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 $50\, cm$. Now, when the $10 \Omega$ resistance is removed the balance point shifts to $40 cm$. The value of $R_{1}$ is (in ohm)

• A. 20
• B. 10
• C. 60
• D. 40

Answer 1

The Correct Option is 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 = 20O $