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AakashIIT-JEE

-Section A : Straight Objective Type

1. Answer (4)

Work done by a paddle wheel on a quantity of liquid is non-zero. While dV = 0 2. Answer (3)

Average momentum is independent of temperature. It is equal to zero) 3. Answer (3)

It is assumed that collision is elastic. Therefore Δp = 2 pcosθ

Δp = 2pcosθ ≤ 2p   ∴ Δp ≠ 3p 4. Answer (3) 2 3 2 3 rot trans _{ } nRT nRT K K . 5. Answer (1) constant 3 1  PV 3 1 .V V nRT _{= C} 3 2  TV = C T ∝ 3 2 V

T3_{∝ V}2_{. As volume is increasing, therefore temperature is increasing.}

6. Answer (3)

P = m.V

or PV–1 _{= m. By comparison with PV}α _{= constant, α = –1}

Now, C = C_v R C_V R R R 2R 2 2 3 1 1 1         . 7. Answer (3) 3 2 3 2 2 2 1 2 V V V V_rms    = 3 ) 12 4 3 ( 2 2 2V2 3 169v2  3 13 rms V V   _{= 7.51 V .}

Heat

UNIT

2

8. Answer (2)

Pressure increases in a direction opposite to that of acceleration. 9. Answer (1) 3 1 2_V P = C 6 1 PV  = C ⇒ α = 6 1 C = C_v – 1  R = C_v – 1 6 1  R = C_v – 6 5  R C > C_v 10. Answer (1) ΔQ = ΔU + W for isobaric process

nC_pΔT = nC_vΔT + nRΔT C_p = C_v + R

C_p – C_v = R.

10a. Answer (2, 4) (IIT-JEE 2009)

Diatomic 7 5 ( ) 6 2 2 P V R R C C    R Monoatomic 5 3 ( ) 4 2 2 P V R R C C    R 2 Diatomic 35 ( . ) 4 P V R C C  2 Monoatomic 15 ( . ) 4 P V R C C  p v C C = γ = 2 1 f

 is smaller for diatomic gas

C_p – C_v = R = constant 11. Answer (1)

Pressure ∝ momentum transferred to the walls.

If T is increased, v is increased, hence Δp will increase. 12. Answer (3) U = nC_VT = n fRT 2 = .( ) 2 f nRT = 2 ) (f PV .

AakashIIT-JEE -2 H V = 2 O V 3 – 10 2 3  RT = –3 10 32 320 3   R 2 3RT = 3R × 10 T = 20 K 14. Answer (3)

At a given temperature, energy of A is more than that of B. So degrees of freedom for A should be more than B. 15. Answer (2) 100 P P 100–x x Area of cylinder = = 76 × 100 × A PV A P P x V x A  = + pressure due to Hg = (76 + ) = (100 – ) atm

⇒ PV = constant for isothermal process 76 × 100 × A = (76 + x) (100 – x) A 7600 = 7600 + 100 x – 76x – x2 x = 24, x = 0 x = 0, 24. 16. Answer (3) Nmax N (no. of molecules) Vm.p V

At most probable speed, number of molecules is maximum 17. Answer (4) M V_mp 1 18. Answer (3) M RT P _

RT M P.   _ρ top = ₂₈₀ 70  R M ρbottom = ₃₀₀ 76  R M 99 . 0 bottom top _    19. Answer (2)

The gas in the tube should have an acceleration towards centre. Therefore it should experience a net force inwards. So P₂ > P₁. 20. Answer (2) As V E P 2 3  E = 2 3 2 3_{PV } nRT

. ∴ E must be translational kinetic energy 21. Answer (4)

Collision is elastic

Hence total kinetic energy and momentum remains conserved. 22. Answer (1)

Let the water rises upto a height of H. A

B ( – )L H0 L0 H P_A = P_B ...(i) P₀L₀ = P_A (L₀– H) ...(ii) P_B = P₀ + ρg (L₀– H) ...(iii)

From equation (ii)

) – ( ₀ 0 0 H L L P P_A  ) – ( ) – ( ₀ 0 0 0 0 _{P g L H} H L L P    ) – ( 1 – ) – ( ₀ 0 0 0 g L H H L L P _       P₀H = ρg (L₀– H)2 ∴ 105_{× H = 10}3_{× 10 (20 × 10}–2_{– H)}2 ⇒ H = 0.38 cm 23. Answer (3)          V V P B – . When P is constant, ΔP = 0.

AakashIIT-JEE -Q₁ = n C_PΔT = 1 × 2 5 R (2T – T) = 2 5 RT T T T V T V 2 2 2 2 1 1 _{  } Q₂ = W = nR (2T) n       2 1 P P 2 2 1 _ P P  = 2 RT n 2 Q_Total = (2.5 + 2 n 2) RT 25. Answer (1) Apply 2 2 1 1 T P T P  ( volume is constant) ⇒ 1 1 2 2 P T P T  26. Answer (1) dQ = dU + dW CΔT = C_vΔT + W. T  T₂– T₁ R V P R V P_{2 2 1 1} –  R V P R V P R V P_{0 0 0 0} 5 _{0 0} – 6 _  W = Area ( 0 2 0) 2 0 3 0 0 2 1 V P V P P     5 3 3 5 5 0 0 0 0 0 0 _{PV C C} R R V P C R V P C   _V     _V  ⇒ 10 21R C  _ CV  R_ 2 3  27. Answer (2) dQ = 3dU C C_V R 2 9 3    ⇒ C = 4.5 R 28. Answer (1) ΔQ = ΔU + W.

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29. Answer (3) PV = RT 2 0 ( ) R P T V V    0 2 0 RT 0 dP R dV    V    ⇒ 0 T V   2 2 0 0 0 2 0 T T T  V T  _ _  T   

Minimum value of pressure is 0

min 0 0 (2 ) 2 R T RT P R T V T      30. Answer (4) As PV = nRT ⇒ U = 2 + 2nRT ⇒ ΔU = 2nRΔT ⇒ CV = 2R

Since C_V = 2R for system, R C R _V 2 5 2 3 _{ }

∴ So it is mixture of monoatomic and diatomic gas 31. Answer (2) As 2 2 1 2 V V P   1 2 2 1    P and 5 8 2  P P₂V₂– P₁V₁ = nRΔT K R T 5 11    32. Answer (2) PVγγγγγ = constant V m   Pρ–γγγγγ _{= constant}   m V 33. Answer (1) ΔW = – Area of cycle = 1 3 6 100 10 200 10 10 2       10 4.17 2.4 W J Q     

AakashIIT-JEE

-ΔQ = ΔW (cyclic process)

ΔQ = 3.14 × 100 × 103 _{× 100 × 10}–6

ΔQ = 31.4 J 35. Answer (2)

W = Area of rectangle = Area of triangle W = (2V₀ × P₀) + 2 1 (2V₀ × 2P₀) = 4 P₀V₀ 36. Answer (4)

As initial and final temperatures are same, ΔU = 0 37. Answer (3) P1 – γ_Tγ _{= constant} 2 1 2 3 2 1 2 3 ) 27 ( 300 P T P  ⇒ T = 900 K 38. Answer (3) Using m RT v   , we get 1.4 273 31 . 8 315 315 10 32 –3 2         RT mv 39. Answer (4) λ = 6.6 cm As v = fλ ⇒ v = 5 × 103_{× 6.6 × 10}–2 _m/s Aslo, M RT v   and 1 –   R C_v 40. Answer (1) C_P = C_V + R 41. Answer (3) For a process PVn _{= k,} n R C C _V – 1   As 2 3R C_v  2 1 – 1 2     n n R R Constant or 2 2 1    PV k P V

42. Answer (1) dT PdV C C  _V  PV = nRT 2e2V_{V = nRT} ) 2 1 ( 2e2 V nR dT dV V _   V nR dT PdV 2 1   (1 2 ) V nR C C V n     ⇒ V R C C _V 2 1   43. Answer (1) T = 300 + 2V R PV = 300 + 2V R V R P  300  2 Work done

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Calculate P₁V₁ and P₂V₂ where V₁ = 2m3

and then nR Δ T = P₂V₂ – P₁V₁ = 400 So ΔU = 3 × 400 _usingU  nR_–₁T_ = 1200 J 45. Answer (3) 1 1 R R C n      Pv k (a constant) 1 1 R R C      1 1 R R O      α = –γ

AakashIIT-JEE -U = V2 U nC T nR T V  _   1 – ⇒ 3RT = V2 3RdT = 2VdV Now, dT dV P C C  V  V R V RT R C 2 3 3   2 2 2 3 3 V T R R C   RT T R R C 3 2 3 3 2     2 3R  R  C = 3.5 R 47. Answer (4) ΔQ = ΔU + W W = 2J nRΔT = 2J ΔU = nCVΔT = ( ) 2 5 T nR = 5 J ∴ ΔQ = 7 J 48. Answer (1) P V 2P0 P0 V0 2V0 A C B 2 3 ) 3 ( 2 1 _{0 0} 0 0 V P V P W_AB   0 0 0 0 0 0 (4 ) 3 1 P V P V U  P V      0 0 0 0 2 9 3 2 3 V P V P Q_AB          0 0 2 1 V P W_net  9 1    AB net Q W

49. Answer (2) T V  1 PV V  Constant PV2 _{= Constant} ⇒ n = 2 n R C C _V – 1    _{⇒ C = C} V – R C = 2R ΔQ = CΔT = 200 R 50. Answer (1) Q₁ : Q₂ : Q₃ 900 : T : 400 900 Q1 Q2 T 400 K Q3 W W Q₂ – Q₁ = Q₃ – Q₂ T – 900 = 400 – T 2T = 1300 T = 650 K 51. Answer (3) ΔW = 25 = PΔV = RΔT ...(i) 3 4 3 1 1 6 2 1 2 1        f R R C_V 3 1 –    ΔU = CVΔT = 3R ΔT ...(ii) or ΔU = 3(RΔT) = 3 × 25 = 75 ΔQ = ΔU + ΔW = 75 + 25 = 100 J 52. Answer (4) Since dW = 0 dV = 0 and dQ = dU > 0 ∴ ΔT > 0 53. Answer (4) PVn _{= constant}

PV = constant is an Isothermal process PV∞ _{= constant}   1 P C V

⇒ V = Constant is an Isochoric process

AakashIIT-JEE -1 2 1 T T    _since 1 2 T T

is same so η will be same. 55. Answer (4) 100 1 – 1 2 1                T T 56. Answer (2) PT = Constant P(PV) = K P2_{V = K} K PV12  2 1 – 1 R C C  V  37.35 = C_V + 2R 2 fR = 37.35 – 2R f = 5 57. Answer (1) W = nRT₀ ln i f V V W = 3R × 300 ln       2 1 W = – 5188 J 58. Answer (3)

Molar heat capacity depend on molecular structure 59. Answer (1) dQ = ΔU + W W = –ΔU = – (U₂ – U₁) = U₁ – U₂ 60. Answer (1) Isochoric process ∴ P ∝ T So, P T P T    _⇒ 1 1 100 T ⇒ T = 100 K

61. Answer (3) θH – θR = k (θR – θ0) θH – 20 = k (20 + 20) θH – 10 = k (10 + 40) or _ _1020  ₅4 H H or 5θ_H – 100 = 4θ_H – 40 θH = 60° 62. Answer (1) i = dx dT kA So k dx dT _1 or K₁ > K₂ 63. Answer (1)

Energy received by earth per second is ₂ 2

2 sun 4 sun 4 R 4 ) T ( e R d _         

Energy radiated by earth per second is [T_earth44R_e2]. Equating the two values, we get T_earth = 290 K 64. Answer (2) P = σ εT4_{A (A = πdl)} 100 = 8 10 0.6 7 22 7 . 0 10 67 . 5  8 T4   5 or T 4₌ 6 . 0 8 22 7 . 0 67 . 5 10 7 100 13       ⇒ T = 2018 K 65. Answer (3)

dQ – dW = dU which is an exact differential, since U does not depend on path

66. Answer (4) PVn _{= constant} ⇒ P1/n _{V = constant} For n = ∞ V = constant 67. Answer (3) W = 0 and Q = 0 So ΔV = 0 or V = constant

AakashIIT-JEE -E₁ = ∈₁σT₁4₍₄πr 1 2₎ E₂ = ∈₂σT₂4₍₄πr 2 2₎ So 2 2 1 4 2 1 2 1 2 1                r r T T E E 69. Answer (4) P = σT 4 _(πr2₎ 450 = σT 4_πr2 P = 2 4 2 ) 2 (         T r 4 1 450  P ⇒ P = 1800 W

69(a). Answer (9) IIT-JEE 2010

As per Wein’s displacement law 3 A B B A T T    

As per Stefan’s law Also, 4 2 4 4 2 (6) (3) 9 (18) A A A B B B P A T P  A T    70. Answer (3) 5 1 0.05 2 10 º C 25 100 A L L T           5 1 0.04 10 º C 40 100 B L L T          ₁ + ₂ = 50 cm ... (i) 1   2 50.03 cm   αA1ΔT + αB2ΔT = 0.03 2 × 10–5 _{× } 1 × 50 + 10–5 × 2 × 50 = 0.03 cm 2₁ + ₂ = 60 cm ... (ii) From (i) and (ii) ₁ = 10 cm 71. Answer (2) k₁x₁ = k₂x₂ and x₁ + x₂ = L α ΔT x₁ = 3x₂ x₁ = ) ( ) ( 2 1 2 2 1 k k k x x   = ) 3 ( 3 K K K T L    = 4 3LT

72. Answer (4)

ρ1 = ρ2 [1 + γ (T2 – T1)]

v₂ = v₁[1 + β(T₂ – T₁)]

The loss in weight at T₁ = v₁ρ₁g ⇒ w₀ – w₁ = v₁ρ₁g

The loss in weight at T₂ = v₂ρ₂g ⇒ w₀ – w₂ = v₂ρ₂g

0 1 2 1 0 2 2 1 1 ( ) 1 ( )          w w T T w w T T 73. Answer (1) 50g + 2g = Vρ₁g 50g + xg = Vρ₂g

where ρ₁ is density of water at 10°C and ρ₂ is density of water at 30°C ⇒ ₅₀52_x = 2 1   = ) 1 ( 1 T    ⇒ 52 – 52 × 10–4 _{× 20 = 50 + x} ⇒ (2 – x) = 52 × 10–4 _{× 20} x = 0.104 kg 74. Answer (1) 30 = (50 – 20)α [where α is constant) mc dt d = (40 – 20)α mc dt d = 30 30 20 ⇒ 20 10 1 . 0 _  mc Heat capacity = mc = 2000 J/ºC 75. Answer (4)

For gas A temperature is maximum and A is polyatomic gas. [Slope is proportional to the γ]

A B V1 V2 P 76. Answer (2) 4Q = nC_PΔT 3Q = nC_VΔT ⇒   ₃4 ⇒ f = 6 77. Answer (1) ( 3 _C) ( _C 3) kA T T kA T T    A B C T3 ( ) ( )T ⇒ 1 3 2 C T T  

AakashIIT-JEE

-Work done for isobaric process is nRΔT

R (T₀ – T₁) + R (T₀ – T₂) = W T₀ = R T T R W 2 ) ( ₁ ₂ 

78(a). Answer (1, 2) IIT-JEE 2010

AB is an isothermal process

⇒ Internal energy at A = Internal energy at B. Applying PV

T = constant between A and B V0

4V0 A B V T0 T C 0 0 0 0 0 0 4 4  B   B P V P V P P T T

As it is not clearly indicated that BC is passing through origin, prediction about point C is not possible. 79. Answer (2) P = σT4_A so 2 2 4 2 2 1 4 1 4 4T r  T r ⇒ 4 1 4 2 2 2 2 1 T T r r  ⇒ 2 1 2 2 1 _       T T r r

79(a). Answer (9) IIT-JEE 2010

As per Wein’s displacement law 3 A B B A T T    

As per Stefan’s law

Also, 4 2 4 4 2 (6) (3) 9 (18) A A A B B B P A T P  A T    80. Answer (2)        _     0 1 2 1 2 2 k t θ1 = 50 θ2 = 40 θ0 = 20 t = 5 ) 20 45 ( 5 10   k       _    20 2 40 5 40 k

     2 25 40 10 10θ = 2000 – 50θ 60θ = 2000 C 3 100    81. Answer (2)

Under steady state constant thermal current flows. 82. Answer (3)

Rate of cooling is proportional to V A or r 1 V₁= 2V₂ 3 2 3 1 3 4 2 3 4 r r   

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Where r is distance from sun

r = diameterofsun  D D _{D } Sun Planet So Solar constant ∝ θ2 84. Answer (1) T  = 2xiˆ2yjˆ2zkˆ ) 2 , 1 , 1 ( ) ( T = 2iˆ2jˆ 85. Answer (3) R_eq = R₁ + 2R₂ = 10_A  _A(2₀(_.0₀₀₁.01)_K)  _AK30      33.4 3 100 100 10 eq Cu R AK T 86. Answer (1) P 2_{T = constant} 3 1 1 2 5    R R C P 3_{V = C C = 4R} 3 1  x _{Q = 4P} 0V0

AakashIIT-JEE

-Heat current per unit area = x T k   88. Answer (1) constant  T k (Weidmann-Franz law) 89. Answer (1)

Black spot is good absorber of heat, so according to Kirchhoff’s law this will be good emitter also. 90. Answer (4)

Fraunhoffer lines are absorption lines which are characteristic of the elements in the sun atmosphere. They are also observed in the emission spectra of the elements. This is based on Kirchhoff’s law.

91. Answer (3)

E_n = nhv [energy of oscillator] 92. Answer (2)

In laboratory, rate of cooling of liquids is observed to measure their heat capacity. 93. Answer (2) and S N O Q P R T T T T T T     A P S B O R N Q So A B ( , , )P Q R ( , , )O N S ⇒ R/3 R/6 R/3 ⇒ 6 5 3 6 3 R R R R R_eq     94. Answer (3)

Air is bad conductor of heat and non conducting medium between two surfaces reduces transfer of heat. 95. Answer (1)

When temperature of two bodies are same then there will be no transfer of heat. 96. Answer (3) x₁ + x₂ + x₃ = LαΔT + 3 2 2 L T L T    Kx1 2 Kx2 2 Kx2 3 Kx3 x₁ = 2x₂ = 3x₃ ∴ x1 + 3 1 3 2 3 2 x x L T     ⇒ 1 9 1 x  LT 2 2 2 2 2 2 1 1 81 81 2 121 242 E  KL T  KL T

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