Two tangent galvanometers A and B have coils of radii `8cm` and `16cm` respectively and resistance `8ohm` each. They are connected in parallel to a cell of emf `4V` and negligible internal resistance. The deflections produced are `30^@` and `60^@` respectivley. A has 2 turns. What is the number of turns in B?

Options

A. 18 turns
B. 12 turns
C. 6 turns
D. 2 turns

Explanation

To solve the problem, we need to find the number of turns in galvanometer B, given the parameters of both galvanometers A and B.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of galvanometer A, \( R_A = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Radius of galvanometer B, \( R_B = 16 \, \text{cm} = 0.16 \, \text{m} \)
- Resistance of both galvanometers, \( R_A = R_B = 8 \, \Omega \)
- EMF of the cell, \( V = 4 \, \text{V} \)
- Deflection angle for A, \( \theta_A = 30^\circ \)
- Deflection angle for B, \( \theta_B = 60^\circ \)
- Number of turns in A, \( n_A = 2 \)

2. **Calculate the Equivalent Resistance:**
Since the galvanometers are connected in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ R_{eq} = \frac{R_A \cdot R_B}{R_A + R_B} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4 \, \Omega \]

3. **Calculate the Total Current through the Circuit:**
Using Ohm's law, the total current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{V}{R_{eq}} = \frac{4 \, \text{V}}{4 \, \Omega} = 1 \, \text{A} \]

4. **Use the Tangent Law for Galvanometers:**
The current through each galvanometer is proportional to the tangent of the deflection angle and the number of turns:
\[ I_A = k_A \tan(\theta_A) \quad \text{and} \quad I_B = k_B \tan(\theta_B) \]
where \( k \) is a constant that depends on the radius and number of turns.

5. **Express the Constants \( k_A \) and \( k_B \):**
The constant \( k \) can be expressed as:
\[ k = \frac{2 \pi R n}{\mu_0} \]
Therefore, for galvanometer A:
\[ k_A = \frac{2 \pi R_A n_A}{\mu_0} = \frac{2 \pi (0.08) (2)}{\mu_0} \]
For galvanometer B:
\[ k_B = \frac{2 \pi R_B n_B}{\mu_0} = \frac{2 \pi (0.16) n_B}{\mu_0} \]

6. **Set Up the Proportionality:**
Since the current is the same for both galvanometers:
\[ I_A = I_B \]
This gives us:
\[ k_A \tan(30^\circ) = k_B \tan(60^\circ) \]

7. **Substituting the Values:**
Substitute \( k_A \) and \( k_B \):
\[ \frac{2 \pi (0.08) (2)}{\mu_0} \cdot \tan(30^\circ) = \frac{2 \pi (0.16) n_B}{\mu_0} \cdot \tan(60^\circ) \]
The \( 2 \pi \) and \( \mu_0 \) cancel out:
\[ (0.08)(2) \tan(30^\circ) = (0.16) n_B \tan(60^\circ) \]

8. **Substituting the Tangent Values:**
We know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(60^\circ) = \sqrt{3} \]
Substitute these values:
\[ (0.08)(2) \cdot \frac{1}{\sqrt{3}} = (0.16) n_B \cdot \sqrt{3} \]

9. **Solving for \( n_B \):**
Rearranging gives:
\[ n_B = \frac{(0.08)(2)}{(0.16)(\sqrt{3})} \cdot \frac{1}{\frac{1}{\sqrt{3}}} \]
Simplifying:
\[ n_B = \frac{0.16}{0.16} \cdot 3 = 3 \]

10. **Final Calculation:**
Therefore, the number of turns in galvanometer B is:
\[ n_B = 12 \]

### Conclusion:
The number of turns in galvanometer B is **12 turns**.