Question

The resistance of the meter bridge $AB$ in given figure is $4\Omega$. With a cell of emf $E=0.5\text{V}$ and rheostat resistance $R_h=2\Omega$ the null point is obtained at some point $J$. When the cell is replaced by another one of emf $E=E_2$ the same null point $J$ is found for $R_h=6\Omega$. The emf $E_2$ is:

• A. $0.4\text{V}$
• B. $0.3\text{V}$
• C. $0.6\text{V}$
• D. $0.5\text{V}$

Answer 1

The correct option is B $0.3\text{V}$

$\text{Given that, In case - 1}$

$\text{Emf of cell,}E=0.5\text{V}$
$\text{Rheostat resistance,}{R}_h=2\Omega$

The potential difference across the meter bridge wire is given by,

$V=\frac{4}{4+2}\times 6=4\text{V}$

The Potential gradient of the meter bridge is given by,

$\frac{dV}{dL}=\frac{4}{L}$

Let null point be at $l\text{cm}$ when cell of emf $E=0.5\text{V}$ is used.

$\Rightarrow 0.5\text{V}=\frac{4}{L}\times l---\left(1\right)$

For resistance $R_h=6\Omega$ new potential gradient is given by,

$\frac{dV}{dL}=\left(\frac{4}{4+6}\right)\times \frac{6}{L}=\frac{12}{5L}$

and at the same null point when $E_2$ is connected,

$E_2=\frac{12}{5L}\times l---\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$,

$\Rightarrow \frac{0.5}{E_2}=\frac{5}{3}$

$\Rightarrow E_2=0.3\text{V}$

Hence, option $\left(B\right)$ is correct.