Question

The resistance of the meter bridge $AB$ is given figure is $4\Omega$ . With a cell of emf $\varepsilon = 0.5 \;V$ 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 $\varepsilon = \varepsilon_2$ the same null point $J$ is found for $R_h = 6 \; \Omega$. The emf $\varepsilon_2$ is;

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

Answer 1

Answer: (C)

0.3 V

Answer 2

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