A magnet makes `40` oscillations per minute at a place having magnetic field intensity of `0.1xx10^(-5)T`. At another place, it takes 2.5 sec to complete one vibrating. The value of earth's horizontal field at that place is

Options

A. `0.25xx10^(-6)T`
B. `0.36xx10^(6)T`
C. `0.66xx10^(-8)T`
D. `1.2xx10^(-6)T`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the frequency and time period at the first location.
Given that the magnet makes 40 oscillations per minute, we can convert this to frequency (f1) in Hertz (Hz):
\[ f_1 = \frac{40 \text{ oscillations}}{1 \text{ minute}} = \frac{40}{60} \text{ Hz} = \frac{2}{3} \text{ Hz} \]

### Step 2: Calculate the time period (T1) at the first location.
The time period (T1) is the reciprocal of the frequency:
\[ T_1 = \frac{1}{f_1} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \text{ seconds} \]

### Step 3: Identify the time period at the second location.
We are given that at the second location, it takes 2.5 seconds to complete one vibration:
\[ T_2 = 2.5 \text{ seconds} \]

### Step 4: Use the relationship between time periods and magnetic fields.
The relationship between the time periods and the magnetic fields is given by the formula:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_{H2}}{B_1}} \]
Where:
- \(B_{H2}\) is the horizontal magnetic field at the second location.
- \(B_1 = 0.1 \times 10^{-5} \text{ T}\) is the magnetic field at the first location.

### Step 5: Substitute the known values into the equation.
Substituting \(T_1\), \(T_2\), and \(B_1\) into the equation:
\[ \frac{\frac{3}{2}}{2.5} = \sqrt{\frac{B_{H2}}{0.1 \times 10^{-5}}} \]

### Step 6: Simplify the left side of the equation.
Calculating the left side:
\[ \frac{3}{2} \div 2.5 = \frac{3}{2} \times \frac{1}{2.5} = \frac{3}{5} = 0.6 \]

### Step 7: Square both sides to eliminate the square root.
\[ (0.6)^2 = \frac{B_{H2}}{0.1 \times 10^{-5}} \]
\[ 0.36 = \frac{B_{H2}}{0.1 \times 10^{-5}} \]

### Step 8: Solve for \(B_{H2}\).
Multiplying both sides by \(0.1 \times 10^{-5}\):
\[ B_{H2} = 0.36 \times 0.1 \times 10^{-5} = 0.36 \times 10^{-6} \text{ T} \]

### Final Answer:
The value of the earth's horizontal field at that place is:
\[ B_{H2} = 0.36 \times 10^{-6} \text{ T} \]