In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1 . 5…

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 x10 V. The value of x is ______.

Solution

We know, the condition where the galvanometer shows a null point is,

E1E2=l1l2

So, 1.5E2=6060+40=610=35

⇒E2=52 V

Hence, the value of x=25.

Asked in: JEE Main 2023 (01 Feb Shift 1)

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