Question:

A wire of length 0.1\(\pi\) m having linear density \( \frac{1}{\pi} \) kg m\(^{-1}\) is bent in the form of a ring. The moment of inertia of the ring about an axis passing through its centre and perpendicular to its plane is

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For a ring, the moment of inertia about an axis through its center and perpendicular to its plane is \( I = m r^2 \). First find the radius using the circumference \( 2\pi r = l \), and the mass using \( m = \lambda l \).
Updated On: May 19, 2025
  • 0.50 \(\pi \times 10^{-3}\) kg m\(^2\)
  • 0.25 \(\pi \times 10^{-2}\) kg m\(^2\)
  • 0.50 \(\times 10^{-3}\) kg m\(^2\)
  • 0.25 \(\times 10^{-3}\) kg m\(^2\)
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The Correct Option is D

Approach Solution - 1

First, calculate the radius of the ring. The wire of length \( l = 0.1\pi \) m is bent into a ring, so the circumference of the ring is: \[ 2\pi r = 0.1\pi \quad \Rightarrow \quad r = \frac{0.1\pi}{2\pi} = 0.05 \, \text{m} \] Next, calculate the mass of the ring. The linear density is \( \lambda = \frac{1}{\pi} \) kg m\(^{-1}\), so: \[ m = \lambda \times l = \frac{1}{\pi} \times 0.1\pi = 0.1 \, \text{kg} \] The moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is: \[ I = m r^2 \] Substituting the values: \[ I = 0.1 \times (0.05)^2 = 0.1 \times 0.0025 = 0.00025 \, \text{kg m}^2 = 0.25 \times 10^{-3} \, \text{kg m}^2 \] So, the moment of inertia of the ring is 0.25 \(\times 10^{-3}\) kg m\(^2\).
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Approach Solution -2

Step 1: Understand the given data
- Length of wire, \( L = 0.1 \pi \, \text{m} \)
- Linear density, \( \lambda = \frac{1}{\pi} \, \text{kg/m} \)

Step 2: Calculate mass of the wire
\[ m = \lambda \times L = \frac{1}{\pi} \times 0.1 \pi = 0.1 \, \text{kg} \]

Step 3: Calculate radius of the ring formed
Since the wire is bent into a ring,
\[ \text{Circumference} = 2 \pi R = L = 0.1 \pi \] \[ \Rightarrow R = \frac{0.1 \pi}{2 \pi} = 0.05 \, \text{m} \]

Step 4: Calculate moment of inertia of the ring about an axis through its centre and perpendicular to its plane
\[ I = m R^2 = 0.1 \times (0.05)^2 = 0.1 \times 0.0025 = 0.00025 \, \text{kg m}^2 \] \[ = 0.25 \times 10^{-3} \, \text{kg m}^2 \]

Final answer:
The moment of inertia of the ring is \(0.25 \times 10^{-3}\) kg m\(^2\).
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