A steel wire of length 2 m and Young's modulus 2.0 \times 10^11 ~N ~m^-2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10^-3 respectively, then the elastic potential energy density of the wire is \underline\hspace2cm \times 10^5 (in SI units).

Properties of Matter PYQ — JEE Main 2025 · 2 Apr · Shift 1 · Q71 with Solution

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Correct answer: 25

We need the elastic potential energy density.

Poisson's ratio is

\nu = -\frac{\text{transverse strain}}{\text{longitudinal strain}}

Taking magnitudes,

\nu = \frac{\varepsilon_t}{\varepsilon_l}

So,

\varepsilon_l = \frac{\varepsilon_t}{\nu} = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}

Y = \frac{\text{stress}}{\text{strain}} \quad \Rightarrow \quad \text{stress} = Y\,\varepsilon_l

Therefore,

\sigma = 2.0 \times 10^{11} \times 5 \times 10^{-3}

\sigma = 1.0 \times 10^9\,\text{N m}^{-2}

Energy density for a stretched wire is

u_e = \frac{1}{2}(\text{stress})(\text{strain})

So,

\nu_e = \frac{1}{2} \times 1.0 \times 10^9 \times 5 \times 10^{-3}

\nu_e = \frac{1}{2} \times 5 \times 10^6

\nu_e = 2.5 \times 10^6\,\text{J m}^{-3}

Given format is

\underline{\hspace{2cm}} \times 10^5

Now,

2.5 \times 10^6 = 25 \times 10^5

So the required integer is

\boxed{25}