A micrometer has a resistance of `100 Omega` and full scale deflection current of `50 muA`. How can it be made to work as an ammeter of range `5mA`?

Explanation

To convert a micrometer (galvanometer) into an ammeter with a specified range, we need to use a shunt resistor. The steps to solve this problem are as follows:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer, \( R_G = 100 \, \Omega \)
- Full scale deflection current of the galvanometer, \( I_G = 50 \, \mu A = 50 \times 10^{-6} \, A \)
- Desired range of the ammeter, \( I = 5 \, mA = 5 \times 10^{-3} \, A \)

2. **Understand the Configuration:**
- When converting a galvanometer into an ammeter, a shunt resistor \( S \) is connected in parallel with the galvanometer. The total current \( I \) splits between the galvanometer and the shunt resistor.

3. **Apply Kirchhoff's Law:**
- The current through the galvanometer is \( I_G \) and the current through the shunt resistor is \( I - I_G \).
- The voltage across the galvanometer and the shunt resistor must be equal:
\[ I_G \cdot R_G = (I - I_G) \cdot S \]

4. **Substitute Known Values:**
- Substitute \( I_G \), \( R_G \), and \( I \) into the equation:
\[ (50 \times 10^{-6}) \cdot 100 = (5 \times 10^{-3} - 50 \times 10^{-6}) \cdot S \]
- Simplifying the left side:
\[ 5 \times 10^{-4} = (5 \times 10^{-3} - 50 \times 10^{-6}) \cdot S \]
- Calculate \( 5 \times 10^{-3} - 50 \times 10^{-6} = 5 \times 10^{-3} - 0.000050 = 0.00495 \, A \).

5. **Rearranging the Equation:**
- The equation now looks like:
\[ 5 \times 10^{-4} = 0.00495 \cdot S \]
- Solving for \( S \):
\[ S = \frac{5 \times 10^{-4}}{0.00495} \approx 0.10101 \, \Omega \]

6. **Final Calculation:**
- The value of \( S \) can be approximated to \( 1 \, \Omega \) for practical purposes.

### Conclusion:
To convert the galvanometer into an ammeter with a range of 5 mA, a shunt resistor of approximately \( 1 \, \Omega \) should be connected in parallel with the galvanometer.