JEE Main202125 Jul 2021Morning ShiftPhysicsCurrent ElectricityActual
In the given figure, there is a circuit of potentiometer of length A B = 10 m . The resistance per unit length is 0 . 1 Ω per cm . Across A B , a battery of emf E and internal resistance r is connected. The maximum value of emf measured by this potentiometer is:
Options
- A5   V
- B2 . 25   V
- C6   V
- D2 . 75   V
Correct answer
A. 5   V
Step-by-step solution
Maximum voltage that can be measured by this potentiometer will be equal to potential drop across A B , R AB = 10 × 0 . 1 × 100 = 100   ohm ∴   V AB = 6 20 + 100 × 100 = 6 × 100 120 = 5   V