NEET Physics Question (2026) — Solution

Step-by-step solution

Let the nodes of the circuit be identified as follows: Node A: Left L-shaped metallic strip. Node B: Right L-shaped metallic strip. Node C: Central metallic strip. Node D: The point on the wire where the jockey touches. The resistances in the four arms of the circuit are: 1. Resistance between A and C is R_1. 2. Resistance between B and C is R_2. 3. Resistance between A and D is R_ AD (left part of the metre bridge wire). 4. Resistance between B and D is R_ DB (right part of the metre bridge wire). In the given figure, the cell E is connected between nodes C and D, and the galvanometer G is connected between nodes A and B. This arrangement forms a Wheatstone bridge. Let the potential at node D be 0 and the potential at node C be V. The current from the cell divides at node C into two parallel branches: C A D and C B D. The potential at node A is given by the voltage divider rule: V_A = V ( R_ AD R_1 + R_ AD ) Similarly, the potential at node B is: V_B = V ( R_ DB R_2 + R_ DB ) The galvanometer shows no deflection when the potential difference across it is zero, i.e., V_A = V_B. R_ AD R_1 + R_ AD = R_ DB R_2 + R_ DB R_1 + R_ AD R_ AD = R_2 + R_ DB R_ DB R_1 R_ AD + 1 = R_2 R_ DB + 1 R_1 R_ AD = R_2 R_ DB This is the standard balance condition for a Wheatstone bridge. Thus, interchanging the cell and the galvanometer does not affect the balance condition. A null point will still be obtained on the wire. If the jockey is moved to the left of the balance point, R_ AD decreases and R_ DB increases, making V_A If the jockey is moved to the right of the balance point, R_ AD increases and R_ DB decreases, making V_A > V_B. Current will flow from A to B, causing a deflection in the opposite direction. Therefore, the galvanometer will show both right-sided and left-sided deflections depending on the jockey's position, and no deflection at the balance point. Answer: Both right-sided and left-sided deflection and at balance point, no deflection