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Question

Physics Question on elastic moduli

A force FF is applied on the wire of radius rr and length LL and change in the length of wire is ll. If the same force FF is applied on the wire of the same material and radius 2r2\, r and length 2L2\, L, then the change in length of the other wire is

A

ll

B

2l2l

C

l2\frac{l}{2}

D

4l4l

Answer

l2\frac{l}{2}

Explanation

Solution

Elongation, l=FLAyl=\frac{F L}{A y} or lLr2l \propto \frac{L}{r^{2}} l2l1=L2L1×(r1r2)2\therefore \frac{l_{2}}{l_{1}} =\frac{L_{2}}{L_{1}} \times\left(\frac{r_{1}}{r_{2}}\right)^{2} =2×(12)2=12=2 \times\left(\frac{1}{2}\right)^{2} =\frac{1}{2} or l2=l12l_{2} =\frac{l_{1}}{2} The change in the length of other wire is l2\frac{l}{2}.