A wire of resistance $x$ ohm is drawn out, so that its length is increased to twice its original length, and its new resistance becomes $20 \,\Omega$, then $x$ will be ........ $\Omega$
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(a)
$x=\frac{\rho l}{A}$
$20=\frac{\rho(2 l)}{(A / 2)}$
$\frac{x}{20}=\frac{1}{4}$
$x=5$
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