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Two similar springs P and Q have spring constant `K_(P)"and "K_(Q)` , such that `K_(P)gtK_(Q)`. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs `W_(P)" and "W_(Q)` are related as , in case (a) and case (b), respectively

A

`W_(P) = W_(Q) , W_(P) gt W_(Q)`

B

`W_(P) = W_(Q) , W_(P) = W_(Q)`

C

`W_(P) gt W_(Q) , W_(Q) gt W_(P)`

D

`W_(P) lt W_(Q) , W_(Q) lt W_(P)`

Text Solution

Verified by Experts

The correct Answer is:
C
(c) Given, `K_(P) gt K_(Q)`
In case (a) , the elongation is same
i.e.., `x_(1)=x_(2)=x`
So, `W_(P)=(1)/(2)K_(P)x^(2)` and `W_(Q)=(1)/(2)K_(Q)x^(2)`
`therefore (W_(P))/(W_(Q))=(K_(P))/(K_(Q)) gt 1" "(K_(P) gt K_(Q))`
`implies W_(P) gt W_(Q)`
In case (b) , the spring force is same
i.e., `F_(1)=F_(2)=F`
So, `x_(1)=(F)/(K_(P)), x_(2)=(F)/(K_(Q))`
`therefore W_(P)=(1)/(2)K_(P)x_(1)^(2)=(1)/(2)K_(P)(F^(2))/(K_(P)^(2))=(1)/(2)(F^(2))/(K_(P))`
and `W_(Q)=(1)/(2)K_(Q)x_(2)^(2) =(1)/(2)K_(Q).(F^(2))/(K_(Q))=(1)/(2)(F^(2))/(K_(Q))`
`therefore (W_(P))/(W_(Q))=(K_(Q))/(K_(P)) lt 1 implies W_(P) lt W_(Q)`
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