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A wire of resistance $$R$$ and radius $$r$$ is stretched till its radius became $$r/2$$. If new resistance of the stretched wire is $$xR$$, then value of $$x$$ is _______
Correct Answer: 16
Volume conserved: $$\pi r^2 l = \pi (r/2)^2 l' \Rightarrow l' = 4l$$.
$$R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\pi(r/2)^2} = \rho \frac{4l}{\pi r^2/4} = 16 \cdot \rho\frac{l}{\pi r^2} = 16R$$.
So $$x = 16$$.
The answer is 16.
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