Given, Emf of cell, ε=0.5v Rheostat resistance, Rh=2Ω Potential gradient is dLdv=(2+46)×L4 Let null point be at ℓcm when cell of emfε=0.5v is used. thus ε1=0.5 V=(2+46)× L4×ℓ...(i) For resistance Rh=6Ω new potential gradient is (4+66)×L4 and at null point (4+66)( L4)×ℓ=ε2...(ii) Dividing equation (i) by (ii) we get ε20.5=610 thus ε2=0.3v
The resistance of the meter bridge AB in given figure is 4 Ω. With a cell of emf ε=0.5 ~V and rheostat resistance R_ h=2 Ω the null point is obtained…
JEE Main 2019 — Physics Mechanics
The resistance of the meter bridge AB in given figure is 4Ω. With a cell of emf ε=0.5 V and rheostat resistance Rh=2Ω 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=6Ω. The emfε2 is: 
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
0.4 V
- B
0.3 V
- C
0.6 V
- D
0.5 V
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