Physics - Practical physics Question with Solution | TestHub

Given, Emf of cell, $\epsilon =0.5\mathrm{v}$ Rheostat resistance, ${\mathrm{R}}_{\mathrm{h}}=2\Omega$ Potential gradient is $\frac{dv}{dL}=\left(\frac{6}{2+4}\right)\times \frac{4}{L}$ Let null point be at $\ell \mathrm{cm}$ when cell of $\mathrm{emf}\epsilon =0.5\mathrm{v}$ is used. thus ${\epsilon }_1=0.5\mathrm{V}=\left(\frac{6}{2+4}\right)\times \frac{4}{\mathrm{L}}\times \ell$...(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 $\left(\frac{6}{4+6}\right)\left(\frac{4}{\mathrm{L}}\right)\times \ell ={\epsilon }_2$...(ii) Dividing equation (i) by (ii) we get $\frac{0.5}{{\epsilon }_2}=\frac{10}{6}$ thus ${\epsilon }_2=0.3\mathrm{v}$