Physics - Current Electricity Question with Solution | TestHub

Let the null point be at a distance $l$ from the left end of the metre bridge wire.

The balancing condition of the metre bridge is given by $\frac{R_1}{R_2}=\frac{l}{100-l}$

Substituting the given values, we get $\frac{40}{60}=\frac{l}{100-l}$

$\frac{2}{3}=\frac{l}{100-l}$

$200-2l=3l$

$5l=200$

$l=40\text{ cm}$

The centre of the metre bridge wire is at $50\text{ cm}$.

The distance of the null point from the centre of the wire towards the left is $50\text{ cm}-40\text{ cm}=10\text{ cm}$.

Answer: 10 cm