Question

When two wires are connected in the two gaps of a meter bridge, the balancing length is 50 cm. When the wire in the right gap is stretched to double its length and again connected in the same gap, then the new balancing length from the left end of the bridge wire is:

• A. 80 cm
• B. 20 cm
• C. 33.3 cm
• D. 66.6 cm

Hint:

Remember, in a meter bridge setup, the product of resistances on either side of the bridge remains constant if the total length of the wire is unchanged.

Answer 1

The Correct Option is B

By stretching the wire to double its original length, the resistance of the wire also doubles (Resistance \( R = \rho \frac{L}{A} \), where \( L \) is length). Originally, the balance condition (50 cm) implies equal resistance on both sides. After doubling the resistance on one side, the bridge is balanced by the inverse ratio of lengths: \[ \frac{l_1}{l_2} = \frac{R_2}{R_1} = \frac{2R}{R} = 2 \] Thus, the new balance length \( l_1 \) from the left end is \( \frac{1}{3} \) of 60 cm = 20 cm.