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InterviewSolution
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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)))` isA. `(1)/(3)`B. `(1)/(3)`C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - B (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)` |
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