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. `(1)/(3)`
B. `(1)/(3)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

B

Explanation

(b) We know that the work done ina stretched wire
`W=(1)/(2)kx^(2)`
Given,`" " k_(A)=k and k_(B)=2k`
So that`" " W_(A)=(1)/(2)kx^(2) and W_(B)=(1)/(2)(2k)x^(2)=kx^(2)`
Hence, the ratio of work done in stretch wire
`(W_(A))/(W_(B))=((1//2)kx^(2))/(kx^(2))rArr(W_(A))/(W_(B))=(1)/(2)`