In a meter bridge experiment, the balance point is obtained at length $\ell_1 \mathrm{~cm}$ from left end…
In a meter bridge experiment, the balance point is obtained at length $\ell_1 \mathrm{~cm}$ from left end when resistances in the left gap and right gap are $5 \Omega$ and $R \Omega$ respectively. When the resistance $R$ is shunted with equal resistance, the new balance point is at $1.6 \ell_1$. The resistance $\mathrm{R}$ in ohm is
25
15
10
20
Solution
$\frac{5}{R}=\frac{\ell_1}{100-\ell_1}$
When $\mathrm{R}$ is shunted,
$\begin{aligned}
& \frac{5 \times 2}{\mathrm{R}}=\frac{1.6 \ell_1}{100-1.6 \ell_1} \\
& \therefore \frac{5}{\mathrm{R}}=\frac{0.8 \ell_1}{100-1.6 \ell_1}
\end{aligned}$
Equating (1) and (2) and solving we get $\ell_1=25 \mathrm{~cm}$ Putting this in Eq. (1), we get $R=15 \Omega$