QJEEM23Phc555: JEE Main 2023 Physics - Class 12 JEEM23Phc555 Solution: A wire of resistance is melted and drawn in a wire of…

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:

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

Substituting the given value of R, we get:

$R' = \frac{160}{16} = 10 ~\Omega$

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