JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Current Electricity

Using the metre-bridge balance condition with a left-gap resistance of 2 Ω and a balance length of 40 cm: $\frac{2}{40}=\frac{X}{60}$, giving $X=3\Omega$. Shunting X with a 2 Ω resistor gives $X^{\prime }=\frac{2\times 3}{2+3}=1.2\Omega$. If the new balance length is $\ell$, then $\frac{2}{\ell }=\frac{1.2}{100-\ell }$, so $200-2\ell =1.2\ell$ and $\ell =\frac{200}{3.2}=62.5$ cm. The balance length therefore changes by $62.5-40=22.5$ cm.