The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA . For a cell of emf 5 V and internal resistance of $20 \Omega$, the null point on it is found to be at 1000 cm . The resistance of whole wire is:

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Current Electricity

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA . For a cell of emf 5 V and internal resistance of

20Ω

, the null point on it is found to be at 1000 cm . The resistance of whole wire is:

Answer: (B)

The potential gradient along the potentiometer wire is

k = \frac{EMF}{balancing length} = \frac{5}{1000}

V/cm. Over the full

1200

cm the potential drop is

V_{P} = k \times 1200 = \frac{5}{1000} \times 1200 = 6

V. By Ohm's law the wire's resistance is

R_{P} = \frac{V _{P}}{I} = \frac{6}{60 \times 1 0 ^{- 3}} = 100 Ω

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option