A battery of emf 10 ~V is connected to a uniform wire A B of 1 ~m length and having a resistance of 10 in series with a 10 resistor as shown in the figure. Two cells of emf 2 ~V and 3 ~V having internal resistance 2 and 3 , respectively are connected as shown in the figure. If the galvanometer shows null deflection at point J on the wire, then the distance of point J from the point B is.

Step-by-step solution

Given r=10 , R=10 , E=10 ~V and L=100 ~cm . Now, voltage drop on the wire, E^ = E (r+R) R E^ = 10 20 10=5 R So, potential gradient on wire, Effective emf of combination in secondary circuit, array rlrl & & V r_ eff & = ₁ r₁ + ₂ r₂ Here, & 1 r_ eff & = 1 2 + 1 3 = 5 6 & V & = 6 5 (1+1)= 12 5 array Now, at balancing point (from point A ) is aligned & l=V / x= 12 5 100 5 & l=48 ~cm aligned So, length from point B is 100-48=52 ~cm