Question

Moving coil galvanometers \(M_1\) and \(M_2\) have resistance, number of turns, area of coil and magnetic field as follows: \(R_1 = 10\Omega,\; R_2 = 14\Omega,\; N_1 = 30,\; N_2 = 42,\) \(A_1 = 3.6 \times 10^{-3}\,\text{m}^2,\; A_2 = 1.8 \times 10^{-2}\,\text{m}^2,\; B_1 = 0.25\,\text{T},\; B_2 = 0.50\,\text{T}\). (Spring constants are same for both materials.) The ratio of (i) current sensitivity and (ii) voltage sensitivity for galvanometer \(M_2\) to \(M_1\) is respectively

• A. \(1:1,\; 1:4:1\)
• B. \(1:1:4,\; 1:1\)
• C. \(4:1,\; 1:1\)
• D. \(1:4:1,\; 1:1\)

Hint:

Current sensitivity depends on \(NBA\), while voltage sensitivity also depends on resistance.

Answer 1

The Correct Option is D

Step 1: Current sensitivity.
Current sensitivity of a moving coil galvanometer is \[ S_I \propto NBA. \] Thus, \[ \frac{S_{I2}}{S_{I1}} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1}. \] Substituting values, \[ \frac{S_{I2}}{S_{I1}} = \frac{42 \times 0.50 \times 1.8\times10^{-2}}{30 \times 0.25 \times 3.6\times10^{-3}} = 4. \] Hence, current sensitivity ratio \(= 4:1\).
Step 2: Voltage sensitivity.
Voltage sensitivity is given by \[ S_V = \frac{S_I}{R}. \] So, \[ \frac{S_{V2}}{S_{V1}} = \frac{S_{I2}/R_2}{S_{I1}/R_1} = \frac{4}{1}\times\frac{10}{14} = 1. \]
Step 3: Conclusion.
The ratios are \(4:1\) for current sensitivity and \(1:1\) for voltage sensitivity.