A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ______ Ω.
Physics · Circular Motion
JEE Mains 29 January 2023 Shift 2 Evening Shift

Question

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ______ Ω.

Physics · Circular Motion

JEE Mains 29 January 2023 Shift 2 Evening Shift

G1 Physics · Accepted Answer

Let the potential gradient be k and emf be E. With shunt R, terminal voltage is ER/(r+R) = kl. Thus E×5/(r+5) = 200k and E×15/(r+15) = 300k. Dividing the two equations gives [5/(r+5)]/[15/(r+15)] = 2/3. Solving gives r = 5 Ω.