DC Pandey Waves And Thermodynamics

6 Waves & Thermodynamics During recede, f2 = f  v  So, we have 6 possibilities.  +   v vs = 1000  320  = 941.18 Hz    320 + 20  |% change in frequency| =  f1 − f2  × 100 ≈ 12%  f1    15. (b) Given, U ∝ T 4 Alternate method V In closed organ pipe, fundamental node U = αT 4 ...(i) V It is also given that, P= 1  U  ⇒ nR 0T = 1 (αT 4 ) 3 V V 3 = 0.85 = λ (R0 = Gas constant) l 4 or VT 3 = 3nR0 = constant α ∴  4 πR 3 T 3 = constant or RT = constant i.e. λ = 0.85 3 4 ∴ T∝1 ⇒ λ = 4 × 0.85 R As we know, ν = c λ 340 16. (b) Average time between two collisions is given by ⇒ 4 × 0.85 = 100 Hz 1 τ= 2 πnvrms d 2 ...(i) Here, n = number of molecules per unit volume = N ∴ Possible frequencies = 100 Hz, 300 Hz, 500 Hz, V 700 Hz, 900 Hz, 1100 Hz below 1250 Hz. 3RT 19. (a) In steady state M and vrms = Substituting these values in Eq.(i) we have, ...(ii) I S =4πR2 τ∝ V T For adiabatic process, TV γ −1 = constant πR2 Radiation Incident V Substituting in Eq. (ii), we have τ ∝ Energy incident per second = Energy radiated per 1 second  V 1  ∴ IπR 2 = σ (T 4 − T04 ) 4 πR 2 γ− ⇒ I = σ (T 4 − T04 ) 4 τ ∝ V1 +  γ − 1  V  1 + γ  ⇒ T 4 − T04 = 40 × 108  2   2  or or τ ∝ ⇒ T 4 − 81 × 108 = 40 × 108 ⇒ T 4 = 121 × 108 17. (a) Entropy is a state functions. Therefore in both cases answer should be same. ⇒ T ≈ 330 K 18. (c) For closed organ pipe = (2 n + 1)v [n = 0, 1, 2 ...... ] 20. (c) In thermal conduction, it is found that in steady 4l state the heat current is directly proportional to the (2 n + 1)v area of cross-section A which is proportional to the 4l < 1250 change in temperature (T1 − T2 ). Then, ∆Q = KA(T1 − T2 ) (2 n + 1) < 1250 × 4 × 0.85 ∆t x 340 (2 n + 1) < 12.5 2 n < 11.50 ⇒ n < 5.25 So, n = 0, 1, 2, 3,...,5

Previous Years’ Questions (2018-13) 7 According to thermal conductivity, we get 23. (c) According to Newton’s cooling law, option (c) is 100°C correct answer. l1 dQ1 24. (b) Heat is extracted from the source means heat is K1 dt given to the system (or gas) or Q is positive. This is positive only along the path ABC. Heat supplied ∴ QABC = ∆U ABC + WABC dQ2 K2 = nC V (Tf − Ti ) + Area under p-V graph dt l2 dQ3 3 K3 dt = n 2 R  (TC − TA ) + 2 p0V0 l3 0°C 0°C = 3 (nRTC − nRTA ) + 2 p0V0 2 i.e. dQ 1 = dQ2 + dQ3 dt dt dt = 3 ( pC VC − pA VA ) + 2 p0V0 0.92 (100 T) 0.26 (T − 0) + 0.12 (T − 0) 2 46 − = 13 12 3 ⇒ T = 40° C = 2 (4 p0V0 − p0V0 ) + 2 p0V0 ∴ dQ1 = 0.92 × 4 (100 − 40) 4.8 cal/s = 13 p0 V0 dt 40 = 2 21. (d) According to first law of thermodynamics, we get 25. (c) In equilibrium, (i) Change in internal energy from A to B i.e. ∆U AB p0 A = Mg …(i) when slightly displaced downwards, ∆U AB = nC V5(TRB( − TA ) = 1× 2 800 − 400) = 1000 R (ii) Change in internal energy from B to C ∆U BC = nC V(TC − TB ) x 5R (600 1× 2 800) 0 = − = − 500 R (iii) ∆Uisothermal = 0 dp = − γ  p0  dV  V0  (iv) Change in internal energy from C to A i.e. ∆UCA   ∆UCA = nC V(TA − TC ) = − 500 R  As in adiabatic process, dp = − γ Vp dV 22. (b) Fundamental frequency of sonometer wire ∴ Restoring force, f = v = 1 T = 1 T F = (dp)A 2l 2l µ 2l Ad = −  γ p0  ( A) ( Ax)   Here, µ = mass per unit length of wire.  V0  Also, Young’s modulus of elasticity Y = Tl F ∝− x A∆l Therefore, motion is simple harmonic comparing with 1 ⇒ T = Y∆l ⇒ f = 2l Y∆l F = − kx we have A l ld k = γ p0 A2 ⇒ l = 1.5 m, ∆l = 0.01 V0 l 1 k d = 7.7 × 103 kg / m3 ∴ f = 2π m ⇒ Y = 2.2 × 1011 N / m2 = 1 γ p0 A2 After substituting the values we get, 2π MV0 f ≈ 178.2 Hz

JEE Advanced 1. One mole of a monoatomic ideal gas cylinders in the steady state is 200K, then undergoes a cyclic process as shown in the figure (where, V is the volume and T is K1/ K2 = .......... . (Numerical Value, 2018) the temperature). Which of the statements below is (are) true ? Insulating material (More than One Correct Option, 2018) T1 K1 K2 T2 T L II L I III 4. In an experiment to measure the speed of sound by a resonating air column, a IV tuning fork of frequency 500 Hz is used. The length of the air column is varied by V changing the level of water in the resonance tube. Two successive resonances are heard (a) Process I is an isochoric process at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (b) In process II, gas absorbs heat (are) true?(More than One Correct Option, 2018) (c) In process IV, gas releases heat (a) The speed of sound determined from this m s−1 (d) Processes I and III are not isobaric experiment is 332 2. Two men are walking along a horizontal straight line in the same direction. The (b) The end correction in this experiment is 0.9 cm main in front walks at a speed 1.0ms−1 and the man behind walks at a speed (c) The wavelength of the sound wave is 66.4 cm 2.0 ms−1. A third man is standing at a height 12 m above the same horizontal (d) The resonance at 50.7 cm corresponds to line such that all three men are in a vertical plane. The two walking men are the fundamental harmonic blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of 5. One mole of a monoatomic ideal gas sound in air 330 ms−1. At the instant, undergoes an adiabatic expansion in when the moving men are 10 m apart, the which its volume becomes eight times its stationary man is equidistant from them. initial value. If the initial temperature of The frequency of beats in Hz, heard by the the gas is 100 K and the universal gas stationary man at this instant, is ............. constant R = 8.0 j mol−1 K −1, the decrease . (Numerical Value, 2018) in its internal energy in joule, is ............ . 3. Two conducting cylinders of equal length but different radii are connected in series (Numerical Value, 2018) between two heat baths kept at temperatures T1 = 300K and T2 = 100K, as 6. One mole of a p II shown in the figure. The radius of the monoatomic ideal 3p0 I IV III bigger cylinder is twice that of the smaller gas undergoes four one and the thermal conductivities of the thermodynamic p0 materials of the smaller and the larger processes as shown cylinders are K1 and K 2, respectively. If the temperature at the junction of the schematically in the V pV-diagram below. V0 3V0 Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-l with the corresponding statements in List-II. (Matching Type Question, 2018)

Previous Years’ Questions (2018-13) 9 List-I List-II 7. Which of the following options is the only correct representation of a process in P. In process I 1. Work done by the gas is zero which ∆U = ∆Q − p∆V ? Q. In process II R. In process III 2. Temperature of the gas (a) (II) (iii) (S) (b) (II) (iii) (P) remains unchanged S. In process IV (c) (III) (iii) (P) (d) (II) (iv) (R) 3. No heat is exchanged between the gas and its 8. Which one of the following options is the surroundings correct combination? 4. Work done by the gas is (a) (II) (iv) (P) (b) (III) (ii) (S) 6 p0 V0 (c) (II) (iv) (R) (d) (IV) (ii) (S) (a) P → 4; Q → 3; R → 1; S → 2 9. A block M hangs vertically at the bottom (b) P → 1; Q → 3; R → 2; S → 4 (c) P → 3; Q → 4; R → 1; S → 2 end of a uniform rope of constant mass (d) P → 3; Q → 4; R → 2; S → 1 per unit length. The top end of the rope is Directions (Q.Nos. 7-8) Matching the attached to a fixed rigid support at O. A information given in the three columns of the following table. transverse wave pulse (Pulse 1) of An ideal gas is undergoing a cyclic wavelength λ 0 is produced at point O on the rope. The pulse takes time TOA to thermodynamic process in different ways as reach point A. If the wave pulse of shown in the corresponding p-V diagrams in wavelength λ 0 is produced at point A (Pulse 2) without disturbing the position column 3 of the table. Consider only the path of M it takes time TAO to reach point O. from state 1 to state 2. W denotes the Which of the following options is/are correct? (More than One Correct Option, 2017) corresponding work done on the system. The O Pulse 1 equations and plots in the table have standards notations and used in thermodynamic processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of Pulse 2 AM moles in the gas is n. (Matching Type, 2017) Column 1 Column 2 Column 3 (a) The time TAO = TOA (i) Isothermal (P) p 1 2 (I) W1→ 2 = 1 (b) The wavelength of Pulse 1 becomes longer − γ 1 when it reaches point A ( p2V2 − p1V1) (c) The velocity of any pulse along the rope is V independent of its frequency and wavelength (II) W1→ 2 (ii) Isochoric (Q) p 1 (d) The velocities of the two pulses (Pulse 1 and (R) p 1 2 = − pV2 + pV1 Pulse 2) are the same at the mid-point of rope V 10. A stationary source emits sound of 2 (III) W1→ 2 = 0 (iii) Isobaric frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms−1. The reflected signal is received by the source V and superposed with the original. What 2 (IV) W1→ 2 = − nRT (iv) Adiabatic (S) p 1 will be the beat frequency of the resulting V ln  V2  signal in Hz? (Given that the speed of   sound in air is 330 ms−1 and the car  V1  reflects the sound at the frequency it has received). (Single Integer Type, 2017)

10 Waves & Thermodynamics 11. Two loudspeakers M and N are located P 3V 5 = constant. Consider another 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car in thermodynamic process that brings the initially at a point P, 1800 m away from system from the same initial state to the the midpoint Q of the line MN and moves same final state in two steps : an isobaric towards Q constantly at 60 km/h along the expansion at Pi followed by an isochoric perpendicular bisector of MN. (isovolumetric) process at volume V f . The It crosses Q and eventually reaches a amount of heat supplied to the system in point R, 1800 m away from Q. Let v(t) the two-step process is approximately represent the beat frequency measured by a person sitting in the car at time t. Let (Single Correct Option, 2016) νP, νQ and νR be the beat frequencies measured at locations P, Q and R (a) 112 J (b) 294 J (c) 588 J (d) 813 J respectively. The speed of sound in air is 330 ms−1. 13. A water cooler of storage capacity 120 Which of the following statement(s) is litres can cool water at a constant rate of (are) true regarding the sound heard by P watts. In a closed circulation system (as the person? shown schematically in the figure), the water from the cooler is used to cool an (More than One Correct Option, 2016) external device that generates constantly 3 kW of heat (thermal load). The (a) The plot below represents schematically the temperature of water fed into the device cannot exceed 30°C and the entire stored variation of beat frequency with time 120 litres of water is initially cooled to 10°C. ν(t) The entire system is thermally insulated. P The minimum value of P (in watts) for νQ Q which the device can be operated for Rt 3 hours is (Single Correct Option, 2016) Cooler Hot Device (b) The rate of change in beat frequency is Cold maximum when the car passes through Q (Specific heat of water is 4.2 kJ kg−1K−1 (c) νP + νR = 2νQ and the density of water is 1000 kg m−3) (d) The plot below represents schematically the variations of beat frequency with time ν(t) P (a) 1600 (b) 2067 (c) 2533 (d) 3933 νQ Q 14. A metal is heated in a furnace where a sensor is kept above the metal surface to Rt read the power radiated (P ) by the metal. The sensor has a scale that displays 12. A gas is enclosed in a cylinder with a log2(P/ P0 ), where P0 is a constant. When the metal surface is at a temperature of movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa 487°C, the sensor shows a value 1. and volume Vi = 10−3 m3 changes to a final state at Pf = (1/ 32) × 105 Pa and Assume that the emissivity of the metallic V f = 8 × 10−3 m3 in an adiabatic surface remains constant. What is the quasi-static process, such that value displayed by the sensor when the temperature of the metal surface is raised to 2767°C? (Single Integer Type, 2016)

Previous Years’ Questions (2018-13) 11 15. The ends Q and R of two thin wires, PQ Ignoring the friction between the piston and RS, are soldered (joined) together. and the cylinder, the correct statements Initially, each of the wire has a length of is/are (More than One Correct Option, 2015) 1 m 10°C. Now, the end P is maintained at 10°C, while the end S is heated and (a) IfV = 2V and T = 3T , then the energy maintained at 400°C. The system is 21 21 thermally insulated from its surroundings. If the thermal conductivity stored in the spring is 1PV of wire PQ is twice that of the wire RS and the coefficient of linear thermal 11 expansion of PQ is 1.2 × 10−5 K−1, the 4 change in length of the wire PQ is (b) IfV = 2V and T = 3T , then the change in (Single Correct Option, 2016) 21 21 internal energy is 3PV 11 (c) IfV = 3V and T = 4T , then the work done by 21 21 the gas is 7 PV 11 3 (d) IfV = 3V and T = 4T , then the heat 21 21 supplied to the gas is 17 PV 11 6 (a) 0.78 mm (b) 0.90 mm 18. Two spherical stars A and B emit (c) 1.56 mm (d) 2.34 mm blackbody radiation. The radius of A is 400 times that of B and A emits 104 times 16. A container of fixed volume has a mixture the power emitted from B. The ratio  λ A  of one mole of hydrogen and one mole of helium in equilibrium at temperature T . λB Assuming the gases are ideal, the correct statements is/are of their wavelengths λ A and λ B at which the peaks occur in their respective (a) The average energy per mole of the gas mixture is 2RT radiation curves is (Single Integer Type, 2015) (b) The ratio of speed of sound in the gas 19. Four harmonic waves of equal frequencies and equal intensities I0 have phase 6 angles 0, π, 2π and π. When they are mixture to that in helium gas is 33 superposed, the intensity of the resulting 5 wave is nI0. The value of n is (c) The ratio of the rms speed of helium atoms to (Single Integer Type, 2015) 1 20. Two vehicles, each moving with speed u that of hydrogen molecules is on the same horizontal straight road, are approaching each other. Wind blows 2 along the road with velocity w. One of these vehicles blows a whistle of (d) The ratio of the rms speed of helium atoms to frequency f1. An observer in the other vehicle hears the frequency of the whistle 1 to be f2. The speed of sound in still air is v. that of hydrogen molecules is The correct statement(s) is (are) 2 (More than One Correct Option, 2014) (More than One Correct Option, 2015) (a) If the wind blows from the observer to the source, f > f . 17. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded 21 piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 (b) If the wind blows from the source to the and volume V1 and the spring is in its observer, f > f relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and 21 volume V2. During this process the piston moves out by a distance x. (c) If the wind blows from the observer to the source, f < f 21 (d) If the wind blows from the source to the observer, f < f 21

12 Waves & Thermodynamics 21. One end of a taut string of length 3 m and Wbf = 100 J respectively. The heat supplied to the system along the path iaf, along the x-axis is fixed at x = 0. The ib and bf are Qiaf , Qib and Qbf speed of the waves in the string is respectively. If the internal energy of the 100 ms−1. The other end of the string is system in the state b is U b = 200 J and Qiaf = 500 J, the ratio Qbf / Qib is vibrating in the y-direction so that (Single Integer Type, 2014) stationary waves are set up in the string. 24. Parallel rays of light of intensity The possible waveform(s) of these I = 912 Wm −2 are incident on a spherical stationary wave is (are) black body kept in surroundings of temperature 300 K. Take Stefan constant (More than One Correct Option, 2014) σ = 5.7 × 10−8 Wm−2K−4 and assume that the energy exchange with the (a) y (t) = A sin πx cos 50πt surroundings is only through radiation. The final steady state temperature of the 63 black body is close to (b) y (t) = A πx 100πt (Single Correct Option 2014) cos sin 33 (c) y (t) = A 5 πx 250πt sin cos 23 (d) y (t) = A 5 πx cos 250πt sin 2 22. A student is performing an experiment (a) 330 K (b) 660 K (c) 990 K (d) 1550 K using a resonance column and a tuning fork of frequency 244 s−1. He is told that the air in the tube has been replaced by Passage (Q. Nos. 25-26) another gas (assume that the column remains filled with the gas). If the In the figure a container is shown to have a minimum height at which resonance movable (without friction) piston on top. The occurs is (0.350 ± 0.005) m, the gas in the container and the piston are all made of tube is (Useful information : perfectly insulating material allowing no heat transfer between outside and inside the 167RT = 640 J1/ 2 mole−1/ 2; container. The container is divided into two 140RT = 590 J1/ 2 mole−1/ 2. The molar compartments by a rigid partition made of a thermally conducting material that allows masses M in grams are given in the options. slow transfer of heat. Take the value of 10/ M for each gas as given there.) (Single Correct Option, 2014) (a) Neon (M = 20, 10 / 20 = 7 / 10) (b) Nitrogen (M = 28, 10 / 28 = 3 / 5) (c) Oxygen (M = 32, 10 / 32 = 9 / 16) (d) Argon (M = 36, 10 / 36 = 17 / 32) 23. A thermodynamic p system is taken a f The lower compartment of the container is from an initial filled with 2 moles of an ideal monoatomic state i with gas at 700 K and the upper compartment is internal energy ib filled with 2 moles of an ideal diatomic gas U i = 100 J to the V at 400 K. The heat capacities per mole of an final state f along =3 =5 ideal monoatomic gas are CV 2 R, C p 2 R, two different paths iaf and ibf, as and those for an ideal diatomic gas are schematically shown in the figure. The CV =5 R, C p = 7 R. 2 2 work done by the system along the paths (Passage Type, 2014) af, ib and bf are Waf = 200 J, Wib = 50 J

Previous Years’ Questions (2018-13) 13 25. Consider the partition to be rigidly fixed p F so that it does not move. When 32p0 equilibrium is achieved, the final temperature of the gases will be (a) 550 K (b) 525 K (c) 513 K (d) 490 K p0 E H G V 26. Now consider the partition to be free to V0 move without friction so that the pressure of gases in both compartments is the same. Match the paths in Column I with the Then total work done by the gases till the magnitudes of the work done in Column time they achieve equilibrium will be II and select the correct answer using the codes given below the lists. (a) 250 R (b) 200 R (c) 100 R (d) −100 R (Matching Type, 2013) 27. Two rectangular blocks, having indentical Column I Column II P. G → E 1. 160 p0V0 ln 2 dimensions, can be arranged either in Q. G → H 2. 36 p0V0 R. F → H 3. 24 p0V0 configuration I or in configuration II as S. F → G 4. 31 p0V0 shown in the figure. One of the blocks has thermal conductivity K and the other 2 K . The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from Codes PQRS the hot end to the cold end in the PQRS (a) 2 3 1 4 (b) 1 2 4 3 configuration I. The time to transport the (c) 4 3 2 1 (d) 2 3 4 1 same amount of heat in the configuration II is (Single Correct Option, 2013) 30. The figure below Configuration II shows the variation Configuration I of specific heat 2K capacity (C) of a C KX K 2K solid as a function (a) 2.0 s (b) 3.0 s (c) 4.5 s (d) 6.0 s of temperature (T ). 100 200 300 400 500 The temperature is T(K) 28. Two non-reactive monoatomic ideal gases increased have their atomic masses in the ratio 2 : 3. continuously from 0 to 500 K at a constant The ratio of their partial pressures, when rate. lgnoring any volume change, the enclosed in a vessel kept at a constant following statement(s) is (are) correct to temperature, is 4 : 3. The ratio of their reasonable approximation. densities is (Single Correct Option, 2013) (More than One Correct Option, 2013) (a) 1: 4 (b) 1: 2 (c) 6 : 9 (d) 8 : 9 (a) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperatureT 29. One mole of a monatomic ideal gas is taken along two cyclic processes (b) heat absorbed in increasing the temperature E → F → G → E and E → F → H → E as from 0-100 K is less than the heat required for shown in the p-V diagram. increasing the temperature from 400-500 K The processes involved are purely isochoric, (c) there is no change in the rate of heat isobaric, isothermal or adiabatic. absorbtion in the range 400-500 K (d) the rate of heat absorption increases in the range 200-300 K

14 Waves & Thermodynamics 31. A horizontal stretched string, fixed at two (a) the number of nodes is 5 ends, is vibrating in its fifth harmonic (b) the length of the string is 0.25 m according to the equation, y(x, t) = (0.01 m) (c) the maximum displacement of the mid-point [sin(62.8 m−1)x] cos[(628 s−1)t]. of the string from its equilibrium position is Assuming π = 3.14, the correct 0.01 m statement(s) is (are) (d) the fundamental frequency is 100 Hz (More than One Correct Option, 2013) Answer with Explanations 1. (b,c,d) (b) Process-II is isothermal expansion, ⇒ K1 = A2 = 4 K 2 A1 ∆U = 0 , W > 0 ∆Q =W > 0 4. (a, c) Let n th harmonic is corresponding to 50.7 cm (c) Process-IV is isothermal compression, and (n + 1)th harmonic is corresponding 83.9 cm. ∆U = 0 , W < 0 ∴ Their difference is λ. ∆Q = W < 0 2 (d) Process-I and III are not isobaric because in isobaric process T ∝ V hence, T-V graph should be ∴ λ = (83.9 − 50.7) cm a straight line passing through origin. 2 2. (5 Hz) or λ = 66.4 cm Observer (stationary) ∴ λ = 16.6 cm 4 13m 13m cos θ = 5 Length corresponding to fundamental mode must be 13 close to λ and 50.7 cm must be an odd multiple of 12m 4 θ θB this length. 16.6 × 3 = 49.8 cm. Therefore, 50.7 is 3rd A 2m/s 5m 5m 1m/s harmonic. If end correction is e, then 1430  330  e + 50.7 = 3λ  4 fA =  330 2 cosθ  −  e = 49.8 − 50.7 = − 0.9 cm  ∴ Speed of sound, v = fλ  1 2 cosθ = 1430   ≈ 1430 1 + 330  ⇒ v = 500 × 66.4 cm/s = 332 m/s 2 cosθ  1 − 330  5  5. (900) Given, n = 1, γ = 3 [from binomial expansion] T-V equation in adiabatic process is fB = 1430  330  TV γ −1 = constant    330 + 1cos θ  ∴ T1V1γ − 1 = T2V2γ −1 ≈ 1430 1 − cos θ   V1  γ −1  1  2 330   V2  8 3 ⇒ T2 = T1 = 100 × 3cosθ Beat frequency = fA − fB = 1430 330  = 13cos θ ⇒ T2 = 25 K  = 13  5  = 5.00 Hz CV = 3 R for monoatomic gas 13 2 3. (4) Rate of heat flow will be same, ∴ ∆U = nC V ∆T = n ×  3R  (T2 − T1) 2 ∴ 300 − 200 = 200 − 100  as dQ T ⋅D  R1 R2 H = dt = R 3 = 1× 2 × 8 × (25 − 100) = − 900 J ∴ R1 = R2 ⇒ L1 = L2 ∴ Decrease in internal energy = 900 J K1 A1 K 2 A2

Previous Years’ Questions (2018-13) 15 6. (c)  dp  = γ  dp  11. (b,c,d) Speed of car, dV dV adiabatic isothermal v = 60 km / h = 500 m / s 3 List-I At a point S, between P and Q (P) Process I ⇒ Adiabatic ⇒ Q = 0 νM C + v cosθ  (Q) Process II ⇒ Isobaric ν ′ = C ; M ∴ W = p∆V = 3 p0[3V0 − V0] = 6 p0V0 νN′ = νN  C + vcos θ  (R) Process III ⇒ Isochoric ⇒ W = 0 C (S) Process (IV) ⇒ Isothermal ⇒ ∆v = (vN − vM)1 + v cos θ  ⇒ Temperature = Constant C 7. (b) ∆U = ∆Q − p∆V Similarly, between Q and R ∆U + p∆V = ∆Q ∆ν = (νN − νM)1 − v cos θ  As ∆U ≠ 0, W ≠ 0, ∆Q ≠ 0. The process represents, C isobaric process d(∆ν) = ± (νN − νM )v sinθ d θ Wgas = − p(∆V ) = − p(V2 − V1) = − pV2 + pV1 dt C dt Graph ‘P’ satisfies isobaric process. θ ≈ 0° at P and R as they are large distance apart. 8. (b) Work done in isochoric process is zero. ⇒ Slope of graph is zero. W12 = 0 as ∆V = 0 at Q, θ = 90° Graph ‘S’ represents isochoric process. sinθ is maximum also value of dθ is maximum 9. (a,c,d or a,c) v = T , so speed at any dt µ as dθ = v , where v is its velocity and r is the length of position will be same for both pulses, therefore time dt r taken by both pulses will be same. the line joining P and S. and r is minimum at Q. λf = v ⇒ λ = v f ⇒ Slope is maximum at Q. since when pulse 1 reaches at A tension and hence At P, νP = ∆ν = (νN − νM)1 + V  speed decreases therefore λ decreases. C (θ ≈ 0°) At R, νR = ∆ν = (νN − νM ) 1 − V  C 10. (6) Car (θ ≈ 0°) At Q, νQ = ∆ν = (νN − νM) vc=2m/s (θ = 90° ) From these equations, we can see that νP + νR = 2 νQ 12. (c) In the first process : piViγ = pf Vfγ p Frequency observed at car pi Vi Vf pf f1 = f0  v + vC  (v = speed of sound) v  Frequency of reflected sound as observed at the source V f2 = f1  v  = f0  v + vC  pi   γ 5    v − vC    3  v − vC    ⇒ = Vf ⇒ 32 = 8γ ⇒ γ = …(i) Beat frequency = f2 − f0 pf  Vi   v + vC − 1 =  2 vC  For the two step process  v − vC   − vC  = f0  f0  v  W = pi(Vf − Vi ) = 105(7 × 10−3) 2 ×2 = 492 × 328 = 6 Hz = 7 × 102 J ⇒ λ ∝v ∝T

16 Waves & Thermodynamics ∆U = f ( pf Vf − piVi ) or 2T − 20 = 400 − T or 3T = 420 2 ∴ T = 140° = γ 1 1  1 × 102 − 102 P 140° S − 4 10°C Q, R 400°C ∆U = − 3 ⋅ 3 × 102 = − 9 × 102 J K 2 4 8 x dx Junction Q − W = ∆U ⇒ Q = 7 × 102 − 9 × 102 Temperature of junction is 140°C 8 Temperature at a distance x from end P is 47 = 8 × 102 J= 588 J Tx = (130 x + 10° ) Change in length dx is suppose dy 13. (b) Heat generated in device in 3 h Then, dy = αdx (Tx − 10) = time × power = 3 × 3600 × 3 × 103 ∆y 1 = 324 × 105 J ∫ dy = ∫ αdx(130 x + 10 − 10) 00 Heat used to heat water ∆y = αx2 × 1 = ms ∆θ = 120 × 1 × 4.2 × 103 × 20 J   2 130 Heat absorbed by coolant 0 = Pt = 324 × 105 − 120 × 1 × 4.2 ×103 × 20 J ∆y = 1.2 × 10−5 × 65 Pt = (325 − 100.8) × 105 J = 223.2 × 105J ∆y = 78.0 × 10−5 m = 0.78 mm 223.2 × 105 2067 W 16. (a, b, d) (a) Total internal energy 3600 P = = U = f1 nRT + f2 nRT 2 2 log 2 p1 1, 14. (9) p0 = (Uave )per mole = U = 1 [5RT + 3RT ] = 2 RT 2n 4 Therefore, p1 = 2 p0 (b) γmix n1C p1 + n2C p2 = n2C v1 + n2C v2 According to Stefan's law, p ∝ T (1)7 R (1) 5R 2 2 p2  T2  4  2767 + 273  4 + 3 p1  T1  487 + 273 (1) 5R + (1) 3R 2 ⇒ = = = 44 = 2 2 = p2 = p2 = 44 ⇒ p2 = 2 × 44 n1M1 + n2M 2 M1 + M2 2 + 4 3 p1 2 p0 p0 n1 + n2 2 2 Mmix = = = = log 2 p2 = log2 [2 × 44 ]= log22 + log2 44 Speed of sound V = γRT p0 M = 1 + log22 8 = 1 + 8 = 9 15. (a) ⇒ V∝ γ P Q, R S M 10°C 2K K 400°C Vmix = γmix × MHe = 3/2 × 4 = 6 VHe γHe Mmix 5/3 3 5 1m 1m 3RT Rate of heat flow from P to Q (d) Vrms = M ⇒ Vrms ∝ 1 , VHe = dQ = 2 KA(T − 10) M VH MH = 2 1 dt 1 MHe 4 2 = Rate of heat flow from Q to S dQ = KA ( 4000 − T) 17. (a, b, c) Note This question can be solved if right hand dt 1 side chamber is assumed open, so that its pressure remains constant even if the piston shifts towards right. At steady state, state rate of heat flow is same ∴ 2 KA(T − 10) = KA ( 400 − T) 1

Previous Years’ Questions (2018-13) 17 p0 = 7 p1V1 + 3  4 p1 ⋅ 3V1 − p1V1  3 2 3 p1= p0 = 7 p1V1 + 9 p1 V1 = 41p1V1 3 2 6 x Note ∆U = 3 ( p2 V2 − p1 V1) has been obtained in 2 p0 = p1 p part (b). kx 18. (2) Power, P = (σT 4 A) = σT 4(4 πR 2) or, P ∝ T 4 R 2 …(i) (a) pV = nRT According to Wien’s law, λ∝1 ⇒ p∝T T V (λ is the wavelength at which peak occurs) Temperature is made three times and volume is ∴ Eq. (i) will become, doubled P ∝ R2 λ4 ⇒ p2 = 3 p1 2 1/ 4  R 2 Further x = ∆V = V2 − V1 = 2 V1 − V1 = V1 or  λ ∝    P  AA AA 3 p1 kx p1 A 1/ 2 1/ 4 p2 = 2 = p1 + A ⇒ kx = 2 λA  R   PB  ⇒ λB =  R A   PA   B    Energy of spring 1/ 4 1 kx2 p1 A p1V1 [400]1/ 2 1  2 = 4 x= 4 = 104  = 2 (b) ∆U = ncv ∆ T = n  3 R  ∆T 19. (3) Let individual amplitudes are A0 each. Amplitudes 2 can be added by vector method. = 3 ( p2 V2 − p1 V1 ) A3 A2 2 = 3   3 p1(2 V1) − p1V1  = 3 p1V1 60° 2  2  60° 60° A4 A1 (c) p2 = 4 p1 3 A1 = A2 = A3 = A4 = A0 ⇒ p2 = 4 p1 = p1 + kx Resultant of A1 and A4 is zero. Resultant of A2 and A3 is 3 A A = A02 + A02 + 2 A0 A0 cos 60° = 3 A0 p1 A ⇒ kx = 3 This is also the net resultant. ⇒ x = ∆V = 2 V1 Now, I ∝ A2 AA ∴ Net intensity will become 3I0. Wgas = ( p0 ∆V + Wspring ) ∴ Answer is 3. = ( p1 Ax + 1 kx ⋅ x) P 2 S 2 V1 1 2 V1 = +  p1 A ⋅ A + 2 ⋅ p1 A ⋅ A  θθ 3 = 2 p1V1 + p1V1 = 7 p1V1 M 3 3 10 m N (d) ∆Q = W + ∆U 18 m = 7 p1V1 + 3 ( p2V2 − p1V1 ) 3 2

18 Waves & Thermodynamics 20. (a,b) When wind blows from S to O So ∆U iaf = Qiaf − Wiaf = 500 J − 200 J = 300 J f2 = f1  v + w + u  v + w −  = Uf − Ui  u  So, U f = U iaf + U i or f2 > f1 = 300 J + 100 J = 400 J when wind blows from O to S f2 = f1  v − w + u  ⇒ f2 > f1 ∆U ib = U b − U i  v − w − u  = 200 J − 100 J = 100 J   21. (a,c,d) There should be a node at x = 0 and antinode at Qib = ∆U ib + Wib = 100 J + 50 J = 150 J x = 3 m. Also, Qibf = ∆U ibf + Wibf v = ω = 100 m/ s. k = ∆U iaf + Wibf = 300 J + 150 J = 450 J ∴ y = 0 at x = 0 and y = ± A at x = 3 m. Only (a), (c) and (d) satisfy the condition. 22. (d) Minimum length = λ So, the required ratio Qbf = Qibf − Qib 4 Qib Qib ⇒ λ = 4l 450 − 150 Now, v = f λ = (244) × 4 × l = 150 = 2 as l = 0.350 ± 0.005 24. (a) In steady state ⇒ v lies between 336.7 m/s to 346.5 m/s Now, v = M γ RT , here M is molecular mass in × 10−3 gram I S =4πR2 = 100 γRT × 10 ⋅ πR2 Radiation M Incident For monoatomic gas, γ = 1.67 Energy incident per second = Energy radiated per ⇒ v = 640 × 10 second ∴ IπR 2 = σ (T 4 − T04 ) 4 πR 2 M ⇒ I = σ (T 4 − T04 ) 4 ⇒ T 4 − T04 = 40 × 108 For diatomic gas, ⇒ T 4 − 81 × 108 = 40 × 108 γ = 1.4 ⇒ v = 590 × 10 M ∴ vNe = 640 × 7 = 448 m/s ⇒ T 4 = 121 × 108 10 v Ar = 640 × 17 = 340 m/s ⇒ T ≈ 330 K 32 25. (d) Let final equilibrium temperature of gases is T vO 2 = 590 × 9 = 331.8 m/s Heat rejected by gas by lower compartment 16 = nC V ∆T 590 × 3 354 m/s 3 vN 2 = 5 = = 2 × 2 R (700 − T) ∴ Only possible answer is Argon. Heat received by the gas in above compartment 23. (2) Wibf = Wib + Wbf = nC p∆T = 50 J + 100 J = 150 J =2 × 7 R (T − 400) Wiaf = Wia + Waf 2 = 0 + 200 J = 200 J Equating the two, we get Qiaf = 500 J 2100 − 3T = 7T − 2800 ⇒ Τ = 490 K

Previous Years’ Questions (2018-13) 19 26. (d) ∆W1 + ∆U1 = ∆Q1 ...(i) nRT ln  Vf  = 32 p0V0 ln  32 V0  ...(ii)    V0  ∆W2 + ∆U 2 = ∆Q2  Vi  ∆Q1 + ∆Q2 = 0 = 32 p0V0 ln 2 5 ∴ (nC p∆T )1 + (nC p∆T )2 = 0 = 160 p0 V0 ln 2 But n1 = n2 = 2 In G → E, ∴ 5 R (T − 700) + 7 R (T − 400) = 0 ∆W = p0∆V 2 2 = p0(31V0) = 31 p0V0 Solving, we get T = 525 K Now, from equations (i) and (ii), we get In G → H work done is less than 31 p0V0 i.e.,24 p0V0 ln F → H work done is 36 p0V0 ∆W1 + ∆W2 = − ∆U1 − ∆U 2 as ∆Q1 + ∆Q2 = 0 30. (b, c, d) Q = mCT ∴ ∆W1 + ∆W2 = − [(nC V ∆T )1 + (nC V ∆T )2] dQ = mc dT dt dt = − 2 × 3 R × (525 − 700) 2 R = rate of absortion of heat = dQ ∝ C dt + 2 × 5 R × (525 − 400) 2 (i) in 0 − 100 K = −100R C increase, so R increases but not linearly 27. (a) RI = R1 + R2 =  l  +  2 l  = 3  l  (ii) ∆Q = mC∆T as C is more in KA KA 2 KA (400 K − 500 K) then 1=1+ 1 (0 − 100 K) so heat is increasing RII R1 R2 (iii) C remains constant so there no change in R from = KA + 2 KA (400K − 500 K) ll (iv) C is increases so R increases in range or RII = l = RI (200 K − 300K) 3KA 4.5 31. (b,c) l Since thermal resistance RII is 4.5 times less than thermal resistance RI . ∴ t II = tI = 9s =2s P 4.5 4.5 28. (d) ρ = pM Number of nodes = 6 RT From the given equation, we can see that ∴ ρ ∝ pM k = 2 π = 62.8 m−1 ρ1  p1   M1   2   4  8 λ or ρ2 =  p2   M2  = 3 3 = 9 2π ∴ λ = 62.8 m = 0.1 m 29. (c) F l = 5λ = 0.25 m p 32T0 2 32p0 Isothermal The mid-point of the string is P, an antinode Isochoric Adiabatic ∴maximum displacement = 0.01 m ω = 2 πf = 628 s−1 p0 E T0 H Isobaric G ∴ f = 628 = 100 Hz 2π V But this is fifth harmonic frequency. V0 32V0 ∴ Fundamental frequency f0 = f = 20 Hz. In F → G work done in isothermal process is 5