A galvanometer coil has a resistance `90 Omega` and full scale deflection current `10 mA`. A `910 Omega` resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is `0.1 V` the number of divisions on its scale is

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given values
- Resistance of the galvanometer (Rg) = 90 Ω
- Full scale deflection current (Ig) = 10 mA = 0.01 A
- Series resistance (Rs) = 910 Ω
- Least count of the voltmeter (LC) = 0.1 V

### Step 2: Calculate the total resistance
The total resistance (R_total) when the galvanometer is connected in series with the resistance is given by:
\[ R_{\text{total}} = R_g + R_s \]
\[ R_{\text{total}} = 90 \, \Omega + 910 \, \Omega = 1000 \, \Omega \]

### Step 3: Calculate the maximum voltage (V_max)
Using Ohm's law, the maximum voltage (V_max) that can be read by the voltmeter is given by:
\[ V_{\text{max}} = I_g \times R_{\text{total}} \]
\[ V_{\text{max}} = 0.01 \, A \times 1000 \, \Omega = 10 \, V \]

### Step 4: Relate the maximum voltage to the number of divisions
Let the number of divisions on the scale be \( x \). The relationship between the maximum voltage, number of divisions, and least count is:
\[ V_{\text{max}} = x \times LC \]
Substituting the known values:
\[ 10 \, V = x \times 0.1 \, V \]

### Step 5: Solve for the number of divisions (x)
Rearranging the equation to find \( x \):
\[ x = \frac{10 \, V}{0.1 \, V} \]
\[ x = 100 \]

### Final Answer
The number of divisions on the scale of the voltmeter is **100**.

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