The resistance per centimeter of a meter bridge wire is r, with X Ω resistance in left gap. Balancing length from left end is at 40 cm with 25 Ω resistance in right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for same settings will be at (a) 10 cm (b) 80 cm (c) 40 cm (d) 20 cm

Answer: (c). Balance: X/25 = (λ l)/(λ(100-l)); the resistance per cm cancels. Doubling the resistance per cm does not change the ratio. So the answer is (c) 40 cm.

The resistance per centimeter of a meter bridge wire is r, with X Ω… | JEE Main 2024 (31 Jan, Shift 2)

Q37JEE Main 2024 · 31 JAN · SHIFT 2Difficult

Wheatstone Bridge & Metre Bridge

The resistance per centimeter of a meter bridge wire is $r$ , with $X\ \Omega$ resistance in left gap. Balancing length from left end is at $40\text{ cm}$ with $25\ \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for same settings will be at

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Answer: (c)

Balance: $\dfrac{X}{25} = \dfrac{\lambda l}{\lambda(100-l)}$ ; the resistance per cm cancels.
Doubling the resistance per cm does not change the ratio.
So the answer is (c) 40 cm.

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