Physics - Current Electricity Question with Solution | TestHub

Here length of potentiometer wire, $l=10\mathrm{m}$ Resistance of potentiometer wire $R=\rho \times \frac{l}{A}$ or $R=\left(\rho \times \frac{10}{A}\right)$ The value of $2.5\mathrm{m}$ length wire $$R^{\prime }=\frac{\rho \times 10}{A\times 10}\times 2.5\Rightarrow R^{\prime }=\left(\frac{2.5\rho }{A\times 10}\right)$$ Potential, $V^{\prime }=I\times R^{\prime }=I\left(\frac{2.5\rho }{A\times 10}\right)$ Again the length of potentiometer wire is increased by $1\mathrm{m}$. Resistance of null position $$R^n=\left(\frac{\rho \times l}{11\times A}\right)$$ $\therefore V^{\prime \prime }=I{R}^{\prime \prime }$ and $V=V^{\prime }$ $$\begin{array}{l}\Rightarrow \frac{I\times 2.5\rho }{A\times 10}=\frac{\rho \times I}{11\times A}\times I\\ \text{ or }\frac{2.5\times 11}{10}=I=2.75\mathrm{m}\end{array}$$