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Question

Physics Question on mechanical properties of solids

If the work done in stretching a wire by 1mm1 \,mm is 2J2 J, the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by 1mm1 \,mm is

A

16J16\, J

B

8J8 \,J

C

4J4 \,J

D

14\frac{1}{4} JJ

Answer

16J16\, J

Explanation

Solution

Stretching force, F=F= Yπr2ΔLL\frac{Y \pi r^{2}\Delta L}{L} where the symbols have their usual meanings. Both the wires are of same material, so YY will be equal, extension in both the wires is same, so ΔL\Delta L will be equal. \therefore\quad FF \propto r2L\frac{r^{2}}{L} \therefore\quad FF\frac{F'}{F} =(2r)2(L/2)=\frac{\left(2r\right)^{2}}{\left(L /2\right)} ×Lr2=8\times\frac{L}{r^{2}}=8 or FF' =8F=8 F (i)\quad\ldots\left(i\right) Work done in stretching a wire, W=12×W=\frac{1}{2}\times F×ΔLF\times\Delta L For same extension WW \propto FF \therefore\quad WW\frac{W'}{W} =FF=8=\frac{F'}{F}=8 [Using(i)]\quad\left[Using \left(i\right)\right] WW' =8W=8×2J=8W=8\times2 \,J =16J=16 \,J