A battery of emf 12 V and internal resistance `2Omega` is connected in series with a tangent galvanometer of resistance `4Omega` . The deflection is `60^@` when the plane of the coil is along the magnetic meridian . To get a deflection of `30^@`, the resistance to be connected in series with the tangent galvanometer is

Options

A. `12Omega`
B. `20Omega`
C. `10Omega`
D. `5Omega`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the situation
We have a battery with an EMF of 12 V and an internal resistance of 2 Ω connected in series with a tangent galvanometer that has a resistance of 4 Ω. The initial deflection is 60°, and we want to find the additional resistance needed to achieve a deflection of 30°.

### Step 2: Calculate the total resistance in the first case
The total resistance when the deflection is 60° can be calculated as:
\[ R_{\text{total1}} = R_{\text{internal}} + R_{\text{galvanometer}} = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega \]

### Step 3: Use the relationship between current and deflection
The current \( I \) through the galvanometer is given by:
\[ I = \frac{E}{R_{\text{total}}} \]
where \( E \) is the EMF of the battery.

For the first case (60° deflection):
\[ I_1 = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A} \]

### Step 4: Set up the relationship for the second case
For the second case (30° deflection), we need to find the new total resistance \( R_{\text{total2}} \). Let \( R \) be the additional resistance we need to add. The new total resistance will be:
\[ R_{\text{total2}} = R_{\text{total1}} + R = 6 \, \Omega + R \]

### Step 5: Use the tangent law relationship
From the tangent galvanometer principle, we know:
\[ \frac{\tan \theta_2}{\tan \theta_1} = \frac{R_1}{R_2} \]
Substituting the known values:
\[ \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{6 \, \Omega}{6 + R} \]

### Step 6: Calculate the tangent values
Using the known values of tangent:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan 60^\circ = \sqrt{3} \]
Substituting these values into the equation:
\[ \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{6}{6 + R} \]
This simplifies to:
\[ \frac{1}{3} = \frac{6}{6 + R} \]

### Step 7: Cross-multiply and solve for R
Cross-multiplying gives:
\[ 1 \cdot (6 + R) = 3 \cdot 6 \]
\[ 6 + R = 18 \]
\[ R = 18 - 6 = 12 \, \Omega \]

### Conclusion
The additional resistance that needs to be connected in series with the tangent galvanometer to achieve a deflection of 30° is **12 Ω**.