1. Current without ammeter

The circuit has a resistance of $50\,\Omega$ connected to a battery of $5\,\text{V}$.

So the true current is

2. Condition for ammeter

When an ammeter is inserted in series, its internal resistance should be very small so that the current does not change appreciably.

The measured current must be within $1\%$ of the original current $0.1\,\text{A}$.

Thus current with ammeter, $I$, should satisfy

3. Maximum allowable ammeter resistance

If the ammeter has resistance $R_A$, then total series resistance becomes

So current becomes

We need

Solving,

Hence,

So the ammeter resistance must be at most about $0.5\,\Omega$.

4. Find equivalent resistance of galvanometer + shunt combination

Given galvanometer resistance

To make an ammeter, a small shunt resistance must be connected in parallel with the galvanometer.

Now check options:

• A: $r_s = 0.5\,\Omega$ in parallel with galvanometer

  $$R_A = \frac{100\times 0.5}{100+0.5} = \frac{50}{100.5} \approx 0.498\,\Omega$$

  This is less than $0.505\,\Omega$, so acceptable.

• B: $0.5\,\Omega$ in series with galvanometer

  $$R_A = 100 + 0.5 = 100.5\,\Omega$$

  Much too large.

• C: $1\,\Omega$ in series with galvanometer

  $$R_A = 101\,\Omega$$

  Much too large.

• D: $1\,\Omega$ in parallel with galvanometer

  $$R_A = \frac{100\times 1}{100+1} = \frac{100}{101} \approx 0.990\,\Omega$$

  This is greater than $0.505\,\Omega$, so not acceptable.

5. Verification for option A

With $R_A \approx 0.498\,\Omega$, total resistance is

Current is

This differs from $0.1\,\text{A}$ by about

So it is within $1\%$.

Therefore, the correct choice is A.