Mechanical Properties of Solids Previous Year Questions — JEE Main Physics
21 past-year Mechanical Properties of Solids questions from JEE Main (Physics).
Q24 (2025)
The length of a light string is $1.4\mathrm{m}$ when the tension on it is $5\mathrm{N}$ . If the tension increases to $7\mathrm{N}$ , the length of the string is $1.56\mathrm{m}$ . The original length of the string is ______ $\mathrm{m}$ .
### Related Formula
By Hooke's Law, the tension ($T$) in a stretched elastic string is proportional to its change in length:
$$T = k \cdot \Delta L = k(L - L_0)$$
where:
$k$ = spring constant of the string
$L$ = current stretched length
$L_0$ = natural original length
### Core Logic
We can set up two algebraic equations using the given conditions:
1. **Case 1:** Tension $T_1 = 5 \ \mathrm{N}$ produces stretched length $L_1 = 1.4 \ \mathrm{m}$:
$$5 = k(1.4 - L_0) \quad \dots (1)$$
2. **Case 2:** Tension $T_2 = 7 \ \mathrm{N}$ produces stretched length $L_2 = 1.56 \ \mathrm{m}$:
$$7 = k(1.56 - L_0) \quad \dots (2)$$
### Step 1: Solve for natural length
Divide equation (1) by equation (2) to eliminate the spring constant $k$:
$$\frac{5}{7} = \frac{1.4 - L_0}{1.56 - L_0}$$
Cross-multiply and solve for $L_0$:
$$5(1.56 - L_0) = 7(1.4 - L_0)$$
$$7.8 - 5 L_0 = 9.8 - 7 L_0$$
Rearrange to isolate the variable:
$$2 L_0 = 9.8 - 7.8 = 2$$
$$L_0 = 1 \ \mathrm{m}$$
Thus, the original length of the string is $1 \mathrm{~m}$.
### Pattern Recognition
Sees: Elastic stretching of a string/wire under variable load.
Trap: Attempting to resolve details like Young's modulus or cross-sectional area. Taking ratios allows the spring constant $k$ to cancel cleanly, saving computational effort.
Shortcut: Use the ratio equation $\frac{T_1}{T_2} = \frac{L_1 - L_0}{L_2 - L_0}$ directly. Solving the linear equation yields $L_0 = 1\text{ m}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q22 (2025)
A steel wire of length $2\mathrm{~m}$ and Young's modulus $2.0\times 10^{11}\mathrm{~N/m}^2$ is stretched by a force. If Poisson's ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is $\text{[value]} \times 10^{5}$ (in SI units)
### Related Formula
$$\sigma = \frac{\epsilon_{\text{transverse}}}{\epsilon_{\text{longitudinal}}}$$
$$u = \frac{1}{2} Y \epsilon_{\text{longitudinal}}^2$$
### Core Logic
Given parameters:
- Young's modulus, $Y = 2.0 \times 10^{11}\mathrm{~N/m}^2$
- Poisson's ratio, $\sigma = 0.2$
- Transverse strain, $\epsilon_{\text{trans}} = 10^{-3}$
First, calculate the longitudinal strain $\epsilon$:
$$\sigma = \frac{\epsilon_{\text{trans}}}{\epsilon} \implies 0.2 = \frac{10^{-3}}{\epsilon}$$
$$\epsilon = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$$
Now, compute the elastic potential energy density $u$:
$$u = \frac{1}{2} Y \epsilon^2 = \frac{1}{2} \times (2.0 \times 10^{11}) \times (5 \times 10^{-3})^2$$
$$u = 10^{11} \times 25 \times 10^{-6} = 25 \times 10^5\mathrm{~J/m}^3$$
Expressing as $x \times 10^5$, we have $x = 25$.
### Step 1: Final Conclusion
The value of the coefficient is 25.
### Pattern Recognition
Poisson's ratio is defined as the ratio of lateral/transverse strain to longitudinal strain. Use this to determine the longitudinal stretch, then apply the basic potential energy density formula $\frac{1}{2} Y \epsilon^2$ directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q19 (2025)
Two wires A and B are made of same material having ratio of lengths $\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}} = \frac{1}{3}$ and their diameters ratio $\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}} = 2$ . If both the wires are stretched using same force, what would be the ratio of their respective elongations?
- $1:6$
- $1:12$
- $3:4$
- $1:3$
### Related Formula
Young's modulus $Y$ is defined as:
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} \implies \Delta L = \frac{F L}{A Y}$$
Area of cross-section of wire with diameter $d$ is:
$$A = \frac{\pi d^2}{4} \implies \Delta L = \frac{4 F L}{\pi d^2 Y}$$$
### Core Logic
Since both wires are made of the same material ($Y_A = Y_B$) and stretched with the same force ($F_A = F_B$):
$$\Delta L \propto \frac{L}{d^2}$$
Set up the ratio for wires A and B:
$$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{L_A}{L_B} \right) \times \left( \frac{d_B}{d_A} \right)^2$$
### Step 1: Substitute Given Ratios
Substitute the ratios $\frac{L_A}{L_B} = \frac{1}{3}$ and $\frac{d_A}{d_B} = 2 \implies \frac{d_B}{d_A} = \frac{1}{2}$:
$$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{1}{3} \right) \times \left( \frac{1}{2} \right)^2 = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$$
Therefore, the ratio of their elongations is $1:12$.
### Pattern Recognition
Sees: Same material, same stretching force.
Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $(1/3) / (2^2) = 1/12$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q3 (2025)
A $3\mathrm{~m}$ long wire of radius $3\mathrm{~mm}$ shows an extension of $0.1\mathrm{~mm}$ when loaded vertically by a mass of $50\mathrm{~kg}$ in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P \times 10^{11}\mathrm{~N}\cdot\mathrm{m}^{-2}$, where the value of $P$ is:
(Take $g = 3\pi\mathrm{~m/s}^{2}$)
- $5$
- $10$
- $25$
- $2.5$
### Related Formula
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{mg L}{\pi r^2 \Delta L}$$
where,
$Y$ = Young's modulus
$F = mg$ = stretching force (load)
$A = \pi r^2$ = cross-sectional area of the wire
$L$ = original length of the wire
$\Delta L$ = extension produced
### Core Logic
Given parameters:
- Original length of wire, $L = 3\mathrm{~m}$
- Radius of wire, $r = 3\mathrm{~mm} = 3 \times 10^{-3}\mathrm{~m}$
- Mass loaded, $m = 50\mathrm{~kg}$
- Acceleration due to gravity, $g = 3\pi\mathrm{~m/s}^{2}$
- Extension, $\Delta L = 0.1\mathrm{~mm} = 10^{-4}\mathrm{~m}$
Substitute the values into the Young's modulus formula:
$$Y = \frac{50 \times (3\pi) \times 3}{\pi \times (3 \times 10^{-3})^2 \times 10^{-4}}$$
### Step 1: Detailed Computation
Simplify the equation:
$$Y = \frac{450\pi}{\pi \times 9 \times 10^{-6} \times 10^{-4}}$$
Cancelling $\pi$ from numerator and denominator:
$$Y = \frac{450}{9 \times 10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11}\mathrm{~N/m}^{2}$$
Comparing this with $P \times 10^{11}\mathrm{~N/m}^{2}$:
$P = 5$
### Pattern Recognition
Sees: Standard Young's modulus measurement.
Trap: Don't forget that the radius is in millimeters and the extension is in millimeters. Convert all units to SI systematically first.
Calculation Tip: The variable $g = 3\pi$ is chosen deliberately to cancel the $\pi$ in the area formula. Keep your eye open for these mathematical shortcuts! ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Elasticity
Q5 (2025)
A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $10^{5}$ 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, $G=10^{10}\text{ N/m}^{2}$)
- $1/160\pi$
- $1/4\pi$
- $1/40\pi$
- $1/2\pi$
### Related Formula
$$G = \frac{\sigma_{\text{shear}}}{\theta} = \frac{F}{A \cdot \theta}$$
where $A = \pi r^2$ is the cross-sectional area.
### Core Logic
Rearranging the formula for $\theta$:
$$\theta = \frac{F}{A \cdot G} = \frac{F}{\pi r^2 \cdot G}$$
Substituting the values:
$$F = 10^5\text{ N}$$
$$r = 4\text{ cm} = 4 \times 10^{-2}\text{ m}$$
$$G = 10^{10}\text{ N/m}^2$$
### Step 1: Calculate Angular Displacement
$$\theta = \frac{10^5}{\pi \times (4 \times 10^{-2})^2 \times 10^{10}}$$
$$\theta = \frac{10^5}{\pi \times 16 \times 10^{-4} \times 10^{10}} = \frac{10^5}{16\pi \times 10^6}$$
$$\theta = \frac{1}{160\pi}\text{ radians}$$
### Pattern Recognition
Shear strain definition is straightforwardly matching the geometry angle of shift. Avoid radius-to-diameter or unit mismatches.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
