When a resistance of 100`Omega` is connected in series with a galvanometer of resistance R, then its range is V. To double its range, a resistance of 1000`Omega` is connected in series. Find the value of R.

Options

A. 700`Omega`
B. 800`Omega`
C. 900`Omega`
D. 100`Omega`

Answer

To solve the problem step by step, we will derive the equations based on the given information about the galvanometer and the resistances in series.

### Step 1: Understand the initial setup
When a resistance of 100 Ω is connected in series with a galvanometer of resistance R, the total resistance in the circuit is \( R + 100 \) Ω. The range of the galvanometer is given as V, which can be expressed in terms of the full-scale deflection current \( I_G \):

\[ V = I_G \times (R + 100) \tag{1} \]

### Step 2: Analyze the second setup
To double the range, a resistance of 1000 Ω is added in series. The new total resistance becomes \( R + 100 + 1000 = R + 1100 \) Ω. The new range is then:

\[ 2V = I_G \times (R + 1100) \tag{2} \]

### Step 3: Substitute equation (1) into equation (2)
From equation (1), we know that \( V = I_G \times (R + 100) \). Therefore, we can substitute this into equation (2):

\[ 2(I_G \times (R + 100)) = I_G \times (R + 1100) \]

### Step 4: Simplify the equation
Now, we can simplify the equation:

\[ 2I_G(R + 100) = I_G(R + 1100) \]

Assuming \( I_G \neq 0 \), we can divide both sides by \( I_G \):

\[ 2(R + 100) = R + 1100 \]

### Step 5: Expand and rearrange the equation
Expanding the left side gives us:

\[ 2R + 200 = R + 1100 \]

Now, rearranging the equation to isolate R:

\[ 2R - R = 1100 - 200 \]

\[ R = 900 \, \Omega \]

### Conclusion
The value of the resistance \( R \) of the galvanometer is \( 900 \, \Omega \).

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