A) 1 J
B) 0.1 J
C) 0.01 J
D) 0.001 J
Correct Answer: B
Solution :
As elastic potential energy \[U=\frac{1}{2}\times stress\times strain\times volume\] \[U=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta l}{l}\times Al=\frac{1}{2}F\Delta l\] \[=\frac{1}{2}\times 200\times {{10}^{-3}}=0.1\,J\]You need to login to perform this action.
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