Question

As shown in the figure, a potentiometer wire of resistance 20 Ω 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 1

Correct Answer: 780

The potential drop per unit length (potential gradient) along the potentiometer wire is calculated using the formula:
ℇ = V/L
where V is the voltage across the potentiometer wire and L is its length.

The total resistance in the circuit is the resistance of the potentiometer wire plus the resistance R from the resistance box:
Total resistance = 20 Ω + R 

The current i flowing through the circuit is given by:
i = ℰ / (20 Ω + R)
with ℰ = 4 V (emf of the standard cell).

The potential drop across the potentiometer wire:
V = i × 20 Ω

Therefore, the potential gradient along the wire is:
Potential gradient = V / 300 cm = (4 V × 20 Ω) / ((20 Ω + R) × 300 cm)

At the null point (60 cm), the potential difference is equal to the emf of the 20 mV cell:
(Potential gradient) × 60 cm = 20 mV

Substitute the potential gradient equation:
((4 × 20) / (20 + R) × 300) × 60 = 0.02

Simplifying:
(80 / (20 + R)) × (60 / 300) = 0.02
(80 / (20 + R)) × (1/5) = 0.02
80 / (20 + R) = 0.1

Cross-multiplying gives:
80 = 0.1 × (20 + R)
800 = 20 + R
R = 780 Ω

Since 780 Ω is within the given range (780,780), the solution is validated.

Answer 2

E= \(\frac{AC}{AB(V_A-V_B)}\)
∴ 20×10⁻³= \(\frac{60}{300}\)× 4×\(\frac{20}{R+20}\)
​∴ R=780 Ω