Question

The circuit diagram given in the figure shows the experimental setup for the measurement of unknown resistance by using a meter bridge. The wire connected between the points $P\&Q$ has non-uniform resistance such that resistance per unit length varies directly as the distance from the point $P.$ Null point is obtained with the jockey $J$ with $R_{1}$ and $R_{2}$ in the given position. On interchanging the positions $R_{1}$ and $R_{2}$ in the gaps the jockey has to be displaced through a distance $\Delta $ from the previous position along the wire to establish the null point. If the ratio of $\frac{R_{1}}{R_{2}}=3,$ find the value of $\Delta $ (in $cm$ ). Ignore any end corrections. [Take $\sqrt{3}=1.7$ ]

Answer 1

Total resistance $=\int\limits_{0}^{L} \lambda x d x=\frac{\lambda}{2} L^{2}$
For balance condition
$\frac{R_{1}}{R_{2}}=\frac{(\lambda / 2) l_{1}^{2}}{(\lambda / 2)\left(L^{2}-l_{1}^{2}\right)}$
Similarly, $\frac{R_{2}}{R_{1}}=\frac{l_{2}^{2}}{L^{2}-l_{2}^{2}}$
$\therefore \frac{R_{1}}{R_{2}}=\frac{l_{1}^{2}}{L^{2}-l_{1}^{2}}=\frac{l_{1}^{2}}{L^{2}-l_{1}^{2}}=n$
Solving, $l_{1}=L \sqrt{\frac{n}{n+1}}$ and $l_{2}=L \sqrt{\frac{1}{n+1}}$
$\therefore \Delta=l_{1}-l_{2}=\frac{L}{\sqrt{n+1}}(\sqrt{n}-1)$
Putting $L=100 \,cm$ and $n=3$
$D=\frac{100}{2}(\sqrt{3}-1)=35 \,cm$