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https://t.me/NEET_StudyMaterial, Chapter, , 8, , Gravitation, Solutions, SECTION - A, Objective Type Questions, 1., , According to Kepler, planets move in, (1) Circular orbits around the sun, (2) Elliptical orbits around the sun with sun at exact centre, (3) Straight lines with constant velocity, (4) Elliptical orbits around the sun with sun at one of its foci, , Sol. Answer (4), Kepler's first law,, Law of Orbits : All planets move in elliptical orbits, with the sun at one of the foci of the ellipse., 2., , The minimum and maximum distances of a planet revolving around sun are r and R. If the minimum speed of, planet on its trajectory is v0, its maximum speed will be, (1), , v 0R, r, , (2), , v 0r, R, , (3), , v 0R 2, r2, , (4), , v 0r 2, R2, , Sol. Answer (1), According to Kepler's second law., Low of Areas : The line that joins any planet to the sun sweeps out, equal areas in equal intervals of time. Thus planets appear to move slower, when they are farther from sun than when they are nearer., Now, for planets moving around the sun in an elliptical orbit, Angular, momentum is conserved., , , , , LP LA, , vmin, , P, , s, , vmax, , r, , A, R, , mvmaxr = mv0R, , v max , , v0R, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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2, 3., , Gravitation, , Solution of Assignment, , A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum and minimum, distances of the planet from the sun are r1 and r2 respectively. The time period of the planet is proportional, to, (1) r13/2, , (2), , r23/2, , (3), , (r1 + r2)3/2, , (4), , (r1– r2)3/2, , Sol. Answer (3), Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the, semi-major axis of the ellipse traced out by the planet., T 2 a3, where,, T = Time period of revolution of a planet., a = Semi-major axis of the elliptical orbit traced by the planet., r1 a ae, r2 a ae, , Alternatively,, 3, , ⎛r r ⎞, ⇒ T 2 ⎜ 1 2 ⎟, ⎝ 2 ⎠, , r1 r2 2a, ⎛ r r ⎞, ⇒ T ⎜⎜ 1 2 ⎟⎟, 2, ⎝, ⎠, 2, , ⇒, , 4., , (–a, 0), , From figure, r1 r2 2a, , 3, , (–ae, 0), s, (0, 0), , r, , (a, 0), , r1, , 3/2, , ⇒ T r1 r2 , , 3/2, , T r1 r2 , , In a satellite if the time of revolution is T, then PE is proportional to, (1) T 1/3, , (2), , T3, , (3), , T –2/3, , (4), , T –4/3, , Sol. Answer (3), According to Kepler's third law,, T2 r3, r = radius of orbit, For a satellite of mass m orbiting in an orbit of radius r around planet of mass M,, Potential energy (PE) =, , PE , , GMm, r, , GMm, T, , 2/3, , PE T–2/3, 5., , The torque on a planet about the centre of sun is, (1) Zero, , (2), , Negative, , (3) Positive, , (4), , Depend on mass of planet, , Sol. Answer (1), Force of gravity is acting on the planet,, , Torque of force of gravity = r Fg rFg sin , , P, Fg, s, , r, (0, 0), , Since = 180°, = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 6., , Gravitation, , 3, , During motion of a planet from perihelion to aphelion the work done by gravitational force of sun on it is, (1) Zero, , (2), , Negative, , (3) Positive, , (4), , May be positive or negative, , Sol. Answer (2), According to Kepler's Law of areas, vA < vP, vA = speed of planet at aphelion, vP = speed of planet at perihelion, Now, work done by gravitational force of sun K.E , , , , 1, 2, m vA, v P2, 2, , , , Wgravitation force is negative., 7., , The gravitational constant depends upon, (1) Size of the bodies, , (2), , Gravitational mass, , (3) Distance between the bodies, , (4), , None of these, , Sol. Answer (4), Gravitational constant 'G' is independent of size of bodies, gravitational mass and distance between the bodies., 8., , Gravitation is the phenomenon of interaction between, (1) Point masses only, , (2), , Any arbitrary shaped masses, , (3) Planets only, , (4), , None of these, , Sol. Answer (2), Gravitation is the phenomenon of interaction between any arbitrary shaped bodies., 9., , Force of gravitation between two masses is found to be F, in vacuum. If both the masses are dipped in water, at same distance then, new force will be, (1) > F, , (2), , <F, , (3), , F, , (4), , Cannot say, , Sol. Answer (3), Force of gravitation is independent of the medium. Force is F when masses are in vacuum. When masses, are dipped in water force will be same., 10. Two point masses m and 4m are separated by a distance d on a line. A third point mass m0 is to be placed, at a point on the line such that the net gravitational force on it is zero., d, , m, , 4m, , The distance of that point from the m mass is, (1), , d, 2, , (2), , d, 4, , (3), , d, 3, , (4), , Sol. Answer (3), Force of gravitation on m0 due to m , , Gmm0, , Force of gravitation on m0 due to 4m , , r2, , (d – r), , r, m0, , F1, , G 4mm0, (d r )2, , d, 5, , m, , F1, , F2, d, , 4m, , F2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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4, , Gravitation, , Solution of Assignment, , Net force = 0, F1 = F2, Gmm0, r, , 4Gmm0, , , , 2, , (d r )2, , (d – r)2 = (2r)2, d – r = 2r, d = 3r, Thus, r , , d, 3, , 11. A large number of identical point masses m are placed along x-axis, at x = 0, 1, 2, 4, ........ The magnitude, of gravitational force on mass at origin (x = 0), will be, , y, x, m........., x = 4 and so on, , m m, m, x=0 x=1 x=2, (1) Gm2, , (2), , 4, Gm 2, 3, , (3), , 2, Gm 2, 3, , (4), , 5, Gm 2, 4, , Sol. Answer (2), Let, F1, F2 , F4, F8 …… be the forces of gravitation due masses 'm' at x = 1, 2, 4, 8 … respectively., F1 Gm, 12, , F2 , F4 , F8 , , 2, , Gm2, 22, Gm2, , m, m, m, m, m, m, x = 0 x = 1 x = 2 x = 4 x = 8 x = 16, , 42, Gm2, 82, , F1 F2 F4 F8 , , 1, 1, ⎛1 1, ⎞, Gm2 ⎜ , , , 1 4 16 64 ⎟⎠, ⎝, , infinite G.P. with common ratio =, , 1, 4, , ⎛ a ⎞, For an infinite G.P, sum = ⎜, ⎟, ⎝ 1 r ⎠, a is the first term, r is the common ratio, Sum , , 1, , ⎛4⎞, ⎜ ⎟, 1 ⎝3⎠, 1, 4, , F1 F2 F4 F8 , , 4, Gm 2, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 5, , 12. Three particles A, B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the, particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A, is, , A, , m, , L, B, , (1), , 3, , Gm 2, L2, , (2), , 2, , m, , L, L, , Gm 2, L2, , C, m, , Sol. Answer (4), , A, , At this moment,, , B, 2, , m, , Gm, L2, , (4), , 3, , Gm, L2, , m, , L, , Forces acting on particle at A can be shown,, where, F Gm, L2, , 2, , (3), , L, L, , C, m, , 60°, , F, , F, , Net force will be resultant of both,, , Fresultant F 2 F 2 2F 2 cos 60 3 F, Fresultant , , a, , F, , m, , 3 Gm2, L2, , 3 Gm, L2, , 13. Two planets have same density but different radii. The acceleration due to gravity would be, (1) Same on both planets, , (2), , Greater on the smaller planet, , (3) Greater on the larger planet, , (4), , Dependent on the distance of planet from the sun, , Sol. Answer (3), Acceleration due to gravity at the surface of a planet, g , , GM, R2, , where M is the mass of planet,, R is the radius of the planet,, Also, M = pV, g, , G, , ⎛4, ⎞, ⎜ GR 3 ⎟, ⎝3, ⎠, R, , Thus, g , , 2, , 4, GR, 3, , Thus g Radius of the planet,, Thus, acceleration due to gravity would be greater on the larger planet., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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6, , Gravitation, , Solution of Assignment, , 14. If the radius of earth shrinks by 1.5% (mass remaining same), then the value of gravitational acceleration, changes by, (1) 2%, , (2), , –2%, , (3), , 3%, , (4), , –3%, , Sol. Answer (3), Alternate method:, g, g' , , GM, R, , g' , , 2, , GM, 2, , (0.985 R ), , g ' (1.0306), , GM, R2, , ⇒ g ' 1.0306 g, , GM, (R R )2, , g ' GM (R R )2, 2, , R ⎞, GM ⎛, 1, 2 ⎜⎝, R ⎟⎠, R, R, 1, we can use binomial and approximately,, for, R, GM ⎛ 2R ⎞, g' , ⎜1 , ⎟, R ⎠, R2 ⎝, , g' , , ⇒ Acceleration changes by, ⇒, g, 100 3%, g, ⇒, , g' g g, , 2R, R, , g 2R, ⎛ 1.5 ⎞ 3, , 2 ⎜, 3%, ⎟, g, R, ⎝ 100 ⎠ 100, , [g ' g g ], , 15. If density of a planet is double that of the earth and the radius 1.5 times that of the earth, the acceleration, due to gravity on the surface of the planet is, (1), , 3, times that on the surface of the earth, 4, , (2), , 3 times that on the surface of the earth, , (3), , 4, times that on the surface of the earth, 3, , (4), , 6 times that on the surface of the earth, , Sol. Answer (2), Acceleration due to gravity on the surface of a planet is, given by, g , , GM, R2, , M Mass of the planet, R Radius of the planet, 4, R 3 , 3, , Also, M , g, , G, R, , 2, , , , 4, 4, R 3 GR, 3, 3, , Density of the planet., Acceleration due to gravity R, , , gplanet, gearth, , , , 2e 1.5 Re, 3, e Re, , Acceleration due to gravity on the surface of planet is 3 times that on the surface of earth., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 7, , 16. At what height above the surface of earth the value of "g" decreases by 2%? [radius of the earth is 6400 km], (1) 32 km, , (2), , 64 km, , (3), , 128 km, , (4), , 1600 km, , Sol. Answer (2), ⎛, 2h ⎞, Acceleration due to gravity above the surface of earth at a height h is given g ' g ⎜ 1 , ⎟, R, e ⎠, ⎝, here, g' = 0.98 g, , 0.98 1 , , , 2h, Re, , 2h, 0.02, Re, , h = 0.01 Re, = 0.01 × 6400 km, = 64 km, 17. During motion of a man from equator to pole of earth, its weight will (neglect the effect of change in the radius, of earth), (1) Increase by 0.34%, , (2), , Decrease by 0.34%, , (3) Increase by 0.52%, , (4), , Decrease by 0.52%, , Sol. Answer (1), weq = mg – m2Re, wp = mg, , w p w eq, w eq, , , , , , m2R, mg m2R, 2R, , [2R 0.0337 m/s2 ], , g 2R, , , , w, 0.0337, , 0.3447 102, w eq 9.81 0.0337, , , , w, 100 0.3447, w eq, , Increases by 0.34%, 18. If earth suddenly stop rotating, then the weight of an object of mass m at equator will [ is angular speed of, earth and R is its radius], (1) Decrease by m2R, , (2), , Increase by m2R, , (3), , Decrease by mR2, , (4), , Increase by mR2, , Sol. Answer (2), At the equator,, Apparent weight, w' = w – m2R, If Earth stops rotating, w' will be equal to ., Thus, the weight of an object of mass m at equator will increase by m2R., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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8, , Gravitation, , Solution of Assignment, , 19. If R is the radius of earth and g is the acceleration due to gravity on the earth’s surface. Then mean density, of earth is, , 4G, 3gR, , (1), , (2), , 3R, 4gG, , (3), , 3g, 4 RG, , (4), , Rg, 12G, , Sol. Answer (3), Acceleration due to gravity at earth's surface is given by,, , GM, , g, , R2, , M, , 4, R 3 , 3, , M Mass of earth, Density of earth, R Radius of earth, g, , , G, R, , 2, , , , 4, R 3, 3, , 3g, 4 GR, , 20. The value of g at the surface of earth is 9.8 m/s2. Then the value of ‘g’ at a place 480 km above the surface, of the earth will be nearly (radius of the earth is 6400 km), (1) 9.8 m/s2, , (2), , 7.2 m/s2, , (3), , 8.5 m/s2, , (4), , 4.2 m/s2, , Sol. Answer (3), ⎡ R ⎤, gh g ⎢, ⎥, ⎣R h ⎦, , 2, , ⎡ 6400 ⎤, 2, gh 9.8 ⎢, ⎥ 8.48 m/s, ⎣ 6400 480 ⎦, 21. If the change in the value of ‘g’ at a height ‘h’ above the surface of the earth is same as at a depth x below, it, then (x and h being much smaller than the radius of the earth), (1) x = h, , (2), , x = 2h, , (3), , x, , h, 2, , (4), , x = h2, , Sol. Answer (2), ⎛, 2h ⎞, gh g ⎜ 1 , ⎟, ⎝ Re ⎠, , ⎛, x ⎞, g x g ⎜1 , ⎟, R, e⎠, ⎝, According to the question,, gh – g = gx – g, ⎛ 2h ⎞, ⎛ x ⎞, g ⎜, ⎟ g⎜, ⎟, ⎝ Re ⎠, ⎝ Re ⎠, , x = 2h, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 9, , 22. As we go from the equator to the poles, value of ‘g’, (1) Remains the same, , (2), , Decreases, , (3) Increases, , (4), , First increase and then decrease, , Sol. Answer (3), At Latitude ,, g' = g0 – 2 R cos2 , at equator, = 0, g' = g0 – 2 R, at poles, = 90°, g' = g0, As we g0 from equator to the poles, value of g' increase., 23. What should be the angular speed with which the earth have to rotate on its axis so that a person on the, equator would weigh, , (1), , 3, th as much as present?, 5, , 2g, 5R, , (2), , 2R, 5g, , (3), , 2 R, , (4), , 5g, , 2g, 5R, , Sol. Answer (1), w' = w – m2R, mge = mg – m2R, mge =, , 3, mg, 5, , 2, m R , , , , 2, mg, 5, , 2g, 5R, , 24. The acceleration due to gravity on a planet is 1.96 m/s2. If it is safe to jump from a height of 3 m on the earth,, the corresponding height on the planet will be, (1) 3 m, , (2), , 6m, , (3), , 9m, , (4), , 15 m, , Sol. Answer (4), It is safer to jump from a height of 3 m on earth,, Corresponding velocity attained =, , 2g1h1, , It will be safer to jump from a height on other planet, If the velocity attained is same =, , , 2g1h1 2g 2 h2, , h1, , h2, , 2g 2 h2, , v 2g1h1, , v 2g2h2, , 9.8 × 3 = 1.96 × h2, h2 = 5 × 3 = 15 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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10, , Gravitation, , Solution of Assignment, , 25. Which of the following graph represents the variations of acceleration due to gravity(g) with distance r from the, centre of earth?, , g, , g, , g, , (1), , (2), , O, , r, , R, , g, , (3), , O, , R, , (4), , r, , O, , r, , R, , r, , Sol. Answer (1), Me Mass of earth, Re Radius of earth, , P, r1, , The acceleration due to gravity at a distance r1, from the centre of earth such that r1 < R,, is given by gr1 , , GM, R3, , Re, , r2, , Q, , r1, , gr, The acceleration due to gravity at a distance, r2 from the centre of earth such that r2 > R,, is given by gr 2 , g, , GM, , g, , , r 22, , 1, r, , 1, r2, , r, , r=R, , r, , 2, , 26. An object is taken to height 2R above the surface of earth, the increase in potential energy is [R is radius of, earth], (1), , mgR, 2, , (2), , mgR, 3, , (3), , 2mgR, 3, , (4), , 2 mgR, , m, , Sol. Answer (3), Potential energy at surface , , GMm, R, , Potential energy at height, 2R , Change in potential energy , , GMm, 3R, , 2R, , R, , GMm GMm, , 3R, R, , , , GMm ⎛ 1 3 ⎞, ⎜, ⎟, R ⎝ 3 ⎠, , , , 2 GMm, 3 R, , , , 2 ⎛ GM ⎞, ⎜, ⎟ mR, 3 ⎝ R2 ⎠, , , , 2, mgR, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 11, , 27. The change in potential energy when a body of mass m is raised to height nR from the earth’s surface is (R, is radius of earth), ⎛ n ⎞, (1) mgR ⎜, ⎟, ⎝ n 1⎠, , (2), , nmgR, , (3), , ⎛ n ⎞, mgR ⎜, ⎟, ⎝ n 1⎠, , (4), , ⎛ n2 ⎞, ⎟, mgR ⎜⎜ 2, ⎟, ⎝ n 1⎠, , Sol. Answer (3), Potential energy at the surface , Potential energy at height, nR , Change in potential energy , , , , GMm, R, , m, , GMm, (n 1)R, , nR, , GMm, GMm, , (n 1)R, R, , R, , GMm ⎛ 1 n 1⎞, ⎜, ⎟, R ⎝ n 1 ⎠, , ⎛ n ⎞ ⎛ GM ⎞, mR, ⎜, ⎝ n 1⎟⎠ ⎜⎝ R 2 ⎟⎠, , ⎛ n ⎞, mgR ⎜, ⎝ n 1⎟⎠, 28. A stationary object is released from a point P at a distance 3R from the centre of the moon which has radius, R and mass M. Which of the following gives the speed of the object on hitting the moon?, , P, , Moon, , R, ⎛ 2GM ⎞, ⎟, (1) ⎜, ⎝ 3R ⎠, , 12, , (2), , ⎛ 4GM ⎞, ⎜, ⎟, ⎝ 3R ⎠, , 3R, , 12, , (3), , ⎛ GM ⎞, ⎟, ⎜, ⎝ 3R ⎠, , 12, , (4), , ⎛ GM ⎞, ⎜, ⎟, ⎝ R ⎠, , 12, , Sol. Answer (2), Conserving mechanical energy between points P and S,, , , , , GMm 1, GMm, mv 2 , R, 3R, 2, , 1, GMm GMm, mv 2 , , R, 2, 3R, , , , GMm ⎛ 1 ⎞, ⎜ 1⎟⎠, R ⎝ 3, , 1, 2GMm, mv 2 , 2, 3R, v, , Moon, , S, , P, , R, 3R, , 4GM, 3R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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12, , Gravitation, , Solution of Assignment, , 29. Four particles A, B, C and D each of mass m are kept at the corners of a square of side L. Now the particle, D is taken to infinity by an external agent keeping the other particles fixed at their respective positions. The, work done by the gravitational force acting on the particle D during its movement is, m, m, A, B, , m, , (1) 2, , Gm 2, L, , (2), , 2, , D L C, , Gm 2, L, , m, , (3), , Gm 2, L, , ⎛ 2 2 1⎞, ⎟, ⎜, ⎟, ⎜, 2, ⎠, ⎝, , (4), , , , Gm 2, L, , ⎛ 2 2 1⎞, ⎜, ⎟, ⎜, ⎟, 2, ⎝, ⎠, , Sol. Answer (4), Work done by the gravitational force acting on the particle D during its movement, = – U, = – (Ufinal – Uinitial), , m, A, , = Uinitial – Ufinal, Now, when the particle is at infinity, U = 0, Ufinal = 0, , D, , m, B, , C, m L m, , Work done = Uinitial, Uinitial , , Gm 2 Gm 2 Gm 2, , , L, L, 2L, , , , Gm 2, L, , 1 ⎞, ⎛, ⎜⎝ 2 , ⎟, 2⎠, , , , Gm 2, L, , ⎛ 2 2 1⎞, ⎜⎝, ⎟, 2 ⎠, , 30. If an object is projected vertically upwards with speed, half the escape speed of earth, then the maximum height, attained by it is [R is radius of earth], (1) R, , (2), , R, 2, , (3), , 2R, , (4), , R, 3, , Sol. Answer (4), , Ve , , 2 GM, R, , M mass of earth, R Radius of earth, Now, conserving potential energy at the surface of earth and highest point,, , , GMm 1 ⎛ 1 2GM ⎞, m⎜, ⎟, R, 2 ⎝2, R ⎠, , , , GMm GMm, GMm, , , R, 4R, r, , 2, , , , GMm, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , , , Gravitation, , 13, , 3GMm, GMm, , 4R, r, , r , , 4R, 3, , Rh, , 4R, 3, , R, h ⎛⎜ ⎞⎟, ⎝ 3⎠, 31. The total mechanical energy of an object of mass m projected from surface of earth with escape speed is, (1) Zero, , (2), , Infinite, , (3), , , , GMm, 2R, , (4), , , , GMm, 3R, , Sol. Answer (1), Total mechanical energy = K.E + P.E, 2GM, R, , Vescape , , Total mechanical energy , , 1, 2 GM GMm, m, , 2, R, R, , =0, 32. A body is thrown with a velocity equal to n times the escape velocity (ve). Velocity of the body at a large, distance away will be, (1) v e n 2 1, , (2), , ve n2 1, , (3), , v e 1 n2, , (4), , None of these, , Sol. Answer (1), At large distance potential energy = 0, Conserving mechanical energy at surface of earth and large distance from earth,, , 1, GMm 1, m(nve ) 2 , mv 2, 2, R, 2, Also, ve , , , , 2GM, R, , 2GMm 2, (n 1) mv 2, R, , v, , 2GM 2, (n 1)1/2, R, , v ve n 2 1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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14, , Gravitation, , Solution of Assignment, , 33. The escape velocity of a body from earth is about 11.2 km/s. Assuming the mass and radius of the earth to, be about 81 and 4 times the mass and radius of the moon, the escape velocity in km/s from the surface of, the moon will be, (1) 0.54, , (2), , 2.48, , (3), , 11, , (4), , 49.5, , Sol. Answer (2), , Vescape , , GM, R, , V escape, , Earth, , Vescape, , moon, , Vmoon , , , , Me Rm, , , Re M m, , 81 ⎛ 9 ⎞, ⎜ ⎟, 4 ⎝2⎠, , 2, 11.2 2.48 km/s, 9, , 34. If M is mass of a planet and R is its radius then in order to become black hole [c is speed of light], GM, c, R, , (1), , (2), , GM, c, 2R, , (3), , 2GM, c, R, , (4), , 2GM, c, R, , Sol. Answer (3), A planet can become a black hole if its mass and radius are such that it has an immense force of gravity, on its surface. The force of attractum has to be so large that even light cannot escape from its surface., Speed of light = c, ve , , 2GM, R, , If ve c, Even light can't escape from the surface of such planet making it appear black., 35. The atmosphere on a planet is possible only if [where vrms is root mean square speed of gas molecules on planet, and ve is escape speed on its surface], (1) vrms = ve, , (2), , vrms > ve, , (3), , vrms ve, , (4), , vrms < ve, , Sol. Answer (4), The atmosphere on a planet is possible only if vrms < ve, If vrms vescape the gas molecules will leave the surroundings of the planet, i.e., will be free from gravitational, pull of the planet., 36. A small satellite is revolving near earth’s surface. Its orbital velocity will be nearly, (1) 8 km/s, , (2), , 11.2 km/s, , (3), , 4 km/s, , (4), , 6 km/s, , Sol. Answer (1), For a satellite revolving near earth's surface,, v0 , , GM e, gR e, Re, , Taking g = 9.81 m/s2 and Re = 6400 km, , v0 , , 9.8, 6400 7.92 km/s 8 km/s, 1000, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 15, , 37. The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular, orbit of radius 4R is, (1) 4T, , (2), , T, 4, , (3), , 8T, , (4), , T, 8, , Sol. Answer (3), Using Kepler's third law,, T2 R3, , , T 2 ⎛ 4R ⎞ 3/2, ⎜, ⎝ R ⎟⎠, T, , , , T2 = T × 23, = 8T, , 38. When speed of a satellite is increased by x percentage, it will escape from its orbit, where the value of x is, (1) 11.2%, , (2), , 41.4%, , (3), , 27.5%, , (4), , 34.4%, , (4), , 4 MJ, , Sol. Answer (2), For a satellite near Earth's surface,, GM e, , ve , Re, , v0 , , 2GM e, Re, , ve 2 v0, ⎛, ⎞, % increase, x = ⎜ 2 1⎟ 100 = 41.4%, ⎝ 1 ⎠, , 39. If potential energy of a satellite is –2MJ, then the binding energy of satellite is, (1) 1 MJ, , (2), , 2 MJ, , (3), , 8 MJ, , Sol. Answer (1), For a satellite of mass m revolving around a planet of mass in a circular orbit of radius r,, P.E , , K.E , , GMm, r, , 1 GM GMm, m, , 2, 2r, r, , T.E , , GMm, 2r, , Binding energy T.E , , , GMm, 2r, , P.E, 1 MJ, 2, , Alternate method,, Binding energy = – T.E, , , , P.E, 2, , = 1 MJ, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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16, , Gravitation, , Solution of Assignment, , 40. The time period of polar satellites is about, (1) 24 hr, , (2), , 100 min, , (3), , 84.6 min, , (4), , 6 hr, , Sol. Answer (2), Time period of polar satellites is about 100 minutes polar satellites are low altitude satellites. (h 500 – 800 km), 41. If a satellite of mass 400 kg revolves around the earth in an orbit with speed 200 m/s then its potential energy, is, (1) –1.2 MJ, , (2), , –8.0 MJ, , (3), , –16 MJ, , (4), , –2.4 MJ, , Sol. Answer (3), For a satellite,, , P.E , , GMm, r, , m = mass of satellite, r = radius of orbit, , 1, GMm, P.E, mv 2 , , 2, 2r, 2, , K.E , , P.E = – mv2, = – 400 × 4 × 104, = – 16 MJ, 42. An artificial satellite revolves around a planet for which gravitational force(F) varies with distance r from its centre as, F r2. If v0 its orbital speed, then, (1) v0 r –1/2, , (2), , v0 r 3/2, , (3), , v0 r –3/2, , (4), , v0 r, , Sol. Answer (2), Gravitational force (F) provides the necessary centripetal force to keep the satellite in orbit,, , , mv 02, F, r, , mv 02, r2, r, v0 Orbital speed, r Radius of orbit, v0 r 3/2, 43. The mean radius of earth is R, and its angular speed on its axis is . What will be the radius of orbit of a, geostationary satellite?, 1/3, , 1/3, , ⎛ Rg ⎞, (1) ⎜ 2 ⎟, ⎝ ⎠, , (2), , ⎛ R 2g ⎞, ⎜⎜ 2 ⎟⎟, ⎝ ⎠, , 1/3, , (3), , ⎛ R 2g ⎞, ⎜⎜, ⎟⎟, ⎝ ⎠, , 1/3, , (4), , ⎛ R 2 2 ⎞, ⎜⎜, ⎟⎟, ⎝ g ⎠, , Sol. Answer (2), Time period of rotation of earth , , 2, , , (Duration of one day), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , Geostationary satellite has same time period, T , , Time period of satellite , , Also, g , , T , , 2, . Let r be the radius of orbit of satellite, , , 2r 3/2, GMe, , GM e, R e2, , 2 r 3/2, g (R e ), , r 3/2 , , , , 17, , Re, , , ⎛ R2, r ⎜ e, ⎝ 2, , , , 2 r 3/2, Re g, , , , 2, , , g, , ⎞, g⎟, ⎠, , 1/2, , 44. A satellite of the earth is revolving in a circular orbit with a uniform speed v. If the gravitational force suddenly, disappears, the satellite will, (1) Continue to move with velocity v along the original orbit, (2) Move with a velocity v, tangentially to the original orbit, (3) Fall down with increasing velocity, (4) Ultimately come to rest somewhere on the original orbit, Sol. Answer (2), For a satellite revolving in a circular orbit, gravitational force provides the necessary centripetal force. If the, gravitational force suddenly disappears, the satellite will move with a velocity v, tangentally to the original orbit., , v, , 45. The relay satellite transmits the television signals continuously from one part of the world to another because, its, (1) Period is greater than the period of rotation of the earth, (2) Period is less than the period of rotation of the earth, (3) Period has no relation with the period of the earth about its axis, (4) Period is equal to the period of rotation of the earth about its axis, Sol. Answer (4), A relay satellite transmits the television signals continuously from one part of the world to another bacause, its period is equal to the period of rotation of the earth about its axis., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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18, , Gravitation, , Solution of Assignment, , 46. If height of a satellite from the surface of earth is increased, then its, (1) Potential energy will increase, , (2), , Kinetic energy will decrease, , (3) Total energy will increase, , (4), , All of these, , Sol. Answer (4), For a satellite orbiting at height h from earth,, P.E , , K.E , , GM e m s, (R e h ), , GM e m s, 2(R e h ), , T.E , , GM e m s, 2(R e h ), , If h is increased,, , P.E increases (becomes less negative), K.E decreases, T.E increases (becomes less negative), , 47. The gravitational force on a body of mass 1.5 kg situated at a point is 45 N. The gravitational field intensity, at that point is, (1) 30 N/kg, , (2), , 67.5 N/kg, , (3), , 46.5 N/kg, , (4), , 43.5 N/kg, , Sol. Answer (1), Gravitation force = mg, g = gravitation field intensity., 45 = 1.5 × g, g, , 45, 30 N/kg, 1.5, , 48. A uniform sphere of mass M and radius R is surrounded by a concentric spherical shell of same mass but, radius 2R. A point mass m is kept at a distance x (>R) in the region bounded by spheres as shown in the, figure. The net gravitational force on the particle is, , m, M, , x, , M, , R, 2R, , (1), , GMm, x2, , (2), , GMmx, R3, , (3), , G(M m ), x2, , (4), , Zero, , Sol. Answer (1), The gravitational force on the point mass m due to uniform sphere , , GMm, x2, , ., , The gravitational force on the point mass due to the outer spherical shell is zero because gravitational force, of attraction on a point mass due to various rejoins of the spherical shell cancels each other completely as, their vector sum is zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 19, , 49. If the gravitational potential on the surface of earth is V0, then potential at a point at height half of the radius, of earth is, (1), , V0, 2, , (2), , 2, V0, 3, , (3), , V0, 3, , (4), , 3V0, 2, , Sol. Answer (2), Gravitational potential on the surface,, V0 , , GM e, Re, , Gravitational potential at height h,, GM e, Re ⎞, ⎛, ⎜⎝ R e , ⎟, 2 ⎠, , Vn , , , , 2 GM e, 3 Re, , 2, V0, 3, 50. Two point masses having mass m and 2m are placed at distance d. The point on the line joining point masses,, where gravitational field intensity is zero will be at distance, 2d, , (1), , (3), , 3 1, , d, 1 2, , from point mass "2m", , (2), , from point mass "m", , (4), , 2d, 3 1, , d, 1 2, , from point mass "2m", , from point mass "m", , Sol. Answer (3), d-r, , r, , 2m, , m, , Gravitational field intensity will be zero,, 2 Gm, , , , Gm, , , , 1, 2, , r d r, , r, , 2, , , , (d r ) 2, , d r 2r, , , , , , r 1 2 d, r , , d, , 1 2 , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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20, , Gravitation, , Solution of Assignment, , SECTION - B, Objective Type Questions, 1., , The ratio of kinetic energy of a planet at perigee and apogee during its motion around the sun in elliptical orbit, of eccentricity e is, (1) 1 : e, , (2), , 1 e, 1 e, , (3), , ⎛ 1 e ⎞, ⎜ 1 e ⎟, ⎝, ⎠, , 2, , ⎛ 1 e ⎞, ⎜ 1 e ⎟, ⎝, ⎠, , (4), , 2, , Sol. Answer (3), K.E of a planet =, , 1, mv2, 2, , K.E at perigee =, , 1, mvP2, 2, , vA, , 1, K.E at apogee = mvA2, 2, , (–a, 0), , P, vP, , Using conservation of angular momentum at P and A, , (–ae, 0) (0, 0), , (a, 0), , rP, , A, , rA, , mvP rP = mvArA, , 2., , , , vP, r, a (1 e ), A , vA, rP, a (1 e ), , , , K.E P v P2 ⎛ 1 e ⎞ 2, , ⎜, ⎟, K.E A vA2 ⎝ 1 e ⎠, , An earth satellite X is revolving around earth in an orbit whose radius is one-fourth of the radius of orbit of a, communication satellite. Time period of revolution of X is, (1) 3 hrs, , (2), , 6 hrs, , (3), , 4 days, , (4), , 72 days, , Sol. Answer (1), Time period of a communication satellite = 24 hours., Using kepler's third law,, T2 r 3, Tc ⎛ r c ⎞, , , T x ⎜⎝ r x ⎟⎠, , , , 3/2, , 24, (4) 3/2, Tx, , Tx , , 24, 3 hrs, 8, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 3., , Gravitation, , 21, , Two satellites of equal mass are revolving around earth in elliptical orbits of different semi-major axis. If their, angular momenta about earth centre are in the ratio 3 : 4 then ratio of their areal velocity is, (1), , 3, 4, , (2), , 2, 3, , (3), , 1, 3, , (4), , 4, 3, , Sol. Answer (1), Areal velocity,, , , A, L vA, t, 2m, , , L is the angular momentum of satellite, m is the mass of satellite,, , 4., , , vA1, L, ⎛ 3⎞, 1 ⎜ ⎟, vA2, L 2 ⎝ 4⎠, , When a satellite moves around the earth in a certain orbit, the quantity which remains constant is, (1) Angular velocity, , (2), , Kinetic energy, , (3), , Areal velocity, , (4), , Potential energy, , Sol. Answer (3), The path of a satellite moving around sun in a certain orbit is not exactly circular but elliptical with low value, of eccentricity, e. Thus only areal velocity is constant., 5., , Consider a planet moving around a star in an elliptical orbit with period T. The area of the elliptical orbit is, proportional to, (1) T, , 4, 3, , 2, , (2), , T, , (3), , Sol. Answer (1), , 1, , T3, , (4), , T2, , Y, , Area of ellipse, A = r1r2, ∵ r1 = a – ae = a(1 – e), and r2 = a + ae = a(1 + e), , (1 – e) (1 + e), , ae, , F1, , A = {a(1 – e)} {a(1 + e)}, =a2, , O, , S, , ae F, 2, , a, r1, , x, , a, r2, , =a2(12 – e2), ∵ e2 << a2 then, A = a2, So, a A1 2, , (1), , According to Keplar’s III law, T2 a3, 1 ⎤, ⎡, T 2 ⎢( A) 2 ⎥, ⎣, ⎦, , 3, , T 2 A3 2, A (T 2 )2 3, , A T4 3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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22, 6., , Gravitation, , Solution of Assignment, , A body weighs 72 N on surface of the earth. When it is taken to a height of h = 2R, where R is radius of earth, it, would weigh, (1) 36 N, , (2), , 18 N, , (3), , 9N, , (4), , 8N, , Sol. Answer (4), Weight on earth = mg m , , GM, R2, , 72 N, , GM, GMm 72, ⎛ GM ⎞, , 8N, Weight at height, h = 2R will be mg m ⎜ 2 ⎟ m , 2 , ⎝ r ⎠, 9, (R 2R ), 9R 2, , 7., , If all objects on the equator of earth feel weightless then the duration of the day will nearly become, (1) 6.2 hr, , (2), , 4.4 hr, , (3), , 2.2 hr, , (4), , 1.41 hr, , Sol. Answer (4), Weq = mg – m2R, mg – m2R = 0, 2R = g, g, R, , , , Time period of rotation, i.e., duration of the day,, , 2, R, 2, 1.41 hr, , g, , , , 8., , A body is projected vertically upwards with a speed of, , GM, (M is mass and R is radius of earth). The body will, R, , attain a height of, (1), , R, 2, , (2), , R, , (3), , 5, R, 4, , (4), , 3R, 2, , Sol. Answer (2), Conserving mechanical energy at earth surface and at the maximum height attained by the body,, P.Ei + K.Ei = P.Ej + K.Ej, , , , , , , , GMm 1 ⎛ GM ⎞, GMm, m⎜, 0, ⎟, R, 2 ⎝ R ⎠, r, GMm, GMm, , 2R, r, r = 2R, , R + h = 2R, h =R, 9., , If the gravitational potential energy of two point masses infinitely away is taken to be zero then gravitational, potential energy of a galaxy is, (1) Zero, , (2), , Positive, , (3), , Negative, , (4), , Can have any value, , Sol. Answer (3), A galaxy is a bounded system, for a bounded system or closed system like planet-sun, satellite Earth, electronnucleus etc. total energy and the potential energy both are negative., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 23, , 10. A particle of mass m is dropped from a height R equal to the radius of the earth above the tunnel dug through the, earth as shown in the figure. Hence the correct statement is, , m, R, , R, C, , (1) Particle will oscillate through the earth to a height h = R on both sides, (2) Motion of the particle is periodic, (3) Motion of the particle is simple harmonic, (4) Both (1) & (2), Sol. Answer (4), When the particle is outside the tunnel force acting on it is , is the distance from the centre of the earth., When the particle is inside the tunnel force acting on it is , , 1, r, , 2, , m, , where, r, , R, , 1, where, r is, r, , R, , the distance from the centre of the earth., In both cases, force is always directed towards the centre of the earth., Thus motion is oscillatory and also periodic but not SHM., 11. The particles A and B of mass m each are separated by a distance r. Another particle C of mass M is placed, at the midpoint of A and B. Find the work done in taking C to a point equidistant r from A and B without, acceleration (G = Gravitational constant and only gravitational interaction between A, B and C is considered), (1), , GMm, r, , (2), , 2GMm, r, , (3), , 3GMm, r, , (4), , 4GMm, r, , r/2, , m, , Sol. Answer (2), Since particle C is moved without any acceleration,, K.E = 0, , m, , Work done by external agent + Wgravitation = 0, , A, , Work done by external agent = – Wg, , r/2, , M, , B, r, , r, , = – (– U), = U, = Uf – Uin, Uf , , GMm GMm, 2GMm, , , r, r, r, , Ui , , GMm GMm, 4GMm, , , r /2, r /2, r, , Work done , , 2GMm, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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24, , Gravitation, , Solution of Assignment, , 12. The magnitude of potential energy per unit mass of an object at the surface of earth is E, then the escape, velocity of the object is, (1), , 2E, , (2), , 4E2, , (3), , E, , (4), , 2E, , Sol. Answer (1), P.E of an object on earth surface , , GMm, R, , ⎛ GM ⎞, E, Magnitude of potential energy per unit mass ⎜, ⎝ R ⎟⎠, Vescape , , 2GM, R, , 2E, , 13. The orbital speed of a satellite revolving around a planet in a circular orbit is v0. If its speed is increased by, 10%, then, (1) It will escape from its orbit, (2) It will start rotating in an elliptical orbit, (3) It will continue to move in the same orbit, (4) It will move in a circular orbit of radius 20% more then radius of initial orbit, Sol. Answer (2), When the orbital speed of a satellite revolving around a planet is increased by 10%, it corresponds to the case, when v0 < v < ve ., 14. If L is the angular momentum of a satellite revolving around earth in a circular orbit of radius r with speed v,, then, (1) L v, , (2), , Lr, , (3), , L r, , (4), , L v, , Sol. Answer (3), Angular momentum of a satellite revolving around earth in a circular orbit, L = mvr, m mass of satellite, v speed of satellite, r radius of orbit, L = mvr, Also, v , , , , GM, r, , Lm, , GM, r, r, , Thus, L r, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 25, , 15. Two satellites of mass m and 2 m are revolving in two circular orbits of radii r and 2r around an imaginary, planet, on the surface of which gravitational force is inversely proportional to distance from its centre. The ratio, of orbital speed of satellites is, (1) 1 : 1, , (2), , 1:2, , (3), , 2:1, , (4), , 1: 2, , Sol. Answer (1), Force of gravitation provides the necessary centripetal force,, , , mv 2 GMm, , r, r, v GM, , Independent of mass of satellite and radius of orbit., , , v1 1, , v2 1, , 16. A satellite of mass m is revolving close to surface of a planet of density d with time period T. The value of, universal gravitational constant on planet is given by, (1) 2d 2T, , (2), , dT 2, , (3), , 1, 2, , d T, , (4), , 3, dT 2, , Sol. Answer (4), Time period of a satellite revolving close to surface,, T , , T2 , , M, , T2 , , G, , 2R 2R 3/2, , v, GM, , 4 2 R 3, GM, 4, R 3 d, 3, 4 2R 3, 4, G R 3 d, 3, , 3, dT 2, , 17. When energy of a satellite-planet system is positive then satellite will, (1) Move around planet in circular orbit, , (2), , Move around planet in elliptical orbit, , (3) Escape out with minimum speed, , (4), , Escape out with speed greater than escape velocity, , Sol. Answer (4), When the energy of a satellite-planet system is positive, satellite escapes away from the gravitational field of, the planet with speed greater than the escape speed., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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26, , Gravitation, , Solution of Assignment, , 18. If radius of an orbiting satellite is decreased, then its kinetic energy, (1) And potential energy decrease, , (2), , And potential energy increase, , (3) Decreases and potential energy increases, , (4), , Increases and potential energy decreases, , Sol. Answer (4), K.E , , GMm, 2r, , P.E , , GMm, r, , M mass of planet, m mass of satellite, r radius of orbit, When r is decreased,, Kinetic energy increases,, Potential energy decreases (becomes more negative)., 19. Two point masses having m and 4m are placed at distance at r. The gravitational potential at a point, where, gravitational field intensity zero is, (1), , 9GM, r, , (2), , 2GM, 3r, , 3GM, r, , (3), , (4), , 6GM, 5r, , Sol. Answer (1), Gravitational field intensity at O is zero,, , , , , Gm, d, , 2, , 4 Gm, , , , (r d ), d2, , d, , (r d ) 2, 2, , m, 4, , r -d, , ×, O, , 4m, r, , r d, 2, d, r – d = ± 2d, d, , r, , r, 3, , (d = – r, not possible), , Taking the +ve value of d,, Calculating gravitational potential at O,, V , , Gm 4Gm, , r /3, 2r /3, , , , 3Gm 6Gm, , r, r, , , , 9Gm, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 27, , 20. If gravitational field intensity is E at distance R/2 outside from then surface of a thin shell of radius R, the, gravitational field intensity at distance R/2 from its centre is, (1) Zero, , (2), , 2E, , 2E, 3, , (3), , (4), , 3E, 2, , Sol. Answer (1), Gravitational field intensity at every point inside a hollow spherical shell of uniform density is zero, because, gravitational field due to various regions of the spherical shell cancels each other completely as their vector, sum is zero., 21. If potential at the surface of earth is assigned zero value, then potential at centre of earth will be, (Mass = M, Radius = R), (1) 0, , , , (2), , GM, 2R, , (3), , , , 3GM, 2R, , (4), , 3GM, 2R, , Sol. Answer (2), The concept involved here is that,, Gravitational potential difference between any two points in a gravitational field is independent of the choice, of reference. When potential at the infinity is assigned zero value,, Potential at the surface , Potential at the centre , , GM, Vs, R, , 3GM, Vc, 2R, , GM 3GM GM, , , R, 2R, 2R, Now, when potential at the surface is assigned zero value,, Vs Vc , , Vs – Vc = Vs' – Vc', , , GM, 0 Vc, 2R, , Vc , , GM, 2R, , Here, Vs' and Vc' are the new values of potential at the sum and centre respectively., 22. An object is projected horizontally with speed, is mass of earth, then object will], , 1 GM, , from a point at height 3 R [where R is radius and M, 2 R, , (1) Fall back on surface of earth by following parabolic path, (2) Fall back on surface of earth by following hyperbolic path, (3) Start rotating around earth in a circular orbit, (4) Escape from gravitational field of earth, Sol. Answer (3), At height 3 R, i.e at distance 4 R from the centre of the earth,, , Vorbital , Here, r = 4 R, , GM, r, V0 , , GM 1 GM, , ,, 4R 2 R, , Thus, an object taken to a height 3 R if projected horizontally with speed, earth in a circular orbit., , 1 GM, , will start rotating around, 2 R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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28, , Gravitation, , Solution of Assignment, , 23. If acceleration due to gravity at distance d[< R] from the centre of earth is , then its value at distance d above, the surface of earth will be [where R is radius of earth], (1), , R 2, , R, 2d, , (2), , (R d )3, , (3), , d, (R d )2, , (4), , R 3, d (R d )2, , Sol. Answer (4), Here, g d , gd , , GM, R3, , d , d < R, , GM, (R d )2, , Using (1), GM , , gd , , ...(1), , , d>R, , ...(2), , R 3, d, , R 3, d (R d )2, , 24. If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at, height h above the surface of earth is [R is radius of earth and M is mass of earth], (1), , GMm, Rh, , GMm, h, , (2), , (3), , GMmh, R (R h ), , (4), , GMmh, h 2R, , Sol. Answer (3), The concept involved here is that,, Gravitational potential energy difference between any two points in a gravitational field is independent of the, choice of reference., When potential at infinity is assigned zero value,, Potential energy of a body of mass m on the surface of earth , Potential energy at height, h , , GMm, Us, R, , GMm, Un, Rh, , 1 ⎞, ⎛ 1, U s U h GMm ⎜ , ⎟, ⎝ R Rh⎠, ⎛ R h R ⎞, GMm ⎜, ⎝ R (R h ) ⎟⎠, , , , GMmh, R (R h ), , Now, when potential at the surface is taken zero, Let Us', Uh' be the new values of potential energy at the surface, and height h respectively,, And, Us – Uh = Us' – Uh', , , GMmh, 0 Uh, R (R h ), , Uh , , GMmh, R (R h ), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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30, , Gravitation, , Solution of Assignment, , SECTION - C, Previous Years Questions, 1., , A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared, to the mass of the earth. Then,, [Re-AIPMT-2015], (1) The acceleration of S is always directed towards the centre of the earth, (2) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains, constant, (3) The total mechanical energy of S varies periodically with time, (4) The linear momentum of S remains constant in magnitude, , Sol. Answer (1), 2., , A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface, of earth. If earth's radius is 6.38 × 106 m and g = 9.8 ms–2, then the orbital speed of the satellite is, [Re-AIPMT-2015], (1) 6.67 km s–1, , (2), , 7.76 km s–1, , (3), , 8.56 km s–1, , (4), , 9.13 km s–1, , Sol. Answer (2), 3., , Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third, If the masses of, power of average distance r between sun and planet, i.e., T2 = Kr3, here K is constant., sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between, them is F , , GMm, r2, , , here G is gravitational constant. The relation between G and K is described as, [AIPMT-2015], , (1) K , , 1, G, , (2), , GK = 42, , (3), , GMK = 42, , (4), , K=G, , Sol. Answer (3), 4., , A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what, approximate radius would earth (mass = 5.98 × 1024 kg) have to be compresed to be a black hole?, [AIPMT-2014], (1) 10–9 m, , (2), , 10–6 m, , (3), , 10–2 m, , (4), , 100 m, , Sol. Answer (3), 5., , , Dependence of intensity of gravitational field ( E ) of earth with distance (r) from centre of earth is correctly, respresented by, [AIPMT-2014], , E, (1), , O, , E, , R, r, , (2), , O, , E, R, , r, , (3), , O, , E, , R, r, , (4), , O, , R, , r, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 6., , Gravitation, , 31, , A body of mass m is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The, change in potential energy of body will be, [NEET-2013], (1), , 2, mgR, 3, , (2), , 3mgR, , (3), , 1, mgR, 3, , (4), , mg2R, , Sol. Answer (1), GMm, R, , P.E at surface of earth , , GM, mR, R2, , , , Uin = –mgR, , GMm, 3R, , P.E at height, h = 2R , , GM, mR, 3R 2, , , , Uf , Uf – Uin = Change in P.E , , , , 7., , GM ⎤, ⎡, ⎢⎣g R 2 ⎥⎦, , mgR, 3, , mgR, mgR, 3, 2mgR, 3, , Infinite number of bodies, each of mass 2 kg are situated on x-axis at distances 1 m, 2 m, 4 m,, 8 m, ....., respectively, from the origin. The resulting gravitational potential due to this system at the origin will, be, [NEET-2013], (1) –, , 8, G, 3, , (2), , –, , 4, G, 3, , (3), , – 4G, , (4), , –G, , Sol. Answer (3), 8., , A spherical planet has a mass Mp and diameter Dp. A particle of mass m falling freely near the surface of this, planet will experience an acceleration due to gravity, equal to, [AIPMT (Prelims)-2012], (1) GMp / Dp2, , (2), , 4GMpm / Dp2, , (3), , 4GMp / Dp2, , (4), , GMpm / Dp2, , Sol. Answer (3), Acceleration due to gravity, near surface , , GM p, R p2, , gp, , Here, Dp = 2Rp, gp , , 4 GM p, D p2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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32, 9., , Gravitation, , Solution of Assignment, , A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the, radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth, is, [AIPMT (Prelims)-2012], (1) 6 2, , (2), , 6, , (3), , 2, , 5, , (4), , 10, , Sol. Answer (1), Time period of a geostationary satellite = 24 hours., Using Keplers third law,, T2 r3, T1, time period of geostationary satellite (6R)3/2 thus T2 (3R)3/2, , , T 2 ⎛ 3R ⎞, ⎜, ⎟, T1 ⎝ 6R ⎠, , T 2 24 , , 3/ 2, , 1, 24, , 6 2 hours., 2 3/ 2, 2 2, , 10. The height at which the weight of a body becomes, , (1) 3R, , (2), , 1, th, its weight on the surface of earth (radius R), is, 16, [AIPMT (Prelims)-2012], , 4R, , (3), , 5R, , (4), , 15R, , Sol. Answer (1), Weight on surface of earth, W = mg, ⎛ GM ⎞, m⎜ 2 ⎟, ⎝ Re ⎠, , Weight at height h from surface, W m, , GM, (R e h ) 2, , Re2, W, 1, , , W 16 (Re h )2, Re + h = 4Re, , , h = 3Re, , 11. If ve is escape velocity and vo is orbital velocity of a satellite for orbit close to the earth’s surface, then these are, related by, [AIPMT (Mains)-2012], (1) v o 2 v e, , (2), , vo = ve, , (3), , ve 2vo, , (4), , ve 2 vo, , Sol. Answer (4), , ve , , 2GM, R, , vo , , GM, R, , v e 2 vo, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 33, , 12. Which one of the following plots represents the variation of gravitational field on a particle with distance r due to, a thin spherical shell of radius R? (r is measured from the centre of the spherical shell), [AIPMT (Mains)-2012], F, , F, , F, , (2), , (1), O, , (3), O, , r, , R, , F, , (4), O, , r, , R, , O, , r, , R, , R, , r, , Sol. Answer (2), For a thin spherical shell gravitational field for r < R is zero., For a thin spherical shell gravitational field for r > R is given by F , , GM, r2, , Thus, most suitable plot is, , F, , r, 13. A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance, v1, of r2. If v1 and v2 are the linear velocities at these points respectively, then the ratio v is, 2, [AIPMT (Prelims)-2011], r1, (1) r, 2, , (2), , ⎛ r1 ⎞, ⎜ ⎟, ⎝ r2 ⎠, , 2, , (3), , r2, r1, , (4), , ⎛ r2 ⎞, ⎜ ⎟, ⎝ r1 ⎠, , 2, , Sol. Answer (3), , P, , 2, A, , s, , 1, r1, , r2, , Using conservation of angular momentum at P and A,, m1r1 = m2r2, 1 r2, , 2 r1, , 14. A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude, of the gravitational potential at a point situated at, , (1), , 4GM, a, , (2), , GM, a, , a, distance from the centre, will be [AIPMT (Mains)-2011], 2, (3), , 2GM, a, , (4), , 3GM, a, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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34, , Gravitation, , Solution of Assignment, , 15. A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius, of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the, surface of the earth. The minimum value of u so that the particle does not return back to earth, is, [AIPMT (Mains)-2011], 2gR 2, , (1), , 2GM, , (2), , R, , 2, , 2GM, R, , (3), , 2gM, , (4), , R2, , Sol. Answer (3), Particle will not return back if it is thrown upwards with escape velocity,, GMm 1, mv e2 0, R, 2, , 2GM, R, , ve , , 16. The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of, satellite A is 3V, then the speed of satellite B will be, [AIPMT (Prelims)-2010], (1), , 3V, 4, , (2), , 6V, , (3), , 12 V, , (4), , 3V, 2, , Sol. Answer (2), V0 , , GMe, r, , r radius of orbit, , , , VA, , VB, , rB, rA, , 3V, , VB, , 1, 4, , VB = 6 V, , 17. A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational, a, distance from the centre, will be, [AIPMT (Prelims)-2010], potential at a point situated at, 2, (1) , , 3GM, a, , (2), , , , 2GM, a, , (3), , , , GM, a, , (4), , , , 4GM, a, , Sol. Answer (1), , a, distance from the centre will be sum of potential due to spherical, Gravitational potential at a point situated at, 2, shell and due to mass M at the centre,, Thus, V = V1 + V2, Vspherical shell , Vmass M , , Vtotal , , GM, V1, a, , GM 2GM, , V2, a /2, a, , a/2, a, , M, , 3GM, a, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 35, , 18. The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to, transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is, ⎛ 1, 1 ⎞, (1) GmM ⎜ 2 2 ⎟, R, R, 2 ⎠, ⎝ 1, , (2), , ⎛ 1, 1 ⎞, GmM ⎜, , ⎟, ⎝ R1 R2 ⎠, , (3), , ⎛ 1, 1 ⎞, 2GmM ⎜, , ⎟, ⎝ R1 R2 ⎠, , [AIPMT (Mains)-2010], , (4), , ⎛ 1, 1, 1 ⎞, , GmM ⎜, ⎟, 2, R, R, ⎝ 1, 2 ⎠, , Sol. Answer (4), To find out the additional kinetic energy to be provided to a satellite of mass m,, We can use conservation of mechanical energy,, K.Ei + P.Ei + K.Eadditional = K.Ef + P.Ef, , 1 GMm GMm, 1 GMm GMm, , K.Eadditional , , 2 R1, R1, 2 R2, R2, , K.Eadditional , , 1 ⎤, GMm ⎡ 1, , 2 ⎢⎣ R1 R2 ⎥⎦, , 19. The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be, a sphere of radius R of uniform density is as shown in figures below, , g, , g, , (a), , (b), , (c), , r, , R, , R, , g, , (d), , r, , R, , The correct figure is, , g, , r, , R, , r, , [AIPMT (Mains)-2010], , (1) (d), , (2), , (a), , (3), , (b), , (4), , (c), , Sol. Answer (1), ginside , , GM, R3, , g outside , , r i.e., for r < R, , GM, r2, , i.e., for r > R, , The suitable graph is,, , , , r, , g, , R, , 1, r2, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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36, , Gravitation, , Solution of Assignment, , 20. The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area, SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then, , m v, C, , B, A, , S, , D, [AIPMT (Prelims)-2009], , (1) t1 = 4t2, , (2), , t1 = 2t2, , (3), , t1 = t2, , (4), , t1 > t2, , Sol. Answer (2), According to Kepler's law of Areas : The line that joins any planet to the sun sweeps out equal areas in equal, intervals of time, , i.e, , A, is constant., t, , Area SCD = 2 × Area SAB, , Using,, , ASCD tSCD ⎛ t1 ⎞, , ⎜ ⎟, ASAB tSAB ⎝ t 2 ⎠, , t1 = 2t2, 21. Two satellites of earth S1 and S2 are moving in the same orbit. The mass of S1 is four times the mass of S2., Which one of the following statements is true?, [AIPMT (Prelims)-2007], (1) The potential energies of earth and satellite in the two cases are equal, (2) S1 and S2 are moving with the same speed, (3) The kinetic energies of the two satellites are equal, (4) The time period of S1 is four times that of S2, Sol. Answer (2), v0 , , GM e, , r is the radius of the orbit., r, , Radius of orbit is same for both S1 and S2,, Thus, v01 = v02, S1 and S2 are moving with the same speed., 22. The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the, Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface, of the Earth. The value of f is, [AIPMT (Prelims)-2006], , (1), , 2, , (2), , 1, 2, , (3), , 1, 3, , (4), , 1, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 37, , Sol. Answer (2), Escape velocity from height, h = R from earth can be evaluated using conservation of mechanical energy,, , GMm 1, m(ve)2 0, 2R, 2, GM, R, , ve , , From surface of earth, v e , , 1, , v e , , 2, , 2GM, R, , v e fv, , 1, , Thus, f , , 2, , 23. Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size., If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g’,, then, [AIPMT (Prelims)-2005], (1) g= 3g, , (2), , g=, , g, 9, , (3), , g= 9g, , (4), , g= 27g, , Sol. Answer (1), , g, , 4GR, 3, , g , , 4G(3R ), 3, , ⎛ 4GR ⎞, 3⎜, ⎟ 3g, ⎝ 3 ⎠, 24. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is :, [AIPMT (Prelims)-2005], (1), , 2, , (2), , 1, 2, , (3), , 1, 2, , (4), , 2, , Sol. Answer (2), , , , P.E , , GMm, R, , K.E , , GMm, 2R, , K.E 1, , P.E 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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38, , Gravitation, , Solution of Assignment, , 25. The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. If VP and, VE are escape velocities of the planet and the earth, respectively, then, (1) VE = 1.5VP, , (2), , VP = 1.5VE, , (3), , VP = 2VE, , (4), , VE = 3VP, , Sol. Answer (3), , ⎛ 8GP, Vescape R ⎜, ⎜, 3, ⎝, , , ⎞, ⎟⎟, ⎠, , VP, 2, VE, VP = 2VE, , 26. A particle of mass 'm' is kept at rest at a height 3R from the surface of earth, where 'R' is radius of earth and, 'M' is mass of earth. The minimum speed with which it should be projected, so that it does not return back,, is (g is acceleration due to gravity on the surface of earth), , ⎛ GM ⎞, (1) ⎜, ⎝ R ⎟⎠, , 1/2, , (2), , ⎛ GM ⎞, ⎜⎝, ⎟, 2R ⎠, , 1/2, , (3), , ⎛ gR ⎞, ⎜⎝, ⎟, 4 ⎠, , 1/2, , (4), , ⎛ 2g ⎞, ⎜⎝ ⎟⎠, R, , 1/2, , Sol. Answer (2), The particle won't return back if it is provided speed such that its total mechanical energy is zero or positive, for minimum speed, we take total energy zero,, , , GMm 1, mv 2 0, 4R, 2, , v, , GM, 2R, , 27. Which of the following graphs shows the variation of acceleration due to gravity g with depth h from the surface, of the earth?, , g, , g, , g, , h, , h, (a), (1) (a), , g, , h, (c), , (b), (2), , (b), , (3), , (c), , h, (d), (4), , (d), , Sol. Answer (3), , h⎞, ⎛, Acceleration due to gravity at depth h from surface of earth, g h g 0 ⎜⎝ 1 ⎟⎠, R, h, g g0 g0, R, g g 0, , h, g0, R, h, , ⎛ g ⎞, g ⎜ 0 ⎟ g0, ⎝ R ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 39, , Comparing with equation of straight line,, y = mx + c, Slope, m is –ve, Intercept c is +ve,, Thus, most appropriate graph is, , g, , h, 28. At what altitude (h) above the earth’s surface would the acceleration due to gravity be one fourth of its value, at the earth’s surface?, (1) h = R, , (2), , h = 4R, , (3), , h = 2R, , (4), , h = 16R, , Sol. Answer (1), At altitude (h) above the earth's surface, g h , gh , , GM, R2, , R 2 (R h ) 2, , gh g , , , , GM, (R h ) 2, , R2, (R h ) 2, , R2, R, 1, 1, , , , 4 (R h ) 2, 2 (R h ), , Using the +ve value,, R + h = 2R, h =R, 29. If the gravitational force between two objects were proportional to 1/R (and not as 1/R2), where R is the, distance between them, then a particle in a circular path (under such a force) would have its orbital speed, v, proportional to, (1) R, (3), , 1, R, , 2, , (2), , R0 (independent of R), , (4), , 1, R, , Sol. Answer (2), Gravitational force provides the necessary centripetal force for a particle to move in the circular path., , , mv 2 K, , R, R, , v, , K ⎤, ⎡, ⎢not 2 ⎥, ⎣, R ⎦, , K, m, , Thus independent of R., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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40, , Gravitation, , Solution of Assignment, , 30. The distance of two planets from the sun are 1013 m and 1012 m respectively. The ratio of time periods of, the planets is, (1), , 10 : 1, , (2), , 10 10 : 1, , (3), , 10 : 1, , (4), , 1:1, , Sol. Answer (2), Using Kepler's third law,, T2 r3, T1 ⎛ r1 ⎞, ⎜ ⎟, T2 ⎝ r2 ⎠, , , , 3/2, , T1, 103/2 10 10, T2, , 31. The radius of earth is about 6400 km and that of Mars is 3200 km. The mass of the earth is about, 10 times the mass of Mars. An object weighs 200 N on the surface of Earth. Its weight on the surface of, mars will be, (1) 20 N, , (2), , 8N, , (3) 80 N, , (4), , 40 N, , Sol. Answer (3), Re = 6400 km, RM = 3200 km, Me, 10, MM, We m , , GMe, , WM m , , mge, , Re2, , GMM, RM2, , mgM, , We ⎛ Me ⎞ ⎛ RM ⎞, ⎜, ⎟⎜, ⎟, WM ⎝ MM ⎠ ⎝ Re ⎠, , 200, ⎛ 1⎞, 10 ⎜ ⎟, WM, ⎝2⎠, , 2, , 2, , WM , , 200 4, 80 N, 10, , 32. The earth (mass = 6 × 1024 kg) revolves around the sun with an angular velocity of 2 × 10–7 rad/s in a, circular orbit of radius 1.5 × 108 km. The force exerted by the sun on the earth, in newtons, is, (1) 36 × 1021, , (2), , 27 × 1039, , (3), , Zero, , (4), , 18 × 1025, , Sol. Answer (1), The force of gravitation exerted by sun provides the necessary centripetal force = m2r, Fg = 6 × 1024 × 4 × 10–14 × 1.5 × 1011, = 36 × 1021 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 41, , 33. Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational, attraction. The speed v of each particle is, , (1), , 1 Gm, 2 R, , 4Gm, R, , (2), , (3), , 1, 1, 2R Gm, , (4), , Gm, 2R, , Sol. Answer (1), , m, , m, R, , R, , Gravitation force provides the necessary centripetal force,, , Gm 2, , , , 2, , (2R ), , mv 2, r, , Where, r is the radius of circular path i.e R, Gm 1 Gm, , 4R, 2 R, , v, , 34. The acceleration due to gravity g and mean density of the earth are related by which of the following, relations? (where G is the gravitational constant and R is the radius of the earth.), (1) , , 3g, 4 GR, , , , (2), , 3g, 4 GR 3, , (3), , , , 4 gR 2, 3G, , (4), , , , (4), , G, , 4 gR 3, 3G, , Sol. Answer (1), g, , M, , GM, R2, 4, R 3 , 3, , Thus, , , ⇒ g, , G, R, , 2, , , , 4, R 3 , 3, , 3g, 4GR, , 35. What will be the formula of mass of the earth in terms of g, R and G?, (1) G, , R2, g, , (2), , g, , R2, G, , (3), , g2, , R, G, , R, g, , Sol. Answer (2), g, , GM, R2, , Mass of earth g, , R2, G, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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42, , Gravitation, , Solution of Assignment, , 36. The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is, how many times greater than that of B from the sun?, (1) 4, , (2), , 5, , (3), , 2, , (4), , 3, , Sol. Answer (1), Using kepler's third law,, T2 r3, TA ⎛ rA ⎞, ⎜ ⎟, TB ⎝ rB ⎠, , 3/2, , ⎛r ⎞, (8)2/3 ⎜ A ⎟, ⎝ rB ⎠, , rA = 4rB, 37. The escape velocity of a body on the surface of the earth is 11.2 km/s. If the earth’s mass increases to twice, its present value and radius of the earth becomes half, the escape velocity becomes, (1) 22.4 km/s, , (2), , 44.8 km/s, , (3), , 5.6 km/s, , (4), , 11.2 km/s, , Sol. Answer (1), , Ve , , 2GM, R, , Ve , , 2G(2M ), R /2, , 2, , 2GM, R, , 22.4 km/s, , 38. The escape velocity of a sphere of mass m from the surface of earth is given by (G = Universal gravitational, constant; M = Mass of the earth and Re = Radius of the earth), (1), , 2GMm, Re, , (2), , 2GM, Re, , (3), , GM, Re, , (4), , 2GM R e, Re, , Sol. Answer (2), ve , , 2GM, Re, , GMm 1, mv e2 0, Re, 2, , ve , , 2GM, , independent of the mass of sphere., Re, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 43, , 39. A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational, force exerted on it will be, (1) 36 N, , (2), , 32 N, , (3), , 144 N, , (4), , 50 N, , Sol. Answer (2), Gravitational force on body mgs , , mGM, R2, , 72 N, , (On the surface of earth), Gravitational force at height, h , , , , R, mGM, mg , 2, 2, ⎛ 3R ⎞, ⎜, ⎟, ⎝ 2 ⎠, mGM, R2, , , , 4 4, 72 32 N, 9 9, , 40. A planet has mass equal to mass of the earth but radius one fourth of radius of the earth. Then escape, velocity at the surface of this planet will be, (1) 11.2 km/s, , (2), , 22.4 km/s, , (3), , 5.6 km/s, , (4), , 44.8 km/s, , Sol. Answer (2), Ve , , GM, 11.2 km/s, R, , Vp , , GM, GM, 2, 22.4 km/s, R /4, R, , 41. With what velocity should a particle be projected so that it attains a height equal to radius of earth?, 1/2, , 1/2, , ⎛ GM ⎞, (1) ⎜, ⎟, ⎝ R ⎠, , (2), , ⎛ 8GM ⎞, ⎜, ⎟, ⎝ R ⎠, , 1/2, , 1/2, , (3), , ⎛ 2GM ⎞, ⎜, ⎟, ⎝ R ⎠, , (4), , ⎛ 4GM ⎞, ⎜, ⎟, ⎝ R ⎠, , Sol. Answer (1), Using conservation of mechanical energy at the surface of earth and at the height, h = R, P.Ei + K.Ei = P.Ej + K.Ej, , GMm 1, GMm, mv 2 , R, 2, 2R, , , 1, GMm ⎛ 1 ⎞, mv 2 , ⎜ 1⎟, 2, R ⎝ 2 ⎠, 1, GMm, mv 2 , 2, 2R, , v, , GM, R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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44, , Gravitation, , Solution of Assignment, , 42. A body of mass m is placed on earth surface which is taken from earth surface to a height of, h = 3R, then change in gravitational potential energy is, (1), , mgR, 4, , (2), , 2, mgR, 3, , (3), , 3, mgR, 4, , (4), , mgR, 2, , Sol. Answer (3), Potential energy of the body at earth surface, , , , , , GMm, R, GM, R2, , Rm, , Ui = –mgR, Potential energy of the body at height,, h = 3R , , Uf , , GMm, 4R, , mgR, 4, , Change in P.E = Uf – Ui, , , , mgR, mgR, 4, , , , 3, mgR, 4, , 43. The acceleration due to gravity on a planet A is 9 times the acceleration due to gravity on planet B. A man, jumps to a height of 2 m on the surface of A. What is the height of jump by the same person on the planet, B?, (1) 2/9 m, , (2), , 18 m, , (3), , 6m, , (4), , 2/3 m, , Sol. Answer (2), Maximum height to which man jumps on A,, , hA , , v2, 2g A, , Height to which man jumps on B,, , hB , , v2, 2gB, , hA gB, , hB g A, , ⇒, , 2, 1, , hB 9, , hB = 18 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 45, , 44. Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space, around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be, (1) 3F, , (2), , F, , (3), , F/3, , (4), , F/9, , Sol. Answer (2), Gravitational force is independent of the medium between the particles, thus force will remain unchanged., 45. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface, of the planet is equal to that at the surface of the earth. If the radius of the earth is R, then the radius of the, planet would be, (1) 2R, , (2), , 4R, , (3), , 1, R, 4, , (4), , 1, R, 2, , Sol. Answer (4), Acceleration due to gravity on the surface of a planet,, g, , , , , , , , GM, R2, G, R, , 2, , , , 4, R 3 , 3, , 4GR, 3, , Ree = Rpp, R × e = Rp × 2e, , Rp , , R, 2, , 46. A ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to, the ball?, (1) It will fall down to the earth gradually, (2) It will go very far in the space, (3) It will continue to move with the same speed along the original orbit of spacecraft, (4) It will move with the same speed, tangentially to the spacecraft, Sol. Answer (3), A ball dropped from a spacecraft revolving around the earth will have zero relative velocity with respect to the, aircraft., But with respect to the centre of the earth its speed will be equal to the speed of the aircraft i.e the orbital, speed., Thus, it will continue to move with same speed along the original orbit and force of gravitation of earth will, provide it the necessary centripetal force., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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46, , Gravitation, , Solution of Assignment, , SECTION - D, Assertion - Reason Type Questions, 1., , A : The gravitational force does not depend on the intervening medium., R : The value of G has same value anywhere in the space., , Sol. Answer (1), Property of gravitational force:, It is independent of the medium between the particles., 2., , A : The acceleration due to gravity for an object is independent from its mass., R : The value of 'g' depends on the mass of planet., , Sol. Answer (2), , g, , GM, R2, , , independent of mass of object., , M Mass of planet., 3., , A : If angular speed of the earth increases, the effective value of g will decrease at all places on earth., R : The value of 'g' at latitude is given by g = g – mR2 cos., , Sol. Answer (4), g' = g – mR2 cos is incorrect., g' = g – 2R cos2 is correct., At poles = 90°, Thus, g' = g, no effect of earth's rotation., 4., , A : The gravitational field intensity is zero everywhere inside a uniform spherical shell., R : The net force on a point mass inside a uniform spherical shell is zero everywhere., , Sol. Answer (1), Gravitational force of attraction on a point mass due to various regions of the spherical shell cancels each other, completely as their vector sum is zero., 5., , A : The value of potential energy depends on the reference taken for zero potential energy., R : The value of change in potential energy is independent from reference level., , Sol. Answer (2), Potential at a point depends on the choice of reference. Potential difference is independent of the choice of, reference., Potential energy is mass times the potential at the point., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 6., , Gravitation, , 47, , A : When a satellite is orbiting then no energy is required to keep moving in its orbit., R : The total mechanical energy of a satellite is conserved., , Sol. Answer (1), Total mechanical energy of the system is conserved, since the dissipative forces are absent or negligible., 7., , A : An astronaut in a satellite may float in the free space outside and inside the satellite., R : An astronaut in a satellite is in weightless state., , Sol. Answer (1), The force of gravitation provides the necessary centripetal force, for an astronant in a satellite, the F.B.D can, be drawn,, , Fg, , F.B.D, N Fg, , Fg N , , mv 02, r, , v 0 orbital speed , , , , , , GMm, r, , 2, , GMm, r, , 2, , GM, r, , ⎛ m ⎞ GM, N ⎜ ⎟, ⎝r ⎠ r, N , , GMm, r2, , Thus, N = 0, making the astronaut feel weightless., 8., , A : The speed of a planet is maximum at perihelion., R : The angular momentum of a planet about centre of sun is conserved., , Sol. Answer (1), , P, , rmin, , A, , Angular momentum of a planet about centre of sun is conserved,, , , Thus, mr v constant, At perihelion r is minimum,, Thus speed of planet is maximum., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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48, 9., , Gravitation, , Solution of Assignment, , A : Kepler's third law of planetary motion is valid only for inverse square forces., R : Only inverse square forces are always central., , Sol. Answer (3), T2 r3, Is valid only for inverse square forces, for a planet going in a circular orbit,, T , , 2r, v, , v is the orbital speed , , GM, r, , T 2r 2 r 3/2, GM, GM, r, Also, it is not true that,, Only inverse square forces are always central., 10. A : Kepler's law cannot be used for asteroids and comets., R : Asteroids and comets do not revolve around sun under its gravitational force., Sol. Answer (4), Kepler's laws can be used for asteroids and comets. All the 3 laws can be proved from the Newton's universal, law of gravitation., Asteroids and comets do revolve around sun., 11. A : During orbital motion of planet around the sun work done by the centripetal force is not zero at all points, on the orbit., R : Planet is revolving around the sun in elliptical orbit., Sol. Answer (1), During motion of a planet around sun, the centripetal force is not always perpendicular to the velocity of planet, in an elliptical orbit. Thus work done is not zero. Although, incase of circular orbits centripetal force is always, perpendicular to velocity., 12. A : Angular momentum of a satellite about a planet is constant., R : Gravitational force is a central force so its torque about the sun is zero., Sol. Answer (1), , r F = rF sin, , = rF sin180°, , P, F, , r, , S, , =0, , , dL, 0, Angular momentum is constant., Also, , dt, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Gravitation, , 49, , 13. A : Gravitational potential is constant everywhere inside a spherical shell., R : Gravitational field inside a spherical shell is zero everywhere., Sol. Answer (1), v, Gravitation field, I , r, , Inside a spherical shell, I = 0, V is constant everywhere., 14. A : Field created by the point mass in its surroundings is a non-uniform gravitational field., R : Since the field is E , , GM, r2, , and it is dependent on r, hence Non-uniform., , Sol. Answer (1), Field due to point mass, E , , GM, r2, , Dependent on distance r from the mass, thus non-uniform., 15. A : If the force of gravitation is inversely proportional to the distance r rather than r2 given by Newton, then, orbital velocity of the satellite around the earth is independent of r., , R:, , GMm mv 2, , r, r, , So,, , v=, , GM, , Hence independent of r., Sol. Answer (1), Force of gravitation provides the necessary centripetal force,, , GMm mv 2, , r, r, v GM, Independent of r., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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50, , Gravitation, , Solution of Assignment, , 16. A : Work done by the gravitational force is positive when the two point masses are brought from infinity to, any two points in space., R : Gravitational potential energy increases during the above process., Sol. Answer (3), Force of gravitation is attractive, thus when masses are brought from infinity to any two points in space,, displacement of masses is in the direction of force., Work done is positive., Also, Wgravity = – U = Ui – Uf, During this process potential energy decreases., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Chapter, , 9, , Mechanical Properties of Solids, Solutions, SECTION - A, Objective Type Questions, 1., , Due to addition of impurities, the modulus of elasticity, (1) Decreases, , (2), , Increases, , (3) Remains constant, , (4), , May increase or decrease, , Sol. Answer (4), It depends on the elastic property of impurities if they themselves more elastic, elasticity will increase. If they, are less elastic, elasticity will decrease., 2., , The shear strain is possible in, (1) Solids, , (2), , Liquids, , (3), , Gases, , (4), , All of these, , Sol. Answer (1), Shear strain is possible in solids only, as only solids have a definite surface., 3., , The ratio of radii of two wires of same material is 2 : 1. If these wires are stretched by equal force, the ratio, of stresses produced in them is, (1) 2 : 1, , (2), , 1:2, , (3), , 1:4, , (4), , 4:1, , Sol. Answer (3), We know,, Force, Stress =, Area, So, Stress × Area = Force, , S×A=F, , ⎧S Stress ⎫, ⎪F Force ⎪, ⎪, ⎪, ⎨, ⎬, ⎪ A Area ⎪, ⎪⎩r radius ⎪⎭, , ∵ (Since) Force applied on the wires is equal we can, relate two conditions as, S1A1 = S2A2, , S1 A2 r22, , , S2 A1 r12, , S1, 1, r2, r2, , , , 2, 2, 4, S2 (2r ), 4r, , ⎧ Where, ⎫, ⎪, ⎪, st, ⎪S1 Stress in 1 wire ⎪, ⎪, ⎪, st, ⎪ A1 Area of 1 wire ⎪, ⎪, ⎪, st, ⎨r1 Radius of 1 wire ⎬, ⎪, ⎪, nd, ⎪S2 Stress in 2 wire ⎪, ⎪, ⎪, nd, ⎪ A2 Area of 2 wire ⎪, ⎪r Radius of 2nd wire ⎪, ⎩2, ⎭, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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52, 4., , Mechanical Properties of Solids, , Solution of Assignment, , A load of 2 kg produces an extension of 1 mm in a wire of 3 m in length and 1 mm in diameter. The Young’s, modulus of wire will be, (1) 3.25 × 1010 Nm–2, , (2), , 7.48 × 1012 Nm2, , (3) 7.48 × 1010 Nm–2, , (4), , 7.48 × 10–10 Nm–2, , Sol. Answer (3), We know, Force Length, Young's Modulus, Area of cross-section elongation, , ⎧⎪F 2 10 N, A (1/2)2 106 m2 ⎫⎪, ⎨, ⎬, , L 1 10 3 m, ⎩⎪L 3 m, ⎭⎪, , F L, Y, A L, Substituting values, 20 3, , , 1, 10 6 1 10 3, 4, , 20 3 4, 3.14 109, , Y, , Y, , 7.48 × 1010 Nm–2 = Y, 5., , Young’s modulus depends upon, (1) Stress applied on material, , (2), , Strain produced in material, , (3) Temperature of material, , (4), , All of these, , Sol. Answer (3), Young's modulus is a material property and it also depends on temperature of material., 6., , The value of Young’s modulus for a perfectly rigid body is, (1) 1, , (2), , Less than 1, , (3), , Zero, , (4), , Infinite, , Sol. Answer (4), For perfectly rigid body the condition is that there should not be any elognation (L = 0) for any value of force, So from the formulae we know, , FL, Y, A L, , If we put L = 0, We get Y as , 7., , The breaking stress of aluminium is 7.5 × 107 Nm–2. The greatest length of aluminium wire that can hang vertically, without breaking is (Density of aluminium is 2.7 × 103 kg m–3), (1) 283 × 103 m, , (2), , 28.3 × 103 m, , (3), , 2.83 × 103 m, , (4), , 0.283 × 103 m, , Sol. Answer (3), Breaking stress = × g × L, Substitute values from the question, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , Breaking stress = 7.5 × 107 Nm–2, , 53, , ⎧ Density of material, ⎫, ⎪, ⎪, , g, Acceleration, due, to, gravity, ⎨, ⎬, ⎪L Length of wire that can hang without breaking⎪, ⎩, ⎭, , = 2.7 × 103 kg m–3, g = 9.8 m/s, 7.5 × 107 = 2.7 × 103 × 9.8 × L, 7.5 107, 2.7 103 9.8, , L, , 2.83 × 103 m = L, The stress strain graphs for two materials A and B are shown in figure. The graphs are drawn to the same, scale. Select the correct statement, , Q, , Strain, Material A, , (0, 0), , Q, , P, , Stress, , P, , Stress, , 8., , (0, 0), , Strain, Material B, , (1) Material A has greater Young’s Modulus, , (2), , Material A is ductile, , (3) Material B is brittle, , (4), , All of these, , (0, 0), , P, A, Strain, (A), , Q, , Stress, , Stress, , Sol. Answer (4), , (0, 0), , P, B, , Q, , Strain, (B), , Slope of stress strain curve (tan ) gives the value of young's modulus for given material, tan = y, And from the graph we can clearly see, tan A > tan B, So material A has greater young's modulus, P to Q distance in material A is greater than P to Q distance in material B, Which implies more deformation is possible in A as compared to B, Hence we can say A is ductile, B is brittle., 9., , A steel wire of diameter 2 mm has a breaking strength of 4 × 105 N. What is the breaking force of similar, steel wire of diameter 1.5 mm?, (1) 2.3 × 105 N, , (2), , 2.6 × 105 N, , (3), , 3 × 105 N, , (4), , 1.5 × 105 N, , Sol. Answer (1), We know, Force Length, elongation, Area young's modulus, , ⎧ FL, ⎫, x ⎬, ⎨, Ay, ⎩, ⎭, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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54, , Mechanical Properties of Solids, , Solution of Assignment, , ⎛ x y ⎞, F ⎜, ⎟A, ⎝ L ⎠, ⎛ x y ⎞ 2, F ⎜, ⎟d, 4⎠, ⎝ L, We can say F d 2, So we can use, , ⎧F1 4 105 N⎫, ⎪, ⎪, ⎪d1 2 mm ⎪, ⎨, ⎬, ⎪F2 ?, ⎪, ⎪d 1.5 mm ⎪, ⎩ 2, ⎭, , F1 d12, , F2 d22, , Substituting values, , 4 105, (2)2, , F2, (1.5)2, F2 = 2.3 × 105 N, 10. A steel wire is 1 m long and 1 mm2 in area of cross-section. If it takes 200 N to stretch this wire by 1 mm,, how much force will be required to stretch a wire of the same material as well as diameter from its normal, length of 10 m to a length of 1002 cm?, (1) 1000 N, , (2), , 200 N, , (3), , 400 N, , (4), , 2000 N, , Sol. Answer (3), , FL, x, AY, Since A, Y remain constant in given case, We can say, FL x, or, , ⎧F1 200 N, ⎪, ⎪x1 1 mm, ⎪, ⎨x2 10.02 m 10 m 0.02 m 20 mm, ⎪L 1 m, ⎪ 1, ⎩⎪L2 10 m, , F1L1, x1, , F2L2 x2, , Substitute the values, 200 1 1 mm, , F2 10 20 mm, , F2 = 400 N, 11. A wire 2 m in length suspended vertically stretches by 10 mm when mass of 10 kg is attached to the lower, end. The elastic potential energy gain by the wire is (take g = 10 m/s2), (1) 0.5 J, , (2), , 5J, , (3), , 50 J, , (4), , 500 J, , Sol. Answer (1), Potential energy per unit volume , , , , 1, stress strain, 2, 1 F L, , 2 A L, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 55, , So, Potential energy = potential energy per unit volume × volume, , , U , , 1 F L, , AL, 2 AL, , 1, F L, 2, , ⎧⎪F 10 10 N, ⎨, 3, ⎪⎩L 10 mm 10 10 m, , Substituting values, U , , {Volume = Length × cross-sectional area}, , 1, 10, 100 , 2, 1000, , U = 0.5 J, 12. A wire of length L and cross-sectional area A is made of material of Young’s modulus Y. The work done in, stretching the wire by an amount x is, , YAx 2, L, , (1), , YAx 2, 2L, , (2), , (3), , 2YAx 2, L, , (4), , 4YAx 2, L, , Sol. Answer (2), , W , , ∵, , 1, Fx, 2, , and Y , , F, , FL, Ax, , YAx, L, , W , , 1 ⎛ YAx ⎞, x, 2 ⎜⎝ L ⎟⎠, , W , , 1 YAx 2, 2 L, , 13. A spherical ball contracts in volume by 0.01% when subjected to a normal uniform pressure of 100 atm. The, Bulk modulus of its material is, (1) 1.01 × 1011 Nm–2, , (2), , 1.01 × 1012 Nm–2, , (3), , 1.01 × 1010 Nm–2, , (4), , 1.0 × 1013 Nm–2, , Sol. Answer (1), We know, , V, P, , V, B, , Substituting values, , , , 0.01, V, 100, 100, , 1.01 105, V, B, , {1 atm = 1.01 × 105 Pa or Nm–2}, , B = 1.01 × 1011 Nm–2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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56, , Mechanical Properties of Solids, , Solution of Assignment, , 14. What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt?, Young’s modulus of elasticity of wire = 12.5 × 1011 dyne/cm2, (1) 0.16%, , (2), , 0.32%, , (3), , , , Percentage increase, , 0.08%, , (4), , 0.12%, , Sol. Answer (1), Y , , FL, AL, , Diameter = 2.5 mm, , Area , , , , , , d 2, 4, , A, , d, , L, F, 100 , 100, L, AY, , 2.5, m, 1000, , ⎧1 dyne 0.1 N ⎫, , ⎨, ⎬, m2 ⎭, ⎩ cm2, , Y = 12.5 × 1011 dyne/cm2, , (2.5)2, 4, , F = 100 × 10 = 1000 N, , 1000 L, 3.14 (2.5)2, 12.5 1011 0.1, 4 (1000)2, , , , L, 100, L, , = 0.16%, 15. Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 2 cm,, then how much is the elongation in steel and copper wire respectively? Given, Ysteel = 20 × 1011 dyne/cm2,, Ycopper = 12 × 1011 dyne/cm2., (1) 1.25 cm; 0.75 cm, , (2), , 0.75 cm; 1.25 cm, , (3), , 1.15 cm; 0.85 cm, , (4), , 0.85 cm; 1.15 cm, , Sol. Answer (2), Let us say that elongation in copper = x, Than elongation in steel = 2 – x, We know, , FL, x, AY, , ∵ F, A, L are same only material is different, We can say, , 1, x, Y, Y2 x1, , Y1 x2, , Substituting values, 20 1011, 11, , 12 10, , , , x, 2x, , ⎧ Where, ⎪, ⎪Y2 Ysteel, ⎪, ⎨Y1 Ycopper, ⎪, ⎪ x1 elongation in copper x, ⎪⎩ x 2 x, 2, , x = 1.25 cm, So xcopper = 1.25 cm, xsteel = 0.75 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 57, , Strain, , Strain, , (4), , Stress, , (3), , Stress, , (2), , Stress, , (1), , Stress, , 16. Which of the following is the graph showing stress-strain variation for elastomers?, , Strain, , Strain, , Sol. Answer (3), , Stress, , In elastomers stress varies exponentially with strain e.g. Rubber, , Strain, 17. A steel rod has a radius 10 mm and a length of 1.0 m. A force stretches it along its length and produces a, strain of 0.32%. Young’s modulus of the steel is 2.0 × 1011 Nm–2. What is the magnitude of the force stretching, the rod?, (1) 100.5 kN, , (2), , 201 kN, , (3), , 78 kN, , (4), , 150 kN, , Sol. Answer (2), Strain = 0.32%, , We know, , , , L, 100 0.32, L, , FL, L, AY, , , , L 0.32, , L, 100, , ⎛ L ⎞, F ⎜, ⎟ A Y, ⎝ L ⎠, , ⎛ 10 ⎞, A = r2 = 3.14 ⎜, ⎟, ⎝ 1000 ⎠, , 2, , Substituting values, , F, , Y = 2 × 1011 Nm2, , 2, , 0.32, ⎛ 10 ⎞, 11, 3.14 ⎜, ⎟ 2 10, 100, ⎝ 1000 ⎠, , F = 201 kN, 18. The proportional limit of steel is 8 × 108 N/m2 and its Young’s modulus is 2 × 1011 N/m2. The maximum, elongation, a one metre long steel wire can be given without exceeding the elastic limit is, (1) 2 mm, , (2), , 4 mm, , (3), , 1 mm, , (4), , 8 mm, , Sol. Answer (2), At proportional limit, Stress strain, Stress = Y × strain, Stress = Y ×, , L, L, , Substituting values, 8 108 1, 2 1011, , {Y = Young's Modulus}, ⎧Stress 8 108 N/m2, ⎪⎪, 11, 2, ⎨Y 2 10 N/m, ⎪L 1 m, ⎪⎩, , L, , 4 mm = L, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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58, , Mechanical Properties of Solids, , Solution of Assignment, , 19. In a series combination of copper and steel wires of same length and same diameter, a force is applied at, one of their ends while the other end is kept fixed. The combined length is increased by, 2 cm. The wires will have, (1) Same stress and same strain, , (2), , Different stress and different strain, , (3) Different stress and same strain, , (4), , Same stress and different strain, , Sol. Answer (4), Stress =, , F, A, , Strain =, , L, L, , Force is same, A is same, , L is same, but due to different young's modulus, , So same stress, , (Material's different), L would be different so strain is different, , 20. Select the correct alternative(s), (1) Elastic forces are not always conservative, (2) Elastic forces are always conservative, (3) Elastic forces are conservative only when Hooke’s law is obeyed, (4) Elastic forces are not conservative, Sol. Answer (1), Since at every value of force material is not able to gain its shape. Therefore elastic forces are not always, conservative., 21. A metallic rod of length l and cross-sectional area A is made of a material of Young’s modulus Y. If the rod, is elongated by an amount y, then the work done is proportional to, (1) y, , (2), , 1, y, , (3), , y2, , (4), , 1, y2, , Sol. Answer (3), Work done = energy stored, W=, , 1, × force × elongation, 2, , W=, , 1 L, , A Y L, 2 L, , W=, , 1 AY, L2, 2 L, , L, ⎧, AY, ⎨Force , L, ⎩, , W L2, 22. The Poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in, the cross-sectional area by 4%. The percentage increase in the length is, (1) 1%, , (2), , 2%, , (3), , 2.5%, , (4), , 4%, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 59, , Sol. Answer (4), Lateral strain, , Longitudinal strain, , ∵ A r2, , r / r, 0.5, l / l, , So, , 4, r, 2, 100, r, , Substitute r/r = 2/100, l, 4, , l, 100, , % increase , , A 2 r, , A, r, , 2, r, , 100, r, , l, 100 4%, l, , Load, , 23. The load versus elongation graph for four wires of same length and the same material is shown in figure. The, thinnest wire is represented by line, , O, (1) OC, , (2), , OD, , D, , C, , B, A, , Elongation, (3), , OA, , (4), , OB, , Sol. Answer (3), For the same load wire with maximum elongation has minimum cross-section area, , FL, x, AY, , F, L, Y are fixed so, , D, , Load (F), , As, , 1, x, A, , O, , OA is the thinnest., , C, , B, A, , Elongation (x), , 24. A rod of uniform cross-sectional area A and length L has a weight W. It is suspended vertically from a fixed, support. If Young’s modulus for rod is Y, then elongation produced in rod is, , W, (1), , WL, YA, , (2), , WL, 2YA, , (3), , WL, 4YA, , (4), , 3WL, 4YA, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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60, , Mechanical Properties of Solids, , Solution of Assignment, , Sol. Answer (2), Center of mass is at, to a, , L, distance from top so it can be assumed for easy calculation that W weight is hanged, 2, , L, length string, 2, , L/2, , FL, L, Now use, AY, L , , L, W, , W L, 2 AY, , 25. If in case A, elongation in wire of length L is l, then for same wire elongation in case B will be, , L, W, , W, , B, W, A, , (1) 4l, , (2), , 2l, , (3), , l, , (4), , l/2, , Sol. Answer (3), Since tension in both cases is same and all other parametrs (Y, A, L) are also same, Elongation will be same in both cases., , Load, , 26. In the given figure, if the dimensions of the two wires are same but materials are different, then Young’s modulus, is, , A, B, Extension, , (1) More for A than B, , (2), , More for B than A, , (3), , Equal for A and B, , (4), , None of these, , Sol. Answer (1), At same value of load, A has less elongation than B, FL, L, AY, , 1, L, So, Y, , A, , ...(1), ⎧∵ L, A are same, ⎫, ⎨, ⎬, ⎩ F Load is also taken same ⎭, , B, , Load, L, EA, , ...(2), , EB, Extension, , Using conditions (1) and (2), We can say, YA > YB, , {Young's modulus of A greater than B}, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 61, , 27. If the Bulk modulus of lead is 8.0 × 109 N/m2 and the initial density of the lead is 11.4 g/cc, then under the, pressure of 2.0 × 108 N/m2, the density of the lead is, (1) 11.3 g/cc, , (2), , 11.5 g/cc, , (3), , 11.6 g/cc, , (4), , 11.7 g/cc, , Sol. Answer (4), We know,, ⎧ Where, ⎪, 8, 2, ⎪P 2 10 N/m, ⎪, 9, 2, ⎨B 8 10 N/m, ⎪ 11.4 g/cc, ⎪ 1, ⎪2 ?, ⎩, , ⎡1, 1⎤, P, ⎢ ⎥, B, ⎣ 2 1 ⎦, , Substitute value's, ⎡1, 1 ⎤, 2 108, , , , ⎢, ⎥, 8 109, ⎣ 2 114 ⎦, , After solving, we get, 2 = 11.7 g/cc, 28. Two wires A and B of same material have radii in the ratio 2 : 1 and lengths in the ratio 4 : 1. The ratio of, the normal forces required to produce the same change in the lengths of these two wires is, (1) 1 : 1, , (2), , 2:1, , (3), , 1:2, , (4), , 1:4, , Sol. Answer (1), From, , FL, x, AY, , We using F , , {∵ x, Y same}, L, L, 2, A r, , F, L r2 ⎛ L ⎞ ⎛r ⎞, So 1 21 2 ⎜ 1 ⎟ ⎜ 2 ⎟, F2 r1 L2 ⎝ L2 ⎠ ⎝ r1 ⎠, , 2, , Substitute the ratio's, We get, , F1 1, , F2 1, , or F1 : F2 : : 1 : 1, , 29. For a given material, the Young’s modulus is 2.4 times its modulus of rigidity. Its Poisson’s ratio is, (1) 0.2, , (2), , 0.4, , (3), , 1.2, , (4), , 2.4, , Sol. Answer (1), Y = 3B [1 – 2], , ...(1), , Y = 2.4B (given), , ...(2), , Using both (1) and (2), 2.4 B, 1 2, 3B, , ⎧ Where, ⎪Y Young's modulus, ⎪, ⎨, ⎪B Bulk's modulus, ⎪⎩ Poisson's ratio, , 0.2 = , 30. When the temperature of a gas is 20°C and pressure is changed from P1 = 1.01 × 105 Pa to P2 = 1.165 ×, 105 Pa, then the volume changes by 10%. The Bulk modulus is, (1) 1.55 × 105 Pa, , (2), , 1.01 × 105 Pa, , (3), , 1.4 × 105 Pa, , (4), , 0.115 × 105 Pa, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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62, , Mechanical Properties of Solids, , Solution of Assignment, , Sol. Answer (1), , V P, , V, B, , ⎧V 10% of V (∵ Pressure increases volume must, ⎪, decreases by 10% so we will use a +ve sign), ⎪, ⎪⎪If V 100 cc, ⎨⇒ V 10 cc, ⎪, ⎪P P2 P1, ⎪, 1.165 105 1.01 105, ⎪⎩, , Substituting the values, , 10 (1.165 106 1.01 106 ), , 100, B, 1 .155 105, , 10, B, , B = 1.55 × 105 Pa, 31. The stress versus strain graph for wires of two materials A and B are as shown in the figure. If YA and YB, are the Young’s moduli of the materials, then, , Stress, , y, A, B, 60° 30°, , x, , Strain, , (1) YB = 2YA, , (2), , YA = 3YB, , (3), , YB = 3YA, , (1) Elastic materials only, , (2), , Plastic materials only, , (3) Elastomers only, , (4), , All of these, , (4), , YA = YB, , Sol. Answer (2), Y = tan , YA = tan 60°,, , YB = tan 30°, , = 3, , 1/ 3, , YA = 3YB, 32. Hooke’s law is applicable for, , Sol. Answer (1), Hooke's law is applicable only for elastic materials as only they follow the stress-strain proportionality., 33. The length of wire, when M1 is hung from it, is l1 and is l2 with both M1 and M2 hanging. The natural length of, wire is, , M2, M1, , (1), , M1, (l1 l 2 ) l1, M2, , (2), , M 2 l1 M1l 2, M1 M 2, , (3), , l1 l 2, 2, , (4), , l1l 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 63, , Sol. Answer (1), Let the natural length of wire be = l, When only M1 hanging, Using l , , FL, AY, , (l1 l ) , , M1 g l, AY, , ...(1), , When both M1, M2 hanging, , (l 2 l ) , , (M1 M2 ) g l, ...(2), AY, , Dividing (1) by (2), l1 l, M1, , l 2 l M1 M2, , Solving this we get, , l, , M1, (l1 l 2 ) l1, M2, , 34. When a load of 10 kg is suspended on a metallic wire, its length increase by 2 mm. The force constant of, the wire is, (1) 3 104 N/m, , (2), , 2.5 103 N/m, , (3), , 5 104 N/m, , (4), , 7.5 103 N/m, , Sol. Answer (3), Force constant (K) , , Force, F, , Elongation x, , ⎧F 10 kg 100 N, ⎨, ⎩ x 2 mm 0.002 m, , Substituting values, K, , 100, 5 10 4 N/m, 0.002, , 35. A rod of length l and radius r is held between two rigid walls so that it is not allowed to expand. If its, temperature is increased, then the force developed in it is proportional to, (1) L, , (2), , 1/L, , (3), , r2, , (4), , r–2, , Sol. Answer (3), (L) Thermal expansion = L Q, Where L = Length original, , radius = r, , = Coeffcient of linear expansion, Q = Change in temperature, , L, , Or we can say, L L, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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64, , Mechanical Properties of Solids, , Solution of Assignment, , And force required to produce similar elongation can be calculated by, , F AY , , So F r2 , , L, L, , ∵ Y is constant , , L, L, , Also L L, So F only proportional to r2, 36. If the temperature of a wire of length 2 m and area of cross-section 1 cm2 is increased from 0° C to 80°C, and is not allowed to increase in length, then force required for it is {Y = 1010 N/m2, = 10–6/°C}, (1) 80 N, , (2), , 160 N, , (3), , 400 N, , (4), , 120 N, , Sol. Answer (1), Thermal expansion would be = L T, Where L = original length, = coefficient of linear expansion, T = Change in temperature, So substituting values, L = 2 × 10–6 × 80, L = 1.6 × 10–4 m, FL, AY, , Now L , , L AY, F, L, , Substitute values, 1.6 10 4 1010 1, F, 2 10000, , 80 N = F, 37. Energy stored per unit volume in a stretched wire having Young’s modulus Y and stress ‘S’ is, (1), , YS, 2, , (2), , S 2Y, 2, , (3), , S2, 2Y, , (4), , S, 2Y, , Sol. Answer (3), U , , 1, stress strain, 2, , , , 1, stress, stress , 2, Y, , , , 1 S2, , 2 Y, , {∵ Stress = Y strain}, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 65, , 38. A wire suspended vertically from one end is stretched by attaching a weight 200 N to the lower end. The weight, stretches the wire by 1 mm. The elastic potential energy gained by the wire is, (1) 0.1 J, , (2), , 0.2 J, , (3), , 0.4 J, , (4), , 10 J, , Sol. Answer (1), 1, × force × elongation, 2, , Elastic potential energy =, , , 1, 1, 200 , 0.1 J, 2, 1000, , 39. Work done by restoring force in a string within elastic limit is –10 J. Maximum amount of heat produced in, the string is, (1) 10 J, , (2), , 20 J, , (3), , 5J, , (4), , 15 J, , Sol. Answer (1), Within elastic limit there is no loss of energy in deforming because no permanent deformation., We can say, Work done by external force = heat produced, Or (–)ve of work done by restoring force = heat produced, –1 × –10 J = H, 10 J = H, 40. The work done per unit volume to stretch the length of area of cross-section 2 mm2 by 2% will be, [Y = 8 1010 N/m2], (1) 40 MJ/m3, , (2), , 16 MJ/m3, , (3), , 64 MJ/m3, , (4), , 32 MJ/m3, , Sol. Answer (4), Work done = Force × elongation, W = F x...., W=, , AY ⎫, ⎧, ⎨F x , ⎬, L ⎭, ⎩, , AY, × x2, L, , Multiply and divide by L, We get, W , , Volume Y, L2, , x 2, , Cross multiply (L A) volume, , W, ⎛ x ⎞, Y ⎜, ⎟, LA, ⎝ L ⎠, , , 2, , Work done per unit volume, Substitute values, , ⎛ 2 ⎞, 8 1010 ⎜, ⎟, ⎝ 100 ⎠, , 2, , = 32 MJ/m3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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66, , Mechanical Properties of Solids, , Solution of Assignment, , 41. Which of the following affects the elasticity of a substance?, (1) Change in temperature, , (2), , Impurity in substance, , (3) Hammering, , (4), , All of these, , Sol. Answer (4), Elasticity is hampered by change in temperature as it changes the structure of grains of the material impurity, if elastic increases the elasticity if plastic increases plasticity., By hammering also grain shape gets changes and effects elasticity., 42. Select the wrong definition, (1) Deforming Force – force that changes configuration of body, (2) Elasticity – property of regaining original configuration, (3) Plastic body – which can be easily melted, (4) Elastic limit – beyond which property of elasticity vanishes, Sol. Answer (3), Plastic body is defined as a body which cannot regain its shape after deforming force is removed., 43. Figure shows graph between stress and strain for a uniform wire at two different temperatures. Then, , Stress, , T2, T1, , (0, 0), (1) T1 > T2, , (2), , T2 > T1, , Strain, (3), , T1 = T2, , (4), , None of these, , Sol. Answer (1), , Stress, , From the graph we can see young's modulus, is less for T1 as compared to T2, (Y = slope of stress-strain curve), As T increases Y decreases, , (0, 0), , So T1 > T2, , T2, T1, , Strain, , Stress, , A, , Stress, , 44. Two different types of rubber are found to have the stress-strain curves as shown. Then, , (0, 0), , Strain, , (0, 0), , B, Strain, , (1) A is suitable for shock absorber, , (2), , B is suitable for shock absorber, , (3) B is suitable for car tyres, , (4), , None of these, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 67, , Sol. Answer (2), One with higher hysterysis loss suitable for shock absorber because high hysterysis loss will lead to dampen, shocks in a easy manner., One with lower hysterysis loss suitable for types because it will have low relaxation time., Area between loop gives amount of hysterysis loss. More area more loss, less area less loss., Therefore, B is suitable for shock absorber and A for types., 45. The ratio of adiabatic to isothermal elasticity of a diatomic gas is, (1) 1.67, , (2), , 1.4, , (3), , 1.33, , (4), , 1.27, , Sol. Answer (2), Kisothermal = Pressure of gas (P), Kadiabatic = × pressure of gas ( P), Ratio , , P, , P, , of diatomic gas , , 7, 1.4, 5, , 46. A uniform cubical block is subjected to volumetric compression, which decreases its each side by 2%. The, Bulk strain produced in it is, (1) 0.03, , (2), , 0.02, , (3), , 0.06, , (4), , 0.12, , Sol. Answer (3), Volume = (side)3, v = (a)3, So, , v 3a, , v, a, , , , v, 3 2, v, , a, ⎧, ⎫, 2%⎬, ⎨given, a, ⎩, ⎭, Side decreases so we used (–)ve sign, , = –6%, We know, , v, P, , v, B, Substituting value of v/v, , , 6, P, , 100, B, , So bulk strain produced is 0.06, 47. When a rubber ball is taken to the bottom of a sea of depth 1400 m, its volume decreases by 2%. The Bulk, modulus of rubber ball is [density of water is 1 g cc and g = 10 m/s2], (1) 7 108 N/m2, , (2), , 6 108 N/m2, , (3), , 14 108 N/m2, , (4), , 9 108 N/m2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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68, , Mechanical Properties of Solids, , Solution of Assignment, , Sol. Answer (1), Pressure at the bottom of sea = wgh = 1000 kg/m3 × 10 m/s2 × 1400 m = 14000000 N/m2, Also we know, , v, P, , v, B, , 2 ⎫, ⎧ v, , ⎨, ⎬, 100 ⎭, ⎩v, , 2, 14000000, , 100, B, , B = 7 × 108 N/m2, 48. A spherical ball contracts in volume by 0.02%, when subjected to a normal uniform pressure of, 50 atmosphere. The Bulk modulus of its material is, (1) 1 × 1011 N/m2, , (2), , 2 × 1010 N/m2, , 2.5 × 1010 N/m2, , (3), , (4), , 1 × 1013 N/m2, , Sol. Answer (3), , v, P, , v, B, , ⎧ v 0.02 ⎫, , ⎨, ⎬, 100 ⎭, ⎩v, , P = 50 atm = 50 × 1.01 × 105 Pa or N/m2, 100, 0.02, , So B = 50 × 1.01 × 105 ×, = 2.5 × 1010 N/m2, , 49. A steel plate of face area 1 cm2 and thickness 4 cm is fixed rigidly at the lower surface. A tangential force, F = 10 kN is applied on the upper surface as shown in the figure. The lateral displacement x of upper surface, w.r.t. the lower surface is (Modulus of rigidity for steel is 8 × 1011 N/m2), , F, , x, , x, 4 cm, , (1) 5 × 10–5 m, , (2), , 5 × 10–6 m, , (3), , 2.5 × 10–3 m, , (4), , 2.5 × 10–4 m, , Sol. Answer (2), Modulus of rigidity (G) , , Force Length, FL, , Area Lateral displacement A x, , F = 10 kN = 10 × 103 N, L = 4 cm = 0.04 m, A = 1 cm2 = 1 × 10–4 m2, G = 8 × 1011 N/m2, Substituting values, 8 1011 , x , , x, , 10 103 0.04, 1 10 4 x, , 10 103 0.04, 1 10 4 8 1011, , F, , A = 1 cm2, h = 4 cm, , 5 10 6 m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 69, , 50. The Poisson's ratio cannot have a value of, (1) 0.7, , (2), , 0.2, , (3), , 0.1, , (4), , 0.5, , Sol. Answer (1), Poisson's ratios value can't be practically more than 1/2 so only value above 1/2 is 0.7, 51. A material has Poisson’s ratio 0.5. If a uniform rod of it suffers a longitudinal strain of 3 × 10–3, what will be, percentage increase in volume?, (1) 2%, , (2), , 3%, , (3), , 5%, , (4), , 0%, , Sol. Answer (4), , Lateral strain, 0.5, Longitudinal strain, r / r 1, , l / l, 2, , Now volume area × length, v r2 L, , 2r l, , r, l, , v 2r L, , , v, r, L, , Magnitute wise both are equal but sign's would, , Substituting value of, , L, L, , v, 0, v, No change in volume, , be different as both quantities cannot increase, , 52. When a uniform metallic wire is stretched the lateral strain produced in it is . If and Y are the Poisson’s, ratio and Young’s modulus for wire, then elastic potential energy density of wire is, , Y2, 2, , (1), , (2), , Y2, 2, , 2, , (3), , Y2, 2, , (4), , Y 2, 2, , Sol. Answer (2), = Strain (lateral), = Poisson's ratio, Y = Young's modulus, Elastic potential energy density =, , Also, , 1, × Y × (strain longitudinal)2, 2, , ...(1), , Lateral strain, Poisson's ratio, Longitudinal strain, , , Longitudinal strain, , Longitudinal strain , , , , , Substituting the value in equation (1), 2, , 1, 1Y 2, ⎛⎞, E.P.E = Y ⎜ ⎟ , 2, 22, ⎝⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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70, , Mechanical Properties of Solids, , Solution of Assignment, , 53. The substances having very short plastic region are, (1) Ductile, , (2), , Brittle, , (3), , Malleable, , (4), , All of these, , Sol. Answer (2), Substances with short plastic region are brittle because less amount of permanent deformation could be done, in them., 54. For an elastic material, (1) Y > , , (2), , Y<, , (3), , Y = 1, , (4), , Y=, , Sol. Answer (1), Y = Young's modulus, = modulus of rigidity, We have a formulae, Y = 2[1 + ], , ⎡ Where poisson's ratio practical⎤, ⎢ value of lies between 0 to 0.5 ⎥, ⎣, ⎦, , 0 < 0.5, Using maximum value of , Y = 3, Y>, 55. Correct pair is, , (1) Change in shape – Longitudinal strain, , (2), , Change in volume – Shear strain, , (3) Change in length – Bulk strain, , (4), , Reciprocal of Bulk modulus – Compressibility, , Sol. Answer (4), B = bulk modulus, and, , 1, is defined as compressibility, B, , SECTION - B, Objective Type Questions, 1., , Two wires of equal length and cross-sectional area are suspended as shown in figure. Their Young's modulii, are Y1 and Y2 respectively. The equivalent Young’s modulii will be, , (1) Y1 + Y2, , (2), , Y1 Y2, Y1 Y2, , (3), , Y1 Y2, 2, , (4), , Y1 Y2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 71, , Sol. Answer (3), Forces acting on both wires would be equal to T1 and T2 respectively by free body diagram, Let equivalent force constant of wire = K, ⎡Crossectional area will double ⎤, ⎢ when both wires taken together ⎥, ⎣, ⎦, , K1 + K2 = K, , Y1, L, A, , Y2, L, A, , AY1 AY2 2 AY, Y Y2, , , ⇒ Y 1, L, L, L, 2, , 2., , T2, , T1, , Mg, , A uniform rod of length L has a mass per unit length and area of cross-section A. If the Young’s modulus, of the rod is Y. Then elongation in the rod due to its own weight is, (1), , 2gL2, AY, , gL2, AY, , (2), , (3), , gL2, 4 AY, , (4), , gL2, 2 AY, , Sol. Answer (4), Total mass can be assumed to be concentrated at center of mass at distance, , L, from top, 2, , M, , L, , L, , M = L, , x , 3., , FL / 2 1 L2 g, , AY, 2 AY, , A,Y, , When a small mass m is suspended at lower end of an elastic wire having upper end fixed with ceiling. There, is loss in gravitational potential energy, let it be x, due to extension of wire, mark correct option, (1) The lost energy can be recovered, (3) Only, , x, amount of energy is recoverable, 2, , (2), , The lost energy is irrecoverable, , (4), , Only, , x, amount of energy is recoverable, 3, , Sol. Answer (3), U (loss in gravitational potential energy) = mg × l, U = x (given), , ⎧ Where, ⎪, ⎨m mass suspended, ⎪l elongation in wire, ⎩, , So x = mg × l, , Elastic potential energy gained , , So only, , 1, Force Elongation, 2, , , , 1, Mg l, 2, , , , 1, Mg l, 2, , , , 1, x, 2, , ∵ Mg l x , , x, amount of energy is recoverable which is stored as elastic potential energy in wire., 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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72, 4., , Mechanical Properties of Solids, , Solution of Assignment, , A solid sphere of radius R made of a material of bulk modulus B surrounded by a liquid in a cylindrical, container. A massless piston of area A floats on the surface of the liquid. Find the fractional decrease in the, , ⎛ R ⎞, radius of the sphere ⎜, ⎟ when a mass M is placed on the piston to compress the liquid, ⎝ R ⎠, Mg, AB, , (1), , Mg, 4 AB, , (2), , (3), , Mg, 3 AB, , (4), , Mg, 2 AB, , Sol. Answer (3), , A, , Pressure increased to weight M, P, , Mg, , Force Mg, , Area, A, , ...(1), , Piston, , R, , And we know,, , V P, , V, B, , V Mg, , V, AB, , Mg ⎤, ⎡, ⎢∵ P A ⎥, ⎣, ⎦, , Volume of a sphere is V , , ...(2), , 4 3 ⎧ Where ⎫, r ⎨, ⎬, 3, ⎩r is radius ⎭, , V, ⎧, ⎪ Where V Fractional decrease in volume, ⎪, P Pressure increased, ⎨, ⎪, B Bulk modulus, ⎪, ⎩, , V 3R, , V, R, , , , Using (2), , , Mg, R, , 3 AB, R, , [(–)ive sign indicates decrease], , Fractional decrease in radius is, 5., , Mg, 3 AB, , A sphere contracts in volume by 0.01% when taken to the bottom of sea 1 km deep. Find Bulk modulus of, the material of sphere, (1) 9.8 × 106 N/m2, , 1.2 × 1010 N/m2, , (2), , (3), , 9.8 × 1010 N/m2, , (4), , 9.8 × 1011 N/m2, , Sol. Answer (3), Pressure at bottom of sea = wgh, w = 1000 kg/m3 = 1 g/cc,, , g = 9.8 m/s2, h = 1000 m, , P = 103 × 9.8 × 1000 N/m2, Now, , V P, , V, B, , ⎧ V 0.01, , (given), ⎨, 100, ⎩ V, , 0.01 103 9.8 1000, , 100, B, , B = 9.8 × 1010 N/m2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 6., , Mechanical Properties of Solids, , 73, , A mild steel wire of length 2l meter cross-sectional area A m2 is fixed horizontally between two pillars. A small, mass m kg is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then, depression at the mid point of wire will be, 1/3, , 1/3, , 1/3, , ⎛ Mg ⎞, (1) ⎜, ⎟, ⎝ YA ⎠, , (2), , ⎛ Mg ⎞, ⎜, ⎟, ⎝ lA ⎠, , (3), , ⎛ Mgl 3 ⎞, ⎜, ⎟, ⎝ YA ⎠, , (4), , Mg, 2YA, , Sol. Answer (3), Let OC = x [depression], , O, , and be small angle, , L, , A, , ∵ x is small, , x, , L (Extension in OB part of wire) = BC – OB, , T, , BC = (L2 + x2)1/2 and OB = L, , B, , , , C T, , L = {(L2 + x2)1/2 – L}, We know F , So F , , mg, , YA L, L, , , , , , YA 2, (L x 2 )1/2 L, L, , 1/2, ⎧⎛, ⎫, x2 ⎞, ⎪, ⎪, YA ⎨⎜ 1 2 ⎟ 1⎬, ⎜, ⎟, L ⎠, ⎪⎭, ⎩⎪⎝, , ⎧Using binomial theorem, ⎪ 2, ⎨ x, ⎪ 2 1, ⎩ L, , ⎧⎪, ⎫⎪, x2, YA ⎨1 2 1⎬, ⎪⎭, ⎩⎪ 2L, , F, , YAx 2, , 1/2, , ⎛, x2 ⎞, So ⎜ 1 2 ⎟, ⎜, L ⎟⎠, ⎝, , 2, , 2L, , 1, , x2, 2L2, , Tension in each part of wire will be equal to 'F', , T sin 2T sin , , By vertical equilibrium, , T, , Mg = 2T sin, {If is small. So sin }, , = 2T , , ∵, , T, , , T cos , , T cos , , T =F, Mg 2 , , Mg 2 , , YAx 2, 2L2, , YAx 2, 2, , 2L, , , , , , x, L, , ⎧in OBC, ⎪, ⎨x, ⎪⎩ L tan , , Mg, , ∵ is small, tan sin , , We get,, ⎛ MgL3, ⎜⎜, ⎝ YA, , 1/3, , ⎞, ⎟⎟, ⎠, , x, , So, , x, , L, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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74, 7., , Mechanical Properties of Solids, , Solution of Assignment, , A rigid bar of mass 15 kg is supported symmetrically by three wire each of 2 m long. These at each end are, of copper and middle one is of steel. Young’s modulus of elasticity for copper and steel are 110 × 109 N/m2, and 190 × 109 N/m2 respectively. If each wire is to have same tension, ratio of their diameters will be, (1), , 11, 19, , (2), , 19, 11, , (3), , 30, 11, , 11, 30, , (4), , Sol. Answer (2), Tension is same (given), , 2 m Copper, , From free body diagram, , Steel, , Copper, , 3T = 150 N, , 15 kg, , T = 50 N, Since the bar has to be supported symmetrically, , T, , Therefore extension in each wire will be same, FL, AY, Compare 1 copper wire with another steel wire, , T, , T, , We know x , , FL, FL, , ACYC ASYS, , , , 150 N, ⎧ Where,, ⎪, ⎪ AC Area of copper wire, ⎪, ⎨YC Young's modulus copper, ⎪ A Area of steel wire, ⎪ S, ⎪⎩YS Young's modulus steel, , AS YC, , AC YS, , Substitutuing value of YC and Ys, , dS2, 110 109, , 9, , 4 dC2 190 10, 4, ⎧dS diameter of steel wire, ⎨, ⎩dC diameter of copper wire, , dS, 11, , 19, dC, dC, 19, , dS, 11, 8., , A solid cube of copper of edge 10 cm subjected to a hydraulic pressure of 7 × 106 pascal. If Bulk modulus, of copper is 140 GPa, then contraction in its volume will be, (1) 5 × 10–8 m3, , (2), , 4 × 10–8 m3, , (3), , 2 × 10–8 m3, , (4), , 108 m3, , Sol. Answer (1), Initial volume V = (side)3 = (10 × 10–2)3, P = 7 × 106 Pa, B = 140 × 109 Pa, We know, , V P, , V, B, V, 10 3, , , , {–V = Contraction in volume}, , 7 106, 140 109, , –V = 5 × 10–8 m3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 9., , Mechanical Properties of Solids, , 75, , Three bars having length l, 2l and 3l and area of cross-section A, 2A and 3A are joined rigidly end to end., Compound rod is subjected to a stretching force F. The increase in length of rod is (Young’s modulus of material, is Y and bars are massless), (1), , 13 Fl, 2 AY, , (2), , Fl, AY, , (3), , 9Fl, AY, , (4), , 3Fl, AY, , Sol. Answer (4), If extension of rod = x, x = x1 + x2 + x3, x1 , , So x , , Where x1, x2, x3 are individual extensions in rod 1, 2, 3, , Fl, 2Fl, 3Fl, , x2 , , x3 , AY, 2 AY, 3 AY, , 3Fl, AY, , 10. A force F is applied along a rod of transverse sectional area A. The normal stress to a section PQ inclined, to transverse section is, P, , F, , Q, (1), , F sin , A, , (2), , F, cos , A, , (3), , F, sin2, 2A, , (4), , F, cos2 , A, , Sol. Answer (4), , P, , F1 = F cos, , P, , , O, , F, , , , , , A, , F, , Q, O, , Q, , Fnormal, F, 1, Area, Area, , Crossectional area PO = A, , , , F cos , A / cos , , Crossectional area of PQ , , , , F, cos2 , A, , Stress , , PO, A, , cos cos , , 11. The strain energy stored in a body of volume V due to shear strain is (shear modulus is ), (1), , 2V, 2, , (2), , V 2, 2, , (3), , 2V, , , (4), , 1 2, V, 2, , Sol. Answer (4), , Shear modulus , , , , Shear stress, Shear stress, , Shear stress, , , = Shear stress, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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76, , Mechanical Properties of Solids, , Strain energy per unit volume =, , , , Strain energy 1, , Volume, 2, , Strain energy =, , Solution of Assignment, , 1, × shear stress × shear strain, 2, (Cross multiply volume), , 1 2, V, 2, , 12. A vertical hanging bar of length l and mass m per unit length carries a load of mass M at lower end, its upper, end is clamped to a rigid support. The tensile stress a distance x from support is (A area of cross-section, of bar), (1), , Mg mg (l x ), A, , Mg, A, , (2), , (3), , Mg mgl, A, , (4), , (M m ) gx, Al, , Sol. Answer (1), Tensile stress =, , Tension at point, Area, , x, , Tension at distance x from top would be the amount of force, acting due to all the weight below it, , l, , = Mass per unit length of rod × length of rod + Mg, = m × (l – x) g + Mg, So Tensile stress , , M, , m(l x )g Mg, A, , 13. A metal wire having Poisson’s ratio 1/4 and Young’s modulus 8 1010 N/m2 is stretched by a force, which, produces a lateral strain of 0.02% in it. The elastic potential energy stored per unit volume in wire is, [in J/m3], (1) 2.56 × 104, , 1.78 × 106, , (2), , (3), , 3.72 × 102, , (4), , 2.18 × 105, , Sol. Answer (1), Lateral strain, Poisson's ratio, Longitudinal strain, , ⎧Y (Young's modulus), ⎪, 10, (given), ⎪⎪ 8 10, ⎨, 1, (given), ⎪Poission's ratio , 4, ⎪, ⎪⎩Lateral strain 0.02% (given), , 0.02 / 100 1, , l / l, 4, l 0.08, , l, 100, , U (Elastic potential energy per unit volume =, , 1, × Y × (Longitudinal strain)), 2, , Substituting values, , U , , 1, ⎛ 0.08 ⎞, 8 1010 ⎜, ⎟, 2, ⎝ 100 ⎠, , 2, , U = 2.56 × 104 J/m3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 77, , 14. An ideal gas has adiabatic exponent . It expands according to the law P = V, where is constant. For, this process, the Bulk modulus of the gas is, (1) P, , (2), , P, , , (3), , P, , (4), , (1 – )P, , Sol. Answer (1), , B, , P, V / V, , B, , V, V, V / V, , ⎧ Where,, ⎪P increase in pressure, ⎪, ⎪⎪, V {∵ P V }, ⎨, ⎪ B Bulk modulus, ⎪V Change in volume, ⎪, ⎪⎩ V Initial volume, , And V = P, So B = P, , 15. Which of the following curve represents the correctly distribution of elongation (y) along heavy rod under its, own weight L length of rod, x distance of point from lower end?, , y, , y, , y, , (1), , (2), , (3), , x, , L, , y, , L, , (4), , x, , L, , x, , L, , x, , Sol. Answer (2), , ⎧ Where,, ⎪x Elongation, ⎪, ⎪ Density of rod, ⎪, ⎨Y Young's modulus, ⎪L Length, ⎪, ⎪g Acceleration due to gravity, ⎪, ⎩ x Distance of point from lower end, , For elongation of rod under its own weight, We know x , , gx 2, 2Y, , We can clearly see that elongation (x2), So graph of x vsx should be a upward parabola., , Elongation, , y, , x, , Length, , 16. The length of a metal wire is l1, when tension in it is T1 and l2 when its tension is T2. The natural length of, the wire is, (1), , l1 l 2, , (2), , l1T2 l 2T1, T2 T1, , (3), , l 2T2 l1T1, T1 T2, , (4), , l1 l 2, 2, , Sol. Answer (2), Let natural length of wire = l, Case I : when tension in wire is T1, , l1 l , , T1l, AY, , ...(1), , FL ⎫, ⎧, ⎨l , ⎬, AY ⎭, ⎩, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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78, , Mechanical Properties of Solids, , Solution of Assignment, , Case II : when tension in wire is T2, l2 l , , T2 l, AY, , ...(2), , Dividing (2) by (1), l 2 l T2, , l1 l T1, , Solving this we get, l, , l1T2 l 2T1, T2 T1, , 17. A wire can sustain a weight of 15 kg. If it cut into four equal parts, then each part can sustain a weight, (1) 5 kg, , (2), , 45 kg, , (3), , 15 kg, , (4), , 30 kg, , Sol. Answer (3), , Stress , , F, A, , So Stress , , 1, A, , Since, we are not reducing the crossectional area of the wire. Therefore each part can still sustain same force, i.e, 15 kg weight., 18. Two wire A and B are stretched by same force. If, for A and B, YA : YB = 1 : 2, rA : rB = 3 : 1 and lA : lB = 4 : 1,, , ⎛ l A ⎞, then ratio of their extension ⎜, ⎟ will be, ⎝ lB ⎠, (1) 10 : 13, , (2), , 11 : 7, , (3), , 8:9, , (4), , 6:5, , Sol. Answer (3), , x , , FL, AY, , For wire A, LA , , For wire B, F LA, , rA2, , LB , , ...(1), , YA, , F LB, , rB2 YB, , ...(2), , Divide (1) by (2), 2, , ⎛r ⎞ Y, F L, L, LA, r 2 Y, 2 A B B A ⎜ B ⎟ B, LB ⎝ rA ⎠ YA, LB, rA YA F LB, Substituting the value of ratio's, 2, , LA 4 ⎛ 1 ⎞, 2 8, ⎜ ⎟ , 1 ⎝3⎠, 1 9, LB, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 79, , 19. A bar is subjected to axial forces as shown. If E is the modulus of elasticity of the bar and A is its crosssection area. Its elongation will be, 3F, , 2F, l, , (1), , Fl, AE, , (2), , 2Fl, AE, , F, l, , 3Fl, AE, , (3), , (4), , 4Fl, AE, , Sol. Answer (4), st, , I part, , nd, , II part, , 3F, , 2F, l, , Elongation in Ist part, x1 , , (3 2)FL, AE, , , , 5FL, AE, , l, Elongation in IInd part, , Net force L, AY, , , , F, , x2 , , (F 2F )L, FL, , AE, AE, , So net elongation, x = x1 + x2, , , , 5FL FL 4FL, , , AE AE, AE, , 20. A metal ring of initial radius r and cross-sectional area A is fitted onto a wooden disc of radius, R > r. If Young’s modulus of metal is Y then tension in the ring is, (1), , AYR, r, , (2), , AY (R r ), r, , (3), , Y ⎛R r⎞, ⎜, ⎟, A⎝ r ⎠, , (4), , Yr, AR, , Sol. Answer (2), r – radius of metal ring, R – radius of wooden disc, Given, R>r, So 2R > 2r, To get the metal ring fitted on wooden disc the circumfrence should be increased by (2R – 2r) of metal ring, L = 2(R – r), F = tension developed in ring, , , 2(R r ) , , T (2r ), AY, , FL ⎞, ⎛, ⎜ L AY ⎟, ⎝, ⎠, , AY (R r ), T, r, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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80, , Mechanical Properties of Solids, , Solution of Assignment, , 21. The normal density of gold is and its modulus is B. The increase in density of piece of gold when pressure, P is applied uniformly from all sides, (1), , P, 2B, , (2), , B, 2P, , (3), , P, BP, , (4), , B, BP, , Sol. Answer (3), We know, , V P, , V, B, , ...(1), , M, V, , And , , ⎧V Change in volume, ⎪, ⎨P Pressure applied, ⎪B Bulk modulus, ⎩, ⎧ Density, ⎪, ⎨M Mass, ⎪V Volume, ⎩, , ...(2), , From (2), , , , M, M, , V V V, , , , M, V, , V V V, , , , , , , , 1, V, 1, V, , [From eq. (2)], , 1, B, 1, P, , [From eq. (1)], , P, BP, , 22. A wire of length 5 m is twisted through 30° at free end. If the radius of wire is 1 mm, the shearing strain in, the wire is, (1) 30°, , (2), , 0.36, , (3), , 1°, , (4), , 0.18°, , Sol. Answer (2), , , r, , L, , , , 1 10 3 30, 5, , ⎧ Where, ⎪ Angle of shear, ⎪⎪, ⎨ Angleof twist, ⎪r Radius of rod, ⎪, ⎪⎩l length or rod, , = 6 × 10–3, = 0.36', , 23. One end of uniform wire of length L and of weight W is attached rigidly to a point in roof and a weight W1 is, suspended from the lower end. If A is area of cross-section of the wire, the stress in the wire at a height, , 3L, 4, , from its lower end is, W1, (1), A, , (2), , W⎞, ⎛, ⎜ W1 , ⎟, ⎝, 4 ⎠, A, , (3), , 3W ⎞, ⎛, ⎜ W1 , ⎟, ⎝, 4 ⎠, A, , (4), , W1 W, A, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 81, , Sol. Answer (3), , Stress , , Tension at point, Area of cross-section, , Tension = force due to weight hanging below the choosen point, , ⎛ 3W, ⎞, W1 ⎟, That is ⎜, ⎝ 4, ⎠, , L, , 3L/4, W1, , 3W / 4 W1, Stress , A, 24. Two wires A and B of same length and of same material have radii r1 and r2 respectively. Their one end is, fixed with a rigid support and at other end equal twisting couple is applied. Then ratio of the angle of twist at, the end of A and the angle of twist at the end of B will be, (1), , r12, , (2), , r22, , r22, r12, , (3), , r24, r14, , (4), , r14, r24, , Sol. Answer (3), r14 1 = r24 2, A (rB )4 ⎛ r2 ⎞, , ⎜ ⎟, B (rA )4 ⎝ r1 ⎠, , 4, , 25. A uniform wire of length L and radius r is twisted by an angle . If modulus of rigidity of the wire is , then, the elastic potential energy stored in wire, is, (1), , r 4 , , (2), , 2L2, , r 4 2, 4L, , (3), , r 4 , 4L2, , (4), , r 4 2, 2L, , Sol. Answer (2), U = work done, We know, Work done , , ⎧ Where,, ⎪, ⎨ Angle of twist , ⎪S Modulus of rigidity = 4, ⎩, , Sr 4 2, 4L, , Substituting values, , U, , r 4 2, 4L, , 26. What is called the ratio of the breaking stress and the working stress?, (1) Elastic fatigue, , (2), , Elastic after effect, , (3), , Yield point, , (4), , Power of safety, , Sol. Answer (4), Breaking stress, n, Working stress, , n = power of safely, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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82, , Mechanical Properties of Solids, , Solution of Assignment, , 27. If is the depression produced in a beam of length L, breadth b and thickness d, when a load is placed at, the mid point, then, (1) L3, , (2), , , , 1, b, , , , (3), , 3, , 1, d, , (4), , All of these, , Sol. Answer (1), Since we know, , , , , WL3, 4Y bd 3, , W, , So we can say L3, , SECTION - C, Previous Years Questions, 1., , The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross, section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of, the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio, of, [Re-AIPMT-2015], (1) 1 : 1, , (2), , 1:2, , (3), , 2:1, , (4), , 4:1, , Sol. Answer (3), 2., , The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10–11 Pa–1 and density, of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?, [AIPMT-2015], (1) 1.4 × 10–2, , (2), , 0.8 × 10–2, , (3), , 1.0 × 10–2, , (4), , 1.2 × 10–2, , Sol. Answer (4), B, , V P, P, ⇒, , V / V, V, B, , V hdg, , V, B, , , , 2700 103 10, 1, 45.4 1011, , = 1.2 × 10–2, 3., , Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension, produced in the wire is l . Which of the following graphs is a straight line?, (1) l versus, , 1, l, , (2), , l versus l 2, , (3), , l versus, , 1, l2, , [AIPMT-2014], (4), , l versus l, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 83, , Sol. Answer (2), V=AL, Y , , FL, FL, ⇒ L , V, AL, Y, L, , L , , FL2, VY, , L L2, Thus, L versus L2 is straight line., 4., , The following four wires of length L and radius r are made of the same material. Which of these will have the, largest extension, when the same tension is applied?, [NEET-2013], (1) L = 400 cm, r = 0.8 mm, , (2), , L = 300 cm, r = 0.6 mm, , (3) L = 200 cm, r = 0.4 mm, , (4), , L = 100 cm, r = 0.2 mm, , Sol. Answer (4), We know, , x , x , , FL, FL, , AY r 2Y, L, r2, , x directly proportional to L, And x inversely proportional to r2, For option (1), L, r2, , , , 400 10, (0.8)2, , 6250, , For option (2), L, r2, , , , 300 10, (0.6)2, , 8333.33, , For option (3), , L, r, , 2, , , , 200 10, (0.4)2, , 12,500, , For option (4), L, r, , 2, , , , 100 10, (0.2)2, , 25,000, , For option (4) we are getting maximum value of, , L, , r2, x also maximum for L = 100 cm and r = 0.2 mm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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84, 5., , Mechanical Properties of Solids, , Solution of Assignment, , A rope 1 cm in diameter breaks, if the tension in it exceeds 500 N. The maximum tension that may be given, to similar rope of diameter 3 cm is, (1) 500 N, , (2), , 3000 N, , (3), , 4500 N, , (4), , 2000 N, , Sol. Answer (3), , FL ⎤, ⎡, ⎢ x 2 ⎥, r Y ⎦, ⎣, , Tension (radius)2, T1 ⎛ r1 ⎞, ⎜ ⎟, T2 ⎝ r2 ⎠, , 2, , So T r2, ⎧T1 500 N⎫, ⎪, ⎪, ⎨r1 1 cm ⎬ given, ⎪r 3 cm ⎪, ⎩2, ⎭, , Substituting values, Let T2 = x, 500 12, 2, x, 3, , https://t.me/NEET_StudyMaterial, , x = 4500 N, T2 = 4500 N, 6., , A wire of length L and radius r fixed at one end and a force F applied to the other end produces an extension, l. The extension produced in another wire of the same material of length 2L and radius 2r by a force 2F, is, (1) l, , (2), , 2l, , (3), , 4l, , (4), , l, 2, , Sol. Answer (1), , L, , FL, r 2Y, , FL ⎤, ⎡, ⎢ x AY ⎥, ⎣, ⎦, , ...(1), , Now, new parameters, F = 2F, L = 2L, r = 2r, Substituting new parameters in eq. (1), L , , , , 2F 2L, (2r )2Y, , FL, r 2Y, , FL, ⎧, ⎫, l from equation 1⎬, ⎨∵, r 2Y, ⎩, ⎭, , L' = L, 7., , The increase in pressure required to decrease the 200 L volume of a liquid by 0.008% in kPa is (Bulk modulus, of the liquid = 2100 MPa is), (1) 8.4, , (2), , 84, , (3), , 92.4, , (4), , 168, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 85, , Sol. Answer (4), V = 200 L, V = –0.008% of 200 L, , , (Decrease in volume so we use (–)ive sign), , 0.008, 200 0.016 L, 100, , B = 2100 MPa = 21 × 108 Pa, We know, , V P, , V, B, , 0.016, P, , 200, 21 108, 168 × 103 Pa = P, 168 kPa = P, 8., , Which of the following relations is true?, (1) Y = 2η(1 – 2σ), , (2), , Y = 2η(1 + 2σ), , (3), , Y = 2η(1 – σ), , (4), , (1 + σ)2η = Y, , Sol. Answer (4), Y = 2(1 + ), , 9., , ⎧ Where,, ⎪Y Young's modulus, ⎪, ⎨, ⎪ Shear modulus, ⎪⎩ Poisson's ratio, , A 5 m long aluminium wire (Y = 7 × 1010 Nm–2) of diameter 3 mm supports a 40 kg mass. In order to have, the same elongation in the copper wire (Y = 12 × 1010 Nm–2) of the same length under the same weight, the, diameter should now be (in mm), (1) 1.75, , (2), , 1.5, , (3), , 2.3, , (4), , 5.0, , Sol. Answer (3), For aluminium wire, , x1 , , FL, 4FL, , AY d 2Y, , Substituting values, x1 , , 4 400 5, (3)2 7 1010, , ...(1), , ⎧F 400 N, ⎪L 5 m, ⎪, ⎨, ⎪d 3 mm, ⎪Y 7 1010 Nm2, ⎩, , For copper wire, Using same formulae, , x2 , , ⎧F 400 N, ⎪L 5 m, ⎪, ⎨, 10, ⎪Y 12 10, ⎪d ?, ⎩, , 4FL, 2, , d Y, , Let diameter be = d, , x2 , , 4 400 5, d 2 12 1010, , ...(2), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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86, , Mechanical Properties of Solids, , Solution of Assignment, , Equating (1) and (2), Because x1 = x2, , [given condition], , 4 400 5, 2, , 10, , (3) 7 10, , , , 4 400 5, d 2 12 1010, , Solving this we get, , 21, 2.3 mm, 2, , d, , 10. Two wires of same material and radius have their lengths in ratio 1 : 2. If these wires are stretched by the, same force, the strain produced in the two wires will be in the ratio, (1) 2 : 1, , (2), , 1:1, , (3), , 1:2, , (4), , 1:4, , Sol. Answer (2), , Strain , , l, l, , We know, l , , FL, AY, , F, F, l, , , l, AY r 2Y, For wire 1, , S1 strain , , l1, F, 2, L, r Y, , ...(1), , l 2, F, , 2L r 2Y, , ...(2), , For wire 2, , S2 strain , Therefore,, Ratio of strains , , S1 F r 2Y 1, , , S2 r 2Y F 1, , 1:1, 11. A steel wire of cross-sectional area 3 10–6 m2 can withstand a maximum strain of 10–3. Young’s modulus, of steel is 2 × 1011 Nm–2. The maximum mass the wire can hold is (take g = 10 ms–2), (1) 40 kg, , (2), , 60 kg, , (3), , 80 kg, , (4), , 100 kg, , Sol. Answer (2), , l, F, , l, AY, Substituting values, , ⎧Given,, ⎪ l, ⎪⎪ 103, ⎨ l, ⎪ A 3 106 m2, ⎪, ⎪⎩Y 2 1011 Nm2, , Strain , , 103 , , F, 3 10, , 6, , 2 1011, , 600 N = F, Therefore maximum mass , , F 600, , 60 kg, g, 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 87, , 12. The hollow shaft is ...... than a solid shaft of same mass, material and length., (1) Less stiff, , (2), , More stiff, , (3), , Equally stiff, , (4), , None of these, , Sol. Answer (2), Let C' = restoring couple per unit twist for hollow cylinder, So C , , S(r24 r14 ), 2L, , ⎧ Where,, ⎫, ⎨, ⎬, r, and, r, are, outer, and, inner, radii, 1, ⎩2, ⎭, , And, C = restoring couple per unit twist for solid cylinder, , C, , S r 4, 2L, , C r24 r14 (r22 r12 )(r22 r12 ), , , , C, r4, r4, , , , C r22 r12, , 1, C r22 r12, , ⎧If mass of both cylinder same than, ⎪ 2, 2, 2, ⎨r L (r2 r1 )L, ⎪, 2, 2, 2, ⎩or r r2 r1, , Hence hollow cylinder more stronger than solid one., , 13. The Bulk modulus for an incompressible liquid is, (1) Zero, , (2), , Unity, , (3), , Infinity, , (4), , Between 0 and 1, , Sol. Answer (3), We know, , V P, , V, B, B, , P V, V, , For incompressible liquid, V (Change in volume) = 0, For every value of pressure applied, Put V = 0, B = (Infinity), 14. A copper rod length L and radius r is suspended from the ceilling by one of its ends. What will be elongation, of the rod due to its own weight when ρ and Y are the density and Young’s modulus of the copper respectively?, (1), , 2 gL2, 2Y, , (2), , gL2, 2Y, , (3), , 2 g 2 L2, 2Y, , (4), , gL, 2Y, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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88, , Mechanical Properties of Solids, , Solution of Assignment, , Sol. Answer (2), Let W be the total weight acting downwards, Let the centre of mass, , L/2, , L, for the top, which is at the distance of, 2, , W, , So it can be assumed that a mass W is hung by a massless, wire of length, , L, , Young's modulus Y,, 2, , L/2, Y, , = density of wire, W = Mg = × r2L × g, Using L , , FL r 2Lg L 1 gL2, , , AY, 2 Y, 2Y, r 2, , W, , 15. Which of the following substances has the highest elasticity?, (1) Steel, , (2), , Copper, , (3), , Rubber, , (4), , Sponge, , Sol. Answer (1), Substance which requires more force for per unit elongation have more elasticity, OR, Less stretchable means more elastic, So, steel is least stretchable, Most elastic., 16. When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is, 0.01 10–3 m. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m2, its, Young’s modulus is, (1) 1.6 × 108 Nm–2, , (2), , 2.5 × 1010 Nm–2, , (3), , 1.25 × 1011 Nm–2, , (4), , 16 × 109 Nm–2, , Sol. Answer (1), Poisson's ratio , l, l, , , , , Lateral strain, Longitudinal strain, , F, AY, , Longitudinal, strain, , So , , Lateral strain, F / AY, , Therefore, , Y , , F, Lateral strain A, , ⎧Given,, ⎪F 100 N, ⎪, ⎪, 3, ⎨Lateral strain 0.01 10 m, ⎪ 0.4, ⎪, ⎪⎩ A 0.025 m2, , Substituting values, , Y , , 0.4 100, 0.01 10, , 3, , 0.025, , 1.6 108 Nm2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 89, , 17. Two wires of length l, radius r and length 2l, radius 2r respectively having same Young’s modulus are hung, with a weight mg. Net elongation is, (1), , 3 mgl, r 2Y, , 2 mgl, 3 r 2Y, , (2), , (3), , 3 mgl, 2r 2Y, , (4), , 3 mgl, 4 r 2Y, , Sol. Answer (3), Tension in both wires will be same, Let elongation in wire 1 be = l1, , l1 , , mgl, , Wire 1, , FL ⎤, ⎡, ⎢ x AY ⎥, ⎣, ⎦, , r 2Y, , Let elongation in wire 2 be = l 2, l 2 , , Wire 2, , mg 2l, , Mg, , (2r 2 )Y, , Net elongation = l1 + l2, , , , , , mgl, 2, , r Y, , , , mgl, 2r 2Y, , 3mgl, 2r 2Y, , 18. A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing, modulus of cube is, (1) 2 109 Nm–2, , (2), , 4 109 Nm–2, , (3), , 8 109 Nm–2, , (4), , 16 109 Nm–2, , Sol. Answer (1), , F h, Ax, Substituting values, Shear modulus , , , , ⎧Given,, ⎪, ⎪F 8 kN 8000 N, ⎪, 6, 2, ⎨ A 40 40 1600 10 m, ⎪, 3, ⎪ x 0.1 mm 0.1 10 m, ⎪h 40 103 m, ⎩, , 8000 40 10 3, 1600 10 6 0.1 10 3, , = 2 × 109 Nm–2, , l, r, and radius, of same material. The free end of, 2, 2, small rod is fixed to a rigid base and the free end of larger rod is given a twist of θ°, the twist angle at the, joint will be, , 19. A rod of length l and radius r is joined to a rod of length, , (1), , , 4, , (2), , , 2, , (3), , 5, 6, , (4), , 8, 9, , Sol. Answer (4), Torque will be same, × Restoring couple per unit twist will be same for both the rods, , Sr 4, , 2L, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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90, , Mechanical Properties of Solids, , Solution of Assignment, , We can use, , Sr14, Sr24, 1 , 2, 2L1, 2L2, , ⎧given, ⎪, ⎪r1 r, ⎪, ⎨r2 r / 2, ⎪l l, ⎪l, ⎩⎪l 2 l / 2, , r14, r4, 1 2 2, L1, L2, Substituting values, , r4, (r / 2)4, 1 , 2, L, l /2, 2, 8, , 1 , , Also 1 + 2 = , , (given), , 2, 2 , 8, 8, 9, , 2 , , 20. The Young’s modulus of the material of a wire is 2 × 1010 Nm–2. If the elongation strain is 1%, then the energy, stored in the wire per unit volume is Jm–3 is, (1) 106, , (2), , 108, , (3), , 2 × 106, , (4), , 2 × 108, , Sol. Answer (1), 1, U Y (strain)2, 2, , (given), , 1 ⎞, ⎛ l, ⎜ l 100 ⎟, ⎝, ⎠, , Substituting values, , U, , 1, ⎛ 1 ⎞, 2 1010 ⎜, ⎟, 2, ⎝ 100 ⎠, , 2, , U = 106, 21. A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire, (1) Remains the same, , (2), , Decreases, , (3) Increases, , (4), , First decreases then increases, , Sol. Answer (3), We know, , L , , FL, AY, , and L = L, , Equating both, , FL, = L F, AY, , ⎧ Where,, ⎪ Change in temperature, ⎪⎪, ⎨ Coefficeince of linear expansion, ⎪L Elongation, ⎪, ⎪⎩L Original length, , So whenever stretching is there will be (+)ive hence temperature will increase., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 91, , 22. A wire of natural length l, Young’s modulus Y and area of cross-section A is extended by x. Then the energy, stored in the wires is given by, (1), , 1 YA 2, x, 2 l, , 1 YA 2, x, 3 l, , (2), , (3), , 1 YI 2, x, 2 A, , (4), , 1 YA 2, x, 2 l2, , Sol. Answer (1), Energy density per unit volume , , 1, (strain)2 Y, 2, , ⎧ Where,, ⎪, x, ⎪, ⎨Strain , l, ⎪, ⎪⎩Y Young's modulus, , Volume = length × area of cross-section, Energy (total) , , , , E, , 1, (strain)2 Y L A, 2, , 1 x2, Y LA, 2 L2, 1 YA 2, x, 2 L, , 23. When a force is applied on a wire of uniform cross-sectional area 3 × 10–6 m2 and length 4 m, the increase, in length is 1 mm. Energy stored in it will be (Y = 2 × 1011 N/m2), (1) 6250 J, , (2), , 0.177 J, , (3), , 0.075 J, , (4), , 0.150 J, , Sol. Answer (3), Energy stored =, , 1, × work done, 2, , =, , 1, × F × x, 2, , =, , 1 YA, , x x, 2 L, , YAx ⎤, ⎡, ⎢F L ⎥, ⎣, ⎦, , Substituting values, , E, , 1 2 1011 3 106 1 103 103, , 2, 4, , E = 0.075 J, 24. If in a wire of Young’s modulus Y, longitudinal strain X is produced then the potential energy stored in its unit, volume will be, (1) 0.5 YX2, , (2), , 0.5 Y2X, , (3), , 2 YX2, , (4), , YX2, , Sol. Answer (1), Potential energy per unit volume , , 1, × (strain)2 × Young's modulus, 2, , Substituting data from question, We get, , U, , 1 2, X Y, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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92, , Mechanical Properties of Solids, , Solution of Assignment, , 25. A uniform metal rod of 2 mm2 cross-section is heated from 0°C to 20°C. The coefficient of the linear expansion, of the rod is 12 × 10–6/°C. Its Young’s modulus of elasticity is 1011 Nm–2. The energy stored per unit volume, of the rod is, (1) 1440 Jm–3, , 15750 Jm–3, , (2), , (3), , 1500 Jm–3, , (4), , 2880 Jm–3, , Sol. Answer (4), We know, L = L, , Also, , L, , L, L, strain, L, , Strain = , , ...(i), , Energy stored per unit volume , , , Substituting values , , 1, (strain)2Y, 2, , 1, 2 2 Y, 2, , [Using eqn. (i)], , 1, (12 106 )2 (20)2 1011, 2, , = 2880 Jm–3, 26. A material has Poisson’s ratio 0.50. If a uniform rod of it suffers a longitudinal strain of 2 × 10–3, then the, percentage change in volume is, (1) 0.6, , (2), , 0.4, , (3), , 0.2, , (4), , Zero, , Sol. Answer (4), Poisson's ratio = 0.50, , Lateral strain, 0.50, Longitudinal strain, , , r / r 1, , L / L 2, 2r L, , r, L, , ...(1), , Volume = area × length, , , V A L, , , V, A, L, , V 2r L, , , V, r, L, , ⎡A r 2, ⎤, ⎢, ⎥, ⎢ A 2 r ⎥, ⎢⎣ A, r ⎥⎦, , Using equation (1), We get, , V, 0, V, So, , V, 100 0%, V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 93, , 27. There is no change in the volume of a wire due to the change in its length on stretching. The Poisson’s ratio, of the material of the wire is, (1) , , 1, 2, , (2), , , , 1, 2, , (3), , , , 1, 4, , (4), , –, , 1, 4, , Sol. Answer (1), V=A×L, V = r2L, , V 2r L, , , V, r, L, V, 0, V, , [given], , 2r L, , r, L, , ...(1), , Poisson's ratio , , r, r, , L, L, , ...(2), , Using equation (1) in (2), , , , r, r, , , , 1, 2, , ⎛ r ⎞, 2⎜, ⎟, ⎝ r ⎠, , 28. If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient η, the Poisson’s, ratio σ will be, (1) , , 2, 3, , (2), , , , 1, 4, , (3), , , , 1, 4, , (4), , , , 2, 3, , Sol. Answer (2), Y , , 3, [given], 2, , =?, And we know, Y = 2(1 + ), , 3, 2(1 ), 2, Solving we get, , , 1, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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94, , Mechanical Properties of Solids, , Solution of Assignment, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Hooke’s law is obeyed only for small values of strain., R : The deformation beyond elastic limit is called plasticity., , Sol. Answer (2), Statement (A) is true, Statement (R) is also true, But (R) is not the correct explaination of (A), Because correct reason is Hooke's law is obeyed in elastic limit only., 2., , A : Strain is a dimensionless quantity., R : Strain is internal force per unit area of a body., , Sol. Answer (3), (A) Is true, Strain =, , L, L, , (R) Is false because strain is change in dimension by original dimension., 3., , A : Diamond is more elastic than rubber., R : When same deforming force is applied diamond deforms less than rubber., , Sol. Answer (1), (A) Is true because modulus of elasticity is more for diamond so less deformation in diamond than rubber when, same deforming force applied, (R) Is true and correct explanation., 4., , A : Bulk modulus for a perfectly plastic body is zero., R : For perfect plastic material, there is no restoring force., , Sol. Answer (1), (A) Is true because a perfectly plastic body cannot regain its shape even when the deforming forces are removed, because restoring forces are absent, (R) Is true and correct explanation for (A), 5., , A : The railway bridges are declared unfit after their use for a long period., R : Due to repeated strain the elasticity of material decreases., , Sol. Answer (1), (A) Is true because after a long use the material weakens and shows dangerous deformation when load is, applied because its elasticity has decreased gradually over the time., (R) Is true and correct explanation for (A), 6., , A : Spring balances show wrong readings after they have been used for a long time., R : Spring in spring balance temporary losses elasticity due to repeated alternating deforming force., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Solids, , 95, , Sol. Answer (1), (A) Is true because after a long use elasticity decreases and small temporary deformation remains these which, in turn tend to be the reason of wrong readings., (R) Is true and correct explanation for (A), 7., , A : Modulus of elasticity is independent of dimensions of the body., R : Modulus of elasticity depends on the material of the body., , Sol. Answer (2), (A) True because modulus of elasticity is a material property, (R) True, But (R) is not the correct explanation because no where it reasons why modulus of elasticity is independent of, dimensions of the body., 8., , A : Adiabatic elasticity of a gas is greater than isothermal elasticity., R:, , Eadiabatic, ., Eisothermal, , Sol. Answer (1), (A) True, Because, , Eadiabatic, and always greater than 1, Eisothermal, , So Eadiabatic is always greater than Eisothermal, (R) True and also correct explanation., 9., , A : When a beam is bent only tensile strain is produced., R : The depression produced in a rectangular beam is directly proportional to its width., , Sol. Answer (4), (A) Is false because strain is there so stress will also be present, , ∵ Stress strain, (R) False depression , , 1, width, , 10. A : To minimise the depression in a beam, it is designed as ‘I’ shape girder., R : The ‘I’ shape girders have large load bearing surface, which decreases the stress., Sol. Answer (1), (A) Is true because having more surface area means less force per unit area i.e. less stress, (R) Is true and correct explanation of (A), 11. A : Iron is more elastic than copper., R : Under a given deforming force, Iron is deformed less than copper., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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96, , Mechanical Properties of Solids, , Solution of Assignment, , Sol. Answer (1), (A) Is true because less deformation under a similar deforming force means more elasticity, (R) Is true and correct explanation of (A), 12. A : Lateral strain is directly proportional to the longitudinal strain within the elastic limit., R : Poisson’s ratio for a given material at a constant temperature is constant., Sol. Answer (1), (A) Is true because,, Lateral strain, , Longitudinal strain, , Lateral strain longitudinal strain, As is constant, (R) Is true and correct explanation of (A), 13. A : Equal amount of work is done when two identical springs of steel and copper are equally stretched., R : Both springs have same spring constant., Sol. Answer (4), (A) Is wrong because amount of work done is not same because the spring constants are different., (R) Is wrong., 14. A : Increase in temperature of a substance decreases the modulus of elasticity., R : The graph between potential energy of molecules and separation between them is asymmetric., Sol. Answer (1), (A) Is true because when we increase the temperature the average distance between the molecules tend to, increase hence decreasing the modulus of elasticity., (R) Is true and correct explanation., 15. A : It is the breaking stress and not the breaking strength which depends on the material., R : Breaking strength =, , Breaking stress, ., Area, , Sol. Answer (4), (A) Is wrong both depend on the material because breaking strength is maximum stress a body can take, (R) Is wrong, Breaking strength = breaking stress × area, , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Chapter, , 10, , Mechanical Properties of Fluids, Solutions, SECTION - A, Objective Type Questions, 1., , The term ‘fluid’ is used for, (1) Liquids only, , (2), , Gases only, , (3) A mixture of liquid and gas only, , (4), , Both liquids and gases, , Sol. Answer (4), Substances that can flow is called fluid. Thus both liquids and gases are fluids., 2., , Select wrong statement about pressure, (1) Pressure is a scalar quantity, , (2), , Pressure is always compressive in nature, , (3) Pressure at a point is same in all directions, , (4), , None of these, , Sol. Answer (4), Pressure is scalar as it is not added vectorially. Pressure is compressive in nature, it is same in all the, directions at a point., 3., , Gauge pressure, (1) May be positive, , (2), , May be negative, , (3), , May be zero, , (4), , All of these, , Sol. Answer (4), Gauge pressure depends on the reference chosen, it can be positive, negative or zero., 4., , Figure shows two containers P and Q with same base area A and each filled upto same height with same, liquid. Select the correct alternative, , A, , A, , px, (P), , (1) px = py, , (2), , px > py, , py, (Q), , (3), , py > px, , (4), , Cannot say, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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98, , Mechanical Properties of Fluids, , Solution of Assignment, , Sol. Answer (1), The level of water above both points is same so by hydrostatic paradox., px = py, 5., , The pressure of confined air is p. If the atmospheric pressure is P, then, , (1) P is equal to p, (2) P is less than p, (3) P is greater than p, (4) P may be less or greater than p depending on the mass of the confined air, Sol. Answer (2), , P, , Pressure at point A = P (Hydrostatic paradox), , h, , and P + wgh = pressure at A = p, , p, , A, , p, , air, , P = p – wgh, , Water, , So, P < p, 6., , Figure shows a container filled with a liquid of density . Four points A, B, C and D lie on the diametrically opposite, points of a circle as shown. Points A and C lie on vertical line and points B and D lie on horizontal line. The, incorrect statement is (pA, pB, pC, pD are absolute pressure at the respective points), A, D, , B, C, , (1) pD = pB, , (2), , pA < pB = pD < pC, , (3), , pD pB , , pC p A, 2, , (4), , pD pB , , pC pA, 2, , Sol. Answer (3), Points at same height have same pressure, points with height difference say 'h' will have difference of gh., Let radius of circle is r, , p0, , pA = p0 + hg, pB = pD = p0 + (h + r)g, pC = p0 + (h + 2r)g, Then,, pC + pA = [p0 + (h + 2r)g] – [p0 + hg], , A, B, , D, C, , h, r, r, , = p0 + hg + 2rg + p0 + hg, = 2[p0 + (h + r)g], , pC pA, p0 (h r )g, 2, pC pA, pB pD, 2, i.e., option (1), (2) and (4) gives correct statement but incorrect statement is (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 7., , Mechanical Properties of Fluids, , 99, , The volume of an air bubble is doubled as it rises from the bottom of lake to its surface. The atmospheric pressure, is 75 cm of mercury. The ratio of density of mercury to that of lake water is, (1) 10, , (2), , 15, , (3), , 40, . The depth of the lake in metre is, 3, , 20, , (4), , 25, , Sol. Answer (1), 2P0 = P0 + gh, P0 = gh, P0 = 75 cm mercury, mercury g , , [Atmospheric pressure], , 75, water g h, 100, , , , m 75, , h, w 100, , , , 40 75, , h, 3 100, , ⎡ m 40, ⎤, , (given)⎥, ⎢∵, 3, ⎣ w, ⎦, , h = 10 m, 8., , A beaker containing a liquid of density moves up with an acceleration ‘a’. The pressure due to the liquid at a, depth h below free surface of the liquid is, (1) hg, , (2), , h (g – a), , (3) h (g + a), , (4), , ⎛g a⎞, 2hg ⎜, ⎟, ⎝g a⎠, , Sol. Answer (3), Due to upward acceleration pseudo force will act downwards so value of acceleration due to gravity will increase, by 'a', g' = (g + a), P = g'h, P = (g + a)h, 9., , (Substitute g'), , A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with some acceleration, the, reading will be, (1) 76 cm, , (2), , > 76 cm, , (3), , < 76 cm, , (4), , Zero, , Sol. Answer (3), When elevator goes up with some acceleration upward, due to pseudo force acting downwards. Value of g, increases to g'., If g increases to g',, , i.e., g' = g + a, , ∵ P = gh = (9 + a)h' = constant, Then, h' < h, i.e., h' < 76 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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100, , Mechanical Properties of Fluids, , Solution of Assignment, , 10. In a hydraulic jack as shown, mass of the car W = 800 kg, A1 = 10 cm2, A2 = 10 m2. The minimum force F required, to lift the car is, W, , F, A2, , A1, , (1) 1 N, , (2), , 0.8 N, , (3), , 8N, , (4), , 16 N, , Sol. Answer (2), Pressure in a liquid is divided equally so we can say pressure at both the pistons should be same, ⎧ Where,, F1 F2, ⎪, , , ⎪F1 F, A1 A2, ⎪, 2, ⎨ A1 10 cm, ⎪, Substituting values,, 2, 4, 2, ⎪ A2 10 m 10 10 cm, ⎪F 8000 N, F, 8000, ⎩ 2, , 4, 10 10 10, ⎪⎧Take, ⎪⎫, ⎨, 2⎬, ⎪⎩g 10 m/s ⎪⎭, , F = 0.8 N, , 11. A wooden cube just floats inside water with a 200 gm mass placed on it. When the mass is removed, the cube, floats with its top surface 2 cm above the water level. What is the side of the cube ?, (1) 6 cm, , (2), , 8 cm, , (3), , 10 cm, , (4), , 12 cm, , Sol. Answer (3), Mass × g = Volume of part of cube × × g, 200 × g = L2 (2 × w × g), 100 = L2, , {∵ w = 1}, , 200 gm, , L, , 10 cm = L, , L, , From the two figures we can see that the 200 gm block, is provided with required buoyant force but a part of, cube which is afloat in 2nd figure., , 2 cm, (L –2) cm, , L, , 12. A block of steel of size 5 × 5 × 5 cm3 is weighed in water. If relative density of steel is 7, its apparent weight is, (1) 6 × 5 × 5 × 5 g wt, , (2), , 4 × 4 × 4 × 7 g wt, , (3) 5 × 5 × 5 × 7 g wt, , (4), , 4 × 4 × 4 × 6 g wt, , Sol. Answer (1), Apparent weight = s vb g – w vb g, = vbg (s – w ), = 5 × 5 × 5 × g × (7 – 1), , ⎡w 1, ⎢, ⎣s 7 (given), , ⎧ Where,, ⎪ density of steel, ⎪ s, ⎨, ⎪w density of water, ⎪v volume of block, ⎩ b, (side side side), , = 6 × 5 × 5 × 5 × g wt, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 13. A block of wood floats in water with, of liquid is (in kg/m3), (1) 750, , (2), , 101, , 4, th of its volume submerged, but it just floats in another liquid. The density, 5, , 800, , (3), , 1000, , (4), , 1250, , Sol. Answer (2), , ⎧ Where,, ⎪v volume of block, ⎪ b, ⎨, 3, ⎪w density of water 1000 kg/m, ⎪ density of block, ⎩ b, , 4, v × w × g = vb × b × g, 5 b, , , , w 5, , b 4, 4, 1000 800 kg/m3, 5, , b , , And when block is put in liquid of density l it just floats, So, vb × b × g = vb × l × g, b = l, So, l = 800 kg/m3, 14. A cubical block is floating in a liquid with one fourth of its volume immersed in the liquid. If whole of the system, accelerates upward with acceleration g/4, the fraction of volume immersed in the liquid will be, (1) 1/4, , (2), , 1/2, , (3), , 3/4, , (4), , 2/3, , Sol. Answer (1), Upward acceleration just causes the acceleration due to gravity increases by some value, but since the term of 'g', gets cancelled out in the buoyancy equation., Volume immersed × w × g = Total volume × cube × g, So, increasing it will not have any effect on the immersed volume., 15. A body of density is dropped from rest from a height h into a lake of density , where > . Neglecting all, dissipative forces, the maximum depth to which the body sinks before returning to float on surface, , (1), , h, , , (2), , h, , , (3), , h, , , (4), , h, , , Sol. Answer (3), Between point 2 and 3, , P0 gh , , 1 2, v gh P0, 2, , gh' + gh = gh', , So, h , , h, , , P, , 1, [P0 = atmospheric pressure], , ⎡∵ v 2gh ⎤, ⎣, ⎦, , h, v 2gh, , 2, , , h, , 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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102, , Mechanical Properties of Fluids, , Solution of Assignment, , 16. A boat carrying a number of stones is floating in a water tank. If the stones are unloaded into water, the water level, in the tank will, (1) Remain unchanged, (2) Rise, (3) Fall, (4) Rise or fall depends on the number of stones unloaded, Sol. Answer (3), Previously when stones are on the boat they are increasing the weight on the boat and to balance this weight boat, needs to generate buoyancy force by displacing more water, but when stones are removed the boat starts displacing, less amount of water hence the level of water in tank falls., , Decrease in level, , 17. A block of ice is floating in a liquid of specific gravity 1.2 contained in a beaker. When the ice melts completely, the, level of liquid in the vessel, (1) Increases, , (2), , Decreases, , (3) Remain unchanged, , (4), , First increases then decreases, , Sol. Answer (1), Density of ice is less than water and density of liquid is more than water. So even when ice melts the level will rise., If liquid > water then level (liquid + water) will rise., 18. Two liquids having densities d1 and d2 are mixed in such a way that both have same mass. The density of the, mixture is, (1), , d1 d 2, 2, , (2), , d1 d 2, d1d 2, , (3), , d1d 2, d1 d 2, , (4), , 2d1d 2, d1 d 2, , Sol. Answer (4), Let each have mass = M and densities d1 and d2, dmix , , Mmix, 2d1d 2, M M, , , Vmix, ⎛ M ⎞ ⎛ M ⎞ d1 d 2, ⎜ ⎟⎜ ⎟, ⎝ d1 ⎠ ⎝ d 2 ⎠, , 19. A barometer tube reads 75 cm of Hg. If tube is gradually inclined at an angle of 30° with horizontal, keeping the, open end in the mercury container, then find the length of mercury column in the barometer tube, (1) 86.7 cm, , (2), , 150 cm, , (3), , 75 cm, , (4), , 92.5 cm, , Sol. Answer (2), , ∵ We are not changing the atmospheric pressure, so, height of Hg from the surface should not change., , 75, cos 60, x, , , x, , 60°, , 75 cm, , 30°, , x = 150 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 103, , 20. A metallic sphere weighing 3 kg in air is held by a string so as to be completely immersed in a liquid of relative, density 0.8. The relative density of metallic is 10. The tension in the string is, (1) 18.7 N, , (2), , 42.5 N, , (3), , 32.7 N, , (4), , 27.6 N, , Sol. Answer (4), , T, L, FB, , ⎧ Where,, ⎪F Force of Buoyancy, ⎪ B, ⎪⎪L Density of liquid, ⎨, ⎪M Density of metal, ⎪M Mass of sphere, ⎪, ⎪⎩T Tension in string, , v,M, , Mg, , T = Mg – Buoyant Force, , M⎞, ⎛, ⎜ M V ⎟, ⎝, ⎠, , T = MVg – LVg, = (M – L) Vg, (10 0.8) , , 3, 10, 10, , T = 9.2 × 3 = 27.6 N, 21. A rectangular block is 10 cm 10 cm 15 cm in size is floating in water with 10 cm side vertical. If it floats with, 15 cm side vertical, then the level of water will, (1) Rise, , (2), , Fall, , (3) Remain same, , (4), , Change according to density of block, , Sol. Answer (3), Mass of block remains same, volume displaced of water will also remain same so level of water will not change., 22. Two cubical blocks identical in dimensions float in water in such a way that 1st block floats with half part immersed, in water and second block floats with 3/4 of its volume inside the water. The ratio of densities of blocks is, (1) 2 : 3, , (2), , 3:4, , (3), , 1:3, , (4), , 1:4, , Sol. Answer (1), , ⎧ Where,, ⎪V Volume of block, ⎪, ⎨, ⎪1 density of liquid 1, ⎪⎩2 density of liquid 2, , 3, 1, V 2 g V 1 g, 4, 2, , , , 2 2, , 1 3, , 23. On putting a capillary tube in a pot filled with water, the level of water rises upto a height of, 4 cm in the tube. If a tube of half the diameter is used instead, the water will rise to a height of nearly, (1) 2 cm, , (2), , 4 cm, , (3), , 8 cm, , (4), , 11 cm, , Sol. Answer (3), For capillary tube, , h, , 2T, r g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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104, , Mechanical Properties of Fluids, , Solution of Assignment, , We can say, 1, 1, or h , r, d, , h, , So,, , h1 d 2, , h2 d1, , , , 4, d, , x 2d, , x = 8 cm, 24. A flat plate of area 0.1 m2 is placed on a flat surface and is separated from it by a film of oil 10–5 m thick whose, coefficient of viscosity is 1.5 N sm–2. The force required to cause the plate to slide on the surface at constant, speed of 1 mm s–1 is, (1) 10 N, , (2), , 15 N, , (3), , 20 N, , (4), , 25 N, , Sol. Answer (2), F A, , v, l, , Substituting values,, 1.5 0.1, , 1 10 3, 10 5, , = 15 N, 25. A ball of density and radius r is dropped on the surface of a liquid of density from certain height. If speed of ball, does not change even on entering in liquid and viscosity of liquid is , then the height from which ball dropped is, ⎡ ( )r ⎤, (1) 2g ⎢, ⎥, ⎣ 9 ⎦, , 2, , (2), , 2g ( )2 r 2, 9, , (3), , 2( )gr 2, 9, , (4), , ⎡ ( )r 2 ⎤, 2g ⎢, ⎥, ⎢⎣ 9 ⎥⎦, , 2, , Sol. Answer (4), The ball has already reached the magnitude of velocity which is equal to its terminal velocity in fluid., , v 2gh, v Terminal , , {Velocity of body fallen form height h 2gh }, 2r 2, ( )g, 9, , Equating both, 2gh , , , , 2r 2, ( )g, 9, , ⎡ ( )r 2 ⎤, h 2g ⎢, ⎥, ⎣⎢ 9 ⎦⎥, , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 105, , Mechanical Properties of Fluids, , 26. Viscous drag force depends on, (1) Size of body, , (2), , Velocity with which it moves, , (3) Viscosity of fluid, , (4), , All of these, , Sol. Answer (4), F A, , V, d, , ⎧ Where,, ⎪F Drag Force, ⎪⎪, ⎨ Viscosity of fluids, ⎪ A Area size of body, ⎪, ⎪⎩V Velocity, , 27. The terminal velocity of a small sized spherical body of radius r falling vertically in a viscous liquid is given by the, proportionality, (1) v , , 1, , (2), , r2, , v r2, , (3), , v, , 1, r, , (4), , vr, , Sol. Answer (2), vT , , 2r 2, [ ] g, 9, , So, vT r 2, 28. A spherical ball is dropped in a long column of viscous liquid. The speed v of the ball varies as function of time as, , v, , v, , v, (2), , (1), , (0, 0), , t, , v, , (3), , (0, 0), , t, , (4), , (0, 0), , t, , (0, 0), , t, , Sol. Answer (2), , VTerm., Velocity does not increases after terminal speed is achieved., , V, (0, 0), , t, , 29. Which of the following is not the property of an ideal fluid?, (1) Fluid flow is irrotational, , (2), , Fluid flow is streamline, , (3) Fluid is incompressible, , (4), , Fluid is viscous, , Sol. Answer (4), An ideal fluid is not viscous., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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106, , Mechanical Properties of Fluids, , Solution of Assignment, , 30. Water is flowing through a channel (lying in a vertical plane) as shown in the figure. Three sections A, B and C are, shown. Sections B and C have equal area of cross section. If PA, PB and PC are the pressures at A, B and C, respectively then, B, A, , h, , C, (1) PA > PB = PC, , (2), , PA < PB < PC, , (3) PA < PB = PC, , (4), , PA > PB > PC, , Sol. Answer (2), Solution by using Bernoulli's principle and equation of continuity, Comparing points A and B, AAvA = ABvB, , {equation of continuity}, , ∵ AA < AB, vA > vB, 1, 1, VA2 gh PB VB2 gh, 2, 2, ∵ vA > vB, PA , , , , {Bernoulli's equation}, , 1, 1, VA2 VB2, 2, 2, , PA < PB, , ...(1), , Now comparing C and B, AB = AC vB = vC, PB , , [equation of continuity], , 1, 1, V 2 ghB PC V 2 ghC, 2, 2, , PB + ghB = PC + ghC, ∵ hB > hC then, , ...(2), , PB < PC, Using (1) and (2), We can say,, , PA < PB < PC, , 31. A liquid flows in the tube from left to right as shown in figure. A1 and A2 are the cross-sections of the portions of the, tube as shown. The ratio of speed, , v1, will be, v2, , v1, A1, (1) A, 2, , (2), , A2, A1, , A2,v2, , A1, , (3), , A2, A1, , (4), , A1, A2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 107, , Sol. Answer (2), By equation of continuity, v1A1 = v2A2, So,, , v1 A2, , v 2 A1, , 32. Water ( = 1000 kg/m3) and kerosene ( = 800kg/m3) are filled in two identical cylindrical vessels. Both vessels, have small holes at their bottom. The speed of the water and kerosene coming out of their holes are v1 and v2, respectively. Select the correct alternative, (1) v1 = v2, , (2), , v1 = 0.8 v2, , (3), , 0.8 v1 = v2, , (4), , v 1 0. 8 v 2, , Sol. Answer (1), Velocity of efflux for small holes 2gh, Which clearly is independent of '' (density), So, v1 = v2, 33. A tank is filled with water to a height H. A hole is made in one of the walls at a depth D below the water surface. The, distance x from the foot of the wall at which the stream of water coming out of the tank strikes the ground is given, by, (1) x = 2 [D (H – D)]1/2, , (2), , x = 2 (gD)1/2, , (3), , x = 2 [D (H + D)]1/2, , (4), , None of these, , Sol. Answer (1), Velocity of efflux 2gD v, Say time taken by water to travel the vertical distance of (H – D) = 't', 1, Using s ut at 2, 2, , ⎧ Where,, ⎪s H D, ⎪, ⎨, ⎪u 0, ⎪⎩a g, , We get,, , 't ' , , 2(H D ), g, , D, , v, , H, x, , Now, x = v × t, Substituting the values, , x 2gD , , 2(H D ), g, , x = 2[D (H – D)]1/2, 34. A tank is filled with water and two holes A and B are made in it. For getting same range, ratio of, h /h is, , B, A, , (1) 2, , (2), , 1, 2, , (3), , 1, 3, , (4), , 1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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108, , Mechanical Properties of Fluids, , Solution of Assignment, , Sol. Answer (4), , For hole 'A', , For hole 'B', , Velocity of efflux 2g ( x h), , Velocity of efflux 2gh, , R = 2[(x + h')h]1/2, , R = 2[h'(x + h)]1/2, , ...(1), , ...(2), , Equating (1) and (2), We get, 2[(x + h')h]1/2 = 2[h'(x + h)]1/2, (x + h')h = h'(x + h), h = h', , , h, 1, h, , 35. Water is filled in a tank upto 3 m height. The base of the tank is at height 1 m above the ground. What should be, the height of a hole made in it, so that water can be sprayed upto maximum horizontal distance on ground?, , 3m, , 1m, (1) 3 m from ground, , (2), , 1.5 m from ground, , (3) 1.5 m from base of tank, , (4), , 2 m from ground, , Sol. Answer (4), Let height of hole from the base of container be h, Velocity of efflux 2g (3 h ), R = 2[(h + 1) (3 – h)]1/2, R = 2(–, , h2, , + 2h +, , [R = Range proved in Q. 33], , 3)1/2, , dR, 2h 2, dh, , If, , dR, 0, then range would be more for corresponding height, dh, , So, 0 = – 2h + 2, , h=1, , Height from the ground = 1 + 1 = 2 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 109, , 36. Soap helps in cleaning clothes, because, (1) It attracts the dirt particles, (2) It decreases the surface tension of water, (3) It increases the cohesive force between water molecules, (4) It increases the angle of contact, Sol. Answer (2), Soap helps cleaning clothes because, it decreases the surface tension of water thus water molecules penetrate, easily into dirt and oil., 37. On increasing temperature of a liquid, its surface tension generally, (1) Increases, , (2), , Decreases, , (3) Remains constant, , (4), , First increases and then decreases, , Sol. Answer (2), On increasing the temperature energy increases hence surface tension decreases. Because surface tension is, nothing but some extra energy required by surface molecules to stay at the place., 38. The raincoats are made water proof by coating it with a material, which, (1) Absorb water, , (2), , Increase surface tension of water, , (3) Increase the angle of contact, , (4), , Decreases the density of water, , Sol. Answer (3), Raincoats are coated with material which increase the angle of contact, so water does not penetrates inside the, layer., 39. An iron needle slowly placed on the surface of water floats because, (1) It displaces water more than its weight, (2) The density of material of needle is less than that of water, (3) Of surface tension, (4) Of its shape, Sol. Answer (3), , F sin, F, O, , F sin, F, , O, , F cos, , F cos, Mg, , Needle floats due to surface tension of water which balances the weight of needle., In equilibrium 2F sin = mg, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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110, , Mechanical Properties of Fluids, , Solution of Assignment, , 40. Two water droplets merge with each other to form a larger droplet. In this process, (1) Energy is liberated, , (2), , Energy is absorbed, , (3) Energy is neither liberated nor absorbed, , (4), , Some mass is converted into energy, , Sol. Answer (1), Work is done when we break a drop into 'n' drops equal to 4R2(n1/3 – 1), So energy will be liberated if we merge back those drops., 41. The radius of a soap bubble is r. The surface tension of soap solution is ‘S’. Keeping temperature constant, the radius, of the soap bubble is doubled. The energy necessary for this will be, (1) 24 r 2S, , 8 r 2S, , (2), , (3), , 16 r 2S, , (4), , 12 r 2S, , Sol. Answer (1), Work done in making a soap bubble of radius r = 4 r2S × 2 = 8 r2S, Energy of bubble = 8, , r2S, , = Er, , ⎧Multiply by 2 due ⎫, ⎨, ⎬, ⎩ to two free surface ⎭, , Work done in making a 2r radius soap bubble = 4 (2r2) S × 2 = 32 r2S, Energy of bubble = 32 r2S = E2r, So energy required to expand a bubble from r to 2r will be equal to E2r – Er, Substituting values, We get,, 32 r2S – 8 r2S = 24 r2S, 42. The surface tension of a liquid is 5 N/m. If a film is held on a ring of area 0.02 m2, its surface energy is about, (1) 5 × 10–2 J, , (2), , 2.5 × 10–2 J, , (3), , 2 × 10–1 J, , (4), , 3 × 10–1 J, , Sol. Answer (3), Surface energy = surface tension × area of film × number of free surface, = 5 × 0.02 × 2, = 2 × 10–1 J, 43. Two soap bubbles having radii 3 cm and 4 cm in vacuum, coalesce under isothermal conditions. The radius of the, new bubble is, (1) 1 cm, , (2), , 5 cm, , (3), , 7 cm, , (4), , 3.5 cm, , Sol. Answer (2), Energy initial = Energy final, 8 (3)2S + 8 (4)2S = 8 (r)2S, , {Surface tension remains constant throughout process, , (3)2 + (4)2 = (r)2, , , 5 cm = r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 111, , 44. The excess pressure in a soap bubble is double that in other one. The ratio of their volume is, (1) 1 : 2, , (2), , 1:8, , (3), , 1:4, , (4), , 1:1, , Sol. Answer (2), Excess pressure in soap bubble =, , 4S, R, , Let for first bubble,, P, , ⎧S Surface tension, ⎨, ⎩R Radius, , 4S, R, , For second bubble,, 2P , , ⎧S Surface tension, ⎨, ⎩ x Radius, , 4S, x, , Substitute value of P, , , , 2, , 4S 4S, , R, x, , x, , R, 2, , Ratio of Radii , , R/2 1, , R, 2, , So, Ratio of volume = (Ratio of Radii)3, 3, , 1, ⎛ 1⎞, ⎜ ⎟ , 8, ⎝2⎠, 45. The work done to break a spherical drop of radius R in n drops of equal size is proportional to, (1), , 1, 1, n 2/3, , 1, 1, n 1/3, , (2), , (3), , n1/3 – 1, , (2), , Because the wick attract s the oil, , (4), , n4/3 – 1, , Sol. Answer (3), , ∵ Volume = constant, i.e.,, , 4, 4, R 3 n r 3, 3, 3, , Radius of each new droplet , , R, , n1/3, Work done to break into 'n' drops = S [n × 4r2 – 4R3], = S × 4R2 [n1/3 – 1], 46. The kerosene oil rises up in the wick of a lamp, (1) Due to high surface tension of oil, , (3) Because wick decreases the surface tension of oil (4), , Due to capillaries formed in the wick, , Sol. Answer (4), Capillary action is responsible., Wick has a lot of capillaries which help the oil rise., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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112, , Mechanical Properties of Fluids, , Solution of Assignment, , 47. Ploughing help to retain water by soil, (1) By creating capillaries, , (2), , By breaking capillaries, , (3) By turning the soil upside down, , (4), , None of these, , Sol. Answer (2), By breaking capillaries as they do not allow water to seep inside., 48. A capillary tube of radius r is immersed in a liquid and mass of liquid, which rises up in it is M. If the radius of tube, is doubled, then the mass of liquid which will rise in capillary tube will be, (1) 2 M, , (2), , M, , (3), , M/2, , (4), , M/4, , Sol. Answer (1), M = v, , ⎧ Area r 2, ⎪, 1, ⎪, ⎨Height , r, ⎪, ⎪ Where r radius of cube, ⎩, , Mass volume of tube, Mass height × Area, Mass , , 1 2, r, r, , ⎛, 2S ⎞, ⎜h , ⎟, rgh, ⎝, ⎠, , Mass r, If radius doubles, mass of liquid that rises up also doubles., 49. A massless inextensible string in the form of a loop is placed on a horizontal film of soap solution of surface, tension T. If film is pierced inside the loop and it convert into a circular loop of diameter d, then the tension, produced in string is, , (1) Td, , Td, , (2), , (3), , d2T, , (4), , d 2T, 4, , Sol. Answer (1), , SF, T cos , T, , SF, T cos , , , T sin T sin , , T, , T cos , , , , , T sin , , , , T cos , , T sin , , By force balancing in vertical direction, SF = 2T sin , SF = 2T , S × 2r × 2 = 2 × T × , S × 2r = Tension, S × d = Tension, , ∵ S=T, So, Tension = Td, , ⎧∵ is small, ⎨, ⎩sin , , ⎧r radius, ⎨, ⎩d diameter, , ⎧ Where,, ⎪S Force due to surface tension, ⎪ F, ⎪, ⎨T Tension in string, ⎪ Small angle, ⎪, ⎪⎩S Surface tension, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 113, , SECTION - B, Objective Type Questions, 1., , The atmospheric pressure at a place is 105 Pa. If tribromomethane (specific gravity = 2.9) be employed as, the barometric liquid, the barometric height is, (1) 3.52 m, , (2), , 1.52 m, , (3), , 4.52 m, , (4), , 2.52 m, , Sol. Answer (1), 1 atm 105 Pa = 76 cm of Hg, Density of Hg = 13.6, Hg × g × hHg = TBM × g × hTBM, Substituting values, 13.6 × 76 = 2.9 × h, , ⎧Hg density of Hg, ⎪, ⎪hHg height of Hg, ⎪, ⎨TBM density of, ⎪, tribromomethane, ⎪, ⎪hTBM height of tribromomethane, ⎩, , 3.52 m = h, 2., , A large vessel of height H, is filled with a liquid of density , upto the brim . A small hole of radius r is made, at the side vertical face, close to the base. The horizontal force is required to stop the gushing of liquid is, (1) (gH)r2, , (2), , gH, , (3), , gHr, , gr2, , (4), , Sol. Answer (1), Pressure close to the base = gH, Force required = pressure × area of hole = gH(r2), 3., , A vertical U-tube of uniform cross-section contains water in both the arms. A 10 cm glycerine column, (R.D. = 1.2) is added to one of the limbs. The level difference between the two free surfaces in the two limbs, will be, (1) 4 cm, , (2), , 2 cm, , (3), , 6 cm, , (4), , 8 cm, , Sol. Answer (2), Let the difference between 2 limbs be x, , x, , Pressure on the line Y Y' should be same below both limbs, so, , = 1.2, , glycerine × g × h = water × g × (h + x), , Y, , 1.2 × 10 = 1 × (h + x), , 4., , h = 10 cm, , P, , P, , Y, , 2 cm = x, , The pressure at the bottom of a water tank is 4 P, where P is atmospheric pressure. If water is drawn out till, the water level decreases by, , (1), , 3P, 8, , (2), , 3, th, then pressure at the bottom of the tank is, 5, 7P, 6, , (3), , 11 P, 5, , (4), , 9P, 4, , Sol. Answer (3), Let height of water in tank be h, So, 4P – P = wgh, , ...(1), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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114, , Mechanical Properties of Fluids, , ∵, , 3, 2, water taken out th water is left to exert pressure, 5, 5, P P , , P P , P , 5., , Solution of Assignment, , 2, w gh, 5, 2, 3P, 5, , [From eq. (1)], , 11 P, 5, , A air bubble rises from bottom of a lake to surface. If its radius increases by 200% and atmospheric pressure, is equal to water coloumn of height H, then depth of lake is, (1) 21 H, , (2), , 8H, , (3), , 9H, , (4), , 26 H, , Sol. Answer (4), Let initial radius be = r, Final radius = r + 200% of r, 3r, , = 3r, Atmospheric pressure = gH, , P = gH, (1), P, , h, , Let depth of the lake be h, r, , So, pressure at the bottom of lake = gH + gh, , P= gH + gh (2), , Using P1V1 = P2V2, gH , , 4, 4, (3r )3 (gH gh ) r 3, 3, 3, , gH , , 4, 4, 4, 27r 3 (gH ) r 3 gh r 3, 3, 3, 3, , Solving this equation we get, 26 H = h, 6., , A piece of gold weighs 10 g in air and 9 g in water. What is the volume of cavity?, (Density of gold = 19.3 g cm–3), (1) 0.182 cc, , (2), , 0.282 cc, , (3), , 0.382 cc, , (4), , 0.482 cc, , Sol. Answer (4), , Vg, , Vc, , When dipped in water, Wapp = Wair – FB, 9 gm × g = 10 gm × g – FB, 1 × g = FB, Now (total volume displaced) × w × g = 1 × g, (Vc + Vg) × 1 = 1, Vc 1 , , ⎧ Where,, ⎪V volume of cavity, ⎪ c, ⎪Vg volume of gold, ⎪, ⎪⎪Wapp 9 gm, ⎨, ⎪Wair 10 gm, ⎪F force of buoyancy, ⎪ B, ⎪w density of water 1, ⎪, ⎪⎩g density of gold 19.3, , Mass of gold in air, 10, 1, 0.482 cc, g, 19.3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 7., , Mechanical Properties of Fluids, , 115, , A block of ice floats in an oil in a vessel when the ice melts, the level of oil will, (1) Go up, , (2), , Go down, , (3) Remain same, , (4), , Go up or down depending on quantity of ice, , Sol. Answer (2), Since block of ice is displacing some oils to stay afloat when the ice block melts level of oil will go down., 8., , An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in, wire reduces by, (1) 3 : 1, , 10, mm. The ratio of relative densities of the object and liquid is, 3, (2) 1 : 3, (3) 1 : 2, , (4), , 2:1, , Sol. Answer (1), , FL, AY, Elongation force and force is due to weight, , ⎧Let density of liquid , ⎪, ⎨Let density of object , ⎪Mass of object M, ⎩, , L , , So elongation weight, L1 weight, , ...(1), , {When not submerged in liquid}, , L2 apparant weight ...(2), , {When submerged in liquid}, , Dividing (1) by (2), Mg, 10, , Mg , 10, Mg , 10 , , 3, , , , 1, 1, 1, 3, , , , 1, 1, , , , , Solving this we get, 1, , 3, So relative densities of object () and liquid () is 3 : 1, , 9., , Water flows in a stream line manner through a capillary tube of radius a. The pressure difference being P and the, a, and the pressure is increased to 4P, then the rate of flow becomes, rate of flow is Q. If the radius is reduced to, 4, (1) 4Q, , (2), , Q, 2, , (3), , Q, , (4), , Q, 64, , Sol. Answer (4), Rate of flow pressure difference × (radius)4, Q P × a4, So,, , Q1 P1 a14, , Q2 P2 a24, , Q1, , Q2, , , , Q2 , , P a4, ⎛a⎞, 4P ⎜ ⎟, ⎝4⎠, , 4, , , , Pr 4 ⎪⎫, ⎪⎧, ⎨∵ Q , ⎬, 8L ⎪⎭, ⎪⎩, , 64, 1, , Q1 Q, , 64 64, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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116, , Mechanical Properties of Fluids, , Solution of Assignment, , 10. A vessel contain a liquid has a constant acceleration 19.6 m/s2 in horizontal direction. The free surface of water, get sloped with horizontal at angle, ⎡ 1⎤, (1) tan1 ⎢ ⎥, ⎣2⎦, , (2), , ⎡ 1 ⎤, sin1 ⎢, ⎥, ⎣ 3⎦, , (3), , tan1 ⎡ 2 ⎤, ⎣ ⎦, , (4), , ⎡ 2 ⎤, sin1 ⎢, ⎥, ⎣ 5⎦, , Sol. Answer (4), , tan , , a 19.6, , 2, g, 9.8, , a, a, , tan = 2, sin , , , , ⎡ 2 ⎤, ⇒ sin1 ⎢, ⎥, ⎣ 5⎦, , 2, 5, , g, , , , 11. A cylinder containing water, stands on a table of height H. A small hole is punched in the side of cylinder at, its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of, water in the cylinder is, (1) H, , (2), , R, , (3), , RH, , (4), , R 2/4H, , Sol. Answer (4), Let depth of water in cylinder be x, So velocity (v) of efflux 2gx, , x, , Time taken (t) by water to travel vertical distance of H, , , , v, , 2H, g, , H, , Range = v × t, , R 2gx , , R, , 2H, g, , Solving this we get, , R2, x, 4H, 12. Air streams horizontally past an air plane. The speed over the top surface is 60 m/s and that under the bottom, surface is 45 m/s. The density of air is 1.293 kg/m3, then the difference in pressure is, (1) 1018 N/m2, , (2), , 516 N/m2, , (3), , 1140 N/m2, , (4), , 2250 N/m2, , Sol. Answer (1), Applying Bernoullis equation, P1 + gh +, , 1 2, 1, v1 = P2 + gh + v22, 2, 2, , 1, × [v12 – v22] = P2 – P1 = P, 2, 1, × 1.293 [(60)2 – (45)2] = P, 2, 1018 N/m2 P, , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 117, , 13. Two water pipes P and Q having diameter 2 × 10–2 m and 4 × 10–2 m respectively are joined in series with, the main supply line of water. The velocity of water flowing in pipe P is, (1) Four times that of Q, (3), , 1, times that of Q, 2, , (2), , Two times that of Q, , (4), , 1, times that of Q, 4, , Sol. Answer (1), Rate of flow through both pipes will be same, i.e., Q1 = Q2, V1 V2, , t, t, , r12 l1 r22 l 2, , t, t, l1, l2, ⎛, ⎞, ⎜ Where t VP and t VQ ⎟, ⎝, ⎠, , , , d12, d 2, VP 2 VQ, 4, 4, 2, , ⎛d ⎞, VP ⎜ 2 ⎟ VQ, ⎝ d1 ⎠, ⎛ 4 10 2, VP ⎜, ⎜ 2 102, ⎝, , 2, , ⎞, ⎟⎟ VQ, ⎠, , VP = 4VQ, 14. At what speed, the velocity head of water is equal to pressure head of 40 cm of mercury?, (1) 2.8 m/s, , (2), , 10.32 m/s, , (3), , 5.6 m/s, , (4), , 8.4 m/s, , Sol. Answer (2), 1, , V 2 = mercury gh, 2 water, , V 2, , mercury, water, , g h, , 2 13.6 9.8 , , 40, 100, , V = 10.32 m/s, 15. If the terminal speed of a sphere of gold (density 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5, kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the same size in the same liquid., (1) 0.2 m/s, , (2), , 0.4 m/s, , (3), , 0.1 m/s, , (4), , 0.133 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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118, , Mechanical Properties of Fluids, , Solution of Assignment, , Sol. Answer (3), Vterminal , , 2a 2, ( ) g, 9, , ⎧ Where, ⎪, ⎨ density of material, ⎪ density of liquid, ⎩, , VT ( – ), , , , , VT1, VT2, , , , gold liquid, , ⎧Given,, ⎪V 0.2 m/s, ⎪ T1, ⎪V V , ?, ⎪ T2, ⎨, 3, ⎪gold 19.5 kg/m, ⎪, 3, ⎪liquid 1.5 kg/m, ⎪, 3, ⎩silver 10.5 kg/m, , silver liquid, , 0.2 19.5 1.5, , V, 10.5 1.5, , V = 0.1 m/s, , 16. Three capillaries of length L,, , L, L, r, r, and, are connected in series. Their radii are r,, and, respectively. Then, 2, 3, 2, 3, , if stream-line flow is to be maintained and the pressure across the first capillary is P, then, (1) The pressure difference across the ends of second capillary is 8P, (2) The pressure difference across the third capillary is 43P, (3) The pressure difference across the ends of second capillary is 16P, (4) The pressure difference across the third capillary is 59P, Sol. Answer (1), , P1, , P2, , P3, , L, , L/2, , L/3, , r, , r/2, , r/3, , P1 = P (given), , ∵ Rate of flow will be same across all pipes, So, pressure across the pipe , , , , , , L /r 4, P1, , P2 ⎛ L /2, ⎜, ⎜ r /2 4, ⎝, , ⎞, ⎟, ⎟, ⎠, , , , ⎧rate of flow of liquid (Q ), ⎪, ⎨, Pr 4, Q, , ⎪, 8 L, ⎩, , length, (radius)4, , 1, 8, , ThenP2 = 8P1, 17. A large open tank has two holes in its wall. One is a square of side a at a depth x from the top and the other, is a circular hole of radius r at depth 4x from the top. When the tank is completely filled with water, the, quantities of water flowing out per second from both holes are the same. Then r is equal to, (1) 2a, , (2), , a, , (3), , a, 2, , (4), , a, , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 119, , Sol. Answer (3), Since quantities of water flowing out of both holes is same, Area of hole × velocity of efflux = constant, , ⎧ A1 Area of square hole, ⎪, ⎪V1 Velocity of efflux from square hole 2gx, ⎪, ⎨ A2 Area of circular hole, ⎪V Velocity of efflux from, ⎪ 2, ⎪, circular hole 2g (4 x ), ⎩, , So, A1 × V1 = A2 × V2, Substituting values., , a2 2gx r 2 8gx, a2 = 2r2, , , a, 2, , r, , 18. If T is the surface tension of a fluid, then the energy needed to break a liquid drop of radius R into 64 equal, drops is, (1) 6R2T, , (2), , R2T, , (3), , 12R2T, , (4), , 8R2T, , Sol. Answer (3), Work done = surface tension × change in area, Since volume will remain equal, Let us assume radius of new drop = r each, , , 4, 4, R 3 64 r 3, 3, 3, , , , R, r, 4, , W = T A, = T [n × 4r2 – 4R2], 2, ⎡, ⎤, ⎛R ⎞, T ⎢ 64 4 ⎜ ⎟ 4R 2 ⎥ 12R 2T, ⎝4⎠, ⎣⎢, ⎦⎥, , 19. The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective, mass ratio is, (1) 1 : 16, , (2), , 1 : 64, , (3), , 1:4, , (4), , 1:8, , Sol. Answer (2), P , , 2T, R, , Pressure , , , , 1, Radius, , ⎧ Where,, ⎪P Excess pressure, ⎪ 0, ⎨, ⎪T Surface tension, ⎪⎩R Radius, , P1 R2, , P2 R1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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120, , Mechanical Properties of Fluids, , , , Solution of Assignment, , 4P R2, , P, R1, , 4R1 = R2, , , R1 1, , R2 4, ⎧ Where,, ⎪M Mass, ⎪, ⎨, ⎪V Volume, ⎪⎩ Density, , M=V×, And V R3 M R3, is same for both, M R3, 3, , So,, , 3, , M1 ⎛ R1 ⎞, 1, ⎛ 1⎞, ⎜, ⎟ ⎜ ⎟ , M2 ⎝ R2 ⎠, 64, ⎝4⎠, , 20. The work done in blowing a soap bubble of 10 cm radius is (surface tension of soap solution is, 0.03 N/m)., (1) 37.68 × 10–4 J, , (2), , 75.36 × 10–4 J, , (3), , 126.82 × 10–4 J, , (4), , 75.36 × 10–3 J, , Sol. Answer (2), Work done = surface tension × change in area × number of free surfaces = S × A × 2, = 0.03 × 4 × (10 × 10–2)2 × 2, = 75.36 × 10–4 J, 21. A glass capillary tube of inner diameter 0.28 mm is lowered vertically into water in a vessel. The pressure to, be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel is, (surface tension of water = 0.07 N/m and atmospheric pressure = 105 N/m2)., (1) 103, , (2), , 99 × 103, , (3), , 100 × 103, , (4), , 101 × 103, , Sol. Answer (4), Height of liquid in capillary , , 2T, h, r g, , ⎧ Where,, ⎪T Surface tension, ⎪⎪, ⎨r Radius of capillary, ⎪ Density of liquid, ⎪, ⎪⎩P0 Atmospheric pressure, , Pressure we need to apply = gh + P0, Substitute value of h, , P g , , P, , 2T, 2T, 4T, P0 , P0 , P0, r g, r, d, , 4 0.07, (0.28 103 ), , ⎧Given,, ⎪, ⎨T 0.07 N/m, ⎪d 0.28 mm, ⎩, , P0 = 1000 Nm–2 + 105 Nm–2, , P = (103 + 105) Nm–2 = 101 × 103 Nm–2, , 22. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same, capillary tube. If the density of mercury is 13.6 kg/m3 and angle of contact is 135°, the ratio of surface tension, for water and mercury is (angle of contact for water and glass is 0°)., (1) 1 : 0.5, , (2), , 1:3, , (3), , 1 : 6.5, , (4), , 1.5 : 1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 121, , Sol. Answer (3), , h, , 2T cos , r g, , For water,, , 10 cm , , 2 Tw cos0, r 1 g, , ...(1), , {Tw – Surface tension of water, , For mercury,, , 3.42 cm , , 2 TM cos135, ...(2), r 13.6 g, , {TM – Surface tension of mercury, , Dividing Eqn (1) by (2), , 2 Tw 1 r 13.6 g, 10, , 1, 3.42, r 1 g 2 TM , 2, , , T, 10, 2 13.6 w, 3.42, TM, , , , T, 10, w, 3.42 1.41 13.6 TM, , , , T, 1, w, 6.5 TM, , 23. A spherical drop of water has 1 mm radius. If the surface tension of water is 75 × 10–3 N/m, then difference, of pressure between inside and outside of the drop is, (1) 35 N/m2, , (2), , 70 N/m2, , (3), , 140 N/m2, , (4), , 150 N/m2, , Sol. Answer (4), Excess pressure , , , ⎧T surface tension, ⎨, ⎩R radius, , 2T, R, 2 75 10 3, 1 10 3, , = 150 N/m2, 24. A capillary tube is dipped in water and it is 20 cm outside water. The water rises upto 8 cm. If the entire, arrangement is put in freely falling elevator the length of water column in the capillary tube will be, (1) 20 cm, , (2), , 4 cm, , (3), , 10 cm, , (4), , 8 cm, , Sol. Answer (1), If entire arrangement is in free fall then the weight of water in capillary will be balanced by pseudo force which, would be equal to the weight of water., Hence, surface tension has no weight to balance so full capillary will be filled with water., 25. If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the surface, tension of soap solution will be (r = 1 cm, density of oil = 0.8 g/cm3), (1) 3.9 N/m, , (2), , 3.9 × 10–2 N/m, , (3), , 3.9 × 10–3 N/m, , (4), , 3.9 × 10–1 N/m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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122, , Mechanical Properties of Fluids, , Solution of Assignment, , Sol. Answer (2), Pressure due to oil column = oil × g × hoil , , 0.08 103, (102 )3, , 9.8 2 103 15.68, , Now, excess pressure = pressure due to oil column, , , , 4T, 15.68, R, , 4 T, 1 102, , 15.68, , T = 3.92 × 10–2 N/m, 26. There is small hole in a hollow sphere. The water enters in it when it is taken to a depth of 40 cm under water., The surface tension of water is 0.07 N/m. The diameter of hole is, (1) 7 mm, , (2), , 0.07 mm, , (3), , 0.0007 mm, , (4), , 0.7 m, , Sol. Answer (2), Let take g = 10 m/s2, For water to enter the sphere, pressure required is = gh, 1 10 , , = 4000, , 40, 1000, 100, N, , m2, , ( = 1000 kg/m3), R, , = excess pressure, , Let the hole have radius = R, Excess pressure , 4000 , , 2T, R, , [One surface air, one surface water], , 2 0.07, R, , , , 2R = 0.07 × 10–3 m, , , , d, , = 0.07 mm, , 27. Two equal drops are falling through air with a steady velocity of 5 cm/second. If two drops coalesce, then new, terminal velocity will be, (1) 5 × (4)1/3 cm/s, , (2), , 5 2 cm/s, , (3), , 5, cm/s, 2, , (4), , 5 × 2 cm/s, , Sol. Answer (1), VTerminal r2, If initial radius = r, let new radius = R, Then 2 ×, , 4 3, 4, r = R3, 3, 3, , (2)1/3 r = R, VT R2, (2)2/3 r 2, , (For bigger drops), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , , , VT, , smaller drop, , VT, , bigger drop, , Mechanical Properties of Fluids, , , , 123, , r2, (2)2/3 r 2, , 5, 1, , x (2)2/3, , 5 × (2)2/3 = x, 5 × (4)1/3 cm/s = x, 28. A small drop of water falls from rest through a large height h in air; the final velocity is, (1) Proportional to, , h, , (3) Inversely proportional to h, , (2), , Proportional to h, , (4), , Almost independent of h, , Sol. Answer (4), Since drop is falling from a large height it achieves its terminal velocity and then there is no further increase, in velocity so v is independent of 'h' if 'h' is very large., 29. A spring balance reads 200 gF when carrying a lump of lead in air. If the lead is now immersed with half of, its volume in brine solution, what will be the new reading of the spring balance? specific gravity of lead and, brine are 11.4 and 1.1 respectively, (1) 190.4 gF, , (2), , 180.4 gF, , (3), , 210 gF, , (4), , 170.4 gF, , Sol. Answer (1), W ' = W – FB, , v g , , ⎧ Where,, ⎪W apparent weight, ⎪⎪, ⎨W read weight = actual weight of body in vaccum, ⎪ density of solution (1.1), ⎪, ⎪⎩ density of material (11.4), , v, g, 2, , ⎞, ⎛, v g ⎜ 1 , ⎟, ⎝ 2 ⎠, 1.1 ⎞, ⎛, W 200 ⎜ 1 , ⎟, ⎝ 11.4 2 ⎠, = 190.35 gF, , 30. A liquid mixture of volume V, has two liquids as its ingredients with densities and . If density of the mixture, is , then mass of the first liquid in mixture is, (1), , V [ 1], [ ], , (2), , V [ ], [ ], , (3), , V ( ), , , (4), , V [1 ], [ ], , Sol. Answer (3), Let mass of liquid with density = M1, Let mass of liquid with density = M2, Total volume = V, Net density of mixture = , Now,, Total mass = M1 + M2, , , V = M1 + M2, , , , M2 = V – M1, , ...(1), , Total mass, ⎡, ⎤, ⎥, ⎢∵, V, ⎣, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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124, , Mechanical Properties of Fluids, , Solution of Assignment, , Now,, , , , (M1 M2 ), Total mass, , Total volume ⎛ M1 ⎞ ⎛ M2 ⎞, ⎜ ⎟⎜ ⎟, ⎝, ⎠ ⎝, ⎠, , Substituting value of M2 from equation (1), , , M1 (V M1 ), M1 (V M1 ), , , , , Solving this we get, , M1 , , V ( ), , , SECTION - C, Previous Years Questions, 1., , The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the, speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is, [Re-AIPMT-2015], V 2R, nr, , (1), , (2), , VR 2, n2r 2, , (3), , VR 2, nr 2, , (4), , VR 2, n3 r 2, , Sol. Answer (3), 2., , Water rises to a height h in capillary tube. If the length of capillary tube above the surface of water is made, less than h, then, [Re-AIPMT-2015], (1) Water does not rise at all, (2) Water rises upto the tip of capillary tube and then starts overflowing like a fountain, (3) Water rises upto the top of capillary tube and stays there without overflowing, (4) Water rises upto a point a little below the top and stays there, , Sol. Answer (3), 3., , A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming, that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the, [AIPMT-2015], direction of the force will be (Pair = 1.2 kg/m3), (1) 2.4 × 105 N, downwards, , (2), , 4.8 × 105 N, downwards, , (3) 4.8 × 105 N, upwards, , (4), , 2.4 × 105 N, upwards, , Sol. Answer (4), , P1 , , 1 2, 1, v 1 P2 v 22, 2, 2, , , , , , , , , , 1, v 22 v 12 P1 P2, 2, , 1, F, v 22 v 12 , 2, A, , , , P2 = P, V2 = 40 m/s, V1 = 0, P1 = P0, , , , 1, F, 1.2 402 02 , 2, 250, , F = 2.4 × 105 N, upward (because P1 > P2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 4., , Mechanical Properties of Fluids, , 125, , A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V., If T is the surface tension of the liquid, then, [AIPMT-2014], , 1 1, (1) Energy = 4VT ⎛⎜ ⎞⎟ is released, ⎝r R⎠, , (2), , ⎛1 1⎞, Energy = 3VT ⎜ ⎟ is absorbed, ⎝r R⎠, , ⎛1 1 ⎞, (3) Energy = 3VT ⎜ ⎟ is released, ⎝r R⎠, , (4), , Energy is neither released nor absorbed, , Sol. Answer (3), Let n drops of radius r coaslece to form a big drop of radius R, n, , 4 3 4, r R 3 V, 3, 3, , n × 4r3 = 4R3 = 3V ...(1), Energy = T.A, = T [n × 4r2 – 4R2], , ⎡ n 4r 3 4R 3 ⎤, T, , ⎥, = ⎢, r, R ⎥⎦, ⎢⎣, , ⎡ 3V 3V ⎤, , = T⎢, R ⎥⎦, ⎣ r, ⎡1 1 ⎤, = 3VT ⎢ ⎥, ⎣r R ⎦, 5., , The wettability of a surface by a liquid depends primarily on :, , [NEET-2013], , (1) Surface tension, (2) Density, (3) Angle of contact between the surface and the liquid, (4) Viscosity, Sol. Answer (3), 6., , The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force, 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is, (1) 30 N, , (2), , 150 N, , (3), , 300 N, , (4), , 600 N, , Sol. Answer (3), Pressure applied on 1 point in a liquid spreads equally, So let P1 be pressure at neck, P2 be pressure at bottom, , 3 cm, P1, , P1 = P2, , , F1 F2, , A1 A2, , , , F2, 12, , 9 225, , F⎤, ⎡, ⎢∵ P A ⎥, ⎣, ⎦, , P2, 15 cm, , 300 N = F2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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126, 7., , Mechanical Properties of Fluids, , Solution of Assignment, , A liquid X of density 3.36 g cm–3 is poured in a U-tube, which contains Hg. Another liquid Y is poured in left, arm with height 8 cm, upper levels of X and Y are same what is density of Y?, , Y, , X, , 8 cm, , (1) 0.8 gcc–1, , (2), , 10 cm, , 1.2 gcc–1, , 1.4 gcc–1, , (3), , 1.6 gcc–1, , (4), , Sol. Answer (1), , Y, , Pressure at 1 and 2 will be same, , X, , 8 cm, , X gHX = Y gHY + Hg g × 2, , 10 cm, , 2 cm, , 3.36 × 10 = Y × 8 + 13.6 × 2, , 2, , 1, , Solving this we get, Y = 0.8 g cc–1, 8., , A wooden ball of density D is immersed in water of density d to a depth h below the surface of water and, then released. Upto what height will the ball jump out of water?, (1), , d, h, D, , (2), , ⎛d, ⎞, ⎜ D 1⎟ h, ⎝, ⎠, , (3), , h, , (4), , Zero, , Sol. Answer (2), Force acting on ball at depth 'h' (i.e. apparent weight), F = Vg [d – D], , h, , Vg [d D], Acceleration (a ) , VD, , v, , [Mass = volume × density], , Vdg, Velocity 2ah v, , [Using, , h' (height above water) , , 9., , v2, , –, , u2, , h, , = 2as], , v 2 2 Vg [d D] h ⎡ d, ⎤, , ⎢ 1⎥ h, 2g, 2 gVD, ⎣D ⎦, , VDg, , A piece of solid weighs 120 g in air, 80 g in water and 60 g in a liquid. The relative density of the solid and, that of the liquid are respectively, (1) 3, 2, , (2), , 2,, , 3, 4, , (3), , 3, ,2, 2, , (4), , 3,, , 3, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 127, , Sol. Answer (4), , ⎡ ⎤, w w ⎢1 ⎥, ⎣ ⎦, , ⎧ Where,, ⎪, ⎨ density of liquid, ⎪ density of body, ⎩, , Inside water, ⎡ , ⎤, 80 120 ⎢1 water ⎥, solid ⎦, ⎣, , solid = 3, , [∵ water = 1], , Inside liquid, ⎡ liquid ⎤, 60 120 ⎢1 , ⎥, ⎣ solid ⎦, , Using solid = 3, We get liquid , , 3, 2, , 10. A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ1 and, ρ2 respectively. If ρ1 < ρ < ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquids is, 2 , (1) , 1, , (2), , 1, 2, , (3), , 2, 1, , (4), , 1 2, , , Sol. Answer (1), ρ1 < ρ < ρ2 (given), , 1(V – x)g, , Let volume of sphere in lower liquid = x, Force of buoyancy by lower liquid = ρ2xg, , 1, , , , Force of buoyancy by upper liquid = ρ1(V – x)g, , 2, , Force of gravity on sphere = Mg = Vρg, Balancing all the forces for vertical equilibrium, , Mg, , We get, , 2 xg, , Vρg = ρ1(V – x) g + ρ2 xg, Solving this we get, x, , So, , V ( 1 ), (2 1 ), , V x 2 , , x, 1, , ⎧ Where,, ⎪, ⎨V x volume of sphere in upper liquid, ⎪ x volume of sphere in lower liquid, ⎩, , 11. Ice pieces are floating in a beaker A containing water and also in a beaker B containing miscible liquid of, specific gravity 1.2. When ice melts, the level of, (1) Water increases in A, , (2), , Water decreases in A, , (3) Liquid in B decreases, , (4), , Liquid in B increases, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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128, , Mechanical Properties of Fluids, , Solution of Assignment, , Sol. Answer (4), For beaker 'A', Ice is floating in water, ice vice g = water vwater displaced g, , ∵ ice water, So we can say, vice vwater displaced, So after the ice melts the level of water will not change., For beaker 'B', Ice is floating in liquid with density 1.2, clearly liquid > ice, So from above analogy, vice > vliquid displaced, So when ice melts the level in beaker 'B' increases., 12. A vessel contains oil (density 0.8 g cm–3) over mercury (density 13.6 g cm–3). A homogenous sphere floats, with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in g, cm–3 is, (1) 12.8, , (2), , 7.2, , (3), , 6.4, , (4), , 3.3, , Sol. Answer (2), Let density of sphere be , , FBoil oil, , And volume be v, , oil, , Balancing forces for vertical equilibrium, Vg , , , , HgVg, 2, , , , V, g, 2, , oil Vg, 2, , Mercury, , 13.6 0.8, , 2, 2, , V, Vg FBHg Hg 2 g, , = 7.2 g cm–3, 13. Two solid pieces, one of steel and the other of aluminium when immersed completely in water have equal, weights. When the solid pieces are weighed in air, (1) The weight of aluminium is half the weight of steel (2), , Steel peice will weigh more, , (3) They have the same weight, , Aluminium piece will weigh more, , (4), , Sol. Answer (4), Apparent weight = weight in air – FBuoyancy, ∵ Apparent weight of steel and aluminium is same, , So weight of aluminium – FB on Aluminium = weight of steel – FB on steel, , ...(1), , Al VAl g – water VAl g = steel Vsteel g – water Vsteel g, steel > Al and water = 1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 129, , So (Al – 1) VAl = (steel – 1) Vsteel, ∵ steel > Al, , (steel – 1) > (Al – 1), So VAl > Vsteel, Also water VAl g > water Vsteel g, Force of buoyancy on Aluminium > Force of buoyancy on steel., Using this condition in equation (1), We get,, weight of Aluminium – weight of steel > 0, weight of Aluminium > weight of steel, 14. A piece of wood is floating in water. When the temperature of water rises, the apparent weight of the wood, will, (1) Increase, , (2), , Decrease, , (3) May increase or decrease, , (4), , Remain same, , Sol. Answer (3), When temperature of water is raised from 0 to 4°C, its density increases and after 4°C density decreases and, apparent weight FBuoyancy density of water, So apparent weight may increase or decrease., 15. A wooden block, with a coin placed on its top, floats in water as shown in the figure. The distances h and l, are shown there. After some time, the coin falls into the water, then, , Coin, l, h, , (1) Both l and h increase, , (2), , Both l and h decrease, , (3) l decreases and h increases, , (4), , l increases and h decreases, , Sol. Answer (2), When coin falls into water block has to displace lesser volume to stay afloat., Implies that block will go up and water will go down., Hence both l and h will decrease., 16. An iceberg is floating in water. The density of ice in the iceberg is 917 kg m–3 and the density of water is, 1024 kg m–3. What percentage fraction of the iceberg would be visible?, (1) 5%, , (2), , 10%, , (3), , 12%, , (4), , 8%, , Sol. Answer (2), ice × volume of ice × g = water × volume of ice inside water × g, 917 × volume of ice = 1024 × volume of ice inside water, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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130, , Mechanical Properties of Fluids, , Solution of Assignment, , Let volume of ice = V, % Volume visible , , V volume inside water, 100, V, , 917V ⎞, ⎛, ⎜ V 1024 ⎟, ⎜, ⎟ 100, ⎝, ⎠, V, , ⎛ 1024 917 ⎞, ⎜ 1024 ⎟V, ⎠ 100, ⎝, V, = 10%, 17. A piece of wax wieghs 18.03 g in air. A piece of metal is found to weigh 17.3 g in water. It is tied to the wax, and both together weigh 15.23 g in water. Then, the specific gravity of wax is, , (1), , 18.03, 17.03, , (2), , 17.03, 18.03, , (3), , 18.03, 19.83, , (4), , 15.03, 17.03, , Sol. Answer (3), Weight of wax in air = 18.03 g, Apparent weight of metal in water = 17.3 g, Apparent weight = weight in air – water Vmetal g, So weight of metal in air = apparent weight + Vmetal g, , [∵ water = 1], , = 17.3 + Vmetal × g, When wax and metal are tied together, Total weight in air = 18.03 + 17.3 + Vmetal × g, And apparent weight in water = 15.23 = weight in air – water Vwax g – water Vmetal g, 15.23 = 18.03 + 17.3 + Vmetal g – Vwax g – Vmetal g, Vwax g = 20.1, , , Mass of wax, g 20.1, density, , M⎤, ⎡, ⎢∵ V ⎥, ⎣, ⎦, , , , 18.03, g 20.1, g , , ⎡, weight ⎤, ⎢Mass , g ⎥⎦, ⎣, , So specific gravity of wax , , 18.03, 0.897 ∼ 0.9, 20.1, , ⎛ 18.03, ⎞, ⎜, 0.9 ⎟, 19.83, ⎝, ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 131, , 18. Eight equal drops of water are falling through air with a steady velocity of 10 cm–1. If the drops combine to, form a single drop big in size, then the terminal velocity of this big drop is, (1) 80 cms–1, , (2), , 30 cms–1, , (3), , 10 cms–1, , (4), , 40 cms–1, , Sol. Answer (4), Let radius of smaller drops be r, and bigger be R, When 8 such drops combine to form a bigger drop the total volume of water remains same, 4 3 4, r R 3, 3, 3, , So, 8 , , 2r = R, And we know,, VTerminal r 2, , , , , VT smaller, , , , VT bigger, , 10, , , , VT, , bigger, , VT, , bigger, , r2, R, , 2, , , , r2, 4r 2, , 1, 4, , [∵ VT, , smaller, , = 10 cms–1 (given)], , = 40 cms–1, , 19. A small spherical ball falling through a viscous medium of negligible density has terminal velocity . Another, ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have, terminal velocity, (3), , , 2, , (1) High density and low viscosity, , (2), , Low density and high viscosity, , (3) High density and high viscosity, , (4), , Low density and low viscosity, , (1) ν, , (2), , , 4, , (4), , 4ν, , Sol. Answer (4), vTerminal r2, vT1, vT2, , , , r12, r22, , Substituting values, , r2, 2, vT2 4r, , ⎧ Where,, ⎪, ⎪vT1 , ⎪, ⎨r1 r, ⎪r 2r, ⎪2, ⎪vT ?, ⎩ 2, , vT = 4, 2, , 20. Streamline flow is more likely for liquid with, , Sol. Answer (2), Streamline flow is more likely for liquid with low density and high viscosity., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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132, , Mechanical Properties of Fluids, , Solution of Assignment, , 21. An air bubble of radius 10–2 m is rising up at a steady rate of 2 × 10–3 ms–1 through a liquid of density 1.5 × 103, kg m–3, the coefficient of viscosity neglecting the density of air, will be (g = 10 ms–2), (1) 23.2 units, , (2), , 83.5 units, , (3), , 334 units, , (4), , 167 units, , Sol. Answer (4), VT , , ⎧ Where given is, ⎪, 3, 1, ⎪VT 2 10 ms, ⎪, 2, ⎪a 10 m, ⎨, 3, 3, ⎪ 1.5 10 kg m, ⎪ ∼ 0, ⎪, ⎪⎩g 10 ms2, , 2a 2, g ( ), 9, , Substituting values, 2 10 3 , , 2 10 4 10 1.5 103, 9, , 167 units, , 22. The flow of liquid is laminar or streamline is determined by, (1) Rate of flow of liquid, , (2), , Density of fluid, , (3) Radius of the tube, , (4), , Coefficient of viscosity of liquid, , Sol. Answer (1), It is decided by rate of flow of liquid, Given by Reynolds number , , vd, n, , 23. A boat carrying a number of large stones is floating in a water tank. What would happen to the water level, if, a few stones are unloaded into water?, (1) Rises, (2) Falls, (3) Remains unchanged, (4) Rises till half the number of stones are unloaded and then begins to fall, Sol. Answer (2), Previously when stones are on the boat they are increasing the weight on the boat and to balance this weight, boat needs to generate buoyancy force by displacing more water, but when stones are removed the boat starts, displacing less amount of water hence the level of water in tank falls., , Decreases in level, , 24. The velocity of a small ball of mass M and density d1 when dropped in a container filled with glycerine becomes, constant after some time. If the density of glycerine is d2, the viscous force acting on the ball is, ⎛ d ⎞, (1) Mg ⎜ 1 2 ⎟, d1 ⎠, ⎝, , (2), , Mg, , d1, d2, , (3), , Sol. Answer (1), ⎛ d ⎞, ⎛ d ⎞, Fv mg Fv vd1g vd 2 g vd1g ⎜ 1 2 ⎟ mg ⎜ 1 2 ⎟, d, d1 ⎠, ⎝, ⎝, 1 ⎠, , mg(d1 – d2), , (4), , mgd1d2, , Fv, , FB, , mg, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 133, , 25. There are two holes one each along the opposite sides of a wide rectangular tank. The cross-section of each, hole is 0.01 m2 and the vertical distance between the holes is one metre. The tank is filled with water. The, net force on the tank in newton when the water flows out of the holes is (density of water = 1000 kgm–3), (1) 100, , (2), , 200, , (3), , 300, , (4), , 400, , Sol. Answer (2), Net force = F2 – F1, = Av22 – Av12, , v1, , h, 1m, , ⎛v ⎞, ⎛v ⎞, F ma v ⎜ ⎟ Ah ⎜ ⎟ Av 2, t, ⎝ ⎠, ⎝t ⎠, , v2, , ⎡v 2gx ⎤, ⎣, ⎦, , = 2g (h + 1) A – 2ghA, = 2gA, = 1000 × 10 × 0.01 × 2, = 200 N, , 26. A hole is made at the bottom of the tank filled with water (density 1000 kg/m3). If the total pressure at the, bottom of the tank is 3 atm (1 atm = 105 N/m2), then the velocity of efflux is, (1), , 200 m/s, , 400 m/s, , (2), , (3), , 500 m/s, , (4), , 800 m/s, , Sol. Answer (2), Apply Bernoulli's theorem, 1 2, P, , gH, , , 2 v constant, , Total pressure, , At point 1, 3 atm + 0 = constant, At point 2, 1 atm +, , ⎧Given, ⎨, ⎩Total pressure 3 atm, , ...(i), , 1 2, v = constant, 2, , 1 2, ...(ii), , v, , Equate (i) and (ii), 3=1+, , 1 2, v, 2, , [Use = 1000 and 1 atm = 105 N/m2], , We get,, , v 400 m/s, 27. A horizontal pipe line carries water in stremline flow. At a point where the cross-sectional area is 10 cm2 the, water velocity is 1 ms–1 and pressure is 2000 Pa. The pressure of water at another point where the crosssectional area is 5 cm2, is, (1) 200 Pa, , (2), , 400 Pa, , (3), , 500 Pa, , (4), , 800 Pa, , Sol. Answer (3), , 10 cm2, –1, , 1 ms, P = 2000 Pa, , 5 cm2, v, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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134, , Mechanical Properties of Fluids, , A1V1 = A2V2, , Solution of Assignment, , (equation of continuity), , 10 × 1 = 5 × v, So, v = 2 ms–1, Apply Bernoulli theorem at both the points, 1, 1, × 1000 × 12 = P + × 1000 × 4, 2, 2, P = 500 Pa, , 2000 +, , 28. A rectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time, will it take to be emptied when half filled with water?, (1) 9 min, , (2), , 7 min, , (3), , 5 min, , (4), , 3 min, , Sol. Answer (2), Let time taken by height 'x' to get reduced by dx = dt, , , dt , , volume, A dx, , efflux speed, 2gx, , T, , h, , ∫0 dt ∫0, , A, , {A is area of cross-section}, , dx, , 2g, , x, , A, , A 2h, T , a g, , dx, , T h, , x, , So we can use, , h1, , T1, , T2, , h2, , 10 min, , t min, t, , a, , 10, 2, , h, h/2, , ∼ 7 min, , 29. A metal plate of area 103 cm2 rests on a layer of oil 6 mm thick. A tangential force 10–2 N is applied on it to, move it with a constant velocity of 6 cms–1. The coefficient of viscosity of the liquid is, (1) 0.1 poise, , (2), , 0.5 poise, , (3), , 0.7 poise, , (4), , 0.9 poise, , Sol. Answer (1), v, F = A, d, , 10 2 (103 10 4 ) , , 6 10 2, 6 10 3, , 0.01 poiseuille, , = 0.1 poise, , ⎧ Where,, ⎪F Force, ⎪, ⎪⎪ Coefficient of viscosity, ⎨, ⎪ A Area, ⎪v Velocity, ⎪, ⎪⎩d Thickness of layer, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 135, , 30. With an increase in temperature, surface tension of liquid (except molten copper and cadmium), (1) Increases, , (2), , Remain same, , (3) Decreases, , (4), , First decreases then increases, , Sol. Answer (3), When we increase the temperature, we are providing energy to the molecules. This increase in potential energy, causes the surface energy to drop and become less negative hence decreasing surface tension because surface, tension is nothing but surface energy per unit area., 31. Determine the energy stored in the surface of a soap bubble of radius 2.1 cm if its tension is 4.5 10–2 Nm–1., (1) 8 mJ, , (2), , 2.46 mJ, , (3), , 4.93 × 10–4 J, , (4), , None of these, , Sol. Answer (3), Energy = surface tension × surface area × number of free surfaces, = (4.5 × 10–2) × 4 × (2.1 × 10–2) × 2, = 4.98 × 10–4 J, 32. A mercury drop of radius 1.0 cm is sprayed into 106 droplets of equal sizes. The energy expended in this, process is (surface tension of mercury is equal to 32 × 10–2 Nm–1), (1) 3.98 × 10–4 J, , (2), , 8.46 × 10–4 J, , (3), , 3.98 × 10–2 J, , (4), , 8.46 × 10–2 J, , Sol. Answer (3), Energy expended = surface tension × increase in area, , (Formulae), , So, volume initially = volume of 106 drops, 3, , , , 4 ⎛ 1 ⎞, 4, ⎜, 106 r 3, ⎟, 3 ⎝ 100 ⎠, 3, , , , ⎡ ⎛ 1 ⎞3, 1 ⎤, 6⎥, ⎢⎜, ⎟, ⎣⎢⎝ 100 ⎠ 10 ⎦⎥, , 1/3, , r, , [Let radius of small drops = r], , 10–4 m = r, 2, ⎡⎛ 1 ⎞2, ⎛ 1 ⎞ ⎤, 6, , 10, , So increase in surface area 4 ⎢⎜, ⎟, ⎜ 100 ⎟ ⎥, ⎝, ⎠ ⎥⎦, ⎢⎣⎝ 10000 ⎠, , 1 ⎤, ⎡ 1, 4 ⎢, , ⎥, ⎣100 10000 ⎦, A 4 0.99, 100, Using this value in formulae, Energy 32 10 2 , , 4 0.99, 100, , [∵ Surface tension = 32 × 10–2 (given)], , = 3.98 × 10–2 J, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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136, , Mechanical Properties of Fluids, , Solution of Assignment, , 33. When a glass capillary tube of radius 0.015 cm is dipped in water, the water rises to a height of, 15 cm within it. Assuming contact angle between water and glass to be 0°, the surface tension of water is, [ρwater = 1000 kg m–3, g = 9.81 ms–2], (1) 0.11 Nm–1, , (2), , 0.7 Nm–1, , (3), , 0.072 Nm–1, , (4), , None of these, , Sol. Answer (1), h, , 2 S cos , rg, , ⎧ Where,, ⎪S surface tension ?, ⎪, ⎪⎪h height of water in capillary 15 cm, ⎨, ⎪r radius of capillary 0.015 cm, ⎪ angle of contact 0, ⎪, 2, ⎪⎩g 9.8 ms, , Substituting values, 15, 2 S 1 100, , 100 1000 0.015 9.81, , S = 0.11 Nm–1, 34. A liquid does not wet the sides of a solid, if the angle of contact is, (1) Obtuse, , (2), , 90°, , (3), , Acute, , (4), , Zero, , Sol. Answer (1), Solid will not get wet if the liquid has high surface tension (example mercury) and liquids with high surface, tension have obtuse angle of contact., 35. Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to a, smaller one?, (1) 21/2 : 1, , (2), , 1:1, , (3), , 22/3 : 1, , (4), , None of these, , Sol. Answer (3), Surface energy = surface tension × surface area, Let the radius of smaller drops be r, And that of bigger drop be R, [∵ Surface tension is same for both], , Then ratio of surface energies = ratio of surface area, = 4R2 : 4r2, = R2 : r2, , ...(1), , ∵ 2 smaller drops are forming 1 big drop so, 2, , So,, , 4 3 4, r R 3, 3, 3, , 21/3r = R, , ...(2), , Using 1 and 2 we can say that ratio of surface energies = 22/3r 2 : r 2 = 22/3 : 1, 36. The excess pressure inside a spherical drop of water is frour times that of another drop. Then their respective, mass ratio is, (1), , 1 : 16, , (2), , 8:1, , (3), , 1:4, , (4), , 1 : 64, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Mechanical Properties of Fluids, , 137, , Sol. Answer (4), Excess pressure , , P , , 2T, r, , {Where, r = radius of drop, , 1, r, , P1 r2 4, , , P2 r1 1, , ⎡, P1 1 ⎤, ⎥, ⎢∵, , P2 4 ⎦, ⎣, , V r3, 3, , 3, , V1 ⎛ r1 ⎞, 1, ⎛ 1⎞, ⎜ ⎟ ⎜ ⎟ , V2 ⎝ r2 ⎠, 64, ⎝4⎠, M = V × then, M1 V1, 1, , , M2 V2 64, , 37. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be, removed from it so that it starts moving up with an acceleration a?, (1), , 2ma, g a, , (2), , 2ma, g a, , (3), , ma, g a, , (4), , ma, g a, , Sol. Answer (1), , N, , m, , a, , mg, , mg – N = ma, , ..... (1), , Let m mass remove from ballon, , N, a, (m –m), , (m –m)g, N – (m – m)g = (m – m)a, , ..... (2), , After addition of equation (1) & (2), then, m' , , 2ma, g a, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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138, , Mechanical Properties of Fluids, , Solution of Assignment, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Hydraulic lift is based on Pascal’s Law., R : Hydrostatic pressure is a scalar quantity, , Sol. Answer (2), A : is true, R : is true, But reason is not the correct explanation. Correct explanation is, change in pressure is transferred, undiminished, from one point to the other., 2., , A : The apparent weight of a body floating on the surface of a liquid is zero., R : The net force on a body floating on the surface of a liquid is zero., , Sol. Answer (1), A : is true, R : is true, And reason is also the correct explanation., 3., , A : It is better to wash cloths in hot water than cold water., R : On increasing temperature surface tension of water decreases., , Sol. Answer (1), A : is true, R : is true and correct explanation., 4., , A : The impurities added to water may increase or decrease surface tension., R : The change in surface tension depends on the nature of impurities., , Sol. Answer (1), A is true, R is true and correct explanation., 5., , A : On increasing temperature the angle of contact generally decreases., R : With rise in temperature, the surface tension of liquid increases., , Sol. Answer (4), A : is false : Angle of contact increases with increase in temperature., R : is false : Surface tension decreases with rise in temperature., 6., , A : If air blows over the roof of a house, the force on the roof is upwards., R : When air blows over the roof, the pressure over it from out side decreases., , Sol. Answer (1), A is true, R is true and correct explanation., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 7., , Mechanical Properties of Fluids, , 139, , A : When rain drops fall through air some distance, they attain a constant velocity., R : The viscous drag of air just balances the weight of rain drops., , Sol. Answer (1), A : is true, R : is true and correct explanation., 8., , A : Bernoulli’s theorem holds good only for non-viscous and incompressible liquid., R : Bernoulli’s theorem is based on the conservation of energy., , Sol. Answer (2), A : is true, R : is true, Correct explanation of 'A' is Bernoulli's equation does not take into account the elastic energy of the fluids., 9., , A : At high altitudes (mountains), it is very difficult to stop bleeding from a cut in the body., R : At high altitude the atmospheric pressure is less than the blood pressure inside the body., , Sol. Answer (1), A : is true, R : is true and correct explanation., 10. A : When liquid drops merge into each other to form a large drop, energy is released., R : When liquid drops merge to form large drop surface tension decreases., Sol. Answer (3), A : is true, R : is false, because when large drop is formed, surface area gets reduced. Hence surface energy gets reduced, due to reduction in surface area not the surface tension., , 11. A : Excess pressure inside a soap bubble is, , 4T, (symbols have their usual meanings)., r, , R : The pressure difference across a curved surface of radius of curvature r is, , 2T, . There are two surfaces in a, r, , soap bubble., Sol. Answer (1), A : is true, R : is true and correct explanation., 12. A : Buoyant force is always vertically upward., R : Buoyant force is always opposite to the direction of acceleration due to gravity., Sol. Answer (4), A : is wrong, Buoyant force is not always vertically upwards, R : is wrong, Buoyant force is always opposite to the direction of effective acceleration., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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140, , Mechanical Properties of Fluids, , Solution of Assignment, , 13. A : Equation of continuity is A1v11 = A2v22 (symbols have their usual meanings)., R : Equation of continuity is valid only for incompressible liquids., Sol. Answer (3), A : is true, R : is false,, 14. A : Atomizer is based on the principle of Bernoulli’s theorem., R : Bernoulli’s theorem is based on the conservation of energy., Sol. Answer (2), A : is true, R : is true, but not the correct explanation, Correct explanation is, decrease in pressure forces the water to move up the tube and get sprayed., 15. A : The spiders and insects can run on the surface of water., R : Buoyant force balances the weight of insects., Sol. Answer (3), A : true, R : is false, surface tension balances the weight of insect., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Chapter, , 11, , Thermal Properties of Matter, Solutions, SECTION - A, Objective Type Questions, 1., , Temperature is a measure of, (1) Hotness or coldness, , (2), , Heat possessed by a body, , (3) Potential energy, , (4), , Thermal energy, , Sol. Answer (1), Temperature is the measure of hotness and coldness., 2., , The readings of a bath on Celsius and Fahrenheit thermometers are in the ratio 2 : 5. The temperature of the, bath is, (1) –26.66°C, , (2), , 40°C, , Sol. Answer (3), , (3), , 45.71°C, , (4), , 26.66°C, , ⎧ where, ⎫, ⎪⎪, ⎪⎪, ⎨TC temperature in Celcius, ⎬, ⎪, ⎪, ⎩⎪TF temperature in Fahrenheit ⎭⎪, , TC, T 32, F, 100, 180, , TC : TF given as 2 : 5, Let TC = 2x, TF = 5x, 2 x 5 x 32, , 10, 18, , x, , 160, 7, , So, 2 x , 3., , 320, ~ 45.7C, 7, , The pressure of a gas filled in the bulb of a constant volume gas thermometer at temperatures 0°C and 100°C are, 27.50 cm and 37.50 cm of Hg respectively. At an unknown temperature the pressure is, 32.45 cm of Hg. Unknown temperature is, (1) 30°C, , (2), , 39°C, , (3), , 49.5°C, , (4), , 29.6°C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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142, , Thermal Properties of Matter, , Solution of Assignment, , Sol. Answer (3), P P0, x x0, T T0, , , P100 P0 x100 x0 T100 T0, 32.45 27.50, T O, , 37.50 27.50 100 0, , T = 49.5°C, 4., , A graph is plotted by taking pressure along y-axis and centigrade temperature along x-axis for an ideal gas at, constant volume. x intercept of the graph is, (1) –273.15ºC, , (2), , –273.15 K, , (3), , –273ºC, , (4), , –273 K, , Pressure, , Sol. Answer (1), , –273.15°C, 5., , Temperature, , A hole is drilled in a copper sheet. The diameter of hole is 4.24 cm at 27.0°C. Diameter of the hole when it, is heated to 35°C is, (1) Less than 4.24 cm, , (2), , Equal to 4.24 cm, , (3) More than 4.24 cm, , (4), , Data insufficient, , Sol. Answer (3), Thermal expansion in this case can be imagined as a photographic enlargement, hence the diameter of hole, will also increase., 6., , The density of water is maximum at, (1) 39.2°F, , (2), , 4°F, , (3), , 0°C, , (4), , 273 K, , Sol. Answer (1), Density of water is maximum at 4°C or 39.2°F., 7., , On heating a uniform metallic cylinder length increases by 3%. The area of cross-section of its base will, increase by, (1) 1.5%, , (2), , 3%, , (3), , 9%, , (4), , 6%, , Sol. Answer (4), = 2, Area of crossection increases by 2 × 3 = 6%, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 8., , Thermal Properties of Matter, , 143, , A circular metallic disc of radius R has a small circular cavity of radius r as shown in figure. On heating the, system, , (1) R increases and r decreases, , (2), , R decreases and r increases, , (3) Both R and r increases, , (4), , Both R and r decreases, , Sol. Answer (3), Both R and r will increase, superficial expansion is always outwards., 9., , If in winter season the surface temperature of lake is 1°C, the temperature at the bottom of lake will be, (1) 1°C, , (2), , 0°C, , (3) 4°C, , (4), , All values less than 1°C are possible, , Sol. Answer (3), Temperature of the surface is the lowest and it should increase as we go down to the bottom., Temperature > 1°C, 10. A uniform copper rod of length 50 cm and diameter 3.0 mm is kept on a frictionless horizontal surface at 20°C. The, coefficient of linear expansion of copper is 2.0 × 10–5 K–1 and Young’s modulus is 1.2 × 1011 N/m2. The copper rod is, heated to 100°C, then the tension developed in the copper rod is, (1) 12 × 103 N, , (2), , 36 × 103 N, , (3), , 18 × 103 N, , (4), , Zero, , Sol. Answer (4), Since the rod is not bounded, No compressive stress, hence no tensions, 11. A seconds pendulum clock has a steel wire. The clock shows correct time at 25°C. How much time does, the clock lose or gain, in one week, when the temperature is increased to 35°C?, (steel = 1.2 × 10–5 / °C), (1) 321.5 s, , (2), , 3.828 s, , (3), , 82.35 s, , (4), , 36.28 s, , Sol. Answer (4), , T 1, , T, 2, , , 1, 1.2 105 10, 2, , T, 6.0 105, T, , Hence time lost in 1 week = 6.0 × 10–5 × T, = 6.0 × 10–5 × 7 × 24 × 3600, = 36.28 s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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144, , Thermal Properties of Matter, , Solution of Assignment, , 12. The apparent coefficient of expansion of a liquid when heated in a brass vessel is X and when heated in a, tin vessel is Y. If is the coefficient of linear expansion for brass, the coefficient of linear expansion of tin is, X Y 3, 3, , (1), , (2), , X 3 Y, 3, , (3), , X Y 2, 3, , (4), , ( X Y 2 ), 2, , Sol. Answer (2), Coefficient of expansion of liquid = [Apparent coefficient of expansion + Coefficient of expansion of vessel], Let coefficient of expansion of liquid be = x, Then, x = X + 3 brass, x = Y + 3 tin, X + 3 brass = Y + 3 tin, tin , , 3 brass Y X 3 Y X, , 3, 3, , 13. The coefficient of volume expansion of glycerin is 49 × 10–5 K–1. The fractional change in the density on a, 30°C rise in temperature is, (1) 1.47 × 10–2, , (2), , 1.47 × 10–3, , (3), , 1.47 × 10–1, , (4), , 1.47 × 10–4, , Sol. Answer (1), 2 = 1(1 – T), 2 1, , 49 105 30 ⇒, 1.47 102, 1, 1, , [∵ = 1 – 2], , 14. A solid cube is first floating in a liquid. The coefficient of linear expansion of cube is and the coefficient of volume, expansion of liquid is . On increasing the temperature of (liquid + cube) system, the cube will sink if, , (1) = 3, , (2), , > 3, , (3), , < 3, , (4), , = 2, , Sol. Answer (2), Cube will sink if expansion in liquid upon heating is more than that of cube, or 3 < , 15. A steel tape is calibrated at 20°C. On a cold day when the temperature is –15°C, percentage error in the tape will, be [steel = 1.2 × 10–5 °C–1], (1) –0.035%, , (2), , –0.042%, , (3), , 0.012%, , (4), , –0.018%, , Sol. Answer (2), % error , , , , L, 100, L, , L T, 100, L, , 1.2 105 35 100 = – 0.042%, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 145, , 16. In engines water is used as coolant, because, (1) It's good conductor of heat energy, , (2), , It has low density, , (3) It has high specific heat, , (4), , It’s bad conductor of heat energy, , Sol. Answer (3), Because it absorbs and gives off heat readily or it has high specific heat., 17. Which of the following material is used to make calorimeter?, (1) Glass, , (2), , Ebonite, , (3), , Metal, , (4), , Superconductor, , (4), , 100 cal/°C, , Sol. Answer (3), Metal is used (copper), 18. The thermal capacity of 100 g of aluminum (specific heat = 0.2 cal/g°C) is, (1) 0.002 cal/°C, , (2), , 20 cal/°C, , (3), , 200 cal/°C, , Sol. Answer (2), Thermal capacity = m.C, = 100 × 0.2 = 20 cal/°C, 19. Select correct statement related to heat, (1) Heat is possessed by a body, (2) Hot water contains more heat as compared to cold water, (3) Heat is a energy which flows due to temperature difference, (4) All of these, Sol. Answer (3), Heat is a energy which flows due to temperatures difference., 20. A block of ice at –12°C is slowly heated and converted into steam at 100°C. Which of the following curves, best represents the event?, , T, , T, , T, (1), , (2), , Q, , T, , (3), , Q, , (4), , Q, , Q, , Sol. Answer (1), , T, 100°C, , 0°C, –12°C, , Q, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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146, , Thermal Properties of Matter, , Solution of Assignment, , 21. The water equivalent of 20 g of aluminium (specific heat 0.2 cal g–1 °C–1 ), is, (1) 40 g, , (2), , 4g, , (3), , 8g, , (4), , 160 g, , Sol. Answer (2), 0.2 × 20 = 4, , [Thermal Capacity], , Water equivalent numerically equal to thermal capacity, So W = 4 g, 22. 100 g of ice (latent heat 80 cal g–1, at 0°C) is mixed with 100 g of water (specific heat 1 cal g–1 °C–1) at 80°C., The final temperature of the mixture will be, (1) 0°C, , (2), , 40°C, , (3), , +, , 100 g water 80°C, , 80°C, , (4), , < 0°C, , Sol. Answer (1), 100 g ice 0°C, , if T = 80 [Qf becomes = 0°], , Qrequired = mL, = 100 × 80, , Qavailable = mCT, , = 8000 cal, , = 100 × 1 × 80 = 8000 cal, , [Required to melt whole ice], Since Qrequired = Qavailable, Whole system will be water at equillibrium and temperature would be 0°C, 23. 200 g of ice at –20°C is mixed with 500 g of water at 20°C in an insulating vessel. Final mass of water in vessel is, (specific heat of ice = 0.5 cal g–1°C–1), (1) 700 g, , (2), , 600 g, , (3), , 400 g, , (4), , 200 g, , Sol. Answer (2), Maximum heat supplied by water, Q1 = 500 × 1 × (20 – 0), = 10,000 cal, Heat required to raise the temperature of ice upto 0°C, Q2 = 200 × 0.5 × 20, = 2000 cal, Q1 – Q2 = 8000 cal, Melts the ice, 8000 = m × 80, m = 100 g, So, mass of water is 500 + 100 = 600 g., 24. Which of the following material is most suitable for cooking utensil?, (1) Low conductivity and low specific heat, , (2), , High conductivity and low specific heat, , (3) Low conductivity and high specific heat, , (4), , High conductivity and high specific heat, , Sol. Answer (2), For cooking high conductivity and low specific heat because we do not want to waste heat energy in heating, up the vessel it self also we want the vessel to absorb as much as heat is available., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 147, , 25. Which of the following factors affect the thermal conductivity of a rod?, (1) Area of cross-section, , (2), , Length of rod, , (3) Material of rod, , (4), , All of these, , Sol. Answer (3), Thermal conductivity is material property., 26. What is the dimensional formula for thermal resistance?, (1) [M –1 L–2 T –1 K], , (2), , [ML2 T –2 K –1], , (3), , [ML–3 T 2 K –1], , (4), , [M –1 L–2 T 3 K], , Sol. Answer (4), , R, , L, KA, , [R] = [M–1 L–2 T3 K], 27. Two ends of a rod of non uniform area of cross-section are maintained at temperature T1 and T2 (T1 > T2) as, shown in the figure, , x, T2, T1, If I is heat current through the cross-section of conductor at distance x from its left face, then the variation of, I with x is best represented by, , I, , I, , I, , (2), , (1), , t, , I, , (3), , (4), , t, , t, , t, , Sol. Answer (1), In steady state rate of heat transfer is constant so heat current through any cross-section remains same with, time., 28. Four rods of same material with different radii r and length l are used to connect two reservoirs of heat at, different temperatures. Which one will conduct maximum heat?, (1) r = 1 cm, l = 1 m, , (2), , r = 2 cm, l = 2 m, , (3) r = 1 cm, l = 1/2 m, , (4), , r = 2 cm, l = 1/2 m, , Sol. Answer (4), H, , K r 2, T2 T1 , L, , H r 2 and H , , 1, L, , So the rod with maximum, , r2, ratio will conduct maximum heat., L, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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148, , Thermal Properties of Matter, , Solution of Assignment, , 29. Two walls of thickness d1 and d2, thermal conductivities K1 and K2 are in contact. In the steady state if the, temperatures at the outer surfaces are T1 and T2, the temperature at the common wall will be, (1), , K 1T1 K 2T2, d1 d 2, , K 1T1d 2 K 2T2d1, K 1d 2 K 2 d1, , (2), , (K 1d1 K 2 d 2 )T1T2, T1 T2, , (3), , (4), , K 1d1T1 K 2d 2T2, K 1d1 K 2 d 2, , Sol. Answer (2), Heat flow across both the walls will be same., , T, , K1A, K A, T T1 2 T2 T , d1, d2, , T1, , K1Td2 K1T1d 2 K 2T2d1 K 2Td1, , K1, , K2, , d1, , K T d K 2T2d1, T 11 2, K1d2 K 2d1, , T2, , d2, , 30. A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner, radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system, are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the, system is in steady state. The effective thermal conductivity of the system is, (1) K1 + K2, , (2), , K 1 3K 2, 4, , (3), , K 1K 2, K1 K 2, , (4), , 3K 1 K 2, 4, , Sol. Answer (2), , R, , L, KA, , [R is thermal resistance], , These two cylinders are like two resistances in a parallel connection., , , 1, 1, 1, , , Reff R1 R2, K eff 2R , , 2, , L, , , , L, 2, , K1 R, , L, , , , K 2 4R 2 R 2, L, , R, , K1, , 2R, , Solving for Keff we get, K, , , , K2, , K1 3K 2, 4, , 31. Four rods of same material and having the same cross section and length have been joined, as shown. The, temperature of the junction of four rods will be, , 0°, , 30°, , , 90°, (1) 20°C, , (2), , 30°C, , 60°, (3), , 45°C, , (4), , 60°C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 149, , Sol. Answer (3), Incoming heat = outgoing heat, (90° – ) + (60° – ) = – 30° + – 0°, 180° = 4, 45°C = , 32. Why it is more hotter for same distance over the top of a candle than it in the side of its flame?, (1) Conduction of heat in air is upward, (2) Heat is maximum radiated in upward direction, (3) Radiation and conduction both contribute in transferring heat upwards, (4) Convection takes more heat in upward direction, Sol. Answer (4), On the sides heating is only due to Radiation but over the top heating is due to Radiation as well as convection., 33. In gravity free space heat transfer is not possible by, (1) Conduction, , (2), , Convection, , (3), , Radiation, , (4), , Both (1) & (3), , Sol. Answer (2), For heat transfer through convention presence of gravity is a must requirement., 34. Which factor does not affect convection?, (1) Temperature difference, , (2), , The rate of movement of carrying medium, , (3) The volumetric specific heat of carrying medium, , (4), , The thermal conductivity of carrying medium, , Sol. Answer (4), The thermal conductivity of carrying medium does not affects convection process., 35. A polished plate with rough black spot is heated to a high temperature and then taken to a dark room, then, (1) Spot will appear brighter than the plate, , (2), , Spot will appear darker than the plate, , (3) Both will appear equally brighter, , (4), , Both will not be visible, , Sol. Answer (1), Rough black spot will act like a black body so radiations absorbed as well as emitted by the spot would be, more than the other parts of the plate., 36. Select the incorrect statement, (1) A body radiates at all temperatures except 0 K, (2) A good reflector is a bad radiator, (3) A colder body can radiate heat to the hotter surroundings, (4) A body does not radiate when its temperature is below 0ºC, Sol. Answer (4), 0°C is 273 K so a body will radiate., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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150, , Thermal Properties of Matter, , Solution of Assignment, , 37. “A good absorber is a good emitter” is explained by, (1) Stefan’s law, , (2), , Wien’s law, , (3) Newton’s law of cooling, , (4), , Kirchhoff’s law, , Sol. Answer (4), Kirchhoff's Law states that "A good absorber is a good emitter"., 38. The rate of radiation of energy from a hot object is maximum, if its surface is, (1) White and smooth, , (2), , Black and rough, , (3), , Black and smooth, , (4), , White and rough, , Sol. Answer (2), Black and rough surfaces absorb maximum amount of radiation and than radiate the same amount of radiations., 39. Two balls of same material and same surface finish have their diameters in the ratio 1 : 2. They are heated to the, same temperature and are left in a room to cool by radiation, then the initial rate of loss of heat, (1) Will be same for the balls, (2) For larger ball is half that of other ball, (3) For larger ball is twice that of other ball, (4) For larger ball is four times that of the other ball, Sol. Answer (4), By Stefan-Boltzmann's law, , , , H eA T 4 T04, , , , H A r2, , f or larger ball is (2)2 that of the smaller ball., 40. A black body, which is at a high temperature T K, thermal radiation emitted at the rate of E W/m2. If the, temperature falls to T/4 K, the thermal radiation emitted in W/m2 will be, (1) E, , (2), , E/4, , (3), , E/64, , (4), , E/256, , Sol. Answer (4), Radiation T4, , E1 T14, , E2 T24, 4, E T 4, , E2, T4, , E2 , , 4, , E, 256, , 41. A sphere, a cube and a thin circular plate, all made of the same mass and finish are heated to a temperature, of 200°C. Which of these objects will cool slowest, when left in air at room temperature?, (1) The sphere, , (2), , The cube, , (3) The circular plate, , (4), , All will cool at same rate, , Sol. Answer (1), Objects with more surface area cool faster [Stefan Boltzmann law]., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 151, , Thermal Properties of Matter, , 42. A body cools down from 80°C to 60°C in 10 minutes when the temperature of surroundings is 30°C. The, temperature of the body after next 10 minutes will be, (1) 30°C, , (2), , 48°C, , (3), , 50°C, , (4), , 52°C, , Sol. Answer (2), Apply newton's law of cooling, ⎛ T T ⎞, ln ⎜ 1 0 ⎟ Kt, ⎝ T2 T0 ⎠, , ⎛ 80 30 ⎞, ln ⎜, ⎟ K 10, ⎝ 60 30 ⎠, , …(1), , Let temperature of body after next 10 minutes = T, So,, , ⎛ 60 30 ⎞, ln ⎜, ⎟ K 10, ⎝ T 30 ⎠, , …(2), , Using equation 1 and 2, ln, , 5, 30, ln, 3, T 30, , 30 × 3 = 5T – 5 × 30, T = 48°C, 43. Two bodies A and B of same mass, area and surface finish with specific heats SA and SB (SA > SB ) are allowed, to cool for given temperature range. Temperature varies with time as, , , , A, , B, , (1), (0, 0), , , , (2), (0,0), , t, , B, , A, , , B, , (3), (0,0), , t, , A, , B, A, , , , (4), (0,0), , t, , t, , Sol. Answer (1), Body with more specific heat takes time to cool if initial temperature is same., 44. Instantaneous temperature difference between cooling body and the surroundings obeying Newton’s law of cooling, is . Which of the following represents the variation of ln with time t?, , ln , , ln , , (1), , ln , , (2), (0,0), , t, , ln , , (3), (0, 0), , t, , (4), (0, 0), , t, , (0, 0), , t, , Sol. Answer (2), ⎛ T T0 ⎞, ln ⎜ f, ⎟ Kt, ⎝ T T0 ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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152, , Thermal Properties of Matter, , Solution of Assignment, , If is the instantaneous temperature than, ⎛ 0 ⎞, ln ⎜ i, ⎟ Kt, ⎝ 0 ⎠, , ⎧ initial temperature, ⎫, ⎪ i, ⎪, ⎨, ⎬, ⎪⎩0 temperature of surrounding⎪⎭, , ln (i – 0) – ln ( – 0) = KT, ln ( – 0) = – Kt + ln (i – 0), Comparing to, y = mx + C, , We get a negative slope, so graph will be a straight line with decreasing slope., , In , , t, , (0, 0), , 45. Two metal spheres have radii r and 2r and they emit thermal radiation with maximum intensities at wavelengths , and 2 respectively. The respective ratio of the radiant energy emitted by them per second will be, (1) 4 : 1, , (2), , 1:4, , (3), , 16 : 1, , (4), , 8:1, , Sol. Answer (1), T , , 1, , , [wien's displacement law], , T1 2 2, So, T 2, 2, 1, and H eAT 4 ⇒ H AT 4, H1, T4 1, 4r 2, 4, , 1 2 4, H2 4 2r 2 T24 4, , H1 : H2 :: 4 : 1, 46. If temperature of sun is decreased by 1 % then the value of solar constant will change by, (1) 2%, , (2), , Sol. Answer (2), Solar constant T, S 4 T, , S, T, , So, , T, 1%, T, , (3), , –2%, , (4), , 4%, , 2, ⎡, ⎤, ⎡R ⎤, ⎢∵ S ⎢ ⎥ T 4 ⎥, ⎣r ⎦, ⎢⎣, ⎥⎦, , 4, , and, , –4%, , [given], , S, 4%, S, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 153, , 47. The value of solar constant is, (1) 2 kcal m–2 minute–1, , (2), , 20 kcal m–2 minute–1, , (3) 2 kWm–2, , (4), , 200 Wm–2, , Sol. Answer (2), 2, , ⎡R ⎤, 4, Value of solar constant ⎢ ⎥ T, ⎣r ⎦, Where,, R = Radius of sun, r = Distance of earth from sun, = Stefan's constant, T = Temperature of sun, , R, 4.65 103 radians, r, = 5.67 × 10–8 Wm–2K–4, T = 5800 K, So substituting values we get, S = (4.65 × 10–3)2 × 5.67 × 10–8 × (5800)4, , 1360 Wm–2, ~ 21 Kcal m–2 min–1, , SECTION - B, Objective Type Questions, 1., , A uniform thermometre scale is at steady state with its 0 cm mark at 20°C and 100 cm mark at 100°C., Temperature of the 60 cm mark is, (1) 48°C, , (2), , 68°C, , (3), , 52°C, , (4), , 58°C, , Sol. Answer (2), T 20, 60 0, , 100 20 100 0, T 20, 60, , 80, 100, , T = 48 + 20 = 68°C, 2., , Two uniform rods AB and BC have Young’s modulii 1.2 × 1011 N/m2 and 1.5 × 1011 N/m2 respectively. If, coefficient of linear expansion of AB is 1.5 × 10–5/°C and both have equal area of cross section, then, coefficient of linear expansion of BC, for which there is no shift of the junction at all temperatures, is, , A, , (1) 1.5 × 10–5/°C, , (2), , 1.2 × 10–5/°C, , B, , (3), , C, , 0.6 × 10–5/°C, , (4), , 0.75 × 10–5/°C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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154, , Thermal Properties of Matter, , Solution of Assignment, , Sol. Answer (2), YAB = 1.2 × 1011 N/m2, YBC = 1.5 × 1011 N/m2, No shift of the junction at all, , , , , , , FL, AY, , 1, Y, , 1 Y2, So, Y, 2, 1, Substituting values, 1.5 105 1.5 1011, , 2, 1.2 1011, , 2 = 1.2 × 10–5/°C, 3., , Coefficient of linear expansion of a vessel completely filled with Hg is 1 × 10–5/°C. If there is no overflow of, Hg on heating the vessel, then coefficient of cubical expansion of Hg is, (1) 4 × 10–5/°C, , (2), , > 3 × 10–5/°C, , (3), , 3 × 10–5/°C, , (4), , Data is insufficient, , Sol. Answer (3), Expansion in Hg volume expansion in container., Volume coefficient of Hg 3 × Linear coefficient of expansion of vessel, 3 × 1 × 10–5/°C, Hg 3 × 10–5/°C, 4., , A metallic tape gives correct value at 25ºC. A piece of wood is being measured by this metallic tape at 10ºC., The reading is 30 cm on the tape, the real length of wooden piece must be, (1) 30 cm, , (2), , > 30 cm, , (3) < 30 cm, , (4), , Data is not sufficient, , Sol. Answer (3), At lesser temperature tape will decrease in length so the reading 30 cm on the tape is lesser than 30 cm in, real., 5., , In a thermostat two metal strips are used, which have different, (1) Length, , (2), , Area of cross-section, , (3) Mass, , (4), , Coefficient of linear expansion, , Sol. Answer (4), Coefficient of linear expansion should be different., 6., , The coefficient of linear expansion of a crystalline substance in one direction is 2 × 10–4/°C and in every direction, perpendicular to it is 3 × 10–4/°C. The coefficient of cubical expansion of crystal is equal to, (1) 5 × 10–4/°C, , (2), , 4 × 10–4/°C, , (3), , 8 × 10–4/°C, , (4), , 7 × 10–4/°C, , Sol. Answer (3), 1 + 2 + 3 = , 2 × 10–4 + 3 × 10–4 + 3 × 10–4 = , 8 × 10–4/°C = , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 7., , Thermal Properties of Matter, , 155, , If Cp and Cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M, where R is the, molar gas constant, (1) Cp – Cv = R/M2, , (2), , Cp – Cv = R, , (3), , Cp – Cv = R/M, , (4), , Cp – Cv = MR, , Sol. Answer (3), R, M, , Cp Cv , , because Cp and Cv are given per unit mass., 8., , The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant, pressure to that at constant volume is, (1), , 9, 7, , 7, 5, , (2), , (3), , 8, 7, , (4), , 5, 7, , Sol. Answer (2), , Cp , , 7, R, 2, , Cv Cp R, , , , , 9., , 5, R, 2, , Cp, Cv, , , , 7, 5, , A bullet of mass 10 g moving with a speed of 20 m/s hits an ice block of mass 990 g kept on a frictionless, floor and gets stuck in it. How much ice will melt if 50% of the lost KE goes to ice? (initial temperature of, the ice block and bullet = 0°C), (1) 0.001 g, , (2), , 0.002 g, , (3), , 0.003 g, , (4), , 0.004 g, , Sol. Answer (3), 50% of lost KE goes to melt ice, , , 1 1 10 20 20, , 1J, 2 2, 1000, , Amount of ice that melts , , 1, 0.003 g, 80 4.2, , 10. 50 g ice at 0°C is dropped into a calorimeter containing 100 g water at 30°C. If thermal capacity of calorimeter, is zero then amount of ice left in the mixture at equilibrium is, (1) 12.5 g, , (2), , 25 g, , (3), , 20 g, , (4), , 10 g, , Sol. Answer (1), Heat required to bring 100 g of water from 30°C to 0°C will be, Q = 100 × 1 × 30 = 3000 cal, Amount of ice that get melted =, , 3000, 37.5 g, 80, , So amount left = 12.5 g, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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156, , Thermal Properties of Matter, , Solution of Assignment, , 11. Heat is being supplied at a constant rate to the sphere of ice which is melting at the rate of 0.1 gm/, s. It melts completely in 100 s. The rate of rise of temperature thereafter will be, (1) 0.4°C/s, , (2), , 2.1°C/s, , (3), , 3.2°C/s, , (4), , 0.8°C/s, , Sol. Answer (4), dQ dm, , L, dt, dt, , ∵ Q mL , , dQ, 0.1 80 8 cal/gs, dt, , also, Q = mst, So, , dQ, dT, ms, dt, dt, , 8 10 1, 8C/s , , dT, dt, , [∵ It melts in 100 s so total mass of sphere = 0.1 × 100 = 10 g], , dT, dt, , 12. In a calorimeter of water equivalent 20 g, water of mass 1.1 kg is taken at 288 K temperature. If steam at, temperature 373 K is passed through it and temperature of water increases by 6.5°C then the mass of steam, condensed is, (1) 17.5 g, , (2), , 11.7 g, , (3), , 15.7 g, , (4), , 18.2 g, , Sol. Answer (2), , Steam at 100°C, , 6.5°C, water at 15°C + calorimeter of W = 20 g, Let mass of steam that gets condenced while the temperature of water is raised by 6.5°C = x g, So, heat released by steam = 540x cal + x × 1 × 78.5 [Q = mL + msT], This heat goes to the water + calorimeter system, Q required by water = 1100 × 1 × 6.5 = 7150 cal, Q required by calorimeter = 20 × 1 × 6.5 = 130 cal, Qreleased = Qrequired, 78.5 x + 540 x = 7150 + 130, x 11.7 g, 13. Heat energy at constant rate is given to two substances P and Q. If variation of temperature (T) of substances, with time (t) is as shown in figure, then select the correct statement., , T, , P, Q, , (1) Specific heat of P is greater than Q, , (2), , t, Specific heat of Q is greater than P, , (3) Both have same specific heat, , (4), , Data is insufficient to predict it, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 157, , Sol. Answer (2), dT, (slope) less means more specific heat., dt, , 14. If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which, the rate of energy production is Q? ( stands for Stefan's constant.), 1/ 4, , 1/ 4, , ⎛ 4R 2Q ⎞, ⎟, (1) ⎜, ⎝ ⎠, , ⎛ Q ⎞, ⎜, 2 ⎟, ⎝ 4R ⎠, , (2), , (3), , Q, 4R 2 , , (4), , ⎛ Q ⎞, ⎜, 2 ⎟, ⎝ 4R ⎠, , (4), , Radiation, , 1/ 2, , Sol. Answer (2), [H = eAT4] for black body e = 1, , H = (4 R2)T4, 1/4, , ⎡ Q ⎤, T ⎢, ⎥, ⎣ 4r 2 ⎦, , 15. Gravitational force is required for, (1) Stirring of liquid, , (2), , Convection, , (3), , Conduction, , (1) Transfer of heat by conduction, , (2), , Transfer of heat by radiation, , (3) Isothermal compression, , (4), , Electrical heating of a nichrome wire, , Sol. Answer (2), Gravity is required for convection., 16. Which of the following processes is reversible?, , Sol. Answer (3), In isothermal compression, Work done = Q (heat supplied), Isothermal compression is reversible, 17. Solar constant (S) depends upon the temperature of the Sun (T) as, (1) S T, , (2), , S T2, , (3), , S T3, , (4), , S T4, , Sol. Answer (4), 2, , ⎡R ⎤, S ⎢ ⎥ T 4, ⎣r ⎦, S T4, 18. Three rods of same dimensions have thermal conductivities 3K, 2K and K. They are arranged as shown, with, their ends at 100°C, 50°C and 0°C. The temperature of their junction is, , 3K, 100°C, , 50°C, , 2K, K, , (1) 75°C, , (2), , 200, °C, 3, , 0°C, (3), , 40°C, , (4), , 100, °C, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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158, , Thermal Properties of Matter, , Solution of Assignment, , Sol. Answer (2), 3K(100 – ) + 2K (50 – ) = K, , 3K, 100°C, , 300K – 3K + 100K – 2K = K, 400 = 6, , 50°C, , 2K, , K, , 0°C, , 400, , 6, 200, , 3, , 19. If wavelength of maximum intensity of radiation emitted by Sun and Moon are 0.5 × 10 –6 m and, 10–4 m respectively, then the ratio of their temperature is, (1), , 1, 10, , (2), , 1, 50, , (3), , 100, , (4), , 200, , Sol. Answer (4), mT1 = Constant, , [Wien's law], , 1T1 = 2T2, 0.5 × 10–6 × T1 = 10–4 × T2, T1, 200, T2, , 20. The three rods shown in figure have identical dimensions. Heat flows from the hot end at a rate of 40 W in, the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b)., (Assume KAl = 200 W/m °C and KCu = 400 W/m °C), , 0°C, , Al, , Cu, , Al, , 100°C, , (a), 0°C, , Al, Cu, Al, , 100°C, , (b), (1) 75 W, , (2), , 200 W, , (3), , 400 W, , (4), , 4W, , Sol. Answer (3), (a) 0°C, , Al, , Cu, , Al, , 100°C, , ∵ The rods have identical dimensions., Let their area of crossection be = A, and length be = L, So each rod would have heat resistance of, R, , L, KA, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 159, , Reff = R1 + R2 + R3, L, L, L, , , K Al A KCu A K Al A, , , , Reff , , H1 , , L, 5, , A 400, , [∵ KAl = 200, KCu = 400], , T, 100, , L, 5, Reff, , A 400, , …(1), , (b) When rods are connected in parallel, , 1, 1, 1, 1, , , , Reff R1 R 2 R3, Reff , H2 , , L, 800, A, , 100, L, 800, A, , …(2), , Dividng (2) by (1), , 40, 400, , ⇒ H2 400 W, H2 800 5, 21. Two bodies A and B of equal masses, area and emissivity cooling under Newton’s law of cooling from same, temperature are represented by the graph. If is the instantaneous temperature of the body and 0 is the, temperature of surroundings, then relationship between their specific heats is, , log(– 0), A, B, t, (1) SA = SB, , (2), , SA > SB, , (3), , SA < SB, , (4), , None of these, , Sol. Answer (2), Body loosing its temperature soon means low specific heat, SA > SB, 22. Two spheres of same material and radius r and 2r are heated to same temperature and are kept in identical, surroundings, ratio of their rate of loss of heat is, (1) 1 : 2, , (2), , 1:4, , (3), , 1:6, , (4), , 1:8, , Sol. Answer (2), Heat loss Area (Radius)2, 2, , So, , 2, H1 ⎡ r1 ⎤, 1, ⎡r ⎤, ⎢ ⎥ ⎢ ⎥ , H2 ⎣ r2 ⎦, 4, ⎣ 2r ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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160, , Thermal Properties of Matter, , Solution of Assignment, , 23. Assume that Solar constant is 1.4 kW/m2, radius of sun is 7 × 105 km and the distance of earth from centre, of sun is 1.5 × 108 km. Stefan’s constant is 5.67 × 10–8 Wm–2 K– 4, find the approximate temperature of sun, (1) 5800 K, , (2), , 16000 K, , (3), , 15500 K, , (4), , 8000 K, , Sol. Answer (1), 2, , ⎡R ⎤, S ⎢ ⎥ T4, ⎣r ⎦, 2, , ⎡ 7 105 ⎤, 4, 8, 1.4 103 ⎢, ⎥ 5.67 10 T, 8, ⎣⎢ 1.5 10 ⎦⎥, T = 5800 K, 24. If a graph is plotted by taking spectral emissive power along y-axis and wavelength along x-axis then the area, below the graph above wavelength axis is, (1) Emissivity, , (2), , Total intensity of radiation, , (3) Diffusivity, , (4), , Solar constant, , Sol. Answer (2), , , ∫0 e d , , area under the graph of e and , , also it gives total radiated average power per unit surface area which is called total intensity of radiation., 25. A spherical black body with radius 12 cm radiates 450 W power at 500 K. If the radius is halved and, temperature is doubled, the power radiated in watt would be, (1) 225, , (2), , 450, , (3), , 900, , (4), , 1800, , Sol. Answer (4), Power radiated r2 T4, 2, , P1 ⎡ r1 ⎤ ⎡ T1 ⎤, ⎢ ⎥ ⎢ ⎥, P2 ⎣ r2 ⎦ ⎣T2 ⎦, , 4, , 2, , 4, 450 ⎡⎢ r ⎤⎥ ⎡ T ⎤, , 1, ⎢ r ⎥ ⎢⎣ 2T ⎥⎦, P2, ⎣ 2⎦, , P2 = 1800 W, 26. Three rods of same material, same area of cross-section but different lengths 10 cm, 20 cm and 30 cm are, connected at a point as shown. What is temperature of junction O?, , 20ºC, 20 cm, cm, 30, , 10, , O, , cm, , 30ºC, (1) 19.2°C, , (2), , 16.4°C, , 10ºC, (3), , 11.5°C, , (4), , 22°C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 161, , Sol. Answer (2), Let temperature of junction be = , Heat flowing to junction = heat out flowing, KA, KA, KA, 30 20 10 , 30, 20, 10, , 30 20 10 , 3, , 30 , 3, , , , , , 2, , , , 1, , 20 2 20, 2, , 60 – 2 = 9 – 120, 180, , 11, , 16.36°C = , 27. If transmission power of a surface is, , (1), , 18, 13, , (2), , 1, 1, , reflective power is , then what is its absorptive power?, 9, 6, , 13, 18, , (3), , 3, 15, , (4), , 15, 3, , Sol. Answer (2), t + r+a=1, a = 1 – (t + r), , ⎛ 1 1⎞, 1 ⎜ ⎟, ⎝9 6⎠, a, , 13, 18, , 28. A solid cylinder of length L and radius r is heat upto same temperature as that of a cube of edge length a. If, both have same material, volume and allowed to cool under similar conditions, then ratio of amount of radiations, radiated will be (Neglect radiation emitted from flat surfaces of the cylinder), (1), , a, 3r, , (2), , 2a, rL, , (3), , a2, rL, , (4), , a2, 2rL, , Sol. Answer (1), ∵ Both have same volume, a3 = r2L, , …(1), , Amount of radiation Surface area, , [∵ Temperature, material are same for both], , Radiation cylinder 2rL 2rL . a, , , Radiation cube, 6a2, 6a3, using equation (1), We get, , Rcylinder, Rcube, , , , a, 3r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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162, , Thermal Properties of Matter, , Solution of Assignment, , 29. A very thin metallic shell of radius r is heated to temperature T and then allowed to cool. The rate of cooling, of shell is proportional to, (1) rT, , (2), , 1, r, , (3), , r2, , (4), , r0, , Sol. Answer (4), Rate of cooling depends on temperature of body, surrounding temperature, not on radius., 30. If an object at absolute temperature (T) radiates energy at rate R, then select correct graph showing the, variation of logeR with loge(T)., logeR, , loge(R), , (1), , logeR, , (2), , logeR, , (3), , (4), , loge(T), , loge(T), , loge(T), , logeT, , Sol. Answer (1), R = eAT4, loge R = eA× 4 loge T, directly proportional., 31. Two diagonally opposite corners of a square made of a four thin rods of same material, same dimensions are at, temperature 40°C and 10°C. If only heat conduction takes place, then the temperature difference between other, two corners will be, (1) 0°C, , (2), , 10°C, , (3), , 25°C, , (4), , 15°C, , Sol. Answer (1), Arrangement is like resistances in wheat stone bridge, No temperature difference between two outer corners., 32. Bottom of a lake is at 0°C and atmospheric temperature is –20°C. If 1 cm ice is formed on the surface in, 24 h, then time taken to form next 1 cm of ice is, (1) 24 h, , (2), , 72 h, , (3), , 48 h, , (4), , 96 h, , Sol. Answer (2), Time intervals to change thickness from 0 to x from x to 2x are in ratio of 1 : 3 : 5 : 7 ……, t1 : t2 = 1 : 3, = 24 : 24 × 3, t2 = 72 hours, 33. The power received at distance d from a small metallic sphere of radius r(<<d) and at absolute temperature, T is P. If temperature is doubled and distance reduced to half of initial value, then the power received at that, point will be, (1) 4p, , (2), , 8p, , (3), , 32p, , (4), , 64p, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 163, , Sol. Answer (4), , Solar constant , , T4, r2, , Solar constant equivalent to power received so, , P1 T14 r22, , , P2 r12 T24, 2, , P T 4 r / 2, , , P2 r 2, (2T )4, P2 = 64p, , SECTION - C, Previous Years Questions, 1., , The value of coefficient of volume expansion of glycerin is 5 × 10–4 K–1. The fractional change in the density, of glycerin for a rise of 40°C in its temperature is, [Re-AIPMT-2015], (1) 0.010, , (2), , 0.015, , (3), , 0.020, , (4), , 0.025, , Sol. Answer (3), 2., , The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the, rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow, will be, [AIPMT-2015], (1) 4.0 J/s, , (2), , 44.0 J/s, , (3), , 16.8 J/s, , (4), , 8.0 J/s, , Sol. Answer (1), 3., , On observing light from three different stars P, Q and R , it was found that intensity of violet colour is maximum, in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red, colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperatures of P, Q, and R then it can be concluded from the above observations that, [AIPMT-2015], (1) TP < TQ < TR, , (2), , TP > TQ > TR, , (3), , TP > TR > TQ, , (4), , TP < TR < TQ, , Sol. Answer (3), 4., , Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass, of water present will be:, [Take specific heat of water = 1 cal g–1°C–1 and latent heat of steam = 540 cal g–1], (1) 24 g, , (2), , 31.5 g, , (3), , 42.5 g, , [AIPMT-2014], (4), , 22.5 g, , Sol. Answer (4), 5., , Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next, 5 minutes. The temperature of the surroundings is, [AIPMT-2014], (1) 45°C, , (2), , 20°C, , (3), , 42°C, , (4), , 10°C, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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164, 6., , Thermal Properties of Matter, , Solution of Assignment, , A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to, white hot. The correct explanation for the above observation is possible by using, [NEET-2013], (1) Wien's displacement law, , (2), , Kirchoff's law, , (3) Newton's law of cooling, , (4), , Stefan's law, , Sol. Answer (1), According to Wien's displacement Law, if temperature rises than decreases., Which explains change of colour., , Time, , (3), Time, , (4), Time, , Temperature, , (2), , Temperature, , (1), , Temperature, , Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which, one of the following graphs represents the variation of temperature with time?, [AIPMT (Prelims)-2012], Temperature, , 7., , Time, , Sol. Answer (3), , Temperature, , Liquid oxygen when heated will observe a rise in temperature as well as change in state one time which can, be represented as, , Time, 8., , If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the, rate of energy production is Q?, [AIPMT (Prelims)-2012], 1/ 4, , 1/ 4, , ⎛ 4R 2Q ⎞, ⎟, (1) ⎜, ⎝ ⎠, , (2), , ⎛ Q ⎞, ⎜, 2 ⎟, ⎝ 4R ⎠, , (3), , Q, 4R 2 , , (4), , ⎛ Q ⎞, ⎜, 2 ⎟, ⎝ 4R ⎠, , 1/ 2, , ( stands for Stefan's constant), Sol. Answer (2), Q = eAT4, For black body e = 1, 1/4, , T ⎛⎜ Q ⎞⎟, ⎝ 4R 2 ⎠, 9., , A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A, block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal, conductivity of slab is (Given, latent heat of fusion of ice = 3.36 × 105 J kg–1), [AIPMT (Mains)-2012], (1) 1.24 J/m/s/°C, , (2), , 1.29 J/m/s/°C, , (3), , 2.05 J/m/s/°C, , (4), , 1.02 J/m/s/°C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 165, , Sol. Answer (1), In 1 hour, 4.8 kg ice melts, Heat supplied = 3.36 × 105 × 4.8 J, Heat supplied per second , , {Q = L m}, , KA, T, L, , 3.36 105 4.8 K 0.36, , 100, 1 3600, 0.1, K = 1.24 J/m/s/°C, 10. When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to, be 80 cal/°C, is, [AIPMT (Prelims)-2011], (1) 293 cal/K, , (2), , 273 cal/K, , (3), , 8 × 104 cal/K, , (4), , 80 cal/K, , Sol. Answer (1), Change in entropy S , S , , Q, T, , S , , 1000 80, 273, , Heat absorbed, Temperature at that instant, , [Q = mL1], , S = 293 cal/K, 11. A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of, heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original, rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs, in time t?, [AIPMT (Prelims)-2010], (1), , Q, 4, , (2), , Q, 16, , (3), , 2Q, , (4), , Q, 2, , Sol. Answer (2), Original volume r 2h, New volume also same as the original volume, But new radius = r/2, Let new height be = h', 2, , ⎛r ⎞, 2, So r h ⎜ ⎟ h, ⎝2⎠, 4h = h', , Q, , KR 2, L, , Q2 , , KR 2, Q, , 4 4 L 16, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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166, , Thermal Properties of Matter, , Solution of Assignment, , 12. The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the, centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by, (where is Stefan’s constant), [AIPMT (Prelims)-2010], (1), , r 2T 4, , (2), , R2, , r 2T 4, 4r, , 2, , (3), , r 4T 4, r4, , (4), , 4r 2T 4, R2, , Sol. Answer (1), , P, , P, , Q, 4R, , 2, , , Q 4r 2 T 4, , r 2T 4, R2, , 13. The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1 and T2, (T1 > T2). The rate of heat transfer,, , dQ, , through the rod in a steady state is given by, dt, , [AIPMT (Prelims)-2009], (1), , dQ k (T1 – T2 ), , dt, LA, , (2), , dQ, kLA(T1 – T2 ), dt, , (3), , dQ kA(T1 – T2 ), , dt, L, , (4), , dQ kL(T1 – T2 ), , dt, A, , Sol. Answer (3), H, , dQ kA(T1 T2 ), , dt, L, , 14. A black body at 227°C radiates heat at the rate of 7 cals/cm2s. At a temperature of 727°C, the rate of heat, radiated in the same units will be, [AIPMT (Prelims)-2009], (1) 50, , (2), , 112, , (3), , 80, , (4), , 60, , Sol. Answer (2), Radiation T 4, R1 ⎛ T1 ⎞, ⎜ ⎟, So, R2 ⎝ T2 ⎠, , 4, , 7 ⎛ 500 ⎞, , x ⎜⎝ 1000 ⎟⎠, , 4, , x = 112 cal/cm2s, 15. On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water, are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature, of 39° C on the Celsius scale?, [AIPMT (Prelims)-2008], (1) 139° W, , (2), , 78° W, , (3), , 117° W, , (4), , 200° W, , Sol. Answer (3), 39 0, T 39, , 100 0 239 39, , 117°W = T, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 167, , 16. Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t°C,, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun, is (where is the Stefan’s constant), [AIPMT (Prelims)-2007], (1), , r 2 (t 273 ) 4, , (2), , R2, , 4r 2 t 4, R2, , (3), , r 2 (t 273 ) 4, 4R 2, , (4), , 16 2 r 2 t 4, R2, , Sol. Answer (1), , P, , P, , Q, 4R 2, , , Q = 4r2(t + 273)4, , r 2 (t 273)4, R2, , 17. A black body is at 727°C. It emits energy at a rate which is proportional to, (1) (727)4, , (2), , (727)2, , (3), , (1000)4, , [AIPMT (Prelims)-2007], (4), , (1000)2, , Sol. Answer (3), E T4, So E (727 + 273)4, (1000)4, 18. A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature, of the body is increased by 1000°C, the maximum intensity will be observed at :, [AIPMT (Prelims)-2006], (1) 4000 Å, , (2), , 5000 Å, , (3), , 6000 Å, , (4), , 3000 Å, , Sol. Answer (4), , 1⎤, ⎡, ⎢ T ⎥, ⎣, ⎦, , Wien's law, , , , T1 2, , T2 1, , 1500 K, , , 2500 K 5000 Å, 3000 Å = , 19. Which of the following circular rods, (given radius r and length l) each made of the same material and whose, ends are maintained at the same temperature will conduct most heat ?, [AIPMT (Prelims)-2005], (1) r = 2r0; l = 2l0, , (2), , r = 2r0; l = l0, , (3), , r = r0; l = l0, , (4), , r = r0; l = 2l0, , Sol. Answer (2), Rod with more, , A, r2, ratio or, ratio will conduct more heat, L, L, , A⎫, ⎧, ⎨∵ H ⎬, L⎭, ⎩, , 20. The coefficients of linear expansion of brass and steel are 1 and 2 respectively. When we take a brass rod, of length l1 and steel rod of length l2 at 0°C, then difference in their lengths (l2 – l1) will remain the same at, all temperatures, if, (1) 12l1 = 22l2, , (2), , 1l2 = 2l1, , (3), , 1l1 = 2l2, , (4), , 1l22 = 2l12, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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168, , Thermal Properties of Matter, , Solution of Assignment, , Sol. Answer (3), Let the temperature difference be = t, l1' = l1 + l11t, l2' = l2 + l22t, , {Where l2' and l2 are increased lengths}, , l2'– l1' = l2 – l1, , {∵ difference of length is same at all temperature}, , l2 + l22t – l1 – l11t = l2 – l1, l22 = l11, 21. The density of water at 20°C is 998 kg/m3 and at 40°C 992 kg/m3. The coefficient of volume expansion of water, is, (1) 10–4/°C, , (2), , 3 × 10–4/°C, , (3), , 2 × 10–4/°C, , (4), , 6 × 10–4/°C, , Sol. Answer (2), 2 = 1 (1 – T), 992 = 998 (1 – × 20), 3 × 10–4/°C, 22. If 1 g of steam at 100°C steam is mixed with 1 g of ice at 0°C, then resultant temperature of the mixture is, (1) 100°C, , (2), , 230°C, , (3), , 270°C, , (4), , 50°C, , Sol. Answer (1), Heat required to convert phase of ice = 80 cal, Heat required to bring water at 0°C to water at 100°C = 100 cal, Total heat required = 180 cal, Heat available = 540 cal, , [Lu of 1 g steam], , Final mixture will have steam + water, and when steam is present in mixture temperature has to be 100°C, 23. Heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio, 1 : 2 and the lengths in the ratio 2 : 1. If the temperature difference between the ends is same, then ratio of, the rate of flow of heat through them will be, (1) 2 : 1, , (2), , 8:1, , (3), , 1:1, , (4), , 1:8, , Sol. Answer (4), , H, , KA, ⎡, ⎤, ⎢H L ( T )⎥, ⎣, ⎦, , d2, L, 2, , H1 ⎡ d1 ⎤ ⎡ L2 ⎤, ⎢ ⎥ ⎢ ⎥, So, H2 ⎣ d 2 ⎦ ⎣ L1 ⎦, 2, , ⎡ 1⎤ 1 1, ⎢ ⎥ , ⎣2⎦ 2 8, H1 1, , H2 8, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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170, , Thermal Properties of Matter, , Solution of Assignment, , 28. Consider a compound slab consisting of two pieces of same length and different materials having equal, thicknesses and thermal conductivities K and 2 K, respectively. The equivalent thermal conductivity of the slab, is, (1) 2/3 K, , (2), , 2 K, , (3), , 3K, , (4), , 4/3 K, , Sol. Answer (4), Connected series way so,, R = R1 + R2, , A, , [For series], , K, , 2K, , l, , l, , A, , 2l, l, l, , , K A KA 2KA, , K = 4/3 K., 29. Gravitational force is required for, (1) Stirring of liquid, , (2), , Convection, , (3), , Conduction, , (4), , Radiation, , Sol. Answer (2), Gravity is the necessary requirement for convection., 30. A black body is at a temperature of 500 K. It emits energy at a rate which is proportional to, (1) (500)3, , (2), , (500)4, , (3), , 500, , (4), , (500)2, , Sol. Answer (2), E T4, So, E (500)4, 31. Which of the following is closest to an ideal black body?, (1) Black lamp, (2) Cavity maintained at constant temperature, (3) Platinum black, (4) A lamp of charcoal heated to high temperature, Sol. Answer (2), Cavity maintained at constant temperature is closest to black body., 32. For a black body at temperature 727°C, its rate of energy loss is 20 watt and temperature of surrounding is 227°C., If temperature of black body is changed to 1227°C then its rate of energy loss will be, (1) 304 W, , (2), , 320, W, 3, , (3), , 240 W, , (4), , 120 W, , Sol. Answer (2), According to Stefan - Boltzmann's Law, dH, T 4 T04, dt, , 20 1000 4 5 4, , x, 1500 4 5 4, , , , x, , 320, W, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermal Properties of Matter, , 171, , 33. A beaker full of hot water is kept in a room. If it cools from 80°C to 75°C in t1 minutes, from 75°C to 70°C in, t2 minutes and from 70°C to 65°C in t3 minutes, then, (1) t1 < t2 < t3, , (2), , t1 > t2 > t3, , (3), , t1 = t 2 = t 3, , (4), , t1 < t2 = t3, , Sol. Answer (1), , 80C , 75C , 70C , 65C, t, t, t, 1, , 2, , 3, , According to newton's law of cooling, , ln, , T1 T0, kt, T2 T0, , Where T0 is temperature of surroundings and T1 and T2 are initial and final temperature so more difference, between T1, T2 and T0 less is the time taken to reach T2 from T1, t3 > t2 > t1, 34. A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns, to white hot. The correct explanation for the above observation is possible by using, (1) Wien's displacement Law, , (2), , Kirchoff's Law, , (3) Newton's Law of cooling, , (4), , Stefan's Law, , Sol. Answer (1), 35. The Wien’s displacement law expresses the relation between, (1) Wavelength corresponding to maximum intensity and temperature, (2) Radiation energy and wavelength, (3) Temperature and wavelength, (4) Colour of light and temperature, Sol. Answer (1), Maximum Temperature = b(constant), 36. We consider the radiation emitted by the human body. Which one of the following statements is correct?, (1) The radiation emitted is in the infra-red region, (2) The radiation is emitted only during the day, (3) The radiation is emitted during the summers and absorbed during the winters, (4) The radiation emitted lies in the ultraviolet region and hence is not visible, Sol. Answer (1), Wavelength lies in infra-red region as temperature of human body is very low., 37. If m denotes the wavelength at which the radiative emission from a black body at a temperature T K is, maximum, then, (1) m T 4, , (2), , m is independent of T (3), , m T, , (4), , m T–1, , Sol. Answer (4), Wien's displacement Law, m T = b, So m , , 1, T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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172, , Thermal Properties of Matter, , Solution of Assignment, , 38. A black body has wavelength corresponding to maximum intensity m at 2000 K. Its corresponding wavelength at, 3000 K will be, (1), , 3, m, 2, , (2), , 2, m, 3, , (3), , 16, m, 81, , (4), , 81, m, 16, , Sol. Answer (2), , , 1, T, , [Wien's Law], , 1 T2, So T, 2, 1, m 3000, , , 2000, 2, m , 3, , 39. The radiant energy from the sun, incident normally at the surface of earth is 20 kcal/m2 min. What would have, been the radiant energy, incident normally on the earth, if the sun had a temperature, twice of the present one?, (1) 320 kcal/m2 min, , (2), , 40 kcal/m2 min, , (3), , 160 kcal/m2 min, , (4), , 80 kcal/m2 min, , Sol. Answer (1), E T4, , 20 ⎛ T ⎞, , E ⎜⎝ 2T ⎟⎠, , 4, , E = 320 kcal/m2 minute, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Density of water is maximum at 4°C., R : Water has both positive and negative temperature coefficients of volumetric expansions depending on the, temperature range., , Sol. Answer (2), A : is true, R : is true, But the correct explanation is that due to structural changes in moleculer of water we observe this anomalous, behaviour., 2., , A : A solid and a hollow sphere of same diameter and same material when heated for the same temperature rise, will, expand by the same amount., R : The change in volume is independent of the original mass but depends on original volume., , Sol. Answer (1), A : is true, R : is true and correct explanation of Assertion., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 3., , Thermal Properties of Matter, , 173, , A : Fahrenheit is the smallest unit for measuring temperature., R : Fahrenheit was the first temperature scale used for measuring temperature., , Sol. Answer (2), A : is true, R : is true, But correct explanation is Fahrenheit has 180 divisions where as other scales have 100 divisions., 4., , A : Material used for making cooking utensils is the one having high specific heat and high conductivity., R : Low conductivity means high specific heat., , Sol. Answer (4), A : is false, material used for making cooking utensils is the one having low specific heat and high conductivity., R : is false., 5., , A : The value of the absorptive power and the emissivity has the same value for a single body at a particular, temperature., R : Value of absorptive power is 1 for a black body., , Sol. Answer (2), A : is true, R : is true, But the correct reason is black body radiate as much as it absorbs., 6., , A : The reflectance of a black body is zero., R : Black body absorbs all radiations incident on it., , Sol. Answer (1), A : is true, R : is true and correct explanation., 7., , A : Evaporation of water is fast on the surface of moon as compared to earth., R : On the surface of moon temperature is much greater than the surface of earth., , Sol. Answer (3), A : is true, R : is false, temperature of surface of moon is much less than that of earth's surface., 8., , A : The internal energy of a solid substance increases during melting., R : Latent heat is required to melt a solid substance., , Sol. Answer (1), A : is true, R : is true and correct explanation., 9., , A : Transmission cables are not tightly fixed on the poles., R : During winters the length of cables decreases due to decrease in temperature, which can damage poles., , Sol. Answer (1), A : True, R : True and correct explanation., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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174, , Thermal Properties of Matter, , Solution of Assignment, , 10. A : The thermal conductivity of a body depends on its material and dimensions., R : Thermal conductivity is proportional to length and inversely proportional to area cross-section of body., Sol. Answer (4), A : False, Thermal conductivity depends only on material, R : False, Thermal conductivity is constant for a given material, 11. A : Eskimos make double wall houses of ice blocks., R : The air trapped between double walls prevents the conduction of heat energy from inside the house to, outside it., Sol. Answer (1), A : True, R : True and correct explanation, 12. A : The rate of growth of ice on the surface of a lake decreases with increase in thickness of ice layer., R : Ice is poor conductor of heat energy., Sol. Answer (1), A : True, R : True and correct explanation., 13. A : Natural convection is not possible in an orbiting satellite., R : Natural convection is not possible in gravity free space., Sol. Answer (1), A : True, R : True and correct explanation., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Chapter, , 12, , Thermodynamics, Solutions, SECTION - A, Objective Type Questions, 1., , In thermodynamics the Zeroth law is related to, (1) Work done, , (2), , Thermal equilibrium, , (3), , Entropy, , (4), , Diffusion, , (3), , W=0, , (4), , Both (1) & (3), , (4), , CV =, , (4), , All of these, , Sol. Answer (2), Zeroth law related to thermal equilibrium., 2., , For a cyclic process, (1) U = 0, , (2), , Q = 0, , Sol. Answer (1), Since initial and final points are at same, temperature so U = 0, 3., , Select the incorrect relation. (Where symbols have their usual meanings), (1) CP =, , R, –1, , (2), , CP – CV = R, , Pf Vf – PV, i i, 1– , , (3), , U =, , (3), , Volume, , R, –1, , Sol. Answer (3), , P V PV, i i, U f f, is the correct relation., 1, 4., , Internal energy of a non-ideal gas depends on, (1) Temperature, , (2), , Pressure, , Sol. Answer (4), Depends on Temperature, Pressure, Volume., 5., , For an adiabatic expansion of an ideal gas the fractional change in its pressure is equal to, (1) – , , V, dV, , (2), , –, , dV, V, , (3), , dV, V, , (4), , –, , dV, V, , Sol. Answer (4), , ∵ PV = constant, Then,, , So, P V– , , P, ⎛ dV ⎞, ⎜, ⎟, P, ⎝ V ⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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176, 6., , Thermodynamics, , Solution of Assignment, , Which of the following laws of thermodynamics defines internal energy?, (1) Zeroth law, , (2), , Second law, , (3), , First law, , (4), , Third law, , Sol. Answer (3), Internal energy is defined in first law, ∵ Q = U + W, So, U = Q – W, 7., , Select the correct statement for work, heat and change in internal energy., (1) Heat supplied and work done depend on initial and final states, (2) Change in internal energy depends on the initial and final states only, (3) Heat and work depend on the path between the two points, (4) All of these, , Sol. Answer (4), All statements are correct., 8., , Morning breakfast gives 5000 cal to a 60 kg person. The efficiency of person is 30%. The height upto which the, person can climb up by using energy obtained from breakfast is, (1) 5 m, , (2), , 10.5 m, , (3), , 15 m, , (4), , 16.5 m, , Sol. Answer (2), ∵ W = JQ, , So, mgh = JQ, , ⎛ 30 ⎞, 4.2 5000, JQ ⎜⎝ 100 ⎟⎠, , h, = 10.5 m, mgh, 60 10, 9., , Select the incorrect statement about the specific heats of a gaseous system., (1) Specific heat at no exchange condition, CA = 0, (3) Specific heat at constant pressure, CP , , R, 1, , (2), , Specific heat at constant temperature, CT = , , (4), , Specific heat at constant volume, CV , , R, , , Sol. Answer (4), The correct value of CV , , R, 1, , 10. Work done in the cyclic process shown in figure is, , P, 3P0, P0, (0, 0), (1) 4P0V0, , (2), , –4P0V0, , V0, , V, , 3V0, (3), , , , 22, P0V0, 7, , (4), , –13P0V0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 177, , Sol. Answer (3), Cyclic process is anticlockwise then, Work done = –(Area of P-V graph), W = –R1R2, , ⎛ 3P P0, ⎜ 0, ⎝, 2, , , ⎞ ⎛ 3V0 V0 ⎞, ⎟⎜, ⎟, ⎠ ⎝, ⎠, 2, , 22, P0V0, 7, , 11. In following figures (a) to (d), variation of volume by change of pressure is shown in figure. The gas is taken, along the path ABCDA. Change in internal energy of the gas will be, (a), , P, , (b), , D, , O, , P, , C, B, , A, , (c), , D, , P, D, , (d), , P, , D, , C, , C, , C, A, O, , V, , B, , A, , V, , O, , A, , B, V, , O, , B, V, , (1) Positive in all cases from (a) to (d), (2) Positive in cases (a), (b) and (c) but zero in case (d), (3) Negative in cases (a), (b) and (c) but zero in case (d), (4) Zero in all the four cases, Sol. Answer (4), U = 0 in all cases because cyclic process., 12. In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas, releases 20 J of heat when 8 J of work was done on the gas. If the initial internal energy of the gas was, 30 J, then the final internal energy will be, (1) 2 J, , (2), , 18 J, , (3), , 42 J, , (4), , 58 J, , Sol. Answer (2), We know by 1st Law of Thermodynamics, Q = U + W, – 20 J = U – 8 J, ∵ U = Ufinal – Uinitial, U = – 12 J, So, Ufinal = Uinitial + U, = 30 + (–12) = 18 J, 13. A perfect gas goes from state A to state B by absorbing 8 × 105 joule and doing 6.5 × 105 joule of external, work. If it is taken from same initial state A to final state B in another process in which it absorbs 105 J of, heat, then in the second process work done, (1) On gas is 105 J, , (2), , On gas is 0.5 × 105 J, , (3) By gas is 105 J, , (4), , By gas is 0.5 × 105 J, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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178, , Thermodynamics, , Solution of Assignment, , Sol. Answer (2), Q = U + W, 8 × 105 = U + 6.5 × 105, 1.5 × 105 J = U, Again using Q = U + W for the second case U will stay the same., Now, 105 = 1.5 × 105 + W, – 0.5 × 105 = W, negative sign indicates work is being done on the gas., 14. Figure shows two processes a and b for a given sample of gas. If Q1, Q2 are the amount of heat absorbed, by the system in the two cases; and U1, U2 are changes in internal energy respectively, then, , P, , a, b, , (0,0), , V, , (1) Q1 = Q2; U1 = U2, , (2), , Q1 > Q2; U1 > U2, , (3) Q1 < Q2; U1 < U2, , (4), , Q1 > Q2; U1 = U2, , Sol. Answer (4), ∵ Initial and final states are same., U1 = U2, , P, , a, b, , Area under 'a' > area under 'b' i.e., W1 > W2, Heat absorbed by a > heat absorbed by b, , (0,0), , Q1 > Q2, , V, , 15. A gas undergoes a change at constant temperature. Which of the following quantities remain fixed?, (1) Pressure, , (2), , Entropy, , (3) Heat exchanged with the system, , (4), , All the above may change, , Sol. Answer (4), When temperature change = 0 then,, P1V1 = P2V2 = constant, Rest may change., 16. Following figure shows P-T graph for four processes A, B, C and D. Select the correct alternative., , P, , D, , C, B, , A, T, , (0, 0), (1) A – Isobaric process, , (2), , B – Adiabatic process, , (3) C – Isochoric process, , (4), , D – Isothermal process, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , Sol. Answer (3), , P, , (A) Temperature is constant – isothermal, , D, , 179, , C, B, , (B) Pressure is constant – Isobaric, (C) Pressure Temperature – Isochoric process, , A, , (D) P1–T = constant – Adiabatic process, , T, , (0, 0), , 17. An ideal gas with adiabatic exponent is heated at constant pressure. It absorbs Q amount of heat. Fraction, of heat absorbed in increasing the temperature is, (1) , , 1, , , (2), , 1, , (3), , 1, , , (4), , 2, , Sol. Answer (2), Heat absorbed in increasing temperature = U = Q – W = nCV T, Fraction of heat absorbed =, , , , Heat absorbed, Total heat, , nCV T, nCP T, , C, 1, V , CP , , 18. A certain amount of an ideal monatomic gas needs 20 J of heat energy to raise its temperature by 10°C at, constant pressure. The heat needed for the same temperature rise at constant volume will be, (1) 30 J, , (2), , 12 J, , (3), , 200 J, , (4), , 215.3 J, , Sol. Answer (2), Q = nCP T, 20 = nCP × 10, , ...(1), , U = nCV T, , U n, , CP, T, , , ∵, , mono 5 / 3, , U = 12 J, 19. Two cylinders contain same amount of ideal monatomic gas. Same amount of heat is given to two cylinders., If temperature rise in cylinder A is T0 then temperature rise in cylinder B will be, , 4, T0, (1), 3, , (2), , 2T0, , Free piston, , Fixed piston, , A, , B, , Heat, , Heat, (3), , T0, 2, , (4), , 5, T0, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 181, , Sol. Answer (2), Process carried out suddenly so process is adiabatic., PV K, P1V1 P2V2, , ⎛V ⎞, P2 P1 ⎜ 1 ⎟, ⎝ V2 ⎠, , , , 7/5, , ⎛ V1 ⎞, P2 (1 atm) ⎜, ⎟, ⎝ V1 / 32 ⎠, P2 = 1 atm × (25)7/5, = 128 atm, , 23. Two samples A and B of a gas initially at the same temperature and pressure, are compressed from volume, V to, , V, (A isothermally and B adiabatically). The final pressure, 2, , (1) PA > PB, , (2), , PA = PB, , (3), , PA < PB, , (4), , PA = 2PB, , (3), , 1.4 105 N/m2, , (4), , 2.75 105 N/m2, , (3), , Q = W, , (4), , Q = –U, , Sol. Answer (3), , P, PB, , B, , PA, , A, , V, 2, , V, , V, , i.e., PA < PB, 24. The adiabatic elasticity of a diatomic gas at NTP is, (1) Zero, , (2), , 1 105 N/m2, , Sol. Answer (3), Adiabatic elasticity = P , , 7, 1.01 105, 5, , 1.414 105 N/m2, 25. For an isometric process, (1) W = –U, , (2), , Q = U, , Sol. Answer (2), For an isometric process, (i.e., isochoric) workdone = zero, So Q = U, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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182, , Thermodynamics, , Solution of Assignment, , 26. A mixture of gases at NTP for which = 1.5 is suddenly compressed to, , 1, th of its original volume. The final, 9, , temperature of mixture is, (1) 300°C, , (2), , 546°C, , (3), , 420°C, , (4), , 872°C, , (4), , Adiabatic, , Sol. Answer (2), TV –1 = constant, T1V1–1 = T2V2–1, , ⎛V ⎞, T2 T1 ⎜ 1 ⎟, ⎝ V2 ⎠, , 1, , 1.5 1, , ⎡ V ⎤, T2 (273 K) ⎢ 1 ⎥, ⎣V / 9 ⎦, T2 = (273 K) × 3, = 819 K, = 546°C, , 27. In which process P-V diagram is a straight line parallel to the volume axis?, (1) Isochoric, , (2), , Isobaric, , Sol. Answer (2), , (3), , Isothermal, , P, , V, Process having a constant pressure, so isobaric process., 28. The P-V plots for two gases during adiabatic processes are shown in the figure. The graphs, 1 and 2 should correspond respectively to, P, 1, 2, , (0, 0), (1) O2 and He, , (2), , He and O2, , (3), , V, , O2 and CO, , (4), , N2 and O2, , Sol. Answer (1), PV = constant [equation of graphs], So for more less the rate of change or slope of graph, , P, , and is less for diatomic., , 1 (diatomic), , So graph 1 for O2, , 2 (monoatomic), V, , Graph 2 for He., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 183, , 29. The pressure and volume of a gas are changed as shown in the P-V diagram in this figure. The temperature, of the gas will, P, A, B, , D, , C, V, , (1) Increase as it goes from A to B, , (2), , Increase as it goes from B to C, , (3) Remain constant during these changes, , (4), , Decrease as it goes from D to A, , Sol. Answer (1), In the process A B, , P, A, , Pressure is constant., , BT, , 1, , PV nRT, , D, , So V T, , C, , and volume is increasing so temperature also increases., , T2, T3, T4, T5, , T5 > T4 > T3 > T2 > T1, V, , 30. The figure shows P-V diagram of a thermodynamic cycle. Which corresponding curve is correct?, , P, B, , C, , A, , D, V, , (0, 0), , P, , B, A, , (1), , O, , P, , C, , D, , A, , (2), , T, , O, , C, , D, , P, , B, , (3), , B, O, , T, , P, , D, , A, , C, T, , D, A, , (4), , O, , C, B, T, , Sol. Answer (1), A B V = constant, , P, , ∵ PV = RT, , R, P T, V, , B, A, , O, , C, , D, T, , Compare with y = mx, P-T graph is a straight line which must passes from origin, A B volume constant, P-increasing, T-increasing., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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184, , Thermodynamics, , Solution of Assignment, , B C pressure constant, volume - increasing, temperature - increasing, B C P = constant, origin P-T graph is a straightline parallel to v-axis, C D V = constant then, P, , R, T, V, , P-T graph is straight line must passes from origin, D A P = constant, P-T graph is a straightline parallel to T-axis., 31. During the thermodynamic process shown in figure for an ideal gas, , P, , V, (1) T = 0, , (2), , Q = 0, , (3), , W<0, , (4), , U > 0, , Sol. Answer (4), For a straight P-V graph line P V, , P, , If pressure increases, volume increases then T also, increases [PV T], So T 0, , V, , Volume increasing so work is positive, W > 0, and temperature also increasing so Q > 0, ∵ Q = U + W, So U > 0, , 32. For P-V diagram of a thermodynamic cycle as shown in figure, process BC and DA are isothermal. Which, of the corresponding graphs is correct?, , P, , A, , B, C, D, , P, , B, , A, , P, , B, , A, , C, , (1), , V, , D, , C, , (2), T, , T, , C, , (0,0), , B, A, , A, , (3), , D, (0,0), , V, , B, , C, , D, (0,0), , V, , (4), T, , D, (0,0), , T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 185, , Sol. Answer (2), From A B, volume increasing, pressure constant, , B C, Pressure , , P, , B, , A, , 1, ⇒ Temperature constant, Volume, , C, D, , Same for D A, , (0,0), , T, , C D pressure decreasing, volume constant, , So P T, 33. Work done for the process shown in the figure is, , V, , B(30 kPa, 25 cc), , A(10 kPa, 10 cc), P, (1) 1 J, , (2), , 1.5 J, , (3), , 4.5 J, , (4), , 0.3 J, , Sol. Answer (4), , v, , Area under graph and V axis = work done, , w, , 25 cc, , 1, (30 10) 103 (25 10) 106, 2, , 10 cc, , = 0.3 J, , B, A, , P, , 34. During which of the following thermodynamic process represented by PV diagram the heat energy absorbed, by system may be equal to area under PV graph?, , P, , P, , A, , (1), , B, O, , P, , A, , (2), , B, , V, , O, , A, , (3), , V, , B, , O, , (4), , All of these, , V, , Sol. Answer (4), , P, , P, , A, BT, , O, , V, , O, , P, , A, , B T, V, , A, , BT, O, , V, , ∵ T is constant in all cases., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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186, , Thermodynamics, , Solution of Assignment, , 35. The specific heat of a gas in a polytropic process is given by, (1), , R, R, , –1 N –1, , (2), , R, R, , 1– 1– N, , (3), , R, R, –, –1 N –1, , (4), , R, R, –, 1– 1– N, , Sol. Answer (3), ∵, , C CV , , R, R, R, , , , , 1, N, 1, 1 N, , 36. For a certain process, pressure of diatomic gas varies according to the relation P = aV 2, where a is constant., What is the molar heat capacity of the gas for this process?, (1), , 17R, 6, , (2), , 6R, 17, , (3), , 13R, 6, , (4), , 16R, 7, , Sol. Answer (1), P = aV2, PV–2 = a, , Compare with PVN = constant then N = – 2, , Polytropic process, ∵, , C CV , , , , , , , , R, 1 N, , , , R, R, , 1 1 N, , of diatomic , , 7, 5, , , , R, R, , ⎛ 7 ⎞ 1 ( 2), ⎜ 1⎟, ⎝5 ⎠, 5R, R, 17R, , , 2 1 2, 6, , 37. In a thermodynamic process two moles of a monatomic ideal gas obeys P V 2 . If temperature of the gas, increases from 300 K to 400 K, then find work done by the gas (where R = universal gas constant)., (1) 200 R, , (2), , –200 R, , (3), , –100 R, , (4), , –400 R, , Sol. Answer (2), P V–2, PV2 = constant, , Compare with PVN = constant then N = 2, , ⎛ R ⎞, W ⎜, ⎟ T, ⎝ 1 N ⎠, , W , , , , R, (T2 T1 ), 1 N, , 2 R(400 300), (1 2), , = –200 R, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 187, , 38. Entropy of a system decreases, (1) When heat is supplied to a system at constant temperature, (2) When heat is taken out from the system at constant temperature, (3) At equilibrium, (4) In any spontaneous process, Sol. Answer (2), Entropy of a system decreases when heat is taken out of the system at constant temperature., 39. If during an adiabatic process the pressure of mixture of gases is found to be proportional to square of its, absolute temperature. The ratio of CP /CV for mixture of gases is, (1) 2, , (2), , 1.5, , (3), , 1.67, , (4), , 2.1, , Sol. Answer (1), P T2, PT–2 = constant, , ⎛ ⎞, compare with PT ⎜, ⎟ constant, ⎝ 1 ⎠, , CP, 2, CV, , 40. If the efficiency of a carnot engine is , then the coefficient of performance of a heat pump working between, the same temperatures will be, (1) 1 – , , (2), , 1 , , , (3), , 1, , , (4), , 1+, , 1, , , Sol. Answer (3), Coefficient of performance of heat pump =, , 1, 1, , , efficiency of Carnot engine, , 41. In a Carnot engine, when heat is absorbed from the source, temperature of source, (1) Increases, , (2), , Decreases, , (3) Remains constant, , (4), , Cannot say, , Sol. Answer (3), Even when heat is taken out temperature stays the same. i.e., heat capacity of surface is infinite., 42. A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. The amount of heat, energy supplied to engine from the source in each cycle is, (1) 800 J, , (2), , 1600 J, , (3), , 3200 J, , (4), , 6400 J, , Sol. Answer (2), W = 800 J, , T, W, 1 2, Q, T1, , 800, 300, 1, Q, 600, 1600 J = Q, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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188, , Thermodynamics, , Solution of Assignment, , 43. An ideal heat engine operates on Carnot cycle between 227°C and 127°C. It absorbs 6 × 104 cal at the higher, temperature. The amount of heat converted into work equals to, (1) 4.8 × 104 cal, , (2), , 3.5 × 104 cal, , (3), , 1.6 × 104 cal, , (4), , 1.2 × 104 cal, , Sol. Answer (4), , W, 6 10, , 4, , 1, , 400, 500, , W = 1.2 × 104 cal, 44. The maximum possible efficiency of a heat engine is, (1) 100%, (2), , T1, T2, , (3), , T1, 1, T2, , ⎛ T ⎞, (4) Dependent upon the temperature of source (T1) and sink (T2) and is equal to ⎜⎜1 2 ⎟⎟, ⎝ T1 ⎠, , Sol. Answer (4), T, 1 2, T1, , So it depends on source and sink temperature., 45. A frictionless heat engine can be 100% efficient only if its exhaust temperature is, (1) Equal to its input temperature, , (2), , Less than its input temperature, , (3) 0 K, , (4), , 0°C, , Sol. Answer (3), , ∵, , T, 1 2, T1, , If exhaust temperature zero kelvin then = 100%., 46. A reversible engine and an irreversible engine are working between the same temperatures. The efficiency of, the, (1) Two engines are same, , (2), , Reversible engine is greater, , (3) Irreversible engine is greater, , (4), , Two engines cannot be compared, , Sol. Answer (2), Efficiency of reversible engine is greater, because there is no loss of heat., 47. Which of the following can be coefficient of performance of refrigerator?, (1) 1, , (2), , 0.5, , (3), , 9, , (4), , All of these, , Sol. Answer (4), , , , 1 , , , , , 1, 1, , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , is less than 1 so, , , Thermodynamics, , 189, , 1, 1, , , 1, 1 0, , , >0, 48. The temperature inside and outside a refrigerator are 273 K and 300 K respectively. Assuming that the, refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly, (1) 11 J, , (2), , 22 J, , (3), , 33 J, , (4), , 50 J, , Sol. Answer (1), T, 273, 9, 1 2 ; 1, , T1, 300 100, , , , 1 100, 91, , 1, ∼ 11 J, , 9, 9, , , , Q, W, , For W = 1 J, Q=, Q = 11 J, 49. By opening the door of a refrigerator placed inside a room you, (1) Can cool the room to certain degree, (2) Can cool it to the temperature inside the refrigerator, (3) Ultimately warm the room slightly, (4) Can neither cool nor warm the room, Sol. Answer (3), Ultimately warm the room because work is being done by the refrigerator., 50. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of, source be increased to as to increase its efficiency by 50% of original efficiency?, (1) 150 K, , (2), , 250 K, , (3), , 300 K, , (4), , 450 K, , Sol. Answer (2), , 40, 300, 1, 50% increase in efficiency, T1, 100, T1 500 K, , 150, 0.4 0.6, 100, new efficiency = 0.6 =, 60, 300, 1, T1, 100, , 60, 100, , T1 750 K, , Difference between 2 Temperatures = 250 K, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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190, , Thermodynamics, , Solution of Assignment, , SECTION - B, Objective Type Questions, 1., , A container is filled with 20 moles of an ideal diatomic gas at absolute temperature T. When heat is supplied, to gas temperature remains constant but 8 moles dissociate into atoms. Heat energy given to gas is, (1) 4RT, , (2), , 6RT, , (3), , 3RT, , (4), , 5RT, , Sol. Answer (1), Heat supplied = U = Ufinal – Uinitial, , Uinitial , U , , 5, 5, 3, 20 RT , Ufinal 20 8 RT 2 8 RT, 2, 2, 2, , , 8 RT, 2, , = 4RT, Heat energy given is 4RT., , Time, , (3), Time, , (4), , Temperature, , (2), , Temperature, , (1), , Temperature, , Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which, one of the following graphs represents the variation of temperature with time?, , Temperature, , 2., , Time, , Time, , Liquid oxygen when heated will observe, a rise in temperature as well as change, in state one time, which can be, represented as, , 3., , Temperature, , Sol. Answer (3), , For an isobaric process, the ratio of Q (amount of heat supplied) to the W (work done by the gas) is, ⎛, C ⎞, ⎜⎜ P ⎟⎟, CV ⎠, ⎝, , (1) , , (2), , – 1, , (3), , , 1, , (4), , , 1, , Sol. Answer (4), For isobaric process Q = nCPT and W = nRT, , So,, , CP, Q CP, , , , W, R, CP CV, , , 1, , C, 1, 1 V, CP, , ⎡ CP, ⎤, ⎥, ⎢∵, ⎣ CV, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 4., , Thermodynamics, , 191, , 3 moles of an ideal gas are contained within a cylinder by a frictionless piston and are initially at temperature, T. The pressure of the gas remains constant while it is heated and its volume doubles. If R is molar gas, constant, the work done by the gas in increasing its volume is, 3, RT ln 2, 2, , (1), , (2), , 3RT ln 2, , (3), , 3, RT, 2, , (4), , 3RT, , Sol. Answer (4), W = PV, = PV, = nRT, = 3RT, 5., , Two moles of a gas at temperature T and volume V are heated to twice its volume at constant pressure. If, Cp, Cv, , (1), , then increase in internal energy of the gas is, , RT, 1, , (2), , 2RT, 1, , (3), , 2RT, 3( 1), , (4), , 2T, 1, , Sol. Answer (2), Q = U + W, Q – W = U, , , W W U, 1, ⎛ 1 ⎞ (P.V ) nRT 2RT, U W ⎜, ⎟ 1 1 1, ⎝ 1⎠, , 6., , A triatomic, diatomic and monatomic gas is supplied same amount of heat at constant pressure, then, (1) Fractional energy used to change internal energy is maximum in monatomic gas, (2) Fractional energy used to change internal energy is maximum in diatomic gas, (3) Fractional energy used to change internal energy is maximum in triatomic gases, (4) Fractional energy used to change internal energy is same in all the three gases, , Sol. Answer (3), C, U nCV T, 1, , V , Q nCP T, CP , 1, 3, ⎛ U ⎞, , , ⎜, ⎟, ⎝ Q ⎠mono mono 5, , 1, 5, ⎛ U ⎞, , ⎜, ⎟ , ⎝ Q ⎠dia dia 7, 1, 3, ⎛ U ⎞, , ⎜, ⎟ , ⎝ Q ⎠tria tria 4, Fractional energy used to change internal energy is maximum in Triatomic gas., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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192, 7., , Thermodynamics, , Solution of Assignment, , 105 calories of heat is required to raise the temperature of 3 moles of an ideal gas at constant pressure from, 30°C to 35°C. The amount of heat required in calories to raise the temperature of the gas through the range, Cp, ⎛, ⎞, (60°C to 65°C) at constant volume is ⎜⎜ , 1.4 ⎟⎟, Cv, ⎝, ⎠, , (1) 50 cal, , (2), , 75 cal, , (3), , 70 cal, , (4), , 90 cal, , Sol. Answer (2), At constant pressure heat absorbed = Q = nCP T, , …(1), , At constant volume heat absorbed = U = nCV T, , …(2), , Dividing (1) by (2),, 105, Q CP, , 1.4 ⇒, 1.4, U CV, U, , UV = 75 cal, 8., , To an ideal triatomic gas 800 cal heat energy is given at constant pressure. If vibrational mode is neglected,, then energy used by gas in work done against surroundings is, (1) 200 cal, , (2), , 300 cal, , (3), , 400 cal, , (4), , 60 cal, , Sol. Answer (1), Heat at constant pressure, Q = nCP T, Heat for doing work, W = nRT, Then, , W, nR T, , Q nCP T, , W ⎛ 1 ⎞, , 800 ⎝⎜ ⎠⎟, , 1, W, 1, , 800, W, 3, 1, 800, 4, , W = 200 cal, 9., , A closed cylindrical vessel contains N moles of an ideal diatomic gas at a temperature T. On supplying heat,, temperature remains same, but n moles get dissociated into atoms. The heat supplied is, (1), , 5, (N n )RT, 2, , (2), , 5, nRT, 2, , (3), , 1, nRT, 2, , (4), , 3, nRT, 2, , Sol. Answer (3), Heat supplied = U = U final Uinitial, , Total internal energy initially =, , 5, NRT, 2, , [Only diatomic gas is present], , Total internal energy when, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 'n' moles get dissociated =, , 5, 3, N n RT 2n RT, 2, 2, , 193, , [diatomic and monoatomic both are present], , 3, ⎧5, ⎫ 5, U ⎨ N n RT 2n RT ⎬ NRT, 2, 2, ⎩, ⎭ 2, Solving this we get, U , , 1, nRT, 2, , Heat supplied is, , 1, nRT ., 2, , 10. Figure shows the isotherms of a fixed mass of an ideal gas at three temperatures TA, TB and TC, then, , 1, V, , A, , B, C, , P, , O, (1) TA > TB > TC, , (2), , TA < TB < TC, , (3), , TB < TA < TC, , (4), , TA = TB = TC, , Sol. Answer (2), , A, , 1, V, , ∵ PV = RT, , RT, P, V, , B, C, , ∵ For constant, , 1, V, , So, P T, , O, , ∵ PC > PB > PA then, , PA PB, , PC, , P, , TC > TB > TA, 11. An ideal monatomic gas at 300 K expands adiabatically to 8 times its volume. What is the final temperature?, (1) 75 K, , (2), , 300 K, , (3), , 560 K, , (4), , 340 K, , Sol. Answer (1), Adiabatic expansion, for monoatomic gas , , 5, 3, , T1V11 T2V2 1, , ⎛V ⎞, T2 T1 ⎜ 1 ⎟, ⎝ V2 ⎠, ⎛V ⎞, 300 ⎜ 1 ⎟, ⎝ 8V1 ⎠, , 1, , 5/3, , , , 300, 75 K, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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194, , Thermodynamics, , Solution of Assignment, , 5, ) is 3 × 105 N/m2. If the same gas is undergoing adiabatic change, 3, then adiabatic elasticity at that instant is, , 12. Slope of isotherm for a gas (having , , (1) 3 × 105 N/m2, , (2), , 5 × 105 N/m2, , (3), , 6 × 105 N/m2, , (4), , 10 × 105 N/m2, , Sol. Answer (2), Adiabatic elasticity = P, , , 5, 3 105 5 105 N/m2, 3, , 13. A gas may expand either adiabatically or isothermally. A number of P–V curves are drawn for the two, processes over different range of pressure and volume. It will be found that, (1) An adiabatic curve and an isothermal curve may intersect, (2) Two adiabatic curves do not intersect, (3) Two isothermal curves do not intersect, (4) All of these, Sol. Answer (4), Slope for isothermal and adiabatic are not same so they will intersect., 14. The variation of pressure P with volume V for an ideal monatomic gas during an adiabatic process is shown, in figure. At point A the magnitude of rate of change of pressure with volume is, , P, , A, , 3 P0, O, (1), , 3 P0, 5 V0, , (2), , V, , 2 V0, , 5 P0, 3 V0, , (3), , 3 P0, 2 V0, , (4), , 5 P0, 2 V0, , Sol. Answer (4), PV = constant, P V–, , P, , dP, dV, , P, V, dP, P, , dV, V, , , , 5 3 P0, , 3 2 V0, , , , 5 P0, 2 V0, , 3 P0, O, , A, 2 V0, , V, , ⎛ dP ⎞ 5 P0, Then ⎜, ⎟, ⎝ dV ⎠ 2V0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 195, , 15. Figure shows, the adiabatic curve on a log T and log V scale performed on ideal gas. The gas is, log T, 5, 4, 3, 2, 1, O, , A, B, , 1 2 3 4 5, , log V, , (1) Monatomic, , (2), , Diatomic, , (3) Polyatomic, , (4), , Mixture of monatomic and diatomic, , Sol. Answer (1), , TV 1 K, , logT 1 logV 0, logT 1 logV, y = – ( – 1) x, , y, 24, 1 slope =, x, 4 1, – ( – 1) = , , , 2, 3, , 5, 3, , Monoatomic., 16. A cyclic process on an ideal monatomic gas is shown in figure. The correct statement is, P, , B, , C, A, V, , (1) Work done by gas in process AB is more than that in the process BC, (2) Net heat energy has been supplied to the system, (3) Temperature of the gas is maximum at state B, (4) In process CA, heat energy is absorbed by system, Sol. Answer (2), It is a cyclic system U = 0, and work done is (+)ive, so heat is supplied to system., 17. A diatomic gas undergoes a process represented by PV1.3 = constant. Choose the incorrect statement, (1) The gas expands by absorbing heat from the surroundings, (2) The gas cools down during expansion, (3) The work done by surroundings during expansion of the gas is negative, (4) None of these, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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196, , Thermodynamics, , Solution of Assignment, , Sol. Answer (4), PV1.3 = K, W , , P2V2 PV, 1 1, 1 N, , ∵ N > 1, so W is negative., , Heat supplied by surrounding heat goes to do work., Down when expands., 18. If a gas is taken from A to C through B then heat absorbed by the gas is 8 J. Heat absorbed by the gas in, taking it from A to C directly is, , P (kPa), 20, , C, , 10, , B, , A, 200, , (1) 8 J, , (2), , 400, , 9J, , (3), , V (cc), 11 J, , (4), , 12 J, , Sol. Answer (2), When taken through ABC [U + work = heat absorbed], Heat absorbed = area under graph + U = 8, U 8 , , 10 200, 6, 1000, , when taken directly to C, W + U = Q, , ⎡10 200 1 2000 ⎤, ⎢ 1000 2 1000 ⎥ 6 Q Q = 9 J, ⎣, ⎦, 19. The process CD is shown in the diagram. As system is taken from C to D, what happens to the temperature, of the system?, P, , 3p0, , C, , p0, , D, v0, , 3v0, , V, , (1) Temperature first decreases and then increases, , (2), , Temperature first increases and then decreases, , (3) Temperature decreases continuously, , (4), , Temperature increases continuously, , Sol. Answer (2), T3 > T2 > T1, , P, , T1T2 T3, , So from C D, Temperature first increases then decreases., , V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 197, , 20. A P-T graph is shown for a cyclic process. Select correct statement regarding this, , P, , C, B, , O, , A, , D, , T, , (1) During process CD, work done by gas is negative, (2) During process AB, work done by the gas is positive, (3) During process BC internal energy of system increases, (4) During process BC internal energy of the system decreases, Sol. Answer (3), In process BC (isochoric process) where T is (+)ive., So U = nCVT, ∵ T is positive U increases, 21. A hydrogen cylinder is designed to withstand an internal pressure of 100 atm. At 27°C, hydrogen is pumped, into the cylinder which exerts a pressure of 20 atm. At what temperature does the danger of explosion first, sets in?, (1) 500 K, , (2), , 1500 K, , (3), , 1000 K, , (4), , 2000 K, , Sol. Answer (2), Constant volume process, PV = nRT, P1 P2, , T1 T2, , 20 100, , 300 T2, T2 = 1500 K, 22. An ideal gas of volume V and pressure P expands isothermally to volume 16 V and then compressed, adiabatically to volume V. The final pressure of gas is [ = 1.5], (1) P, , (2), , 3P, , (3), , 4P, , (4), , 6P, , Sol. Answer (3), Isothermal expansion, P1V1 = P2V2, PV = 16 V × P', P, P', 16, P1V1 P2V2, , [adiabatic compression], , P'(16 V)1.5 = P" (V)1.5, P, 161.5 P " 4P, 16, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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198, , Thermodynamics, , Solution of Assignment, , 23. The pressure P of an ideal diatomic gas varies with its absolute temperature T as shown in figure. The molar, heat capacity of gas during this process is [R is gas constant], P, , T, , (1) 1.7 R, , (2), , 3.25 R, , (3), , 2.5 R, , (4), , 4.2 R, , Sol. Answer (3), , CV of diatomic =, , 5, R, 2, , 24. An ideal gas expands according to the law P 2V = constant. The internal energy of the gas, (1) Increases continuously, , (2), , Decreases continuously, , (3) Remain constant, , (4), , First increases and then decreases, , Sol. Answer (1), P2V = K, or PV–2 = K, N = –2, , C CV , , R, positive quantity, 1 N, , C>0, W>0, ∵, , PV–2, , [gas is expanding], = K so TV–3 = constant, , T will increases if V increases., T > 0, So U = CT > 0, It will increase continuously., 25. The variation of pressure P with volume V for an ideal diatomic gas is parabolic as shown in the figure. The molar, specific heat of the gas during this process is, , P, , V, , O, (1), , 9R, 5, , (2), , 17R, 6, , (3), , 3R, 4, , (4), , 8R, 5, , Sol. Answer (2), P = aV–2, , C, , R, R, , 1 1 N, , C, , 17R, 6, , So, PV2 = constant then N = 2, ∵ = 1.4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 199, , 26. Neon gas of a given mass expands isothermally to double volume. What should be the further fractional, decrease in pressure, so that the gas when adiabatically compressed from that state, reaches the original, state?, (1) 1 – 2–2/3, , (2), , 1 – 31/3, , 21/3, , (3), , (4), , 32/3, , Sol. Answer (1), , PV, 1 1 P2V2, , [for isothermal], , PV P ' 2V, P, P', 2, P1V1 P2V2, , [for adiabatic], , P, 5/3, 5/3, 2V , P2 V , 2, , [ for neon = 5/3 ], , P P2 . 2 , , 2/3, , P P P2 P2 . 2 , , Fractional decrease 2, P2, P2, , 2/3, , 1 22/3, , 27. When 1 kg of ice at 0°C melts to water at 0° C, the resulting change in its entropy, taking latent heat of ice, to be 80 cal/°C is, (1) 293 cal/K, , (2), , 273 cal/K, , 8 × 104 cal/K, , (3), , (4), , 80 cal/K, , Sol. Answer (1), ∵, , ⎛ Q, Entropy ⎜, ⎝ T, y , , ⎞ ⎛ mlf ⎞, ⎟⎜, ⎟, ⎠ ⎝ T ⎠, , 1000 80, 293 cal/K, 273, , 28. Carnot cycle is plotted in P-V graph. Which portion represents an isothermal expansion?, , P A, , B, C, , D, , V, (1) AB, , (2), , BC, , (3), , CD, , (4), , DA, , Sol. Answer (1), AB is isothermal expansion. BC is adiabatic expansion, CD is isothermal compression, A = adiabatic compression., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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200, , Thermodynamics, , Solution of Assignment, , 29. Efficiency of a heat engine working between a given source and sink is 0.5. Coefficient of performance of the, refrigerator working between the same source and the sink will be, (1) 1, , (2), , 0.5, , (3), , 1.5, , (4), , 2, , Sol. Answer (1), , , , 1, 1 , , 0.5 , , 1, 1 , , =1, 30. A heat engine rejects 600 cal to the sink at 27°C. Amount of work done by the engine will be, (Temperature of source is 227°C & J = 4.2 J/cal), (1) 1680 J, , (2), , 840 J, , (3), , 2520 J, , (4), , None of these, , Sol. Answer (1), T, W, 1 2 , T1 Q1, 1, , 300 W, , 500 Q1, , W 2, , Q1 5, Q1 , , 5W, 2, , ∵ W = Q1 – Q2, Then Q2 = Q1 – W, Q2 , , 5W, 3W, W , 2, 2, , Then W , , 2Q2 2 600, = 400 cal = 400 × 4.2 J = 1680 J, 3, 3, , SECTION - C, Previous Years Questions, 1., , 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0, JK–1 mol–1. If the speed of sound in this gas at NTP is 952 ms–1, then the heat capacity at constant pressure, is (Take gas constant R = 8.3 JK–1 mol–1), [Re-AIPMT-2015], (1) 8.5 JK–1 mol–1, , (2), , 8.0 JK–1 mol–1, , (3), , 7.5 JK–1 mol–1, , (4), , 7.0 JK–1 mol–1, , Sol. Answer (2), 2., , The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is –20°C, the temperature, of the surroundings to which it rejects heat is, [Re-AIPMT-2015], (1) 21°C, , (2), , 31°C, , (3), , 41°C, , (4), , 11°C, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 3., , Thermodynamics, , 201, , An ideal gas is compressed to half its initial volume by means of several processes. Which of the process, results in the maximum work done on the gas?, [Re-AIPMT-2015], (1) Isothermal, , (2), , Adiabatic, , (3), , Isobaric, , (4), , Isochoric, , Sol. Answer (2), 4., , One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, , P (in kPa), 5, , A, B, , 2, 4, , 6, 3, , V (in m ), The change in internal energy of the gas during the transition is, (1) –12 kJ, , (2), , 20 kJ, , (3), , [AIPMT-2015], , –20 kJ, , (4), , 20 J, , Sol. Answer (3), U = nCVT, ⎛ R ⎞, n⎜, ⎟ (T2 T1 ), ⎝ 1⎠, , P (kPa), 5, , n[RT2 RT1 ], , 1, , , , A, B, , 2, , n(P2V2 PV, 1 1), 1, , O, , , , 1(2 6 103 5 4 103 ), 7 / 5 1, , , , 8 103, 20 103, 2/5, , 4, , 6, , 3, , V (m ), , = –20 kJ, , 5., , 1, as heat engine, is used as a refrigerator. If the work done, 10, on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is, , A Carnot engine, having an efficiency of =, , [AIPMT-2015], (1) 1 J, , (2), , 100 J, , (3), , 99 J, , (4), , 90 J, , Sol. Answer (4), ∵, , , , Q, 1, 1 2, , W, , Q, 1, 1 2, (1/ 10), 10, (10 1) , , Q2, 10, , Q2 = 90 J, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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202, 6., , Thermodynamics, , Solution of Assignment, , Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB,, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat, absorbed by the system in the process AC will be, [AIPMT-2015], , P, B, , 4, , 6 × 10 Pa, , C, , A, , 4, , 2 × 10 Pa, , –3, , 2 × 10 m, , (1) 300 J, , (2), , 380 J, , 3, , (3), , –3, , 4 × 10 m, , 500 J, , 3, , V, (4), , 460 J, , Sol. Answer (4), In process ABC, ∵ Q = U + W, So, U = Q – W, U = (400 + 100) – (6 × 104 × 2 × 10–3), U = 500 – 120, U = 380 J, In process AC, Q = U + W, ⎡1, ⎤, 380 ⎢ (2 104 6 104 ) 2 103 ⎥, ⎣2, ⎦, , = 380 + 80, = 460 J, 7., , A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically, to a volume 16V. the final pressure of the gas is: (take = 5/3), [AIPMT-2014], (1) 64P, , (2), , 32P, , (3), , P/64, , (4), , 16P, , Sol. Answer (3), In isothermal process, P1V1 = P2V2, PV = P2(2V), , P, ...(1), 2, In adiabatic process, P2 , , , , , , P2V2 = P3V3, , ⎛P ⎞, , , ⎜ ⎟ (2V ) P3 (16V ), ⎝2⎠, , P3 , , P ⎛ 2v ⎞, ⎜, ⎟, 2 ⎝ 16V ⎠, , , , , , P ⎛ 1⎞, ⎜ ⎟, 2 ⎝8⎠, , 5/3, , , , P, 64, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , 8., , Thermodynamics, , 203, , A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system, in the cycle is, [AIPMT-2014], , P, , C, , 3 P0, , B, , 2 P0, P0, , (1) P0 V0, , (2), , D, , A, V0, , 2P0 V0, , 2V0 V, P0V0, 2, , (3), , (4), , Zero, , Sol. Answer (4), W = area enclosed by AODA + by area enclosed OBCO, , P, 3 P0, , ⎡1, ⎤ ⎡ 1, ⎤, ⎢ (2V0 V0 ) P0 ⎥ ⎢ (2V0 V0 )P0 ⎥, ⎣2, ⎦ ⎣ 2, ⎦, , 9., , P0, , B, O, , 2 P0, , ⎧ AODA is a clockwise, ⎨, ⎩ while DBCO is anticlockwise, , =0, , C, , A, V0, , D, 2V0 V, , A gas is taken through the cycle A B C A, as shown. What is the net work done by the gas?, , P(105 Pa), 7, 6, 5, 4, 3, 2, 1, 0, , B, A, 1 2, , C, 4, , 6, , V(10–3 m3), , 8, , [NEET-2013], (1) 1000 J, , (2), , Zero, , (3), , – 2000 J, , (4), , 2000 J, , Sol. Answer (1), ∵ Cyclic curve is clockwise i.e., W = +ve, W = area enclosed, , 1, 5 103 4 105, 2, = 1000 J, , , 10. The molar specific heats of an ideal gas at constant pressure and volume are denoted by Cp and Cv respectively., If , , (1), , Cp, Cv, , and R is the universal gas constant, then Cv is equal to, , R, ( 1), , (2), , ( 1), R, , (3), , R, , [NEET-2013], , (4), , 1 , 1 , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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204, , Thermodynamics, , Solution of Assignment, , Sol. Answer (1), , , , Cp, Cv, , We know Cp – Cv = R, , R, 1, , So Cv , , 11. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature., Cp, The ratio of, for the gas is:, [NEET-2013], Cv, (1) 2, , 5, 3, , (2), , (3), , 3, 2, , (4), , 4, 3, , Sol. Answer (3), P T3, PT–3 = constant, , Compare with, , Then,, , ⎛ ⎞, ⎜, ⎟, PT ⎝ 1 ⎠, , constant, , , 3, 1 , , , , 3, 2, , 12. In the given (V – T) diagram, what is the relation between pressures P1 and P2?, , V, , [NEET-2013], , P2, P1, 2, 1, , (1) P2 > P1, , (2), , P2 < P1, , (3), , T, Cannot be predicted, , (4), , P2 = P1, , Sol. Answer (2), ∵ PV = RT, , ⎧∵ Slope of V -T graph ⎫, ⎪, ⎪, ⎨, ⎬, V, m, , tan, , , ⎪⎩, ⎪⎭, T, , V R, tan , T P, i.e., P , , 1, tan , , 2 > 1 so tan2 > tan1, P2 < P1, , then P2 < P1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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206, , Thermodynamics, , Solution of Assignment, , 16. During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies, that, [AIPMT (Prelims)-2011], (1) 150 J of heat has been added to the gas, (2) 150 J of heat has been removed from the gas, (3) 300 J of heat has been added to the gas, (4) No heat is transferred because the process is isothermal, Sol. Answer (2), It implies 150 J heat has been removed from the gas., 17. A mass of diatomic gas ( = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its, temperature rises from 27°C to 927°C. The pressure of the gas in the final state is, [AIPMT (Mains)-2011], (1) 256 atm, , (2), , 8 atm, , (3), , 28 atm, , (4), , 68.7 atm, , Sol. Answer (1), P, , 1, , ⎛ ⎞, ⎜, ⎟, , , , then, PT ⎝ 1 ⎠ constant, , T C, , P2 ⎛ T1 ⎞, ⎜ ⎟, P1 ⎝ T2 ⎠, , , , 1, , 1.4, , P2 ⎛ 300 ⎞, , 2 ⎜⎝ 1200 ⎟⎠, P2 ⎛ 1 ⎞, , 2 ⎜⎝ 4 ⎟⎠, , 7, , 11.4, , 2, , P2 26 256 atm, 18. If U and W represent the increase in internal energy and work done by the system respectively in a, thermodynamical process, which of the following is true?, [AIPMT (Prelims)-2010], (1) U = – W, in a isothermal process, , (2), , U = – W, in a adiabatic process, , (3) U = W, in a isothermal process, , (4), , U = W, in a adiabatic process, , Sol. Answer (2), In adiabatic process Q = 0, So U = – W, , [∵ Q = W + U], , 19. If Cp and Cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M, where R is the, molar gas constant, [AIPMT (Mains)-2010], (1) Cp – Cv = R/M2, , (2), , Cp – Cv = R, , (3), , Cp – Cv = R/M, , (4), , Cp – Cv = MR, , Sol. Answer (3), , Cp Cv , , R, M, , Because Cp & Cv are given per unit mass, And Cp – Cv = R is for 1 mole, So here we use R/M where M is molecular mass., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 20. A monoatomic gas at pressure P1 and V1 is compressed adiabatically to, final pressure of the gas?, (1) 64 P1, , (2), , P1, , (3), , 207, , 1, th its original volume. What is the, 8, [AIPMT (Mains)-2010], , 16 P1, , (4), , 32 P1, , Sol. Answer (4), ∵ PV = constant, , , 5, ⎛V ⎞, P2 P1 ⎜ 1 ⎟ P 8 3 32P1, ⎝ V2 ⎠, , 21. In thermodynamic processes which of the following statements is not true?, , [AIPMT (Prelims)-2009], , (1) In an isochoric process pressure remains constant, (2) In an isothermal process the temperature remains constant, (3) In an adiabatic process PV = constant, (4) In an adiabatic process the system is insulated from the surroundings, Sol. Answer (1), In isochoric processes volume remains constant., 22. The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is, [AIPMT (Prelims)-2009], (1) 6400 J, , (2), , 5400 J, , (3), , 7900 J, , (4), , 8900 J, , Sol. Answer (3), 2 × 4.2 × 1000 = dU + 500, dU = 7900 J, 23. If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed, cycle process, then, [AIPMT (Prelims)-2008], (1) Q = 0, , (2), , W=0, , (3), , Q=W=0, , (4), , E=0, , Sol. Answer (4), E = change in U, and in cyclic process U = 0, E=0, 24. At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110°C this ratio, is, [AIPMT (Prelims)-2008], (1), , 283, x, 383, , (2), , x, , (3), , 383, x, 283, , (4), , 10, x, 110, , Sol. Answer (1), , , x, P, , at 10°C, , M, x, PV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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208, , Thermodynamics, , Solution of Assignment, , Molecular mass × number of moles, x, R T, 1, x, , T, 383 x, , 283 x , x , , 283, x, 383, , 25. An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency is doubled., Temperature of the source is, [AIPMT (Prelims)-2007], (1) 99°C, , (2), , 124°C, , (3), , 37°C, , (4), , 62°C, , Sol. Answer (1), , T 62, 1, 1 L, 3, TH, ⎡ TL 5 ⎤, ⎥, ⎢, ⎣TH 6 ⎦, , 1, 5 62, 1 , 3, 6 TH, , TH = 372° K = 99°C, TL = 37°C, 26. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of, source be increased so as to increase its efficiency by 50% of original efficiency? [AIPMT (Prelims)-2006], (1) 275 K, , (2), , 325 K, , (3), , 250 K, , (4), , 380 K, , Sol. Answer (3), Where T2 Sink Temperature, , T, 1 2, T1, , T1 Source Temperature, , Temperature of sink is given to be 300 K., = 0.4, So 0.4 1 , , 300, T1, , T1 = 500 K, Now, is increased by 50%., , , 150, 15, , 0.4 0.6, 100, 10, , To maintain same sink temperature new source temperature is, , 0.6 1 , , 300, T1, , T1 = 750 K, Increase in temperature = 750 – 500 = 250 K, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 27. The molar specific heat at constant pressure of an ideal gas is, pressure to that at constant volume is :, , (1), , 7, 5, , (2), , 8, 7, , (3), , 209, , 7, R . The ratio of specific heat at constant, 2, [AIPMT (Prelims)-2006], , 5, 7, , (4), , 9, 7, , Sol. Answer (1), , CP , , , , , , 7, R, R, , 2, , 7, 5, , 28. Which of the following processes is reversible ?, , [AIPMT (Prelims)-2005], , (1) Transfer of heat by radiation, , (2), , Electrical heating of a nichrome wire, , (3) Transfer of heat by conduction, , (4), , Isothermal compression, , Sol. Answer (4), Isothermal compression takes place slowly at constant pressure, also U is zero so it is a reversible process., 29. An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 104 cal of heat at, higher temperature. Amount of heat converted to work is, (1) 2.4 × 104 cal, , (2), , 6 × 104 cal, , (3), , [AIPMT (Prelims)-2005], 1.2 × 104 cal, , (4), , 4.8 × 104 cal, , Sol. Answer (3), 30. A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy, at a is Ua = 10 J. Along the path adc the amount of heat absorbed Q1 = 50 J and the work obtained W1 =, 20 J whereas along the path abc the heat absorbed Q2 = 36 J. The amount of work along the path abc is, , d, , c, , a, , b, , P, , V, (1) 6 J, , (2), , 10 J, , (3), , 12 J, , (4), , 36 J, , Sol. Answer (1), dQ1 adc = 50 = dUadc + 20 dUadc = 30 = dUabc, dQabc = 36 = 30 + dWabc, dWabc = 6 J, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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210, , Thermodynamics, , Solution of Assignment, , 31. Consider two insulated chambers (A, B) of same volume connected by a closed knob, S. 1 mole of perfect, gas is confined in chamber A. What is the change in entropy of gas when knob S is opened? R = 8.31 J, mol–1K–1., , S, A, , (1) 1.46 J/K, , (2), , 3.46 J/K, , B, , (3), , 5.46 J/K, , (4), , 7.46 J/K, , Sol. Answer (3), V, S 2.303 nR loge 2, V1, , If initially volume is taken as V, then final volume = ZV, as volume of both chambers is given to be same., S 2.303 1 8.31 loge, , 2V, V, , S 5.46 J/K, , 32. A Carnot engine has efficiency 25%. It operates between reservoirs of constant temperatures with temperature, difference of 80°C. What is the temperature of the low-temperature reservoir?, (1) –25°C, , (2), , 25°C, , (3), , –33°C, , (4), , 33°C, , Sol. Answer (3), , 1, , TL, TH, , T, 1, 1 L, 4, TH, TH , , 4, TL, 3, , also TH – TL = 80, TL = 240 K = –33°C, 33. In an adiabatic change, the pressure and temperature of a monatomic gas are related as, P Tc, where c equals, (1), , 3, 5, , (2), , 5, 3, , (3), , 2, 5, , (4), , 5, 2, , Sol. Answer (4), P T C ⇒ PT C K, , And compare with, Then, C , C, , ⎛ ⎞, ⎜, ⎟, PT ⎝ 1 ⎠, , constant, , [condition from adiabatic process], , , 1 , , 5/3, 5/3, 5, , , 1 5 / 3, 2 / 3 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 211, , 34. An ideal Carnot engine, whose efficiency is 40%, receives heat at 500 K. If its efficiency is 50%, then the, intake temperature for the same exhaust temperature is, (1) 800 K, , (2), , 900 K, , (3), , 600 K, , (4), , 700 K, , Sol. Answer (3), , 1, , T2, T1, , T, 40, 1 2, 100, 500, , T2 = 300 K, If = 50%, , 50, 300, 1, 100, T1, T1 = 600 K, 35. A monatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The, temperature after compression will be, (1) 1164 K, , (2), , 144 K, , (3), , 18 K, , (4), , 887.4 K, , Sol. Answer (1), ∵ TV–1 = constant, ⎛V ⎞, T1 ⎜ 1 ⎟, ⎝ V2 ⎠, , 1, , T2, , 291 × (8)2/3 = T2, T2 = 291 × 4 = 1164 K, 36. An ideal gas, undergoing adiabatic change, has which of the following pressure temperature relationship?, (1) PT1– = constant, , (2), , P1–T = constant, , (3), , P–1T = constant, , (4), , PT–1 = constant, , Sol. Answer (2), PV constant, , ⎧∵ PV RT, ⎪, ⎨, ⎛ RT, V ⎜, ⎪, ⎝ P, ⎩, , , , ⎛T ⎞, P ⎜ ⎟ constant, ⎝P ⎠, , ⎫, ⎪, ⎞⎬, ⎟⎪, ⎠⎭, , P 1 . T constant, 37. A sample of gas expands from volume V1 to V2. The amount of work done by the gas is greatest, when the, expansion is, (1) Adiabatic, , (2), , Equal in all cases, , (3), , Isothermal, , (4), , Isobaric, , Sol. Answer (4), Work done is maximum in isobaric process, W = P.V = P(V2 – V1) = nR(T2 – T1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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212, , Thermodynamics, , Solution of Assignment, , 38. The efficiency of a Carnot engine operating with reservoir temperature of 100°C and – 23°C will be, (1), , 373 250, 373, , (2), , 373 250, 373, , (3), , 100 23, 100, , (4), , 100 23, 100, , Sol. Answer (2), ⎡ Where, ⎤, ⎢, ⎥, ⎢T2 sink temperature, ⎥, ⎢⎣T1 reservoir temperature ⎥⎦, , T, 1 2, T1, , , , 373 250, 373, , 39. We consider a thermodynamic system. If U represents the increase in its internal energy and W the work, done by the system, which of the following statements is true?, (1) U = –W in an isothermal process, , (2), , U = W in an isothermal process, , (3) U = – W in an adiabatic process, , (4), , U = W in an adiabatic process, , Sol. Answer (3), As Q = zero for adiabatic process, So U = –W for adiabatic process, 40. If the ratio of specific heat of a gas at constant pressure to that at constant volume is , the change in internal, energy of a mass of gas, when the volume changes from V to 2V at constant pressure P, is, (1), , PV, ( 1), , (2), , PV, , (3), , R, ( 1), , (4), , PV, ( 1), , Sol. Answer (1), nR(T2 T1 ) nRT2 nRT1, ⎛ R ⎞, U nCV T n ⎜, T , , ⎟, 1, 1, ⎝ 1⎠, , , , P (2V V ) PV, , 1, 1, , 41. An ideal gas at 27°C is compressed adiabatically to 8/27 of its original volume. The rise in temperature is (Take, = 5/3), (1) 275 K, , (2), , 375 K, , (3), , 475 K, , (4), , 175 K, , Sol. Answer (2), ∵ TV –1 = constant, T1V11 T2V21, , ⎡V ⎤, T1 ⎢ 1 ⎥, ⎣V2 ⎦, , 1, , T2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , ⎡ 27 ⎤, 300 ⎢ ⎥, ⎣8 ⎦, , 5 1, 3, , Thermodynamics, , 213, , T2, , 2, , ⎡3 ⎤, 300 ⎢ ⎥ T2, ⎣2⎦, , 300 , , 9, T2, 4, , 675 K = T2, T = 675 – 300 = 375 K, 42. Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature, T1 and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature, T and rejects to its sink at temperature T2. For what value of T the efficiencies of the two engines are equal?, (1), , T1 T2, 2, , (2), , T1 T2, 2, , (3), , T1T2, , (4), , T1T2, , Sol. Answer (4), A = B, ⎛ TL ⎞, ⎛ TL ⎞, ⎜, ⎟ ⎜, ⎟, ⎝ TH ⎠ A ⎝ TH ⎠B, , T T2, , T1 T, T 2 T1T2 ⇒ T T1T2, , 43. The (W/Q) of a Carnot engine is 1/6. Now the temperature of sink is reduced by 62°C, then this ratio becomes, twice, therefore the initial temperature of the sink and source are respectively, (1) 33°C, 67°C, , (2), , 37°C, 99°C, , (3), , 67°C, 33°C, , (4), , 97K, 37K, , Sol. Answer (2), , T, 1, 1 L ...(1), 6, TH, , ⎛ T 62 ⎞, ⎛ 1⎞, 2 ⎜ ⎟ 1 ⎜ L, ⎟, ⎝6⎠, ⎝ TH ⎠, , T 62 , 1, 1 2, 3, TH, 1, 5 62, 1 , 3, 6 TH, , TH = 372°K = 99°C, TL = 37°C, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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214, , Thermodynamics, , Solution of Assignment, , 44. A scientist says that the efficiency of his heat engine which works at source temperature 127°C and sink, temperature 27°C is 26%, then, (1) It is impossible, , (2), , It is possible but less probable, , (3) It is quite probable, , (4), , Data are incomplete, , Sol. Answer (1), , 1, , TL, 300, 1, 25%, 400, TH, , 45. The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept, constant and its efficiency raised to 60%, then the required temperature of sink will be, (1) 100 K, , (2), , 600 K, , (3), , 400 K, , (4), , 500 K, , Sol. Answer (3), 1, 500, 1, TH, 2, , T, 6, 1 L, 10, 103, , TH 103, , TL 4 102 400 K, , 46. An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C. It absorbs 6 kcal at the higher, temperature. The amount of heat (in kcal) converted into work is equal to, (1) 4.8, , (2), , 3.5, , (3), , 1.6, , (4), , 1.2, , Sol. Answer (4), , T, W, 1 L, Q1, TH, W, 400, 1, 1.2 J, 6, 500, , 47. One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of, specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas, will be, (1) (T + 2.4) K, , (2), , (T – 2.4) K, , (3), , (T + 4) K, , (4), , (T – 4) K, , Sol. Answer (4), W , , nR T2 T1 , 1, , 5⎤, ⎡, ⎢ 3 ⎥, ⎣, ⎦, , dU, , Tfinal T 4 K, 48. The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1 K to T2 K is, , 3, Na kB (T2 T1 ), 2, , (1), , (2), , 3, Na kB (T2 T1 ), 4, , (3), , ⎛T ⎞, 3, Na kB ⎜ 2 ⎟, 4, ⎝ T1 ⎠, , (4), , 3, Na k B (T2 T1 ), 8, , Sol. Answer (4), , Q ., , , f, R dT , 2, , 3, Na K B T2 T1 Q, 8, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 215, , 49. Which of the following relations does not give the equation of an adiabatic process, where terms have their, usual meaning?, (1) P.T1– = constant, , (2), , P1–T = constant, , (3) PV = constant, , (4), , TV–1 = constant, , Sol. Answer (1), It is P1–T = K, 50. According to C.E. van der Waal, the interatomic potential varies with the average interatomic distance (R) as, (1) R–1, , (2), , R–2, , (3), , R–4, , (4), , R–6, , Sol. Answer (4), According to van der Waal's formulae, interatomic potential is inversely proportion to R6., So, U R –6, 51. In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled,, then the resultant pressure will be, (1) 4P, , (2), , 2P, , (3), , P, , (4), , P/2, , Sol. Answer (2), P, , 1, MnV 2, 3, , P' , , 1 M, 1, 2, n 2V 2 MnV 2 2P, 3 2, 3, , 52. The mean free path of collision of gas molecules varies with its diameter (d) of the molecules as, (1) d–1, , (2), , d–2, , (3), , d–3, , (4), , d–4, , (4), , Potential energy, , Sol. Answer (2), , , , 1, d2, , 53. At 0 K, which of the following properties of a gas will be zero?, (1) Volume, , (2), , Density, , (3), , Kinetic energy, , Sol. Answer (3), at 0 K Vrms = 0 so K.E. = 0, 54. The value of critical temperature in terms of van der Waals’ constants a and b is given by, (1) TC , , 8a, 27Rb, , (2), , TC , , 27a, 8Rb, , (3), , TC , , a, 2Rb, , (4), , TC , , a, 27Rb, , Sol. Answer (1), TC , , 8a, 27Rb, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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216, , Thermodynamics, , Solution of Assignment, , 55. The degrees of freedom of a triatomic gas is, (Consider moderate temperature), (1) 6, , (2), , 4, , (3), , 2, , (4), , 8, , Sol. Answer (1), Degree of freedom = 3 rotational + 3 translational + 0 vibrational [T is moderate] = 6, 56. To find out degree of freedom, the expression is, (1) f , , 2, 1, , (2), , f , , 1, 2, , (3), , f , , 2, 1, , (4), , f , , 1, 1, , Sol. Answer (1), ∵, , CV , , fR, 2, , Then, f , , 2CV, 2CV, 2, 2, , , , CP, R, CP CV, 1, 1, CV, , 57. The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will, be (where R is the gas constant), (1) PV =, , 5, RT, 32, , (2), , PV = 5RT, , (3), , PV =, , 5, RT, 2, , (4), , PV =, , 5, RT, 16, , Sol. Answer (1), ∵, , ⎛m⎞, PV nRT ⎜ ⎟ RT, ⎝M ⎠, PV , , 5, RT, 32, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Work done by a gas in isothermal expansion is more than the work done by the gas in the same, expansion adiabatically., R : Temperature remains constant in isothermal expansion and not in adiabatic expansion., , Sol. Answer (2), A : is true, R : is true, but not correct explanation, correct explanation is, in isothermal expansion., ∵ T = 0 so U = 0, Q = W, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment, , Thermodynamics, , 217, , all the heat goes in doing work., Whereas in adiabatic process, Heat goes to work as well as in increasing internal energy., Wisothermal > Wadiabatic, 2., , A : Efficiency of heat engine can never be 100%., R : Second law of thermodynamics puts a limitation on the efficiency of a heat engine., , Sol. Answer (1), A : is true, R : is true, and correct explanation, 3., , A : Heat absorbed in a cyclic process is zero., R : Work done in a cyclic process is zero., , Sol. Answer (4), A : is false, in cyclic process only U = 0, Q = W., R : is false, work done is not zero only change in internal energy is zero., 4., , A : Coefficient of performance of a refrigerator is always greater than 1., R : Efficiency of heat engine is greater than 1., , Sol. Answer (4), A : is false., R : is false, Because efficiency of heat engine can never be equal to greater to 1., 1, , ∵ all the heat cannot be converted to work., and coefficient of performane of refrigerator, , 5., , , , 1 1, 1, , , , ∵, , 1 so may be less than 1., , A : Adiabatic expansion causes cooling., R : In adiabatic expansion, internal energy is used up in doing work., , Sol. Answer (1), A : is true, R : is true, and correct explanation, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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218, 6., , Thermodynamics, , Solution of Assignment, , A : The specific heat of an ideal gas is zero in an adiabatic process., R : Specific heat of a gas is process independent., , Sol. Answer (3), A : is true, R : is false, Because specific heat depends on the process., 7., , A : The change in internal energy does not depend on the path of process., R : The internal energy of an ideal gas is independent of the configuration of its molecules., , Sol. Answer (2), A : is true, R : is true, but not the correct explanation, because internal energy depends on the temperature of the gas., 8., , A : Heat supplied to a gaseous system in an isothermal process is used to do work against surroundings., R : During isothermal process there is no change in internal energy of the system., , Sol. Answer (1), A : true, R : true and correct explanation, 9., , A : In nature all thermodynamic processes are irreversible., R : During a thermodynamic process it is not possible to eliminate dissipative effects., , Sol. Answer (1), A : is true, R : is true and correct explanation, 10. A : During a cyclic process work done by the system is zero., R : Heat supplied to a system in the cyclic process converts into internal energy of the system., Sol. Answer (4), A : is false, in cyclic process, work done is not zero, internal energy change is zero., R : is false, heat supplied converts to work as initial state is equal to final state., No change in internal energy., , , , , , , , https://t.me/NEET_StudyMaterial, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
