Question

In the meter bridge experiment, the length AB of the wire is $1m$. The resistors $X$ and $Y$ have values $5\Omega$ and $2\Omega$ respectively. When a shunt resistance $S$ is connected to $X$, the balancing point is found to be $0.625m$ from A. Then, the resistance of the shunt is

• A. $5\Omega$
• B. $10\Omega$
• C. $7.5\Omega$
• D. $12.5\Omega$

Answer 1

Answer: (B)

$10\Omega$

In the balanced wheatstone's bridge condition,

$\frac{R_{X||S}}{R_{AJ}}=\frac{R_Y}{R_{JB}}$

Since $AB$ is a uniform wire, its resistance is proportional to length

Hence, $\frac{R_{X||S}}{0.625}=\frac{2}{0.375}$.

or, $R_{X||S}=\frac{10}{3}\Omega =\frac{5S}{5+S}$.

or, $50+10S=15S$

or, $S=10\Omega$