JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Current Electricity

At balance the meter bridge compares the two gap resistances through the lengths: $\frac{R}{S}=\frac{\ell }{100-\ell }=\frac{25}{75}=\frac{1}{3}$, so $S=3R$. Replacing the wire by one of half the length and half the diameter changes its resistance through $R\propto \frac{\text{length}}{d^2}$: halving the length divides by two while halving the diameter multiplies by four, giving a net $R^{\prime }=2R$. The new balance satisfies $\frac{R^{\prime }}{S}=\frac{{\ell }^{\prime }}{100-{\ell }^{\prime }}$, i.e. $\frac{2R}{3R}=\frac{{\ell }^{\prime }}{100-{\ell }^{\prime }}$, so $\frac{2}{3}=\frac{{\ell }^{\prime }}{100-{\ell }^{\prime }}$. Then $3{\ell }^{\prime }=2(100-{\ell }^{\prime })$, and ${\ell }^{\prime }=40\text{cm}$.