Mechanical Properties of Solids NEET PYQs (2005–2026) with Solutions

10 solved Mechanical Properties of Solids PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyElastic behaviour, Stress-strain relationship, Hooke's Law, Young's modulus, Bulk modulus

   Match List I with List II: Choose the correct answer from the options given below:

   1. AOption A: A-IV, B-I, C-II, D-III
   2. BOption B: A-III, B-II, C-I, D-IV
   3. COption C: A-I, B-IV, C-III, D-II
   4. DOption D: A-II, B-III, C-IV, D-I

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   Correct answer: (D) A-II, B-III, C-IV, D-I

   Explanation

   Young’s modulus = Stress/Strain = FL/[A(ΔL)] → II Compressibility = 1/(Bulk modulus) = -(1/ΔP)(ΔV/V) → III Bulk modulus = -P(V/ΔV) → IV Poisson’s ratio = Lateral strain/Longitudinal strain = Δd/ΔL × (L/d) → I Hence the correct matching is A-II, B-III, C-IV, D-I (Option D).

2. Question 2 (NEET 2024)

   NEET 2024MediumHooke's Law & Elastic Moduli

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

   1. AOption A: 40 mm
   2. BOption B: 8 mm
   3. COption C: 4 mm
   4. DOption D: 0.4 mm

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   Correct answer: (C) 4 mm

   Explanation

   Young's modulus is: $$Y = \frac{\text{stress}}{\text{strain}}$$ Therefore, $$\text{strain} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$$ Since, $$\text{strain} = \frac{\Delta L}{L}$$ $$\Delta L = 4 \times 10^{-3} \times 1 = 4 \times 10^{-3}\,m$$ $$\Delta L = 4\,mm$$ Hence, option (C) is correct.

3. Question 3 (NEET 2024)

   NEET 2024MediumThermal Expansion and Stress

   A metallic bar of Young’s modulus 0.5 × 10¹¹ N m⁻² and coefficient of linear 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:

   1. AOption A: 100 × 10³ N
   2. BOption B: 2 × 10³ N
   3. COption C: 5 × 10³ N
   4. DOption D: 50 × 10³ N

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   Correct answer: (D) 50 × 10³ N

   Explanation

   Thermal stress developed when expansion is prevented: $$\text{stress} = Y \alpha \Delta T$$ Force developed: $$F = \text{stress} \times A$$ Substituting values: $$F = (0.5 \times 10^{11})(10^{-5})(100)(10^{-3})$$ $$F = 0.5 \times 10^5$$ $$F = 50 \times 10^3\,N$$ Hence, option (D) is correct.

4. Question 4 (NEET 2023)

   NEET 2023EasyStress & Strain

   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 :

   1. AOption A: 2W/A
   2. BOption B: W/A
   3. COption C: W/2A
   4. DOption D: Zero

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   Correct answer: (B) W/A

   Explanation

   Longitudinal stress is defined as force per unit area. Hence stress = W/A.

5. Question 5 (NEET 2022)

   NEET 2022MediumHooke's Law & Elastic Moduli

   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.

   1. AOption A: (A) is false but (R) is true
   2. BOption B: Both (A) and (R) are true and (R) is the correct explanation of (A)
   3. COption C: Both (A) and (R) are true and (R) is not the correct explanation of (A)
   4. DOption D: (A) is true but (R) is false

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   Correct answer: (D) (A) is true but (R) is false

   Explanation

   Extension in a coil spring depends on the shear modulus (modulus of rigidity) of the material, so Assertion (A) is true. Copper has lower tensile strength than steel, hence Reason (R) is false. Therefore option (D) is correct.

6. Question 6 (NEET 2020)

   NEET 2020MediumHooke's Law & Elastic Moduli

   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:

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (D) Option (4)

   Explanation

   Young’s modulus, Y = (Stress)/(Strain) = (F/A)/((ΔL)/L) = (Mg/A)/((L₁ − L)/L) = MgL / A(L₁ − L). Hence, option (D) is correct.

7. Question 7 (NEET 2019)

   NEET 2019EasyStress-Strain Curve

   The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively),

   1. AOption A: Plastic and ductile
   2. BOption B: Ductile and brittle
   3. COption C: Brittle and ductile
   4. DOption D: Brittle and plastic

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   Correct answer: (C) Brittle and ductile

   Explanation

   Since the fracture point and ultimate strength point are very close for material X, it behaves as a brittle material. For material Y, these two points are far apart, indicating large plastic deformation before fracture, hence Y is ductile.

8. Question 8 (NEET 2018)

   NEET 2018MediumYoung's modulus and elastic deformation

   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?

   1. AOption A: 9F
   2. BOption B: 6F
   3. COption C: F
   4. DOption D: 4F

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   Correct answer: (A) 9F

   Explanation

   Extension in a wire is given by: $$\Delta l = \frac{Fl}{AY}$$ where $$Y$$ is Young's modulus. Since both wires have the same volume: For wire 1: $$A_1 = A, \quad l_1 = 3l$$ So, $$\Delta l = \frac{F(3l)}{AY} \quad ...(1)$$ For wire 2: $$A_2 = 3A, \quad l_2 = l$$ If force required is $$F'$$: $$\Delta l = \frac{F'l}{3AY} \quad ...(2)$$ Equating (1) and (2): $$\frac{3Fl}{AY} = \frac{F'l}{3AY}$$ $$F' = 9F$$ Hence, option (A) is correct.

9. Question 9 (NEET 2017)

   NEET 2017MediumHooke's Law & Elastic Moduli

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

   1. AOption A: $\dfrac{p}{3B}$
   2. BOption B: $\dfrac{p}{B}$
   3. COption C: $\dfrac{B}{3p}$
   4. DOption D: $\dfrac{3p}{B}$

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   Correct answer: (A) $\dfrac{p}{3B}$

   Explanation

   Bulk modulus is defined as: $$B=\frac{p}{\Delta V/V}$$ For a sphere: $$V=\frac{4}{3}\pi r^3$$ Hence, $$\frac{\Delta V}{V}=3\frac{\Delta r}{r}$$ Therefore, $$B=\frac{p}{3\Delta r/r}$$ So the fractional decrease in radius is: $$\frac{\Delta r}{r}=\frac{p}{3B}$$

10. Question 10 (AIPMT 2015)

    AIPMT 2015MediumHooke's Law & Elastic Moduli

    The Young's modulus of steel is twice that of brass. Two wires of the same length and same cross-sectional area, one of steel and another of brass, are suspended from the same roof. If the lower ends of the wires are to be at the same level, the weights added to the steel and brass wires must be in the ratio:

    1. AOption A: 1 : 1
    2. BOption B: 1 : 2
    3. COption C: 2 : 1
    4. DOption D: 4 : 1

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    Correct answer: (C) 2 : 1

    Explanation

    For equal extensions: $$\Delta l=\frac{Fl}{AY}$$ Since the lower ends must remain at the same level, $$\Delta l_{steel}=\Delta l_{brass}$$ $$\frac{W_s l}{AY_s}=\frac{W_b l}{AY_b}$$ $$\frac{W_s}{W_b}=\frac{Y_s}{Y_b}$$ Given: $$Y_s=2Y_b$$ Therefore, $$\frac{W_s}{W_b}=2$$ Hence the ratio is $2:1$.