A compose needle makes `10` oscillations per minute in the earths horizontal field. `A` bar magnet deflects the needle by `60^(@)` from the magnetic meridian. The frequency of oscillaiton in the deflected position in oscillations per minute is (field due to magnet is perpendicular to `B_(H)`)

Options

A. `5sqrt2`
B. `20sqrt2`
C. `10sqrt2`
D. `10`

Explanation

To solve the problem, we need to determine the frequency of oscillation of a compass needle when it is deflected by a bar magnet. Here are the steps to arrive at the solution:

### Step 1: Understand the Initial Conditions
The compass needle makes 10 oscillations per minute in the Earth's horizontal magnetic field. This means the initial frequency \( N_1 \) is:
\[ N_1 = 10 \text{ oscillations per minute} \]

### Step 2: Analyze the Effect of the Bar Magnet
When a bar magnet is introduced, it deflects the needle by \( 60^\circ \) from the magnetic meridian. The magnetic field due to the bar magnet is perpendicular to the Earth's horizontal magnetic field \( B_H \).

### Step 3: Determine the Resultant Magnetic Field
The resultant magnetic field \( B_R \) when the bar magnet is present can be calculated using vector addition of the Earth's magnetic field \( B_H \) and the magnetic field \( B_M \) due to the bar magnet. Since the angle between them is \( 60^\circ \), we can use the cosine rule:
\[ B_R = \sqrt{B_H^2 + B_M^2 + 2 B_H B_M \cos(60^\circ)} \]
Since \( \cos(60^\circ) = \frac{1}{2} \), this simplifies to:
\[ B_R = \sqrt{B_H^2 + B_M^2 + B_H B_M} \]

### Step 4: Relate Frequency to Magnetic Field
The frequency of oscillation \( N \) is directly proportional to the square root of the magnetic field:
\[ N \propto \sqrt{B} \]
Thus, we can express the new frequency \( N_2 \) in terms of the original frequency \( N_1 \):
\[ \frac{N_2}{N_1} = \sqrt{\frac{B_R}{B_H}} \]

### Step 5: Calculate the Ratio of the Magnetic Fields
From the deflection angle \( 60^\circ \):
\[ \frac{B_M}{B_H} = \tan(60^\circ) = \sqrt{3} \]
Using this, we can express \( B_R \) in terms of \( B_H \):
\[ B_R = B_H \sqrt{1 + \sqrt{3} + 1} = B_H \sqrt{2 + \sqrt{3}} \]

### Step 6: Substitute into the Frequency Equation
Now substituting back into the frequency ratio:
\[ \frac{N_2}{10} = \sqrt{\frac{B_R}{B_H}} = \sqrt{1 + \sqrt{3}} \]
Thus, we find:
\[ N_2 = 10 \sqrt{1 + \sqrt{3}} \]

### Step 7: Calculate the Final Frequency
Now we can calculate \( N_2 \):
\[ N_2 = 10 \sqrt{2} \text{ oscillations per minute} \]

### Conclusion
The frequency of oscillation in the deflected position is:
\[ N_2 = 10 \sqrt{2} \text{ oscillations per minute} \approx 14.14 \text{ oscillations per minute} \]