JEE Main20231 Feb 2023Morning ShiftPhysicsCurrent ElectricityActual
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1 . 5 V is found to be 60 cm . If this cell is replaced by another cell of emf E . the length-of null point increases by 40 cm . The value of E is x 10 V . The value of x is ______.
Correct answer
0
Step-by-step solution
We know, the condition where the galvanometer shows a null point is, E 1 E 2 = l 1 l 2 So, 1 . 5 E 2 = 60 60 + 40 = 6 10 = 3 5 ⇒ E 2 = 5 2   V Hence, the value of x = 25 .