A ring is suspended at a point on its rim and it behaves as a second’s pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be (g = `pi^(2))`

Options

A. 0.5 m
B. 1.0 m
C. 0.67 m
D. 1.5 m

Answer

To solve the problem, we need to determine the radius of a ring suspended at a point on its rim, which behaves as a second's pendulum. The time period of a second's pendulum is given as 2 seconds. Let's go through the solution step by step.

### Step 1: Understand the Time Period of a Physical Pendulum
The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgL_{cm}}} \]
where:
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the pendulum,
- \( g \) is the acceleration due to gravity,
- \( L_{cm} \) is the distance from the pivot point to the center of mass.

### Step 2: Identify the Moment of Inertia
For a ring of radius \( r \) and mass \( m \), the moment of inertia about its center of mass is:
\[ I_{cm} = mr^2 \]
Using the parallel axis theorem, the moment of inertia about the pivot point (which is on the rim) is:
\[ I = I_{cm} + md^2 \]
where \( d = r \) (the distance from the center of mass to the pivot point). Thus,
\[ I = mr^2 + mr^2 = 2mr^2 \]

### Step 3: Calculate the Distance to Center of Mass
The distance from the pivot point to the center of mass \( L_{cm} \) is simply the radius \( r \):
\[ L_{cm} = r \]

### Step 4: Substitute Values into the Time Period Formula
Substituting \( I \) and \( L_{cm} \) into the time period formula gives:
\[ T = 2\pi \sqrt{\frac{2mr^2}{mgr}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{2r}{g}} \]

### Step 5: Set the Time Period Equal to 2 Seconds
Since the time period \( T \) is given as 2 seconds:
\[ 2 = 2\pi \sqrt{\frac{2r}{g}} \]

### Step 6: Solve for \( r \)
Squaring both sides:
\[ 4 = 4\pi^2 \frac{2r}{g} \]
This simplifies to:
\[ 1 = \pi^2 \frac{2r}{g} \]
Rearranging gives:
\[ g = 2\pi^2 r \]

### Step 7: Substitute the Value of \( g \)
Given that \( g = \pi^2 \):
\[ \pi^2 = 2\pi^2 r \]
Dividing both sides by \( \pi^2 \):
\[ 1 = 2r \]
Thus:
\[ r = \frac{1}{2} \text{ meters} = 0.5 \text{ meters} \]

### Final Answer
The radius of the ring is \( 0.5 \) meters.

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