Physics - Current Electricity Question with Solution | TestHub

Let $l$ be the balancing length from the left end of the metre bridge.

Using the principle of the metre bridge, $\frac{R_1}{R_2}=\frac{l}{100-l}$

Substituting the given values, $\frac{20}{60}=\frac{l}{100-l}$

$\frac{1}{3}=\frac{l}{100-l}$

$100-l=3l$

$4l=100\Rightarrow l=25$ cm

The centre of the metre bridge wire is at $50$ cm.

Distance of the null point from the centre of the wire $=50-25=25$ cm.

Answer: $25$ cm