In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that the variation (dR)/(dl) of its resistance R with length l is (dR)/(dl)∝1/(√l). Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is…

Question

In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that the variation (dR)/(dl) of its resistance R with length l is (dR)/(dl)∝1/(√l). Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP? [Figure: metre bridge wire AB of 1 m; equal resistances R' in the left and right gaps; a cell across the outer ends of the gaps; galvanometer G from the junction of the two R' to a jockey at P; AP=l, PB=1-l.]

Accepted Answer

(3) 0.25 m. (dR)/(dl)=k/(√l) gives R(0→ l)=2k√l and R(l→1)=2k(1-√l). Equal gap resistances mean the two wire segments must have equal resistance. 2k√l=2k(1-√l), so √l=1/2. AP=l=0.25 m.

Step-by-step solution

Why the other options are wrong