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

Options

A. `W_(P)=W_(Q),W_(P)gtW_(Q)`
B. `W_(P)=W_(Q),W_(P)=W_(Q)`
C. `W_(P)gtW_(Q),W_(Q)gtW_(P)`
D. `W_(P)ltW_(Q),W_(Q)tW_(P)`

Answer

c

Explanation

(c )Given, `K_(P)gtK_(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))gt1impliesW_(P)gtW_(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)^(2))=1/(2)(F^(2))/(K_(Q))`
`therefore(W_(P))/(W_(Q))=(K_(Q))/(K_(P))lt1`
`implies W_(P)ltW_(Q)`