The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 150 ms. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles toghether. The time period of this combination (in ms) will be

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understanding the Initial Conditions
The original magnet has a time period \( T = 150 \, \text{ms} \). The time period of a magnet in a vibration magnetometer is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{M g}} \]
where \( I \) is the moment of inertia, \( M \) is the mass of the magnet, and \( g \) is the acceleration due to gravity.

### Step 2: Cutting the Magnet
The magnet is cut into three equal parts. When we cut the magnet along its length, each part will have:
- Mass \( M' = \frac{M}{3} \)
- Length \( L' = \frac{L}{3} \)

### Step 3: Calculating the Moment of Inertia of Each Part
The moment of inertia of a magnet about its end is given by:
\[ I = \frac{1}{3} m L^2 \]
For each part, the moment of inertia \( I' \) will be:
\[ I' = \frac{1}{3} \left(\frac{M}{3}\right) \left(\frac{L}{3}\right)^2 = \frac{1}{3} \cdot \frac{M}{3} \cdot \frac{L^2}{9} = \frac{M L^2}{81} \]

### Step 4: Total Moment of Inertia for the Combined Magnet
When the three parts are stacked together with like poles together, the total moment of inertia \( I_{total} \) will be:
\[ I_{total} = 3 \cdot I' = 3 \cdot \frac{M L^2}{81} = \frac{M L^2}{27} \]

### Step 5: Finding the New Time Period
Now, we can find the new time period \( T' \) using the formula for the time period:
\[ T' = 2\pi \sqrt{\frac{I_{total}}{M g}} = 2\pi \sqrt{\frac{\frac{M L^2}{27}}{M g}} = 2\pi \sqrt{\frac{L^2}{27g}} = \frac{2\pi L}{\sqrt{27g}} \]
Since the original time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{I}{M g}} = 2\pi \sqrt{\frac{\frac{1}{3} M L^2}{M g}} = 2\pi \sqrt{\frac{L^2}{3g}} \]

### Step 6: Relating the Two Time Periods
Now, we can relate \( T' \) and \( T \):
\[ \frac{T'}{T} = \sqrt{\frac{I_{total}}{I}} = \sqrt{\frac{\frac{M L^2}{27}}{\frac{1}{3} M L^2}} = \sqrt{\frac{3}{27}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]
Thus, we have:
\[ T' = \frac{T}{3} = \frac{150 \, \text{ms}}{3} = 50 \, \text{ms} \]

### Final Answer
The time period of the combination will be \( 50 \, \text{ms} \).

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