JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Current Electricity

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In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is

3 : 2

, the difference in the balancing length of the potentiometer wire in above two cases will be ______ cm.

Answer: 25

Solution

In a potentiometer, the emf of a cell is proportional to the balancing length:

\frac{ε _{1}}{ε _{2}} = \frac{ℓ _{1}}{ℓ _{2}}

Given:

\frac{ε _{1}}{ε _{2}} = \frac{3}{2}

and

ℓ_{1} = 75

cm.

Therefore:

\frac{3}{2} = \frac{75}{ℓ _{2}}

ℓ_{2} = \frac{75 \times 2}{3} = 50 cm

The difference in balancing lengths:

Δ ℓ = ℓ_{1} - ℓ_{2} = 75 - 50 = 25 cm

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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