Question

The resistance of the meter bridge $AB$ is given figure is $4\Omega$ . With a cell of emf $\epsilon =0.5V$ 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 $\epsilon ={\epsilon }_2$ the same null point $J$ is found for $R_h=6\Omega$. The emf ${\epsilon }_2$ is;

• A. 0.6 V
• B. 0.5 V
• C. 0.3 V
• D. 0.4 V

Answer 1

Answer: (C)

Potential gradient with $R_h=2\Omega$ is $\left(\frac{6}{2+4}\right)\times \frac{4}{L}=\frac{dV}{dL};L=100cm$
Let null point be at $\ell$ cm
thus ${\epsilon }_1=0.5V=\left(\frac{6}{2+4}\right)\times \frac{4}{L}\times \ell$ .....(1)
Now with $R_h=6\Omega$ new potential gradient is
$\left(\frac{6}{4+6}\right)\times \frac{4}{L}$ and at null point
$\left(\frac{6}{4+6}\right)\left(\frac{4}{L}\right)\times \ell ={\epsilon }_2$ ....(2)
dividing equation (1) by (2) we get
$\frac{0.5}{{\epsilon }_2}=\frac{10}{6}{\epsilon }_2=0.3$