Question

When an unknown resistance is connected in the left gap and a resistance of 15 $\Omega$ is connected in the right gap of a meter bridge, the balancing point is found to be at 40 cm from the left end of the bridge wire. The emf of the battery of negligible internal resistance used in the circuit is 2 V. If the current drawn from the battery is 280 mA, the resistance of the bridge wire per unit length is

• A. $0.4 \Omega cm^{-1}$
• B. $0.3 \Omega cm^{-1}$
• C. $0.1 \Omega cm^{-1}$
• D. $0.2 \Omega cm^{-1}$

Hint:

Understand the circuit diagram: The battery is connected across the ends of the 1m wire. The resistors X and R are connected in series with each other, and this combination is in parallel with the wire.

Answer 1

The Correct Option is C

Step 1: Calculate Unknown Resistance X:
Using Meter Bridge principle: $\frac{X}{R} = \frac{l}{100-l}$. $R = 15 \Omega$, $l = 40$ cm. \[ \frac{X}{15} = \frac{40}{60} = \frac{2}{3} \implies X = 10 \Omega \]
Step 2: Calculate Total Circuit Resistance:
The battery supplies current to the entire bridge circuit. Usually, the wire and the resistor gaps are in parallel across the battery in standard diagrams where the battery is connected to the ends of the wire A and C. Current $I = 280$ mA = 0.28 A. Voltage $V = 2$ V. Total External Resistance $R_{ext} = \frac{V}{I} = \frac{2}{0.28} = \frac{200}{28} \approx 7.14 \Omega$.
Step 3: Determine Wire Resistance ($R_w$):
The circuit consists of the wire resistance $R_w$ in parallel with the series combination of gap resistors $(X + R)$. $X + R = 10 + 15 = 25 \Omega$. \[ \frac{1}{R_{ext}} = \frac{1}{R_w} + \frac{1}{X+R} \] \[ \frac{1}{7.14} = \frac{1}{R_w} + \frac{1}{25} \] Or simpler: $\frac{1}{R_w} = \frac{I}{V} - \frac{1}{25} = \frac{0.28}{2} - 0.04 = 0.14 - 0.04 = 0.1$. \[ R_w = \frac{1}{0.1} = 10 \Omega \]
Step 4: Calculate Resistance per unit length ($\lambda$):
Length of wire = 100 cm. \[ \lambda = \frac{R_w}{100} = \frac{10}{100} = 0.1 \, \Omega cm^{-1} \]
Step 5: Final Answer:
The resistance per unit length is $0.1 \Omega cm^{-1}$.