The length of potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance 10 Ω is balanced on this potentiometer the balance length is found to be 500 cm . The resistance of potentiometer wire will be

Question

The length of potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance 10 Ω is balanced on this potentiometer the balance length is found to be 500 cm . The resistance of potentiometer wire will be (A) 20 Ω (B) 40 Ω (C) 60 Ω (D) 80 Ω

Tardigrade Admin · Accepted Answer

Potential difference of external circuit = Potential gradient × Balance length. 2=(I R/L)× l0 R=(2 L/I l0)=(2 × 600/40 × 1 0- 3 × 500) R=(12 × 1 03/200) R=60 Ω

Q. The length of potentiometer wire is $600 c m$ and a current of $40 m A$ is flowing in it. When a cell of emf $2 V$ and internal resistance $10 Ω$ is balanced on this potentiometer the balance length is found to be $500 c m$ . The resistance of potentiometer wire will be

$20 Ω$

$40 Ω$

$60 Ω$

$80 Ω$

Potential difference of external circuit $=$ Potential gradient $\times$ Balance length.
$2 = \frac{I R}{L} \times l_{0}$
$R = \frac{2 L}{I l _{0}} = \frac{2 \times 600}{40 \times 1 0 ^{- 3} \times 500}$
$R = \frac{12 \times 1 0 ^{3}}{200}$
$R = 60 Ω$

Q. The length of potentiometer wire is 600cm and a current of 40mA is flowing in it. When a cell of emf 2V and internal resistance 10Ω is balanced on this potentiometer the balance length is found to be 500cm . The resistance of potentiometer wire will be

20Ω

40Ω

60Ω

80Ω