Question

In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: 

• A. 54.6 \( \Omega \)
• B. 116.0 \( \Omega \)
• C. 45.0 \( \Omega \)
• D. 103.0 \( \Omega \)

Hint:

To find the figure of merit of the galvanometer, use the relation involving the deflection and the resistance in the circuit. Higher deflections indicate better performance of the galvanometer.

Answer 1

The Correct Option is B

In the first case when \( K_1 \) is closed and \( K_2 \) is open, the deflection is 20 div. When \( K_2 \) is closed, and a resistance of \( 90 \, \Omega \) is removed, the deflection becomes 13 div. The figure of merit \( G \) of the galvanometer is calculated using the relation: \[ G = \frac{R_1 \cdot \Delta \theta_2}{\Delta \theta_1} \] where \( R_1 \) is the resistance in the circuit, \( \Delta \theta_1 \) is the first deflection, and \( \Delta \theta_2 \) is the second deflection. From the given data: \[ \Delta \theta_1 = 20 \, \text{div}, \, \Delta \theta_2 = 13 \, \text{div}, \, R_1 = 520 \, \Omega. \] The deflection decreases as the additional resistance is removed, and the relationship gives us: \[ G = \frac{520 \cdot 13}{20} = 116.0 \, \Omega. \] Thus, the resistance of the galvanometer is \( 116.0 \, \Omega \).