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 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

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Current Electricity

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

25Ω

resistance in right gap. Now the wire is replaced by another wire of

2 r

resistance per centimeter. The new balancing length for same settings will be at

(A)20 cm
(B)10 cm
(C)80 cm
(D)40 cm✓ Correct

Answer: (D)

Solution

In a meter bridge at balance,

\frac{X}{R} = \frac{ℓ _{1} r}{ℓ _{2} r} = \frac{ℓ _{1}}{ℓ _{2}}

, where r is the resistance per unit length of the wire.

This factor r appears identically in both segments and cancels out of the balance condition. So replacing the wire's resistance per centimetre with 2r leaves the ratio

ℓ_{1} / ℓ_{2}

unchanged for the same X and R.

The balancing length stays at 40 cm.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Current Electricity →

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Current Electricity

Solution

More JEE Main questions on Current Electricity