Question

Two unknown resistances are connected in two gaps of a meter bridge. The null point is obtained at 40 cm from the left end. A 30 \(\Omega\) resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of the smaller resistance is

• A. 8 \(\Omega\)
• B. 24 \(\Omega\)
• C. 32 \(\Omega\)
• D. 16 \(\Omega\)

Hint:

The value of the unknown resistance in a meter bridge can be calculated by applying the ratio of the lengths of the bridge and using the principle of the Wheatstone bridge.

Answer 1

The Correct Option is B

Step 1: Understanding the meter bridge principle.
In a meter bridge, the null point is the point where the ratio of resistances on both sides of the bridge is equal: \[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \] where \( R_1 \) and \( R_2 \) are the resistances in the two gaps, and \( L_1 \) and \( L_2 \) are the distances from the left and right ends of the bridge, respectively. Step 2: Apply the conditions.
Initially, the null point is at 40 cm, so: \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \] After connecting the 30 \(\Omega\) resistor in series with the smaller resistance, the null point shifts to 60 cm, so: \[ \frac{R_1 + 30}{R_2} = \frac{60}{40} = \frac{3}{2} \] Step 3: Solving the system of equations.
From the first equation: \[ \frac{R_1}{R_2} = \frac{2}{3} \quad \Rightarrow \quad R_1 = \frac{2}{3} R_2 \] Substitute this into the second equation: \[ \frac{\frac{2}{3} R_2 + 30}{R_2} = \frac{3}{2} \] Solving for \( R_2 \): \[ \frac{\frac{2}{3} R_2 + 30}{R_2} = \frac{3}{2} \quad \Rightarrow \quad \frac{2}{3} + \frac{30}{R_2} = \frac{3}{2} \] Multiply through by 3: \[ 2 + \frac{90}{R_2} = \frac{9}{2} \] Simplifying: \[ \frac{90}{R_2} = \frac{5}{2} \quad \Rightarrow \quad R_2 = \frac{90 \times 2}{5} = 36 \, \Omega \] Step 4: Conclusion.
Thus, the value of the smaller resistance is 24 \(\Omega\), which corresponds to option (B).