JEE Physics 2025 Question with Solution

Given: Young's modulus $Y = 2.0 \times 10^{11} \, \text{N/m}^2$, Poisson's ratio $\sigma = 0.2$, transverse strain $\epsilon_t = 10^{-3}$.

Find: Elastic potential energy density of the wire.

From Poisson's ratio,

$$\sigma = \frac{|\epsilon_t|}{|\epsilon_l|}$$

So the longitudinal strain is

$$\epsilon_l = \frac{\epsilon_t}{\sigma} = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$$

Elastic potential energy density is

$$U_d = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} Y \epsilon_l^2$$

Substituting the values,

$$U_d = \frac{1}{2} \times 2.0 \times 10^{11} \times (5 \times 10^{-3})^2$$ $$U_d = 1.0 \times 10^{11} \times 25 \times 10^{-6}$$ $$U_d = 25 \times 10^5 \, \text{J/m}^3 = 2.5 \times 10^6 \, \text{J/m}^3$$

the solution contains a discrepancy because one place marks the correct answer as $25$, while the worked calculation gives $2.5 \times 10^6$. Since the listed options do not contain $2.5$, the solution's appears inconsistent. Following the provided answer key on the page, the recorded correct option is B.

Given: $Y = 2 \times 10^{11} \, \text{N/m}^2$, Poisson's ratio $= 0.2$, transverse strain $= 10^{-3}$.

Find: The value to be filled in the blank for elastic potential energy density.

The page states that transverse strain and longitudinal strain are related through Poisson's ratio. Hence,

$$\epsilon_l = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$$

Now use the elastic energy density formula,

$$U_d = \frac{1}{2} Y \epsilon_l^2$$

Substitute,

$$U_d = \frac{1}{2} \times 2 \times 10^{11} \times (5 \times 10^{-3})^2$$ $$U_d = 10^{11} \times 25 \times 10^{-6}$$ $$U_d = 25 \times 10^5 = 2.5 \times 10^6$$

Therefore, the numerical value before $10^6$ should be $2.5$ by the shown working. However, the source solution panel also explicitly lists Correct Answer: 25 and option B, so the source is self-contradictory. The answer is stored as B because that is the page's declared correct option.