Question

When the right gap of a meter bridge consists of two equal resistors in series, the balancing point is at 50 cm. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap, the balancing point is at

• A. 60 cm
• B. 33.3 cm
• C. 25 cm
• D. 40 cm

Hint:

For a meter bridge, $\frac{R_L}{R_R} = \frac{l}{100-l}$. If the ratio of resistances is known, solving for $l$ is straightforward.

Answer 1

The Correct Option is D

Step 1: Analyze First Case:
Left gap resistance = $X$. Right gap resistance = $R + R = 2R$. Balancing length $l_1 = 50$ cm. Wheatstone bridge principle: $\frac{X}{2R} = \frac{50}{100-50} = 1$. $\implies X = 2R$.
Step 2: Analyze Second Case:
One resistor $R$ is removed from the right gap. Right gap resistance = $R$. This removed resistor is connected in parallel to the left gap resistor ($X$). New Left gap resistance $X' = \frac{X \cdot R}{X + R}$. Since $X = 2R$, $X' = \frac{2R \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2R}{3}$. New Right gap resistance = $R$.
Step 3: Find New Balancing Point ($l$):
Let the new balancing length be $l$. \[ \frac{X'}{R} = \frac{l}{100 - l} \] Substitute $X' = \frac{2R}{3}$: \[ \frac{2R/3}{R} = \frac{l}{100 - l} \] \[ \frac{2}{3} = \frac{l}{100 - l} \] \[ 2(100 - l) = 3l \] \[ 200 - 2l = 3l \implies 5l = 200 \implies l = 40 \text{ cm} \]