Question

The resistance of 2$\Omega$ is connected across one gap of a meter-bridge (length of the wire is 100 cm) and an unknown resistance greater than 2$\Omega$ is connected across the other gap. When the resistance are interchanged, the balance point shifts by 20cm. Neglecting any corrections, the unknown resistance is: (in $\Omega$)

Answer 1

Let the unknown resistance be $R$ and the balance point be at $l$ cm initially.

Using $\frac{R_1}{R_2}=\frac{l^{\prime }}{100-l^{\prime }}$

Case 1: $R_1=2$ and $R_2=R$ and $l^{\prime }=l$

$\therefore$ $\frac{2}{R}=\frac{l}{100-l}$ ..........(1)

Case 2: $R_1=R$ and $R_2=2$ and $l^{\prime }=l+20$

$\therefore$ $\frac{R}{2}=\frac{l+20}{80-l}$ $⟹$ $\frac{2}{R}=\frac{80-l}{l+20}$ ........(2)

Equating (1) and (2) we get $\frac{l}{100-l}=\frac{80-l}{l+20}$ $⟹$ $l=40$ cm

From (1), $\frac{2}{R}=\frac{40}{100-40}$ $⟹R=3\Omega$