Question 12

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

Name: 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
Uploaded: 2026-05-13T16:19:28.493522+05:30
Duration: 1 min 43 s
Description: 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

Solution

$$R = \rho \frac{l}{A}$$

$$A = \frac{V}{l} \implies R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V} \implies R \propto l^2$$

$$\frac{R'}{R} = \left(\frac{l'}{l}\right)^2$$

$$\frac{R'}{160} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \implies R' = \frac{160}{16} = 10\,\Omega$$

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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

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