To solve the problem, we need to determine the resistance that must be added in series with the tangent galvanometer to achieve a deflection of 30 degrees, given that the initial deflection is 60 degrees.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- EMF of the battery, \( E = 12 \, \text{V} \)
- Internal resistance of the battery, \( r = 2 \, \Omega \)
- Resistance of the tangent galvanometer, \( R_g = 4 \, \Omega \)
- Initial deflection angle, \( \theta_1 = 60^\circ \)
- Required deflection angle, \( \theta_2 = 30^\circ \)
2. **Calculate the Initial Total Resistance:**
The total resistance in the circuit when the galvanometer is connected is:
\[
R_{\text{total}} = r + R_g = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega
\]
3. **Relate Current to Deflection:**
The current through the galvanometer is proportional to the tangent of the angle of deflection:
\[
I \propto \tan(\theta)
\]
Therefore, we can write:
\[
\frac{I_1}{I_2} = \frac{\tan(\theta_1)}{\tan(\theta_2)}
\]
4. **Calculate the Tangents of the Angles:**
- For \( \theta_1 = 60^\circ \):
\[
\tan(60^\circ) = \sqrt{3}
\]
- For \( \theta_2 = 30^\circ \):
\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\]
5. **Set Up the Current Ratio:**
Using the values of the tangents:
\[
\frac{I_1}{I_2} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = 3
\]
This means that the current \( I_1 \) is three times the current \( I_2 \).
6. **Relate Currents to Resistances:**
The current can also be expressed in terms of EMF and resistance:
\[
I_1 = \frac{E}{R_{\text{total}}} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}
\]
Therefore, for \( I_2 \):
\[
I_2 = \frac{I_1}{3} = \frac{2 \, \text{A}}{3} = \frac{2}{3} \, \text{A}
\]
7. **Calculate the New Total Resistance for \( I_2 \):**
Using Ohm's law:
\[
I_2 = \frac{E}{R_{\text{total}}'} \implies R_{\text{total}}' = \frac{E}{I_2} = \frac{12 \, \text{V}}{\frac{2}{3} \, \text{A}} = 18 \, \Omega
\]
8. **Determine the Additional Resistance Needed:**
The new total resistance \( R_{\text{total}}' \) includes the additional resistance \( R \):
\[
R_{\text{total}}' = R_{\text{total}} + R \implies 18 \, \Omega = 6 \, \Omega + R
\]
Solving for \( R \):
\[
R = 18 \, \Omega - 6 \, \Omega = 12 \, \Omega
\]
### Final Answer:
The resistance to be connected in series with the tangent galvanometer is \( 12 \, \Omega \).
