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

Answer: 35

Total resistance $=\underset{0}{\overset{L}{\int }}\lambda xdx=\frac{\lambda }{2}{L}^2$
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 $L=100cm$ and $n=3$
$D=\frac{100}{2}(\sqrt{3}-1)=35cm$