A galvanometer of resistance $$100\Omega$$ when connected in series with $$400\Omega$$ measures a voltage of upto $$10$$ V. The value of resistance required to convert the galvanometer into ammeter to read upto $$10$$ A is $$x \times 10^{-2} \Omega$$. The value of $$x$$ is :

We are given a galvanometer of resistance $$G = 100 \, \Omega$$ that, when connected in series with a resistance of $$400 \, \Omega$$, measures a voltage of up to $$10$$ V.

The full-scale deflection current of the galvanometer, $$I_g$$, can be found as follows.

When the galvanometer is used as a voltmeter with a series resistance $$R_s = 400 \, \Omega$$, the total resistance in the circuit is:

$$R_{total} = G + R_s = 100 + 400 = 500 \, \Omega$$

Using Ohm's law, the full-scale deflection current is:

$$I_g = \frac{V}{R_{total}} = \frac{10}{500} = 0.02 \text{ A}$$

To convert a galvanometer into an ammeter, a low resistance (called a shunt, $$S$$) is connected in parallel with the galvanometer. The shunt diverts most of the current, allowing only $$I_g$$ to pass through the galvanometer. The formula for the shunt resistance is:

$$S = \frac{I_g \cdot G}{I - I_g}$$

This comes from the condition that the voltage across the galvanometer equals the voltage across the shunt: $$I_g \cdot G = (I - I_g) \cdot S$$.

Here, $$I = 10$$ A (the desired full-scale reading of the ammeter), $$I_g = 0.02$$ A, and $$G = 100 \, \Omega$$:

$$S = \frac{0.02 \times 100}{10 - 0.02} = \frac{2}{9.98} \approx 0.2004 \, \Omega$$

Expressed in the required form, $$S \approx 0.2 \, \Omega = 20 \times 10^{-2} \, \Omega$$.

Therefore, $$x = 20$$.

The correct answer is Option (3): 20.