1. Given data

• Galvanometer resistance: $G = 30\,\Omega$
• Full-scale deflection current: $I_g = 20\text{ mA} = 0.02\text{ A}$
• Maximum current to be measured: $I = 3\text{ A}$

2. Current through shunt

The galvanometer carries only $I_g = 0.02\text{ A}$ at full scale, so the remaining current goes through the shunt:

$I_s = I - I_g = 3 - 0.02 = 2.98\text{ A}$

3. Use equal potential drop across galvanometer and shunt

Since galvanometer and shunt are in parallel,

$I_g G = I_s S$

where $S$ is the shunt resistance.

Substituting values:

$0.02 \times 30 = 2.98 \times S$

$0.6 = 2.98S$

$S = \frac{0.6}{2.98} = \frac{60}{298} = \frac{30}{149}\,\Omega$

4. Compare with the given form

Given,

$S = \frac{30}{X}\,\Omega$

So,

$\frac{30}{X} = \frac{30}{149}$

Hence,

$X = 149$

5. Option check

• A: $447$ ❌
• B: $298$ ❌
• C: $149$ ✅
• D: $596$ ❌

Therefore, the correct answer is Option C.