JEE Physics Thermal Properties Of Matter MCQs Set 24

Answer: (d) the temperature will be 0°C and 20 g of ice will be left

Explanation:
1. Heat given by 30 g water cooling to 0°C = 30 × 80 = 2400 cal.
2. Ice melted = 2400 / 80 = 30 g, which is less than 50 g.
3. So 20 g of ice remains and the temperature stays at 0°C.

Teacher's Note:
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Answer: (a) 71.25 cm and 41.25 cm

Explanation:
1. Constant difference needs equal expansion: \( L_s\alpha_s = L_b\alpha_b \), so \( L_s/L_b = 19/11 \).
2. \( L_s - L_b = 30 \Rightarrow L_b\left(\frac{19}{11}-1\right) = 30 \Rightarrow L_b = 41.25 \) cm and \( L_s = 71.25 \) cm.

Teacher's Note:
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Answer: (c) c

Explanation:
1. \( \alpha = \frac{\Delta L}{L\Delta T} \): a = 2 × 10⁻⁵, b = 2 × 10⁻⁵, c = 4 × 10⁻⁵, d = 2 × 10⁻⁵ per °C.
2. Rod c has the greatest value.

Teacher's Note:
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45. .

Answer: (c) \( \frac3T \)

Explanation:
1. \( V = \frac{nRT}{P} \) and \( P \propto T^{-2} \), so \( V \propto T^3 \).
2. \( \gamma = \frac1V\frac{dV}{dT} = \frac3T \).

Teacher's Note:
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Answer: (c) 2.6 × 10⁻³/°C

Explanation:
1. At 30°C: V₃₀ = 20 g / 1.5 g cc⁻¹ = 13.33 cc. At 40°C: V₄₀ = 18 g / 1.25 g cc⁻¹ = 14.4 cc.
2. \( \gamma = \frac{\Delta V}{V\Delta T} = \frac{1.067}{13.33\times10} = 8\times10^{-3} \)/°C.
3. \( \alpha = \gamma/3 \approx 2.6\times10^{-3} \)/°C.

Teacher's Note:
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Answer: (b) \( \frac32T_0 \)

Explanation:
1. At constant volume, \( C_v \) (N₂) = \( \frac52R \) and \( C_v \) (He) = \( \frac32R \).
2. \( \frac52R(T_f-T_0) = \frac32R\left(\frac73T_0-T_f\right) \Rightarrow 4T_f = 6T_0 \Rightarrow T_f=\frac32T_0 \).

Teacher's Note:
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70. .

Answer: (b) \( \frac{Pt}{m} \)

Explanation:
1. The heat lost to the surroundings per second equals P.
2. Heat released on solidifying = \( mL = Pt \Rightarrow L = \frac{Pt}{m} \).

Teacher's Note:
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Answer: (b) 6 kg

Explanation:
1. Heat to warm the ice to 0°C = 2 × 0.5 × 20 = 20 kcal.
2. Heat available from water = 5 × 1 × 20 = 100 kcal, so 80 kcal remains for melting.
3. Ice melted = 80 / 80 = 1 kg, so water = 5 + 1 = 6 kg.

Teacher's Note:
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Answer: (d) α₁ = 4α₂

Explanation:
1. \( OR^2 = l^2(1+\alpha_2t)^2 - \frac{l^2}{4}(1+\alpha_1t)^2 \).
2. For small t: \( OR^2 \approx \frac34l^2 + l^2t\left(2\alpha_2 - \frac{\alpha_1}{2}\right) \). Constant OR needs \( 2\alpha_2 = \frac{\alpha_1}{2} \), so \( \alpha_1 = 4\alpha_2 \).

Teacher's Note:
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Answer: (b) 2 and 4 only

Explanation:
1. The strip bends because of the difference in expansion: \( \frac1R \propto |\alpha_B-\alpha_C|\Delta T \).
2. So R is inversely proportional to both ΔT and |αB − αC|, which are statements 2 and 4.

Teacher's Note:
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108. .

Answer: (b) 1, 3 and 4

Explanation:
1. Linear rise 0.2% in every dimension: area rises 0.4% (1 true), thickness rises 0.2% (2 false), volume rises 0.6% (3 true).
2. \( \alpha = \frac{0.002}{80} = 2.5\times10^{-5} = 0.25\times10^{-4} \)/°C (4 true).

Teacher's Note:
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99. .

Answer: (a) 1 and 4

Explanation:
1. Unoccupied space constant means both change by the same volume: \( V_v\gamma_v\Delta T = V_L\gamma_L\Delta T \).
2. So \( \frac{\gamma_v}{\gamma_L} = \frac{V_L}{V_v} \) (4). Since \( V_L < V_v \), \( \gamma_L > \gamma_v \) (1).

Teacher's Note:
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Answer: (b) 2 and 3

Explanation:
1. \( \beta = 2\alpha \) and \( \gamma = 3\alpha \).
2. \( \beta/\alpha = 2 \) (1 false), \( \beta/\gamma = 2/3 \) (2 true), \( \gamma/\alpha = 3 \) (3 true), \( \beta/\alpha = 2 \neq \gamma/\beta = 3/2 \) (4 false).

Teacher's Note:
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Answer: (a) 1 and 4

Explanation:
1. \( KE = \frac12\cdot\frac25MR^2(2\pi n)^2 = \frac45\pi^2n^2MR^2 \); half of it: \( \frac25\pi^2n^2MR^2 \) (1 true).
2. \( MC\Delta T = \frac25\pi^2n^2MR^2 \Rightarrow \Delta T = \frac{2\pi^2n^2R^2}{5C} \) (4 true).

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Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Metal expands more than glass for the same rise in temperature, so the lid loosens.
2. The reason correctly explains the assertion.

Teacher's Note:
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Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Increase of pressure favours the denser phase (water), lowering the melting point.
2. The contraction on melting explains this.

Teacher's Note:
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Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Setting F = C in the formula gives C = −40.
2. So the reason explains the assertion.

Teacher's Note:
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Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Land heats and cools faster than water, creating pressure differences and winds.
2. The reason explains the assertion.

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48. p
49. e
50. r
51. a
52. t
53. u
54. r
55. e
56.
57. s
58. l
59. o
60. w
61. l
62. y
63. .

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Final temperature = (m₁s₁T₁ + m₂s₂T₂)/(m₁s₁ + m₂s₂), which equals the mean only when the thermal capacities are equal.
2. The reason explains the assertion.

Teacher's Note:
1. T
2. h
3. e
4.
5. f
6. i
7. n
8. a
9. l
10.
11. t
12. e
13. m
14. p
15. e
16. r
17. a
18. t
19. u
20. r
21. e
22.
23. i
24. s
25.
26. a
27.
28. w
29. e
30. i
31. g
32. h
33. t
34. e
35. d
36.
37. a
38. v
39. e
40. r
41. a
42. g
43. e
44. ,
45.
46. w
47. e
48. i
49. g
50. h
51. t
52. e
53. d
54.
55. b
56. y
57.
58. t
59. h
60. e
61. r
62. m
63. a
64. l
65.
66. c
67. a
68. p
69. a
70. c
71. i
72. t
73. y
74. .

Answer: (d) A is false, but R is true.

Explanation:
1. Heat can be converted to work and vice versa, so heat alone is not conserved: the assertion is false.
2. Total energy of an isolated system is conserved, so the reason is true.

Teacher's Note:
1. O
2. n
3. l
4. y
5.
6. t
7. o
8. t
9. a
10. l
11.
12. e
13. n
14. e
15. r
16. g
17. y
18.
19. i
20. s
21.
22. c
23. o
24. n
25. s
26. e
27. r
28. v
29. e
30. d
31. ,
32.
33. n
34. o
35. t
36.
37. h
38. e
39. a
40. t
41.
42. b
43. y
44.
45. i
46. t
47. s
48. e
49. l
50. f
51. .

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. During a change of state heat goes into breaking molecular bonds, raising potential energy rather than kinetic energy.
2. The temperature depends on kinetic energy, so it stays constant.

Teacher's Note:
1. L
2. a
3. t
4. e
5. n
6. t
7.
8. h
9. e
10. a
11. t
12.
13. i
14. n
15. c
16. r
17. e
18. a
19. s
20. e
21. s
22.
23. p
24. o
25. t
26. e
27. n
28. t
29. i
30. a
31. l
32.
33. e
34. n
35. e
36. r
37. g
38. y
39. ,
40.
41. n
42. o
43. t
44.
45. t
46. e
47. m
48. p
49. e
50. r
51. a
52. t
53. u
54. r
55. e
56. .