JEE Main 2023 — Current Electricity Question with Solution

Let the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:

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

where $$\rho$$ is the resistivity of the material of the wire.

When the wire is melted and drawn into a wire of one-fourth of its length, its new length is L/4 and its new cross-sectional area is 4A (since the same amount of material is now spread over a longer length). Therefore, its new resistance R' is given by:

$$R^{\prime }=\frac{\rho (L/4)}{4A}=\frac{R}{16}$$

Substituting the given value of R, we get:

$$R^{\prime }=\frac{160}{16}=10\Omega$$

Therefore, the new resistance of the wire is $$10\Omega$$.