A galvanometer of resistance `25Omega` is connected to a battery of 2V along with a resistance of `3000Omega`. In this case, a full - scale deflection of 30 units is obtained in the galvanometer. In order or reduce this deflection to 10 units, how much more resistance `("in "Omega)` should be added to the circuit in series?

Answer

To solve the problem step by step, we will use the relationship between the current through the galvanometer and the deflection it produces. The deflection is directly proportional to the current flowing through the galvanometer.

### Step 1: Define the initial conditions
- Given:
- Resistance of galvanometer, \( R_g = 25 \, \Omega \)
- External resistance, \( R_1 = 3000 \, \Omega \)
- Voltage of the battery, \( V = 2 \, V \)
- Full-scale deflection (initial), \( D_1 = 30 \) units

### Step 2: Calculate the total resistance for the initial condition
The total resistance in the circuit when the galvanometer shows full-scale deflection is:
\[ R_{\text{total}_1} = R_g + R_1 = 25 \, \Omega + 3000 \, \Omega = 3025 \, \Omega \]

### Step 3: Calculate the current for the initial condition
Using Ohm's law, the current \( I_1 \) through the circuit when the deflection is 30 units is given by:
\[ I_1 = \frac{V}{R_{\text{total}_1}} = \frac{2 \, V}{3025 \, \Omega} \approx 0.000661 \, A \]

### Step 4: Define the new conditions
- New deflection (target), \( D_2 = 10 \) units

### Step 5: Set up the equation for the new condition
Let \( R \) be the additional resistance needed to achieve the new deflection. The total resistance in this case will be:
\[ R_{\text{total}_2} = R_g + R_1 + R = 25 \, \Omega + 3000 \, \Omega + R = 3025 \, \Omega + R \]

### Step 6: Relate the new current to the new deflection
Since the deflection is directly proportional to the current, we can set up the ratio:
\[ \frac{D_1}{D_2} = \frac{I_1}{I_2} \]
Where \( I_2 \) is the current for the new deflection. Thus:
\[ \frac{30}{10} = \frac{I_1}{I_2} \implies I_2 = \frac{I_1}{3} \]

### Step 7: Calculate the new current
Substituting \( I_1 \):
\[ I_2 = \frac{0.000661 \, A}{3} \approx 0.0002203 \, A \]

### Step 8: Set up the equation for the new current
Using Ohm's law for the new condition:
\[ I_2 = \frac{V}{R_{\text{total}_2}} \implies 0.0002203 \, A = \frac{2 \, V}{3025 \, \Omega + R} \]

### Step 9: Solve for \( R \)
Rearranging gives:
\[ 3025 \, \Omega + R = \frac{2 \, V}{0.0002203 \, A} \]
Calculating the right side:
\[ 3025 + R = \frac{2}{0.0002203} \approx 9076.5 \, \Omega \]
Thus:
\[ R \approx 9076.5 - 3025 \approx 6051.5 \, \Omega \]

### Step 10: Conclusion
The additional resistance required to reduce the deflection from 30 units to 10 units is approximately \( 6051.5 \, \Omega \).

### Final Answer:
The additional resistance needed is approximately \( 6052 \, \Omega \) (rounded to the nearest whole number).

---