Question

In the metre bridge arrangement, an unknown resistance $R_1$ is connected in series with resistance of $10\Omega$. When the combination is connecetd to left gap of the meter bridge and $R_2$ is connected to right gap, balance point from left end is at $50\text{cm}$. When $10\Omega$ resistor is removed, the balance point shifts to $40\text{cm}$. The value of $R_1$ is

• A. $20\Omega$
• B. $10\Omega$
• C. $60\Omega$
• D. $40\Omega$

Answer 1

The correct option is A $20\Omega$

$\frac{R_1+10}{R_2}=\frac{50}{50}$

$\Rightarrow R_1+10=R_2...\left(1\right)$

Now, when $10\Omega$ is removed, then

$\frac{R_1}{R_2}=\frac{40}{60}$

$\Rightarrow \frac{R_1}{R_2}=\frac{2}{3}$

$\Rightarrow R_2=\frac{3}{2}{R}_1...\left(2\right)$

Put $\left(2\right)$ in $\left(1\right)$, we get

$R_1+10=\frac{3}{2}{R}_1$

$\Rightarrow 2R_1+20=3R_1$

$\therefore R_1=20\Omega$

Hence, the correct answer is (A).