Question

In the experimental setup of metre bridge shown in the figure, the null point is obtained at a distance of $40\text{cm}$ from $\text{A}$. If a $10\Omega$ resistor is connected in series with $R_1\Omega$ , the null point is observed to shift by $10\text{cm}$ from $\text{A}$. Find the resistance that should be connected in parallel with $(R_1+10)$$\Omega$ such that the null point shifts back to its initial position is

• A. $120\Omega$
• B. $90\Omega$
• C. $60\Omega$
• D. $30\Omega$

Answer 1

The correct option is C $60\Omega$

Step-$\left(\text{I}\right):$

$\frac{R_1}{R_2}=\frac{l}{100-l}=\frac{40}{60}=\frac{2}{3}\Rightarrow R_1=\frac{2}{3}{R}_2$

Step-$\left(\text{II}\right):$

When $10\Omega$ is connected in series with $R_1$ the balancing length is $40+10=50\text{cm}$.

$\frac{R_1+10}{R_2}=\frac{50}{100-50}=1\Rightarrow R_1+10=R_2$

$\frac{2}{3}{R}_2+10=R_2\Rightarrow 10=\frac{R_2}{3}$

$\Rightarrow R_2=30\Omega$ and $R_1=20\Omega$

Step-$\left(\text{III}\right):$

To bring the null point to the initial position, a resistance $x$ is connected parallel to $R_1+10=30\Omega$

ie $\frac{\left(\frac{30x}{30+x}\right)}{R_2}=\frac{40}{60}\Rightarrow \frac{30x}{(30+x)30}=\frac{2}{3}$

$\therefore 3x=60+2x$

$\Rightarrow x=60\Omega$

Hence, option (c) is the correct answer.