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Question:

The resistance of the meter bridge AB is 4 Ω. With a cell of emf ε=0.5 V and rheostat resistance Rh=2 Ω, the null point is obtained at some point J. When the cell is replaced by another one of emf ε=ε2, the same null point J is found for Rh=6 Ω. The emf ε is:

0.5 V

0.3 V

0.6 V

0.4 V

Solution:

Potential gradient with Rh=2Ω is (6/(2+4)) × 4/L = dV/dL; L=100cm
Let null point be at l cm
thus ε1=0.5V = (6/(2+4)) × 4/L × l (1)
Now with Rh=6Ω new potential gradient is (6/(4+6)) × 4/L
and at null point (6/(4+6))(4/L) × l = ε2.. (2)
dividing equation(1) and (2) we get
0.5/ε2 = 10/6
thus ε2 = 0.3