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NEET Properties of Solids and Liquids MCQs with answers

429 questions on Properties of Solids and Liquids for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

mediumNEET 2022

1. Excess pressure inside soap bubble of radius r, surface tension T:

  1. T/r
  2. 2T/r
  3. 4T/r
  4. T/2r
Show answer and explanation

Answer: (C) 4T/r

Two surfaces → 4T/r. Single surface (drop): 2T/r.

Quick tip: Bubble (2 surfaces): 4T/r. Drop/air bubble in liquid (1 surface): 2T/r.

NCERT Class 11, Chapter 10, page 242

easy

2. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 250 g of water by 30 °C is:

  1. 31.5 J
  2. 1050 J
  3. 31500 J
  4. 7500 J
Show answer and explanation

Answer: (C) 31500 J

Heat absorbed without a change of state is Q = mcΔT = 250 × 4.2 × 30 = 31500 J.

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3. Two spherical stars A and B radiate maximum energy at wavelengths 400 nm and 600 nm respectively. The ratio of the surface area of star A to that of star B is 1 : 4. The ratio of total power radiated by star A to that by star B is:

  1. 27 : 16
  2. 81 : 64
  3. 9 : 4
  4. 3 : 2
Show answer and explanation

Answer: (B) 81 : 64

By Wien's law, T is inversely proportional to λ_max, so T_A / T_B = λ_B / λ_A = 600/400 = 3/2. Total radiated power is P = A_surface σ T⁴. Therefore, P_A / P_B = (A_A / A_B) × (T_A / T_B)⁴ = (1/4) × (3/2)⁴ = (1/4) × (81/16) = 81/64.

hard

4. A cylindrical vessel of cross-sectional area A has a tiny hole of area a at its bottom (where a ≪ A). When the vessel is filled with liquid up to height H, the total time required for the vessel to empty completely through the hole is:

  1. (A/(2a)) √(2H/g)
  2. (A/a) √(H/(2g))
  3. (2A/a) √(H/g)
  4. (A/a) √(2H/g)
Show answer and explanation

Answer: (D) (A/a) √(2H/g)

The rate of decrease of water level is -A(dh/dt) = a√(2gh). Rearranging gives -dh/√h = (a/A)√(2g) dt. Integrating h from H to 0 gives 2√H = (a/A)√(2g) t. Solving for t yields t = (A/a) √(2H/g).

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5. A wire of length 1.5 m and cross-sectional area 1 mm² stretches by 2 mm under a load of 150 N. The Young's modulus of its material is:

  1. 1.13 × 10¹¹ N/m²
  2. 7.5 × 10¹⁰ N/m²
  3. 1.13 × 10¹⁴ N/m²
  4. 1.13 × 10⁸ N/m²
Show answer and explanation

Answer: (A) 1.13 × 10¹¹ N/m²

Y = FL/(AΔL) = 150 × 1.5/(1 × 10⁻⁶ × 2 × 10⁻³) = 1.13 × 10¹¹ N/m².

easy

6. An astronaut takes an open beaker filled with a liquid to an effective depth h inside a spaceship. If the spaceship is in a stable circular orbit around the Earth, the gauge pressure at the bottom of the beaker is:

  1. Zero
  2. ρgh
  3. 2ρgh
  4. Dependent on the orbital speed of the spaceship
Show answer and explanation

Answer: (A) Zero

Inside an orbiting spaceship, the effective acceleration due to gravity is zero (state of weightlessness, g_eff = 0). Since gauge pressure is given by P_gauge = ρ(g_eff)h, the gauge pressure at the bottom is zero.

hard

7. A glass tube of length 100 cm is sealed at one end and held vertically with its open end downwards. It is slowly immersed in water until the water level inside the tube rises to half the tube's length. Assuming temperature remains constant and atmospheric pressure equals 10 m of water column, what is the depth of the open end of the tube from the water surface?

  1. 10.5 m
  2. 10.0 m
  3. 9.5 m
  4. 11.0 m
Show answer and explanation

Answer: (A) 10.5 m

By Boyle's law (P₁V₁ = P₂V₂), initially P₁ = P₀, V₁ = L·A. When water rises to half the tube, V₂ = (L/2)·A, so P₂ = 2P₀. The pressure at the internal water-air boundary is P₂ = P₀ + ρg(H - L/2), where H is the depth of the open end. Thus, ρg(H - L/2) = P₀ = ρg(10 m). Hence, H = 10 m + L/2 = 10 m + 0.5 m = 10.5 m.

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8. The thermal radiation emitted by a black body at 2000 K has maximum spectral intensity at 1.5 μm. If the absolute temperature of the black body is increased to 3000 K, what is the wavelength at which maximum emission occurs, and by what factor does the total emissive power increase?

  1. 2.25 μm and 2.25
  2. 1.0 μm and 5.0625
  3. 1.0 μm and 2.25
  4. 0.67 μm and 5.0625
Show answer and explanation

Answer: (B) 1.0 μm and 5.0625

By Wien's displacement law, λ_max T = constant. Thus, λ' = 1.5 × (2000/3000) = 1.0 μm. By Stefan-Boltzmann law, total emissive power E ∝ T⁴, so E'/E = (3000/2000)⁴ = (1.5)⁴ = 5.0625.

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9. A soap bubble of radius r₁ is blown inside another soap bubble of radius r₂ (where r₂ > r₁) in air at atmospheric pressure P₀. The surface tension of the soap solution is T. The excess pressure inside the inner bubble relative to the atmospheric pressure P₀ is:

  1. 2T/r₁ + 2T/r₂
  2. 4T/r₁ + 4T/r₂
  3. 4T(1/r₁ - 1/r₂)
  4. 4T/r₁
Show answer and explanation

Answer: (B) 4T/r₁ + 4T/r₂

Pressure inside the outer bubble is P₁ = P₀ + 4T/r₂. The inner bubble is surrounded by the air inside the outer bubble, so the pressure inside the inner bubble is P₂ = P₁ + 4T/r₁ = P₀ + 4T/r₂ + 4T/r₁. Therefore, the excess pressure inside the inner bubble relative to the outside atmosphere is P₂ - P₀ = 4T/r₁ + 4T/r₂.

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10. A metal rod of length 1.0 m, cross-sectional area 2.0 × 10⁻⁴ m², and coefficient of linear expansion 1.2 × 10⁻⁵ K⁻¹ is prevented from expanding along its length while its temperature is raised by 50 K. If the Young's modulus of the rod is 2.0 × 10¹¹ N m⁻², the thermal force developed in the rod is:

  1. 1.2 × 10⁴ N
  2. 2.4 × 10⁴ N
  3. 4.8 × 10⁴ N
  4. 6.0 × 10³ N
Show answer and explanation

Answer: (B) 2.4 × 10⁴ N

Thermal strain developed when expansion is fully prevented is ΔL/L = α ΔT = (1.2 × 10⁻⁵)(50) = 6.0 × 10⁻⁴. Thermal stress is σ = Y (ΔL/L) = (2.0 × 10¹¹)(6.0 × 10⁻⁴) = 1.2 × 10⁸ N m⁻². The compressive force is F = σ × A = (1.2 × 10⁸)(2.0 × 10⁻⁴) = 2.4 × 10⁴ N.

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11. A metal sphere cools at the rate of 1.6 °C/min when its temperature is 70 °C, in surroundings at 30 °C. According to Newton's law of cooling, its rate of cooling when its temperature falls to 50 °C in the same surroundings will be:

  1. 0.4 °C/min
  2. 1.2 °C/min
  3. 0.8 °C/min
  4. 1.0 °C/min
Show answer and explanation

Answer: (C) 0.8 °C/min

According to Newton's law of cooling, rate of cooling dT/dt = -k(T - T₀). At 70 °C, 1.6 = k(70 - 30) = 40k, which gives k = 0.04 min⁻¹. When the temperature falls to 50 °C, dT/dt = k(50 - 30) = 0.04 × 20 = 0.8 °C/min.

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12. An open manometer tube has one arm connected to a gas container and the other arm open to the atmosphere. The manometric liquid is oil of density 850 kg/m³. If the oil level in the open arm is 40 cm higher than that in the arm connected to the container, what is the gauge pressure of the gas? (Take g = 9.8 m/s²)

  1. 3920 Pa
  2. 3400 Pa
  3. 3332 Pa
  4. 4160 Pa
Show answer and explanation

Answer: (C) 3332 Pa

Gauge pressure is given by P_gauge = P_gas - P_atm = ρ_oil × g × h = 850 kg/m³ × 9.8 m/s² × 0.40 m = 3332 Pa.

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13. Consider the following two statements regarding Pascal's principle: Statement I: A change in pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the containing vessel. Statement II: Pascal's principle is strictly valid only when the effect of gravity is completely neglected. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is the direct statement of Pascal's principle. Statement II is incorrect because Pascal's principle deals with the transmission of an applied external pressure change (ΔP); even when gravity is present, an increase in pressure at one point increases the pressure at every other point by the exact same amount ΔP.

easy

14. When a load of 40 N is suspended from a vertical wire, its length becomes L₁. When the load is increased to 60 N, its length becomes L₂. Assuming Hooke's law is obeyed, the natural (unstretched) length of the wire is:

  1. 3L₂ - 2L₁
  2. 2L₁ - L₂
  3. 3L₁ - 2L₂
  4. (L₁ + L₂) / 2
Show answer and explanation

Answer: (C) 3L₁ - 2L₂

Let the original length be L₀. By Hooke's law, L₁ - L₀ = 40/k and L₂ - L₀ = 60/k, where k is the force constant of the wire. Dividing the two equations gives (L₁ - L₀) / (L₂ - L₀) = 40/60 = 2/3. Cross-multiplying: 3(L₁ - L₀) = 2(L₂ - L₀) ⇒ 3L₁ - 3L₀ = 2L₂ - 2L₀ ⇒ L₀ = 3L₁ - 2L₂.

easy

15. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 500 g of water by 10 °C is:

  1. 21000 J
  2. 5000 J
  3. 21 J
  4. 2100 J
Show answer and explanation

Answer: (A) 21000 J

Heat absorbed without a change of state is Q = mcΔT = 500 × 4.2 × 10 = 21000 J.

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