A galvanometer of resistance `25Omega` is connected to a battery of 2 volt along with a resistance in series. When the value of this resistance is `3000Omega`, a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will be

Options

A. `4512 Omega`
B. `5413 Omega`
C. `2000 Omega`
D. `6000 Omega`

Answer

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

### Step 1: Understand the circuit
We have a galvanometer with a resistance \( G = 25 \Omega \) connected in series with a resistance \( R = 3000 \Omega \) and a battery of voltage \( V = 2 \, \text{V} \). The galvanometer shows a full-scale deflection of 30 units.

### Step 2: Calculate the current for full-scale deflection
The total resistance in the circuit is:
\[ R_{\text{total}} = R + G = 3000 \Omega + 25 \Omega = 3025 \Omega \]
Using Ohm's law, the current \( I \) through the circuit when the galvanometer shows full-scale deflection is:
\[ I = \frac{V}{R_{\text{total}}} = \frac{2 \, \text{V}}{3025 \, \Omega} \approx 0.000661 \, \text{A} \quad (\text{or } 661 \, \mu A) \]

### Step 3: Establish the relationship between current and deflection
Let \( K \) be the figure of merit of the galvanometer. The current \( I_g \) corresponding to full-scale deflection (30 units) can be expressed as:
\[ I_g = K \cdot 30 \]
From the previous step, we have:
\[ I_g = I \quad \text{(for full-scale deflection)} \]
Thus,
\[ K \cdot 30 = 0.000661 \, \text{A} \]
From this, we can find \( K \):
\[ K = \frac{0.000661}{30} \approx 0.00002203 \, \text{A/unit} \]

### Step 4: Calculate the current for the new deflection
Now, we want to reduce the deflection to 20 units. The new current \( I_g' \) will be:
\[ I_g' = K \cdot 20 = 0.00002203 \cdot 20 \approx 0.0004406 \, \text{A} \quad (\text{or } 440.6 \, \mu A) \]

### Step 5: Set up the equation for the new resistance
Using Ohm's law again, we can express the new total resistance \( R' \) needed to achieve this new current:
\[ I_g' = \frac{V}{R' + G} \]
Substituting the values we have:
\[ 0.0004406 = \frac{2}{R' + 25} \]

### Step 6: Solve for the new resistance \( R' \)
Rearranging the equation gives:
\[ R' + 25 = \frac{2}{0.0004406} \]
Calculating the right side:
\[ R' + 25 \approx 4537.5 \, \Omega \]
Thus,
\[ R' \approx 4537.5 - 25 = 4512.5 \, \Omega \]

### Step 7: Conclusion
The new resistance required in series to reduce the deflection to 20 units is approximately:
\[ R' \approx 4512.5 \, \Omega \]