At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will be

Options

A. `4 sec`
B. `2 sec`
C. `1/2 sec`
D. `sqrt(2) sec`

Explanation

To solve the problem, we need to determine the time period of a magnet oscillating in a magnetic field that has been doubled. Let's break down the solution step by step.

### Step 1: Understand the relationship between frequency and time period
The frequency (f) of oscillation is related to the time period (T) by the formula:
\[ f = \frac{1}{T} \]
Given that the magnet makes 30 oscillations per minute, we first convert this to oscillations per second (Hz):
\[ f = \frac{30 \text{ oscillations}}{60 \text{ seconds}} = 0.5 \text{ Hz} \]

### Step 2: Calculate the initial time period (T1)
Using the relationship between frequency and time period:
\[ T_1 = \frac{1}{f} = \frac{1}{0.5} = 2 \text{ seconds} \]

### Step 3: Use the formula for the time period of oscillation in a magnetic field
The time period of a magnet oscillating in a magnetic field is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where:
- \(I\) is the moment of inertia,
- \(m\) is the magnetic moment,
- \(B\) is the magnetic field strength.

### Step 4: Establish the relationship between the time periods in different magnetic fields
Let \(T_2\) be the time period in the new magnetic field where the magnetic field is doubled (\(B_2 = 2B_1\)). The relationship between the time periods can be expressed as:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} = \sqrt{\frac{2B_1}{B_1}} = \sqrt{2} \]
This implies:
\[ T_2 = T_1 \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ seconds} \]

### Step 5: Final answer
Thus, the time period of the magnet when the magnetic field is doubled is:
\[ T_2 = \sqrt{2} \text{ seconds} \]