A magnet oscillating in a horizontal plane has a time period of 2 seconds at a place where the angle of dip is `30^(@)` and 3 seconds at another place where the angle of dip is `60^(@)`. The retio of resultant magnetic field at the two places is

Options

A. `(4sqrt(3))/(7)`
B. `(4)/(9sqrt(3))`
C. `(9)/(4sqrt(3))`
D. `(9)/(sqrt(3))`

Answer

To solve the problem, we need to find the ratio of the resultant magnetic fields at two different places where a magnet oscillates with given time periods and angles of dip.

### Step 1: Understand the relationship between time period and magnetic field
The time period \( T \) of a magnet oscillating in a magnetic field is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{M B \cos \theta}} \]

where:
- \( I \) is the moment of inertia,
- \( M \) is the mass of the magnet,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle of dip.

From this formula, we can see that the time period \( T \) is inversely proportional to the square root of \( B \cos \theta \).

### Step 2: Set up the ratio of time periods
Let \( T_1 = 2 \) seconds at angle of dip \( \theta_1 = 30^\circ \) and \( T_2 = 3 \) seconds at angle of dip \( \theta_2 = 60^\circ \).

Using the relationship established, we can write:

\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2 \cos \theta_2}{B_1 \cos \theta_1}} \]

### Step 3: Substitute the known values
Substituting the values of \( T_1 \), \( T_2 \), \( \theta_1 \), and \( \theta_2 \):

\[ \frac{2}{3} = \sqrt{\frac{B_2 \cos 60^\circ}{B_1 \cos 30^\circ}} \]

### Step 4: Square both sides
Squaring both sides gives:

\[ \left(\frac{2}{3}\right)^2 = \frac{B_2 \cos 60^\circ}{B_1 \cos 30^\circ} \]

This simplifies to:

\[ \frac{4}{9} = \frac{B_2 \cdot \frac{1}{2}}{B_1 \cdot \frac{\sqrt{3}}{2}} \]

### Step 5: Simplify the equation
This can be simplified to:

\[ \frac{4}{9} = \frac{B_2}{B_1} \cdot \frac{1}{\sqrt{3}} \]

### Step 6: Rearranging the equation
Rearranging gives:

\[ \frac{B_2}{B_1} = \frac{4}{9} \cdot \sqrt{3} \]

### Step 7: Finding the ratio \( \frac{B_1}{B_2} \)
Taking the reciprocal to find \( \frac{B_1}{B_2} \):

\[ \frac{B_1}{B_2} = \frac{9}{4\sqrt{3}} \]

### Final Answer
Thus, the ratio of the resultant magnetic field at the two places is:

\[ \frac{B_1}{B_2} = \frac{9}{4\sqrt{3}} \]