Force constant of two wires `A` and `B` of the same material are `K` and `2K` respectively. If the two wires are stretched equally, then the ratio of work done in stretching `((W_(A))/(W_(B)))` is

Options

A. `w_(1) = w_(2)`
B. `w_(1) =2 w_(2)`
C. `w_(1)= 1//2 w_(2)`
D. `w_(1) = 1//4w_(2)`

Answer

B

Explanation

For a spring, `F= - Kx therefore x =|(F)/(K)|`
For the first spring, `w_(1)=(1)/(K_(1)x_(1)^(2)`
but `K_(1)= K and x_(1)=(F)/(K)`
`therefore w_(1)=(1)/(2)K(F^(2))/(K^(2))=(1)/(2)(F^(2))/(K)`
"sin"ce the same force is used
Similarly, `w_(2)=(1)/(2)(2K)((F^(2))/(4K^(2)))=(1)/(2)((F^(2))/(2K))`
`therefore (w_(1))/(w_(2))=2 or w_(1)=2w_(2)`