A steel rod has a radius of 20 mm and a length of 2.0 m. A force… — Solution | JEE Physics

NVAEasyJEE 2023Young's Modulus, Bulk & Rigidity Modulus

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

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: radius $r = 20 \, \text{mm}$ , length $L = 2.0 \, \text{m}$ , force $F = 62.8 \, \text{kN}$ , Young’s modulus $Y = 2.0 \times 10^{11} \, \text{N/m}^2$ .

Find: the longitudinal strain in the form $P \times 10^{-5}$ .

From the solution:

\Delta L = \frac{FL}{AY} = \frac{62.8 \times 1000 \times 2}{3.14 \times 20 \times 20 \times 10^{-6} \times 2 \times 10^{11}} = 50 \times 10^{-5} \, \text{m}

Hence, the value of $P = 50$ .

Therefore, the extracted the solution indicates the answer as $50$ , although the provided correct answer field states $25$ .

Common mistakes

Using radius and diameter interchangeably is incorrect because the cross-sectional area depends on $A = \pi r^2$ . Use the given radius directly unless the question explicitly gives diameter.
Failing to convert $20 \, \text{mm}$ into SI units leads to a wrong area and wrong deformation. Convert all lengths to metres before substitution.
Confusing elongation $\Delta L$ with strain is incorrect because strain is dimensionless and equals $\frac{\Delta L}{L}$ . After finding extension, divide by the original length to get strain.

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Answer

Step-by-step solution

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