Question

An unknown resistance $X$ is connected in series with a resistance of $10\Omega$. This combination connected in the left gap of the metre-bridge is balanced against another resistance $Y$ connected in the other gap. The balance point is found to be $0.50m$. · When the $10\Omega$ resistance is removed in the combination, the balance point is $0.4m$. The value of $X$ in $\Omega$ is

• A. 60
• B. 40
• C. 20
• D. 30

Answer 1

Answer: (C)

Given situation is shown in the figure.
$\frac{X+10}{Y}=\frac{50}{50}$
or, $X+10=Y$ ...(i)
Now, if $10\Omega$ resistance is removed in the combination, the balance point is $0.4m(=40cm)$ .
So, $\frac{X}{Y}=\frac{40}{(100-40)}=\frac{2}{3}$
or, $Y=\frac{3}{2}X$ ...(ii)
From eqn (i) and (ii)
$X=10=\frac{3}{2}X$ or, $X=20\Omega$