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📘 Mechanical Properties of Solids — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Mechanical Properties of Solids for JEE Main with previous year questions on stress, strain, and Young's modulus. Each solution is explained in a simple, step-by-step manner to boost your score.
Mechanical Properties of Solids — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 1)
Two wires A and B are made of same material having ratio of lengths \(L_A/L_B = 1/3\) and their diameters ratio \(d_A/d_B = 2\). If both the wires are stretched using same force, what would be the ratio of their respective elongations?
(1) 1 : 6 (2) 1 : 12
(3) 3 : 4 (4) 1 : 3
(1) 1 : 6 (2) 1 : 12
(3) 3 : 4 (4) 1 : 3
✅ Answer & Explanation:
Correct Option: (2)
- Elongation \(\Delta L = \frac{FL}{AY} = \frac{FL}{(\pi d^2 / 4) Y}\).
- Since \(F\) and \(Y\) are same: \(\Delta L \propto \frac{L}{d^2}\).
- \(\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \left(\frac{d_B}{d_A}\right)^2 = \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right)^2 = \frac{1}{12}\).
Q2. JEE Main 2025 (4 April Shift 2)
A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of \(10^5 \, \text{N}\) at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement \(\theta\) of the rod axis from its original position would be : (shear moduli, \(G = 10^{10} \, \text{N/m}^2\))
(1) \(1/160\pi\) (2) \(1/4\pi\)
(3) \(1/40\pi\) (4) \(1/2\pi\)
(1) \(1/160\pi\) (2) \(1/4\pi\)
(3) \(1/40\pi\) (4) \(1/2\pi\)
✅ Answer & Explanation:
Correct Option: (1)
- Shear modulus \(G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{\frac{F}{A}}{\theta}\).
- \(\theta = \frac{F}{AG} = \frac{10^5}{\pi (0.04)^2 \times 10^{10}} = \frac{10^5}{\pi (16 \times 10^{-4}) \times 10^{10}} = \frac{1}{160\pi}\).
Q3. JEE Main 2024 (05 Apr Shift 2)
Match List-I with List-II:
Choose the correct answer from the options given below :
(1) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
| List-I | List-II |
|---|---|
| (A) A force that restores an elastic body of unit area to its original state | (I) Bulk modulus |
| (B) Two equal and opposite forces parallel to opposite faces | (II) Young's modulus |
| (C) Forces perpendicular everywhere to the surface per unit area same everywhere | (III) Stress |
| (D) Two equal and opposite forces perpendicular to opposite faces | (IV) Shear modulus |
Choose the correct answer from the options given below :
(1) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
✅ Answer & Explanation:
Correct Option: (2)
- (A) Restoring force per unit area is Stress. [III]
- (B) Parallel forces relate to Shear modulus. [IV]
- (C) Uniformly perpendicular forces relate to Bulk modulus. [I]
- (D) Axial perpendicular forces relate to Young's modulus. [II]
Q4. JEE Main 2023 (10 Apr Shift 1)
Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is _____ \(10^{-6} \, m\). [Young's modulus for steel \(= 2 \times 10^{11} \, N \, m^{-2}\) and \(g = 10 \, m \, s^{-2}\)]
✅ Answer & Explanation:
Correct Answer: 20
- Elongation \(\Delta L = \frac{F L}{A Y}\).
- For steel wire, the force \(F\) is the total weight of both blocks \(= (2 + 1.14) \times 10 = 31.4 \, \text{N}\).
- Area \(A = \pi (0.2 \times 10^{-2})^2 = \pi \times 4 \times 10^{-6} \, \text{m}^2\).
- \(L = 1.6 \, \text{m}\).
- \(\Delta L = \frac{31.4 \times 1.6}{\pi \times 4 \times 10^{-6} \times 2 \times 10^{11}} \approx 20 \times 10^{-6} \, \text{m}\).
Q5. JEE Main 2022 (27 Jul Shift 2)
A steel wire of length 3.2 m (\(Y_S = 2.0 \times 10^{11} \, N \, m^{-2}\)) and a copper wire of length 4.4 m (\(Y_C = 1.1 \times 10^{11} \, N \, m^{-2}\)), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: (Given \(\pi = 22/7\))
(1) 360 (2) 180 (3) 1080 (4) 154
(1) 360 (2) 180 (3) 1080 (4) 154
✅ Answer & Explanation:
Correct Option: (4)
- Total elongation \(\Delta L_{net} = \Delta L_s + \Delta L_c = F \left( \frac{L_s}{A Y_s} + \frac{L_c}{A Y_c} \right)\).
- Substituting the radii, lengths, and moduli:
- \(1.4 \times 10^{-3} = \frac{F}{\pi (1.4 \times 10^{-3})^2} \left[ \frac{3.2}{2 \times 10^{11}} + \frac{4.4}{1.1 \times 10^{11}} \right]\).
- Solving for \(F\) results in \(154 \, \text{N}\).
Q6. JEE Main 2020 (07 Jan Shift 1)
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section \(1.0 \, mm^2\)) is \(90 \, m \, s^{-1}\). If the Young's modulus of wire is \(16 \times 10^{11} \, N \, m^{-2}\), the extension of wire over its natural length is:
(1) 0.03 mm (2) 0.02 mm
(3) 0.04 mm (4) 0.01 mm
(1) 0.03 mm (2) 0.02 mm
(3) 0.04 mm (4) 0.01 mm
✅ Answer & Explanation:
Correct Option: (1)
- Velocity \(v = \sqrt{T/\mu}\). Mass per unit length \(\mu = 6 \times 10^{-3} / 0.6 = 0.01 \, \text{kg/m}\).
- \(90 = \sqrt{\frac{T}{0.01}} \implies T = 81 \, \text{N}\).
- Elongation \(\Delta L = \frac{T L}{A Y} = \frac{81 \times 0.6}{(10^{-6}) \times (16 \times 10^{11})} = \frac{48.6}{16 \times 10^5} \approx 3 \times 10^{-5} \, \text{m} = 0.03 \, \text{mm}\).
Q7. JEE Main 2025 (8 April Shift 2)
A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to change of volume of \(0.8 \, \text{cm}^3\). If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was _______ litre. (Take \(1 \, \text{atm} = 10^5 \, \text{Pa}\))
✅ Answer & Explanation:
Correct Answer: 4
- Using the formula for Bulk Modulus: \(B = \frac{\Delta P}{\Delta V / V}\)
- Rearranging for initial volume \(V\): \(V = \frac{B \cdot \Delta V}{\Delta P}\)
- \(\Delta P = 5 - 1 = 4 \, \text{atm} = 4 \times 10^5 \, \text{Pa}\)
- \(\Delta V = 0.8 \, \text{cm}^3 = 0.8 \times 10^{-6} \, \text{m}^3\)
- \(B = 2 \, \text{GPa} = 2 \times 10^9 \, \text{Pa}\)
- \(V = \frac{2 \times 10^9 \times 0.8 \times 10^{-6}}{4 \times 10^5} = \frac{1600}{4 \times 10^5} = 4 \times 10^{-3} \, \text{m}^3 = 4 \, \text{litres}\).
Q8. JEE Main 2025 (29 Jan Shift 1)
The fractional compression \((\Delta V/V)\) of water at the depth of 2.5 km below the sea level is ______ %. Given, the Bulk modulus of water \(= 2 \times 10^9 \, \text{N m}^{-2}\), density of water \(= 10^3 \, \text{kg m}^{-3}\), acceleration due to gravity \(= 10 \, \text{m s}^{-2}\).
(1) 1.25 (2) 1.0
(3) 1.75 (4) 1.5
(1) 1.25 (2) 1.0
(3) 1.75 (4) 1.5
✅ Answer & Explanation:
Correct Option: (1)
- Pressure at depth \(h\): \(\Delta P = \rho gh = 10^3 \times 10 \times 2500 = 2.5 \times 10^7 \, \text{Pa}\).
- Fractional compression \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{2.5 \times 10^7}{2 \times 10^9} = 1.25 \times 10^{-2}\).
- Percentage fractional compression \(= 1.25 \times 10^{-2} \times 100 = 1.25\%\).
Q9. JEE Main 2025 (28 Jan Shift 2)
The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of \(7 \times 10^6 \, \text{Pa}\), would be ____ \(mm^3\). (Given bulk modulus of copper \(= 1.4 \times 10^{11} \, \text{N m}^{-2}\))
✅ Answer & Explanation:
Correct Answer: 50
- Initial volume \(V = (10 \, \text{cm})^3 = 10^{-3} \, \text{m}^3\).
- \(\Delta V = \frac{V \cdot \Delta P}{B} = \frac{10^{-3} \times 7 \times 10^6}{1.4 \times 10^{11}} = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-8} \, \text{m}^3\).
- Convert to \(mm^3\): \(5 \times 10^{-8} \times (10^3)^3 = 5 \times 10^{-8} \times 10^9 = 50 \, \text{mm}^3\).
Q10. JEE Main 2024 (27 Jan Shift 1)
If average depth of an ocean is 4000 m and the bulk modulus of water is \(2 \times 10^9 \, \text{N m}^{-2}\), then fractional compression \((\Delta V/V)\) of water at the bottom of ocean is \(\alpha \times 10^{-2}\). The value of \(\alpha\) is _______, (Given, \(g = 10 \, \text{m s}^{-2}, \rho = 1000 \, \text{kg m}^{-3}\))
✅ Answer & Explanation:
Correct Answer: 2
- \(\Delta P = \rho gh = 1000 \times 10 \times 4000 = 4 \times 10^7 \, \text{Pa}\).
- \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{4 \times 10^7}{2 \times 10^9} = 2 \times 10^{-2}\).
- Comparing with \(\alpha \times 10^{-2}\), we find \(\alpha = 2\).
Q11. JEE Main 2023 (13 Apr Shift 1)
Under isothermal condition, the pressure of a gas is given by \(P = aV^{-3}\), where \(a\) is a constant and \(V\) is the volume of the gas. The bulk modulus at constant temperature is equal to :
(1) \(3P\) (2) \(P\) (3) \(2P\) (4) \(P/2\)
(1) \(3P\) (2) \(P\) (3) \(2P\) (4) \(P/2\)
✅ Answer & Explanation:
Correct Option: (1)
- Bulk Modulus \(B = -V \frac{dP}{dV}\).
- Given \(P = aV^{-3}\), then \(\frac{dP}{dV} = -3aV^{-4}\).
- \(B = -V (-3aV^{-4}) = 3aV^{-3}\).
- Since \(P = aV^{-3}\), then \(B = 3P\).
Q12. JEE Main 2022 (27 Jul Shift 2)
In an experiment to determine the Young's modulus, steel wires of five different lengths but of same cross-section (\(2 \, mm^2\)) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is \(x \times 10^{11} \, N \, m^{-2}\), then the value of \(x\) is _____.
✅ Answer & Explanation:
Correct Answer: 2
- Slope \(S = \frac{\Delta L}{F} = \frac{L}{AY}\).
- From the graph, slope of the *slope vs Length* line is \(\frac{1}{AY}\).
- From graph: \(\text{slope of line} = \frac{1.25 \times 10^{-5} - 0}{5 - 0} = 0.25 \times 10^{-5}\).
- \(\frac{1}{AY} = 0.25 \times 10^{-5} \implies Y = \frac{1}{2 \times 10^{-6} \times 0.25 \times 10^{-5}} = \frac{1}{0.5 \times 10^{-11}} = 2 \times 10^{11} \, \text{N/m}^2\).
- Thus \(x = 2\).
Q13. JEE Main 2022 (26 Jun Shift 1)
The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of \(5 \times 10^{-4}\) is _____ \(kJ \, m^{-3}\).
✅ Answer & Explanation:
Correct Answer: 25
- Energy Density \(u = \frac{1}{2} \times \text{Stress} \times \text{Strain}\).
- From graph, for strain \(3 \times 10^{-10}\), stress is \(60 \, \text{Pa}\).
- Young's modulus \(Y = \frac{60}{3 \times 10^{-10}} = 20 \times 10^{10} \, \text{Pa}\).
- For strain \(\epsilon = 5 \times 10^{-4}\), Stress \(= Y\epsilon = 20 \times 10^{10} \times 5 \times 10^{-4} = 10^8 \, \text{Pa}\).
- \(u = \frac{1}{2} \times 10^8 \times 5 \times 10^{-4} = 2.5 \times 10^4 \, \text{J/m}^3 = 25 \, \text{kJ/m}^3\).
📘 Exam Preparation Tip for Mechanical Properties of Solids
- Stress and Strain: Stress is the restoring force per unit area (\(F/A\)). Strain is the fractional change in dimension. Three types: Longitudinal (Young's modulus), Volume (Bulk modulus), and Shear (Shear modulus).
- Hooke's Law: Within the elastic limit, stress is proportional to strain. \(Y = \frac{\text{Stress}}{\text{Strain}}\) for linear deformation. For volume deformation, \(B = \frac{\Delta P}{\Delta V / V}\).
- Young's Modulus: \(Y = \frac{FL}{A\Delta L}\). Useful for wires, rods, and beams. Remember that elongation \(\Delta L = \frac{FL}{AY}\).
- Bulk Modulus: \(B = \frac{\Delta P}{\Delta V / V}\). The negative sign indicates volume decreases with increasing pressure. Compressibility is the reciprocal of bulk modulus (\(K = 1/B\)).
- Shear Modulus: \(G = \frac{F/A}{\theta}\). Applies when forces are tangential to the surface. For small angles, \(\theta\) is the angular displacement.
- Energy Density: The elastic potential energy per unit volume is \(u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \epsilon^2\).
- Graphs: For \(Y\) determination, the slope of a stress-strain graph gives \(Y\). For \(\Delta L\) vs \(F\) graphs, the slope is \(L/(AY)\).