Question

An unknown resistance $R_1$ is connected in series with a resistance of $10\Omega$ . This combination is connected in one gap of a meter bridge, while a resistance $R_2$ is connected in the other gap. The balance point is obtained at $50\text{cm}$ mark. When the $10\Omega$ resistance is removed, the balance point shifts to $40\text{cm}$ mark. The value of $R_1$ is -

• A. $20\Omega$
• B. $30\Omega$
• C. $10\Omega$
• D. $40\Omega$

Answer 1

The correct option is A $20\Omega$

Case $\left(1\right):$


So,

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

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

Case $\left(2\right):$


So,

$\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$

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

Now, from equations $\left(1\right)$ and $\left(2\right)$,

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

$\Rightarrow R_1=20\Omega$

Hence, option $\left(A\right)$ is the correct answer.