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