1. Find the full-scale current of the galvanometer

A galvanometer of resistance $G = 100\,\Omega$ is connected in series with $R = 400\,\Omega$ and the combination measures up to $V = 10\,\text{V}.$

So the total resistance is $R_{\text{total}} = 100 + 400 = 500\,\Omega.$

Hence the current for full-scale deflection is $I_g = \frac{V}{R_{\text{total}}} = \frac{10}{500} = 0.02\,\text{A}.$

2. Convert galvanometer into ammeter

To make an ammeter of range $I = 10\,\text{A},$ a shunt resistance $S$ is connected in parallel with the galvanometer.

At full-scale reading:

• current through galvanometer = $I_g = 0.02\,\text{A}$
• current through shunt = $I_s = I - I_g = 10 - 0.02 = 9.98\,\text{A}.$

Since galvanometer and shunt are in parallel, potential difference across them is same: $I_g G = I_s S.$

So, $S = \frac{I_g G}{I_s} = \frac{0.02 \times 100}{9.98} = \frac{2}{9.98} \approx 0.2004\,\Omega.$

Thus, $S \approx 0.2\,\Omega.$

3. Match with the given form

Given, $S = x \times 10^{-2}\,\Omega.$

So, $0.2 = x \times 10^{-2}$ $x = 20.$

4. Correct option

$\boxed{\text{Option B: } 20}$