The scale of a galvanometer of resistance `100 ohms` contains 25 divisions. It gives a defelction of one division on passing a current of `4xx 10 ^(-4)` amperes. The resistance in ohms to be added to it, so that it may become a voltmeter of range `2.5` volts is

Options

A. 150
B. 170
C. 110
D. 220

Answer

To solve the problem, we need to determine the resistance to be added to a galvanometer to convert it into a voltmeter with a specified range. Here’s the step-by-step solution:

### Step 1: Understand the given data
- Resistance of the galvanometer (G) = 100 ohms
- Number of divisions on the galvanometer scale = 25
- Current for 1 division = \(4 \times 10^{-4}\) A
- Voltage range of the voltmeter (V) = 2.5 V

### Step 2: Calculate the maximum current (I_max) through the galvanometer
The maximum current that can be measured by the galvanometer when all divisions are used is:
\[ I_{\text{max}} = \text{Number of divisions} \times \text{Current per division} \]
\[ I_{\text{max}} = 25 \times 4 \times 10^{-4} = 100 \times 10^{-4} = 10^{-2} \text{ A} = 0.01 \text{ A} \]

### Step 3: Write the formula for the voltmeter
When the galvanometer is converted into a voltmeter, the total voltage across the galvanometer and the added resistance (R) is given by:
\[ V = I \times R_{\text{eq}} \]
where \(R_{\text{eq}} = G + R\) (the total resistance in the circuit).

### Step 4: Substitute known values into the formula
We know:
- \(V = 2.5 \text{ V}\)
- \(I = I_{\text{max}} = 0.01 \text{ A}\)
- \(G = 100 \text{ ohms}\)

Substituting these values into the voltage formula:
\[ 2.5 = 0.01 \times (100 + R) \]

### Step 5: Solve for R
First, rearranging the equation:
\[ 2.5 = 0.01 \times (100 + R) \]
\[ \Rightarrow 2.5 = 1 + 0.01R \]
\[ \Rightarrow 2.5 - 1 = 0.01R \]
\[ \Rightarrow 1.5 = 0.01R \]
\[ \Rightarrow R = \frac{1.5}{0.01} = 150 \text{ ohms} \]

### Final Answer
The resistance to be added to the galvanometer to convert it into a voltmeter of range 2.5 volts is **150 ohms**.

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