A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is `30^(@)` and 15 oscillations minute at a place where dip angle is `60^(@)`. The ratio of total earth's magnetic field at the two places is

Options

A. `3sqrt(3):8`
B. `16:9sqrt(3)`
C. `4:9`
D. `2sqrt(3):9`

Explanation

To solve the problem, we will use the formula for the frequency of oscillation of a magnetic dipole in a magnetic field. The frequency \( f \) of oscillation of a magnet in a magnetic field \( B \) is given by:

\[ f = \frac{1}{2\pi} \sqrt{\frac{mB_h}{I}} \]

where:
- \( m \) is the magnetic moment of the magnet,
- \( B_h \) is the horizontal component of the Earth's magnetic field,
- \( I \) is the moment of inertia of the magnet.

However, since the moment of inertia \( I \) and the magnetic moment \( m \) remain constant for the same magnet, we can express the frequency in terms of the horizontal component of the magnetic field \( B_h \) at two different locations.

### Step 1: Write the frequency equations for both locations
Let \( B_{h1} \) be the horizontal component of the Earth's magnetic field at the first location (dip angle \( 30^\circ \)), and \( B_{h2} \) be the horizontal component at the second location (dip angle \( 60^\circ \)).

From the problem:
- At dip angle \( 30^\circ \), the frequency \( f_1 = 20 \) oscillations per minute = \( \frac{20}{60} \) Hz = \( \frac{1}{3} \) Hz.
- At dip angle \( 60^\circ \), the frequency \( f_2 = 15 \) oscillations per minute = \( \frac{15}{60} \) Hz = \( \frac{1}{4} \) Hz.

### Step 2: Relate the frequencies to the horizontal components of the magnetic field
Using the frequency equation:

\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{mB_{h1}}{I}} \quad \text{and} \quad f_2 = \frac{1}{2\pi} \sqrt{\frac{mB_{h2}}{I}} \]

### Step 3: Square both frequency equations
Squaring both equations gives:

\[ f_1^2 = \frac{mB_{h1}}{4\pi^2 I} \quad \text{and} \quad f_2^2 = \frac{mB_{h2}}{4\pi^2 I} \]

### Step 4: Set up the ratio of the two frequencies
Taking the ratio of the two equations:

\[ \frac{f_1^2}{f_2^2} = \frac{B_{h1}}{B_{h2}} \]

### Step 5: Substitute the values of frequencies
Substituting the values of \( f_1 \) and \( f_2 \):

\[ \frac{\left(\frac{1}{3}\right)^2}{\left(\frac{1}{4}\right)^2} = \frac{B_{h1}}{B_{h2}} \]

Calculating the left side:

\[ \frac{\frac{1}{9}}{\frac{1}{16}} = \frac{16}{9} \]

### Step 6: Conclusion
Thus, the ratio of the horizontal components of the Earth's magnetic field at the two locations is:

\[ \frac{B_{h1}}{B_{h2}} = \frac{16}{9} \]

### Final Answer
The ratio of the total Earth's magnetic field at the two places is \( \frac{16}{9} \).

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