JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Current Electricity

As shown in the figure, a potentiometer wire of resistance

20Ω

and length 300 cm is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance '

R

' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of '

R

' is _____

Ω

Answer: 780

Solution

In a potentiometer, the potential difference across a length is proportional to that length. The potential gradient along the wire is

k = \frac{V _{A B}}{L _{A B}}

, where

V_{A B}

is the potential difference across the wire.

The current through the potentiometer wire is

I = \frac{4}{R + 20}

.

Potential difference across the wire:

V_{A B} = I \times 20 = \frac{4 \times 20}{R + 20} = \frac{80}{R + 20}

V.

At null point (60 cm), the emf of the test cell equals the potential drop:

E = \frac{l}{L} \times V_{A B}

20 \times 1 0^{- 3} = \frac{60}{300} \times \frac{80}{R + 20}

0.02 = 0.2 \times \frac{80}{R + 20}

0.02 (R + 20) = 16

R + 20 = 800

R = 780 Ω

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer