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 doneA. `W_(2)=0.5 W_(1)`B. `W_(2)=W_(1)`C. `W_(2)=2W_(1)`D. `W_(2)=2W_(1)^(2)`

The force constant of a wire is k and that of another wire is . 2k Whe

1.	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 doneA. `W_(2)=0.5 W_(1)`B. `W_(2)=W_(1)`C. `W_(2)=2W_(1)`D. `W_(2)=2W_(1)^(2)`
Answer» Correct Answer - 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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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 doneA. `W_(2)=0.5 W_(1)`B. `W_(2)=W_(1)`C. `W_(2)=2W_(1)`D. `W_(2)=2W_(1)^(2)`

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