Question

In a meter bridge experiment, $S$ is a standard resistance and $R$ is a resistance wire. It is found that balancing length is $l=25\text{cm}$. If $R$ is replaced by a wire of half the length and half diameter, those of $R$ (of the same material), then the balancing distance $l^{\prime }$ (in cm) will now be

Answer 1

For the given meter bridge,

$\frac{R}{S}=\frac{l_1}{100-l_1}$

Where, $l_1\to$ balancing length
$\Rightarrow \frac{R}{S}=\frac{25}{75}=\frac{1}{3}...\left(i\right)$

With new resistance,

$R^{\prime }=\frac{\rho \frac{l}{2}}{\frac{A}{4}}=\rho \frac{l\times 2}{A}$

$\because R=\rho \frac{l}{A}$

$\Rightarrow R^{\prime }=2R$

$\frac{R^{\prime }}{S}=\frac{l_2}{100-l_2}$

$\Rightarrow \frac{2R}{S}=\frac{l_2}{100-l_2}$

$\Rightarrow 2\times \frac{1}{3}=\frac{l_2}{100-l_2}\text{Using (i)}$

$\Rightarrow l_2=40\text{cm}$