JEE Main · 2008 · 27 Apr 2008 · Offline · Current Electricity

In a balanced meter bridge the galvanometer carries no current, so the bridge is a balanced Wheatstone network in which the ratio of the resistances in the two gaps equals the ratio of the balancing lengths of the wire. With the known resistance of $55\Omega$ and the null point dividing the wire in the ratio $20:80$, the balance relation reads $\frac{55}{R}=\frac{20}{80}$. Cross-multiplying, $R=\frac{55\times 80}{20}=220\Omega$, so the unknown resistor measures $220\Omega$.