The maximum current in a galvanometer can be 10 mA. Its resistance is `10 Omega`. To convert it into an ammeter of 1 A, what resistance should be connected in parallel with galvanometer (in `10^(-1)Omega)`?

Explanation

To solve the problem of converting a galvanometer into an ammeter, we need to determine the resistance that should be connected in parallel with the galvanometer. Here’s a step-by-step solution:

### Step 1: Understand the Given Values
- Maximum current in the galvanometer, \( I_g = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)
- Resistance of the galvanometer, \( R_g = 10 \, \Omega \)
- Desired maximum current for the ammeter, \( I = 1 \, \text{A} \)

### Step 2: Use the Formula for Shunt Resistance
When a galvanometer is converted into an ammeter, a shunt resistance \( S \) is connected in parallel. The relationship between the currents is given by:
\[ I = I_g + I_s \]
where \( I_s \) is the current through the shunt. The potential difference across both the galvanometer and the shunt is the same:
\[ I_g R_g = I_s S \]

### Step 3: Express \( I_s \) in Terms of \( I \) and \( I_g \)
From the first equation, we can express \( I_s \):
\[ I_s = I - I_g \]

### Step 4: Substitute \( I_s \) into the Voltage Equation
Substituting \( I_s \) into the voltage equation gives:
\[ I_g R_g = (I - I_g) S \]

### Step 5: Rearrange to Solve for \( S \)
Rearranging the equation to solve for \( S \):
\[ S = \frac{I_g R_g}{I - I_g} \]

### Step 6: Substitute the Known Values
Now, substitute \( I_g = 10 \times 10^{-3} \, \text{A} \), \( R_g = 10 \, \Omega \), and \( I = 1 \, \text{A} \):
\[ S = \frac{(10 \times 10^{-3}) \times 10}{1 - (10 \times 10^{-3})} \]

### Step 7: Calculate the Values
Calculating the numerator:
\[ (10 \times 10^{-3}) \times 10 = 0.1 \]
Calculating the denominator:
\[ 1 - 10 \times 10^{-3} = 1 - 0.01 = 0.99 \]
Thus,
\[ S = \frac{0.1}{0.99} \approx 0.10101 \, \Omega \]

### Step 8: Convert to the Required Format
Since the problem asks for the answer in \( 10^{-1} \, \Omega \):
\[ S \approx 0.10101 \, \Omega \approx 10^{0} \, \Omega \text{ (approximately)} \]
Thus, we can express it as:
\[ S \approx 1.01 \times 10^{-1} \, \Omega \]

### Final Answer
To convert the galvanometer into an ammeter of 1 A, the resistance that should be connected in parallel with the galvanometer is approximately:
\[ S \approx 10^{-1} \, \Omega \]