Q. If work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material, but wire double the radius and half length 1 mm joule is
A) $1/4$
B) 4
C) 8
D) 16
Explanation
(d) Work done = $\frac{1}{2} \times Y \times (strain)^{2} \times volume\ $
2 = $\frac{1}{2} \times Y \times \left( \frac{\mathrm{\Delta}L}{L} \right)^{2} = \frac{YA\mathrm{\Delta}L^{2}}{2L}\ $
New work done, $W^{'} = \ \frac{Y(4A)\mathrm{\Delta}L^{2}}{2(L/2)}$
= $8\left\lbrack \frac{YA\mathrm{\Delta}L^{2}}{2L} \right\rbrack 8 \times 2 = 16\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.