Question

In the meter bridge experiment, the length $A B$ of the wire is $1\, m$. 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 $0.625\, m$ 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)

Here in given condition, we have
$\frac{\frac{b x}{b+ x}}{2}=\frac{0.625}{0.375}$
$\frac{b x}{(b +x) 2} =\frac{25}{15}$
$\frac{5 b}{(b+5) 2} =\frac{5}{3}$
$\frac{b}{2 b+10} =\frac{1}{3}$
$\Rightarrow 3 b-2 b =10$
$b=10 \Omega$