MHT CET · Physics · Mechanical Properties of Solids

A wire of length ‘L’ and area of cross section ‘A’ is made of material of Young’s modulus ‘Y’. It is stretched by an amount ‘x’. The work done in stretching the wire is

A $\frac{Y x^{2} A}{2 L}$
B $\frac{2 Y x^{2} A}{L}$
C $\frac{Y x A}{2 L}$
D $\frac{Y x^{2} A}{2}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{Y x^{2} A}{2 L}$

Step-by-step Solution

Detailed explanation

If a force F is applied along the length L of wire for stretching by an amount x, then Young’s modulus is given by

Y = \frac{s t r e s s}{s t r a i n} = \frac{F L}{A x}

where, A = area of cross-sectional

\Rightarrow F = \frac{Y A}{L} x

The work done is stretching the wire is given by

W = \int_{0}^{x} F . d x = {\int_{0}^{x} \frac{Y A}{L} x = \frac{Y A}{L} [\frac{x^{2}}{2}]}_{0} = \frac{Y A x^{2}}{2 L}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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