A thin magnetic needle vibrates in the horizontal plane with a period of 4s. The needle is cut into two halves by a plane normal to the magnetic axis of the needle. Then, the period of vibration of each half needle is approximately

Options

A. 4 s
B. 2 s
C. 8 s
D. 1 s

Answer

To solve the problem step by step, we will analyze the situation involving the magnetic needle and its properties before and after it is cut.

### Step 1: Understand the Initial Conditions
The initial period of vibration (T) of the magnetic needle is given as 4 seconds.

### Step 2: Define the Length of the Needle
Let the length of the magnetic needle be \( L \). After cutting the needle into two halves, the length of each half will be \( L/2 \).

### Step 3: Moment of Inertia Calculation
The moment of inertia \( I \) of a thin rod about its end is given by the formula:
\[ I = \frac{1}{3} m L^2 \]
However, since we are considering the magnetic needle as vibrating about its center, we can also use:
\[ I = \frac{1}{12} m L^2 \]
For the full needle, the moment of inertia is:
\[ I = \frac{1}{12} m L^2 \]

### Step 4: Moment of Inertia of Each Half
When the needle is cut into two halves, the mass of each half becomes \( m/2 \) and the length becomes \( L/2 \). The moment of inertia \( I' \) for each half is:
\[ I' = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \cdot \frac{m}{2} \cdot \frac{L^2}{4} = \frac{m L^2}{96} \]

### Step 5: New Period of Vibration
The period of vibration \( T' \) for each half can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I'}{M'B}} \]
Where \( M' = \frac{m}{2} \) is the new mass and \( B \) is the magnetic field (which remains constant).

Substituting \( I' \) and \( M' \):
\[ T' = 2\pi \sqrt{\frac{\frac{m L^2}{96}}{\frac{m}{2} B}} = 2\pi \sqrt{\frac{L^2}{48B}} \]

### Step 6: Relate New Period to Original Period
The original period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{I}{MB}} = 2\pi \sqrt{\frac{\frac{1}{12} m L^2}{m B}} = 2\pi \sqrt{\frac{L^2}{12B}} \]

Now, we can see that:
\[ T' = 2\pi \sqrt{\frac{L^2}{48B}} = \frac{T}{2} \]
since \( \frac{L^2}{48B} = \frac{1}{4} \cdot \frac{L^2}{12B} \).

### Step 7: Calculate the New Period
Given that the original period \( T = 4 \) seconds, we find:
\[ T' = \frac{4}{2} = 2 \text{ seconds} \]

### Conclusion
The period of vibration of each half of the needle is approximately **2 seconds**.