There are two massless springs A and B spring constant `K_(A)` and `K_(B)` respectively and `K_(A)ltK_(B)` if `W_(A)` and `W_(B)` be denoted as work done on A and work done on B respectvely. Then

Options

A. if they are are compressed to same distance `W_(A)gtW_(B)`
B. if they are compressed by same force (upto equilibrium state) `W_(A)ltW_(B)`
C. if they are compressed by ame distance `W_(A)=W_(B)`
D. if they are compressed by same force (upto equilibrium state) `W_(A)gtW_(B)`

Answer

B

Explanation

If the springs are compressed to same amount:
`W_(A)=(1)/(2)K_(A)x^(2),W_(B)=(1)/(2)K_(B)x^(2)`
`becauseK_(A)gtK_(B)impliesW_(A)gtW_(B)`
If the spring are compressed by same force,
`F=K_(A)x_(A)=K_(B)x_(B),x_(A)=(F)/(K_(A)),x_(B)=(F)/(K_(S))`
`(W_(A))/(W_(B))=((1)/(2)K_(A).(F^(2))/(K_(A)^(2)))/((1)/(2)K_(S)(F^(2))/(K_(B)^(2)))=(K_(B))/(K_(A))`
Hence `W_(A)ltW_(B)`