Q 12-03-141JEE MainJEE Main 2023 (12 Apr, Shift 1)Easy
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
Answer: (B) $10\ \Omega$
Volume is constant, so $R\propto l^2$: $R'=\dfrac{160}{16}=10\ \Omega$.
Solution by Sreeraj P, M.Sc Physics