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A wire of resistance $$160 \ \Omega$$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
$$R = \rho \frac{l}{A}$$
$$A = \frac{V}{l} \implies R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V} \implies R \propto l^2$$
$$\frac{R'}{R} = \left(\frac{l'}{l}\right)^2$$
$$\frac{R'}{160} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \implies R' = \frac{160}{16} = 10\,\Omega$$
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