Question

An unknown resistance $R_1$ is connected in series with a resistance of $10\Omega .$ This combination is connected to one gap of a metre bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50cm$. Now, when the $10\Omega$ resistance is removed the balance point shifts to $40cm$. The value of $R_1$ is (in ohm)

• A. 20
• B. 10
• C. 60
• D. 40

Answer 1

Answer: (A)

The balance condition of a metre bridge experiment
$\frac{R}{S}=\frac{l_1}{(100-l_1)}$
Here, $R=R_1,S=R_2$
$\therefore \frac{R_1}{R_2}=\frac{l_1}{(100-l_1)}$
1st Case $\frac{R_1+10}{R_2}=\frac{50}{50}$
$\Rightarrow R_1+10=R_2........(i)$
2nd Case $\frac{R_1}{R_2}=\frac{40}{60}$
$R_2=\frac{60}{40}$
$\Rightarrow R_2=\frac{60}{40}{R}_1.....(ii)$
So, Eqs. (i) and (ii) give
$R_1+10=\frac{60}{40}{R}_1$
$\Rightarrow \frac{60}{40}{R}_1-R_2=10$
$\Rightarrow \frac{20}{40}{R}_1=\frac{10\times 40}{20}$
$\therefore R_1=20\Omega$