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 $Rh=2\Omega$ 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 $\epsilon$ is,:

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

Answer 1

Answer: (C)

$0.3V$

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