NEET Mechanical Properties of Solids PYQs & Practice Questions

10 free NEET Physics questions on Mechanical Properties of Solids, starting with previous year questions (2017–2026). Tap “Show answer” to check each one. No login needed.

Topics: Stress and Strain · Hooke's Law and Young's Modulus · Elastic Moduli · Bulk Modulus · Stress-Strain Curve · Poisson's Ratio · Shear Modulus · Elastic Potential Energy · Comparison of Elastic Properties · Properties of Materials · Breaking Stress · Young's Modulus and Temperature

1. Q1NEET 2026Matching elastic moduli

   Match List I with List II: List I — A. Young's Modulus, B. Compressibility, C. Bulk Modulus, D. Poisson's Ratio. List II — I. (Δd/d)/(ΔL/L), II. FL/A(ΔL), III. -(1/ΔP)(ΔV/V), IV. -P(ΔV/V).

   • A. A-III, B-II, C-I, D-IV
   • B. A-II, B-III, C-IV, D-I
   • C. A-I, B-IV, C-III, D-II
   • D. A-IV, B-I, C-II, D-III
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   Answer: A — A-III, B-II, C-I, D-IV

   Young's modulus=FL/A(ΔL) (II); compressibility=-(1/ΔP)(ΔV/V) (III); Bulk modulus=-P(ΔV/V) (IV); Poisson's ratio=(Δd/d)/(ΔL/L) (I). So A-II, B-III, C-IV, D-I.

2. Q2NEET 2024Young's Modulus - Maximum Elongation

   The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8×10⁸ N m⁻² and 2×10¹¹ N m⁻², is:

   • A. 40 mm
   • B. 8 mm
   • C. 4 mm
   • D. 0.4 mm
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   Answer: B — 8 mm

   ΔL = FL/AY, so (ΔL·Y)/L = F/A = 8×10⁸ (elastic limit stress). ΔL × 2×10¹¹ / 1 = 8×10⁸, so ΔL = 8×10⁸ / 2×10¹¹ = 4×10⁻³ m = 4 mm.

3. Q3NEET 2024Thermal Stress - Compressive Force in Constrained Bar

   A metallic bar of Young's modulus, 0.5×10¹¹ N m⁻² and coefficient of linear thermal expansion 10⁻⁵ °C⁻¹, length 1 m and area of cross-section 10⁻³ m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is:

   • A. 100×10³ N
   • B. 2×10³ N
   • C. 5×10³ N
   • D. 50×10³ N
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   Answer: C — 5×10³ N

   F = YAαΔT = 0.5×10¹¹ × 10⁻³ × 10⁻⁵ × 100 = 50×10³ N.

4. Q4NEET 2023Longitudinal Stress

   Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is:

   • A. W/A
   • B. W/2A
   • C. Zero
   • D. 2W/A
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   Answer: —

   Longitudinal stress is defined as force per unit cross-sectional area. Since the weight W is the force stretching the wire, stress = W/A at any cross-section along the wire.

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5. Q5NEET 2022Shear Modulus and Tensile Strength

   Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both (A) and (R) are true and (R) is the correct explanation of (A)
   • B. Both (A) and (R) are true and (R) is not the correct explanation of (A)
   • C. (A) is true but (R) is false
   • D. (A) is false but (R) is true
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   Answer: B — Both (A) and (R) are true and (R) is not the correct explanation of (A)

   The stretching/twisting behavior of a coil spring is indeed governed by the shear modulus of its material, making Assertion (A) true. However, steel has greater tensile strength than copper of the same dimensions, making Reason (R) false.

6. Q6NEET 2020Young's modulus formula derivation

   A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young's modulus is:

   • A. MgL1/AL
   • B. Mg(L1-L)/AL
   • C. MgL/AL1
   • D. MgL/(A(L1-L))
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   Answer: C — MgL/AL1

   Young's modulus Y = (Stress)/(Strain) = (Mg/A)/((L1-L)/L) = MgL/(A(L1-L)).

7. Q7NEET 2020Dimensional formula of stress

   Dimensions of stress are:

   • A. [MLT^-2]
   • B. [ML²T^-2]
   • C. [ML⁰T^-2]
   • D. [ML^-1T^-2]
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   Answer: C — [ML⁰T^-2]

   Stress = Force/Area = [MLT^-2]/[L²] = [ML^-1T^-2].

8. Q8NEET 2019Elastic Potential Energy

   When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L+l). The elastic potential energy stored in the extended wire is:

   • A. (1/2)Mgl
   • B. (1/2)MgL
   • C. Mgl
   • D. MgL
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   Answer: —

   At equilibrium, kl = Mg, so k = Mg/l. Elastic PE = (1/2)kl^2 = (1/2)(Mg/l)(l^2) = (1/2)Mgl.

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9. Q9NEET 2018Young's modulus: force for equal extension in wires of different cross-section

   Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?

   • A. 9 F
   • B. F
   • C. 4 F
   • D. 6 F
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   Answer: —

   Since volume is constant, if area is 3A, length is L/3 for the second wire. Young's modulus Y = FL/(AΔl) is the same for both. For wire 2: Y = F2(L/3)/(3A×Δl). Equating with wire 1's Y=FL/(AΔl): F2 = 9F.

10. Q10NEET 2017Bulk modulus

    The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', the fractional decrease in radius is:

    • A. B/(3p)
    • B. 3p/B
    • C. p/(3B)
    • D. p/B
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    Answer: B — 3p/B

    ΔV/V = 3ΔR/R (for a sphere). B = −p/(ΔV/V) ⟹ ΔV/V = p/B. So ΔR/R = (1/3)(ΔV/V) = p/(3B).