JEE Physics Mechanical Properties Of Fluids MCQs Set 23

Answer: (a) P increases faster than linearly with h (its slope grows with depth)

Explanation:
1. The slope of the pressure-depth graph is \( \frac{dP}{dh}=\rho g \).
2. Because \( \rho \) increases with depth, the slope increases too, so P rises faster than linearly: the curve bends upward.

Teacher's Note:
1. P
2. r
3. e
4. s
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113. .

Answer: (b) \( T=\frac{rh\rho g}{4} \)

Explanation:
1. A soap bubble has two surfaces, so the excess pressure inside is \( \frac{4T}{r} \).
2. The manometer measures this: \( \frac{4T}{r}=h\rho g \), so \( T=\frac{rh\rho g}{4} \).

Teacher's Note:
1. F
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Answer: (c) \( 2000\ \text{kg/m}^3 \)

Explanation:
1. Take the pressure at the horizontal level B. In the tube it is \( P_0+\rho_{oil}g(0.20) \).
2. In the container the same level is 5 cm below the free surface: \( P_0+\rho g(0.05) \).
3. Equating: \( \rho=\frac{500\times0.20}{0.05}=2000 \) kg/m³.

Teacher's Note:
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91. .

Answer: (d) \( 4\sqrt5\ \text{m/s} \)

Explanation:
1. Torricelli's law: \( v=\sqrt{2gh} \).
2. \( v=\sqrt{2\times10\times4}=\sqrt{80}=4\sqrt5 \) m/s. The density and pipe diameter do not matter.

Teacher's Note:
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Answer: (a) 9 : 4

Explanation:
1. Continuity: \( A_1v_1=A_2v_2 \), so \( \frac{v_1}{v_2}=\frac{d_2^2}{d_1^2} \).
2. \( \frac{v_1}{v_2}=\left(\frac{3.75}{2.5}\right)^2=\frac94 \).

Teacher's Note:
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103. .

Answer: (b) decreases

Explanation:
1. While the coin is on the block, the total weight of block plus coin is balanced by buoyancy, so the displaced volume is \( \frac{W_b+W_c}{\rho_wg} \).
2. After the coin sinks, the block displaces \( \frac{W_b}{\rho_wg} \) and the coin displaces only its own volume, which is less than \( \frac{W_c}{\rho_wg} \) since the coin is denser than water. The total displaced volume decreases, so the water level falls.

Teacher's Note:
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100. .

Answer: (c) \( \phi^2 \)

Explanation:
1. The maximum level is reached when the inflow equals the outflow: \( \phi=a\sqrt{2gh} \).
2. \( h=\frac{\phi^2}{2ga^2}\propto\phi^2 \). (It is also proportional to \( \frac{1}{a^2} \).)

Teacher's Note:
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82. .

Answer: (d) \( \frac v2 \)

Explanation:
1. With negligible buoyancy, \( mg=6\pi\eta rv_t \), so \( v_t\propto\frac mr \).
2. The same mass and double the radius gives \( v_t'=\frac v2 \).

Teacher's Note:
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Answer: (a) \( 6.4\times10^{-2}\ \text{m}^3 \)

Explanation:
1. Take torques about B. The weight of the rod (240 N) acts 6 m from B, the lead weight (120 N) acts 12 m from B, and the buoyancy F acts at the middle of the submerged half, 9 m from B (same angle factor in all terms).
2. \( 240\times6+120\times12=F\times9 \), so \( F=320 \) N. The tension is \( 360-320=40 \) N.
3. Submerged volume \( =\frac{320}{1000\times10}=0.032 \) m³, so the whole rod has volume \( 0.064 \) m³ \( =6.4\times10^{-2} \) m³.

Teacher's Note:
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98. .

Answer: (b) \( 8P_1+\frac{24T}{r} \)

Explanation:
1. The pressure of the air inside the bubble is \( P+\frac{4T}{r} \). Initially \( p_{in,1}=P_1+\frac{4T}{r} \); finally \( p_{in,2}=P_2+\frac{8T}{r} \).
2. The volume of the bubble falls to one eighth, and isothermally \( p_{in,2}=8p_{in,1} \).
3. \( P_2+\frac{8T}{r}=8P_1+\frac{32T}{r} \), so \( P_2=8P_1+\frac{24T}{r} \).

Teacher's Note:
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99. .

Answer: (c) \( \sqrt{920\times980}\ \text{cm/s} \)

Explanation:
1. Pressure at the bottom above atmospheric: \( p=(1\times100+0.9\times400)g=460g \) (in g/cm² units of weight).
2. The escaping liquid is water: \( \frac12\rho_wv^2=p \), so \( v=\sqrt{2\times460\times980}=\sqrt{920\times980} \) cm/s.

Teacher's Note:
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102. .

Answer: (d) \( \frac{hdg}{2} \)

Explanation:
1. The pressure increases linearly from 0 at the top to \( hdg \) at the bottom.
2. The mean over the wall is \( \frac{0+hdg}{2}=\frac{hdg}{2} \).

Teacher's Note:
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71. .

Answer: (a) \( 1:2^{1/3} \)

Explanation:
1. Volume conserved: \( R=2^{1/3}r \).
2. Final energy \( =4\pi R^2T=2^{2/3}\,4\pi r^2T \); initial \( =2\times4\pi r^2T \). Ratio \( =\frac{2^{2/3}}{2}=2^{-1/3} \), i.e. \( 1:2^{1/3} \).

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Answer: (b) 3

Explanation:
1. Buoyant force \( =60-40=20 \) N.
2. Specific gravity \( =\frac{\text{weight in air}}{\text{buoyant force}}=\frac{60}{20}=3 \).

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Answer: (c) \( a^2 \)

Explanation:
1. At terminal velocity \( \frac43\pi a^3(\sigma-\rho)g=6\pi\eta av \).
2. \( v=\frac{2a^2(\sigma-\rho)g}{9\eta}\propto a^2 \).

Teacher's Note:
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8. s
9. c
10. a
11. l
12. e
13. s
14.
15. a
16. s
17.
18. a
19.
20. c
21. u
22. b
23. e
24. d
25.
26. w
27. h
28. i
29. l
30. e
31.
32. t
33. h
34. e
35.
36. d
37. r
38. a
39. g
40.
41. s
42. c
43. a
44. l
45. e
46. s
47.
48. a
49. s
50.
51. a
52. ,
53.
54. s
55. o
56.
57. t
58. h
59. e
60.
61. t
62. e
63. r
64. m
65. i
66. n
67. a
68. l
69.
70. s
71. p
72. e
73. e
74. d
75.
76. s
77. c
78. a
79. l
80. e
81. s
82.
83. a
84. s
85.
86. a
87.
88. s
89. q
90. u
91. a
92. r
93. e
94. d
95. .

Answer: (d) \( \left(\frac dD-1\right)h \)

Explanation:
1. Net upward force \( =(d-D)Vg \), so \( a=\frac{(d-D)g}{D} \). At the surface \( v^2=2ah=2gh\frac{d-D}{D} \).
2. Above the surface the ball rises \( H=\frac{v^2}{2g}=\left(\frac dD-1\right)h \).

Teacher's Note:
1. U
2. s
3. e
4.
5. e
6. n
7. e
8. r
9. g
10. y
11. :
12.
13. t
14. h
15. e
16.
17. n
18. e
19. t
20.
21. f
22. o
23. r
24. c
25. e
26.
27. w
28. h
29. i
30. l
31. e
32.
33. s
34. u
35. b
36. m
37. e
38. r
39. g
40. e
41. d
42.
43. d
44. o
45. e
46. s
47.
48. w
49. o
50. r
51. k
52.
53. o
54. v
55. e
56. r
57.
58. a
59.
60. d
61. i
62. s
63. t
64. a
65. n
66. c
67. e
68.
69. h
70. ,
71.
72. a
73. n
74. d
75.
76. t
77. h
78. e
79. n
80.
81. i
82. t
83.
84. a
85. l
86. l
87.
88. c
89. o
90. n
91. v
92. e
93. r
94. t
95. s
96.
97. t
98. o
99.
100. h
101. e
102. i
103. g
104. h
105. t
106.
107. i
108. n
109.
110. a
111. i
112. r
113. .

Answer: (a) increases

Explanation:
1. The ice displaces its weight of the denser solution: \( V_d=\frac{m}{\rho_{sol}} \).
2. The melted ice occupies \( \frac{m}{\rho_w} \), which is larger because \( \rho_w\lt \rho_{sol} \). So the liquid level rises.

Teacher's Note:
1. F
2. o
3. r
4.
5. f
6. l
7. o
8. a
9. t
10. i
11. n
12. g
13.
14. i
15. c
16. e
17.
18. i
19. n
20.
21. p
22. u
23. r
24. e
25.
26. w
27. a
28. t
29. e
30. r
31.
32. t
33. h
34. e
35.
36. l
37. e
38. v
39. e
40. l
41.
42. s
43. t
44. a
45. y
46. s
47.
48. t
49. h
50. e
51.
52. s
53. a
54. m
55. e
56. .
57.
58. I
59. n
60.
61. a
62.
63. d
64. e
65. n
66. s
67. e
68. r
69.
70. l
71. i
72. q
73. u
74. i
75. d
76. ,
77.
78. m
79. e
80. l
81. t
82. i
83. n
84. g
85.
86. a
87. d
88. d
89. s
90.
91. m
92. o
93. r
94. e
95.
96. v
97. o
98. l
99. u
100. m
101. e
102.
103. t
104. h
105. a
106. n
107.
108. t
109. h
110. e
111.
112. i
113. c
114. e
115.
116. d
117. i
118. s
119. p
120. l
121. a
122. c
123. e
124. d
125. .

Answer: (b) 27 : 8

Explanation:
1. Terminal velocity \( \propto r^2 \), so \( \frac{r_1}{r_2}=\sqrt{\frac94}=\frac32 \).
2. Volume \( \propto r^3 \): \( \frac{V_1}{V_2}=\frac{27}{8} \).

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

Answer: (c) \( \rho_1\lt \rho_3\lt \rho_2 \)

Explanation:
1. The lighter liquid is on top, so \( \rho_1\lt \rho_2 \).
2. The ball sinks through liquid 1 (so \( \rho_3\gt \rho_1 \)) but floats on liquid 2 (so \( \rho_3\lt \rho_2 \)). Hence \( \rho_1\lt \rho_3\lt \rho_2 \).

Teacher's Note:
1. A
2.
3. b
4. o
5. d
6. y
7.
8. f
9. l
10. o
11. a
12. t
13. s
14.
15. a
16. t
17.
18. t
19. h
20. e
21.
22. b
23. o
24. u
25. n
26. d
27. a
28. r
29. y
30.
31. b
32. e
33. t
34. w
35. e
36. e
37. n
38.
39. t
40. w
41. o
42.
43. l
44. i
45. q
46. u
47. i
48. d
49. s
50.
51. i
52. f
53.
54. i
55. t
56. s
57.
58. d
59. e
60. n
61. s
62. i
63. t
64. y
65.
66. i
67. s
68.
69. i
70. n
71. t
72. e
73. r
74. m
75. e
76. d
77. i
78. a
79. t
80. e
81.
82. b
83. e
84. t
85. w
86. e
87. e
88. n
89.
90. t
91. h
92. e
93. i
94. r
95. s
96. .

Answer: (d) \( \sqrt{400}\ \text{m/s} \)

Explanation:
1. Pressure difference between the inside at the hole and the outside atmosphere \( =3-1=2 \) atm \( =2\times10^5 \) Pa.
2. \( \frac12\rho v^2=2\times10^5 \), so \( v=\sqrt{\frac{4\times10^5}{1000}}=\sqrt{400} \) m/s.

Teacher's Note:
1. O
2. n
3. l
4. y
5.
6. t
7. h
8. e
9.
10. g
11. a
12. u
13. g
14. e
15.
16. p
17. r
18. e
19. s
20. s
21. u
22. r
23. e
24.
25. d
26. r
27. i
28. v
29. e
30. s
31.
32. t
33. h
34. e
35.
36. e
37. f
38. f
39. l
40. u
41. x
42. ,
43.
44. s
45. i
46. n
47. c
48. e
49.
50. t
51. h
52. e
53.
54. o
55. u
56. t
57. s
58. i
59. d
60. e
61.
62. i
63. s
64.
65. a
66. t
67.
68. a
69. t
70. m
71. o
72. s
73. p
74. h
75. e
76. r
77. i
78. c
79.
80. p
81. r
82. e
83. s
84. s
85. u
86. r
87. e
88. .

Answer: (a) rise and fall

Explanation:
1. The ice displaces weight-equivalents of both water and oil: \( m=\rho_wV_w+\rho_oV_o \). After melting, the water formed has volume \( \frac{m}{\rho_w}=V_w+\frac{\rho_o}{\rho_w}V_o \).
2. The melt water exceeds the volume of water displaced, \( V_w \), so the interface rises. It is less than \( V_w+V_o \), the total volume displaced, so the top level of the oil falls.

Teacher's Note:
1. C
2. o
3. m
4. p
5. a
6. r
7. e
8.
9. t
10. h
11. e
12.
13. n
14. e
15. w
16.
17. v
18. o
19. l
20. u
21. m
22. e
23.
24. o
25. f
26.
27. t
28. h
29. e
30.
31. m
32. e
33. l
34. t
35.
36. w
37. a
38. t
39. e
40. r
41.
42. w
43. i
44. t
45. h
46.
47. t
48. h
49. e
50.
51. v
52. o
53. l
54. u
55. m
56. e
57.
58. t
59. h
60. a
61. t
62.
63. t
64. h
65. e
66.
67. i
68. c
69. e
70.
71. d
72. i
73. s
74. p
75. l
76. a
77. c
78. e
79. d
80.
81. i
82. n
83.
84. e
85. a
86. c
87. h
88.
89. l
90. i
91. q
92. u
93. i
94. d
95. .

Answer: (b) \( R^2\sqrt{\frac{2\rho_wg}{3T}} \)

Explanation:
1. The surface tension force holds the bubble down: vertical component \( 2\pi rT\sin\phi \) with \( \sin\phi=\frac rR \), so \( F_T=\frac{2\pi r^2T}{R} \).
2. At detachment it equals the buoyancy: \( \frac43\pi R^3\rho_wg=\frac{2\pi r^2T}{R} \).
3. \( r^2=\frac{2R^4\rho_wg}{3T} \), so \( r=R^2\sqrt{\frac{2\rho_wg}{3T}} \).

Teacher's Note:
1. R
2. e
3. s
4. o
5. l
6. v
7. e
8.
9. t
10. h
11. e
12.
13. s
14. u
15. r
16. f
17. a
18. c
19. e
20.
21. t
22. e
23. n
24. s
25. i
26. o
27. n
28.
29. f
30. o
31. r
32. c
33. e
34.
35. a
36. l
37. o
38. n
39. g
40.
41. t
42. h
43. e
44.
45. v
46. e
47. r
48. t
49. i
50. c
51. a
52. l
53.
54. a
55. t
56.
57. t
58. h
59. e
60.
61. c
62. i
63. r
64. c
65. l
66. e
67.
68. o
69. f
70.
71. c
72. o
73. n
74. t
75. a
76. c
77. t
78. ,
79.
80. a
81. n
82. d
83.
84. e
85. q
86. u
87. a
88. t
89. e
90.
91. w
92. i
93. t
94. h
95.
96. t
97. h
98. e
99.
100. b
101. u
102. o
103. y
104. a
105. n
106. c
107. y
108. .