The volume of a wire remains constant, when subjected to tensile stress is applied on it. If the percentage change in lateral strain is 2 %, then percentage change in longitudinal strain is
A
2 %
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B
4 %
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C
6 %
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D
8 %
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Solution
The correct option is B4 % Let us suppose length of wire is L and radius is r then it's volume is V=πr2L On differentiating both side ΔV=π(2rΔr)L+πr2ΔL As, V= constant, then ΔV=0 ⇒2πrLΔr+πr2ΔL=0 ⇒2πrLΔr=−πr2ΔL ⇒Δrr=−12×ΔLL ⇒ΔLL=−2Δrr ⇒ΔLL×100=−2Δrr×100
⇒ΔLL×100=−2×−2 [Δrr×100=−2%, given, Negative sign in percentage lateral strain shows radius will decrease due to tensile stress.] ⇒ΔLL×100=+4 % Percentage change in longitudnal strain is +4 %