A galvanometer of resistance `1000Omega` gives full-scale deflection for a current of 1mA. To measure a P.D of 10V, the resistance to be connected with the galvanometer is

Options

A. `9kOmega` in series
B. `9kOmega` is parallel
C. `10kOmega` in series
D. `1kOmega` parallel.

Answer

To solve the problem, we need to determine the resistance that should be connected in series with the galvanometer to measure a potential difference (P.D.) of 10V. Here are the steps to arrive at the solution:

### Step 1: Understand the given values
- Resistance of the galvanometer, \( R_g = 1000 \, \Omega \)
- Full-scale deflection current, \( I_g = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)
- Desired potential difference, \( V = 10 \, \text{V} \)

### Step 2: Set up the equation for the total resistance
When we connect a resistance \( R \) in series with the galvanometer, the total resistance \( R_{total} \) is given by:
\[ R_{total} = R_g + R \]

### Step 3: Apply Ohm's Law
According to Ohm's Law, the potential difference across the total resistance is:
\[ V = I \times R_{total} \]
Substituting the known values:
\[ 10 = (1 \times 10^{-3}) \times (1000 + R) \]

### Step 4: Solve for \( R \)
Rearranging the equation gives:
\[ 10 = 1 \times 10^{-3} \times (1000 + R) \]
Dividing both sides by \( 1 \times 10^{-3} \):
\[ 10 \times 10^{3} = 1000 + R \]
\[ 10000 = 1000 + R \]
Now, isolate \( R \):
\[ R = 10000 - 1000 \]
\[ R = 9000 \, \Omega \]

### Step 5: Convert to kilo-ohms
\[ R = 9 \, k\Omega \]

### Conclusion
The resistance to be connected with the galvanometer is \( 9 \, k\Omega \).