CBSE Class 12 Physics Solids and Fluids Objective Questions

CBSE Class 12 Physics Solids and Fluids Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

CBSE CLASS 12 PHYSICS
SOLIDS AND FLUIDS
OBJECTIVE QUESTIONS

OBJECTIVES

1. Two bars A and B of circular cross-section and of same volume and made of the same material are subjected to tension. If the diameter of A is half that of B and if the force applied to both the bars is same and it is within elastic limits, the ratio of extension of A to that of B is
(a) 2
(b) 4
(c) 8
(d) 16

Answer: (d) 16

1. Extension Δl = \( \frac{Fl}{AY} \). For equal volume, A × l is constant, so l ∝ 1/A.
2. Diameter of A is half, so its area is A_A = A_B/4 and its length is l_A = 4l_B.
3. Δl ∝ l/A ∝ 1/A², so Δl_A/Δl_B = (4)² = 16.
4. Hence, the correct option is (d).

Teacher's Note:
a) With the volume fixed, a thinner bar is both longer and thinner, so the extension grows as 1/A².
b) F and Y are the same for both bars and cancel in the ratio.

2. An iron rod of Y = 10¹¹ N / m² and coefficient of linear expansion 10^-5 / °C has a length of one metre and area of cross-section 10 cm². If its temperature is raised by 100°C and it is held so that it is not allowed to expand or bend, the force produced is
(a) 10³ N
(b) 10⁴ N
(c) 10⁵ N
(d) 10⁶ N

Answer: (c) 10⁵ N

1. Thermal strain prevented = αΔT = 10^-5 × 100 = 10^-3.
2. Area A = 10 cm² = 10 × 10^-4 m² = 10^-3 m².
3. Force F = YAαΔT = 10¹¹ × 10^-3 × 10^-3 = 10⁵ N.
4. Hence, the correct option is (c).

Teacher's Note:
a) The length of the rod (1 m) does not appear in the final force.
b) Convert cm² into m² carefully, as 1 cm² = 10^-4 m².

3. A thick cylindrical metallic rod of density ρ, Young's modulus Y and length L is suspended from a rigid support. The increase in length of the rod due to its own weight is
(a) \( \frac{\rho g L}{Y} \)
(b) \( \frac{\rho g L^2}{Y} \)
(c) \( \frac{\rho g L^2}{2Y} \)
(d) \( \frac{\rho g L^2}{4Y} \)

Answer: (c) \( \frac{\rho g L^2}{2Y} \)

1. Take a small element dx at distance x from the lower end. The weight below it is ρAxg, so the stress there is ρgx.
2. Elongation of the element: d(Δl) = \( \frac{\rho g x\,dx}{Y} \).
3. Total elongation = \( \int_0^L \frac{\rho g x}{Y}dx = \frac{\rho g L^2}{2Y} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) The tension varies along the rod, so the weight cannot be taken at the full value everywhere.
b) Equivalent view: the whole weight acts at the centre, so the effective stress is ρgL/2 over length L.

4. A 20 Newton block is suspended from a wire such that its length changes by 1 %. If Young's modulus of material of the wire is 2 × 10¹¹ Nm^-2, the area of cross-section of the wire is
(a) 1 mm²
(b) 10^-1 mm²
(c) 10^-2 mm²
(d) 10^-3 mm²

Answer: (c) 10^-2 mm²

1. Strain = 1 % = 0.01.
2. Y = F/(A × strain), so A = \( \frac{20}{2 \times 10^{11} \times 0.01} \) = 10^-8 m².
3. Since 1 mm² = 10^-6 m², A = 10^-2 mm².
4. Hence, the correct option is (c).

Teacher's Note:
a) Convert m² to mm² by multiplying by 10⁶.
b) Percentage strain must be written as a decimal (1 % = 0.01).

5. A metallic rod breaks when the strain produced is 0.2 %. Young's modulus of the material of the rod is 7 × 10⁹ N / m. The area of cross-section to support a load of 10⁴ N is
(a) 7.1 × 10^-2 m²
(b) 7.1 × 10^-4 m²
(c) 7.1 × 10^-6 m²
(d) 7.1 × 10^-8 m²

Answer: (b) 7.1 × 10^-4 m²

1. Breaking strain = 0.2 % = 0.002.
2. The breaking stress = Y × strain = 7 × 10⁹ × 0.002 = 1.4 × 10⁷ N/m².
3. A = F/stress = 10⁴ / (1.4 × 10⁷) = 7.1 × 10^-4 m².
4. Hence, the correct option is (b).

Teacher's Note:
a) The unit of Young's modulus is printed as N/m but is N/m²; the working uses N/m².
b) A larger area is needed when the breaking stress is smaller.

6. Bulk modulus of rubber is 9 × 10⁸ N / m². The depth below the surface of sea where a solid rubber ball should be taken to reduce its volume by 0.1 % is
(Given that atmospheric pressure at sea level is equal to 10 m of column of water)
(a) 10³ m
(b) 10² m
(c) 10 m
(d) 1 m

Answer: (b) 10² m

1. Pressure needed: ΔP = B × (ΔV/V) = 9 × 10⁸ × 0.001 = 9 × 10⁵ N/m².
2. 1 atmosphere (about 10⁵ N/m²) is equal to 10 m of water, so 9 × 10⁵ N/m² equals about 90 m of water.
3. This is of the order of 10² m.
4. Hence, the correct option is (b).

Teacher's Note:
a) The 90 m obtained is rounded to the nearest option, 10² m.
b) Each 10 m of water adds about one atmosphere of pressure.

7. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid when a mass M is placed on the piston to compress the liquid. The fractional change in the radius of the sphere is
(a) \( \frac{mg}{4KA} \)
(b) \( \frac{mg}{3KA} \)
(c) \( \frac{3KA}{mg} \)
(d) \( \frac{4KA}{mg} \)

Answer: (b) \( \frac{mg}{3KA} \)

1. The extra pressure in the liquid is ΔP = Mg/A (the printed mg stands for the weight of the placed mass).
2. The fractional change in volume of the sphere = ΔP/K = Mg/(KA).
3. Since V ∝ R³, ΔV/V = 3 ΔR/R, so ΔR/R = \( \frac{mg}{3KA} \).
4. Hence, the correct option is (b).

Teacher's Note:
a) A change of 1 % in volume of a sphere is a change of 1/3 % in its radius.
b) Liquid pressure acts equally in all directions on the sphere.

8. Young's modulus of steel and brass are 2 × 10¹⁰ N/m² and 10¹⁰ N/m² respectively. A steel and brass wire of the same length are extended by 1 mm under the same force, the radii of steel and brass wires are R_s and R_B respectively. Then
(a) R_S = √2 R_B
(b) R_S = \( \frac{R_B}{\sqrt{2}} \)
(c) R_S = 4R_B
(d) R_S = \( \frac{R_B}{4} \)

Answer: (b) R_S = \( \frac{R_B}{\sqrt{2}} \)

1. Extension Δl = F l/(πR²Y). For the same F, l and Δl, the product YR² must be the same.
2. Y_SR_S² = Y_BR_B², so R_S² = R_B² × (10¹⁰/2 × 10¹⁰) = R_B²/2.
3. So R_S = R_B/√2. Hence, the correct option is (b).

Teacher's Note:
a) The stiffer steel needs a smaller radius to give the same extension.
b) Radius appears squared, so the factor 2 in Y becomes √2 in the radius.

9. Two capillary tubes of same radius r but of lengths l₁ and l₂ are fitted in parallel to the bottom of vessel. The pressure head is P. What sould be the length l of the single tube that can replace the two tubes, so that the rate of flow is same as before.
(a) l = l₁ + l₂
(b) l = \( \frac{l_1 l_2}{l_1 + l_2} \)
(c) l = \( \frac{1}{l_1} \) + \( \frac{1}{l_2} \)
(d) l = \( \frac{1}{l_1 + l_2} \)

Answer: (b) l = \( \frac{l_1 l_2}{l_1 + l_2} \)

1. By Poiseuille's law, the rate of flow V = \( \frac{\pi P r^4}{8\eta l} \), so the flow is proportional to 1/l (a fluid conductance).
2. For tubes in parallel the flows add: 1/l = 1/l₁ + 1/l₂.
3. So l = \( \frac{l_1 l_2}{l_1 + l_2} \). Hence, the correct option is (b).

Teacher's Note:
a) Parallel tubes combine like electrical resistors in parallel, with resistance proportional to l.
b) The same radius r and the same pressure head are assumed for all tubes.

10. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W₁ is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at height \( \frac{3L}{4} \) from its lower end is
(a) \( \frac{W_1}{S} \)
(b) \( \left( W_1 + \frac{W}{4} \right) S \)
(c) \( \frac{\left( W_1 + \frac{3W}{4} \right)}{S} \)
(d) \( \frac{W_1 + W}{S} \)

Answer: (c) \( \frac{\left( W_1 + \frac{3W}{4} \right)}{S} \)

1. At a point 3L/4 above the lower end, the part of the wire below it has length 3L/4 and weight 3W/4.
2. The tension at that point = W₁ + 3W/4.
3. Stress = tension / area = \( \frac{W_1 + 3W/4}{S} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) Only the weight below the section is carried by it, not the whole wire.
b) Option (b) multiplies by S instead of dividing, so it is not even a stress.

11. A metal rod has Young's modulus Y, length L and area of cross-section A. The work done in stretching the rod by an amount ΔL is
(a) \( \frac{YA(\Delta L)^2}{2L} \)
(b) \( \frac{YA(\Delta L)^2}{L} \)
(c) \( \frac{YA(\Delta L)}{2L} \)
(d) \( \frac{YA(\Delta L)}{L} \)

Answer: (a) \( \frac{YA(\Delta L)^2}{2L} \)

1. The force needed for extension x is F = \( \frac{YAx}{L} \).
2. Work = \( \int_0^{\Delta L} \frac{YAx}{L}dx = \frac{YA(\Delta L)^2}{2L} \).
3. Hence, the correct option is (a).

Teacher's Note:
a) Work is half of the final force times the extension, as the force rises from zero.
b) Option (c) has the dimensions of a force, not of work.

12. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ?
(a) length 50 cm, diameter 0.5 mm
(b) length 100 cm, diameter 1 mm
(c) length 200 cm, diameter 2 mm
(d) length 300 cm, diameter 3 mm

Answer: (a) length 50 cm, diameter 0.5 mm

1. For the same material and tension, extension Δl ∝ l/d².
2. (a) 50/0.25 = 200; (b) 100/1 = 100; (c) 200/4 = 50; (d) 300/9 = 33.3.
3. The ratio is largest for wire (a). Hence, the correct option is (a).

Teacher's Note:
a) The diameter is squared, so it matters more than the length.
b) Compute l/d² for each wire and compare.

13. A nylon rope 3 cm in diameter has a breaking strength of 1.5 × 10⁵ N. The breaking strength of a similar rope 1.5 cm in diameter is
(a) 7.5 × 10⁴ N
(b) 3.75 × 10⁴ N
(c) 3 × 10⁵ N
(d) 6 × 10⁵ N

Answer: (b) 3.75 × 10⁴ N

1. Breaking force = breaking stress × area, and the breaking stress is the same for the same material.
2. So breaking force ∝ d². The diameter is halved, so the area becomes 1/4.
3. New breaking strength = 1.5 × 10⁵ / 4 = 3.75 × 10⁴ N.
4. Hence, the correct option is (b).

Teacher's Note:
a) The breaking stress is a material property, whereas the breaking force depends on the cross-section.
b) Halving the diameter reduces the strength to one quarter, not one half.

14. A highly rigid cubical block A of small mass and side L is fixed rigidly on to another cubical box B of the same dimensions and of low modulus of rigidity η such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side face of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by
(a) \( 2\pi\sqrt{M\eta L} \)
(b) \( 2\pi\sqrt{\frac{M\eta}{L}} \)
(c) \( 2\pi\sqrt{\frac{ML}{\eta}} \)
(d) \( 2\pi\sqrt{\frac{M}{\eta L}} \)

Answer: (d) \( 2\pi\sqrt{\frac{M}{\eta L}} \)

1. The top face of B is displaced by x through shear, so the shear angle is θ = x/L.
2. Restoring force = η × θ × area = η (x/L) L² = ηLx, so the spring constant is k = ηL.
3. Time period T = \( 2\pi\sqrt{\frac{M}{k}} = 2\pi\sqrt{\frac{M}{\eta L}} \), where M is the mass of block A.
4. Hence, the correct option is (d).

Teacher's Note:
a) The shear modulus plays the role of the spring constant through k = ηL.
b) The mass M used in the options is the small mass of block A.

15. On end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to
(a) \( 2\pi\sqrt{\frac{m}{K}} \)
(b) \( 2\pi\sqrt{\frac{m(YA + KL)}{YAK}} \)
(c) \( 2\pi\sqrt{\frac{mYA}{KL}} \)
(d) \( 2\pi\sqrt{\frac{mL}{YA}} \)

Answer: (b) \( 2\pi\sqrt{\frac{m(YA + KL)}{YAK}} \)

1. The wire behaves as a spring of constant k₁ = YA/L and it is in series with the spring K.
2. For springs in series, 1/k = 1/k₁ + 1/K = L/(YA) + 1/K = (KL + YA)/(YAK).
3. T = \( 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{m(YA + KL)}{YAK}} \).
4. Hence, the correct option is (b).

Teacher's Note:
a) The same force acts on the wire and on the spring, which is the sign of a series combination.
b) If the wire were rigid (Y very large), the result reduces to option (a).

16. A wire of area of cross-section 3.0 mm² and natural length 50 cm is fixed at one end and a mass of 2.1 kg is hung from the other end. Young's modulus of the material of the wire is 1.9 × 10¹¹ N / m² and g = 10 ms^-2. The potential energy stored in the wire in steady state is
(a) 1.9 × 10^-5 J
(b) 1.9 × 10^-4 J
(c) 9.1 × 10^-4 J
(d) 9.1 × 10^-5 J

Answer: (b) 1.9 × 10^-4 J

1. Force F = mg = 2.1 × 10 = 21 N and A = 3.0 × 10^-6 m².
2. Extension Δl = \( \frac{FL}{AY} = \frac{21 \times 0.5}{3 \times 10^{-6} \times 1.9 \times 10^{11}} \) = 1.84 × 10^-5 m.
3. Energy stored U = ½ F Δl = ½ × 21 × 1.84 × 10^-5 = 1.93 × 10^-4 J ≈ 1.9 × 10^-4 J.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use half of F × Δl because the force builds up from zero.
b) Convert mm² to m² (1 mm² = 10^-6 m²) and cm to m.

17. If the work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of the same material but double the radius and half the length is
(a) 0.25 J
(b) 4 J
(c) 8 J
(d) 16 J

Answer: (d) 16 J

1. Work done for the same extension x is W = \( \frac{YAx^2}{2l} \), so W ∝ A/l.
2. Double the radius makes the area 4 times, and half the length doubles 1/l, so the ratio is 4 × 2 = 8.
3. W = 2 × 8 = 16 J (the second wire is also stretched by 1 mm).
4. Hence, the correct option is (d).

Teacher's Note:
a) The question is assumed to stretch the second wire also by 1 mm, as in the first case.
b) Area depends on the square of the radius, so the factor 2 becomes 4.

18. Two solids A and B float in water. A floats with half its volume immersed and B floats with 2/3rd of its volume immersed. The densities of A and B are in the ratio of
(a) 2 : 3
(b) 4 : 3
(c) 3 : 4
(d) 3 : 2

Answer: (c) 3 : 4

1. A floating body has density = (fraction immersed) × density of water.
2. ρ_A = ½ ρ_w and ρ_B = ⅔ ρ_w.
3. ρ_A : ρ_B = ½ : ⅔ = 3 : 4.
4. Hence, the correct option is (c).

Teacher's Note:
a) Weight of the body equals the weight of the liquid displaced when it floats.
b) A body with the larger immersed fraction is denser.

19. A solid floats in liquid A with half its volume immersed and in liquid B with 2/3rd of its volume immersed. The densities of liquid A and B are in the ratio of
(a) 4 : 3
(b) 3 : 2
(c) 3 : 4
(d) 1 : 3

Answer: (a) 4 : 3

1. For the same solid, ρ_s = ½ ρ_A = ⅔ ρ_B.
2. So ρ_A : ρ_B = ⅔ : ½ = 4 : 3.
3. Hence, the correct option is (a).

Teacher's Note:
a) The liquid with the smaller immersed fraction is the denser liquid.
b) Contrast this with Question 18, where the liquid is fixed and the solids differ.

20. A solid weighs 5 N in air, 4 N in water and 4.5 N in some other liquid. The specific gravity of liquid is
(a) 0.5
(b) 0.6
(c) 0.9
(d) 1.5

Answer: (a) 0.5

1. Loss of weight in water = 5 - 4 = 1 N, and in the liquid = 5 - 4.5 = 0.5 N.
2. Specific gravity of the liquid = (loss in liquid)/(loss in water) = 0.5/1 = 0.5.
3. Hence, the correct option is (a).

Teacher's Note:
a) The loss of weight equals the upthrust, which is proportional to the liquid density for the same body.
b) The ratio of the two losses gives the relative density directly.

21. A vessel contains oil of density 0.8 gm/cm³ over mercury of density 13.6 gm/cm³. A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of material of the sphere is
(a) 3.3
(b) 6.4
(c) 7.2
(d) 12.8

Answer: (c) 7.2

1. Weight of sphere = upthrust from mercury + upthrust from oil.
2. ρVg = (V/2)(13.6)g + (V/2)(0.8)g.
3. ρ = (13.6 + 0.8)/2 = 7.2 gm/cm³.
4. Hence, the correct option is (c).

Teacher's Note:
a) The density of the sphere is the average of the two liquid densities because equal halves are immersed.
b) The answer is in gm/cm³, the same units as the data.

22. A body of volume V and density ρ is initially submerged in a liquid of density ρ'. If it is lifted through a height h in the liquid, its potential energy will
(a) increase by hV(ρ-ρ')g
(b) decrease by hV(ρ-ρ')g
(c) increase by hVρg / ρ'
(d) decrease by hVρg/ρ'

Answer: (a) increase by hV(ρ-ρ')g

1. When the body rises by h, a volume V of liquid of density ρ' goes down by h to take its place.
2. Change in potential energy = ρVgh (body rises) - ρ'Vgh (liquid falls) = hV(ρ - ρ')g.
3. For ρ greater than ρ' this is an increase. Hence, the correct option is (a).

Teacher's Note:
a) The potential energy of the body-plus-liquid system is considered, not of the body alone.
b) If ρ were less than ρ' the same expression would be negative, meaning a decrease.

23. A sphere has a hollow portion which is one third of its total volume. It floats in water with four fifth of its volume immersed. The specific gravity of the material is
(a) 0.9
(b) 1.2
(c) 1.8
(d) 2.4

Answer: (b) 1.2

1. Let the total volume be V. The material occupies (2/3)V.
2. Weight = upthrust: ρ × (2V/3) × g = 1 × (4V/5) × g.
3. ρ = (4/5) × (3/2) = 1.2.
4. Hence, the correct option is (b).

Teacher's Note:
a) Only the solid part contributes to the mass; the hollow part has none.
b) Specific gravity above 1 is possible because the hollow part makes the whole sphere float.

24. A body A floats in water with half its volume immersed. Another body B floats in a liquid of specific gravity 1.5 with 2/3rd of its volume immersed. The ratio of density of A to that of B is
(a) 1 : 2
(b) 2 : 1
(c) 2 : 3
(d) 3 : 2

Answer: (a) 1 : 2

1. Density of A = ½ × 1 = 0.5 (relative to water).
2. Density of B = ⅔ × 1.5 = 1.0.
3. Ratio ρ_A : ρ_B = 0.5 : 1 = 1 : 2.
4. Hence, the correct option is (a).

Teacher's Note:
a) Use the specific gravity of the liquid as the density of the liquid in the same units.
b) Always multiply the immersed fraction by the density of the liquid in which the body floats.

25. A balloon of total mass 1000 kg floats motionless over the earth's surface. If 100 kg of sand is thrown overboard the balloon, the balloon starts rising with an acceleration of (g = 9.8 m/s²)
(a) 10 m/s²
(b) 9.8 m/s²
(c) 1.09 m/s²
(d) 8.4 m/s²

Answer: (c) 1.09 m/s²

1. Floating motionless means the upthrust equals the weight: B = 1000 g.
2. After throwing the sand the mass is 900 kg, so the net upward force = 1000g - 900g = 100g.
3. a = 100g/900 = g/9 = 9.8/9 = 1.09 m/s².
4. Hence, the correct option is (c).

Teacher's Note:
a) The upthrust stays the same because the balloon volume does not change.
b) The acceleration is the extra force divided by the new mass, not by the old mass.

26. Water stands at a height of H in a large tank whose sides are vertical. A hole is made in one of the walls of the tank at a depth h below the surface of water. The distance R from the foot of the wall where the emerging stream of water strikes the floor is
(a) \( \sqrt{h(H - h)} \)
(b) \( \sqrt{hH} \)
(c) \( 2\sqrt{h(H - h)} \)
(d) \( \sqrt{2hH} \)

Answer: (c) \( 2\sqrt{h(H - h)} \)

1. Speed of efflux v = √(2gh) (Torricelli).
2. The hole is at height (H - h) above the floor, so the time of fall is t = √(2(H - h)/g).
3. R = vt = \( \sqrt{2gh} \times \sqrt{\frac{2(H - h)}{g}} = 2\sqrt{h(H - h)} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) The horizontal range is the same for holes at depth h and at depth (H - h).
b) The range is maximum (equal to H) when the hole is at the middle, h = H/2.

27. The density of ice is 920 kg/m³. The percentage of volume of an iceberg submerged in sea of density 1030 kg/m³ is
(a) 9.20
(b) 10.3
(c) 8.9
(d) 19.3

Answer: (c) 8.9

1. For a floating iceberg, the fraction submerged = ρ_ice/ρ_sea = 920/1030 = 0.893.
2. So about 89.3 % of the volume is submerged.
3. The digits 8.9 of the printed option (c) match this value, so (c) is the intended answer.
4. Hence, the correct option is (c).

Teacher's Note:
a) The actual percentage is about 89 %; option (c) has lost a factor of 10 in printing, but it is the only option with the right digits.
b) About 10.7 % of the iceberg stays above the water.

28. A body of mass 2 kg is floating in water with half its volume submerged. The force required to fully submerge it into water is
(a) 2 N
(b) 9.8 N
(c) 19.6 N
(d) 4.9 N

Answer: (c) 19.6 N

1. At half immersion the upthrust equals the weight: mg = 2 × 9.8 = 19.6 N.
2. Fully submerged, the upthrust becomes 2 × 19.6 = 39.2 N.
3. Extra force needed = 39.2 - 19.6 = 19.6 N (downward).
4. Hence, the correct option is (c).

Teacher's Note:
a) The extra upthrust comes only from the additional half of the volume going under.
b) The answer equals the weight of the body, 19.6 N, using g = 9.8 m/s².

29. A wooden block of volume 1000 cc is suspended from a spring balance. Its weight is 12 N in air. It is suspended in water such that half of the block is below the surface of water. The reading of spring balance is
(a) 10 N
(b) 9 N
(c) 8 N
(d) 7 N

Answer: (d) 7 N

1. Volume immersed = 500 cc = 5 × 10^-4 m³.
2. Upthrust = 5 × 10^-4 × 1000 × 10 = 5 N (taking g = 10 m/s²).
3. Reading of the spring balance = 12 - 5 = 7 N.
4. Hence, the correct option is (d).

Teacher's Note:
a) The balance reads the weight minus the upthrust on the immersed part only.
b) With g = 9.8 m/s² the reading would be 7.1 N, so g = 10 m/s² is the value intended.

30. Density of ice is ρ and that of water is σ. The decrease in volume when a mass M melts is
(a) \( \frac{M}{\sigma - \rho} \)
(b) \( \frac{\sigma - \rho}{M} \)
(c) \( M\left[\frac{1}{\sigma} - \frac{1}{\rho}\right] \)
(d) \( M\left[\frac{1}{\rho} - \frac{1}{\sigma}\right] \)

Answer: (d) \( M\left[\frac{1}{\rho} - \frac{1}{\sigma}\right] \)

1. Volume of ice of mass M = M/ρ and volume of water formed = M/σ.
2. Decrease in volume = M/ρ - M/σ = \( M\left[\frac{1}{\rho} - \frac{1}{\sigma}\right] \).
3. Since σ is greater than ρ, this is positive. Hence, the correct option is (d).

Teacher's Note:
a) Mass stays the same on melting; only the volume changes.
b) Density of ice is less than that of water, which is why the volume decreases.

31. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on the boat. The mass of the man is
(a) 60 kg
(b) 62 kg
(c) 72 kg
(d) 128 kg

Answer: (a) 60 kg

1. Extra volume of water displaced = 3 × 2 × 0.01 = 0.06 m³.
2. Mass of this water = 0.06 × 1000 = 60 kg.
3. By the law of floating, this equals the mass of the man.
4. Hence, the correct option is (a).

Teacher's Note:
a) The extra weight carried equals the extra weight of displaced water.
b) The sides of the boat are assumed vertical near the waterline.

32. If A denotes the area of the free surface of a liquid and h is the depth of the orifice of area of cross-section a below the liquid surface, the velocity v of flow through the orifice is given by
(a) \( \sqrt{2gh} \)
(b) \( \sqrt{2gh}\sqrt{\frac{A^2}{A^2 - a^2}} \)
(c) \( \sqrt{2gh}\sqrt{\frac{A}{A - a}} \)
(d) \( \sqrt{2gh}\sqrt{\frac{A^2 - a^2}{A^2}} \)

Answer: (b) \( \sqrt{2gh}\sqrt{\frac{A^2}{A^2 - a^2}} \)

1. Continuity: A v_A = a v, so v_A = av/A at the free surface.
2. Bernoulli: P₀ + ½ρv_A² + ρgh = P₀ + ½ρv².
3. So v² - (a²/A²) v² = 2gh, giving v² = \( \frac{2gh}{1 - a^2/A^2} \).
4. v = \( \sqrt{2gh}\sqrt{\frac{A^2}{A^2 - a^2}} \). Hence, the correct option is (b).

Teacher's Note:
a) When A is very large compared with a, the result reduces to Torricelli's √(2gh).
b) Do not neglect the speed of the free surface unless A is very large.

33. A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow is given by the following value if h₁ is the thickness of water layer at the bottom and h₂ is the thickness of kerosene layer at the top and density of water and kerosene is ρ₁ and ρ₂ gm / cc respectively.
(a) \( \sqrt{2g(h_1 + h_2)} \)
(b) \( \sqrt{2g(h_1\rho_1 + h_2\rho_2)} \)
(c) \( \sqrt{2g\left[h_1 + h_2\left(\frac{\rho_2}{\rho_1}\right)\right]} \)
(d) \( \sqrt{2g\left[h_1\left(\frac{\rho_1}{\rho_2}\right) + h_2\right]} \)

Answer: (c) \( \sqrt{2g\left[h_1 + h_2\left(\frac{\rho_2}{\rho_1}\right)\right]} \)

1. Pressure at the hole (gauge) = ρ₂gh₂ + ρ₁gh₁.
2. For water leaving the hole, ½ρ₁v² = ρ₁gh₁ + ρ₂gh₂.
3. So v = \( \sqrt{2g\left[h_1 + h_2\frac{\rho_2}{\rho_1}\right]} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) The kerosene layer acts like a shorter equivalent column of water of height h₂ρ₂/ρ₁.
b) Since ρ₂ is less than ρ₁, the speed is less than √(2g(h₁ + h₂)).

34. A raft of wood density 600 kg / m³ of mass 129 kg floats in water. How much weight can be put on the raft to make it just sink.
(a) 120 kg
(b) 200 kg
(c) 40 kg
(d) 81 kg

Answer: (d) 81 kg

1. Volume of the raft = 129/600 = 0.215 m³.
2. When it is just about to sink, the whole volume is submerged: the displaced water has mass 0.215 × 1000 = 215 kg.
3. Extra mass that can be carried = 215 - 129 = 86 kg, so the nearest listed option is 81 kg.
4. Hence, the correct option is (d).

Teacher's Note:
a) The exact working with the printed numbers gives 86 kg; the key (d) 81 kg is the nearest option and corresponds to a raft mass of about 121 kg in the original problem.
b) General rule: extra load = m(ρ_w/ρ - 1) = 129 × (1000/600 - 1).

35. The spring balance A reads 2 kg with a block of mass m suspended from it. When a beaker with a liquid is put on the pan B of the balance, it reads 5 kg. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation,
(a) the balance A will read more than 2 kg.
(b) the balance B will read more than 5 kg.
(c) the balance A will read less than 2 kg and B will read more than 5 kg.
(d) the balance A and B will read 2 kg and 5 kg respectively.

[Figure: A spring balance A hangs from a hatched ceiling support, with a coiled spring and a block of mass m hanging inside a beaker of liquid which stands on the pan of a second balance B resting on a table; the dial of B is drawn below the beaker.]

Answer: (b) and (c) (both are correct)

1. When the block is immersed, the liquid exerts an upward upthrust on it, so the tension in the spring of A falls and balance A reads less than 2 kg.
2. By Newton's third law, the block pushes the liquid down with an equal force, so balance B reads more than 5 kg by the same amount.
3. So statement (b) is true and statement (c) is true; (a) and (d) are false.
4. Hence, the correct options are (b) and (c).

Teacher's Note:
a) The total of the two readings stays 2 + 5 = 7 kg, because the weight of the block and the liquid is just shared differently.
b) This is a question with more than one correct option, as the printed key shows (b, c).

36. A sphere of radius R has a concentric cavity of radius r. The relative density of the material of the sphere is σ. It just floats when it is placed in a tank full of water. The Ratio R/r is
(a) \( \left(\frac{\sigma - 1}{\sigma}\right)^{1/3} \)
(b) \( \left(\frac{\sigma}{\sigma - 1}\right)^{1/3} \)
(c) \( \sigma^{1/3} \)
(d) \( \frac{1}{\sigma^{1/3}} \)

Answer: (b) \( \left(\frac{\sigma}{\sigma - 1}\right)^{1/3} \)

1. The sphere just floats, so it is fully submerged and its weight equals the upthrust.
2. Weight = σ × (4/3)π(R³ - r³) × g and upthrust = 1 × (4/3)πR³ × g.
3. σ(R³ - r³) = R³, so R³(σ - 1) = σr³.
4. R/r = \( \left(\frac{\sigma}{\sigma - 1}\right)^{1/3} \). Hence, the correct option is (b).

Teacher's Note:
a) Relative density σ is the density of the material compared with water.
b) The ratio is greater than 1, as it must be since the cavity is inside the sphere.

37. A horizontal pipeline carries water in a streamline flow. The velocity of water is 1 m/s and the pressure is 2000 Pa at a point along the pipe where cross-sectional area is 10 cm². The pressure of water at another point, where the cross-sectional area is 5 cm², is (1 Pa = 1 N/m²)
(a) 300 Pa
(b) 400 Pa
(c) 500 Pa
(d) 600 Pa

Answer: (c) 500 Pa

1. Continuity: 10 × 1 = 5 × v₂, so v₂ = 2 m/s.
2. Bernoulli for a horizontal pipe: P₂ = P₁ + ½ρ(v₁² - v₂²) = 2000 + ½ × 1000 × (1 - 4).
3. P₂ = 2000 - 1500 = 500 Pa.
4. Hence, the correct option is (c).

Teacher's Note:
a) The pressure drops where the water speeds up.
b) Use ρ = 1000 kg/m³ for water.

38. A small spherical solid ball is dropped in a viscous liquid. Its journey in the liquid is best described in the figure by
(a) Curve A
(b) Curve B
(c) Curve C
(d) Curve D

[Figure: A graph with Distance on the x axis and Velocity in liquid on the y axis. All four curves start at the origin O. Curve A rises slowly then steeply upward. Curve B is a straight line rising to the upper right. Curve C rises steeply, bends over and becomes nearly flat. Curve D rises and then turns downward. The curves cross near the middle of the graph.]

Answer: (c) Curve C

1. The ball speeds up at first, but as the speed grows the viscous drag grows until it balances the net downward force.
2. So the velocity increases with distance at a decreasing rate and then becomes constant (terminal velocity).
3. Curve C rises steeply and then flattens; A keeps rising faster, B is a straight line (constant acceleration) and D falls after the peak, which are not possible.
4. Hence, the correct option is (c).

Teacher's Note:
a) A falling ball never slows down in a long viscous column; it approaches a constant terminal velocity.
b) Curve D shows the speed dropping, which would need a force opposing motion that is larger than the weight.

39. From amongst the following curves, which one shown the variation of the velocity V with the t for a small sized spherical body falling vertically in a long column of viscous liquid.
(a)
(b)
(c)
(d)

[Figure: Four small v-t graphs, each with v on the vertical axis and t on the horizontal axis, given as options. (a) A horizontal line. (b) A straight line rising from a point above the origin. (c) A curve starting high, dipping slightly and then rising a little. (d) A curve starting above the origin, rising steeply at first and then flattening out.]

Answer: (d)

1. The sphere starts with some speed and is accelerated by gravity, so the speed first rises.
2. As the speed grows, the viscous drag (6πηrv) rises until weight = upthrust + drag, and the acceleration becomes zero.
3. So v rises with a decreasing slope and levels off at the terminal velocity, as in curve (d).
4. Hence, the correct option is (d).

Teacher's Note:
a) The options are graphs; their shapes are described in the figure line above the answer.
b) A straight rising line (b) would need a constant acceleration, which stops once the drag builds up.

40. Two equal drops are falling through air with a steady velocity of 5 cm/sec. If the drop coalesce, the new terminal velocity will be
(a) 5 × 2 cm/sec
(b) 5 × √2 cm/sec
(c) 5 × ( 4)^1/3 cm/sec
(d) 5/√2 cm/sec

Answer: (c) 5 × ( 4)^1/3 cm/sec

1. Terminal velocity v ∝ r². Two drops of radius r coalesce to form a drop of radius R with (4/3)πR³ = 2 × (4/3)πr³, so R = 2^1/3 r.
2. v' = v (R/r)² = 5 × (2^1/3)² = 5 × 2^2/3 = 5 × (4)^1/3 cm/s.
3. Hence, the correct option is (c).

Teacher's Note:
a) Terminal velocity depends on the square of the radius, so the bigger drop falls faster.
b) 2^2/3 = 4^1/3 ≈ 1.59, so the new speed is about 7.9 cm/s.

41. A large no. of liquid drops each of radius r coalesce to form a single drop of radius R. The energy released in the brocess is converted into the kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of liquid is T, density of liquid is ρ)
(a) \( \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)} \)
(b) \( \sqrt{\frac{4T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)} \)
(c) \( \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} + \frac{1}{R}\right)} \)
(d) \( \sqrt{\frac{4T}{\rho}\left(\frac{1}{r} + \frac{1}{R}\right)} \)

Answer: (a) \( \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)} \)

1. Let n drops form the big drop: nr³ = R³. Surface energy released = T × 4π(nr² - R²) = \( 4\pi T\left(\frac{R^3}{r} - R^2\right) \).
2. Kinetic energy gained = ½ × (4/3)πR³ρ × v².
3. Equating: \( 4\pi T R^3\left(\frac{1}{r} - \frac{1}{R}\right) = \frac{2}{3}\pi R^3 \rho v^2 \), so v² = \( \frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right) \).
4. Hence, the correct option is (a).

Teacher's Note:
a) The surface area decreases on coalescing, so energy is released and not absorbed.
b) Check that the factor under the root has a minus sign, since 1/r is greater than 1/R.

42. Water rises in a capillary tube to a height h. If the tube is inclined to the liquid surface at 45°, the liquid will rise in the tube upto its length equal to
(a) \( \frac{h}{\sqrt{2}} \)
(b) h
(c) \( \sqrt{2} \) h
(d) 2 h

Answer: (c) \( \sqrt{2} \) h

1. The vertical height of the water column stays fixed at h, whatever the inclination.
2. If the tube makes an angle of 45° with the vertical, the length l of water along the tube satisfies l cos 45° = h.
3. l = h / cos 45° = √2 h.
4. Hence, the correct option is (c).

Teacher's Note:
a) Only the vertical height h is determined by surface tension, so the length along the tube increases when the tube is tilted.
b) Because of the symmetry of 45°, the angle with the surface and with the vertical are the same.

43. Drops of liquid of density d are floating half immersed in a liquid of density ρ. If the surface tension of liquid is T, then radius of the drop will be
(a) \( \sqrt{\frac{3T}{g(2d - \rho)}} \)
(b) \( \sqrt{\frac{3T}{g(d - \rho)}} \)
(c) \( \sqrt{\frac{6T}{g(2d - \rho)}} \)
(d) \( \sqrt{\frac{6T}{g(d - \rho)}} \)

Answer: (a) \( \sqrt{\frac{3T}{g(2d - \rho)}} \)

1. The weight of the drop (4/3)πr³dg is balanced by the upthrust on its immersed half, ½ × (4/3)πr³ρg, and the surface tension force T × 2πr acting along the line of contact.
2. (4/3)πr³g(d - ρ/2) = 2πrT.
3. r² = \( \frac{3T}{2g(d - \rho/2)} = \frac{3T}{g(2d - \rho)} \).
4. Hence, the correct option is (a).

Teacher's Note:
a) The surface tension force acts along the circular line where the drop meets the surface, of length 2πr.
b) The condition is only possible if the drop is denser than half the liquid, 2d greater than ρ.

44. A metallic wire of density d floats horizontal in water. The maximum radius of the wire may not sink, will be (surface tension of water = T).
(a) \( \sqrt{\frac{2T}{\pi d g}} \)
(b) \( \sqrt{2\pi T d g} \)
(c) \( \sqrt{\frac{2\pi T}{dg}} \)
(d) \( \sqrt{\frac{dg}{2\pi T}} \)

Answer: (a) \( \sqrt{\frac{2T}{\pi d g}} \)

1. Take a length l of the wire. Its weight = πr²l dg.
2. Surface tension acts on both sides of the wire, so the maximum supporting force = 2Tl.
3. πr²ldg = 2Tl, so r² = \( \frac{2T}{\pi d g} \).
4. r = \( \sqrt{\frac{2T}{\pi d g}} \). Hence, the correct option is (a).

Teacher's Note:
a) The wire is supported by the surface tension on its two sides, hence the factor 2T per unit length.
b) A larger radius means more weight per unit length, so the wire would sink.