A steel rod has a radius of 20 ~mm and a length of 2.0 ~m . A force of 62.8 ~kN stretches it along its length. Young's modulus of steel is 2.0 \times 10^11 ~N / m^2 . The longitudinal strain produced in the wire is \underline\hspace2cm\times 10^-5

Properties of Matter PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q64 with Solution

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Correct answer: 25

Young's modulus is

Y = \frac{\text{stress}}{\text{strain}}

So,

\text{strain} = \frac{\text{stress}}{Y} = \frac{F/A}{Y} = \frac{F}{AY}

r = 20\,\text{mm} = 20 \times 10^{-3} = 2.0 \times 10^{-2}\,\text{m}

F = 62.8\,\text{kN} = 62.8 \times 10^3\,\text{N}

Y = 2.0 \times 10^{11}\,\text{N/m}^2

A = \pi r^2 = \pi (2.0 \times 10^{-2})^2

A = \pi \times 4.0 \times 10^{-4}

A \approx 1.256 \times 10^{-3}\,\text{m}^2

\text{stress} = \frac{F}{A} = \frac{62.8 \times 10^3}{1.256 \times 10^{-3}}

\text{stress} = 5.0 \times 10^7\,\text{N/m}^2

\text{strain} = \frac{5.0 \times 10^7}{2.0 \times 10^{11}}

\text{strain} = 2.5 \times 10^{-4}

Now write it in the form:

\text{strain} = \underline{\hspace{1cm}} \times 10^{-5}

Since

2.5 \times 10^{-4} = 25 \times 10^{-5}

The required integer is

\boxed{25}

Stored correct answer = 25, which matches the derived answer.