JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Current Electricity

A potentiometer wire of length 300 cm is connected in series with a resistance

780Ω

and a standard cell of emf 4 V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is ______

Ω

Answer: 20

Solution

Let the resistance of the potentiometer wire be

R

.

Current through the circuit:

i = \frac{4}{R + 780}

Potential difference across 60 cm length:

V_{A C} = \frac{4 R \times 60}{( R + 780 ) \times 300} = \frac{4 R}{5 ( R + 780 )}

This equals 20 mV:

\frac{4 R}{5 ( R + 780 )} = 0.02

4 R = 0.1 (R + 780)

4 R = 0.1 R + 78

3.9 R = 78

R = 20Ω

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer

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