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. 1100
B. 1000
C. 900
D. 800

Answer

To solve the problem, we need to analyze the situation step by step.

### 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:
\[ R_{\text{total1}} = 100 + R \]

### Step 2: Write the Expression for Current
Using Ohm's Law, the current (I) flowing through the circuit when the voltage (V) is applied can be expressed as:
\[ I = \frac{V}{R_{\text{total1}}} = \frac{V}{100 + R} \]
Let’s call this Equation (1).

### Step 3: Analyze the Second Setup
To double the range, a resistance of 1000 Ω is connected in series with the galvanometer. The total resistance in this case becomes:
\[ R_{\text{total2}} = 1000 + 100 + R = 1100 + R \]

### Step 4: Write the Expression for Current in the Second Setup
In this case, the new voltage range is 2V, so the current can be expressed as:
\[ I = \frac{2V}{R_{\text{total2}}} = \frac{2V}{1100 + R} \]
Let’s call this Equation (2).

### Step 5: Set the Two Current Equations Equal
Since the current remains the same in both cases (because they are in series), we can set the two equations for current equal to each other:
\[ \frac{V}{100 + R} = \frac{2V}{1100 + R} \]

### Step 6: Cancel V from Both Sides
Assuming V is not zero, we can cancel it from both sides:
\[ \frac{1}{100 + R} = \frac{2}{1100 + R} \]

### Step 7: Cross-Multiply to Solve for R
Cross-multiplying gives us:
\[ 1100 + R = 2(100 + R) \]
Expanding the right side:
\[ 1100 + R = 200 + 2R \]

### Step 8: Rearranging the Equation
Now, we rearrange the equation to isolate R:
\[ 1100 - 200 = 2R - R \]
This simplifies to:
\[ 900 = R \]

### Step 9: Conclusion
Thus, the value of R is:
\[ R = 900 \, \Omega \]