A magnet makes 40 oscillations per minute at a place having magnetic field of `0.1 xx 10^(-5) T`. At another place, " it take `2.5 s` to complete one vibration. If the value of the earth's horizontal field at that place is `y xx 10^(-6) T`. then find `y`.

Answer

To solve the problem, we need to find the value of \( y \) given the oscillation frequencies and magnetic fields at two different locations. Here’s the step-by-step solution:

### Step 1: Understand the given data
- At the first location, the magnet makes 40 oscillations per minute.
- The magnetic field at this location, \( B_1 = 0.1 \times 10^{-5} \, T \).
- At the second location, the time taken for one vibration is \( t_2 = 2.5 \, s \).
- We need to find the horizontal magnetic field at the second location, represented as \( B_2 = y \times 10^{-6} \, T \).

### Step 2: Convert oscillations per minute to time period
The frequency of oscillation at the first location can be calculated as:
\[ f_1 = \frac{40 \, \text{oscillations}}{1 \, \text{minute}} = \frac{40}{60} \, \text{oscillations per second} = \frac{2}{3} \, \text{Hz} \]
The time period \( T_1 \) is the reciprocal of frequency:
\[ T_1 = \frac{1}{f_1} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \, s = 1.5 \, s \]

### Step 3: Use the relationship between time periods and magnetic fields
The relationship between the time periods and magnetic fields is given by:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \]
Substituting the known values:
\[ \frac{1.5}{2.5} = \sqrt{\frac{B_2}{0.1 \times 10^{-5}}} \]

### Step 4: Simplify the left side
Calculating the left side:
\[ \frac{1.5}{2.5} = \frac{3}{5} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \left(\frac{3}{5}\right)^2 = \frac{B_2}{0.1 \times 10^{-5}} \]
\[ \frac{9}{25} = \frac{B_2}{0.1 \times 10^{-5}} \]

### Step 6: Solve for \( B_2 \)
Rearranging gives:
\[ B_2 = 0.1 \times 10^{-5} \times \frac{9}{25} \]
Calculating \( B_2 \):
\[ B_2 = 0.1 \times 9 \times 10^{-5} \div 25 = \frac{0.9 \times 10^{-5}}{25} = 0.036 \times 10^{-5} \, T \]

### Step 7: Express \( B_2 \) in terms of \( y \)
We know that \( B_2 = y \times 10^{-6} \, T \). Thus:
\[ 0.036 \times 10^{-5} = y \times 10^{-6} \]
Converting \( 0.036 \times 10^{-5} \) to \( 10^{-6} \):
\[ 0.036 \times 10^{-5} = 0.36 \times 10^{-6} \]
So, we have:
\[ y = 0.36 \]

### Final Answer
The value of \( y \) is:
\[ \boxed{0.36} \]