The resistance of the meter bridge AB in given figure is 4 . With a...

Given, Emf of cell, $\varepsilon=0.5 \mathrm{v}$ Rheostat resistance, $\mathrm{R}_{\mathrm{h}}=2 \Omega$ Potential gradient is $\frac{d v}{d L}=\left(\frac{6}{2+4}\right) \times \frac{4}{L}$ Let null point be at $\ell \mathrm{cm}$ when cell of $\mathrm{emf} \varepsilon=0.5 \mathrm{v}$ is used. thus $\varepsilon_{1}=0.5 \mathrm{~V}=\left(\frac{6}{2+4}\right) \times \frac{4}{\mathrm{~L}} \times \ell$...(i) For resistance $\mathrm{R}_{\mathrm{h}}=6 \Omega$ new potential gradient is \left(\frac{6}{4+6}\right) \times \frac{4}{L} and at null point $\left(\frac{6}{4+6}\right)\left(\frac{4}{\mathrm{~L}}\right) \times \ell=\varepsilon_{2}$...(ii) Dividing equation (i) by (ii) we get $\frac{0.5}{\varepsilon_{2}}=\frac{10}{6}$ thus $\varepsilon_{2}=0.3 \mathrm{v}$