Question

When a wire is connected in the left gap of a metre bridge, the balancing point is at \( 40 \) cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap, then the balancing point from the left end of the bridge wire is:

• A. \( \frac{300}{11} \) cm
• B. \( \frac{800}{11} \) cm
• C. \( \frac{400}{11} \) cm
• D. \( \frac{700}{11} \) cm

Hint:

For stretched wires in metre bridge, resistance scales as \( R \propto L^2 \), affecting the balancing ratio.

Answer 1

The Correct Option is C

Step 1: Understanding Resistance Change The resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where \( \rho \) is resistivity, \( L \) is length, and \( A \) is the cross-sectional area. Since the wire is stretched its length doubles, but cross-sectional area reduces, making its resistance increase by a factor of 4. Step 2: Applying Metre Bridge Formula The metre bridge follows the principle of Wheatstone's bridge: \[ \frac{R_1}{R_2} = \frac{l}{(100 - l)} \] Given initial balancing condition: \[ \frac{R_1}{R_2} = \frac{40}{60} \] After stretching, the resistance increases by a factor of 4: \[ \frac{4R_1}{R_2} = \frac{l'}{(100 - l')} \] Solving, \[ \frac{4 \times 40}{60} = \frac{l'}{100 - l'} \] \[ \frac{160}{60} = \frac{l'}{100 - l'} \] Solving for \( l' \), \[ l' = \frac{400}{11} \text{ cm} \] Conclusion Thus, the correct answer is: \[ \frac{400}{11} \text{ cm} \]