Page 1 :
https://t.me/NEET_StudyMaterial, Chapter, , 13, , Kinetic Theory, Solutions, SECTION - A, Objective Type Questions, 1., , Select the appropriate property of an ideal gas, (1) Its molecules are infinitesimally small, (2) There are no forces of interaction between its molecules, (3) It strictly obeys the ideal gas laws, (4) All of these, , Sol. Answer (4), All of the statements are true for an ideal gas., 2., , A real gas behaves as an ideal gas at, (1) Very low pressure and high temperature, , (2) High pressure and low temperature, , (3) High pressure and high temperature, , (4) Low pressure and low temperature, , Sol. Answer (1), At very low pressure the force of interaction between particles may be considered negligible. Also at high, temperature the force of inter molecular interaction decreases, 3., , A gas at a pressure P0 is contained in a vessel. If the masses of all the molecules are halved and their velocities, doubled, then the resulting pressure P will be, (1) 4P0, , (2) 2P0, , (3) P0, , (4), , P0, 2, , Sol. Answer (4), P0 =, , 1 mn 2, , v, 3, v, , P =, , P, 1 mn, , (2v )2  0, 3 2v, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 2 :
2, 4., , Kinetic Theory, , Solution of Assignment, , If E is the energy density in an ideal gas, then the pressure of the ideal gas is, (1) P , , 2, E, 3, , (2) P , , 3, E, 2, , (3) P , , 5, E, 2, , (4) P , , 2, E, 5, , Sol. Answer (1), , 5., , E=, , 1, n, mv 2 , v, 2, , P=, , 1 mN 2, v, 3 v, , P=, , 2, E, 3, , A sample of gas in a box is at pressure P0 and temperature T0. If number of molecules is doubled and total, kinetic energy of the gas is kept constant, then final temperature and pressure will be, (1) T0 , P0, , (2) T0 , 2P0, , (3), , T0, , 2P0, 2, , (4), , T0, , P0, 2, , Sol. Answer (4), P0 =, , 1 mN 2,  v rms, 3 v, , If E0 is initial KE of one molecule, nE0 = E   2n ⇒ E  , , E, v, ⎛1, ⎞1 1, ⇒ ⎜ mv 2 ⎟  mv 2  v  , 2, ⎝2, ⎠2 2, 2, , Thus KE of every molecule becomes half. Hence temperature becomes, P , , 2, , 1 m2N ⎛ v ⎞,  P0, 3 v ⎜⎝ 2 ⎟⎠, , Thus T  , 6., , T0, ., 2, , T0, , P   P0, 2, , By increasing temperature of a gas by 6°C its pressure increases by 0.4 %, at constant volume. Then initial, temperature of gas is, (1) 1000 K, , (2) 1500 K, , (3) 2000 K, , (4) 750 K, , Sol. Answer (2), , P2 T2, , P1 T1, , T1 = T, T2 = T + 6, , P2, T, 1 2 1, P1, T1, , P2  P1, T 6 ⎞,  100  ⎛⎜,  1⎟100, P1, ⎝ T, ⎠, 0.4 =, , 600,  T = 1500 K, T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 3 :
Solution of Assignment, , 7., , Kinetic Theory, , 3, , Boyle’s law is obeyed by, (1) Real gas of constant mass and temperature, (2) Ideal gas of constant mass and temperature, (3) Both ideal and real gases at constant temperature and variable mass, (4) Both ideal and real gases of constant mass and variable temperature, , Sol. Answer (2), Boyle's law states that if m and T are constant., , 1, P, and gas laws are only valid for ideal gases., V, , 8., , For an ideal gas the fractional change in its volume per degree rise in temperature at constant pressure is, equal to [T is absolute temperature of gas], (1) T 0, , (2) T, , (3) T –1, , (4) T 2, , Sol. Answer (3), PV = n RT, , ... (1), , P dv = n R dT, , ... (2), , Dividing (2) by (1), dV dT, , V, T, Fractional change in volume per degree rise in temperature, , dV, V  1, dT T, 9., , The raise in the temperature of a given mass of an ideal gas at constant pressure and at temperature 27° to, double its volume is, (1) 327°C, , (2) 54°C, , (3) 300°C, , (4) 600°C, , Sol. Answer (1), PV = n RT, Initial temperature T0 = 300 K, V0  T0, 2V0  2T0, 2T0 = 600 K,  T = 600 – 300 = 300 K, 10. The average velocity of gas molecules is, (1) Proportional to T, , (2) Proportional to T, , (3) Zero, , (4) Not possible to determine, , Sol. Answer (3), , v =, , , 3RT, M, , v T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 4 :
4, , Kinetic Theory, , Solution of Assignment, , 11. Which of the following methods will enable the volume of ideal gas to be increased four times?, (1) Double the temperature and reduce the pressure to half, (2) Double the temperature and also double the pressure, (3) Reduce the temperature to half and double the pressure, (4) Reduce the temperature to half and reduce the pressure to half, Sol. Answer (1), V0 = k, , T0, P0, , 4V0 = 4k, , T0, P0, , = k, , 2T0, P0 / 2, ,  Temperature is doubled and pressure halved., 12. A container has N molecules at absolute temperature T. If the number of molecules is doubled but kinetic energy, in the box remain the same as before, the absolute temperature of the gas is, (1) T, , (2), , T, 2, , (3) 3T, , (4) 4T, , Sol. Answer (2), Initial energy of gas = Final energy, Let K.E. of each molecule initially be E0.,  Total kinetic energy = E0 × n, Let final kinetic energy of each molecule be Ef., E0 × n = Ef × 2n, Ef =, , E0, 2, , Since temperature is the average kinetic energy of molecules., T0 = KE0, , Tf =, ,  Temperature becomes Tf =, , KE0, 2, T0, 2, , 13. During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at, temperature T and volume V, when it expands to volume 2V, the resulting temperature is, , (1), , T, 2, , (2) 2T, , (3), , 2T, , (4), , T, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 6 :
6, , Kinetic Theory, , Solution of Assignment, , 16. The average speed of gas molecules is v at pressure P. If by keeping temperature constant the pressure of, gas is doubled, then average speed will become, (1), , 2v, , (2) v, , (3) 2v, , (4), , v, 2, , Sol. Answer (2), vav , , T, , Since temperature is constant vav is constant., 17. Four molecules of a gas have speeds 1, 2, 3 and 4 km/s. The value of the r.m.s. speed of the gas molecules, is, , (1), , 1, 15 km/s, 2, , (2), , 1, 10 km/s, 2, , (3) 2.5 km/s, , (4), , 15, km/s, 2, , Sol. Answer (4), , R.M.S. speed =, , v12  v 22  v 32 ...v n2, n, , v =, , 12  22  32  42, 4, , v =, , 30, 4, , v =, , 15, 2, , 18. The r.m.s. speed of the molecules of enclosed gas is V. What will be the r.m.s. speed if pressure is doubled, keeping the temperature same?, (1), , V, 2, , (2) V, , (3) 2V, , (4) 4V, , Sol. Answer (2), Temperature is a quantities which denotes a value which gives the average kinetic energy of molecules in a, gas. This depends on velocities of gas molecules and vice-versa. If temperature does not change Vr.m.s. will, also not change., 19. The effect of temperature on Maxwell’s speed distribution is correctly shown by, , Nv, , (1), , Nv, , T2 T1 > T2, T1, , v, , (2), , T1 > T2, T2, , Nv, T1, , (3), v, , T1 > T2, , Nv, , T1 T2, , (4), v, , T1 T1 > T2, T2, v, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 7 :
Solution of Assignment, , Kinetic Theory, , 7, , Sol. Answer (1), The Maxwell’s distribution curve heats at the most probable speed, which depends on temperature., vprobable , , T, , Hence T1 > T2 is correctly shown in option (1) as it shows with peaks of the curve at higher temperature,, furthers along the x-axis., 20. Select the incorrect statement about Maxwell’s speed distribution, (1) The distribution function depends only on the absolute temperature, (2) Vrms  Vav  Vmp, (3) The area under the distribution curve gives total number of molecules of the gas, (4) The distribution curve is symmetric about the most probable speed, Sol. Answer (4), The Maxwell’s speed distribution is asymmetric due to the fact that the lowest speed possible is zero. While, the highest speed possible is infinity., 21. The ratio of number of collisions per second at the walls of containers by He and O2 gas molecules kept at, same volume and temperature, is (assume normal incidence on walls), (1) 2 : 1, , (2) 1 : 2, , (4) 1 : 2 2, , (3) 2 2 : 1, , Sol. Answer (3), ⎛ nO2, ⎜⎜, ⎝ nHe, , 2, , ⎞, MHe, ⎟⎟ , M O2, ⎠, , MO2 = 32, , MHe = 4, , nO2, nHe, , , , , 1, 2 2, , nHe : nO2  2 2 : 1, , 22. An ant is moving on a plane horizontal surface. The number of degrees of freedom of the ant will be, (1) 1, , (2) 2, , (3) 3, , (4) 6, , Sol. Answer (2), The number of degrees of freedom of movement of ant is 2 as it can move only in two independent directions, in the plane surface., , 23. If a gas has ‘f ’ degree of freedom, the ratio of the specific heats of the gases, , (1), , 1 f, 2, , (2) 1 , , f, 2, , (3) 1, , 1, f, , Cp, Cv, , is, , (4) 1, , 2, f, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 8 :
8, , Kinetic Theory, , Solution of Assignment, , Sol. Answer (4), Cp = Cv + R, and Cv =, Cp, Cv, Cp, Cv, , Cp, Cv, Cp, Cv, , f, R, 2, f, R R, 2, =, f, 2, =, , f 2, f, , =, , 2f, f, , = 1, , 2, f, , 24. Molar specific heat at constant volume, for a non-linear triatomic gas is (vibration mode neglected), (1) 3R, , (2) 4R, , (3) 2R, , (4) R, , Sol. Answer (1), Molar heat capacities for a gas is given by Cv =, , f, RT, 2, , Where f = 6 in triatomic molecules, Cv = 3 RT, 25. A mixture of ideal gases has 2 moles of He, 4 moles of oxygen and 1 mole of ozone at absolute temperature T. The, internal energy of mixture is, (1) 13 RT, , (2) 11 RT, , (3) 16 RT, , (4) 14 RT, , Sol. Answer (3), Degrees of freedom of He (fHe) = 3, Degrees of freedom of O2 (fO ) = 5, 2, , Degrees of freedom of O3 (fO ) = 6, 3, , nHe = 2,, , nO2 = 4, , nO3 = 1, , Energy of mixture = Sum of individual energies, = nHefHe  nO2 fO2  nO3 fO3 , =  2  3  4  5  1 6 , , RT, 2, , RT, 2, , = (3 + 10 + 3) RT, = 16 RT, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 9 :
Solution of Assignment, , Kinetic Theory, , 9, , 26. E0 and EH respectively represents the average kinetic energy of a molecule of oxygen and hydrogen. If the two, gases are at the same temperature, which of the following statement is true?, (1) E0 > EH, , (2) E0 = EH, , (3) E0 < EH, , (4) Data insufficient, , Sol. Answer (2), Temperature is an approximate value which refers to average kinetic energy per molecule. If temperature of both, is same, average energy will be same according to definition., 27. 14 g of CO at 27°C is mixed with 16 g of O2 at 47°C. The temperature of mixture is (vibration mode neglected), (1) –5°C, , (2) 32°C, , (3) 37°C, , (4) 27°C, , Sol. Answer (3), 1 mole of CO and 1 mole of O2 are mixed., Net internal energy =, , f1, f, RTCO  2 RTO2, 2, 2, , =, , 5, 5, R 300  R 350, 2, 2, , =, , 5, R(650), 2, , = 5 R(325), = 1625 R, 1625 =, , 5, RTfinal  nfinal, 2, , 1625  2, = Tfinal × nfinal, 5, , 325 × 2 = Tfinal × 2, Tfinal = 325 K, Tfinal = 37°C, 28. When one mole of monatomic gas is mixed with one mole of a diatomic gas, then the equivalent value of  for, the mixture will be (vibration mode neglected), (1) 1.33, , (2) 1.40, , (3) 1.50, , 2 5,  = 1, 3 3, , n, n, n,  1  2,   1 1  1  2  1, , (4) 1.6, , Sol. Answer (3), Y for monatomic gas = 1 , , Y for diatomic gas = 1 , , 2 7,  =, 2, 5 5, , 2, 1, 1, , , 7,  1 5, 1, 1, 3, 5, , Solving, we get  = 3/2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 10 :
10, , Kinetic Theory, , Solution of Assignment, , 29. A box of negligible mass containing 2 moles of an ideal gas of molar mass M and adiabatic exponent  moves, with constant speed v on a smooth horizontal surface. If the box suddenly stops, then change in temperature, of gas will be, (1), , (   1)Mv 2, 4R, , (2), , Mv 2, 2R, , (3), , Mv 2, 2(   1)R, , (4), , (   1)Mv 2, 2R, , Sol. Answer (4), Mass of gas in the box = 2 M, Initial kinetic energy =, , 1, nfRT, 2, , Mv2 =,  T =, , 2Mv 2, nfR, , Substitution f =, T =, , 1,  2M  v 2  Mv 2, 2, , 2, and n = 2, 1 r, , (   1) Mv 2, 2R, , 30. On increasing number density for a gas in a vessel, mean free path of a gas, (1) Decreases, , (2) Increases, , (3) Remains same, , (4) Becomes double, , Sol. Answer (1), Mean free path of a substance is the average distance a molecule may travel without collision., If the number of molecules per unit volume increases it increases the frequency of collision and decreases, the mean free path., , SECTION - B, Objective Type Questions, 1., , At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1920 m/s. The, gas is, (2) F2, , (1) H2, , (3) Cl2, , (4) O2, , Sol. Answer (1), Vr.m.s. =, , 1920 =, , M=, , 3RT, M, , 3  8.314  300, M, , 3  8.314  300, 1920, , M = 0.00202 kg, On molar weight = 2.02 g., Hence it is hydrogen., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 11 :
Solution of Assignment, , 2., , Kinetic Theory, , 11, , An ideal gas is enclosed in a container of volume V at a pressure P. It is being pumped out of the container by, using a pump with stroke volume v. What is final pressure in container after n-stroke of the pump? (assume, temperature remains same), ⎛ V ⎞, (1) P ⎜, ⎟, ⎝V  v ⎠, , n, , (2), , PV, (V  v )n, , (3) P, , Vn, , ⎛ V ⎞, (4) P ⎜, ⎟, ⎝V  v ⎠, , vn, , n, , Sol. Answer (1), After stroke PV = constant, PV = P1(V + v), P1 =, , PV, (V  v ), , V, v, , Similarly after 2nd stroke, , ⎛ V ⎞, P2 = P ⎜, ⎟, ⎝V  v ⎠, , 2, , After nth stroke, , V ⎞, Pn = P ⎛⎜, ⎟, ⎝V  v ⎠, 3., , n, , Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. If number of molecules of the gases, are n1, n2 and n3 respectively then temperature of mixture will be (assume no loss of energy), , (1), , T1  T2  T3, 3, , (2), , n12T1  n22T2  n32T3, n1  n2  n3, , (3), , n1T1  n2T2  n3T3, n1  n2  n3, , T1  T2  T3, (4) n  n  n, 1, 2, 3, , Sol. Answer (3), Absolute temperature = T1, T2, T3., Internal energy of gases =, , Average temperature =, , 4., , n1RT1 n2RT2 n3 RT3, , , 2, 2, 2, , Internal energy of mixture, R, (n1  n2  n3 ), 2, , Variation of atmospheric pressure, with height from earth is, (1) Linear, , (2) Parabolic, , (3) Exponential, , (4) Hyperbolic, , Sol. Answer (3), Variation of atmospheric pressure due to height is given by the barometric formula, Ph = P0 e–mgh/RT, Hence the decrease will be exponential., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 12 :
12, 5., , Kinetic Theory, , Solution of Assignment, , A narrow glass tube, 80 cm long and opens at both ends, is half immersed in mercury, now the top of the, tube is closed and is taken out of mercury. A column of mercury 20 cm long remains in the tube. Find, atmospheric pressure, (1) 20 cm of air column, , (2) 60 cm of Hg column, , (3) 60 cm of air column, , (4) 20 cm of Hg column, , Sol. Answer (2), PV = constant, P1V1 = P2V2, , [P1 = P0 atmospheric pressure], , P0 × 40 = P1 × 60, , ... (i), , P1 + 20 = P0, , ... (ii), , From (i), , 2P0, 3, , P1 =, From (ii), , 2P0, + 20 = P0  P0 = 60 cm of Hg., 3, 6., , One mole of monatomic gas and three moles of diatomic gas are put together in a container. The molar specific, heat (in JK–1 mol–1) at constant volume is (Let R = 8 JK–1 mol–1), (1) 18, , (2) 19, , (3) 20, , (4) 21, , Sol. Answer (1), n1 = 1 mole, , f1 = 3, , n2 = 3 moles, , f2 = 5, , R = 8 JK–1 mol–1, Molar specific heat are given by the weighted means of the gases., , Cvf =, , Cv f =, , n1  Cv1  n2  Cv 2, n1  n2, , 3, 5, 1 R  3  R, 2, 2, 4, , =, , 3, 15, R R, 8, 8, , =, , 18, R, 8, , Cvf = 18, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 13 :
Solution of Assignment, , 7., , Kinetic Theory, , 13, , Two closed containers of equal volume filled with air at pressure P0 and temperature T0. Both are connected, by a narrow tube. If one of the container is maintained at temperature T0 and other at temperature T, then, new pressure in the containers will be, , 2P0T, (1) T  T, 0, , P0T, (2) T  T, 0, , P0T, (3) 2(T  T ), 0, , (4), , T  T0, P0, , Sol. Answer (1), , P, = constant, T, , Initially, , P0 P0 2P0, , , T0 T0, T0, , For two containers, P0 P 2P0,  , T0 T, T0, 2P0  T  T0, P = T (T  T ), 0, 0, , 2P0T, or P = (T  T ), 0, 8., , The temperature of a gas is –68°C. At what temperature will the average kinetic energy of its molecules be, twice that of –68°C?, (1) 137°C, , (2) 127°C, , (3) 100°C, , (4) 105°C, , Sol. Answer (1), Average kinetic energy =, , f, RT, 2, , or K.E.avg  T, For K.E. energy to be doubles that of K.E. at – 68°C or 205 K., The temperature must be 2 T or 410 K, When converted to °C = 410 – 273 = 137°C, 9., , An ideal gas is filled in a closed container and container is moving with uniform acceleration in horizontal, direction. Neglect gravity. Pressure inside the container is, (1) Uniform everywhere, , (2) Less in front, , (3) Less at back, , (4) Less at top, , Sol. Answer (2), Each particle closes experience a pseudo force initials, themselves to give low pressure every where. This is, because of Pascal's law., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 14 :
14, , Kinetic Theory, , Solution of Assignment, , 10. One kg of a diatomic gas is at pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m3. The energy of, the gas due to its thermal motion will be, (1) 3 × 104 J, , (2) 5 × 104 J, , (3) 6 × 104 J, , (4) 7 × 104 J, , Sol. Answer (2), PV = n RT, f=4, , ... (1), , kg/m3, ,  v = 0.25 m3, 8 × 104 ×, , 1, PV, 4, , ... (2), , f, RT  n, 2, , Energy of gas is given by =, =, , 5,  PV, 2, , =, , [From (1)], , 5,  2  10 4  5  104 J, 2, , 11. A container contains 32 g of O2 at a temperature T. The pressure of the gas is P. An identical container containing, 4 g of H2 at a temperature 2T has a pressure of, (1) 8P, , (2) 4P, , (3) P, , (4) P18, , Sol. Answer (2), Given, 1 mole of O2 at temperature T, pressure P, and 2 moles of H2 at a temperature 2T, P1 =, , n  RT, V, , P0 =, , RT, P, V, , PH =, , 4 RT,  4P, V, , 12. An ideal gas is expanding such that PT = constant. The coefficient of volume expansion of the gas is, (1), , 1, T, , (2), , 2, T, , (3), , 3, T, , (4), , 4, T, , Sol. Answer (2), PT = Constant, or, , T2, = Constant [PV = nRT]  T2 = KV, V, , ... (i), , Differentiating w.r.t. T, we get, , 2T K dV, 2T, dV, , ⇒, , V V dT, VK VdT, , , dV, 2TV 2, , , VdT VT 2 T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 15 :
Solution of Assignment, , Kinetic Theory, , 15, , 13. 50 cal of heat is required to raise the temperature of 1 mole of an ideal gas from 20°C to 25°C, while the, pressure of the gas is kept constant. The amount of heat required to raise the temperature of the same gas, through same temperature range at constant volume is (R = 2 cal/mol/K), (1) 70 cal, , (2) 60 cal, , (3) 40 cal, , (4) 50 cal, , Sol. Answer (3), , 50, = 10 cal K–1 mol–1, AT, , CP =, , CP = CV + R, CV = CP = R, CV = 8 cal K–1 mol–1, 14. Pressure versus temperature graph of an ideal gas is as shown in figure. Density of the gas at point A is 0., Density at point B will be, , P, B, , 3P0, P0, , A, T0, , (1), , 3, 0, 4, , (2), , 3, 0, 2, , 2T0, (3), , T, 4, 0, 3, , (4) 20, , Sol. Answer (2), PV = nRT, , P=, , , RT, M, , Initially  = 0, P = P0, T = T0, P0 =, , 0 R, T0 initially, M, , ... (i), , 3P0 =, , x R, 2T0 final, M, , ... (ii), , Dividing (ii) by (i), , x, 3=  2, 0, 3, 0   x, 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 16 :
16, , Kinetic Theory, , Solution of Assignment, , 15. The energy (in eV) possessed by a neon atom at 27°C is, (2) 4.75  10–4, , (1) 1.72  10–3, , (3) 3.88  10–2, , (4) 3.27  10–5, , Sol. Answer (3), Neon is a monoatomic gas., So, at 300 K its internal energy is given by, , For one molecule, , 1, fkT, 2, , 3, kT, 2, , 16. If heat energy is given to an ideal gas at constant pressure, then select the graph which best represents the, variation of VT with temperature (T)., , VT, , VT, , VT, , (2), , (1), , VT, , (3), , T, , (4), , T, , T, , T, , Sol. Answer (1), PV = nRT, P, T, , nR V, , and, , V nR, , = constant (K), T, P, , VT = KT2, Assuming VT = y and x = T, y = Kx2, Which is equation of a parabola will focus on y-axis > facing upwards., 17. If hydrogen gas is heated to a very high temperature, then the fraction of energy possessed by gas molecules, correspond to rotational motion, (1), , 3, 5, , (2), , 2, 7, , (3), , 3, 7, , (4), , 2, 5, , Sol. Answer (2), Hydrogen is a diatomic molecules and if vibrational degrees of freedom are increased the degrees of freedom, will be 3 translation 2 rotational and two vibrational.,  So total 7 degree of freedom., Fraction of energy possessed due to rotational motion : Degree of freedom due to rotation total degree of, freedom =, , 2, 7, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 17 :
Solution of Assignment, , Kinetic Theory, , 17, , 18. The temperature (T) of one mole of an ideal gas varies with its volume (V) as T = –V 3 + V 2, where  and,  are positive constants. The maximum pressure of gas during this process is, (1), , , 2R, , (2), , 2 R, 4, , (3), , (   )R, 22, , (4), ,  2R, 2, , Sol. Answer (2), T = –V 3 + V 2, and PV = nRT, , ... (i), ... (ii), , n=1, So, P =, , Multiplying, , RT, V, , R, in (i), V, , RT, = (–V 2 + V )R, V, , or, , P = (–V 2 + V )R ... (iii), dP, = (– 2V + )R, dV, , Maxima is when, , d 2P, dP, = 0 and, in negative, so, dV, dV 2, , O = (–2V + )R, V=, , , 2, , Put in value of V in equation (iii), , ⎛, 2, 2 ⎞, 2R, ⎟R  P =, P = ⎜  2 , 2 ⎠, ⎝ 4, 4, 19. Nitrogen gas is filled in an insulated container. If  fraction of moles dissociates without exchange of any energy,, then the fractional change in its temperature is, (1), , –, 5, , (2), , , 3, , (3), , –3, 2, , (4), , 5, 2  3, , Sol. Answer (1), Degree of freedom of diatomic nitrogen = 5, Degree of freedom of monoatomic nitrogen = 3, Let initial number of moles be n and  fraction dissociated., So fraction dissociated = n fraction remaining = n – n., nbreak into two so new atoms formed is actually 2n., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 18 :
18, , Kinetic Theory, , Solution of Assignment, , Initial energy is given by = n , , f, 5,  RT = n   RT, 2, 2, , 5, 3, Final energy = (n  n ) RT2  2n  RT2, 2, 2, =, , 5, 5, nRT2  nRT2  n3RT2, 2, 2, , =, , nRT2, 5, nRT2 , 2, 2, , =, , (5  2) nRT2, 2, , Change in energy is given on zero., (5   ) nRT2, 5 nRT, =, 2, 2, 5T, = T2, 5, , T = T2 – T, or T =, , , 5T, T, T =, 5, 5, , Fractional change in temperature =, , T, , or , T, 5, , 20. An ideal gas undergoes a polytropic given by equation PVn = constant. If molar heat capacity of gas during, this process is arithmetic mean of its molar heat capacity at constant pressure and constant volume then value, of n is, (1) Zero, , (2) –1, , (3) +1, , (4) , , Sol. Answer (2), Polytropic process, PVn = constant, Given heat capacities is average of CP and CV. So, C=, , CP  CV, 2, , or C =, , 2CV  R, 2, , or C =, , CV  R, 2, , ... (i), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 19 :
Solution of Assignment, , Kinetic Theory, , 19, , Now formula for specific heat of polytropic process is given by, C=, , or, , R, R, , y  1 1 n, , ... (ii), , R, R, R, R, R, as CV ,  , , y  1 2 y  1 1 n, y 1, R, R, , 2 1 n, , or n = –1, 21. If  moles of a monoatomic gas are mixed with  moles of a polyatomic gas and mixture behaves like diatomic, gas, then [neglect the vibrational mode of freedom], (1) 2 = , , (2)  = 2, , (3)  = –3, , (4) 3 = –, , Sol. Answer (1), 22. If different ideal gases are at the same temperature, pressure and have same volume, then all gases have same, (1) Density, , (2) Number of molecules, , (3) Most probable speed, , (4) Internal energy per mole, , Sol. Answer (2), PV = nRT, or, , PV, n, RT, , At the same pressure volume and temperature each molecule will have same number of moles i.e. same, number of molecules of gas., 23. The internal energy of 10 g of nitrogen at N.T.P. is about, (1) 2575 J, , (2) 2025 J, , (3) 3721 J, , (4) 4051 J, , Sol. Answer (2), Number of moles of N2 =, , U=, , =, , 10, 28, , f, nRT, 2, , 5 5,   R  273, 2 14, , = 2025 J, 24. The mean free path of a molecule of He gas is . Its mean free path along any arbitrary coordinate axis will, be, (1) , , (2), , , 3, , (3), , , 3, , (4) 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 20 :
20, , Kinetic Theory, , Solution of Assignment, , Sol. Answer (3), Mean free path of a molecule is the resultant of path along three separate axis and they will be equal.,  2x   2y   22, , =, , So,, , 2, where x =  y  2 = (say) a, , a2  a2  a2, , =, or, , , , a=, , 3, , 25. 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), Interatomic potential varies with average interatomic distance as R–6 which is a fact., 26. 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), Critical temperature is given as :, TC =, , 8a, 27Rb, , 27. To find out degree of freedom, the expression is, 2, (1) f    1, , (2) f , ,  1, 2, , 2, (3) f    1, , 1, (4) f    1, , Sol. Answer (1), CV =, , f, R, 2, , CP =, , f, f, (f  2), R, CV + R = R  R , 0, 2, 2, , CP,  = C, V, =, , (f  2)R  2, 2f R, ,  = 1, , 2, f, , –1=, , f =, , 2, f, , 2,  1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 21 :
Solution of Assignment, , Kinetic Theory, , 21, , 28. Nitrogen gas N2 of mass 28 g is kept in a vessel at pressure of 10 atm and temperature 57ºC. Due to leakage, of N2 gas its pressure falls to 5 atm and temperature to 27ºC. The amount of N2 gas leaked out is, (1), , 5, g, 63, , 63, g, 5, , (2), , (3), , 28, g, 63, , (4), , 63, g, 28, , Sol. Answer (2), Mass = 28 g, Pi = 10 atm, , Ti = 57°C = 330 K, , Pf = 5 atm, , Tf = 27°C = 300 K, , Volume is kept constant., Pi = K × ni Ti, , ... (i), , Pf = K × nf Tf, , ... (ii), , Dividing (i) by (ii), Pi, n T,  i i, Pf, nf Tf, ni, 10 300, , nf = 5 330, , or, , ni, 10, 2, nf =, 11, ni, 20, nf = 11, , Now ni = 1 mole of N2, nf =, , 11, moles, 20, , or Mass of N2 left =, , 11, × 28, 20, ,  Quantity released = 28 –, , =, , 11, × 28, 20, , 63, 9, g,  28 =, 5, 20, , 29. A diatomic gas of molecular mass 40 g/mol is filled in a rigid container at temperature 30ºC. It is moving with, velocity 200 m/s. If it is suddenly stopped, the rise in the temperature of the gas is, (1), , 32, ºC, R, , (2), , 320, ºC, R, , (3), , 3200, ºC, R, , (4), , 3.2, ºC, R, , Sol. Answer (2), Let there by n moles of gas., Mass of gas = 40ng or, , 40n, or 0.04n kg, 1000, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 22 :
22, , Kinetic Theory, , Solution of Assignment, , K.E. of gas in container =, , 1, × 0.04 n × (200)2, 2, , = 0.02 × n × 4 × 104, = 8 × 102 × n J, f, Now heat capacity of gas (C) = nR, 2, , or C =, , 5, R×n, 2, , or CT = 8 × 102 × n, or, , 5, × R × n T = 8 × 102 × n, 2, T =, , 8  102 2, , R, 5, , T =, , 16, 320,  102 =, °C, 5, R, , 30. The ratio of average translatory kinetic energy of He gas molecules to O2 gas molecules is, (1), , 25, 21, , (2), , 21, 25, , (3), , 3, 2, , (4) 1, , Sol. Answer (4), Both He and O2 have 3 translatory degrees of freedom. At the same temperature, energy is divided equally, in all degrees of freedom. Hence ratio of the translatory kinetic energy is one., , SECTION - C, Previous Years Questions, 1., , Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being, twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio, of molecular weight of A and B is, [Re-AIPMT-2015], (1), , 1, 2, , (2), , 2, 3, , (3), , 3, 4, , (4) 2, , Sol. Answer (3), , A, , B, , 2P, , P, , 1.5 d, , d, , We known,, PV = nRT, PV =, , m, RT, M, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 23 :
Solution of Assignment, , Kinetic Theory, , 23, , ⎛m⎞ 1, P  ⎜ ⎟ RT, ⎝V ⎠M, P, , d, RT, M, , P1 d1 M2, So, P  M  d, 2, 1, 2, 2P 1.5 d M2, , , P, M1, d, M1 15 3, , , M 2 20 4, , 2., , The ratio of the specific heats, , CP,   in terms of degrees of freedom (n) is given by, Cv, , 1⎞, ⎛, (2) ⎜ 1  ⎟, ⎝ n⎠, , ⎛ n⎞, (1) ⎜ 1  ⎟, ⎝ 2⎠, , ⎛ n⎞, (3) ⎜ 1  ⎟, ⎝ 3⎠, , [AIPMT-2015], , ⎛ 2⎞, (4) ⎜ 1  ⎟, ⎝ n⎠, , Sol. Answer (4), We known, , Cv , , n, R, 2, , So, CP  R  Cv  R , , ...(i), n, R, 2, , ⎛ n⎞, = R ⎜1  ⎟, ⎝ 2⎠, , ...(ii), , On dividing equation (ii) by (i), , CP, , Cv, , n⎞, ⎛, R ⎜1  ⎟, 2⎠, ⎝, Y, n, R, 2, , n, ⎛ n⎞, So, ⎜ 1  ⎟  Y, 2, 2, ⎝, ⎠, Y  1, , 3., , 2, n, , The mean free path of molecules of a gas, (radius r) is inversely proportional to, (1) r3, , (2) r2, , (3) r, , [AIPMT-2014], (4), , r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 24 :
24, , Kinetic Theory, , Solution of Assignment, , Sol. Answer (2), The mean free path of molecules of gas is given by, , 1, , , , where, , 2d 2n, , d = diameter of molecule, d = 2r, , So,  , 4., , 1, r2, , The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1K to T2K is, [NEET-2013], , (1), , 3, N k (T – T1), 2 a B 2, , (2), , 3, N k (T – T1), 4 a B 2, , (3), , 3, ⎛T ⎞, NakB ⎜ 2 ⎟, 4, ⎝ T1 ⎠, , (4), , 3, N k (T – T1), 8 a B 2, , Sol. Answer (4), Mole of helium is 4 g, So, number of moles of helium =, , 1, moles, 4, , NTP there is constant pressure, CP of gas =, , 3R, f, R=, 2, 2, , Q = CP × n T, , 5., , =, , 3, 1, R   (T2  T1 ), 2, 4, , =, , 3R, (T2  T1 ), 8, , Two container A and B are partly filled with water and closed. The volume of A is twice that of B and it contains, half the amount of water in B. If both are at same temperature, the water vapour in the container will have, pressure in the ratio of, (1) 1 : 2, , (2) 1 : 1, , (3) 2 : 1, , (4) 4 : 1, , Sol. Answer (2), Vapour pressure for the same liquid is always the same. So the ratio will be P : P or 1 : 1., 6., , At constant volume, temperature is increased then, (1) Collision on walls will be less, (2) Number of collisions per unit time will increase, (3) Collisions will be in straight lines, (4) Collisions will not change, , Sol. Answer (2), At constant volume if temperature is increased pressure will increase. Since pressure is increased due to, collisions of particles will the wall of the container. So collisions per unit time will increase., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 25 :
Solution of Assignment, , 7., , Kinetic Theory, , 25, , A polyatomic gas with n degree of freedom has a mean energy per molecule given by, (1), , nKT, N, , (2), , nKT, 2N, , (3), , nKT, 2, , (4), , nKT, 4, , Sol. Answer (3), Numbers of degrees of freedom = n., Internal energy of gas =, , K=, , n, RT, 2, , R, where N is the Avogadro's number., N, , or NK = R, Internal energy =, , n, N KT, 2, , Internal energy per molecule =, , 8., , n N KT, nKT, or, 2 N, 2, , For a certain gas the ratio of specific heats is given to be  = 1.5. For this gas, (1) Cv , , 3R, J, , (2) C p , , 3R, J, , (3) C p , , 5R, J, , (4) Cv , , 5R, J, , Sol. Answer (2), For a certain gas, , Cp, Cv, ,  1.5, , Cp = Cv + R, , 1, , R,  1.5, Cv, , R=, , Cv, 2, , or Cv = 2R, Cp = 3R, 9., , According to kinetic theory of gases, at absolute zero temperature, (1) Water freezes, , (2) Liquid helium freezes, , (3) Molecular motion stops, , (4) Liquid hydrogen freezes, , Sol. Answer (3), According to kinetic energy of gases at absolute temperature molecular motion stops as for ideal gases only, in kinetic energy of gases is considered which is given by, K.E. =, , f, nKT, 2, , So, T = 0 K and motion will be zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 26 :
26, , Kinetic Theory, , Solution of Assignment, , 10. One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at, constant pressure. If the same gas is heated at constant volume to raise the temperature by the same, 10 K, the heat required is [Given the gas constant R = 8.3 J/mol-K], (1) 198.7 J, , (2) 29 J, , (3) 215.3 J, , (4) 124 J, , Sol. Answer (4), 207, = 20.7 J/mol-K, 10, , CP =, , R = 8.3, and CV = CP – R, = 12.4 J/mol-K, Q = CV T, = 124 J, 11. Relation between pressure (P) and average kinetic energy per unit volume of gas (E) is, (1) P , , 2, E, 3, , (2) P , , 1, E, 3, , (3) P , , 1, E, 2, , (4) P = 3E, , Sol. Answer (1), Relation between pressure (P) and average kinetic energy is given by, P=, , 1, mNv 2, 3, , ... (i), , E=, , 1, mv 2, 2, , ... (ii), , Using (i) and (ii), P=, , 2, E, 3, , 12. If Cs be the velocity of sound in air and C be the rms velocity, then, 1/2, , (1) Cs < C, , (2) Cs =C, , , (3) Cs  C ⎛⎜ ⎞⎟, ⎝3⎠, , (4) None of these, , Sol. Answer (3), Velocity of sound in air is given by, , Cs , , P, or, , , C=, , 3RT, P, M, , Cs = C, ,  RT, M, , , 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 27 :
Solution of Assignment, , Kinetic Theory, , 27, , 13. The temperature of gas is raised from 27°C to 927°C. The rms speed is, , (1), , 927, times the earlier value, 27, , (2) Remain the same, , (3) Gets halved, , (4) Get doubled, , Sol. Answer (4), , V=, , 3RT, M, , V= K T, V1 = K 300, V2 = K 1200, , V1, 300, , V2, 1200, , 2 300, 300, , V1  V2, , or V2 = 2V1, 14. The equation of state, corresponding to 8 g of O2 is, (1) PV = 8RT, , (2) PV , , RT, 4, , (3) PV = RT, , (4) PV , , RT, 2, , Sol. Answer (2), , 8 g of O2 =, , 1, moles, 4, , PV = nRT, , PV =, , 1, RT, 4, , 1⎤, ⎡, ⎢⎣as n  4 ⎥⎦, , 15. At 0 K, which of the following properties of a gas will be zero?, (1) Kinetic energy, , (2) Potential energy, , (3) Density, , (4) Mass, , Sol. Answer (1), By definition at absolute zero the kinetic energy of a gas is zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 28 :
28, , Kinetic Theory, , Solution of Assignment, , 16. At 10oC the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110oC, this, ratio is, (1), , 283, x, 383, , (2) x, , (3), , 383, x, 283, , (4), , 10, x, 110, , Sol. Answer (1), PV = nRT, , P, RT, =, , M, x=, , RT, M, , T1 = 283 K, x1 =, , T2 = 383 K, , R,  283, M, , x2 =, , R,  383, M, , x1, 283, x2 = 383, , x2 =, , 383, 283, , 17. The degrees of freedom of a triatomic gas is (consider moderate temperature), (1) 6, , (2) 4, , (3) 2, , (4) 8, , Sol. Answer (1), A non-linear triatomic gas has 3 translators and 3 rotatory degrees of freedom., 18. 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), Number of moles (n) =, , 5, 32, , –PV = nRT, PV =, , 5, RT, 32, , SECTION - D, Assertion-Reason Type Questions, 1., , A : For a real gas internal energy depends on its temperature as well as volume also., R : For a real gas interatomic potential energy depends on volume and kinetic energy depends on temperature., , Sol. Answer (1), The assertion is correct and the reason is the 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

Page 29 :
Solution of Assignment, , 2., , Kinetic Theory, , 29, , A : The gravitational force between the gas molecules is ineffective due to extremely small size and very high, speed., R : No force of interaction acts between molecules of an ideal gas., , Sol. Answer (2), Both the statements are true but reason is the incorrect explanation of assertion. The assertion happens to, be self explanatory., 3., , A : Average velocity of gas molecules is zero., R : Due to random motion of gas molecules, velocities of different molecules cancel each other., , Sol. Answer (1), The assertion is true but reason is correct explanation of assertion., 4., , A : At constant volume on increasing temperature the collision frequency increases., R : Collision frequency  temperature of gas., , Sol. Answer (3), The assertion is true but collision frequency does not increase linearly with temperature., 5., , A : Two gases with the same average translational kinetic energy have same temperature even if one has greater, rotational energy as compared to other., R : Only average translational kinetic energy of a gas contributes to its temperature., , Sol. Answer (1), The assertion is true and reason is correct explanation of assertion., 6., , A : All molecular motion ceases at –273.15°C., R : Temperature below –273.15°C cannot be attained., , Sol. Answer (2), Both the statements are correct but reason is not the correct explanation for assertion., 7., , A : Magnitude of mean velocity of the gas molecules is same as their mean speed., R : The only difference between mean velocity and mean speed is that mean velocity is a vector and mean, speed is a scalar., , Sol. Answer (4), Mean velocity of a gas is not the same as mean speed. Hence both the statements are correct., 8., , A : Mean free path of gas molecules varies inversely as density of the gas., R : Mean free path varies inversely as pressure of the gas., , Sol. Answer (2), Mean free path is given by =, , V, 2N d2, , Where, N is total number of molecules., V is volume., d is the diameter of molecule., N, is the number velocity of gas hence assertion is true. But the mean free path does not depend on pressure., V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 30 :
30, 9., , Kinetic Theory, , Solution of Assignment, , A : Number of air molecules in a room in winter is more than the number of molecules in the same room in, summer., R : At a given pressure and volume, the number of molecules of a given mass of a gas is directly proportional, to the absolute temperature., , Sol. Answer (3), The assertion is true as at a lower temperature there is a higher density., According to PV = nRT, , , , 1, n, , Hence reason is false., 10. A : Evaporation occurs at any temperature whereas the boiling point depends on the external pressure., R : Evaporation of a liquid occurs from the surface of a liquid at all temperature whereas boiling takes place, at a temperature determined by the external pressure., Sol. Answer (1), The assertion is true and the reason is the correct explanation of the assertion., , , , , , , , 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

Page 31 :
Chapter, , 14, , Oscillations, Solutions, SECTION - A, Objective Type Questions, 1., , If a particle is executing simple harmonic motion, then acceleration of particle, (1) Is uniform, , (2) Varies linearly with time, , (3) Is non uniform, , (4) Both (2) & (3), , Sol. Answer (3), If a particle is executing S.H.M., a – 2x, Hence it is not uniform and depends on x rather than time., Hence answer is (3)., 2., , What is the phase difference between acceleration and velocity of a particle executing simple harmonic motion?, (1) Zero, , (2), , , 2, , (3) , , (4) 2, , Sol. Answer (2), v = A cos (t + ), and, , a = – A2 sin (t + ), , ∵, , cos(t +  + /2) = – sin (t + ), a = A2 cos (t +  + /2), , Hence velocity lags /2 with acceleration., 3., , The shape of graph plotted between velocity and position of a particle executing simple harmonic motion is, (1) A straight line, , (2) An ellipse, , (3) A parabola, , (4) A hyperbola, , Sol. Answer (2), 2, 2, v =  A x, , v2,  x 2  A2, 2, , , , , v2, x2, , 1, 2 A2 A2, , This is the equation of an ellipse., Hence answer is (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 32 :
32, 4., , Oscillations, , Solution of Assignment, , If particle is executing simple harmonic motion with time period T, then the time period of its total mechanical, energy is, , (1) Zero, , (2), , T, 2, , (3) 2T, , (4) Infinite, , Sol. Answer (4), The total mechanical energy doesn't change in an undamped S.H.M.  frequency = 0., Hence time period is infinite., 5., , Identify the correct definition, (1) If after every certain interval of time, particle repeats its motion then motion is called periodic motion, (2) To and fro motion of a particle over the same path about its mean position in certain time interval is called, oscillatory motion, (3) Oscillatory motion described in terms of single sine and cosine functions is called simple harmonic motion, (4) All of these, , Sol. Answer (4), All the above definition are true., Hence answer is (4), 6., , The equation of simple harmonic motion may not be expressed as (each term has its usual meaning), (1) x = Asin(t + ), , (2) x = Acos(t – ), , (3) x = asin t + bcos t, , (4) x = Asin (t + ) + Bsin (2t + ), , Sol. Answer (4), The fourth option is a superposition of two S.H.M.'s will different frequencies and time periods., Hence it is not a true S.H.M., 7., , Select wrong statement about simple harmonic motion, (1) The body is uniformly accelerated, (2) The velocity of the body changes smoothly at all instants, (3) The amplitude of oscillation is symmetric about the equilibrium position, (4) The frequency of oscillation is independent of amplitude, , Sol. Answer (1), In S.H.M.., a = – 2x, Thus acceleration varies linearly with time., 8., , The displacement of a particle executing S.H.M. is given by x = 0.01 sin 100(t + 0.05). The time period is, (1) 0.01 s, , (2) 0.02 s, , (3) 0.1 s, , (4) 0.2 s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 33 :
Solution of Assignment, , Oscillations, , 33, , Sol. Answer (2), x = 0.01 sin 100(t + 0.05), Here  = 100 , T=, , 9., , 2, 2, 1, , , = 0.02 s,  100 50, , A particle moves under force F = –5(x – 2)3. Motion of the particle is, (1) Translatory, , (2) Oscillatory, , (3) S.H.M., , (4) All of these, , Sol. Answer (2), F = –5(x – 2)3, Acceleration is negative when (x – 2) is positive and positive when (x – 2) is negative., Hence it is an oscillatory motion about x = 2., 10. For a particle showing motion under the force F = –5(x – 2)2, the motion is, (1) Translatory, , (2) Oscillatory, , (3) S.H.M., , (4) All of these, , Sol. Answer (1), F = –5(x – 2)2, The motion depicts a non uniform translatory motion as the acceleration just keeps increasing in the negative, direction. This is because (x – 2) is always positive., 11. For a particle showing motion under the force F = –5(x – 2), the motion is, (1) Translatory, , (2) Oscillatory, , (3) S.H.M., , (4) Both (2) & (3), , Sol. Answer (4), Force varies linearly with time with respect to – (x – 2), F  – (x – 2), Hence motion of the particle is an S.H.M., 12. A boy is swinging in a swing. If he stands, the time period will, (1) First decrease, then increase, , (2) Decrease, , (3) Increase, , (4) Remain same, , Sol. Answer (2), A swing is like a pendulum. So, , T = 2, , l, g, , When the boy stands the C.O.M. will become higher. Thus ‘l’ will become shorter and so according to the, equation of time period. Time period will decrease., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 34 :
34, , Oscillations, , Solution of Assignment, , 13. Time period of a simple pendulum in a freely falling lift will be, (1) Finite, , (2) Infinite, , (3) Zero, , (4) All of these, , Sol. Answer (2), , T, , l, g, , In a freely falling lift g = 0., Hence T = infinite., 14. If effective length of a simple pendulum is equal to radius of earth (R), time period will be, (1) T  , , R, g, , (2) T  2, , 2R, g, , (3) T  2, , R, g, , (4) T  2, , R, 2g, , Sol. Answer (4), T = 2, , 1, 1 1, g ⎛⎜  ⎞⎟, ⎝R l ⎠, , ∵ l = R  T = 2, , R, 2g, , 15. A particle is executing S.H.M. with time period T. If time period of its total mechanical energy is T then, , (1) 2, , (2), , 1, 2, , (3) Zero, , T, is, T, , (4) Infinite, , Sol. Answer (4), , , Total mechanical energy will never change so T     T  , T, 16. A body executing S.H.M. along a straight line has a velocity of 3 ms–1 when it is at a distance of 4 m, from its mean position and 4 ms–1 when it is at a distance of 3 m from its mean position. Its angular frequency, and amplitude are, (1) 2 rad s–1 & 5 m, , (2) 1 rad s–1 & 10 m, , (3) 2 rad s–1 & 10 m, , (4) 1 rad s–1 & 5 m, , Sol. Answer (4), 2, 2, v =  A x, , v1 = 3 m/s, , x1 = 4 m, , v2 = 4 m/s, , x2 = 3 m, , 3 =  A2  42, , ... (i), , 4 =  A2  32, , ... (ii), , Solving (i) and (ii), we get, A = 5 m and  = 1 rad/s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 35 :
Solution of Assignment, , Oscillations, , 35, , 17. Two identical springs have the same force constant 73.5 Nm–1. The elongation produced in each spring in three, cases shown in Figure-1, Figure-2 and Figure-3 are (g = 9.8 ms–2), , (1), , 1, 2, 1, m, m, m, 6, 3, 3, , (2), , 5 kg, , 5 kg, , 5 kg, , Figure 1, , Figure 2, , Figure 3, , 1, 1, 1, m, m, m, 3, 3, 3, , (3), , 2, 1, 1, m, m, m, 3, 3, 6, , (4), , 1, 4, 2, m, m, m, 3, 3, 3, , Sol. Answer (4), k = 73.5 Nm–1, , Force = 5 × 9.8, , In figure (1), 5 × 9.8 = (2 k) x1,  x1 =, , 5  9.8, 1, , 2  73.5 3, , In figure (2), , 5 × 9.8 =, , or 5 × 9.8 =, , x2 =, , k k,  x2, k k, k,  x2, 2, , 98, 4, , 73.5 3, , In figure (3), 5 × 9.8 = kx3, x3 =, , 5  9.8 2, , 73.5, 3, , 18. The frequency of oscillation of a mass m suspended by a spring is 1. If length of spring is cut to one third, then the same mass oscillates with frequency 2, then, (1) 2 = 31, , (2) 32 = 1, , (3)  2  3 1, , (4), , 3  2  1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 36 :
36, , Oscillations, , Solution of Assignment, , Sol. Answer (3), old =, , kold, m, , When divided into 3 parts the spring constant of smaller parts,  kfinal = 3kold,  final =, , 3 old, ,  = 2 , Hence final =, , 3 old ⇒ 2  31, , 19. Two particles executing S.H.M. of same frequency, meet at x = +A/2, while moving in opposite directions. Phase, difference between the particles is, (1), , , 6, , (2), , , 3, , (3), , 5, 6, , (4), , 2, 3, , Sol. Answer (4), x = A sin t, When displacement, x=, , A, 2, , A, = A sin (t + ), 2, , sin–1, , 1, = t + , 2, , t + = 30° or 150°, When particles are in opposite direction at one lime phase is 30° and at the other 150°. So phase difference, is 120°., 20. A particle is executing S.H.M. with time period T. Starting from mean position, time taken by it to complete, 5, oscillations, is, 8, (1), , T, 12, , (2), , T, 6, , (3), , 5T, 12, , (4), , 7T, 12, , Sol. Answer (4), Total distance covered by particle = 4 A, T, 5, 1, of oscillation means that it has completed, the oscillation taking, seconds. Now it has to cover, 2, 8, 2, 1, A, A, oscillation more. The whole path may be divided into 8 parts of, hence it has to travel, distance from, 8, 2, 2, mean position., , For, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 37 :
Solution of Assignment, , Oscillations, , 37, , A, = A sin t, 2, , = 8 t, 6, , 2 ⎤, ⎡, ⎢⎣Putting   T ⎥⎦, , T, 12, , t=, , Total time =, , T T, 7T, , , 2 12 12, , 21. A particle is executing S.H.M.. between x = ± A. The time taken to go from 0 to, , A, A, is T1 and to go from, 2, 2, , to A is T2 ; then, (1) T1 < T2, , (2) T1 > T2, , (3) T1 = T2, , (4) T1 = 2T2, , Sol. Answer (1), The velocity is greater closer to the mean position so it will take less time gains from 0 to, , A, A, than from, 2, 2, , to A., ⎛, ⎞, 22. The displacements of two particles executing S.H.M. on the same line are given as y 1  a sin ⎜ t   ⎟ and, 2, ⎝, ⎠, ⎛ 2, ⎞, y 2  b sin ⎜, t   ⎟. The phase difference between them at t = 1 s is, ⎝ 3, ⎠, , (1) , , (2), , , 2, , (3), , , 4, , (4), , , 6, , Sol. Answer (4), Phase difference between them is just difference in the angular values., , ⎛ 2, ⎞ ⎛, ⎞ ,  ⎟  ⎜  ⎟ , Phase difference = ⎜, ⎠ 6, ⎝ 3, ⎠ ⎝2, 23. For a particle executing simple harmonic motion, the amplitude is A and time period is T. The maximum speed, will be, (1) 4AT, , (2), , 2A, T, , (3) 2, , A, T, , (4), , 2A, T, , Sol. Answer (4), Maximum speed is given by, v = A, and, , =, , Hence v =, , 2, T, 2, A, T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 38 :
38, , Oscillations, , Solution of Assignment, , 24. A particle is executing S.H.M.. with amplitude A and has maximum velocity v0. Its speed at displacement, , 3A, 4, , will be, , (1), , 7, v0, 4, , (2), , v0, , (3) v0, , 2, , (4), , 3, v0, 2, , Sol. Answer (1), v =  A2  x 2, x=, , 3, A, 4, , v =  A2 , , v = A, , 9 A2, 16, , 7, 16, 7, as (v0 = A), 4, , or v = v 0, , 25. A particle executes simple harmonic motion according to equation 4, , d 2x, dt 2, , + 320x = 0. Its time period of, , oscillation is, (1), , 2, , s, , 5 3, , (2), , , 3 2, , s, , (3), , , 2 5, , s, , (4), , 2, 3, , s, , Sol. Answer (3), , 4, , d2x, dt 2, , + 320x = 0, , 4 a = – 320 x, a = – 80 x, Since a = –2 x in S.H.M., 80 = 2, 16  5 = , , or  = 4 5, , T=, , , 2, 2, , , s,  4 3 2 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 39 :
Solution of Assignment, , Oscillations, , 39, , 26. The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in figure., The time period of oscillation of particle is, , v(m/s), 0.4, , –10, , 0, , 10, , x(cm), , –0.4, (1), , , s, 2, , (2)  s, , (3) 2 s, , (4) 3 s, , Sol. Answer (1), A = 10 cm, , A = 0.4 m/s, , = 0.1 m,   = 4 rad/s, , T=, , 2 ,  s, 4, 2, , 27. A particle of mass 10 g is undergoing S.H.M. of amplitude 10 cm and period 0.1 s. The maximum value of, force on particle is about, (1) 5.6 N, , (2) 2.75 N, , (3) 3.5 N, , (4) 4 N, , Sol. Answer (4), T = 0.1, , m = 0.01 kg, ,   = 20  rad/s, Amplitude A = 0.1 m, a = – 2x, Maximum acceleration = –2 A, Maximum force = – m2 A, Fmax = – 0.01 × (20 )2 × 0.1, Fmax = – 0.001 × 400 2, = –3.95 N, or –4 N approximately, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 40 :
40, , Oscillations, , Solution of Assignment, , 28. Two identical pendulums oscillate with a constant phase difference, , , and same amplitude. If the maximum, 4, , velocity of one is v, the maximum velocity of the other will be, (1) v, , (2), , 2v, , (3) 2v, , (4), , v, 2, , Sol. Answer (1), If the phase difference is constant, they are moving with same frequency and ., Since maximum velocity A = v (given), Maximum velocity of other will still be v., 29. Which of the following graphs best represents the variation of acceleration ‘a’ with displacement x?, , a, x, , (1), , (2), , a, , a, , a, x, , x, , (3), , (4), , x, , Sol. Answer (2), a = – kx is an S.H.M., Hence it will be a straight line with negative slope as in option (2), 30. A body executes S.H.M. with an amplitude A. At what displacement from the mean position, is the potential, energy of the body one-fourth of its total energy?, (1), , A, 4, , (2), , (3), , 3A, 4, , (4) Some other fraction of A, , A, 2, , Sol. Answer (2), Potential energy at displacement x from mean position is given by, P.E. =, , 1 2, kx, 2, , Let total energy be E =, , 1 2, kA, 2, , 1 2 E 1 2, kx =,  kA, 2, 4 8, , x=, , A, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 41 :
Solution of Assignment, , Oscillations, , 41, , 31. A particle of mass 4 kg moves simple harmonically such that its PE(U) varies with position x, as shown. The, period of oscillations is, joule, , (1), , 2, s, 25, , (2), ,  2, s, 5, , (3), , 4, s, 5, , (4), , 2 2, s, 5, , Sol. Answer (4), Mass = 4 kg, , joule, , 1 2, Maximum P.E. = kA, 2, , 1=, , 1,  k  (0.2)2, 2, , 2, = k, 0.04, , k = 50 N/m, T = 2, , m, k, , T = 2, , 4, 50, , =, , 2 2, s, 5, , 32. The kinetic energy and potential energy of a particle executing S.H.M. are equal, when displacement in terms, of amplitude ‘A’ is, (1), , A, 2, , (2), , A, 2, , (3), , A 2, 3, , (4) A 2, , Sol. Answer (2), Total energy =, , 1 2, kA, 2, , When P.E. is half of total energy P.E. = K.E., , , 1 2 1, kx  k ( A2  x 2 ), 2, 2, , x2 = A2 – x2  2x2 = A2, x=, , A, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 42 :
42, , Oscillations, , Solution of Assignment, , 33. A block is resting on a piston which executes simple harmonic motion with a period 2.0 s. The maximum, velocity of the piston, at an amplitude just sufficient for the block to separate from the piston is, (1) 1.57 ms–1, , (2) 3.12 ms–1, , (3) 2.0 ms–1, , (4) 6.42 ms–1, , Sol. Answer (2), Period (T) = 2 s, =, , 2, =  rad/s, 2, , When block just represent from a piston, maximum acceleration must be equal to g., g = – 2x, Acceleration is maximum when x = A, g = – 2A, or A =, , 9.8, 2, , Maximum velocity = A, =, , 9.8, , 2, , =, , 9.8, m/s, , , = 3.119 m/s = 3.12 m/s, 34. A simple pendulum suspended from the ceiling of a stationary lift has period T0. When the lift descends at, steady speed, the period is T1, and when it descends with constant downward acceleration, the period is T2., Which one of the following is true?, (1) T0 = T1 = T2, , (2) T0 = T1 < T2, , (3) T0 = T1 > T2, , (4) T0 < T1 < T2, , Sol. Answer (2), Pseudo force only when there is an acceleration., Hence T0 = T1 as there is uniform downward motion., When it moves downward with a steady acceleration then pseudo force acts upwards, reducing net ‘g’, Since T2 = 2, , l, gnet, , When g reduces time period T2 increases, T0 = T1 < T2, 35. If a Second’s pendulum is moved to a planet where acceleration due to gravity is 4 times, the length of the, second’s pendulum on the planet should be made, (1) 2 times, , (2) 4 times, , (3) 8 times, , (4) 15 times, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 43 :
Solution of Assignment, , Oscillations, , 43, , Sol. Answer (2), Time period of a pendulum, T = 2, , l, l, or T2, g, g, , If g becomes 4 times. l must also be increased by 4 times to keep T constant., 36. A simple pendulum with a metallic bob has a time period T. The bob is now immersed in a, non-viscous liquid and oscillated. If the density of the liquid is 1/4 that of metal, the time period of the same, pendulum will be, (1), , T, , (2), , 3, , 2T, , (3), , 3, , 4, T, 3, , (4), , 2, T, 3, , Sol. Answer (2), , l, g, , Normal time period T = 2, , When immersed in a liquid. It experiences an upthrust., , × volume g, 4, , Upthrust =, , Upward acceleration = Upward force/mass of ball =, T = 2, , g, 4, , l, geff, , geff = g , , g 3,  g, 4 4, , l, 2T, 4 , 3g, 3, , T = 2 , , 37. Two pendulums of length 1.21 m and 1.0 m start vibrating. At some instant, the two are in the mean position, in same phase. After how many vibrations of the longer pendulum, the two will be in phase?, (1) 10, , (2) 11, , (3) 20, , (4) 21, , Sol. Answer (1), Lengths l1 = 1.21 m, T1 = 2, , T1 =, , l1, g, , l2 = 1 m, T2 = 2, , l2, g, , 11, T, 10 2, , or 10 T1 = 11T2, Hence it oscillations of longer pendulum is equal to 11 oscillation of shorter one., Hence the will be in phase again after 10 oscillations of longer pendulum., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 44 :
44, , Oscillations, , Solution of Assignment, , 38. The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44%. then its, new time period of oscillation will be, (1) 96 s, , (2) 58 s, , (3) 82 s, , (4) 72 s, , Sol. Answer (4), Let initial length be l1, Final length l2 = l1 , , 144, 100, , T1 = 2, , l1, g, , T2 = 2, , l1 144, , g 100, , or T2 = 1.2 T1, T1 = 60 s, So T2 = 72 s, 39. If the length of a clock pendulum increases by 0.2% due to atmospheric temperature rise, then the loss in, time of clock per day is, (1) 86.4 s, , (2) 43.2 s, , (3) 72.5 s, , (4) 32.5 s, , Sol. Answer (1), Time period = 2, , l, g, , T l, T, , T, , T = T, , l, l, l  l  , l, , ⎛ 1, ⎞, T = T ⎜1    ⎟ [  0.002], ⎝ 2, ⎠, T = T – T =, , 1, T    T  0.001, 2, , Time lost in time t is, T =, , 1, 2, , t = 1 day = 24 × 3600 s = 86400 s, , ⎛ T ⎞  t, ⎟, T = ⎜, ⎝ T ⎠, T = 0.001 × 86400, T = 86.4 s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 45 :
Solution of Assignment, , Oscillations, , 45, , 40. A simple pendulum is oscillating in a trolley moving on a horizontal straight road with constant acceleration, a. If direction of motion of trolley is taken as positive x direction and vertical upward direction as positive, y direction then the mean position of pendulum makes an angle, , ⎛g⎞, (1) tan1 ⎜ ⎟ with y axis in +x direction, ⎝a⎠, , ⎛a⎞, (2) tan1 ⎜ ⎟ with y axis in –x direction, ⎝g⎠, , ⎛a⎞, (3) tan1 ⎜ ⎟ with y axis in +x direction, ⎝g⎠, , ⎛g⎞, (4) tan1 ⎜ ⎟ with y axis in –x direction, ⎝a⎠, , Sol. Answer (2), T sin = ma, , ... (i), , T cos = mg, , ... (ii), , , , Dividing (i) and (ii), , T cos =, , mg, , a, g, , mg, , 1 ⎛ a ⎞,  = tan ⎜ ⎟, ⎝g ⎠, , 41. The time period of oscillations of a second’s pendulum on the surface of a planet having mass and radius double, those of earth is, (1) 4 s, , (2) 1 s, , (3), , 2s, , (4) 2 2 s, , Sol. Answer (4), , g1 =, , Gm, R2, , g2 =, , G  2m g1, , 2, 4R 2, , l, T1 = 2 g, 1, , l, T2 = 2 g, 2, T2 =, , 2T1, , Since T1 is time period of seconds pendulum T1 = 2., Hence T2 = 2 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 46 :
46, , Oscillations, , Solution of Assignment, , 42. The shape of graph between time period of a simple pendulum and its length is, (1) Straight line, , (2) Parabolic, , (3) Hyperbolic, , (4) Elliptical, , Sol. Answer (2), 43. A hollow metal sphere is filled with water through a small hole in it. It is hung by a long thread, and is made to oscillate. Water slowly flows out of the hole at the bottom. Select the correct variation of its, time period, (1) The period will go on increasing till the sphere is empty, (2) The period will go on decreasing till the sphere is empty, (3) The period will not be affected at all, (4) The period will increase first, then decrease to initial value till the sphere is empty, Sol. Answer (4), , P, , As the water level goes down, the distance of C.O.M., From point of oscillation P keeps increasing., Since T = 2, , l, g, , The time period of pendulum keep increasing., 44. A uniform rod of mass m and length l is suspended about its end. Time period of small angular oscillations, is, , m,l, , (1) 2, , l, g, , (2) 2, , 2l, g, , (3) 2, , 2l, 3g, , (4) 2, , l, 3g, , Sol. Answer (3), This is the case of a physical pendulum., T = 2, , Lcom =, , T = 2, , l com, mg Lcom, L, 2, , Icom =, , mL2, 3, , 2l, 3g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 47 :
Solution of Assignment, , Oscillations, , 47, , 45. A uniform disc of mass M and radius R is suspended in vertical plane from a point on its periphery. Its time, period of oscillation is, (1) 2, , 3R, g, , (2) 2, , R, 3g, , (3) 2, , 2R, 3g, , (4) 2, , 3R, 2g, , Sol. Answer (4), It is the case of a physical pendulum., , l c.o.m., T = 2 mg L, com, Icom =, , 3, MR 2, 2,  MR 2 = MR, 2, 2, , Lcom = R, T = 2, , 3R, 2g, , 46. A solid cylinder of density 0 , cross-section area A and length l floats in a liquid of density  (> 0) with, its axis vertical, as shown. If it is slightly displaced downward and released, the time period will be, , , x, , (1) 2, , l, g, , (2) 2, , 0 l, g, , (3) 2, , l, 0 g, , (4) 2, , l, 2g, , Sol. Answer (2), The time period of a floating uniform cylinder is simply given as R = l., 47. A block of mass m hangs from three springs having same spring constant k. If the mass is slightly displaced, downwards, the time period of oscillation will be, , k, , k, k, m, (1) 2, , m, 3k, , (2) 2, , 3m, 2k, , (3) 2, , 2m, 3k, , (4) 2, , 3k, m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 48 :
48, , Oscillations, , Solution of Assignment, , Sol. Answer (2), The first two springs are in parallel., So, keq of 1st 2 will be = 2k, Then it becomes, The springs 2k and k are in series., , 2k, , 2k  k, 2k  k, , So, keq =, , k, , 2 k k 2,  k, 3k, 3, , =, , m, , m, T = 2 k, eq, , , , T  2, , 3m, 2k, , 48. Two masses m1 = 1 kg and m2 = 0.5 kg are suspended together by a massless spring of spring constant, 12.5 Nm–1. When masses are in equilibrium m1 is removed without disturbing the system. New amplitude of, oscillation will be, (1) 30 cm, , (2) 50 cm, , (3) 80 cm, , (4) 60 cm, , Sol. Answer (3), Points of equilibrium of the spring will be when no force acts on it., or kx = (m1 + m2)g, x=, , (m1  m2 )g, k, , The new equilibrium position which will be the mean position of S.H.M. will be simply, , New amplitude will be maximum displacement from, , A=, , m2g, k, , m2g, which is :, k, , (m1  m2 )g m2g, –, k, k, , or A =, , m1g, k, , or A =, , 1 10, 12.5, , or A =, , 4, m, 5, ,  A = 0.8 m or 80 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 49 :
Solution of Assignment, , Oscillations, , 49, , 49. A mass m is attached to two springs of same force constant K, as shown in following four arrangements. If, T1, T2, T3 and T4 respectively be the time periods of oscillation in the following arrangements, in which case, time period is maximum?, , K, , (a), , K, , K, (b), , T1, , (c), , T2, , m, (1) (a), , K, , K, , m, , K, , K, , (d), , K, , m, , m, , (2) (b), , (3) (c), , (4) (d), , Sol. Answer (2), , m, K, , T = 2, , Time period is maximum when K is minimum., In (a), (c) and (d) the spring constants are in parallel so the Keq = 2K., Only in case (b) springs are in series., K, 2, , So, Keq =, , Hence time period in this case will be maximum., 50. A clock S is based on oscillations of a spring and a clock P is based on pendulum motion. Both clocks run at, the same rate on earth. On a planet having same density as earth but twice the radius then, (1) S will run faster than P, , (2) P will run faster than S, , (3) Both run at same rate, , (4) Both run at same rate but different than earth, , Sol. Answer (2), Time period of spring = 2, , k, m, , Time period of pendulum = 2, , l, g, , Time period of spring will not be affected by gravitational acceleration., Let mass of earth be m, 4, 3, Mass of new planet =   (2R ) = 8 m, 3, , g2 =, , GM2, , 2, , (R2 ), , T2 = 2, , T2 =, , , , G  8M, =2g, (2R )2, , l, 2g, , T, 2, , Hence P will move faster., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 50 :
50, , Oscillations, , Solution of Assignment, , 51. A 100 g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass, must be attached to the spring if the period of vibration is to be 6.28 s?, (2) 105 g, , (1) 1000 g, , (3) 107 g, , (4) 104 g, , Sol. Answer (4), At point of equilibrium kx = mg, k × 9.8 × 10–2 = 100 × 10–3 × 9.8, k = 100 × 10–1, k = 10 N/m, Period of vibration needed = 6.28 s, , T = 2, , m, k, , 6.28 = 2 × 3.14, , 1=, , m, 10, , m, 10, , m = 10 kg or 104 g, 52. An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. At this, instant, the springs are relaxed. The left mass is displaced to the left and the right mass is displaced to the, right by same distance and released. The resulting collision is elastic. The time period of the oscillations of, system is, , k, , (1) 2, , 2m, k, , (2) 2, , m, 2k, , m, , k, , m, , (3) , , m, k, , (4) 2, , m, k, , Sol. Answer (3), If there was no collision each spring will oscillate with period, T = 2, , m, k, , Because of collisions the springs are only compressed but cannot extend beyond their natural length. Hence, the perform only half oscillation., Hence, , T = 2, , or T = , , m, 2, k, , m, k, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 52 :
52, , Oscillations, , Solution of Assignment, , 55. In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half, of its maximum value in 1 s, then after 2 s amplitude will be (A0– initial amplitude), , (1), , 1, A0, 4, , (2), , 1, A0, 2, , (3) A0, , (4), , 3 A0, 2, , Sol. Answer (1), A = A0 e–bt, Amplitude becomes half hence, A0, A e–bt [t = 1], 2 0, ,  e–b =, , 1, 2, ,  In two seconds, ⎛ 1⎞, A = A0 ⎜ ⎟, ⎝2⎠, , A=, , 2, , A0, 4, , 56. In forced oscillations, a particle oscillates simple harmonically with a frequency equal to, (1) Frequency of driving force, (2) Natural frequency of body, (3) Difference of frequency of driving force and natural frequency, (4) Mean of frequency of driving force and natural frequency, Sol. Answer (1), In forced oscillations a particle oscillator simple harmonically with a frequency equal to driving frequency., 57. Resonance is a special case of, (1) Forced oscillations, , (2) Damped oscillations, , (3) Undamped oscillations, , (4) Coupled oscillations, , Sol. Answer (1), Resonance is a special case of force oscillation due to which oscillation to place with greater amplitude., 58. The S.H.M. of a particle is given by the equations = 2 sin t + 4 cos t. Its amplitude of oscillation is, (1) 4 units, , (2) 2 units, , (3) 6 units, , (4) 2 5 units, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 53 :
Solution of Assignment, , Oscillations, , 53, , Sol. Answer (4), x = 2 sin t + 4 cos t, It can also be written as, x=, , 2, 4, ⎛, ⎞, sin t, cos t ⎟, 22  42 ⎜, 2, 2, 2, 2, 2 4, ⎝ 2 4, ⎠, , x=, , 20 sin(t  ), , 20 = Amplitude, , or Amplitude = 2 5, 59. If two mutually perpendicular simple harmonic motion of same amplitude, frequency and having zero phase, difference superimpose on a particle, then its resultant path will be, (1) A straight line, , (2) A circle, , (3) An ellipse, , (4) A hyperbola, , Sol. Answer (1), If A – B is the path followed by one particle superimpose, then the result will be as shown and the particle, will oscillate diagonally., , A, , B, , A, B, , 60. Which of the following represents a S.H.M.?, (1) sin t – cos t, , (2) sin t + cos t, , (3) sin t + 2 cos t, , (4) All of these, , Sol. Answer (4), All of them are superposition of two S.H.M. in the same phase and hence they all represent S.H.M's., , SECTION - B, Objective Type Questions, 1., , The circular motion of a particle with constant speed is, (1) Periodic but not simple harmonic, , (2) Simple harmonic but not periodic, , (3) Period and simple harmonic, , (4) Neither periodic nor simple harmonic, , Sol. Answer (1), The motion repeats itself after same intervals hence it is periodic. Put since acceleration is not proportional, to – x, the motion is periodic but not simple harmonic., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 54 :
54, 2., , Oscillations, , Solution of Assignment, , A body of mass 0.01 kg executes simple harmonic motion about x = 0 under the influence of a force as shown, in figure. The time period of S.H.M. is, F(N), , 80, –0.2, –80, , (1) 1.05 s, , (2) 0.52 s, , 0.2, x(m), , (3) 0.25 s, , (4) 0.03 s, , Sol. Answer (4), Maximum restoring force on particle, F = 80 N, x = – 0.2, Since F = – kx, 80 = k × 0.2, 400 = k, T = 2, , m, k, , = 2, , 0.01, kg, 400, , = 2, , 1, 40000, , = 2, , 1, 2  102, , =, , 1, 2  102, ,  0.03 s, 3., , A 1.00 × 10–20 kg particle is vibrating under simple harmonic motion with a period of 1.00 × 10–5 s and with, a maximum speed of 1.00 × 103 m/s. The maximum displacement of particle from mean position is, (1) 1.59 mm, , (2) 1.00 m, , (3) 10 m, , (4) 3.18 mm, , Sol. Answer (1), m = 1 × 10–20 kg, , T = 1 × 10–5 kg, , Maximum speed = A = 1 × 103 m/s, =, , ... (i), , 2, = 2 × 105 rad/s, T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 55 :
Solution of Assignment, , Oscillations, , 55, , Putting value of  in (i), A × 2 × 105 = 1 × 103, , A=, , 1, 2  102, , A = 1.59 mm, 4., , The equation of an S.H.M. with amplitude A and angular frequency  in which all the distances are measured, from one extreme position and time is taken to be zero at the other extreme position is, (1) x = A sin t, , (2) x = A (cos t + sin t), , (3) x = A – A cos t, , (4) x = A + A cos t, , Sol. Answer (4), At t = 0 the distance from 1 extreme is 2 A, At t = , , x=0, , Hence by resulting values we can get equation for S.H.M.. from S.H.M.., 5., , A body oscillates with S.H.M. according to the equation x = (5.0 m) cos [(2 rad s–1)t + /4], At t = 1.5 s, its acceleration is, (1) – 139.56 m/s2, , (2) 139.56 m/s2, , (3) 69.78 m/s2, , (4) – 69.78 m/s2, , Sol. Answer (2), x = 5 cos (2t +/4), , t =, , 3, s, 2, , x = 5 cos (3 +/4), , ⎛ 13 ⎞, ⎟, x = 5 cos ⎜, ⎝ 4 ⎠, , x = – 5 cos, , x=–, , , 4, , 5, 2, , Acceleration, a = – 2x, a = – 42x –, , a, , 5, 2, ,  139.56 m/s2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 56 :
56, 6., , Oscillations, , Solution of Assignment, , The time period of a particle executing S.H.M. is 8 s. At t = 0 it is at the mean position. The ratio of distance, covered by the particle in 1st second to the 2nd second is, , (1), , , , , , 2 –1 s, , (2), , 2s, , (3), , , , , , 2 1 s, , (4), , 1, 2, , s, , Sol. Answer (3), =, , T=8s, , x1 = A sin, , , A, , 4, 2, , x2 = A sin, , A, , , A, ( 2  1), × 2 – A sin, =A–, =, 4, 4, 2, 2, , x1, =, x2, , 7., , 2 , , 8, 4, , 1, 2 1, ,  2 1, 2 1, 2 1, , Two particle executing S.H.M. of same amplitude of 20 cm with same period along the same line about same, equilibrium position. The maximum distance between the two is 20 cm. Their phase difference in radian is equal, to, , (1), , , 3, , (2), , , 2, , (3), , 2, 3, , (4), , 4, 5, , Sol. Answer (1), x1 = A sin (t + 1), x2 = A sin (t + 2), x1 – x2 = A sin (t + 1) – A sin (t + 2), ⎡, ⎛  ⎞, ⎛    2 ⎞⎤, 20 = 2 × 20 sin ⎜ 1 2 ⎟  cos ⎢t  ⎜ 1, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠⎥⎦, ⎣, , 1, ⎛,  , ⎛   2 ⎞, ⎛ 1  2 ⎞ ⎞, , ,  sin⎜ 1, ⎟  cos ⎜ t  ⎜, ⎟ ⎟ for maximum value. ⇒ 1 2  ⇒ 1  2 , 2, ⎝ 2 ⎠, ⎝ 2 ⎠⎠, 2, 6, 3, ⎝, , 8., , A particle execute S.H.M. along a straight line. The amplitude of oscillation is 2 cm. When displacement of, particle from the mean position is 1 cm, the magnitude of its acceleration is equal to magnitude of its velocity., The time period of oscillation is, , (1), , 2, 2, , (2), , 2, 2, , (3), , 2, 3, , (4), , 3, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 57 :
Solution of Assignment, , Oscillations, , 57, , Sol. Answer (3), A = 2 cm = 2 × 10–2, a = –2x, 2, 2, and v =  A  x, , 4, 2 × 1 × 10–2 =  (4  1)10, ,  × 1 × 10–2 = 3 × 10–2, = 3, T=, , 9., , 2 2, , , 3, , Figure shows the position-time graph of an object in S.H.M. The correct equation representing this motion is, , x(cm), 4, 2, 0, , 5, , 11, , t(s), , –4, ⎞, ⎛ 2, (1) 2 sin⎜ t  ⎟, 6⎠, ⎝ 5, , ⎞, ⎛, (2) 4 sin⎜ t  ⎟, 6⎠, ⎝5, , ⎞, ⎛, (3) 4 sin⎜ t  ⎟, 3⎠, ⎝6, , ⎞, ⎛, (4) 4 sin⎜ t  ⎟, 6⎠, ⎝6, , Sol. Answer (4), Time period is 12 s from diagram., =, , 2 , , 12 6, , Amplitude A = 4, Initial phase is determined by putting known values in the equation., , ⎛, ⎞, 2 = 4 sin⎜ t   ⎟, ⎝6, ⎠, 1, sin1 =  [t = 0], 2, , = , 6, , ⎞, ⎛, Hence equation is x = ⎜ t  ⎟, 6⎠, ⎝6, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 58 :
58, , Oscillations, , Solution of Assignment, , ⎤, ⎡, 10. A particle executes S.H.M. according to equation x = 10 (cm) cos ⎢2t  ⎥ , where t is in second. The, 2⎦, ⎣, magnitude of the velocity of the particle at t =, (1) 24.7 cm/s, , (2) 20.5 cm/s, , 1, s will be, 6, , (3) 28.3 cm/s, , (4) 31.4 cm/s, , Sol. Answer (4), , ⎤, ⎡, x = 10 cos ⎢2t  ⎥, ⎣, 2⎦, , At t =, , 1, s, 6, , ⎡ ⎤, x = 10 cos ⎢  ⎥, ⎣2 3 ⎦, , x = –10 sin, , , 3, , x = 5 3, v =  A2  x 2, v = 2 100  75, v = 10 , or v = 31.4 cm/s, 11. A particle executes S.H.M. and its position varies with time as x = A sint. Its average speed during its motion, from mean position to mid-point of mean and extreme position is, (1) Zero, , (2), , 3A, , , (3), , A, 2, , (4), , 2A, , , Sol. Answer (2), Phase at mean position = 0, Phase at mid point, , A, = A sin , 2,  =, , , 6, , Time it takes to travel a phase difference of , t=, , 2 , ,  2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 59 :
Solution of Assignment, , or t =, , , , , or t =, , , 6, , Average speed =, , Oscillations, , 59, , Total distance, Time taken, , =, , A/2,  / 6, , =, , 3A, , , 12. A particle of mass m in a unidirectional potential field have potential energy U(x) =  + 2x2, where  and , are positive constants. Find its time period of oscillation., , (1) 2, , 2, m, , (2) 2, , m, 2, , (3) , , m, , , (4) , , , m, , Sol. Answer (3), U(x) =  + 2x2, , F=–, , dU ( x ), dx, , F = – 4x, , T = 2, , m, k, , T = 2, , m, 4, , T= , , cos [k = ], , m, , , 13. A particle is executing S.H.M. and its velocity v is related to its position (x) as v2 + ax2 = b, where a and b, are positive constants. The frequency of oscillation of particle is, , (1), , 1 b, 2 a, , (2), , a, 2, , (3), , b, 2, , (4), , 1 a, 2 b, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 60 :
60, , Oscillations, , Solution of Assignment, , Sol. Answer (2), v2 + ax2 = b, v2 = b – ax2, , ⎛b, ⎞, v2 = a ⎜  x 2 ⎟, ⎝a, ⎠, Comparing it to equation, v2 = 2 (A2 – x2), =, , f=, , a, , , a, , 2 2, , 14. A loaded vertical spring executes S.H.M. with a time period of 4 s. The difference between the kinetic energy, and potential energy of this system varies with a period of, (1) 2 s, , (2) 1 s, , (3) 8 s, , (4) 4 s, , Sol. Answer (1), 2 , , rad/s, T, 2, , T=4s, , =, , x = A sin t, , v = A cos t, , K.E. =, , 1, mv2, 2, , 1, mA22 cos2t, 2, , ... (i), , P.E. =, , 1 2, kx, 2, , 1 2, kA sin2t, 2, , ... (ii), , =, , k, m, , or k = m2 putting this value in (ii)., , K.E. – P.E. =, , 1, m2A2 (cos2t – sin2t), 2, , K.E. – P.E. = Emax. (cos 2t), Hence time period of difference of K.E. and P.E. is, , T=, , 2, 2, , or T = 2 s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 61 :
Solution of Assignment, , 61, , Oscillations, , 15. As a body performs S.H.M., its potential energy U varies with time t as indicated in, , U, , U, t, , (1), , U, , U, t, , (2), , t, , (3), , (4), , t, , Sol. Answer (1), , 1 2, 1, Potential energy = kx = A22sin2(t +), 2, 2, The graph for sin2(t +) is given by (1)., 16. A particle is performing S.H.M. with energy of vibration 90 J and amplitude 6 cm. When the particle reaches, at distance 4 cm from mean position, it is stopped for a moment and then released. The new energy of, vibration will be, (1) 40 J, , (2) 50 J, , (3) 90 J, , (4) 60 J, , Sol. Answer (1), Energy = 90 J, Maximum energy =, , or m2 =, , Amplitude = 6 cm, 1, mA22 = 90, 2, , 180, 36  10 4, ,  m2 = 30 × 102, When particle is stopped the point where it is stopped is the new amplitude but angular velocity will remain, same., E=, , 1, mA222, 2, , or E = 3000 A22, A2 = 4 × 10–2, A2 = 3000, 17. The variations of potential energy (U) with position x for three simple harmonic oscillators A, B and C are shown, in figure. The oscillators have same mass. The time period of oscillation is greatest for, , U, , A, (1) A, , (2) B, , U, , x, , B, , U, , x, (3) C, , C, , x, (4) Same for all, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 62 :
62, , Oscillations, , Solution of Assignment, , Sol. Answer (3), , or, , U=, , 1 2, kx, 2, , x2 =, , 2U, k, , x, , 1, (Since U is constant), k, , Also T = 2, , or, , T, , m, k, , 1, k, , Therefore x  T, Hence the oscillation with maximum x will have the maximum time period., 18. If the particle repeats its motion after a fixed time interval of 8 s then after how much time its maximum value, of PE will be attained after attaining its minimum value?, (1) 2 s, , (2) 4 s, , (3) 8 s, , (4) 1 s, , Sol. Answer (1), T=8s, Maximum value of potential energy is reached two times per oscillation which is, , T, time away from mean, 4, , position which has minimum value at position., 19. A particle is executing S.H.M. with total mechanical energy 90 J and amplitude 6 cm. If its energy is somehow, decreased to 40 J then its amplitude will become, (1) 2 cm, , (2) 4 cm, , (3), , 8, cm, 3, , (4), , 4, cm, 3, , Sol. Answer (2), 1, mA22 = 90 J, 2, , m and  remaining same energy is reduced to 40 J., A12, A22, , or, , , , 9, 4, , A1 3, , A2 2, , or A2 = 4 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 63 :
Solution of Assignment, , Oscillations, , 63, , 20. A linear harmonic oscillator of force constant 6 × 105 N/m and amplitude 4 cm, has a total energy 600 J., Select the correct statement., (1) Maximum potential energy is 600 J, , (2) Maximum kinetic energy is 480 J, , (3) Minimum potential energy is 120 J, , (4) All of these, , Sol. Answer (4), k = 6 × 105 N/m, Amplitude = 4 cm = 4 × 10–2 m, , 1 2, 1, kx = × 6 × 105 × (4 × 10–2)2, 2, 2, or E = 480 J, Since energy of S.H.M. is 480 J and there is 600 J provided to the oscillation there must be 600 – 120 = 480 J, stored as energy., 21. A second’s pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is, (1) Moving down with uniform acceleration, (2) Moving around the earth in geostationary orbit, (3) Moving up with uniform velocity, (4) Moving up with uniform acceleration, Sol. Answer (1), T = 2, , l, g, , When the rocket is moving down with acceleration Pseudo force is acting upwards reducing effective gravitational, acceleration (g)., 22. The curve between square of frequency of oscillation and length of the simple pendulum is, (1) Straight line, , (2) Parabola, , (3) Ellipse, , (4) Hyperbola, , Sol. Answer (4), l = kT2, T = 2, , f=, , or f2 =, , l, g, , 1 g, 2 l, k, l, , This is the equation of hyperbola between f2 and l., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 64 :
64, , Oscillations, , Solution of Assignment, , 23. A simple pendulum of mass m executes S.H.M. with total energy E. If at an instant it is at one of extreme, positions, then its linear momentum after a phase shift of, , (1), , 2mE, , (2), , 3mE, 2, , , rad will be, 3, , (3) 2 mE, , (4), , 2mE, 3, , Sol. Answer (2), Energy = E0, After a phase shift of, , E = E0 cos2, , , 3, , , 3, , P 2 E0 3, , 2m, 4, 3 mE, 2, , P=, , 24. A small iron ball of mass m is suspended with the help of a massless rod of length L and is free to oscillate, in vertical plane. Its time period of oscillation is, (1) 2, , mL, 2g, , (2) 2, , mL, g, , (3) 2, , L, g, , (4) 2, , m, gL, , Sol. Answer (3), Since this is the case of a massless rod, the condition is same as that of a pendulum., T = 2, , L, g, , 25. A rectangular block of mass m and area of cross-section A floats in a liquid of density . If it is given a small, vertical displacement from equilibrium it undergoes oscillations with a time period T, then, (1) T , , 1, , (2) T  , , m, , (3) T , , 1, A, , (4) T , , 1, , , Sol. Answer (3), Time period of a floating cylinder is given by, T = 2, , L s, L g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 65 :
Solution of Assignment, , Oscillations, , 65, , When s is density of limit, L3 s, , T = 2, , L2L g, , Ms, AL g, , T = 2, , T , , 1, A, , 26. A body of mass 5 kg hangs from a spring and oscillates with a time period of 2 second. If the body is, removed, the length of the spring will decrease by, (1) glk metre, , (2) klg metre, , (3) 2 metre, , (4) g metre, , Sol. Answer (4), T = 2 sec, , Mass = 5 kg, , Spring constant = k,  = 1 rad/sec, Now  =, , k, m, , So, k = 5, Equilibrium position when it is oscillating is at, kx = mg, or, , x =, , mg, k, , When the mass is removed the spring will return to its natural length, which is x upwards., Since m = 5 and k = 5, x = g metre, 27. In the figure shown, there is friction between the blocks P and Q but the contact between the block Q and, lower surface is frictionless. Initially the block Q with block P over it lies at x = 0, with spring at its natural, length. The block Q is pulled to right and then released. As the spring - blocks system undergoes S.H.M., with amplitude A, the block P tends to slip over Q. P is more likely to slip at, , P, Q, , x = –A, (1) x = 0, , (2) x = +A, , x=0, , frictionless, x, x = +A, (3) x = , , A, 2, , (4) x = , , A, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 66 :
66, , Oscillations, , Solution of Assignment, , Sol. Answer (2), The block is most likely to slip when there is maximum acceleration., This happens when the blocks are at the extremities where the displacement is either +A or –A., , P, Q, , x, , x=0, 28. A flat horizontal board moves up and down under S.H.M. vertically with amplitude A. The shortest permissible, time period of the vibration such that an object placed on the board may not lose contact with the board is, , (1) 2, , g, A, , (2) 2, , A, g, , (3) 2, , 2A, g, , (4), , , 2, , A, g, , Sol. Answer (2), Maximum acceleration of the system (amax) = –2A, For a block to escape the board the acceleration must be equal to 9 at the top-most point., g = 2A, g, A, , =, , Time period =, , 2, A, =, , g, , 29. A simple pendulum with iron bob has a time period T. The bob is now immersed in a non-viscous liquid and, oscillated. If the density of liquid is, , (1) T, , 8, 7, , (2) T, , 1, th that of iron, then new time period will be, 12, 12, 13, , (3) T, , 12, 11, , (4) T, , 6, 5, , Sol. Answer (3), When bob is inserted, in liquid effective of is reduced because of the force of upthrust.,  is density of solid., S,  is density of liquid., L, V is volume of solid, and T = 2, , l, g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 67 :
Solution of Assignment, , 67, , l, , Tnew = 2, , gnew, l, , g Lg, S, , Tnew = 2, , l  12, g  11, , Tnew = 2, , or, , Oscillations, , 12, T, 11, , Tnew =, , 30. When a mass m is attached to a spring it oscillates with period 4 s. When an additional mass of 2 kg is, attached to a spring, time period increases by 1 s. The value of m is, (1) 3.5 kg, , (2) 8.2 kg, , (3) 4.7 kg, , (4) 2.6 kg, , Sol. Answer (1), 1 =, , k, m, , 2 =, , k, m2, , 1, 2 =, , m2, m, , 2, T, , Since  =, , T2, m2, =, T1, m, 2, , ⎛5⎞ = m  2, ⎜ ⎟, m, ⎝4⎠, , 25 m  2, =, 16, m, 25 m = 16 m + 32, 9 m = 32, m= 3, , 5, kg, 9, , m  3.5 kg, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 69 :
Solution of Assignment, , Oscillations, , 69, , (i) – (ii), So, V12  V22  2 ⎡ x22  x12 ⎤, ⎣, ⎦, 2 , , V12  V22, , , , V12  V22, , x22  x12, , x22, , T  2, , 3., , , , x12, , , , 2, T, , x22  x12, , V12  V22, , When two displacements represented by y1 = a sin(t) and y2 = b cos(t) are superimposed the motion is, [AIPMT - 2015], (1) Simple harmonic with amplitude, , (a  b ), 2, , (2) Not a simple harmonic, , (3) Simple harmonic with amplitude, , a, b, , (4) Simple harmonic with amplitude, , a2  b2, , Sol. Answer (4), , b, , Angle between a and b can be given 90º, so their resultant is, R  a2  b2, , a, , Frequency of y1 and y2 are similar so motion of particle is S.H.M., 4., , The oscillation of a body on a smooth horizontal surface is represented by the equation, x = Acos(t), where x is, displacement at time t and  is frequency of oscillation. Which one of the following graphs shows correctly the, variation a with t? (Here a = acceleration at time t and T = time period), [AIPMT - 2014], , a, (1), , a, , O, , T, , t, , (2), , a, (3), , O, , T, , t, , T, , t, , a, , O, , T, , t, , (4), , O, , Sol. Answer (3), X = Acost, v, , a, , dx,   A sin t, dt, , d 2x, dt 2, , a, O, , t, ,   A 2 cos t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 70 :
70, 5., , Oscillations, , Solution of Assignment, , Out of the following functions representing motion of a particle which represents SHM., (A) y = sin t – cos t, , (B) y = sin3 t, , ⎛ 3, ⎞,  3 t ⎟, (C) y  5cos ⎜, ⎝ 4, ⎠, , (D) y = 1 + t + 2t2, , (1) Only (A) and (B), , (2) Only (A), , (3) Only (D) does not represent SHM, , (4) Only (A) and (C), , [AIPMT (Prelims)-2011], , Sol. Answer (4), Only (A) and (C) are of the form, x = A sin (t + ), Hence they are the only ones which represents an S.H.M., 6., , Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and, amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude., The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The, phase difference is, [AIPMT (Mains)-2011], (1) , , (2), , , 6, , (3) Zero, , (4), , 2, 3, , Sol. Answer (4), v1 = A cos t, , ... (i), , v2 = A cos t + , , ... (ii), , According to equation v1 = – v2 when, x=, , A, 2, , A, = A sin t, 2, , t =, , A, , when x =, 6, 2, , cos t = – cos(t + ) equating (i) and (ii), , cos1, , 3 ,  , 2, 6, , 5 ,  , 6 6,  =, 7., , 2, 3, , The displacement of a particle along the x-axis is given by x = a sin2t. The motion of the particle corresponds, to, [AIPMT (Prelims)-2010], (1) Simple harmonic motion of frequency, (3) Non simple harmonic motion, , , , , (2) Simple harmonic motion of frequency, , 3, 2, , (4) Simple harmonic motion of frequency, , , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 71 :
Solution of Assignment, , Oscillations, , 71, , Sol. Answer (1), x = a sin2 t, x=, , a(1  cos 2t ), 2, , 2x – a = – cos 2t, or cos 2t = a – 2x, , 2 , , 2 , , The period of this function is T =, Hence frequency =, 8., , 1 , , T , , The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another, mass M is also suspended, the period of oscillation will now be, [AIPMT (Prelims)-2010], (1) T, , (2), , T, 2, , (3) 2T, , (4), , 2T, , Sol. Answer (4), T m, , 9., , T, , T, , 2M, ⇒ T   2T, M, , A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The, a, will be, [AIPMT (Prelims)-2009], speed of the pendulum at x =, 2, (1), , a, T, , (2), , 32a, T, , (3), , a 3, T, , (4), , a 3, 2T, , Sol. Answer (3), v = A cos t, x=, , , , A, 2, , A, = A sin t, 2, , , = t, 6, , v = Acos, , v = A, , , 6, , 3, 2, , v=, , 2 3, T 2, , v=, ,  3, A, T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 72 :
72, , Oscillations, , Solution of Assignment, , 10. Which one of the following equations of motion represents simple harmonic motion? (Where k, k0, k1 and a are, all positive.), [AIPMT (Prelims)-2009], (1) Acceleration = –k(x + a), , (2) Acceleration = k(x + a), , (3) Acceleration = kx, , (4) Acceleration = –k0x + k1x2, , Sol. Answer (1), According to equation of S.H.M.., a = –2x., The only option of the same form is the third one k0., Acceleration = – k(x + a), Hence answer is (3), 11. Two simple harmonic motions of angular frequency 100 and 1000 rad s–1 have the same displacement amplitude., The ratio of their maximum accelerations is, [AIPMT (Prelims)-2008], (1) 1 : 104, , (2) 1 : 10, , (3) 1 : 102, , (4) 1 : 103, , Sol. Answer (3), Maximum acceleration occurs at the extreme points of an S.H.M.. motion., a = –2x, a = –2A, , At x = A, a1 = (100)2 A, a2 = (1000)2 A, , a1, 1, =, a2 100, a1 : a2 = 1 : 102, 12. A point performs simple harmonic oscillation of period T and the equation of motion is given by, , ⎞, ⎛, x  a sin ⎜ t  ⎟ . After the elapse of what fraction of the time period the velocity of the point will be equal to, 6, ⎝, ⎠, half of its maximum velocity?, [AIPMT (Prelims)-2008], (1), , T, 12, , (2), , T, 8, , (3), , T, 6, , (4), , T, 3, , Sol. Answer (1), Maximum velocity = a, v = a cos t +, , Let time where v =, , Let, , , 6, , a, 2, , a, ,  a cos t , 2, 6, , ,  t , 3, 6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 73 :
Solution of Assignment, , t , , or, , 73, , , 6, , 2 ⎤, ⎡, ⎢⎣ = T ⎥⎦, , 2 , =, T, 6, , t=, , Oscillations, , T, 12, , 13. A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure., The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple, harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that, the mass gets detached from the pan? [Take g = 10 m/s2], m, , [AIPMT (Prelims)-2007], (1) 10.0 cm, , (2) Any value less than 12.0 cm, , (3) 4.0 cm, , (4) 8.0 cm, , Sol. Answer (1), When it disconnects from plates acceleration is maximum for minimum amplitude. Acceleration is maximum, at the extremities., When block leaves from a = g, g = –2x, g = –2a, g= –, –g, , k, a, m, , m, =a, k, , |a|=, , 10 × 2, 200, , a = 10 cm, 14. The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic, motion is, [AIPMT (Prelims)-2007], (1) Zero, , (2) 0.5 , , (3) , , (4) 0.707 , , Sol. Answer (2), v = A cos( + ), a = –A2 sin (t + ), Now, , ⎞, ⎛, a = +A2 cos ⎜ t    ⎟, 2⎠, ⎝, , ⎞, ⎛, , Hence A = ⎜ t    ⎟  (t  ) , 2⎠, ⎝, 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 74 :
74, , Oscillations, , Solution of Assignment, , 15. The particle executing simple harmonic motion has a kinetic energy K0cos2 t. The maximum values of the, potential energy and the total energy are respectively, [AIPMT (Prelims)-2007], (1) K0 and K0, , (2) 0 and 2K0, , (3), , K0, and K0, 2, , (4) K0 and 2K0, , Sol. Answer (1), Let kinetic energy at 4 point be K = K0cos2t, At maximum value of K cos2t = 1 and K = K0, Maximum value of K.E. = Maximum value of P.E. = Total mechanical energy as mechanical energy is, converted in S.H.M., 16. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum, time taken by the particle to travel half of the amplitude from the equilibrium position is, [AIPMT (Prelims)-2007], (1), , T, 2, , (2), , T, 4, , (3), , T, 8, , (4), , T, 12, , Sol. Answer (4), x = a sin t, a, = a sin t, 2, ,  t =, , , 6, , t=, ,  1  T,   , 6  6 2, , t=, , T, 12, , 17. A rectangular block of mass m and area of cross-section A floats in a liquid of density ñ. If it is given a small, vertical displacement from equilibrium it undergoes oscillation with a time period T. Then, [AIPMT (Prelims)-2006], (1) T , , , , (2) T , , 1, A, , (3) T , , 1, , , (4) T , , 1, m, , Sol. Answer (2), 18. The circular motion of a particle with constant speed is:, , [AIPMT (Prelims)-2005], , (1) Simple harmonic but not periodic, , (2) Periodic and simple harmonic, , (3) Neither periodic nor simple harmonic, , (4) Periodic but not simple harmonic, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 75 :
Solution of Assignment, , Oscillations, , 75, , 19. A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The, frequency of its oscillation is, (1) 3 Hz, , (2) 2 Hz, , [AIPMT (Prelims)-2005], (3) 4 Hz, , (4) 1 Hz, , Sol. Answer (4), A = 5 cm, Maximum speed (A) = 31.4, =, , 31.4, 5, , or 2f = 31.4, , f=, , 31.4, 10  3.14, , f = 1 Hz, 20. Which of the following is simple harmonic motion?, (1) Particle moving in a circle with uniform speed, (2) Wave moving through a string fixed at both ends, (3) Earth spinning about its axis, (4) Ball bouncing between two rigid vertical walls, Sol. Answer (2), A wave on a string is an example of simple harmonic motion as the displacement of particles in the motion, may be described by, x = A sin t, 21. A particle executes S.H.M.. along x-axis. The force acting on it is given by, (1) A cos (kx), , (2) Ae–kx, , (3) kx, , (4) – kx, , Sol. Answer (4), Force acts along – kx according to theory of S.H.M.., 22. Which one of the following statements is true for the speed v and the acceleration a of a particle executing, simple harmonic motion?, (1) When v is maximum, a is maximum, (2) Value of a is zero, whatever may be the value of v, (3) When v is zero, a is zero, (4) When v is maximum, a is zero, Sol. Answer (4), Acceleration (a) is zero is the mean position where velocity is maximum., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 76 :
76, , Oscillations, , Solution of Assignment, , 23. A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the, displacement of the mass from the origin is small. Which graph correctly depicts the position of the particle, as a function of time?, , v(x), , m, (x), , 0, x(t), , x(t), (1), , t, , 0, , (2), , 0, , x(t), t, , (3), , 0, , x(t), , t, , (4), , 0, , t, , Sol. Answer (2), Particle of mass m is released in a parabolic path. It will perform S.H.M.. just like a pendulum.', The particle is released from amplitude., Option (2) represents graph of an S.H.M.. starting from amplitude position., 24. In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy, is kinetic?, (1), , 1, 2, , (2), , 3, 4, , (3) Zero, , (4), , 1, 4, , Sol. Answer (2), , A⎞, ⎛, K.E. of half of amplitude = ⎜ x  ⎟, 2⎠, ⎝, , K.E. of, , K.E. =, , 2⎞, 1, 1 ⎛, k ( A2  x 2 )  k ⎜ A2  A ⎟, 2, 2 ⎝, 4 ⎠, , 1 ⎛ 3 A2 ⎞, k, 2 ⎜⎝ 4 ⎟⎠, , 1 ⎛ 3 A2 ⎞, k, K .E. 2 ⎜⎝ 4 ⎟⎠ 3, , , Fraction of the total energy is kinetic energy =, 1 2, T .E., 4, kA, 2, , 25. A linear harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01 m has a total mechanical, energy of 160 J. Its, (1) Maximum P.E. is 160 J, , (2) Maximum P.E. is zero, , (3) Maximum P.E. is 100 J, , (4) Maximum P.E. is 120 J, , Sol. Answer (1), At maximum potential energy all mechanical energy is stored as potential energy., Hence maximum P.E. = Total mechanical energy, = 160 J, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 77 :
Solution of Assignment, , Oscillations, , 77, , 26. Displacement between maximum potential energy position and maximum kinetic energy position for a particle, executing simple harmonic motion is, (1) , , a, 2, , (2) + a, , (3) ± a, , (4) – 1, , Sol. Answer (3), Maximum P.E. is at extremities and maximum K.E. is at mean position., Hence the difference in between the two position is ± a., 27. A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position, being O. Its potential energy U is plotted. It will be as given below in the graph, , U, , U, (2), , (1), , x1, , O, , x2, , x1, , U, (3) x1, , O, , O, , x2, , U, x2, , (4) x1, , O, , x2, , Sol. Answer (1), In S.H.M.. potential energy is minimum at mean position and maximum at the extremities. Also graph will be, parabolic as, , U=, , 1 2, kx, 2, , Hence answer is (1)., 28. The potential energy of a simple harmonic oscillator when the particle is half way to its end point is, (1), , 2, E, 3, , (2), , 1, E, 8, , (3), , 1, E, 4, , (4), , 1, E, 2, , Sol. Answer (3), E=, , 1 2, kA [where E is total energy], 2, , When x =, , Ex =, or Ex =, , A, 2, , 1 2 1 kA2 kA2, kx , , 2, 2 4, 8, E, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 78 :
78, , Oscillations, , Solution of Assignment, , 29. If the length of a simple pendulum is increased by 2%, then the time period, (1) Increases by 1%, , (2) Decreases by 1%, , (3) Increases by 2%, , (4) Decreases by 2%, , Sol. Answer (1), T = 2, , l, g, , If length is increased by 2%, T2 = 2, , l 102, , g 100, , or T2 = 2, , l, (1  0.02), g, , or T2 = 2, , l,  1.01, g, , or T2 = T +, , T, 100, , [By binomial theorem], ,  T2 is 1% more than T., 30. Two simple pendulums of length 5 m and 20 m respectively are given small linear displacements in one, direction at the same time. They will again be in the same phase when the pendulum of shorter length has, completed ______ oscillations., (1) 2, , (2) 1, , (3) 5, , (4) 3, , Sol. Answer (1), l1 = 5 m, T1 = 2, , l2 = 20 m, , 5, g, , T2 = 2, , 20, g, , T1 1, , T2 2, 2T1 = T2, Hence they will be in phase again when shorter one has completed 2 oscillation., 31. Two masses MA and MB are hung from two strings of length lA and lB respectively. They are executing S.H.M., with frequency relation fA = 2fB, then relation, (1) I A , , IB, , does not depend on mass, 4, , (3) IA = 2IB and MA = 2MB, , (2) IA = 4IB, does not depend on mass, (4) I A , , IB, MB, and M A , 2, 2, , Sol. Answer (1), Mass MA and MB, Length lA and lB, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 79 :
Solution of Assignment, , If, , Oscillations, , 79, , fA = 2fB, TB = 2TA, , 2, , [as f =, , 2, given], T, , lA, l,  4 B, g, g, , lA, = 4lB, g, or lB =, , lA, which does not depend on mass., 4, , 32. A mass m is vertically suspended from a spring of negligible mass, the system oscillates with a frequency, n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?, (1), , n, 2, , (2) 4n, , (3), , n, 4, , (4) 2n, , Sol. Answer (1),  = 2n =, , = n, , k, m, , 1, m, , When m becomes 4 m hence n is halved, , n, 2, , 33. A mass is suspended separately by two different springs in successive order then time periods is, t1 and t2 respectively. If it is connected by both the springs as shown in figure then time period is t0, the, correct relation is, , K1, , K2, m, , (1) t 02  t 12  t 22, , (2) t 0 2  t 1 2  t 2 2, , (3) t 01  t 11  t 21, , (4) t0 = t1 + t2, , Sol. Answer (2), t1 = 2, , m, k1, , t2 = 2, , m, k2, , The springs in parallel have keq = k1 + k2, t0 = 2, , m, k1  k2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 80 :
80, , Oscillations, , m, ⎛m m⎞, 4 2 ⎜ 2  2 ⎟, ⎝ t1 t2 ⎠, , t0 = 2, , t0 =, , Solution of Assignment, , t12 t22, , t12  t22, , 1 t12  t 22, =, t02 t12  t22, , 1, 1 1, = , t02 t 22 t12, , 34. The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same, mass is suspended from one of the parts, then the new time period will be, (1), , T, 4, , (2) T, , (3), , T, 2, , (4) 2T, , Sol. Answer (3), T = 2, , m, k, , The k of spring becomes 4 k when cut., T = 2, , m, 4k, , or T =, , T, 2, , 35. A particle, with restoring force proportional to displacement and resisting force proportional to velocity is, subjected to a force F sint. If the amplitude of the particle is maximum for  = 1, and the energy of the, particle maximum for  = 2, then, (1) 1  0 and 2  0, , (2) 1  0 and 2  0, , (3) 1  0 and 2  0, , (4) 1  0 and 2  0, , Sol. Answer (2), F = F0 sin t is the equation of forces vibration., In the case of resonance  is the resonating frequency., At the same amplitude and energy is maximum., Hence 1 = 2 = , , 36. When an oscillator completes 100 oscillations its amplitude reduced to, , 1, of initial value. What will be its, 3, , amplitude, when it completes 200 oscillations?, (1), , 1, 8, , (2), , 2, 3, , A=, , A0, 3, , (3), , 1, 6, , (4), , 1, 9, , Sol. Answer (4), A = A0e–bt, When t = 200 T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 81 :
Solution of Assignment, , Oscillations, , 81, , A0, = A0e–b×100 T, 3, e–100bT =, , 1, 3, , A = A0(e–b×200 T), A = A0(e–b×100 T)2, ⎛ 1⎞, A = A0 ⎜ ⎟, ⎝3⎠, , A =, , 2, , A0, 9, , 37. In case of a forced vibrations, the resonance wave becomes very sharp when the, (1) Damping force is small, , (2) Restoring force is small, , (3) Applied periodic force is small, , (4) Quality factor is small, , Sol. Answer (1), When restoring force is very small there is very little dissipation of energy and the driving force can deliver, maximum amplitude., 38. Two S.H.M.’s with same amplitude and time period, when acting together in perpendicular directions with a, phase difference of, , , , give rise to, 2, , (1) Straight motion, , (2) Elliptical motion, , (3) Circular motion, , (4) None of these, , Sol. Answer (3), Let x be x = a sin t, and y be y = a sin (t + ) or y = a cos t, If they are perpendicular x2 + y2 = a2 which is the equation of circle., , 39. The equations of two S.H.M.’s is given as x = a cos (t + ) and y = a cos (t + ), where     , the, 2, resultant of the two S.H.M.’s represents, , (1) A hyperbola, , (2) A circle, , (3) An ellipse, , (4) None of these, , Sol. Answer (2), x = a cos (t + ), y = a cos (t + ), =+, , ... (i), , , 2, , a – a sin (t + ), , ... (ii), , Squaring both (i) and (ii) it is of the form., x2 + y2 = a2, Hence it represents a circle., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 82 :
82, , Oscillations, , Solution of Assignment, , 40. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of, proportionality are, (1) kg ms–1, , (2) kg ms–2, , (3) kg s–1, , (4) kg s, , Sol. Answer (3), F = kv, , F, v, [k] = [F] [v]–1, k =, , https://t.me/NEET_StudyMaterial, , [k] = [kg ms–2 m–1s], [k] = kg s–1, 41. A wave has S.H.M. whose period is 4 s while another wave which also possess S.H.M. has its period 3 s., If both are combined, then the resultant wave will have the period equal to, (1) 4 s, , (2) 5 s, , (3) 12 s, , (4) 3 s, , Sol. Answer (3), T1 = 4 s, T2 = 3 s, The resultant wave will have a time period equal to LCM of the two waves., LCM of 4 and 3 is 12, Hence T = 12 s, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Simple harmonic motion is not a uniform motion., R : Simple harmonic motion can be regarded as the projection of uniform circular motion., , Sol. Answer (2), Simple harmonic motion is not a uniform motion as acceleration velocity and displacement are all variable with, time simple harmonic motion can be regarded as a project ion of uniform circular motion but this is not the, correct explanation of assertion., 2., , A : In simple harmonic motion, the velocity is maximum when the displacement is minimum., R : Displacement and velocity of S.H.M. differ in phase by /2., , Sol. Answer (1), The assertion is true and the reason is the correct explanation of the given assertion., 3., , A : In reality the amplitude of a freely oscillating pendulum decreases gradually with time., R : The frequency of the pendulum decreases with time., , Sol. Answer (3), Assertion is true as damping and dissipative forces are present reducing the energy and hence amplitude of, the pendulum., The frequency positive remains constant., 4., , A : The graph of velocity as a function of displacement for a particle executing S.H.M. is an ellipse., R : The velocity and displacement are related as v   A 2  x 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 83 :
Solution of Assignment, , Oscillations, , 83, , Sol. Answer (1), The assertion is true and the explanation gives the equation of velocity vs displacement (x) which is in the form, of an ellipse., 5., , A : The phase difference between the two particles shown below is . (Assuming both particles have same, time periods and same amplitudes)., , mean position, , –a, , a, , R : If the particles cross each other while they move in the opposite direction, they have a phase difference, of  radian., Sol. Answer (4), The phase difference between two particles is  only when they pass the mean point with opposite velocities, at the same point., Both the assertion and reason are hence false., 6., , A : All trigonometric functions are periodic., R : All trigonometric functions can represent S.H.M., , Sol. Answer (3), All trigonometric function may be periodic but not all can represent S.H.M. Examples of exception include tan,, cot, cosec etc. Hence answer is (3), 7., , A : In a S.H.M. both kinetic energy and potential energy oscillates with double the frequency of S.H.M., R : Frequency of oscillation of total energy in S.H.M. is infinite, , Sol. Answer (3), K.E. =, , 1, mA22cos2(t + ), 2, , P.E. =, , 1, mA22sin2(t + ), 2, , cos2 and sin2 may both be written in the form of doubles angles. Hence they represent S.H.M. with angular, frequency = 2., Total energy of an S.H.M. is always constant. Constant values are said to have infinite time period and zero, frequency., 8., , A : If a clock based on simple pendulum is taken to hill it will become slower., R : With increase of height above surface of earth g decreases so T will increase., , Sol. Answer (1), T = 2, , l, g, , The assertion is the correct explanation of the reason., 9., , A : If a spring block system, oscillating in a vertical plane is made to oscillate on a horizontal surface, the, time period will remain same., R : The time period of spring block system does not depend on g., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 84 :
84, , Oscillations, , Solution of Assignment, , Sol. Answer (2), T = 2, , m, k, , Gives the time period of spring block system any time period remains the same in both cases., Time period of spring block system doesn't depend on g but that explanation alone is not enough to explain, the non-variation time period of S.H.M., 10. A : In a forced oscillator the energy transferred from driving force to damped oscillator is maximum in resonance, state., R : The amplitude of forced oscillator depends on the frequency of external force., Sol. Answer (2), The assertion and reason are both true. But the reason does not exsplain why maximum energy transfer occurs, in that state. It merely says that the amplitude depends on frequency of external force., 11. A : If length of a spring is halved, then its force constant becomes double., R : The spring constant is inversely proportional to length of spring., Sol. Answer (1), Suppose a spring of natural length l into made of two parts l1 and l2., The entire spring is displaced by a distance x., , l1, , x = x1 + x2, , l2, , Now x1 : x2 = l1 : l2, , F F, : = l1 : l2, :, k1 k2, k2 l1, =, k1 l 2, k, , 1, l, , So if length is halved keq is doubled., 12. A : When soldier cross a bridge, they are asked to break steps., R : If they do not break steps, then they will apply large force on bridge simultaneously., Sol. Answer (3), When soldiers cross a bridge, they are asked to break steps. So the assertion is true., They are told to do this to avoid forced vibration to be created on the bridge. If it is the same as natural, frequency of the bridge, it may cause resonance. The bridge may oscillate with a higher amplitude and the, bridge may fall., 13. A : In S.H.M. the change in velocity is not uniform., R : In S.H.M. the acceleration of body varies linearly with its displacement., Sol. Answer (1), Both the assertion and reason are correct and reason the correct explanation of assertion., a = –2x, , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 85 :
Chapter, , 15, , Waves, Solutions, SECTION - A, Objective Type Questions, 1., , Which of the following phenomenon cannot take place with sound waves?, (1) Polarisation, , (2) Refraction, , (3) Diffraction, , (4) Reflection, , Sol. Answer (1), Polarisation is a phenomenon which cannot take place with mechanical waves., 2., , The speed of sound in air is independent from its, (1) Amplitude, , (2) Frequency, , (3) Phase, , (4) All of these, , Sol. Answer (4), , Speed of sound in air (v) =, , P, , , All the other options are function of the wave but speed of sound depends on the medium. Hence, answer, will be (4)., 3., , The waves which cannot travel without medium are, (1) X-rays, , (2) Radio waves, , (3) Light waves, , (4) Sound waves, , Sol. Answer (4), Sound waves being mechanical waves need a mechanical medium for their propagation., 4., , When a wave propagating through a medium encounters a change in medium, then which of the following, property remains same?, (1) Speed, , (2) Amplitude, , (3) Frequency, , (4) Wavelength, , Sol. Answer (3), The energy carried by a wave is a function of its frequency. If there is no loss of energy in the charge of medium, the frequency still remains the same., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 86 :
86, 5., , Waves, , Solution of Assignment, , A transverse wave travels along x-axis. The particles of medium move, (1) Along x-axis, , (2) Along y-axis, , (3) Along z-axis, , (4) Either along y-axis or z-axis, , Sol. Answer (4), In a transverse wave the energy may transfer along x-direction but particles are displaced perpendicular to the, transfer of energy., 6., , The phenomenon of sound propagation in air is, (1) An isothermal process, , (2) An adiabatic process, , (3) An isobaric process, , (4) An isochoric process, , Sol. Answer (2), The phenomenon of sound propagation was found to be almost an adiabatic process by Laplace's theory., 7., , If at STP, velocity of sound in a gas ( = 1.5) is 600 m/s, the r.m.s. velocity of the gas molecules at STP, will be, (1) 400 m/s, , (2) 600 m/s, , (3) 600 2 m/s, , (4) 300 2 m/s, , Sol. Answer (3), , Velocity of sound =, , RMS speed =, , vs, , rrms, , , 3, , vs, , rrms, , 1, 2, ,  RT, M, , 3RT, M, , vrms = 600 2 m/s, 8., , In a stretched string,, (1) Only transverse waves can exist, , (2) Only longitudinal waves can exist, , (3) Both transverse and longitudinal waves can exist, , (4) None of these, , Sol. Answer (3), In a solid both transverse and longitudinal waves may pass., 9., , Two strings of same material are stretched to the same tension. If their radii are in the ratio 1 : 2, then, respective wave velocities in them will be in ratio, (1) 4 : 1, , (2) 2 : 1, , (3) 1 : 2, , (4) 1 : 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 87 :
Solution of Assignment, , Waves, , 87, , Sol. Answer (2), Let density of material be , Areas of cross section A1 = r12 , A2 = r22 where r2 = 2r1,  1 = r12  , 2 = 4r12 , T, ,  v1 =, , r12 , , 1 T, and v2 = 2, r12, ,  v1 : v2 = 2 : 1, 10. A pulse is generated at lower end of a hanging rope of uniform density and length L. The speed of the pulse, when it reaches the mid point of rope is, , L, , (1), , 2gL, , (2), , gL, , (3), , gL, 2, , (4), , gL, 2, , Sol. Answer (3), , v=, , T, , , Weight of the string below the rope stretching it =, , L, g, 2, , This is equal to T at the middle.,  Velocity at middle =, , =, , Lg, 2, gL, 2, , 11. Which of the following equations represents a transverse wave travelling along –y axis?, (1) x = A sin (t – ky), , (2) x = A sin (t + ky), , (3) y0 = A sin (t – kx) (4) y0 = A sin (t + kx), , Sol. Answer (2), Velocity of wave = , , , k, , Also since the wave is travelling in the y direction, displacement of the particles will be in x-direction., So, equation will be of the form x = A sin (t + ky), Putting positive values of  and k in the velocity equation will give negative value of velocity. Hence, wave is, travelling in –ve direction., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 88 :
88, , Waves, , Solution of Assignment, , y⎞, ⎛, 12. A wave is represented by x  4 cos⎜ 8t  ⎟ , where x and y are in metre and t in second. The frequency of, 2⎠, ⎝, , the wave (in s–1) is, (1), , 4, , , (2), , 8, , , (3), , 2, , , (4), , , 4, , Sol. Answer (1), , y⎞, ⎛, x  4cos⎜ 8t  ⎟, 2⎠, ⎝, Comparing with x = A sin (kx – t), =8, and  = 2f, f=, , 8, 2, , f=, , 4, , ,  ⎞, ⎛, 13. A wave is represented by the equation y = A sin ⎜ 10x  15t  ⎟ where x is in metre and t in second. The, 6⎠, ⎝, expression represents, (1) A wave travelling in negative x-direction with a velocity of 1.5 m/s, (2) A wave travelling in positive x direction with a velocity of 1.5 m/s, (3) A wave traveling in positive x-direction with wavelength 0.2 m, (4) A wave travelling in negative x-direction with a velocity of 150 m/s, Sol. Answer (1), ,  ⎞, ⎛, y = A sin ⎜ 10x  15t  ⎟, 6⎠, ⎝, v= , , , k, , v= , , 15, 10, ,  v = –1.5 m/s, Hence, velocity is 1.5 m/s in negative x-direction.,  ⎞, ⎛, 14. A travelling wave in a string is represented by y  3 sin ⎜ t  x ⎟ . The phase difference between two particles, 4 ⎠, ⎝2, , separated by a distance 4 cm is (Take x and y in cm and t in seconds), (1), , , rad, 2, , (2), , , rad, 4, , (3)  rad, , (4) 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 89 :
Solution of Assignment, , Waves, , 89, , Sol. Answer (3),  ⎞, ⎛, y  3 sin ⎜ t  x ⎟, 4 ⎠, ⎝2, , Take x = 0 and x = 4 cm, , , y1 = 3sin t and y 2  3sin t  , 2, 2, , Phase difference in between the two particles is ., 15. A transverse wave is described by the equation y  A sin 2(nt  x /  0 ) . The maximum particle velocity is equal, to 3 times the wave velocity if, (1)  0 , , A, 3, , (2)  0 , , 2A, 3, , (3)  0  A, , (4)  0  3A, , Sol. Answer (2), y  A sin 2(nt  x /  0 ), , Particle velocity is, , dy, and maximum particle velocity is A., dt, ,  = 2n,  A = 2nA, Wave velocity (v) =, , 2n, , = 2 / , k, 0, , 2nA = 30n, 0 , , 2A, 3, , , , 16. If u is instantaneous velocity of particle and v is velocity of wave, then, , , , , (2) u is parallel to v, (1) u is perpendicular to v, , , , , (3) | u | is equal to | v |, (4) | u |  (slope of wave form)| v |, , Sol. Answer (4), , , u may be perpendicular or parallel to v depending on whether it is a transverse or longitudinal wave., y = Asinkx – t, dy, = Acoskx – t, dt, dy, =v, dt, , u = –Acoskx – t, , |v | =, , , k, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 90 :
90, , Waves, , Solution of Assignment, , Slope of y = Asinkx – t, dy, = Akcoskx – t, dx, , dy, dy, | v |, dx, dx, u = | v | (Slope of waveform), 17. In a simple harmonic wave, minimum distance between particles in same phase always having same speed,, is, (1) /4, , (2) /3, , (3) /2, , (4) , , Sol. Answer (1), y = Asin(t – kx), dy, = Acos(t – kx), dt, , Take x = 0 and x = x1, v1 = Acos(t), v2 = Acos(t + kx), |v1||v2|, |cos(t)| = |cos(t + kx)|, , 2, ,  kx =, , or x =, , , 4, , 18. The sound intensity level at a point 4 m from the point source is 10 dB, then the sound level at a distance, 2 m from the same source will be, (1) 26 dB, , (2) 16 dB, , (3) 23 dB, , (4) 32 dB, , Sol. Answer (2), I, , 1, r2, ,  If at two m intensity is I1 at 4 m, intensity is, , I1, 4, , 19. The tones that are separated by three octaves have a frequency ratio of, (1) 3, , (2) 6, , (3) 8, , (4) 16, , Sol. Answer (3), Each octaves is double the frequency of the previous octave., If initial frequency is f., Final frequency = 2 × 2 × 2 f or 23 f or 8 f,  Frequency ratio = 8 : 1., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 91 :
Solution of Assignment, , Waves, , 91, , ⎞, ⎛, 20. On the superposition of the two waves given as y 1  A0 sin(t  kx ) and y 2  A0 cos⎜ t  kx  ⎟ , the, 6⎠, ⎝, , resultant amplitude of oscillations will be, (1), , 3A0, , A0, 2, , (2), , (3) A0, , (4), , 3, A0, 2, , Sol. Answer (3), Resultant amplitude is given by Ar2 = A12 + A22 + 2A1A2 cos, y 1  A0 sin(t  kx ), , ⎞, ⎛, y 2  A0 cos⎜ t  kx  ⎟, 6⎠, ⎝, , or, ,  ⎞, ⎛, y 2  A0 sin⎜ t  kx   ⎟, 6 2⎠, ⎝, Ar2 = A02 + A02 + 2A02 cos, Ar2 = 2A02 + 2A02 ×, , 2, 3, , 1, 2, ,  Ar = A0, 21. Two waves of amplitudes A0 and xA0 pass through a region. If x > 1, the difference in the maximum and, minimum resultant amplitude possible is, (1) (x + 1)A0, , (2) (x – 1)A0, , (3) 2xA0, , (4) 2A0, , Sol. Answer (4), Amplitudes are A0 and xA0.,  Maximum amplitude where they are in phase are, , A0 + xA0., , Minimum amplitude xA0 – A0, Difference between the two = 2A0., 22. Which of the following represents a standing wave?, (1) y = A sin(t – kx), , (2) y = Ae–bx sin(t – kx + ), , (3) y = A sin kx sin(t – ), , (4) y = (ax + b) sin(t – kx), , Sol. Answer (3), In a standing wave, the amplitude is an oscillating function of x in :, y = A sin kx sin(t + ), A sin kx represents the amplitude and is of the correct form to be a standing wave., y = 3 sin, , x, cos 40 t, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 92 :
92, , Waves, , Solution of Assignment, , ⎛ x ⎞, 23. The equation of standing wave in a stretched string is given by y  5 sin⎜ ⎟ cos( 40 t ) , where x and y are, ⎝ 3 ⎠, , in cm and t in second. The separation between two consecutive nodes is (in cm), (1) 1.5, , (2) 3, , (3) 6, , (4) 4, , Sol. Answer (2), , y  5sin, , x, cos40t, 3, , At the nodes the values of x is such that y = 0., Taking position 1 at x = 0 to make sine function zero; the value of sine function will be zero node for, , x, =., 3, ,  x=3, Hence, shortest distance between two nodes is 3 cm., 24. In a stationary wave, (1) Strain is maximum at nodes, , (2) Strain is minimum at nodes, , (3) Strain is maximum at antinodes, , (4) Amplitude is zero at all points, , Sol. Answer (1), Where the slope of wave is higher the strain will be higher., 25. Select the correct statement about the reflection and refraction of a wave at the interface between the medium, 1 and 2 the, (1) Reflected wave has a phase change of , (2) Wavelength of reflected wave is less than that of incident wave, (3) Frequency of transmitted wave is same as that of incident wave, (4) Frequency of wave changes as per nature of boundary, Sol. Answer (3), It is not said which of waves 1 and 2 is denser. But frequency of a wave remains same in all media. Hence, option (3) is correct., 26. In the standing wave shown, particles at the positions A and B have a phase difference of, , A, , (1) 0, , (2), , , 2, , B, , (3), , 5, 6, , (4) , , Sol. Answer (4), Let, yA = A sin t, yA = A sin t + , To convert sin to – sin,  must be ., Hence phase difference between is ., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 93 :
Solution of Assignment, , Waves, , 93, , 27. A 12 m long vibrating string has the speed of wave 48 m/s. To what frequency it will resonate?, (1) 2 cps, , (2) 4 cps, , (3) 6 cps, , (4) All of these, , Sol. Answer (4), v = 48 m/s, l = 12 m, f =, , nv, 2l, , f = 2n (where n = 1, 2, 3 .....), Hence answer is (4)., 28. A certain string will resonant to several frequencies, the lowest of which is 200 cps. What are the next three, higher frequencies to which it resonants?, (1) 400, 600, 800, , (2) 300, 400, 500, , (3) 100, 150, 200, , (4) 200, 250, 300, , Sol. Answer (1), Resonant frequency occurs according to, , Hence, , nv, and when n = 1 frequency is minimum., 2l, , v, = 200 [n = 1], 2l, , Other resonant frequencies will be simply multiples of fundamental frequency., Hence (1) is the answer., 29. The tension in a wire is decreased by 19%. The percentage decrease in frequency will be, (1) 0.19%, , (2) 10%, , (3) 19%, , (4) 0.9%, , Sol. Answer (2), , Velocity in a wave =, , T, , , Fundamental frequency of waves, , f =, , T 1, ×,  2l, , v, 2l, , ... (i), , If T decreases by 19% value of T will be T – 0.19 T, Putting this value in (i), , f , , T (1  0.19)1/2, , 2l, , ⎛ 1, ⎞, f   f ⎜1  × 0.19 ⎟, ⎝ 2, ⎠, , [Binomial theorem], , f  = f – 0.1 f, , Hence, the frequency decreases by 0.1 f are 10% of initial value., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 94 :
94, , Waves, , Solution of Assignment, , 30. The length of a sonometer wire is 0.75 m and density 9 × 10 3 Kg/m 3 . It can bear a stress of, 8.1 × 108 N/m2 without exceeding the elastic limit. The fundamental frequency that can be produced in the, wire, is, (1) 200 Hz, , (2) 150 Hz, , (3) 600 Hz, , (4) 450 Hz, , Sol. Answer (1),  = 9 × 103 Kg/m3, , l = 0.75 m, , Limiting stress = 8.1 × 108 N/m2, Let area be equal to A., Tension (T) = Stress × Area, = 8.1 × 108 × A, Mass / length() = A, , Velocity (v) =, , T 8.1 108  A, , , 9  103  A, , =, , 8.1 105, 9, , =, , 2, 9  10 4 = 3 × 10 m/s, , Maximum fundamental frequency =, , v, 2l, , =, , 300, = 200 Hz, 2  0.75, , 31. An aluminium rod having a length 100 cm is clamped at its middle point and set into longitudinal vibrations., Let the rod vibrate in its fundamental mode. The density of aluminium is 2600 kg/m3 and its Young’s modulus, is 7.8 × 1010 N/m2. The frequency of the sound produced is, (1) 1250 Hz, , (2) 2740 Hz, , (3) 2350 Hz, , (4) 1685 Hz, , Sol. Answer (2), , v=, , y, , , , 7.8  1010, = 5480 ml, 2600, , Since rod is clamped at the middle, the middle point is a pressure antinode and free ends are nodes. In the, fundamental mode there are no other nodes and antinodes. The length of the rod is therefore half the, wavelength., So,  = 2l = 2m, Frequency =, , v 5480, , = 2740 Hz, , 2, , 32. The string of a violin has a frequency of 440 cps. If the violin string is shortened by one fifth, its frequency, will be changed to, (1) 440 cps, , (2) 880 cps, , (3) 550 cps, , (4) 2200 cps, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 95 :
Solution of Assignment, , Waves, , 95, , Sol. Answer (4), Fundamental frequency = 440 cps =, , If l is made one fifth, , Since, , 5, , v, 2l, , 5v, v, or, 2l, 2l / 5, , v, = 440, 2l, v,  5  440  2200, 2l, , 33. A wire of length one metre under a certain initial tension emits a sound of fundamental frequency 256 Hz. When, the tension is increased by 1 kg wt, the frequency of the fundamental node increases to 320 Hz. The initial, tension is, (1) 3/4 kg wt, , (2) 4/3 kg wt, , (3) 16/9 kg wt, , (4) 20/9 kg wt, , Sol. Answer (3), Let tension be T, , f1 =, , T 1, , = 256,  2l, , ... (i), , f2 =, , T  10 1, , = 320, , 2l, , ... (ii), , 256, T, =, T  10 320, , or, , T, (16)2, =, T  10 (16)2  (20)2, , or, , T, 162, =, T  10 (20)2, , [Dividing (i) by (ii)], , or 400 T = 256 T2 + 2560, or 144 T = 2560, or T =, , 2560, 144, , or T =, , 2560, 16  9, , or T =, , 160, Newton, 9, , =, , 16, kg-wt, 9, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 96 :
96, , Waves, , Solution of Assignment, , 34. In case of closed pipe which harmonic the pth overtone will be, (1) 2p + 1, , (2) 2p – 1, , (3) p + 1, , (4) p – 1, , Sol. Answer (1), In case of closed pipe the pth overtone will be 2p + 1 harmonic., 35. The pitch of an organ pipe is highest when the pipe is filled with, (1) Air, , (2) Hydrogen, , (3) Oxygen, , (4) Carbon dioxide, , Sol. Answer (2), v=f, = is constant for an organ pipe, v=, , v=, ,  RT, M, , 1, M, , v is maximum when M is least.Thus, velocity of sound in air is maximum when M is least., Since f  v so f , , 1, M, , Hence, frequency is maximum when M is least., Hence, among options answer is H2., 36. A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in, water so that half of it is in water. The fundamental frequency of the air column is now, (1), , f, 2, , (2), , 3f, 4, , (3) f, , (4) 2 f, , Sol. Answer (3), Fundamental frequency of open organ pipe is f =, , If closed organ pipe, , v, 2l, , v, 4l, , Initial length l1 = l,  f1 =, , v, 2l, , Final length of air column l2 =, , f2 =, , =, , l, 2, , v, 4l / 2, v, 2l, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 97 :
Solution of Assignment, , Waves, , 97, , 37. For a certain organ pipe, three successive resonance frequencies are observed at 425, 595, and 765, Hz, respectively, Taking the speed of sound in air to be 340 m/s the fundamental frequency of the pipe (in Hz) is, (1) 425, , (2) 170, , (3) 85, , (4) 245, , Sol. Answer (3), Frequencies are 425, 595, 765., Among the options the HCF is here 85., Since resonant frequencies are odd multiples of fundamental frequency., The fundamental frequency is 85 Hz., 38. A closed pipe of length 10 cm has its fundamental frequency half that of the second overtone of an open pipe., The length of the open pipe, (1) 10 cm, , (2) 20 cm, , (3) 30 cm, , (4) 40 cm, , Sol. Answer (3), l = 0.1 m, Fundamental frequency of pipe (f1) =, , v, v, or f1 =, 4l, 0.4, , Frequency of 2nd overtone of open pipe 2 :, , 3v, 2l, , f2 =, , 2f1 = f2, , 2×, , v, 3v, =, 0.4 2l, , l = 0.3 m, 39. The vibrations of four air columns under identical conditions are represented in the figure below. The ratio of, frequencies np : nq : nr : ns will be, , p, (1) 12 : 6 : 3 : 4, , q, , (2) 1 : 2 : 4 : 3, , r, , s, , (3) 4 : 2 : 3 : 1, , (4) 6 : 2 : 3 : 4, , Sol. Answer (2), P, S, will be 1st and 2nd harmonics of closed tube., P=, , v, v, ,S=, 4l, 4l, , Q, R are the 1st and 2nd harmonics of open pipe., Q=, , v, v, ,R=, 2l, l, ,  P:Q:R:S=1:2:4:3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 98 :
98, , Waves, , Solution of Assignment, , 40. In resonance tube two successive positions of resonance are obtained at 15 cm and 48 cm. If the frequency, of the fork is 500 cps, the velocity of sound is, (1) 330 m/s, , (2) 300 m/s, , (3) 1000 m/s, , (4) 360 m/s, , Sol. Answer (1), l1 + e =, , , 4, , 3, l2 + e =, 4, , l2 – l1 =, , l1, , e, l2, , , 2, , 0.48 – 0.15 =, , , 2, , 0  = 0.66, Velocity = f  = 500 × 0.66 = 330 m/s, 41. During superposition of two waves of nearly equal frequencies, beats frequency is defined as the, (1) Sum of frequencies of interfering waves, (2) Number of times the resultant intensity becomes maximum or minimum in one second, (3) Average of frequencies of interfering waves, (4) Number of times the resultant amplitude becomes maximum or minimum in one second, Sol. Answer (2), Beats is the difference in the frequencies of two superimposing waves. Beats are noticed when the resultant, wave acquire its higher amplitudes, due to both waves, sounding higher than the regular sound., The beats would occur f1 – f2 = f times per second. Hence answer is (2)., 42. A tuning fork of unknown frequency produces 4 beats per second when sounded with another tuning fork, of frequency 254 Hz. It gives the same number of beats per second when unknown tuning fork loaded with, wax. The unknown frequency before loading with wax is, (1) 258, , (2) 254, , (3) 250, , (4) Can’t be determined, , Sol. Answer (1), Since there are 4 beats. We know that the unknown frequency (f)., f = 254 + 4, Since in the second case after loading with wax frequency of 2nd fork must have reduced., Initial frequency must be, f = 254 + 4 = 258 Hz, 43. Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning, forks produce 4 beats per second. The highest frequency is twice that of the lowest. Possible highest and, lowest frequencies (in Hz) are, (1) 80 and 40, , (2) 100 and 50, , (3) 44 and 22, , (4) 72 and 36, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 99 :
Solution of Assignment, , Waves, , 99, , Sol. Answer (4), Let frequencies be f1, f2 ... f10., So, f10 = f1 + 9 × (number of beats), f10 = f1 + 36, Hence, answer must be (4)., 44. The displacement at a point due to two waves are y1 = 4sin(500t) and y2 = 2sin(506t). The result due to, their superposition will be, (1) 3 beats per second with intensity relation between maxima and minima equal to 2, (2) 3 beats per second with intensity relation between maxima and minima equal to 9, (3) 6 beats per second with intensity relation between maxima and minima equal to 2, (4) 6 beats per second with intensity relation between maxima and minima equal to 9, Sol. Answer (2), y1 = 4sin500t, y2 = 2sin506t, Frequency of y1(f1 ) , ,  500, , = 250 Hz, 2, 2, , Frequency of y 2 (f2 ) , ,  506, , = 253 Hz, 2, 2, , Amax ( A1  A2 )2, , Intensity relation A, ( A1  A2 )2, min, , =, , 36, =9, 4, , 45. A tuning fork vibrating with a sonometer having a wire of length 20 cm produces 5 beats per second. The beats, frequency does not change if the length of the wire is changed to 21 cm. The frequency of the tuning fork must, be, (1) 200 Hz, , (2) 210 Hz, , (3) 205 Hz, , (4) 215 Hz, , Sol. Answer (3), Let frequency of tuning fork be f., , ⎛v ⎞, From question we know the frequencies of sonometer wire ⎜ ⎟ are, ⎝ 2l ⎠, v, v, and, 2  0.2, 2  0.21, , Now, , v, –f=8, 0.4, , ... (i), , v, =5, 0.42, , ... (ii), , f–, , Solving (i) and (ii), answer is (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 100 :
100, , Waves, , Solution of Assignment, , 46. A tuning fork and an air column whose temperature is 51°C produce 4 beats in one second, when sounded, together. When the temperature of air column decreases the number of beats per second decreases. When, the temperature remains 16°C only one beat per second is produced. The frequency of the tuning fork is, (1) 100 Hz, , (2) 75 Hz, , (3) 150 Hz, , (4) 50 Hz, , Sol. Answer (4), F 4, , F 1, , 273  51, 273  16, , f 4, , f 1, , 324, 289, , f  4 18, , f  1 17, f  50 Hz, , 47. A man standing on a platform observes that the frequency of the sound of a whistle emitted by a train drops, by 140 Hz. If the velocity of sound in air is 330 m/s and the speed of the train is 70 m/s, the frequency of, the whistle is, (1) 571 Hz, , (2) 800 Hz, , (3) 400 Hz, , (4) 260 Hz, , Sol. Answer (2), Velocity of sound in air = 330 m/s., Speed of train = 70 m/s, Let initial frequency be f0., v, f = f0 × v  v, s, , f0 – f = 140, , ⎛ 330 ⎞, f0 ⎜1 , ⎟  140, ⎝ 400 ⎠, f0 = 800 Hz, 48. A sound source is moving towards a stationary observer with (1/10) of the speed of sound. The ratio of apparent, to real frequency is, , 10, (1), 9, , 11, (2), 10, , ⎛ 11 ⎞, (3) ⎜ ⎟, ⎝ 10 ⎠, , 2, , ⎛ 9 ⎞, (4) ⎜ ⎟, ⎝ 10 ⎠, , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 101 :
Solution of Assignment, , Waves, , 101, , Sol. Answer (1), , v, f = f0 v  v, s, f = f0, , v, v  v / 10, , 10 f0, 9, , f=, ,  f : f0 =, , 10, 9, , 49. A train moves towards a stationary observer with a speed 34 m/s. The train sounds a whistle and its frequency, registered by the observer is f1. If the speed of the train is reduced to 17 m/s, the frequency registered is f2., If the speed of sound is 340 m/s then the ratio, , (1), , 18, 19, , (2), , 1, 2, , f1, is, f2, , (3) 2, , (4), , 19, 18, , Sol. Answer (4), , v, f = f0 v  v, s, f1 = f0, , 340, 340  34, , f2 = f0, , 340, 340  17, , f1 =, , 10, f0, 9, , f2 =, , 20, 19, , f, 19, f2 = 18, , 50. An observer is approaching with a speed v, towards a stationary source emitting sound waves of wavelength, 0. The wavelength shift detected by the observer is (Take c = speed of sound), (1), ,  0v, c, , (2), ,  0c, v, , (3), ,  0v 2, c2, , (4) Zero, , Sol. Answer (4), In Doppler effect only change in frequency is observed and not change in wavelength if observer approaches, the source., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 102 :
102, , Waves, , Solution of Assignment, , SECTION - B, Objective Type Questions, 1., , The longitudinal wave can be observed in, (1) Elastic media, , (2) Inelastic media, , (3) Both (1) & (2), , (4) None of these, , Sol. Answer (1), Longitudinal waves travel due to compressions and rarefactions. If medium is inelastic it does not allow, constituent compressions and rarefactions., 2., , A transverse pulse is shown in the figure, on which 4 points are shown at any instant. Which of the following, points are in a state to move upwards in subsequent time?, , A, , B, D, , C, (1) A, B, , (2) A, D, , (3) B, C, , (4) B, D, , Sol. Answer (3), y = A0 sin (t – kx), , dy, = VP = A0 cos(t – kx), dt, Values for VP at B and C are positive force they are in a state to move upwards., 3., , A rope of length L and mass M hangs freely from the ceiling. If the time taken by a transverse wave to travel, from the bottom to the top of the rope is T, then time to cover first half length is, , ⎛ 2  1⎞, ⎟, (2) T ⎜⎜, 2 ⎟⎠, ⎝, , (1) T, , (3), , T, 2, , (4), , T, 2, , Sol. Answer (3), , v=, , N, , , The tension N in the string varies as :, N=, , Mg,  x where x is length from the ground., L, , dx, dt = v and vx =, x, T, , ∫, 0, , L, , dt , , ∫, 0, , Mgx,  gx, LM / L, , dx, gx, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 103 :
Solution of Assignment, , Waves, , 103, , L, , ∫2, , T=, , x dx, , 0, , L, , ∫2, , T=, , Lg, , ... (i), , 0, , If time to cover half length is T2., , 2Lg, , T2 =, , T, 2, 4., , [By putting limits 0 to L/2 in equation (i)], ,  T2, , A uniform rope having some mass hangs vertically from a rigid support. A transverse wave pulse is produced, at the lower end. The speed (u) of the wave pulse varies with height (h) from the lower end as, , u, , u, , u, , (2), , (1), , u, , (3), , h, , (4), , h, , h, , h, , Sol. Answer (2), Velocity of wave at a distance h from lower end is, v=, , gh, , or v2 = gh, This is of the form of a parabola (y = kx2)., Hence answer is (2), 5., , A transverse pulse generated at the bottom of a uniform rope of length L, travels in upward direction. The time, taken by it to travel the full length of rope will be, (1), , L, 2g, , (2), , 2L, g, , (3), , L, g, , (4), , 4L, g, , Sol. Answer (4), Velocity of wave at a distance x from lower end is, vx =, , 2gx, , Time dt =, , dx, 2gx, , Integrating for 0 to L we get, , 4l, g, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 104 :
104, 6., , Waves, , Solution of Assignment, , Which one of the following represents a wave?, (1) y = A sin (t – kx), (2) y = A cos2 (at – bx + c) + A sin2(at – bx + c), (3) y = A sin kx, (4) y = A sin t, , Sol. Answer (1), d 2y, 1 d 2y,  2, is satisfied only by functions of the form f(kx – t), 2, dx, v dt 2, , The only equation in the option which satisfies the above equation., 7., , Which of the following functions for y can never represent a travelling wave?, ⎧ ( x vt ) ⎫, ⎨, ⎬, x, e⎩ 0 ⎭, , (a) (x2 – vt)2, , ⎡ ( x  vt ) ⎤, ⎥, (b) log⎢ x, 0, ⎣, ⎦, , (c), , (1) Only (a), , (2) (b) & (c), , (3) (c) & (d), , 2, , (d), , 1, x  vt, , (4) Only (c), , Sol. Answer (1), The only function in the given function which is not of the form f(kx – t) is the first one (a), Hence option (1) is correct., 8., , In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero, displacement is 0.170 s. The frequency of wave is, (1) 0.73 Hz, , (2) 0.36 Hz, , (3) 1.47 Hz, , (4) 2.94 Hz, , Sol. Answer (3), Time period from maximum displacement T0 zero displacement is, , T, = 0.170 s, 4, ,  Total time = 4 × 0.17 = 0.688, Frequency =, , 9., , 1, = 1.47 Hz, T, , Figure shows a snapshot for a travelling sine wave along a string. Four elemental portions a, b, c and d are, indicated on the string. The elemental portion which has maximum potential energy is/are, , d, c, , a, , x, b, , (1) a, , (2) b, , (3) c, , (4) b and d, , Sol. Answer (1), Maximum potential energy is stored where kinetic energy, i.e., velocity is zero as the total energy of an, element of a wave is constant in its oscillating direction. All the energy is stored as potential energy of the, wire at points b and d., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 105 :
Solution of Assignment, , Waves, , 105, , 10. A sinusoidal wave of frequency 500 Hz has a speed of 350 m/s. The phase difference between two, displacements at a certain point at times 1 m apart is, (1), , , 4, , (2), , , 2, , (3) , , (4), , 3, 2, , Sol. Answer (3), v = 350 m/s, v=, , , k, , k =, , 2f, v, , k =, , 2  500, 350, , k =, , f = 500 Hz, , 2, 0.7, ,   = 0.7 m, Let equation of wave be, y = A sin (kx – t), Let x1 = 0, , x2 = 1 m, , y1 = A sin (t), , ⎛ 2, ⎞,  1  t ⎟, y2 = A sin ⎜, ⎝ 0.7, ⎠, , Hence phase difference =, , 20, or approximately, 7, , Phase difference  3, Which is same as  = , x⎞, ⎛, 11. The equation of travelling wave is y  a sin 2⎜ pt  ⎟ . Then the ratio of maximum particle velocity to wave, 5⎠, ⎝, , velocity is, , (1), , a, 5, , (2) 2 5a, , (3), , 2a, 5, , (4), , 2a, 5, , Sol. Answer (3), Equation of travelling wave, , x⎞, ⎛, y = a sin 2 ⎜ pt  ⎟, 5⎠, ⎝, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 106 :
106, , Waves, , Solution of Assignment, , Maximum particle velocity (vy) = A or vy = a × 2p, Wave velocity (v) =, , , , vy, v, , , ,  2 p, ,  5p, k 2 / 5, , 2a, 5, , 12. A travelling wave pulse is given by y , , 4, 2, , 2, , 3 x  48t  24 xt  2, , where x and y are in metre and t is in second., , The velocity of wave is, (1) 4 m/s, , (2) 2 m/s, , (3) 8 m/s, , (4) 12 m/s, , Sol. Answer (1), , 4, , y, , 2, , 2, , 3 x  48t  24 xt  2, , We need to convert it into the form of f(kx – t), 4, , y, , 2, , 3( x  16t 2  8 x )  2, , y=, , 4, 3( x  4t )2  2, , v=, , , k, , Hence v =, , 4, = 4 m/s, 1, , 13. The ratio of maximum particle velocity to wave velocity is [where symbols have their usual meanings], (1) k A, , (2) A, , (3) k , , (4), , , k, , Sol. Answer (1), Maximum particle velocity (vy) = A, Wave velocity (v) =, , , , vy, v, , , k, ,  Ak, , 14. What is the phase difference between the displacement wave and pressure wave in sound wave?, (1) Zero, , (2), , , 2, , (3) , , (4), , , 4, , Sol. Answer (2), Phase difference between the pressure and displacement wave will always be, , , ., 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 107 :
Solution of Assignment, , Waves, , 107, , 15. The driver of a car travelling with speed 30 m/s towards a hill sounds a horn of frequency 600 Hz. If the velocity, of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is, (1) 500 Hz, , (2) 550 Hz, , (3) 555.5 Hz, , (4) 720 Hz, , Sol. Answer (4), Velocity of sound = 330 m/s, Frequency of incident sound on wall (f1) =, , v, f, v  vs, , Frequency of sound observed by driver (f2 ) , , v  vs,  f1, v, , v  vs, or f2 = v  v × 600, s, , or f2 =, , 360,  600, 300, , or f2 = 720 Hz, 16. The ratio of intensities between two coherent sound sources is 4 : 1. The difference of loudness in dB between, maximum and minimum intensities when they interfere in the space is, (1) 20 log10(3), , (2) 10 log10(2), , (3) 20 log10(2), , (4) 10 log10(3), , Sol. Answer (3), Loudness in dB = 10 log10, , Let L1 = 10 log10, , I, I0, , L2 = 10 log10, , 4I, I0, , I, I0, , L2 – L1 = 10 log10 4, = 20 log10 2, 17. The intensity of sound reduces by 20% on passing through a glass slab. If sound of intensity I is made to, cross through two such slabs, then the intensity of emergent sound will be, (1) 0.36 I, , (2) 0.64 I, , (3) 0.4 I, , (4) 0.8 I, , Sol. Answer (2), Let intensity of sound initially be = I0, After passing through glass slab intensity is =, , 80, I = 0.8I0, 100 0, , After passing through 2nd slab it further reduces by 20% and only 80% is left., Hence final intensity =, , 80, × 0.8I0 = 0.64I0, 100, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 108 :
108, , Waves, , Solution of Assignment, , 18. Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The, sum of the maximum and minimum intensities is, (1) 2(I1 + I2), , (2) I1 + I2, , (3) ( I1 , , I2 )2, , (4) ( I1 , , I2 )2, , Sol. Answer (1), Imax =, , , , I1  I2, , Imin =, , , , I1  I2, , , , , , 2, , 2, ,  I1  I2  2 I1I2,  I1  I2  2 I1I2, , Imax + Imin = 2(I1 + I2), 19. A stationary wave is represented by y = A sin (100t) cos (0.01x), where y and A are in millimetres, t is in, second and x is in metre. The velocity of the constituent wave is, (1) 104 m/s, , (2) Not derivable, , (3) 1 m/s, , (4) 102 m/s, , Sol. Answer (1), y = A sin 100t cos 0.01x,  = 100, v=, , k = 0.01, , , k, , or v = 104 m/s, 20. The length of a sonometer wire AB is 110 cm. Where should the two bridges be placed from A to divide the, wire in 3 segments whose fundamental frequencies are in the ratio of 1 : 2 : 3?, (1) 60 cm and 90 cm, , (2) 30 cm and 60 cm, , (3) 30 cm and 90 cm, , (4) 40 cm and 80 cm, , Sol. Answer (1), LAB = 110 cm, Fundamental frequencies =, , Hence, l2 =, Also, , v, v, v, :, :, are given as 1 : 2 : 3, 2l1 2l 2 2l3, , l, l1, and l3 = 1, 3, 2, , l1 + l2 + l3 = 110, , l1 , , l l1,   110, 2 3, , 6l1 + 3l1 + 2l1 = 660, l1 =, , 660, 11, ,  l1 = 60 cm,  l2 = 30 cm,  l3 = 20 cm, Hence answer is (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 109 :
Solution of Assignment, , Waves, , 109, , 21. Standing waves are produced in 10 m long stretched string fixed at both ends. If the string vibrates in, 5 segments and wave velocity is 20 m/s, the frequency is, (1) 5 Hz, , (2) 10 Hz, , (3) 2 Hz, , (4) 4 Hz, , Sol. Answer (1), The question refers to the 5th harmonic of a vibrating wave., Frequency of 5th harmonic is =, , nv, 5  20, =, = 5 Hz, 2l, 2  10, , 22. If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become, (1), , n, 4, , 2n, , (2), , (3) n, , (4), , n, 2, , Sol. Answer (3), Fundamental frequency (n) =, , or n =, , T 1, ,  2l, , v, 2l, , ... (i), , Tension becomes 2T, , ... (ii), , v = A per metre, or v = r2, Now,  = (2r)2, or, ,  = 2r2, , or, ,  = 2, , , 2, , ... (iii), , Putting (ii) and (iii) in equation (i),, n = n, 23. Two waves have equations x1 = a sin (t + 1) and x2 = a sin (t + 2). If in the resultant wave the, frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between, them is, (1), , , 6, , (2), , 2, 3, , (3), , , 4, , (4), , , 3, , Sol. Answer (2), x1 = a sin (t + 1), x2 = a sin (t + 2), x = x1 + x2, = a[sin(t + 1) + sin(t + 2)], 1   2 ⎞, ⎛, ⎛ 1   2 ⎞, = 2a sin ⎜ t , ⎟ cos ⎜, ⎟, 2 ⎠, ⎝, ⎝ 2 ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 110 :
110, , Waves, , Solution of Assignment, , Now as given in question, 2a cos, , 1   2, 2, , a, , ⎛   2 ⎞ 1, cos ⎜ 1, ⎟, ⎝ 2 ⎠ 2, 1   2, 2, , , , 1   2 , , , 3, , 2, 3, , 24. The standing wave in a medium is expressed as y = 0.2 sin (0.8x) cos (3000t) m. The distance between any, two consecutive points of minimum or maximum displacement is, (1), , , m, 2, , (2), , , m, 4, , (3), , , m, 6, , (4) None of these, , Sol. Answer (4), y = 0.2 sin 0.8x cos 3000t, , ⎛⎞, The distance between any two points of minimum or maximum displacement is simply wavelength ⎜ ⎟ of the, ⎝4⎠, wave., k =, , 2, = 0.8, , , or  =, , 2, 0.8, ,  =, , 5,  5, m, m ⇒ , 2, 4 8, , Hence answer must be (4)., 25. The equation of a standing wave in a string fixed at both ends is given as y = 2A sin kx cos t, The amplitude and frequency of a particle vibrating at the mid of an antinode and a node are respectively, (1) A,, , , 2, , (2), , , 2 2, , A, , ,, , (3) A,, , , , , (4), , 2 A,, , , 2, , Sol. Answer (4), y = 2A sin kx . cos t, In a standing waves the function of amplitude (Ay) is given by, Ay = 2A sin kx, At mid-point of node and antinode x =, , Ay = 2A sin, , , 8, , 2  ⎡, 2 ⎤, , k ⎥, ⎢,  8 ⎣, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 111 :
Solution of Assignment, , or Ay =, , Waves, , 111, , 2A, 2, ,  Ay =, , 2A, , Frequency is same at all points =, , , 2, , 26. Two sinusoidal waves given below are superposed, y1 = A sin (kx – t +, , , ) ,, 6, , y2 = A sin (kx – t –, , , ), 6, , The equation of resultant wave is, (1) y , , A, 3, , sin (kx – t), , (2) y = A 3 sin (kx – t), , (3) y = A 3 sin (kx – t –, , , ), 3, , (4) y =, , A, 3, , sin (kx – t –, , , ), 3, , Sol. Answer (2), y = y1 + y2, or y = A sin (kx – t +, , , , ) + A sin (kx – t –, ), 6, 6, , ⎛⎛  ⎞⎞, , or y = 2A sin (kx – t) . cos ⎜ ⎜⎝ 6 6 ⎟⎠ ⎟, ⎜, ⎟, 2 ⎠, ⎝, or y = 2 A, , 3, sin(kx  t ), 2, ,  y = A 3 sin(kx  t ), 27. For a particular resonance tube, following are four of the six harmonics below 1000 Hz;, 300, 600, 750 and 900 Hz, The two missing harmonics are, (1) 75, 150, , (2) 150, 450, , (3) 400, 800, , (4) 250, 400, , Sol. Answer (2), The 6th harmonic (f6) =, , or, , 6v,  900, 2l, , v, = 150, 2l, , Hence fundamental frequency of string, , v, or 150 Hz is missing., 2l, , 3rd multiple of 150 which is 450 is also missing., Hence required frequencies are 150, 450., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 112 :
112, , Waves, , Solution of Assignment, , 28. A second harmonic has to be generated in a string of length l stretched between two rigid supports. The points, where the string has to be plucked and touched are respectively, (1), , l l, ,, 4 2, , (2), , l 3l, ,, 4 4, , (3), , l l, ,, 2 2, , (4), , l 3l, ,, 2 4, , Sol. Answer (1), In the 2nd harmonic the shape of wave looks like., , Point of, plucking, Point of, touching, B, , A, , The point of touching forms a node point of plucking forms an antinode. Hence, for 2nd harmonic, we need to, need wire touched at, , l, l, and pluck it at ., 2, 4, , 29. In a standing wave, all particles of the medium cross the mean position with, (1) Different speeds at different instants, , (2) Different speeds at same instant, , (3) Same speed at different instants, , (4) Same speed at same instant, , Sol. Answer (2), In a standing waves all particles of same wave have same v and hence vibrate in phase passing the mean, point at same time. The amplitude of the waves is different at different points hence the speed is different., 30. Two waves are represented by, y1 = 5 sin 2(75t – 0.25x), y2 = 10 sin 2(150t – 0.50x), The intensity ratio, (1) 1 : 2, , I1, of the two waves is, I2, , (2) 1 : 4, , (3) 1 : 8, , (4) 1 : 16, , Sol. Answer (4), A1 = 5,, , A2 = 10, , I1 kA12 ⎛ 5 ⎞2, ,  ⎜ ⎟ 1: 4, I2 kA22 ⎝ 10 ⎠, , 31. In a closed organ pipe of length 105 cm, standing waves are set up corresponding to third overtone. What, distance from the closed end, a pressure node is formed?, (1) 5 cm, , (2) 15 cm, , (3) 25 cm, , (4) 30 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 113 :
Solution of Assignment, , Waves, , 113, , Sol. Answer (2), In a closed pipe f =, , (2P  1), v, 4l, , For third overtone (f) =, , =, , v, v, ,  4l, f 7v, , =, , 4l, 7, , =, , 4  105, 7, , 7v, 4l, , /4, ,  = 60 cm, First pressure node will be formed, , , 4, , 32. A uniform string resonates with a tuning fork, at a maximum tension of 32 N. If it is divided into two segments, by placing a wedge at a distance one-fourth of length from one end, then to resonance with same frequency, the maximum value of tension for string will be, (1) 2 N, , (2) 4 N, , (3) 8 N, , (4) 16 N, , Sol. Answer (1), Let length be l., , f=, , T 1, , ... (i),  2l, , f=, , T 4, , ... (ii),  2l, , or f =, , T 4, , ... (iii),  6l, , Equating (i) & (ii) and (i) & (iii), T, =4 &, T1, , T, 4, , T2 3, , Put T = 32 N, 32, = T1, 16, , 9, × 32 = T2, 16, , T1 = 2 N, , T2 = 18 N, , of the options on T1 is right., 33. If in a stationary wave the amplitude corresponding to antinode is 4 cm, then the amplitude corresponding to, a particle of medium located exactly midway between a node and an antinode is, (1) 2 cm, , (2) 2 2 cm, , (3), , 2 cm, , (4) 1.5 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 115 :
Solution of Assignment, , Waves, , 115, , 36. The two waves of the same frequency moving in the same direction give rise to, (1) Beats, , (2) Interference, , (3) Stationary waves, , (4) None of these, , Sol. Answer (2), The two waves of same frequency moving in the same direction give rise to interference., 37. The string of a violin emits a note of 205 Hz at its correct tension. The string is tightened slightly and then it, produces six beats in two seconds with a tuning fork of frequency 205 Hz. The frequency of the note emitted, by the taut string is, (1) 211 Hz, , (2) 199 Hz, , (3) 208 Hz, , (4) 202 Hz, , Sol. Answer (3), Initial frequency = 205 Hz, String tightned so frequency is increased = 205 + f, Final frequency of string – frequency of tuning fork = 3 beats, 205 + f – 205 = 3,  f=3,  Final frequency = 205 + 3 = 208 Hz, 38. When two tuning forks (fork 1 and fork 2) are sounded together, 4 beats per second are heard. Now some, tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second, are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?, (1) 204 Hz, , (2) 196 Hz, , (3) 202 Hz, , (4) 200 Hz, , Sol. Answer (2), Frequency of fork 1 = 200 Hz, Frequency of fork 2 = 200 ± 4, When tape is added frequency of fork 2 decreases., When frequency of fork 2 decreases number of beats increases., Hence we know frequency of fork 2 is, f2 = 200 – 4 = 196 Hz, 39. A train moving at a speed of 220 ms –1 towards a stationary object, emits a sound of frequency, 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of, the echo as detected by the driver of the train is (Speed of sound in air is 330 ms–1), (1) 3500 Hz, , (2) 4000 Hz, , (3) 5000 Hz, , (4) 3000 Hz, , Sol. Answer (3), v = 220 ms–1, f = 1000 Hz, Frequency of echo =, , fe =, , 300  220, f, 330  220, , 550, f, 110, , or fe = 5 f, or fe = 5000 Hz, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 116 :
116, , Waves, , Solution of Assignment, , 40. A source of frequency  gives 5 beats/second when sounded with a source of frequency 200 Hz. The second, harmonic of frequency 2 of source gives 10 beats/second when sounded with a source of frequency 420 Hz., The value of  is, (1) 205 Hz, , (2) 195 Hz, , (3) 200 Hz, , (4) 210 Hz, , Sol. Answer (1),  = 200 ± 5, , ... (i), , 2 = 420 ± 10,  = 210 ± 5, , ... (ii), , Common value for (i) & (ii) is 205 Hz., Hence  = 205 Hz, 41. A vibrating tuning fork is moving slowly and uniformly in a horizontal circular path of radius 8 m. The shortest, distance of an observer in same plane from the tuning fork is 9 m. The distance between the tuning fork and, observer at the instant when apparent frequency becomes maximum is, (1) 9 m, , (2) 25 m, , (3) 15 m, , (4), , 353 m, , Sol. Answer (3), The apparent frequency is maximum when relative velocity of approach of tuning fork with respect to observer, is maximum., , vs, , P, 8, , O, , OP =, , 9, , 8, , 172  82 = 15 m, , 42. The frequency changes by 10% as a sound source approaches a stationary observer with constant speed Vs., What would be percentage change in the frequency as the source recedes the observer with same speed, (Vs < V)?, (1) 10.5%, , (2) 8.5%, , (3) 4.5%, , (4) 1.5%, , Sol. Answer (2), , f , , v, f0, v  vs, , f is such that f =, , When, , 110, f0, 100, , v, 110, , v  v s 100, , 100 v = 110 v – 110 vs, v = 11 vs, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 117 :
Solution of Assignment, , Waves, , 117, , When source is received, v, x, f0 , f0, v  vs, 100, , Putting v = 11 vs, , 11, × 100 = x, 12, x = 91.66%, % change = 100 – 91.66  8.5%, 43. A train blowing its whistle moves with constant speed on a straight track towards observer and then crosses, him. If the ratio of difference between the actual and apparent frequencies be 3 : 2 in the two cases, then, the speed of train is [v is speed of sound], (1), , 2v, 3, , (2), , v, 5, , (3), , v, 3, , (4), , 3v, 2, , Sol. Answer (2), v, fA = v  v f0, s, , (Frequency of approach), , v, fR = v  v f0, s, , (Frequency of departing), , ⎛ v, ⎞, fA = ⎜ v  v  1⎟ f0, s, ⎝, ⎠, , v, fA = v  v f0, s, v ⎞, ⎛, fR = ⎜1  v  v ⎟ f0, s⎠, ⎝, , v, fR = v  v f0, s, fA v  v s 3, , , fR v  v s 2, 2 v = 2 vs = 3 v – 3 vs, 5 vs = v, vs =, , v, 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 118 :
118, , Waves, , Solution of Assignment, , 44. A man is standing on a railway platform listening to the whistle of an engine that passes the man at constant, speed without stopping. If the engine passes the man at time instant t0, how does the frequency f of the whistle, as heard by the man changes with time?, , f, , f, , (1), , f, , (2), , t0, , f, , (3), , t, , (4), , t, , t0, , t0, , t, , t0, , t, , Sol. Answer (1), When the train is approaching with speed vs. The apparent frequency is constant and fA , , When it is departing it is again constant, suddenly changing after t0 to f0 , , v, f0, v  vs, , v, f0 ., v  vs, , Hence graph will be that of option (1)., 45. A whistle ‘S’ of frequency f revolves in a circle of radius R at a constant speed v. What is the ratio of maximum, and minimum frequency detected by a detector D at rest at a distance 2R from the center of circle as shown, in figure? (take ‘c’ as speed of sound), , D, , R, , S, , 2R, ⎛c v ⎞, ⎟, (1) ⎜, ⎝c v ⎠, , (2), , ⎛c v ⎞, 2⎜, ⎟, ⎝c v ⎠, , (3), , 2, , (4), , (c  v ), c 2, , Sol. Answer (1), c = Speed of sound, At maximum frequency sources directly approaches observes with speed vs.,  fA =, , c, f0, c v, , At minimum frequency sources recedes with vs, fR , , c, f0, c v, , fA c  v, , fR c  v, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 119 :
Solution of Assignment, , Waves, , 119, , SECTION - C, Previous Years Questions, 1., , A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies, of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant, frequency for this string is, [Re-AIPMT-2015], (1) 105 Hz, , (2) 155 Hz, , (3) 205 Hz, , (4) 10.5 Hz, , Sol. Answer (1), ( n  1) 420, , n, 315, n  1 84, , n, 63, , 63 n + 63 = 84 n, 63 = 21 n, n=3, 1, of 315 is fundamental., 3, , f = 105, 2., , A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance, from each other. The source is moving with a speed of 19.4 ms–1 at an angle of 60° with the source observer, line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity, of sound in air 330 ms–1), is, [Re-AIPMT-2015], VS, , 60°, (1) 97 Hz, , S, , O, , (2) 100 Hz, , (3) 103 Hz, , (4) 106 Hz, , Sol. Answer (3), ⎛, ⎞, V,    ⎜, ⎟, ⎝ V  Vs cos  ⎠, , Vs, , ⎛, ⎞, ⎜, ⎟, 330, = 100  ⎜, ⎟, ⎜ 330  15.4  1 ⎟, ⎝, 2⎠, , ⎛ 330 ⎞, = 100  ⎜, ⎟, ⎝ 330  9.7 ⎠, =, ,  = 60º, S, , V0 = 0, O, , 33000,  103 Hz, 3  20  3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 120 :
120, 3., , Waves, , Solution of Assignment, , The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ, pipe open at both the ends. The length of organ pipe open at both the ends is, [AIPMT-2015], (1) 140 cm, , (2) 80 cm, , (3) 100 cm, , (4) 120 cm, , Sol. Answer (4), c , , V, 4l c, , ⎛V ⎞, c  3 ⎜, ⎟ = 2nd overtone, ⎝ 2l0 ⎠, 2V, V, , 2l0 4l c, ,  6lc = l0, 6 × 20 = l0, 120 = l0, , 4., , If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original, fundamental frequency n of the string is given by, [AIPMT-2014], , 1, , 1 1, 1, 1, (1) n  n  n  n, 1, 2, 3, , (2), , n  n1  n2  n3, , (3), , n, , , , 1, n1, , , , 1, n2, , , , 1, n3, , (4) n = n1 + n2 + n3, , Sol. Answer (1), , n1 , , 1 T, 1 T, ⇒ l1 , 2n1 , nl1 , , l  l1  l 2  l3, , n2 , , 1 T, 1 T, ⇒ l2 , 2l 2 , 2n2 , , 1 T, 1 T, 1 T, 1 T, , , , 2n  2n1  2n2  2n3 , , n3 , , 1 T, 1 T, ⇒ l3 , 2l3 , 2n3 , , l1, , l2, , l3, , 1 1, 1, 1, , , , n n1 n2 n3, , 5., , The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose, [AIPMT-2014], frequencies lie below 1250 Hz are (velocity of sound = 340 ms–1), (1) 4, , (2) 5, , (3) 7, , (4) 6, , Sol. Answer (4), The frequency which is integral multiple of fundamental frequency is called overtones, Now, n0 , , v, 340, ,  100, 4l0 4  0.85, , Now, combintation of frequencies = 100, 300, 500, 700, 900, 1200, i.e., 6 frequencies., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 121 :
Solution of Assignment, , 6., , Waves, , 121, , A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has, eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of, sound is 343 m/s, the frequency of the honk as heard by him will be, [AIPMT-2014], (1) 1332 Hz, , (2) 1372 Hz, , Sol. Answer (3), , (3) 1412 Hz, , (4) 1454 Hz, , N 0 = 343, v 0 = 10 m/s, , v s = 5 m/s, , ⎛ v  v0 ⎞, ⎛ 343  10 ⎞, N   N0 ⎜, ⎟  1392  ⎜, ⎟  1412 Hz, ⎝ 343  5 ⎠, ⎝ v  vs ⎠, , 7., , If we study the vibration of a pipe open at both ends, then the following statement is not true:, (1), , Odd harmonics of the fundamental frequency will be generated, , (2), , All harmonics of he fundamental frequency will be generated, , (3), , Pressure change will be maximum at both ends, , (4), , Open end will be antinode, , [NEET-2013], , Sol. Answer (3), 8., , A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The, second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source, of frequency 513 Hz. The unknown frequency is, [NEET-2013], (1) 246 Hz, , (2) 240 Hz, , (3) 260 Hz, , (4) 254 Hz, , Sol. Answer (4), 9., , A wave travelling in the positive x-direction having displacement along y-direction as 1 m, wavelength 2m and, 1, [NEET-2013], frequency of Hz is represented by, , (1) y = sin(2x – 2t), (2) y = sin(10x – 20t), (3) y = sin(2x + 2t) (4) y = sin(x – 2t), , Sol. Answer (4), 10. Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4 sin600t and, y2 = 5 sin 608t. An observer located near these two sources of sound will hear, [AIPMT (Prelims)-2012], (1), , 8 beats per second with intensity ratio 81 : 1 between waxing and waning, , (2), , 4 beats per second with intensity ratio 81 : 1 between waxing and waning, , (3), , 4 beats per second with intensity ratio 25 : 16 between waxing and waning, , (4), , 8 beats per second with intensity ratio 25 : 16 between waxing and waning, , Sol. Answer (2), 1 = 600  = 2f1, 2 = 608  = 2f2, f1 = 300 Hz, f2 = 304 Hz,  Beats heard will be 304 – 300 = 4 beats / s, Imax 81, Imax ( A1  A2 )2, , , 2 or I, 1, Imin ( A1  A2 ), min, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 122 :
122, , Waves, , Solution of Assignment, , 11. When a string is divided into three segments of length l1, l2 and l3, the fundamental frequencies of these three, segments are v1, v2 and v3 respectively. The original fundamental frequency (v) of the string is, [AIPMT (Prelims)-2012], , 1 1, 1, 1, (1) v  v  v  v, 1, 2, 3, (3), , (2), , v  v1  v 2  v 3, , 1, v, , , , 1, v1, , , , 1, v2, , , , 1, v3, , (4) v = v1 + v2 + v3, , Sol. Answer (1), Let original fundamental frequency be =, l = l1 + l2 + l3, , , f1 = 2l, 1, Similarly, l 2 , , v, 2l, , ... (i), , , or l1 = 2f, 1, v, 2f2, , l3 , , v, 2f3, , l, , v, 2f, , Putting values for l, l1, l2, l3 in (i), v, v, v, v, , , , 2f 2f1 2f2 2f3, 1 1 1 1,   , f f1 f2 f3, , , (50t – x), where x and y are in metres and t is, 2, in seconds. The ratio of maximum particle velocity to the wave velocity is, [AIPMT (Mains)-2012], , 12. The equation of a simple harmonic wave is given byy = 3sin, , (1) 2 , , (2), , 3, , 2, , (3) 3 , , (4), , 2, , 3, , Sol. Answer (2), 13. A train moving at a speed of 220 ms–1 towards a stationary object, emits a sound of frequency 1000 Hz. Some, of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected, [AIPMT (Mains)-2012], by the driver of the train is: (Speed of sound in air is 330 ms–1), (1) 3500 Hz, , (2) 4000 Hz, , (3) 5000 Hz, , (4) 3000 Hz, , Sol. Answer (3), 14. Two waves are represented by the equations, y1= a sin (t + kx + 0.57) m and, y2 = a cos (t + kx) m, where x is in meter and t in sec. The phase difference between them is, (1) 0.57 radian, , (2) 1.0 radian, , (3) 1.25 radian, , [AIPMT (Prelims)-2011], (4) 1.57 radian, , Sol. Answer (2), Phase difference is simply difference in the argument in sine function., We can write y2 = a sin [t + kx + /2],   = t + kx + /2 – (t + kx + 0.57), = 1.57 – 0.57 = 1 radian, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 123 :
Solution of Assignment, , Waves, , 123, , 15. Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz, acoustic wave as it enters brass from warm air, [AIPMT (Prelims)-2011], (1) Decreases by a factor 20, , (2) Decreases by a factor 10, , (3) Increases by a factor 20, , (4) Increases by a factor 10, , Sol. Answer (4), Frequency will remain constant, va = Velocity in air, vb = Velocity in brass, va = 700 × a, vb = 700 × b, a =, , 350, 700, , a = 0.5 m/s, , b =, , 3500, 700, , b = 50 m, ,  b = 10 a, 16. Two identical piano wires, kept under the same tension T have a fundamental frequency of 600 Hz. The fractional, increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate, together would be, [AIPMT (Mains)-2011], (1) 0.04, , (2) 0.01, , (3) 0.02, , (4) 0.03, , Sol. Answer (3), Beats = 6 Hz, New frequency of one of the wires = 600 + 6 = 606, f=, , v, T 1, , , 2l,  2l, 2, , T  T, T, ⎛ 606 ⎞, or, = 0.0201  0.02, ⎜, ⎟ , T, T, ⎝ 600 ⎠, , 17. A transverse wave is represented by y = A sin(t – kx). For what value of the wavelength is the wave velocity, equal to the maximum particle velocity?, [AIPMT (Prelims)-2010], , A, 2, Sol. Answer (3), (1), , (2) A, , (3) 2A, , (4) A, , 18. A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat, frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency, of the piano string before increasing the tension was, [AIPMT (Prelims)-2010], (1) 508 Hz, , (2) 510 Hz, , (3) 514 Hz, , (4) 516 Hz, , Sol. Answer (1), Tuning fork of frequency = 512 Hz, Frequency of wire = 512 ± 4, Since frequency beats decrease when tension is increased, frequency of tuning fork must be greater than string, initially.,  Frequency of piano = 508 Hz, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 124 :
124, , Waves, , Solution of Assignment, , 19. Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit, length of both the strings is same and equal to 1 g/m. When both strings vibrate simultaneously the number of, beats is, [AIPMT (Prelims)-2009], (1) 7, , (2) 8, , (3) 3, , l2 = 49.1 cm, , T = 20 N, , (4) 5, , Sol. Answer (1), l1 = 51.6 cm, ,  = 1 g/m = 0.001 kg/m, v, 20, 1, =, , 2l1, 0.001 2  0.516, , ⎡, T⎤, ⎢v , ⎥, ⎦, ⎣, ,  137 Hz, Similarly,, , v, = 144 Hz, 2l 2, Number of beats = 144 – 137 = 7 Hz], 20. The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. If the, velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is:, [AIPMT (Prelims)-2009], (1) 555.5 Hz, , (2) 720 Hz, , (3) 500 Hz, , (4) 550 Hz, , Sol. Answer (2), 21. A wave in a string has an amplitude of 2 cm. The wave travels in the positive direction of x axis with a speed, of 128 m/s and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the, wave is, [AIPMT (Prelims)-2009], (1) y = (0.02) m sin (15.7x – 2010t), , (2) y = (0.02) m sin (15.7x + 2010t), , (3) y = (0.02) m sin (7.85x – 1005t), , (4) y = (0.02) m sin (7.85x + 1005t), , Sol. Answer (3), =, , 4, = 0.8 m, 5, , Velocity = 128 m/s, =, , 2v 2  128, , = 320  = 1005, , 0.8, , k=, , 2, = 7.85, 0.8, ,  Equation of the form y = A sin (kx – t) is the first option., 22. The wave described by y = 0.25 sin(10x – 2t), where x and y are in meters and t in seconds, is a wave, travelling along the, [AIPMT (Prelims)-2008], (1) –x-direction with amplitude 0.25 m and wavelength  = 0.2 m, (2) –x-direction with frequency 1 Hz, (3) +x-direction with frequency  Hz and wavelength  = 0.2 m, (4) +x-direction with frequency 1 Hz and wavelength  = 0.2 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 125 :
Solution of Assignment, , Waves, , 125, , Sol. Answer (4), y = 0.25 sin(10 x – 2 t) ... (i), Frequency (f) =, , , = 1 Hz, 2, , Wavelength () =, , 2 1,  m, k, 5, , Equation (i) is the equation of wave travelling in +ve x-direction., Hence answer is (0.2) option (4), 23. Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation, is 0.05 sec and the velocity of the wave is 300 m/sec. What is the phase difference between the oscillations of, two points?, [AIPMT (Prelims)-2008], (1), , , 6, , (2), , , 3, , (3), , 2, 3, , (4) , , Sol. Answer (3), T = 0.05 s,  f=, , Velocity (v) = 300 m/s, , 1, = 20 Hz, T, , = 2f = 40 , , k, , Since velocity =, 300 =, , k =, , 40 , k, , 2, 15, ,  = 15 m, Put x = 10 and x = 15 in equation, y = A sin (kx – t), , ⎛ 2, ⎞, = A sin⎜ x  40t ⎟, ⎝ 15, ⎠, Time (t) is same., , 2, y1 = A sin⎛⎜ 10  t ⎞⎟, ⎝ 15, ⎠, y2 = A sin (2 – t), , ⎛ 4, ⎞,  t ⎟,  = (2 – t) – ⎜, ⎝ 3, ⎠, or  = 2 –, or  =, , 4, 3, , 2, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 126 :
126, , Waves, , Solution of Assignment, , 24. Two sound waves with wavelengths 5 m and 5.5 m respectively, each propagate in a gas with velocity 330 m/, s. We expect the following number of beats per second, [AIPMT (Prelims)-2006], (1) 12, , (2) 0, , (3) 1, , (4) 6, , Sol. Answer (4), v 330, f1 =  = 5 = 66Hz, 1, v, 330, f2 =  = 5.5 = 60Hz, 2, , Beats = f2 – f1 = 6 beats/s, , 25. A transverse wave propagating along x-axis is represented by: y(x, t) = 8 sin(0.5x – 4t –, where x is in metres and t is in seconds. The speed of the wave is, (1) 4 m/s, , (2) 0.5 m/s, , (3), , , m/s, 4, , , ), 4, , [AIPMT (Prelims)-2006], (4) 8 m/s, , Sol. Answer (4), v=, , , k, , v=, , 4,  /2, , v = 8 m/s, 26. The time of reverberation of a room-A is one second. What will be the time (in seconds) of reverberation of a, room, having all the dimensions double of those of room-A?, [AIPMT (Prelims)-2006], (1) 2, , (2) 4, , (3), , 1, 2, , (4) 1, , Sol. Answer (1), Sabini's formula for reverberation time is, T=, , 0.16V,  as, , Where V is volume of hall in m3, , T, V s, , =, T, s V, T, 23, = 2 or T = 2 s, T, 2, , 27. Which one of the following statements is true?, (1), , Both light and sound waves in air are transverse, , (2), , The sound waves in air are longitudinal while the light waves are transverse, , (3), , Both light and sound waves in air are longitudinal, , (4), , Both light and sound waves can travel in vacuum, , [AIPMT (Prelims)-2006], , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 127 :
Solution of Assignment, , Waves, , 127, , 28. A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at, distance of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is, [AIPMT (Prelims)-2005], (1) 9 : 4, , (2) 2 : 3, , (3) 3 : 2, , (4) 4 : 9, , Sol. Answer (1), I, , 1, r2, k, 22, , IP =, , IQ =, , k, 32, , IP = IQ = 9 : 4, 29. Two vibrating tuning forks produce progressive waves given by y1 = 4sin 500t and y2 = 2sin 506t. Number of, beats produced per minute is, [AIPMT (Prelims)-2005], (1) 360, , (2) 180, , (3) 3, , (4) 60, , Sol. Answer (2), y1 = 4 sin 500 t and y2 = 2 sin 506 t, 1 = 2f1 = 500 , f1 = 250 Hz, Similarly,, , 2f2 = 506 , f2 = 253 Hz, , Beats = f2 – f1 = 3 Hz, 3 beats in 1 second then number of beats in 1 minute = 3 × 60 = 180, 30. Two waves are represented by the equations, y1 = asin(t + kx + 0.57)m and, y2 = acos(t + kx) m,, where x is in metre and t in second. The phase difference between them is, (1) 0.57 radian, , (2) 1.0 radian, , (3) 1.25 radian, , (4) 1.57 radian, , Sol. Answer (2), 31. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner, is 4.2 MHz. The speed of sound in a tissue is 1.7 km/s. The wavelength of sound in the tissue is close to, (1) 4 × 10–3 m, , (2) 8 × 10–3 m, , (3) 4 × 10–4 m, , (4) 8 × 10–4 m, , Sol. Answer (3), f = 4.2 × 106 Hz, v = 1700 m/s, =, =, , v, f, 1700, = 0.404 × 10–3, 42  105, , = 4 × 10–4, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 128 :
128, , Waves, , Solution of Assignment, , 32. Two sound waves having a phase difference of 60° have path difference of, (1), , , 6, , (2), , , 3, , (3) 2, , (4), , , 2, , Sol. Answer (1), Path difference =, , , × Phase difference., 2, ,  Path difference =, , , m, 6, , 33. A transverse wave is represented by the equation y  y 0 sin, , 2, (vt  x ) . For what value of  the maximum, , , particle velocity is equal to two times the wave velocity?, (1)  , , y 0, 2, , (2)  , , y 0, 3, , (3)  = 2y0, , (4)  = y0, , Sol. Answer (4), y  y 0 sin, , 2, (vt  x ), , , dy, v, 2, = y 0 2 cos (vt  x ), dt, , , , Maximum particle velocity = y 0, , 2v, , , Wave velocity = v, y 0 2v = 2 v, , y = , 0, , 34. A wave travelling in positive x-direction with A = 0.2 m, velocity = 360 m/s and  = 60 m, then correct, expression for the wave is, ⎡ ⎛, x ⎞⎤, (1) y  0.2 sin⎢2⎜ 6t  60 ⎟⎥, ⎝, ⎠⎦, ⎣, , ⎡ ⎛, x ⎞⎤, (2) y  0.2 sin⎢⎜ 6t  60 ⎟⎥, ⎝, ⎠⎦, ⎣, , ⎡ ⎛, x ⎞⎤, (3) y  0.2 sin⎢2⎜ 6t  60 ⎟⎥, ⎝, ⎠⎦, ⎣, , ⎡ ⎛, x ⎞⎤, (4) y  0.2 sin⎢⎜ 6t  60 ⎟⎥, ⎝, ⎠⎦, ⎣, , Sol. Answer (3), A = 0.2 m, , Velocity = 360 m/s and  = 60 m, , We are looking for a waves of the form y = A sin t – kx, =, , 2v 2  360, ,  12, , 60, , k =, , 2  2, =,  60, , Correct value of  and k are found in option (3)., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 129 :
Solution of Assignment, , Waves, , 129, , 35. The phase difference between two waves, represented by y 1 = 10 –6sin[100t + (x/50) + 0.5] m and, y2 = 10–6cos[100t + (x/50] m, where x is expressed in metre and t is expressed in second, is approximately, (1) 1.07 radian, , (2) 2.07 radian, , (3) 0.5 radian, , (4) 1.5 radian, , Sol. Answer (1), , x, ⎛, ⎞, y1 = 106 sin ⎜100t ,  0.5 ⎟, 50, ⎝, ⎠, x ⎞, y2 = 106 cos ⎛⎜100t , ⎟, 50 ⎠, ⎝, x ⎞, ⎛, y2 = 106 sin ⎜100t ,  ⎟, 50 2 ⎠, ⎝, Phase differences, , x ⎞ ⎛, x, ⎛, ⎞,  ⎟  ⎜100t ,  0.5 ⎟, ⎜100t , 50 2 ⎠ ⎝, 50, ⎝, ⎠, , =, , 3.14 1, , 2, 2, , = 1.07 radian, 36. A wave of frequency 100 Hz travels along a string towards its fixed end. When this wave travels back, after, reflection, a node is formed at a distance of 10 cm from the fixed end. The speed of the wave (incident and, reflected) is, (1) 20 m/s, , (2) 40 m/s, , (3) 5 m/s, , (4) 10 m/s, , Sol. Answer (1), Node is formed at 10 cm,  Wavelength = 2 × 10 = 20 cm,  = 0.2 cm, v = f  = 0.2 × 100 = 20 m/s, 37. A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21 Å, between them. The wavelength of the standing wave is, (1) 6.05 Å, , (2) 2.42 Å, , (3) 1.21 Å, , (4) 3.63 Å, , Sol. Answer (3), Total distance 1.21 Å, , 1.21 Å, According to description the wave must like above diagram wavelength () = 1.21 Å, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 130 :
130, , Waves, , Solution of Assignment, , 38. Two waves of wavelengths 50 cm and 51 cm produced 12 beats per second. The velocity of sound is, (1) 340 m/s, , (2) 331 m/s, , (3) 306 m/s, , (4) 360 m/s, , Sol. Answer (3), v, f1 = , 1, , v, f2 = , 2, , f1 – f2 = 12, v, v, , = 12, 0.5 0.51, , Solving for v, we get v = 306 m/s., 39. Two stationary sources each emit waves of wavelength . An observer moves from one source to another with, velocity u. Then number of beats heard by him, (1), , 2u, , , (2), , u, , , (3), , u, , (4), , u, 2, , Sol. Answer (1), Let both have the same frequencies 0., Let the initial source be s1 and the source is approaching be s2., f0 =, , v, , , fs1 , , v  v0, 1,  f0  (v  v 0 ), v, , , fs2 , , v  v0, 1,  f0  (v  v 0 ), v, , , v0 = u and fs2  fs1  ?, , ⎡(v  u ) (v  u ) ⎤, ⎢⎣    ⎥⎦ = number of beats., ⎡v  u  v  u ⎤, ⎢⎣, ⎥⎦ = number of beats., , 2u, = number of beats., , , 40. Two vibrating tuning forks produce progressive waves given by y1 = 4sin 500 t and y2 = 2sin 506 t. Number, of beats produced per minute is, (1) 360, , (2) 180, , (3) 60, , (4) 3, , Sol. Answer (2), 41. A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the, straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency, n + n1. Then (if the sound velocity in air is 300 m/s), (1) n1 = 0.1 n, , (2) n1 = 0, , (3) n1 = 10 n, , (4) n1 = – 0.1 n, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 131 :
Solution of Assignment, , Waves, , 131, , Sol. Answer (2), If the object and source are not approaching each other in any direction there is no change in frequency due, to Doppler effect., Length = 50 cm, 42. A whistle revolves in a circle with angular speed  = 20 rad/s using a string of length 50 cm. If the frequency, of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer who is far, away from the centre (vsound = 340 m/s)?, (1) 385 Hz, , (2) 374 Hz, , (3) 394 Hz, , (4) 333 Hz, , Sol. Answer (2), vs = r = 20 × 0.5 = 10 m/s, Observed frequency is minimum when source moves from observer., Apparent frequency =, , v,  f0, v  vs, , =, , 340,  385, 340  10, , =, , 340,  385, 350, , v + vs, , = 34 × 11 = 374 Hz, 43. An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The, wavelength and frequency of the source emitted are  and f respectively. The apparent frequency and, wavelength recorded by the observer are respectively, (1) 1.2f, 1.2 , , (2) 1.2f, , , (3) f, 1.2 , , (4) 0.8f, 0.8 , , Sol. Answer (2), Wavelength of wave does not change due to the Doppler effect., Apparent frequency =, , v  vs, 6,  f =  f or 12 f, v, 5, , Wavelength = , 44. A car is moving towards a high cliff. The driver sounds a horn of frequency f. The reflected sound heard by, the driver has frequency 2f if v the velocity of sound, then the velocity of the car, in the same velocity units,, will be, (1), , v, , (2), , 2, , v, 3, , (3), , v, 4, , (4), , v, 2, , Sol. Answer (2), Frequency with which sound hits the wall (f1) , , v, f, v  vc, , Frequency with which man hears the sound again (f2 ) , or, , f2 , , vc =, , v  vc, f, v  vc, , or, , 2f , , v  vc,  f1, v, , v  vc,  f or 2 v – 2 vc = v + vc, v  vc, , v, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 132 :
132, , Waves, , Solution of Assignment, , , 45. The equation of a simple harmonic wave is given by y  3 sin (50t  x ) , where x and y are in metres and t, 2, is in seconds. The ratio of maximum particle velocity to the wave velocity is, , (1) 2, , (2), , 3, , 2, , (3) 3, , (4), , 2, , 3, , Sol. Answer (2), , , y  3 sin (50t  x ), 2, Maximum particle velocity = A, or 3 × 25  = 75 , Wave velocity =, , Ratio =, ,  50, ,  50m/s, k, 1, , 75 3, , 50, 2, , 46. Which one of the following statements is true?, (1) The sound waves in air are longitudinal while the light waves are transverse, (2) Both light and sound waves in air are longitudinal, (3) Both light and sound waves can travel in vacuum, (4) Both light and sound waves in air are transverse, Sol. Answer (1), Sound is a mechanical wave and travels longitudinally in air., Light being electromagnetic will be all transversely irrespective of the medium., , SECTION - D, Assertion-Reason Type Questions, 1., , A : Doppler’s effect in sound is asymmetric but in light, it is symmetric., R : In sound, change in frequency depends on the individual velocity of both the source as well as the observer., In light, change in frequency depends on the relative velocity between source and observer., , Sol. Answer (2), Doppler’s effect in sound is different when the object is moving towards source and when source moves towards, observer., Hence it is asymmetric. Light is symmetric as it just depends on relative motion between the source and the, object., 2., , A : The propagation of sound in air should be an isothermal process., R : As air is bad conductor of heat, its temperature does not change by compression or expansion., , Sol. Answer (4), The propagation of sound in air is an adiabatic process as very little sound energy get dissipated as heat., Air may be a bad conductor of heat but its temperature does change when work is performed on it., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 133 :
Solution of Assignment, , 3., , Waves, , 133, , A : Velocity of sound in air increases with increase in humidity., R : Velocity of sound doesn’t depend upon medium., , Sol. Answer (3), , v=, , P, , , The  or density reduces because of presence of water vapour as both N2 and O2 are heavier than H2O., Velocity of sound, thus, depends largely on the medium., 4., , A : Intensity of sound wave does not change when the listener moves towards or away from the stationary, source., R : The motion of listener towards a stationary source causes an apparent change in wavelength of sound., , Sol. Answer (4), Intensity of sound changes when observer moves towards are away from source because of change in frequency., Motion of observers only causes change in frequency of sound and not the wavelength., 5., , A : A vibrating tuning fork sounds louder when its stem is pressed against desk top., R : When a wave reaches another denser medium, part of the wave is reflected., , Sol. Answer (2), When a vibrating tuning fork is held in hand only air is set into vibration. When its stem is placed in contact, with the labels the entire labels is set into forced vibrations., 6., , A : Longitudinal waves do not exhibit the phenomenon of polarisation., R : In longitudinal waves medium particle vibrate in direction normal to the wave propagation., , Sol. Answer (3), Polarisation means redirecting a wave to propagate in only one plane., The molecules in a longitudinal wave vibrate along the direction of propagation of wave and hence cannot be, redirected by any material., 7., , A : If a wave moving in a rarer medium, gets reflected at the boundary of a denser medium, then it encounters, a sudden change in phase of ., R : If a wave propagating in a denser medium, gets reflected from rarer medium, then there will be no abrupt, phase change., , Sol. Answer (2), Both the statements are true but (2) does not give the correct explanation for (1)., 8., , A : Speed of sound in moist air is more than its speed in dry air., R : Dry air is denser than moist air at atmospheric pressure., , Sol. Answer (1), , Speed of sound in air is v =, , P, , , When the air is humid, the density of air reduces as O2 and N2 are heavier than H2O., Hence speed of sound decreases., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 134 :
134, 9., , Waves, , Solution of Assignment, , A : Sound travels faster in solids as compared to liquids and gases., R : Solids are more elastic than liquids and gases., , Sol. Answer (1), , v, , B, , , Hence since the Bulk molecules of solids is higher the speed of sound is higher in solids., Since ‘B’ is refer to bulk molecule., Speed depends on elasticity and reason is correct explanation of assertion., 10. A : There is no energy transferred by standing waves., R : The total energy of standing waves is twice the energy of each of incident and reflected wave., Sol. Answer (2), Both statements are true but reason is not the correct explanation of each other., 11. A : In Doppler’s effect the value of apparent frequency depends on the relative motion between source and, observer., R : The change in frequency in Doppler effect is independent from the distance between source and observer., Sol. Answer (2), Both the statements are true but reason does not explain the cause for assertion., 12. A : The pitch of female voice is higher than the pitch of male voice., R : Pitch distinguishes between a shrill and a grave sound., Sol. Answer (2), The assertion is a true fact and so is the reason. But the reason offers no explanation for the assertion., , , , , , , , 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