1. Use torque balance at full-scale deflection

For a moving coil galvanometer,

$N B I_g A = C\,\theta$

where:

• $N = 50$
• $B = 0.02\,\text{T}$
• $A = 2\times 10^{-4}\,\text{m}^2$
• $C = 10^{-4}\,\text{N m rad}^{-1}$
• $\theta = 0.2\,\text{rad}$

So,

$I_g = \frac{C\theta}{N B A}$

Substitute values:

$I_g = \frac{10^{-4}\times 0.2}{50\times 0.02\times 2\times 10^{-4}}$

First compute numerator:

$10^{-4}\times 0.2 = 2\times 10^{-5}$

Now denominator:

$50\times 0.02 = 1$

so

$1\times 2\times 10^{-4} = 2\times 10^{-4}$

Thus,

$I_g = \frac{2\times 10^{-5}}{2\times 10^{-4}} = 0.1\,\text{A}$

So the galvanometer gives full-scale deflection at

$I_g = 0.1\,\text{A}$

2. Convert galvanometer into 1 A ammeter

Desired ammeter range:

$I = 1.0\,\text{A}$

Hence current through shunt will be

$I_s = I - I_g = 1.0 - 0.1 = 0.9\,\text{A}$

Galvanometer resistance is

$R_g = 50\,\Omega$

Since galvanometer and shunt are in parallel, voltage across both is same:

$I_g R_g = I_s R_s$

Therefore,

$R_s = \frac{I_g R_g}{I_s}$

Substitute values:

$R_s = \frac{0.1\times 50}{0.9} = \frac{5}{0.9} = 5.56\,\Omega$

3. Final answer

$\boxed{5.56\,\Omega}$