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Q.

A wire is stretched by 0.01 m by a certain force F. Another wire of the same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be

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a

0.005 m

b

0.01 m

c

0.02 m

d

0.002 m

answer is A.

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Detailed Solution

Given that elongation l = 0.01 m

Force applied F, length = L

$s t r e s s = \frac{F}{{πr}^{2}}$

$s t r a i n = \frac{l}{L}$

$l = \frac{F L}{{πr}^{2} Y}$

$∴ l \propto \frac{L}{r^{2}} (Y a n d F a r e c o n s \tan t s)$ $h e r e \frac{l_{2}}{l_{1}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} = (2) \times {(\frac{1}{2})}^{2} = \frac{1}{2}$ $\Rightarrow l_{2} = \frac{l_{1}}{2} = \frac{0.01 m}{2} = 0.005 m$

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