A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R.B. = 2400 $$\Omega$$. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 $$\Omega$$. Then we can conclude:

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In a galvanometer, the deflection ($$\theta$$) is directly proportional to the current ($$I$$) passing through it $$I = k\theta$$

$$I = \frac{V}{R + G}$$,$$V$$ is the battery voltage ($$2\ \text{V}$$). $$R$$ is the resistance from the Resistance Box (R.B.). $$G$$ is the internal resistance of the galvanometer.

Case 1: $$R_1 = 2400\ \Omega$$ resulting in a deflection $$\theta_1 = 40$$ divisions.

$$I_1 = \frac{2}{2400 + G} = 40k \quad \text{--- (Eq. 1)}$$

Case 2: $$R_2 = 4900\ \Omega$$ resulting in a deflection $$\theta_2 = 20$$ divisions.

$$I_2 = \frac{2}{4900 + G} = 20k \quad \text{--- (Eq. 2)}$$

Divide (Eq. 1) by (Eq. 2), $$\frac{I_1}{I_2} = \frac{4900 + G}{2400 + G} = \frac{40k}{20k}$$ $$\implies \frac{4900 + G}{2400 + G} = 2$$

$$4900 + G = 4800 + 2G$$ $$\implies G = 100\ \Omega$$

$$20k = \frac{2}{4900 + 100}$$ $$\implies 20k = \frac{2}{5000}$$

$$k = \frac{1}{20 \times 2500} = \frac{1}{50000}\ \text{A/division}$$ $$\implies k = 20\ \mu\text{A/division}$$

Option A: Current sensitivity is $$20\ \mu\text{A/division}$$. (Correct)

Option B: $$G = 100\ \Omega$$, not $$200\ \Omega$$. (Incorrect)

Option C: For $$\theta = 10$$ div, $$10(20 \times 10^{-6}) = \frac{2}{R + 100} \implies R = 9900\ \Omega$$. (Incorrect)

Option D: Full scale current ($$50$$ divisions) $$= 50 \times 20\ \mu\text{A} = 1\ \text{mA}$$. (Incorrect)