The length of a potentiometer wire is 'L'. A cell of e.m.f. E is balanced at a length L 5 from the positive end of the wire. If the length of the wire is increased by L 2 , the same cell will give balance point at distance ' x '. The value of ' x ' is

Step-by-step solution

The initial potential gradient for potentiometer wire length L with applied voltage V₀ is k₁ = V₀ L . With the cell of e.m.f. E balanced at L 5 , potentiometer principle gives E = k₁ L 5 = V₀ 5 . When the wire length increases to 3L 2 with unchanged V₀ , the new potential gradient becomes k₂ = V₀ 3L 2 = 2V₀ 3L . For the same cell e.m.f. balanced at length x : E = k₂ x = 2V₀ 3L x . Equating both expressions for E : V₀ 5 = 2V₀ 3L x Canceling V₀ and solving yields x = 3L 10 . The new balance point occurs at 3L 10 . Co