Mechanical Properties of Solids

JEE Main level, single correct, medium

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Question 1 of 118

A wire of length LL and radius rr is stretched by length ll under tension FF. A wire of same material of length 2L2L and radius 2r2r under tension 2F2F will be stretched by:

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Answer: (B) ll

ΔL=FLπr2Y\Delta L = \frac{F L}{\pi r^2 Y}. For new wire: ΔL′=(2F)(2L)π(2r)2Y=4FL4πr2Y=l\Delta L' = \frac{(2F)(2L)}{\pi (2r)^2 Y} = \frac{4FL}{4\pi r^2 Y} = l.

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