Question

When the load on a wire is slowly increased from 3 to 5 kg wt, the elongation increases from 0.61 to 1.02 mm. The work done during the extension of wire is

• A. 0.16 J
• B. 0.016 J
• C. 1.6 J
• D. 16 J

Answer 1

The correct option is B

0.016 J

Work done = $\frac{1}{2}$ $\times$ stretching force $\times$ extension.

Therefore, net work done in increasing the length from 0.61 mm to 1.02 mm is

$W_2-W_1=\frac{1}{2}\times 5\times 9.8\times 1.02\times {10}^{-3}-\frac{1}{2}\times 3\times 9.8\times 0.61\times {10}^{-3}=0.016J$