JEE Main201911 Jan 2019Morning ShiftPhysicsExperimental PhysicsActual
The resistance of the meter bridge AB in given figure is 4 . With a cell of emf =0.5 ~V and rheostat resistance R _ h =2 the null point is obtained at some point J. When the cell is replaced by another one of emf = ₂ the same null point J is found for R _ h =6 . The emf ₂ is:
Options
- A0.4 ~V
- B0.3 ~V
- C0.6 ~V
- D0.5 ~V
Correct answer
B. 0.3 ~V
Step-by-step solution
Given, Emf of cell, =0.5 v Rheostat resistance, R _ h =2 Potential gradient is d v d L = ( 6 2+4 ) 4 L Let null point be at cm when cell of emf =0.5 v is used. thus ₁=0.5 ~V = ( 6 2+4 ) 4 ~L ...(i) For resistance R _ h =6 new potential gradient is ( 6 4+6 ) 4 L and at null point ( 6 4+6 ) ( 4 ~L ) = ₂ ...(ii) Dividing equation (i) by (ii) we get 0.5 ₂ = 10 6 thus ₂=0.3 v