Physics - Current Electricity Question with Solution | TestHub

In a meter bridge the ratio of two resistances is

$\frac{R}{R^{\prime }}=\frac{l}{l^{\prime }}$

Where$l$and$l\prime$are balancing lengths.

Resistance$R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$

In material remains same$\rho ={\rho }^{\prime }$

Given,$l^{\prime }=2l$

$r^{\prime }=\frac{r}{2}$

$\therefore R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}=\frac{\rho 2l}{\pi {\left(\frac{r}{2}\right)}^2}=\frac{8\rho l}{\pi r^2}$

$R^{\prime }=8R$

Therefore, the new balancing point is expected to be 8$l$.