A) 4 J
B) 8 J
C) 12 J
D) 16 J
Correct Answer: D
Solution :
Total work done in increasing the length by \[l\] \[W=\frac{1}{2}\frac{YA}{L}{{t}^{2}}\] or \[W\propto {{l}^{2}}\] \[\therefore \] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{l_{2}^{2}}{l_{2}^{2}}\] Let the initial length be 10. \[\frac{4}{{{W}_{2}}}=\frac{10\times 10}{20\times 20}\] \[{{W}_{2}}=16J\]
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