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Chapter, , 1, , Physics and Measurement, UNITS, Measurement of any physical quantity involves its comparison with a certain basic, reference standard called, unit., Measurement = nu, Here, n is numerical value and u is unit. The numerical value is inversely proportional to the size of unit., n × u = constant, n, , 1, u, , DIMENSIONS OF PHYSICAL QUANTITIES, All the physical quantities represented by derived units can be expressed in terms of some combination of, seven fundamental quantities. These seven fundamental quantities are called seven dimensions of the physical, world. They are denoted with square brackets [ ]., Physical quantity, , S. No., 1., , 2., 3., 4., , 5., 6., 7., 8., , Absolute permittivity ( 0 ), , Dimensional formula, –1 –3 4, , M L T A, –2, , –2, , –2, , –2, , Useful result, , 2, , Absolute permeability ( 0), , MLT A, , Resistance R, , MLT A, , Inductance L, , ML2T –2 A–2, , Capacitance C, , M –1 L–2T4 A2, , q1q 2, , 1, , F, , 4, , F, l, , 0, , 0i1i 2, 2 r, 2, , P=IR, U, , 1 2, LI, 2, q2, 2C, , U, , L/R (time constant), , [M0 L0T 1], , –, , RC (time constant), , 0 0, , 1, , [M L T ], , –, , 0 0, , 1, , –, , ML –1 T–2, , –, , LC, , 9., , Stress, Pressure, Energy, 1, 2, 2, density,, 0E , B / 2 0, 2, , 10., , Heat capacity, Boltzmann, constant, , MLT, , 2, , ML T, , –2, , K, , –1, , r2, , –

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2, , Physics and Measurement, , Rapid Revision & Formula Bank, , DIMENSIONAL ANALYSIS AND ITS APPLICATIONS, Principle of Homogeneity of dimensions : It is based on the simple fact that length can be added to length., It states that in a correct equation, the dimensions of each term added or subtracted must be same. If two, quantities are being added or subtracted, they must be of same dimensional formula. Every correct equation, must have same dimensions on both sides of the equation., Note : Although torque and work done by a force have same dimensional formula yet they cannot be added, as their nature is different., Conversion of units : The numerical value of a physical quantity in a system of units can be changed to, another system of units using the equation n[u] = constant i.e., n1[u1] = n2[u2] where n is the numerical value, and u is the unit., a, , n2, , b, , M  L   T , n1  1   1   1 ,  M2   L 2   T2 , , c, , where the dimensional formula of the physical quantity is [MaLbTc]., , To find a relation among the physical quantities. If one knows the quantities on which a particular physical, quantity depends and guesses that this dependence is of product type, method of dimensions are helpful in, deducing their relation., Suppose we want to find the relation between force, mass and acceleration. Let force depends on mass and, acceleration as follows., F = Kmbac when K = dimensionless constant, b and c are powers of mass and acceleration., According to principle of homogeneity,, [F] = [K] [m]b [a]c, [MLT–2] = [M0L0T0] [M]b [LT–2]c, [MLT–2] = MbLc T–2c, Equating the dimension on both sides we get 1 = b, 1 = c, –2c = –2., b = 1 and c = 1., , ACCURACY AND PRECISION, Accuracy, The closeness of the measured value to the true value of the physical quantity is known as the accuracy of, the measurement., , Precision, It is the measure of the extent to which successive measurements of a physical quantity differ from one, another., Suppose the true value of a measurement is 35.75 and two measured values are 35.73 and 35.725. Here 35.73, is closest to 35.75, so its accuracy is more than 35.725 but 35.725 is more precise than 35.73 because 35.725, is measured upto 3 decimal places., , SIGNIFICANT FIGURES, The number of digits in the measured value about the correctness of which we are sure plus one more digit, are called significant figures., , Rules for counting the significant figures, Rule I : All non-zero digits are significant., Rule II : All zeros occurring between the non zero digits are significant. For example 230089 contains six, significant figures., Rule III : All zeros to the left of non zero digit are not significant. For example 0.0023 contains two significant, figures.

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Rapid Revision & Formula Bank, , Physics and Measurement, , 3, , Rule IV : If a number ends in zeros that are not to the right of a decimal, the zeros are not significant., For example, number of significant figures in, 1500 (Two), 1.5 × 103 (Two), 1.50 × 103 (Three), 1.500 × 103 (Four), Note : Length of an object may be represented in many ways say 5 m, 5.0 m, 500 cm, 5.00 m,, 5 × 102 cm. Here 5.00 m is most precise as it contains 3 significant figures., , Rules for Arithmetic Operations with Significant Figures, Rule I : In addition or subtraction, the final result should retain as many decimal places as there are in the, number with the least decimal places., Rule II : In multiplication or division, the final result should retain as many significant figures as are there in, the original number with the least significant figures., , Rounding Off of Uncertain Digits, Rule I : The preceding digit is raised by 1 if the insignificant digit to be removed is more than 5 and is left, unchanged if the later is less than 5., Rule II : When the insignificant digit to be removed is 5 and the uncertain digit is even, 5 is simply dropped, and if it is odd, then the preceding digit is raised by 1., , ERRORS IN MEASUREMENT, 1. Mean Absolute Error :- If a1, a2, a3, ........ an are n measurements then, , am, , a1 a2 ...... an, is taken as the true value of a quantity, if the same is not known., n, , a1 = am – a1, a2 = am – a2, ....................., an = am – an, Mean absolute error,, , a, , | a1 |, , | a2 | ...... | an |, n, , Final result of measurement may be written as :, a = am ±, , a, , 2. Relative Error or Fractional Error : It is given by, , a, am, , Mean absolute Error, Mean value of measuremen t, , 3. Percentage Error, , a, am, , 100%, , 4. Combination of Errors :, (i), , In Sum : If Z = A + B, then maximum absolute error in Z is given by,, , Z =, , A+, , B, maximum, , Z, A, B, i.e., when two physical quantities are added then, Z, A B A B, the maximum absolute error in the result is the sum of the absolute errors of the individual quantities., , fractional error in this case

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4, , Physics and Measurement, , (ii), , Rapid Revision & Formula Bank, , In Difference : If Z = A – B, then maximum absolute error is Z = A + B and maximum fractional, Z, A, B, ., error in this case, Z, A B A B, , (iii) In Product : If Z = AB, then the maximum fractional error,, fractional error., (iv) In Division : If Z = A/B, then maximum fractional error is, (v), , Z, Z, , In Power : If Z = An then, , A pB q, , In more general form if Z, , Z, Z, , A, A, , p, , B, B, , q, , r, , Cr, , Z, Z, , A, A, , B, Z, where, is known as, B, Z, , Z, Z, , A, A, , B, B, , A, A, , n, , then the maximum fractional error in Z is, , C, C, , Applications :, 1. For a simple pendulum, T, , l1/2, , T, T, , 1 l, 2 l, , 2. For a sphere, surface area and volume are given by, A, , 4 3, r, 3, , 4 r 2, V, , A, A, , 2., , r, V, and, r, V, , 3., , r, r, , 3. When two resistors R1 and R2 are connected, (a), , In series, Rs = R1 + R2, , Rs = R1 + R2, , (b), , ∆Rs, Rs, , R1, R1, , R2, R2, , Rp, , R1, , R2, , In parallel,, , 1, RP, Also,, , 1, R1, , Rp, Rp, , 1, R2, , R1, R1, , 4. If x = 2a – 3b then,, , R p2, , R2, R2, , R12, , R22, , R1, R2, R1 R2, , x=2 a+3 b, , LEAST COUNT OF MEASURING INSTRUMENTS, The smallest measurement that can be taken by an instrument is equal to least count of the instrument., For example, a meter scale has smallest division 1 mm. This represents the least count (and also the absolute, error) in the measurement., Let a length measured by the meter scale = 56.0 cm, This implies that x = 56.0 cm, Absolute error x = 1 mm = 0.1 cm, Relative error =, , x, x, , 0 .1, 56.0

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Rapid Revision & Formula Bank, , Physics and Measurement, , 5, , Vernier Callipers, It consists of two scales viz main scale and vernier scale. Vernier scale moves on the main scale. The least, count of the instrument is the smallest distance between two consecutive divisions and it is equal to, 1 MSD – 1 VSD., VS, , 0.9 cm, 1, , 0, In the figure shown,, , 1 cm, , 1 MSD = 0.1 cm, 1 VSD = 0.09 cm, , Least count = 1 MSD – 1 VSD = 0.01 cm, , Screw Gauge, It contains a main scale and a circular scale. The circular scale is divided into a number of divisions. In other, words, the complete rotation of circular scale is divided into a number of parts. The least count of a screw, gauge is pitch divided by no. of circular scale divisions., Least count of spherometer and Screw Gauge =, , Pitch, No. of CSD, , Total reading of screw gauge = Main scale reading + [(Circular scale reading) × Least count], Table : SI Units and Dimensions of Some Important Physical Quantities, S.No., , Quantity, , SI Unit, , Dimensional Formula, , 1., , Volume, , m3, , [M0L3T0], , 2., , Density, , kg m–3, , [M1L–3T0 ], , 3., , Velocity, , ms–1, , [M0L1T–1], , 4., , Acceleration, , ms–2, , [M0L1T–2], , 5., , Angular Velocity, , rad s–1, , [M0L0T–1], , 6., , Frequency, , s–1 or hertz (Hz), , [M0L0T–1], , 7., , Momentum, , kg ms–1, , [M1L1T–1], , 8., , Force, , kg ms–2 or newton (N), , [M1L1T–2], , 9., , Work, Energy, , kg m2s–2 or joule (J), , [M1L2T–2], , 10., , Power, , kg m2 s–3 or Js–1, or watt (W), , [M1L2T–3], , 11., , Pressure, Stress, , Nm–2 or pascal (Pa), , [M1L–1T–2], , 12., , Modulus of Elasticity, , Nm–2, , [M1L–1T–2], , 13., , Moment of Inertia, , kg m2, , [M1L2T0], , 14., , Torque, , Nm, , [M1L2T–2], , 15., , Angular Momentum, , kg m2 s–1 or J.s, , [M1L2T–1], , 16., , Impulse, , Ns, , [M1L1T–1], , 17., , Coefficient of Viscosity, , kg m–1 s–1, , [M1L–1T–1], , 18., , Surface Tension, , Nm–1, , [M1L0T–2]

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2, , Chapter, , Kinematics, DISTANCE TRAVERSED AND SPEED, Distance traversed (Path length), 1. The total length of actual path traversed by the body between initial and final positions is called distance., 2. It has no direction and is always positive., 3. Distance covered by particle never decreases., 4. Its SI unit is metre (m) and dimensional formula is [M0L1T0]., , EQUATIONS OF MOTION, General equations of motion :, v, , dx,  dx, dt, , vdt  dx, , a, , dv,  dv, dt, , adt  dv, , a, , vdv,  vdv, dx, , , ,  vdt, , , ,  adt = area enclosed by acceleration-time graph, , , , adx  vdv, , = area enclosed by velocity-time graph, ,  adx = area enclosed by acceleration-position graph, , Equations of motion of a particle moving with uniform acceleration in straight line :, 1. v = u + at, , 2., , S, , v u, 1 2, t, at = ,  2 , 2, , ut, , 3. v2 = u2 + 2aS, 4., , Sn th, , 5., , x, , u, , x0, , 1, a(2n 1), 2, ut, , 1 2, at, 2, , vt, , 1 2, at, 2

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8, , Kinematics, , Rapid Revision & Formula Bank, , Here,, u = velocity of particle at t = 0, S = Displacement of particle between 0 to t, = x – x0 (x0 = position of particle at t = 0, x = position of particle at time t), a = uniform acceleration, v = velocity of particle at time t, , Sn th = Displacement of the particle in nth second, , GRAPHS, The important properties of various graphs are given below :, 1. Slope of the tangent at a point on the position-time graph gives the instantaneous velocity at that point., , Position, , (x - t) curve, , (x), , P, tan, , dx, dt, , v (Instantaneous Velocity at point P), , Time (t), , Position (x), , 2. Slope of a chord joining two points on the Position-time graph gives the average velocity during the time, interval between those points., , (x - t) curve, xi, xf, , tan, , ti, , tf, , xf, tf, , xi, ti, , v av, , Time (t ), , 3. Slope of the tangent at a point on the velocity-time graph gives the instantaneous acceleration at that point., , Tangent, , v, , (v-t graph), , P, dv, dt, , tan, , a (Instantaneous acceleration at P), , Time (t), 4. Slope of the chord joining two points on the velocity-time graph gives the average acceleration during the, time interval between those points., , vf, v, , vi, , tan, , ti, , tf, , t, , vf, tf, , vi, ti, , aav, , Average acceleration in time interval tf – ti

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Rapid Revision & Formula Bank, , Kinematics, , 9, , 5. The area under the acceleration-time graph between ti and tf gives the change in velocity (vf – vi) between, the two instants., , a, Shaded area = vf – vi = change in velocity during interval ti to tf, , ti, , tf, , t, , 6. The area under speed-time graph between ti and tf gives distance covered by particle in the interval tf – ti., , v, speed, Shaded area = distance covered in time (tf – ti), , ti, , tf t (time), , 7. The area under the velocity-time graph between ti and tf gives the displacement (xf – xi) between the two, instants., , Velocity, , v, , A1, time, A2, , Shaded area (A1 – A2) = Displacement in time (tf – ti), Also, A1 + A2 = Distance covered in time (tf – ti), , 8. In velocity-position graph, the acceleration of particle of any position x0 is given as., , v, , a, , v 0 tan, , v0, , dv, dx, , (v - t graph), , v0, x x0, , O, , x, , x0, , 9. The position-time graph cannot be symmetric about the time-axis because at an instant a particle cannot, have two displacements., 10. The distance-time graph is always an increasing curve for a moving body., 11. The displacement-time graph does not show the trajectory of the particle., , Applications, 1. If a particle is moving with uniform acceleration on a straight line and have velocity vA at A and vB at B,, then velocity of particle midway on line AB is v, 2. If a body starts from rest with acceleration, time of journey is T, then, , v A2, , 2, , v B2, , ., , and then retards to rest with retardation , such that total

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10, , Kinematics, , Rapid Revision & Formula Bank, , , , , , (a) Maximum velocity during the trip vmax., , (b) Length of the journey L, , 1, , 2 , , x1, x2, , vmax, ,  2, T, , , v max ., (c) Average velocity of the trip =, 2, (d), , v, , , .T, , , x1, , T, 2(, , x2, , 1, , 2, , t1, , t2, , T, , t, , ), , t1, ., t2, , MOTION UNDER GRAVITY, If height of object is very small as compared to radius of earth, motion of object will be uniformly accelerated., Equation of motion can be applied with proper sign convention., Following are the important cases of interest., , t=0, O u=0, , 1. Object is released from a height h., Time taken to reach ground, , h, , 0, , 1 2, gT (taking up as positive), 2, , [email protected], , 2h, g, , T, , h, , Velocity of ball when it reaches ground, , v, , 0 gT, , g, , 2h ', g, , 2gh, , ‘–’ sign indicate that velocity will be in downward direction., 2. A particle is projected from ground with velocity u in vertically upward direction then, (a) Time of ascent = Time of descent =, , (b) Maximum height attained =, , Time of flight, 2, , T, 2, , u, g, , u2, 2g, , (c) Speed of particle when it hits the ground = u, (d) Graphs, , +, , +, Velocity, , a, , (g = const), O, , t, , u, , O, , –g, , –u, , –, , –, , u, g, , 2u, g, , Time

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Rapid Revision & Formula Bank, , Kinematics, , 11, , Displacement, , Speed, , +, u, O, , 2 u Time, g, , (Parabolic), , O, , u, g, , 2u Time, g, , Distance, , u, g, , 2, , u, 2g, , O, , u, g, , 2u Time, g, average velocity vav = 0, , (e) Displacement of particle in complete journey = zero, (f) Distance covered by particle in complete journey =, , u2, g, , u, 2, , Average speed in complete journey =, , 3. A body is thrown upward such that it takes t seconds to reach its highest point., (a), , Distance travelled in (t)th second = distance travelled in (t + 1)th second., , (b), , Distance travelled in (t – 1)th second = distance travelled in (t + 2)th second., , (c), , Distance travelled in (t – r)th second = distance travelled in (t + r + 1)th second., , 4. A body is projected upward from certain height h with initial speed u., (a), , Its speed when it acquires the same level is u., , (b), , Its speed at the ground level is, , v, (c), , x, , 2gh, , The time required to attain same level is, T=, , (d), , u2, , A t=, , t =0, –, , 2u, g, , Total time of flight (T') is obtained by solving, h, , uT ', , 1, gT '2 or T ', 2, , u, , u2, g, , u, h, , 0, , T, 2, , v = u2+2 gh, C t =T, , 2gh, , t =0, , 5. A body is projected from a certain height h with initial, speed u downward., (a), , Its speed at ground level is v, , (b), , Time of flight (T), T, , u, , u2, g, , u2, , Here x = 0, particle follows, same path during, ascent and descent, , u, t =T, B, , 2gh, , u, , g, , h, , t=T, , 2gh, , v

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12, , Kinematics, , Rapid Revision & Formula Bank, , Parallelogram Law of Vector Addition : If two vectors having common origin are represented both in, magnitude and direction as the two adjacent sides of a parallelogram, then the diagonal which originates from, the common origin represents the resultant of these two vectors. The results are listed below, (a), , R, , (b), , | R | ( A2, , (c), , tan, , (d), , A B, , B, B2, , 2 AB cos )1/ 2, , B sin, , tan, A B cos, , R, , A sin, B A cos, , A, , If | A | | B | x (say) , then R = x 2(1 cos ) = 2 x cos, , 2, , and, , 2, , i.e., resultant bisect angle, , between A and B ., (e), , If | A | | B | then, , (f), , Rmax = A + B, when, , (g), , R2, , (h), , If | A | | B | | R | , then, , (i), , If R is perpendicular to A , then cos, , (j), , For n coplanar vectors of same magnitude acting at a point such that angle between consecutive vectors, , A2, , B 2 , if, , <, = 0 and Rmin = |A – B| when, , = 180°., , = 90° i.e., A and B are perpendicular., = 120°., A, and A2 + R2 = B2., B, ,  360 ,  , the resultant is zero., are equal ,  n , , VECTOR SUBTRACTION, Subtraction of vector B from vector A is simply addition of vector B with A i.e., A B, Using parallelogram law,, , B, , –B, , O, ,  , Result : R = | A B |, , A2, , B2, , ( – ), , 2 AB cos , tan, , A–B, , A, , B sin, A B cos(, , Note : If | A | | B | x (say) , then R = x 2(1 cos ) = 2 x sin, , ), , 2, , ., , B sin, A B cos, , A ( B)

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Rapid Revision & Formula Bank, , 13, , Kinematics, , RESOLUTION OF VECTORS, Any vector V can be represented as a sum of two vectors P and Q which are in same plane as, V, , P, , Q , where, , are two real numbers. We say that V has been resolved in two component, , and, , P and Q along P and Q respectively., , vector, , Rectangular components in two dimensions :, , , V, , , Vx, ,  , V y,V, , Vx iˆ Vy ˆj , V, , Vx2, , Vy2, , Y, V, , V x and V y are rectangular component of vector in 2-dimension., , Vy, , Vx = V cos, Vy = V sin, , = V cos(90 – ), , Vz = zero., V, , O, , V cos iˆ V sin ˆj, , Vx, , , Note : Unit vector along V is cos iˆ sin jˆ, , SCALAR AND VECTOR PRODUCTS, , , Scalar (dot) Product of Two Vectors : The scalar product of two vectors A and B is defined as,  , A . B AB cos, B, ,  , A.B, AB, , cos, , , ,  , If A and B are perpendicular, then A . B, , If, ,  , < 90°, then A . B, , 0 and if, , 0, ,  , > 90° then A . B, , 0., , ,   B, , , Projection of vector A on B is ( A. B ) 2 ., B, A2, , A.A, , iˆ . iˆ, , jˆ . ˆj, , kˆ . kˆ, , 1., , Scalar product is commutative i.e., A . B, , B.A ., ,   , Scalar product is distributive i.e., A (B C ), ,  , A B, ,  , A C, , Vector Product of two Vectors :, Mathematically, if, , A B, , AB sin nˆ, , is the angle between vectors A and B , then, …(i), , A, , X

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14, , Kinematics, , Rapid Revision & Formula Bank, , A ×B, , A ×B, , A, B A, , (a), , B, , (b), , The direction of vector A B is the same as that of unit vector n̂ . It is decided by any of the following two rules :, (a), , Right handed screw rule : Rotate a right handed screw from vector A to B through the smaller angle, between them; then the direction of motion of screw gives the direction of vector A B (Fig. a)., , (b), , Right hand thumb rule : Bend the finger of the right hand in such a way that they point in the direction, of rotation from vector A to B through the smaller angle between them; then the thumb points in the, direction of vector A B (Fig. b)., , RELATIVE MOTION IN TWO DIMENSIONS, Relative velocity :, , Velocity of object A w.r.t. object B is v AB, , 1., , , vA, ,  , v B , v BA, , , vB, , , vA, , Direction of Umbrella : A person moving on straight road has to hold his umbrella opposite to direction, of relative velocity of rain. The angle, , vM, with vertical in forward direction., vR, , is given by tan, , Umbrella, vRM, vR, , vR, vM, , 2., , , , –vM, , Closest approach : Two objects A and B having velocities vA and v B at separation x are shown in figure, , vA, , vB, , A, , B, , x

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Rapid Revision & Formula Bank, , The relative velocity of A with respect to B is given by, v AB, , vA, , tan, , vA, vB, , -vB, , vB, , vAB, , vA, , -vB, , The above situation is similar to figure given below., , (D, ire, c, of tio, A no, w. f m, r.t o, . B ti, ) on, , y is the distance of closest approach., y, x, , Now, sin, , vAB, , y = x sin, y, , 15, , Kinematics, , x tan, , xVA, , 1 tan 2, , VB2 VA2, , y, , A, , v=0, B, , x, , 3. Crossing a river :, , y, , v = velocity of the man in still water., = angle at which man swims w.r.t. normal to bank such that, , D, , B, , vx = – v sin , vy = v cos, , v, , Time taken to cross the river is given by, d, d, v y v cos, Velocity along the river, t, , vx, , v sin, , v cos, A, , u v sin, , Distance drifted along the river D, , t vx, , d, (u v sin ), v cos, , D, , Case I : (Shortest time), The Minimum time to cross the river is given by, d, (when cos, v, Distance drifted is given by, min, , = 1,, , = 0°, u, , v), , d, u, v, Case II : (Shortest path), D, , To cross the river straight, drift D = 0, , u – v sin, , =0, , u, sin, provided v > u, v, Time to cross the river straight across is given by, , t, , d, v cos, , d, v, , 2, , u2, , v, , v2 – u2, , u, , u, , d, , x

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16, , Kinematics, , Rapid Revision & Formula Bank, , PROJECTILE MOTION, An object moving in space under the influence of gravity is called projectile. Two important cases of interest, are discussed below :, 1. Horizontal projection :, A body of mass m is projected horizontally with a speed u from a height h at the moment t = 0. The path, followed by it is a parabola., , It hits the ground at the moment t = T, with a velocity v such that, , T, , 2H, g, , , v, , u2, , t=0, 2gH, , uiˆ gTˆj, , y-axis, , y, , t = t0, , H, , The position at any instant t0 is given by, , x, , x = ut0, y, , y, , 1 2, gt 0, 2, gx 2, 2u 2, , x-axis, , O, , R, , (trajectory of particle), , t=T, v, , The velocity at any instant t0 is given by, v0, , uiˆ gt 0 ˆj, , The range R will be given by R, , 2H, g, , u, , 2. Oblique projection : A body of mass m is projected from ground with speed u at an angle, horizontal at the moment t = 0., It hits the ground at a horizontal distance R at the moment t = T., , v, , uy, , T, t= 2, , u, , H, , t=0, ux, , 1. Time of flight, , T, , 2. Maximum height, , H, , 3. Horizontal range, , R, , 4. Equation of trajectory: y, or y, , 2u y, g, , u y2, 2g, ux T, x tan, , R, 2u sin, g, , u 2 sin2, 2g, 2u x u y, g, gx 2, 2u cos 2, , , x tan 1, , , 2, , x, , R, , u 2 sin 2, g, , u cos, , = ux, t=T, , above

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Rapid Revision & Formula Bank, , Kinematics, , u2, , 5. Instantaneous velocity v, , (gt )2, , 17, , 2u(gt ) sin, , and direction of motion is such that, tan, , v sin, , u sin – gt, u cos, , v, v cos, , u cos, cos, , (a), , v, , (b), , v sin = u sin – gt, , [∵ Horizontal component is same everywhere], , , (c) When v (velocity at any instant ‘t’) is perpendicular to u (initial velocity), = – (90° – ), (i), , v, , u cos, cos(90, , (ii), , t, , u, g sin, , u, ), , u cot, , v, , Applications :, 1. The height attained by the particle is largest when, and range is minimum (zero)., , = 90°. In this situation, time of flight is maximum, , 2. When R is range, T is time of flight and H is maximum height, then, (a) tan, , gT 2, 2R, , (b) tan, , 4H, R, , 3. When horizontal range is maximum, H, , Rmax, 4, , 4. The horizontal range is same for complimentary angles like ( , 90° – ) or (45° + , 45° – ). It is maximum, for = 45°., 5. If A and B are two points at same level such that the object passes A at t = t1 and B at T = t2, then, y, , u, , t = t1, h, , t=0, , T, , 2u sin, g, , (ii) h, , 1, gt1t 2, 2, , (i), , t1, , t = t2, , A, , t2, , (iii) Average velocity in the interval AB is, vav = u cos [∵ vertical displacement is zero], , B, , h, , t=T, x

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18, , Kinematics, , Rapid Revision & Formula Bank, , 6. If a projectile is projected from one vertex of a triangle such that it grazes second vertex and finally fall, down on 3rd vertex of the triangle on the same horizontal level, then tan, , tan, , tan, , ., , u, , 7. A projectile has same range for two angle of projection. If time of flight in two cases are T1 and T2,, maximum height is H1 and H2 and the horizontal range is R. Then, , 1, gT1T2, 2, , (i) Range of projectile is R, , (ii) Velocity of projection of projectile is u, , 1/2, 1  2, g T1 T22 , 2 , , (iii) R = 4 H1H2, , CIRCULAR MOTION, An object of mass m is moving on a circular track of radius r. At t = 0, it was at A. At any moment of time, AOB, , ‘t’, it has moved to B, such that, , . Let its speed at this instant be v and direction is along the, , tangent. In a small time dt, it moves to B such that, , The angular velocity vector is, , d ., , d kˆ, , The angular displacement vector is d, , , B OB, , y, v+dv, B, , d ˆ, k., dt, , B, d, , At B , the speed of the object has become v + dv., dv, dt, , The tangential acceleration is at, , r, , The radial (centripetal acceleration) is ac, d, dt, , The angular acceleration is, , Relations among various quantities., , , r, , 1., , , v, , , , 2., , , a, , , dv, dt, , 3., , ac, , 4., , at, , , , , , , v, , , , , r, , , dr, dt, , , d, dt, , , r, , ac, , at, , O, , v2, r, , 2, , r, , r, A, , x

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Rapid Revision & Formula Bank, , Kinematics, , 19, , Uniform Circular Motion :, 1. In uniform circular motion, the speed (v) of particle is remain constant ( = constant), 2., , aT, , d, dt, , 0 and, , 0, , y, v, a, , v, , x, , a, , a, , v, , v, 3. Only centripetal acceleration (also called normal acceleration) exists in uniform circular motion, ac = r, , 2, , v2, r, , a, , , 4. In uniform circular motion v, , Nonuniform Circular Motion :, 1. In nonuniform circular motion the speed (v) and angular velocity ( ) change w.r.t. time., 2. Net acceleration of particle in non-uniform circular motion., , y, at, a, , a, , ac 2, , at 2, , v2 ,  r , , , , 2, , r, , ac, , 2, , ‰ ‰ ‰, , x

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Chapter, , 3, , Laws of Motion, Equilibrium of Concurrent Forces, If three forces P , Q and R are acting on an object such that forces are concurrent and the object is in, equilibrium then, , P, sin, , Q, sin, , R, ., sin, P, Q, , R, , APPLICATIONS OF NEWTON’S LAWS OF MOTION, The strings connected to pulley are considered as ideal. Their length is fixed, so the ends of string follow a, fixed relation between displacement velocity and acceleration. These relations are called constraint relation., Case I :, , When the middle end is fixed, , x1 = x2, v1 = v2, a1 = a2, x2, , x1, , Case II : When the side end is fixed, , x1, , x2, OR, x2, , x1, , x2 = 2x1, v2 = 2v1, a2 = 2a1

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Rapid Revision & Formula Bank, , Laws of Motion, , 21, , Case III : When all the three ends are free to move, , x1, , x1, , v1, x3, , x2, , a1, , x2, , x3, 2, , v2, , 2, , a2, , v3, a3, , 2, , Note : In all the above relations downward direction is taken as positive. If any of the direction is, upward in any case then –ve sign must be incorporated in the corresponding equation., 1., , A machine gun fires n bullet per second with speed u and mass of each bullet is m., , m,u, M, The Force required to keep the gun stationary is, , F, , , nmv, , 2. Bullets moving with a speed v hit a wall normally., (i), , If the bullets come to rest in wall, , m, , Force on wall Fwall = nmv, (Here n is number of bullets hitting the wall in one second), (ii), , v, , If the bullets rebound elastically,, Fwall = 2nmv, , 3. Liquid jet of area A moving with speed v hits a wall, , v, , A, (i), , Force required by a pump to move the liquid with this speed is, F, , (ii), , v, , dm, = v × Av = Av2, dt, , As jet hits a vertical wall and does not rebound, the force exerted by it on the wall is, Fwall = Av2, , (iii) When water rebounds elastically, Fwall = 2 Av2, (iv), , For oblique impact as shown, Fwall = 2 Av2 cos, , Liquid jet, v, , v, , wall

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24, , Laws of Motion, , aB, (iv), , Rapid Revision & Formula Bank, , T, , B, , T, T, , 2T, aA, , A, MAg, 2aA = aB, , …(i), , MAg – 2T = MAaA, , …(ii), , T = MBaB, , …(iii), , Two block system :, Case - I :, Let ‘m’ does not slide down relative to wedge ‘M’, , a, , The force required is given by, , m, , F = (M + m)g tan, a = g tan, , M, , (in horizontal direction w.r.t. ground), , Wedge, , Contact force R between m and M is, R, , F, , smooth, , mg, cos, , Case - II :, , m, , Minimum value of F so that ‘m’ falls freely is given by, F = Mgcot, , F, M, , Wedge M moves with acceleration = gcot, , Wedge, , The block falls vertically with acceleration ‘g’., , smooth, , Contact force between M and m is zero., , Angle of Repose, Consider a situation in which a block is placed on an inclined plane with co-efficient of friction ‘ ’. The, maximum value of angle of inclined plane for which the block can remain at rest is defined as angle of repose., , N, N=, , s, , mg sin, , s, , mg cos, , mg cos, , Two blocks placed one above the other on smooth ground, , tan 1(, , s, , )

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26, , Laws of Motion, , Rapid Revision & Formula Bank, , DYNAMICS OF CIRCULAR MOTION, Neglecting Gravity, , axis, , O, , T, , r, , v, , m, , T = Centripetal force =, , mv 2, r, , m, , 2, , r, , Considering gravity (Conical pendulum), , O, T cos, , l, , h, , T, T sin, , C, , r, mg, , mg, Tsin = m, , 2r, , …(1), , Tcos = mg, T, , (a) For, , mg, cos, , …(2), , to be 90º (i.e., string to be horizontal), T=, It is not possible., , (b) Tsin = m, , T, , 2r, , m, , =m, 2, , 2lsin, , l, , (c) Time period = 2, , h, g, , (h = l cos ), , Vehicle negotiating a curve on a banked road, The maximum velocity with which a vehicle can safely negotiate a curve of radius r on a rough inclined road is, , Vmax, , rg (, 1, , tan ), tan

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Rapid Revision & Formula Bank, , Laws of Motion, , N cos, , N, , N sin, f cos, f, , mg, f sin, , mg, Special Cases :, For a smooth inclined surface, For a horizontal rough surface,, , =0, =0, , v max, , rg tan, , v max, , rg, , ‰ ‰ ‰, , 27

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30, , Work, Energy and Power, , 4., , Rapid Revision & Formula Bank, , Oblique elastic collision, A body of mass m collides with a stationary body of same mass., , v1, (a), , +, , (b), , v1cos, , (c), , v1sin, , (d), , u2, , =, , =, , …(1), , 2, , = u …(2), , + v2cos, = v2sin, , v12, , +, , u, m, , …(4), , After, collision, , 90°, , Before, collision, , …(3), , v22, , u=0, m, , v2, , MOTION IN A VERTICAL CIRCLE, A particle of mass m is tied to a string of length l whose other end is fixed. The particle can revolve about O, in a vertical circle. when it is at position L (lowest point), it is given a speed VL horizontal. Following results, are useful in describing its motion., 1., , aT = g sin, , 2., , aC, , 3., , 4., 5., 6., , Tp, , TL, vL, , 2gl, , Tp, , mg cos, m, , mv p2, l, , mg, , vH, , …(1), , H, TH, mg, , v p2, l, , M, mg cos, , …(2), , O, l, , mv L2, , TL, , L, , l, , vL, , mg, , 2gl , it oscillates between M and M, , vL, , 5gl , it will leave the circular path somewhere between M and H., , 7., , When v L, , 5gl , it completes vertical circle (Also vH, , 8., , TH, , mv H2, l, , 9., , TL – TH = 6 mg (always), , mg, , ‰ ‰ ‰, , gl ), , M, vP, P, mg cos, mg sin, , TP

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5, , Chapter, , Rotational Motion, CENTRE OF MASS OF A RIGID BODY (CONTINUOUS MASS DISTRIBUTION), Mathematically position coordinates of the centre of mass of rigid body are given by, , x cm, ,  xdm, ;,  dm, ,  ydm, ;,  dm, , y cm, ,  zdm,  dm, , zcm, , Centre of mass of some commonly used objects., 1., , Semi circular wire of radius R., 2R, , OC, , Semicircular, wire, , C, , , where C is centre of mass, O, , 2., , Semi circular disc of radius R, 4R, 3, , OC, , 3., , Non-uniform rod of length L. The linear mass density, varies linearly from zero at O to maximum at B., , O, , 2L/3, , 2L, 3, , OC, , O, , VELOCITY AND ACCELERATION OF CENTRE OF MASS, Velocity of Centre of Mass, The instantaneous velocity of centre of mass is given by, , v cm, , , m1v 1, , , m2 v 2, , Where P, , , ....mn v n, , n, , m, i 1, , system, , Semicircular, disc, , C, , ;, , or, , , v cm, , , P system, M system, , i, , is the total linear momentum of the system of particles., , L, , C, , B

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32, , Rotational Motion, , Rapid Revision & Formula Bank, , Acceleration of Centre of Mass, Differentiating v, , m1 a 1, , , a cm, , cm, , w.r.t. time we get a cm as, , , m2 a 2, n, , , ....mn a n, , ;, , or, ,  mi, , , a cm, , , F ext, Msystem, , i 1, , Where, , Fext is the vector sum of forces acting on the particles of system., , MOMENT OF INERTIA OF DIFFERENT OBJECTS, For an axis perpendicular to the plane of the ring, , A hollow cylinder, 2, , Ic = MR, , 2, , I = MR, , R, , M, , R, , The axis perpendicular to the plane of the disc., , I cm, , A solid cylinder, , MR 2, 2, , A thin rod, , MR 2, 2, , I cm, A plate, , M, , M, L, L, , L, , Ic, , ML2, 12, , Ic, , ML2, 12

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Rotational Motion, , Rapid Revision & Formula Bank, , A thin rod about a perpendicular, , A plate about one edge, , axis through its end, , M, L, , L, I, , ML2, 3, , ML2, 3, , I, , z, , A Rectangular Plate, (a) I xx, , Mb 2, 12, , (b) I yy, , Ml 2, 12, , O, b, x, , M (l 2 b 2 ), 12, , The axis is perpendicular to the rod and passing through the centre of mass, , A, , B, , ML2, 12, , MR 2, ; IBB, 4, , ML2, 3, , R, , MR 2, 4, , About its diameter, , , , M, , L, , O, B, , A Solid Sphere, , A, , A Hollow Sphere, About its diameter, , M, R, , Icm, , 2, MR 2, 5, , y, z, , A Thick Rod (Solid cylinder), , I AA, , x, , l, , (c) Izz = Ixx + Iyy, (d) I zz, , Mass = M, , y, , I cm, , 2, MR 2, 3, , 33

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34, , Rotational Motion, , Rapid Revision & Formula Bank, , RIGID BODY ROTATION, In this section, rotation of a body about a stationary fixed axis has been discussed, 1., , Rotating Disc, A tangential force F is applied at the periphery, as a result disc is rotating above an axis passing through, its CM, normal to plane of disc, A, = F × R [about O], F, R, 1, R, I, mR 2, B, 2, r O, C, 2F, I, , MR, , (1) Tangential acceleration of A is a A, , R, , 2F, (along horizontal), M, , (2) Tangential acceleration of B is aB, , R, , 2F, (vertically downwards), M, , (3) Tangential acceleration of C is aC, , r, , 2Fr, (along horizontal, opposite to the direction of tangential, MR, , acceleration of A), 2., , Hinged Rod, The rod is released from rest from horizontal position, mg, , I, , N, , L, (about A), 2, , A, , ML2, 3, , I, , C, L, 2, , 3g, 2L, L, 2, , (1) Linear acceleration of COM C is acm, (2) Linear acceleration of point B is aB, , L, , 3g, mg, . Also, N = mg – macm =, 4, 4, 3g, 2, , (3) The rod is released from unstable equilibrium position {from position A}, (a) When at B, Mg, , L, (1, 2, , cos ), , = 90°,, , A, , u=0, , P, , O, , = 0°, , L, , 6g, l, , (c) at P,, , 2, , L, , 6g, cos, L, 2, , (b) at C,, , 1  ML2 , 2  3 , , 3g, l, , B, , C, , L, 2, B, , mg

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36, , Rotational Motion, , Rapid Revision & Formula Bank, , Kinetic energy of the body during pure rolling (E), E = Translational KE + Rotational KE, = ET + ER, , 1, 2, mv cm, 2, , =, , 1, = 2 m, , E, , 1, I cm, 2, , 2, ,  K2 , 1, 2 , , mv cm, 1,  R 2  (where k is radius of gyration), 2, , , , I cm  2, v cm, R2 , ,  K2 , , ET 1,  R2 , , , ,  R2 , , E R 1,  K2 , , , , Similarly, E, , Fraction of total energy, , Fraction of total energy, 1, , 1, , Y, X, , Type of body, , K, , 1. Ring or hollow, , R, , 1, 2, , 0.5, , 50%, , 1, 2, , 0.5, , 50%, , 1: 1, , 2, R, 3, R, , 3, 5, , 0.6, , 60%, , 2, 5, , 0.4, , 40%, , 2:3, , cylinder, 2. Spherical Shell, 3. Disc or solid, cylinder, 4. Solid sphere, , translational X, , , K ,  1, , 2 , R, , , 2, , rotational Y, , , R2 ,  1, , K 2 , , , 2, , 2, 3, , 0.666, , 66.67%, , 1, 3, , 0.333, , 33.33%, , 1: 2, , 2, R, 5, , 5, 7, , 0.714, , 71.4%, , 2, 7, , 0.286, , 28.6%, , 2:5, , Note : Above values X and Y are independent of mass and radius of the body. They only depend on the, type of body., , Applications, 1. A force is applied at the distance h from centre of mass as shown in figure, , h, , F1, , , R, , K2 , M1, R 2 , , , aC.M, , 2, , fr, , F (K, K2, , If, , h, , hR ), R2, , h, fr, , N (must be less than mg for Rolling), , K2, friction is backward, R, , K2, friction become forward, R, 2. If force is applied at centre of mass then (h = 0), h, , So, a, , F,  K2 , , M 1, 2 ,  R , , and fr, , FK 2, K 2 R2, , F, , R2 , 1, , K 2 , , Rough ground, , F

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Rotational Motion, , Rapid Revision & Formula Bank, , 37, , 3. If force is applied at highest point (h = R), , aC.M., , fr, , 2F, , K2 , M 1, , R2 , , ,  R2 K 2 ,  K 2 R2  F , forward direction, , 4. (i), , (ii), , C, , R, , F, R, , fr, , F, , Rough horizontal, surface, , Rough inclined plane, , in, gs, m, =, , a, R, , R, (iii), , fr, , T, m, , a, R, , (iv), , T, , T, R, a, , a, , F = mg, , F=mg, , For all the four situations shown above,, a=, , F, I, m c.m.2, R, , fr or T =, , Ic.m.a, R2, , In the situations described above, the linear acceleration of the moving object can be calculated by same, formula, the value of F, and moment of inertia will depend on the kind of problem., Also consider the following situation., , F, , a=, , (m1, , T2 – T1 =, , m2 ), , I, R2, , R, T1, , T2, , Ia, R2, , Here, F = (m2 – m1) g = Net pulling force, , m1, m2

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38, , Rotational Motion, , Rapid Revision & Formula Bank, , Rolling of a Body on an Inclined Plane, mgh, , By conservation of energy,, , (Total energy), , 1, , Let, , 1., , Icm, MR, , = 1, , 2, , K2, , 2gh, , 2., , v cm, , 3., , Time, , 1, ., sin, , 2h, g, , (Rotatory), , 1, 2, mv cm, 2, , (Translatory), , R, g sin, , fr, mg, , 2gh 1, , =, , 2, , (where k is radius of gyration), , R2, , mg sin, g sin, =, Icm, K2, m, 1, 2, R, R2, , acm, , 1, I, 2, , sin, , (, gh, rou, , s, co, g, m, , vcm, , K2, R2, , ∵ Force of friction fr, , mg sin, R2, 1, K2, , i.e., t, 4. Force of friction fr, , mg sin, R2, 1, K2, , 5. Instantaneous power P = (mg sin )v, 6. Maximum angle of inclination for pure rolling,, Ring, , :, , max, , = tan–1 (2 ),, , Spherical Shell, , :, , max, , = tan–1(2.5 ), , Disc, , :, , max, , = tan–1 (3 ), , Solid sphere, , :, , max, , = tan–1(3.5 )., , max, , , tan 1 , , , ,  R2  ,  1, 2 ,  ,  K , , ANGULAR MOMENTUM, The general formula for angular momentum about any point is, , , , L Mv cmR I, Case - I : Pure translation, , | LO | Mv cm h, LC = 0, , z, M, C, h, O, , no, , ), ing, p, p, i, sl, , vcm, , x, , h

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6, , Chapter, , Gravitation, VARIATION IN THE VALUE OF g, 1., , At height h (Above earth’s surface), g, , g, , 1, , , If h = Re , g, , h , , Re , , 2, , g, 4, , If h < < Re then g, 2., , 2h , Re , , At depth “x” (below earth’s surface), g, , 3., , , g1, , , , g 1, , , x , , Re  , at the centre of earth g = 0, weight = 0, , Due to Rotation of Earth :, Apparent value of acceleration due to gravity., g = g – Re, , 2cos2, , angle of latitude, , GRAVITATIONAL FIELD INTENSITY AND POTENTIAL (V), 1., , Gravitational field intensity, I, , 2., , GM, , r2, , rˆ, , Gravitational potential, V, , W, V, m, , GM, (units J/kg), r

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Gravitation, , Rapid Revision & Formula Bank, , Variation of Intensity and Potential, 1., , For a spherical shell of mass M and radius R, Case-I :, , r < R (internal point), GM, R, , Ii = 0, Vi =, Case-II :, , r = R (on the surface), Is =, , Case-III :, , GM, R, , GM, R, , , Vs, , 2, , r > R (outside the shell), Io = GM , Vo, r2, , GM, r, , V, , |I|, O, , O, , r, , r=R, , r=R, , r, , GM, R, 2., , For Uniform solid Sphere, Case-I :, , r < R (internal point), Ii =, , Case-II :, , R, , GM, , , Vi, , 3, , 2R, , 3, , (3R 2, , r 2 ) At centre Vc, , 3GM, 2R, , 3, VSurface, 2, , r = R (on the surface), , GM, , Vs, R2, , Is, Case-III :, , GMr, , GM, R, , r > R (outside the surface), Io, , GM, , r, , 2, , , Vo, , GM, r, , |I|, |Imax|, , V, I r, , I, , 1, r2, , O, r, , O, r=R, , 3, Vs, 2, , VS, Vc, , r=R, , r, hyperbolic, parabolic, , `, , 41

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42, , Rapid Revision & Formula Bank, , Gravitation, , 3., , Gravitational intensity and potential on the axis of uniform ring of mass M radius R at distance x from, centre., , GMx, , I, , (R, , 3, x2 ) 2, , 2, , ,V, , GM, R, , 2, , x2, R, , At centre I = 0. I is maximum at x, , 2GM, , ; Imax =, , 2, , 3 3R 2, , R, , R, , x, , –, , 2, 4., , 2, , V, , |I|, , x, , GM, R, , Neutral point : The point P at which gravitational field is zero between two massive bodies, is called, neutral point., , M1, r12, , M1, , M2, r22, , P, r1, , M2, , r2, , GRAVITATIONAL POTENTIAL ENERGY, At earth surface U, , GM e m, , At height h, U, Re, , Energy required to escape = Escape energy = +, , GM e m, Re h, , GM e m, = Binding energy., Re, , ESCAPE VELOCITY, ve, , 2GM e, Re, , 8, 2, GRe, 3, , 2gRe, , At earth surface, ve = 11.2 km/s, , KEPLER’S LAWS, (1), , All planets revolve around the Sun in elliptical orbit having the Sun at one focus., If e = eccentricity of ellipse then distance of the planet from the Sun at perigee is, rp = (1 – e)a, and distance of the planet from the Sun at apogee is, ra = (1 + e)a, , (a = semi major axis), , Ratio of orbital speeds at apogee and perigee is, , va, vp, , rp, ra, , 1 e, 1 e

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43, , Gravitation, , Rapid Revision & Formula Bank, , Ratio of angular velocities at apogee and perigee is,  rp ,  ,  ra , , a, p, , 2, ,  1 e, , ,  1 e, , 2, , vp, Apogee, , rp, , ea, , ra, a, , Perigee, , Sun, , va, , 2a, (2), , A planet sweeps out equal area in equal time interval i.e., Areal speed of the planet is constant, 1, vr, 2, , dA, dt, , (3), , L, = constant (L represents angular momentum of planet about the Sun), 2m, , Square of time period is proportional to cube of semi-major axis of the elliptical orbit of the planet., i.e., T, , 2, , a3, , SATELLITES, Important results regarding satellite motion in circular orbit., 1., , Orbital Velocity (v0) : Gravitational attraction of planet gives necessary centripetal force., GM, r, , v0, , v0, , ve, 2., , gRe2, Re h, , GM, r, , (h = height above the surface of earth), , 2v 0, , Time Period : The period of revolution of a satellite is, , T, , 2 r, v0, , 2 r, , r, GM, , 2, , r3, GM, , For a satellite orbiting close to the earth’s surface (r ~ Re), the time period is minimum and is given by, , Tmin, , 2, , Re3, GM, , 2, , Re, g, , For earth Re = 6400 km,, g = 9.8 m/s2, Tmin = 84.6 min.= 1.4 h, Thus, for any satellite orbiting around the earth, its time period must be more than 2, 84.6 minutes., , R, or, g

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Chapter, , 7, , Properties of Solids and Liquids, , Stress, , Strain, (1), , Longitudinal strain =, , l, , ‘A’ (area), , (1), , l, , l, , F, l, , Normal Stress (Tensile) = F/A, , (2), , V, V, , Volumetric strain =, , (2), r, , 2, , A=4 r, , Normal Stress (Compressive) = P (pressure), , (3), , Shear strain =, , A(area), , L, L, , (3), , L, , F, , L, , F, Tangential Stress or Shear Stress = A, , MODULI OF ELASTICITY, (1), , Young’s modulus of elasticity Y, , (2), , Bulk modulus of elasticity, Compressibility =, , (3), , Tensile stress, Longitudinal strain, , Fl, A l, , Normal or compressive stress, Volumetric strain, , V, , P or,, V, , 1, , Modulus of rigidity or shear modulus, , or G, , Shear stress, Shear strain, , F, A, , FL, A L, , V, , dP, dV

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46, , Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , Applications :, 1., , For a wire Y, , Fl, A l, , YA, l, l, , F, , i.e. a wire behaves like a spring of spring constant (k), , 2., , ,  i .e., k, , , YA, l, , k, , When this wire is stretched by applying an external force F, and l is extension produced, then, (a), , Work done by external force = F l, , (b), , Work done by restoring force =, , (c), , Heat produced =, , (d), , Elastic potential energy stored =, , 1, F l, 2, , 1, F l, 2, , Energy density U, , 3., , 1, , l, , 1 F l, 2 volume, , 1, F l, 2, 1F l, 2 Al, , =, , 1, stress × strain, 2, , =, , 1 (stress)2, 2, Y, , 1, Y (strain)2, 2, , A rod of mass m and length l hangs from a support, Area of cross-section = A, Extension produced due to its own weight,, , l, , l, , Mgl, gl 2, =, ( = density of wire), 2 AY, 2Y, , Thermal Stress : Rod Fixed between Rigid Support, , l, , F, , If, , = Rise in temperature, , Compressive strain =, , Y, A, , F, , Heated, , l, l, , Compressive stress = Y × strain = Y, F=Y, , ×A, , Note : If the rod is placed on horizontal frictionless surface, then stress developed on heating is zero.

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Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , Poisson’s Ratio, Consider a uniform bar being stretched by applying two forces at its ends., , R, , F, , R– R, , F, , l, , Longitudinal strain =, , l+ l, , l, l, , R, R, , Lateral strain =, , R /R, l /l, , Poisson’s ratio,, (a), , Theoretically – 1, , 0.5, , (b), , Practically 0, , (c), , When density of material is constant, , 0.5, = 0.5, , Equilibrium of Different Liquids in a U tube, 1. PA = PB, , (as A & B are at same level in same liquid), , P0 + h1 1g = P0 + h2 2g, , (where P0 is atmospheric pressure), , h1 1g = h2 2g, 1h1, , =, , 2h2, , P0, , P0, h2, , h1, , A, , 1, , B, , 2, , 2. When the U tube accelerates horizontally, difference of levels of liquid satisfies the relation,, , tan, , a, g, , h, L, , L, , h, , a, , 47

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48, , Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , 3. When U-tube is rotated about on limb, , h, , L, Here h, , 2 2, , L, and tan, 2g, , h,  tan, L, , Excess pressure, If, , Po = Atmospheric pressure, Pi = Inside pressure, , then Pi – Po = Excess pressure, , Liquid drop, PO, (1), , r, , Pi, , Pi = Po +, , 2T, r, , Soap bubble, , (2), , PO, , Pi, , Pi – Po =, , 4T, r, , Air bubble, , (3), , r, , Po, , Pi, , Pi = Po +, , 2T, r, , 2, , L, ., 2g

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Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , (4), , 49, , Capillary tube, concave meniscus, , O, , R, r, , PO, Pi, , Capillary tube,, Concave Meniscus, , Fc, , (b) Fa, (5), , 2T, R, , Po, , (a) Pi, , 2, , Capillary tube, convex meniscus., PO, r, , Pi, , R, , o, , Convex Meniscus, , (a) Pi, , Po, , 2T, R, , Fc, , (b) Fa, , 2, , Combining of Bubbles, 1., , If the soap bubble coalesce in vacuum, then Po = 0, r2 = r12 + r22, , 2., , If two soap bubbles come in contact to form a double bubble then, r = radius of interface, r1 > r2, 1, r, , 1, r2, , 1, r1, , The interface will be convex towards larger bubble and concave towards smaller bubble because, P2 > P1 > P0., r2, P2, P0, , r1, P1, Radius ‘r’

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50, , Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , CAPILLARY ACTION, Rise or fall of liquid in a tube of fine diameter., , O, , Ascent formula, , h, , 2T, R g, , R, , r, , 2T cos, r g, , Energy of a Liquid, Various energies per unit mass :, 1., , Potential energy/mass = gh, , 2., , Kinetic energy/mass =, , 3., , Pressure energy/mass =, , 1 2, v, 2, , P, , Energy Heads, Various energy heads per unit mass :, 1., , Gravitational head = h, , 2., , Velocity head =, , 3., , Pressure head =, , v2, 2g, , P, g, , BERNOULLI’S THEOREM, It is based on conservation of energy., For an ideal, non-viscous and incompressible liquid,, P1, , v 12, 2, , P2, , gh1, , v 22, 2, , gh2 = constant, , Applications of Bernoulli’s Theorem, (1) To find rate of flow of liquid Q = av. Value of Q in various cases has been given below, Case - (a) :, Q, , a1a2, , P1, a1, v1, , 2(P1, (a12, , P2 ), , a22 ), , P2, a2, v2

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52, , Properties of Solids and Liquids, (e), , Rapid Revision & Formula Bank, , Range of liquid, R, , 2 h( H, , h), , For a given value of total height (H), H, 2, , Rmax = H when h, , h, H, , R, , h, (f), , H, h, R = 2 h(H - h) for both holes, , (3), , If, , A0 = area of cross-section of mouth of tap, A = area of cross-section of water jet at a depth h, A0v0 = Av = Q [rate of flow], , By Bernoulli’s theorem, , v 02, 2, , gh, , v 12, 2, , [∵ pressure is atmospheric at both points], , Reynold’s Number, NR, , vD, , Inertial Force, Viscous force, , Value of NR for various cases :, (1), , NR < 2000, flow is streamline, , (2), , NR > 3000, flow is turbulent, , (3), , 2000 < NR < 3000, flow is unstable, , (4), , When NR = 2000, flow is critical, , vD, , 2000, , vc, , 2000, , D, , (Critical velocity), , A0, h, , A

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Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , 53, , Viscosity and Viscous Force, , F, , Plate, , v, , y, , Stationary plate, , Fluid at rest, , Viscous force is given in this case by,, , F, , A, , dv, dy, 1 Pa.s = 10 poise = 1 decapoise, , : SI, , Units of, , 1 dyne/cm2-s = 1 poise, , C.G.S, , Poiseuille’s Equation, Volume flow rate across a tube with pressure difference between its ends is,, Pr 4, 8 l, , dV, dt, , Q, , Series combination of two tubes, Two tubes of radius r1, length l1 and radius r2, length l2 are connected in series across a pressure difference, of P. Length of a single tube that can replace the two tubes is found using,, l1, , l, r4, , l2, , r14, , r24, , STOKES LAW, When a small spherical body of radius r is moving with velocity v through a perfectly homogeneous medium, having coefficient of viscosity , it experiences a retarding force given by, rv., , F=6, , Important case :, (1), , A body of radius r released from rest in a fluid, = density of body, , If, , = density of liquid or fluid, Terminal velocity is given by,, vT, , 2 r 2g, (, 9, , ), , Thermal Expansion, When the temperature of a body increases, its all dimensions (length, area, volume) increase., (1) Coefficient of linear expansion is given by, , L, L, L = L0 (1 +, , ),, , is the change in temperature in °C or K & L0 is initial length

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54, , Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , (2) Coefficient of superficial expansion is given by, , A, A, A = A0 (1 +, , ), , (3) Coefficient of cubical expansion is given by, , V, , or, , V, m, , m, , θ, , 0, , V = V0 (1 +, , ), , (1, , =, , 0(1, , ), , –, , ), , An Isotropic body expands equally in all directions and we can obtain the following relations, or, , =2, , =3 ,, , 1, , 2, , 3, , ,, , =, , 3, 2, , CALORIMETRY, (1) Specific heat capacity or Gram specific heat (c) : If Q heat is given to a substance of mass ‘m’, and, rise in temperature is , then, c=, , Q, m, , (cal/g°C), , (2) Molar heat capacity (C) :, C = Molar mass (M) × specific heat capacity, C = Mc =, C, , Q, n, , M. Q, Q, =, m, n, , :n=, , m, is number of moles, M, , (3) Heat capacity of an object is defined as product of mass and specific heat., (4) In general if Q heat is given to a substance of mass ‘m’ which increases its temperature by, , or, , Q, , mc, , c is specific heat capacity, , Q, , nC, , C is molar heat capacity, n is number of moles of the substance., , then, , Specific Latent heat, (1) Latent heat of fusion Lf, , Q, m, , (2) Latent heat of vaporisation Lv, During phase change (liquid, energy changes., , Q, m, solid or liquid, , vapours) temperature remains constant, but internal, , Water, Specific heat, , C = 1 cal/gm/°C = 4.2 J/gm/°C = 4200 J/kg/°C, Lf = 80 cal/gm = 336 J/gm, Lv = 540 cal/gm = 2268 J/gm, , For ice : Cice = 0.5 cal/g °C = 2100 J/kg °C

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Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , 57, , Newton’s Law of Cooling, If the temperature T of a body is not much different from surrounding temperature T0, then rate of cooling of, a liquid is directly proportional to the difference in the temperature of liquid T and temperature of surroundings, (T0) i.e., , Rate of, cooling, , – dT, dt, , – dT, dt, , T, , T – T0, , kT0, , , , Rate of cooling, , –, , dT , , dt , , (T, , T0 ), , dT, = (T – T0), dt, , Results, (1) Tf = T0 + (Ti – T0)e– t, where Ti is initial temperature, Tf is temperature after time t., t, , (2) Another form, , (3), , T03, , 4 A, mc, , dT, dt, , log, , (T, , Ti, Tf, , T0, T0, , Ti, , , , m mass of body , , , T0 ),  c specific heat ,  A surface area , , , emissivity, , , , T0, , t, , (4) Another approximate formula is, T1, , t, , T2, ,  T1 T2, ,  2, , , T0 , , , ln(T1 – T0), , t, Above formula gives time ‘t’ taken by the body to cool down from T1 to T2. T0 is temperature of surrounding., (5) If temperature of a body changes from, 2, 1, , –, –, , 0, 0, , =, , 3, 2, , –, –, , 0, 0, , (, , 0, , 1, , to, , 2, , in time ‘t’ and changes from, , 2, , to, , 3, , in next time then, , = temperature of environment), , (6) If equal masses of two liquids having same surface area and finish, cool from same initial temperature to same, final temperature with same surrounding, then, c1, t1, k2, =, =, ;, c2, t2, k1, , c1 & c2 are specific heats

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58, , Properties of Solids and Liquids, , Rapid Revision & Formula Bank, , WIEN’S DISPLACEMENT LAW, This law states that the wavelength corresponding to maximum intensity for a black body is inversely proportional, to the absolute temperature of the body, m, , b, T, , where b is a constant known as Wien's constant, Results, , Spectral emissive, power (e ), A1, 1, , T1, , 2, , A1, A2, , T2, T1, , T14, T24, , T2, 1, , (1), , max, , A2, , 2, , T=b, , (2) b = 2.898 × 10–3 m-K, (3) Area under e – graph = T 4 (Total emissive power), (4) If the temperature of the black body is made two fold,, , max, , becomes half, while area becomes 16 times., , (5) Temperature of the Sun,, If T = temperature of sun, then total energy radiated by sun per second = T4 (4 R2), The Sun, , The Earth, , R, , r, , Intensity at distance r from the sun (i.e., on earth), I = (Power radiated)(Area) =, , So, T, ,  r 2,  2,  R, , S,  ,  ,  , , 14, , T 4R 2, , r2, , = S, where S is called solar constant [S = 1.4 kW / m2], ,   1.5 108  2, , 7 105 ,  , , 1, , 4, 1.4 103 , , 8, 5.67 10 , , , ‰ ‰ ‰, , 5800 K

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Chapter, , 8, , Kinetic Theory of Gases, and Thermodynamics, Pressure Exerted by the Gas, The pressure of the gas is due to continuous bombardment of the gas molecules against the walls of the container., According to kinetic theory, the pressure exerted by an ideal gas is given by, P, , 1M 2, v, 3V, , M = Mass of the enclosed gas, , V = Volume of the container, v 2 = Mean square speed of molecules, 1 M 2, v rm s, 3 V, , Vrms = Root mean square velocity, , 1 Nm 2, v, 3 V, , N = Number of molecules, , or P =, , or, , P, , P, , 3, , 2, v rms, , = density of gas, m = Mass of the molecule, , v 2 = Mean square speed of molecules, Speeds of gas molecules :, (i) Root mean square speed,, , vrms=, , 3RT, Mw =, , 3P, , v12, , v22, , v32, n, , ....... v n2, , Here, MW is molecular weight in kg., vrms =, , (ii) Average speed vavg =, , 8RT, Mw, , 3 kT, , k = Boltzmann’s constant, m = mass of one molecule in kg, m, 8P, , v1 v 2, , v3, , ......... v n, n

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60, , Kinetic Theory of Gases and Thermodynamics, , Rapid Revision & Formula Bank, , (iii) vmp = Most probable speed is defined as the speed corresponding to which there are maximum number of, molecules., 2RT, Mw, , v mp, , 2P, , =, , = Density of gas, Mw = Molecular weight, R = Gas constant, P = Pressure of gas, m = Mass of one molecule, , 2 kT, m, , Order of magnitude : vrms > vavg > vmp, , v rms : v av : v mp, , 3 :, , 8, , :, , ~, 2 –, , 3 : 2 .5 : 2, , Relation between CP & CV :, CP – CV, , (i), , R, , (CP > CV), , CP, (ii) C, V, f, R, 2, , (iii) Cv, , f, nR T, 2, , U, n T, , CP, CV, , Gas, , Degrees of, freedom (f), , Monoatomic, , 3, (Translational), , 3, nR T, 2, , 3, R, 2, , 5, R, 2, , 5, 3, , Diatomic, , 3(Trans) +, 2(Rot), , 5, nR T, 2, , 5, R, 2, , 7, R, 2, , 7, 5, , Non-Linear, Poly atomic, , 3 (Trans) + 3, (Rot), , 3nR T, , 3R, , 4R, , 4, 3, , U, , CV, , CP=CV + R, , For a mixture of two gases A and B containing nA and nB number of moles., (i), , (ii), , ( n A fA, nA, , fmix, , 1, , mix, , nB fB ), nB, , 2, fmix, , Cvmix, , fmix R, 2, , Cpmix, , Cvmix, , R, , Thermodynamic Process, (1) Melting process : (Change of state, solid to liquid), Q= U+W, mLf = U + 0, , [W = 0 as volume remains nearly constant], , (2) Boiling process : (Change of state, liquid to vapours), mLv = U + P[V2 – V1], V2 = volume of vapours, V1 = volume of liquid, When 1 g of water vapourises isobarically at atmospheric pressure. U = 2091 J, P = 1.01 × 105 Pa,, V1 = 1 cm3, V2 = 1671 m3.

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Rapid Revision & Formula Bank, , Kinetic Theory of Gases and Thermodynamics, , (3) Isochoric process : Volume is constant, [dV = change in volume], , W=0, , dV = 0, , Q = nCV T = U, , U, n T, , CV, , (4) Isobaric process : Pressure is constant, P = constant, dW = PdV, W = P V = nR T, Q = nCP T = U + W, nCP T = nCV T + nR T, CP = CV + R, U, f, , W, 2, , Q, f, , 2, , U, 1, , or, , W, , Q, 1, , Fraction of total heat converted to internal energy =, , Fraction of total heat converted to work is,, , W, Q, , U, Q, , 1, , 1, , (5) Isothermal process : Temperature is constant, PV = K, , dT = 0, , dU = 0, C =, , as PV = nRT, So P, , (Constant), , nRT, V, , Work done in isothermal process, W, , Q, , V , nRT loge  2 ,  V1 , , V ,  , 2.303 nRT log10  2  = 2.303 nRT log  P1 , 10,  V1 ,  P2 , , (6) Adiabatic process : Heat exchanged (Q) is zero, PV = K, , [Equation of adiabatic process], , As Q = 0,, , nC T = 0 or C = 0, , Also, 0 = nCV T + W, , [by first law of thermodynamics], , Now, W, , W = – nCV T = –, , U, , i.e., , (7) Polytropic Process, PVx = Constant, , W, , nR T, 1 x, , Molar heat capacity, , C, , CV, , R, 1 x, , nR, (T – T1), –1 2, , 61

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Chapter, , 9, , Oscillations and Waves, Periodic Function, If f(t + T) = f(t) then function ‘f’ is periodic with period T., , Harmonic Motion, When oscillatory motion of a particle can be expressed in terms of sine or cosine functions, it is said to be, a harmonic motion., , SIMPLE HARMONIC MOTION, When a motion can be expressed in terms of a single sine or cosine (sinusoidal) function, the motion is said, to be Simple Harmonic Motion (SHM). For SHM, force -(displacement), –x, , F, , F = – kx [Restoring Force], , k, x, m, , a, d 2x, dt, , k, x, m, , 2, , 0, , or, , d 2x, , 2, , dt 2, , x, , 0, , (This equation represents the differential equation of S.H.M. ), , Velocity and acceleration of a particle executing S.H.M., If, , or,, , x = A sin t, A is amplitude (maximum displacement from mean position), v, , dx, dt, , v, , , A sin t, , , A cos t, , ,  , maximum speed = A, 2, , i.e., velocity leads displacement by, (a), , 2, , . (This is always true in SHM), , Dependence of velocity v with displacement from mean position (x), , v, , A2, , x2

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64, , Oscillations and Waves, , (b), , Rapid Revision & Formula Bank, , Acceleration, dv, dt, , a, , a=A, , A, 2sin(, , 2, , sin t , maximum acceleration = A, , t+ ), , i.e., acceleration leads velocity by, (c), , 2, , 2, , . Acceleration and displacement are in opposite phase., , Dependence of acceleration with position, is a = –, , 2x, , Graphical representation of variation of position, velocity and acceleration, t), , (For x = A sin, x, , v, , A, O, , A, T, 2, , T, , t, , 3T, 2, , x, , A, , –A, x2, A2, , –A, , –A, v, , v2, ( A )2, , 1, , a, , +A, O, , T, 2, , T, , x, , t, , 3T, 2, , slope = –, 2, tan =, , –A, a, , a, , 2, , +A, , A, O, , –A, , T, , T, 2, , t, , 3T, 2, , 2, , Salient points regarding energy in SHM :, , Oscillating quantity Time period Frequency, Displacement, , T, , f, , KE, , T/2, , 2f, , PE, , T/2, , 2f, , |KE ~ PE|, , T/4, , 4f, 0, , v, , A, v2, ( A )2, , 2, , Energy in SHM, , Total Energy, , 2, , a2, ( A 2 )2, , 1

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Rapid Revision & Formula Bank, , Oscillations and Waves, , 1., , KEavg, , 1, m, 4, , 2., , KEmax =, , 1, m, 2, , 3., , KEmin = zero at extreme position., , 4., , PEavg =, , 1, m, 4, , 5., , PEmax =, , 1, m, 2, , 6., , Total energy, E =, , 7., , Both kinetic and potential energy vary parabolically with x., , 2, , 65, , A2 ., , 2, , A 2 at mean position., , 2, , A2 ., , 2, , A 2 at extreme position., 1, m, 2, , 2A2, , which is constant i.e. doesn’t depend on x, , E, 1, m, 2, , 2, , A2, , TE, PE, , TE = Constant, TE = (1/2)m 2A2, , KE, x, , 8., , PE = KE at x, , x = –A, A, 2, , and t, , x=A, , x, , T, . (Starting from mean position towards +x)., 8, , SIMPLE PENDULUM, Time period of oscillation of simple pendulum of length l for small angular amplitude is given by T, , 2, , l, g, , where g is the effective acceleration due to gravity, directed along the length of pendulum when it is at mean, position., SOME IMPORTANT POINTS :, On changing various factors, T changes as :, 1., , If length ‘l’ is changed,, , T, T, , 2., , If gravity ‘g’ is changed,, , T, T, , 1 l, ., 2 l, , 1 g, ., 2 g, , Simple Pendulum in Lift, Effective g = | g, , , , a | , where a is pseudo acceleration. T, , 2, , l, g

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66, , Oscillations and Waves, , Rapid Revision & Formula Bank, , Simple Pendulum of Length Comparable to the Radius of Earth, Time period of such a pendulum is given by,, , T, , 1, , 2, , 1, g , l, , 1, Re, , , , , , OSCILLATION OF SPRING, Horizontal Oscillations, The spring is pulled/pushed from x = 0 to x = x0 and released., , k, , Smooth, , M, , M, , x = x0, , x=0, , The block executes SHM, (1), , Amplitude of oscillation = x0, , (2), , Time period T, , 2, , M, k, , COMBINATIONS OF SPRINGS, Series Combination :, , k1, , k2, M, , (1), , , , 1, k, , 1, k1, , 1, k, k 2 or Effective spring constant,, , 1, k, , 1, k1, , 1, k2, , Effective spring constant, k, , T, , 2, , M, k, , M, Smooth, , Smooth, , (2), , k, , k1k 2, k1 k 2, , k1k 2, T, k1 k 2 ,, , 2, , k1, , M, k, , , , k2, M, , k, , M

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Rapid Revision & Formula Bank, , Oscillations and Waves, , 67, , Parallel Combination, (1), , T, , Effective spring constant, k = k1 + k2 ,, , 2, , k1, , M, k, , k, , k2, , , , M, , M, Smooth, , Smooth, (2), , Effective spring constant, k = k1 + k2 T, , 2, , k2, , k1, , M, k, , k, , , , M, , M, Smooth, , Smooth, , Physical Pendulum, Figure shows an extended body (called physical pendulum) pivoted about point O, which is at a distance d, from its centre of mass., , O, d, cm, , mg, Time period of oscillation, T, , 2, , I, mgd, , I = moment of inertia of the body about pivoted point., d is the distance of centre of mass from suspension point, , Oscillation of a Floating Cylinder, If, , = density of cylinder material, = density of fluid ( < ), , then, T = 2, , L, g, , 2, , h, g, , Oscillations of a Liquid in a Tube, T, , 2, , l, g (sin, , sin ), , l = total length of liquid column, , h, , L

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68, , Oscillations and Waves, , Rapid Revision & Formula Bank, , Superposition of SHMs, Consider two SHMs along the same line, If, , y1 = a1 sin, , a2, , t, , y2 = a2 sin ( t + ), then, equation of resultant SHM is given by,, , a1, , y = y1 + y2 = A sin ( t + ), , a12, , where, A, , a22, , 2a1a2 cos, ,  a2 sin, tan 1,  a1 a2 cos, , &, , A, , , , , , Damped Oscillations, If there is any dissipative force like viscous force in SHM, then the amplitude of the particle decreases with, time such type of oscillations are known as damped oscillations., , x, , t, , (i), , Differential equation for damped oscillation m, , d 2x, dt, , 2, , b, , dx, dt, , kx, , 0,, , where b = Coefficient of damping., (ii), , Displacement-time equation, x = A(t) sin( t + ), , (iii), , Amplitude of damped oscillation, A(t ), , (iv), , Angular frequency of damped oscillation,, , A0 e, , b, t, 2m, , , where A0 = Initial amplitude., , 2, 0, , 2, ,  b ,  2m  , where, , , , 0, , k, =natural frequency., m, , Speed of mechanical waves, (a), , Transverse wave in a stretched string, T = tension in the string, = mass per unit length, D = diameter of string, , W, , = density, v, , v=, , T, , T, A, , stress, density, , 2, D, , T, , A, , D2, 4

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Rapid Revision & Formula Bank, , (b), , 69, , Transverse wave in a long bar is given by, , Y, , v, (c), , Oscillations and Waves, , Y = Young’s modulus,, , where, , = Density of material, , Longitudinal Waves, (i), , In liquid v, , = bulk modulus of elasticity, = density, , (ii), , In gases v, , ., , is bulk modulus of the gas. This value is not a fixed value for gases, , Case - I : Suggested by Newton, Taking isothermal process, , = 1.23 kg/m3, , Put P = 1 atm,, , v, , =P, , P, , v = 280 m/s (more than 15% error), , Case - II : Corrected by Laplace, For Adiabatic, , P, , v, , = P, = 1.4, we get, For air v, , . Taking, , 20 T, , v = 330 m/s, Note : Propagation of sound in air is adiabatic., , Factors affecting speed of sound, P, , RT, =, M, , v=, (1), , v is independent of pressure (If temperature is kept constant), , (2), , v, , (3), , Velocity of a wave depends on medium, not on the frequency of source, , (4), , v, , (5), , Velocity of sound in humid air is more because its density is less than that of dry air., , (6), , Velocity of sound in humid hydrogen is less than in dry hydrogen due to similar reason., , 1, , or v, , 1, M, , (If temperature is kept constant), , T, , SOUND WAVES, These are mechanical and longitudinal waves. They propagate in form of compressions and rarefactions., Particle displacements can be represented by wave function, S = A sin( t – kx), As particles oscillate, pressure variation takes place according to the wave function., P = P0 cos( t – kx), P0 = maximum pressure variation

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70, , Oscillations and Waves, , Rapid Revision & Formula Bank, , Characteristic of Sound, Loudness : Sensation of sound produced in human ear is due to amplitude. It depends upon intensity, density, of medium, presence of surrounding bodies,, (a), , Intensity of Wave, I = 2 2f 2A2 v, , 1, 2, , =, I, (b), , f, , 2, , 2, , A2 v, A2, , and I, , Intensity Level or (Sound Level) ( ), , I 0 minimum intensity of audible sound 10, , I measured intensity, , I , 10log10  ,  I0 , , 12, , W/m 2 , , , , Sonometer : In this case, transverse stationary waves are formed., , T = Mg, (tension in wire), l, , Mg, , The wire vibrates in n loops, then, n, or, 2, , l, , T, , Velocity v, , n, , v, , Pipe, length l, Open, , Closed, , 2l, n, , where ‘ ’ is mass per unit length of wire., , nv, 2l, , n T, 2l, , Fundamental, Mode, , Ist Overtone, , V, 2l, st, I Harmonic, V, 4l, st, I Harmonic, , V, l, 2 Harmonic, nd, , 3V, , th, , (n – 1) overtone, n, , V, , 2l, nth Harmonic, , V, ( 2n 1), 4l, 4l, th, rd, 3 Harmonic (2n –1) Harmonic, , 1:2:3:4, , 1:3:5:7, , Note : Even numbered (i.e., 2nd, 4th .....) harmonics do not exist in close organ pipe.

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Rapid Revision & Formula Bank, , Oscillations and Waves, , 71, , End correction (e) :, The antinodes are formed slightly out side the open end. The distance of antinode from open end of the pipe, is called end correction. It depends on radius of pipe. (e = 0.6 r), Thus, we have,, For open organ pipe, , For closed organ pipe, , e, e, , l + 2e, , l+e, , e, (2n, 4(l, , nV, 2(l 2e ), , 1)V, e), , Resonance Tube:, , ‘ ’, , If resonance is obtained first at length l1., , l1, , then at length l2, then, , l2, , = 2(l2 – l1), distance between two successive lengths is, , 2, , Interference, Consider two waves of same frequency and wavelength,, y1 = a1 sin ( t – kx), I1 = Ca12, y2 = a2 sin ( t – kx + ), I2 = Ca22, Equation of resultant wave is,, y = y1 + y2 = A sin ( t – kx + ), where A, , a12, , a22, , 2a1a2 cos, , and, ,  a2 sin, tan 1,  a1 a2 cos, , , , , , Resultant Intensity is given by, I, , I1 I 2, , 2 I1I 2 cos, , DOPPLER’S EFFECT, If a wave source and a observer are moving relative to each other, the frequency observed by the receiver (f) is, different from the actual source frequency (f0) given by,, f, ,  v v0, f0 ,  v ∓ vs, , , , , , , where v = speed of sound, v0 = speed of observer, vs = speed of source, , ‰ ‰ ‰