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The force constant of a wire is k and th...

The force constant of a wire is k and that of another wire is . 2k When both the wires are stretched through same distance, then the work done

A

`W_(2)=0.5 W_(1)`

B

`W_(2)=W_(1)`

C

`W_(2)=2W_(1)`

D

`W_(2)=2W_(1)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

`(W_(2))/(W_(1))=((1)/(2)k_(2)x^(2))/((1)/(2)k_(1)x^(2))`
`=(k_(2))/(k_(1))=(2k)/(k)`
`(W_(2))/(W_(1))=2 " " therefore W_(2)=2W_(1)`.
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