A ring of diameter `2m` oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equibvalent length of the simple pendulum is

Options

A. `2m`
B. `4m`
C. `1.5m`
D. `3m`

Explanation

To find the equivalent length of the simple pendulum for the given ring oscillating as a compound pendulum, we can follow these steps:

### Step 1: Understand the System
We have a ring with a diameter of 2 meters, which means its radius \( r \) is 1 meter. The ring oscillates about a horizontal axis passing through a point at its rim.

### Step 2: Calculate the Moment of Inertia
The moment of inertia \( I \) of a ring about an axis through its center is given by:
\[ I_{center} = m r^2 \]
However, since the axis of rotation is at the rim of the ring, we will use the parallel axis theorem:
\[ I_{effective} = I_{center} + m d^2 \]
where \( d \) is the distance from the center of mass to the new axis. Here, \( d = r = 1 \, m \).

Thus, we have:
\[ I_{effective} = m r^2 + m r^2 = 2m r^2 \]

### Step 3: Substitute the Values
Substituting \( r = 1 \, m \):
\[ I_{effective} = 2m (1^2) = 2m \]

### Step 4: Determine the Effective Length
The effective length \( l_{effective} \) for the oscillation is simply the distance from the axis of rotation to the center of mass, which is equal to the radius \( r \):
\[ l_{effective} = r = 1 \, m \]

### Step 5: Calculate the Time Period
The time period \( T \) of a compound pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I_{effective}}{m g l_{effective}}} \]
Substituting the values we have:
\[ T = 2\pi \sqrt{\frac{2m}{m g (1)}} = 2\pi \sqrt{\frac{2}{g}} \]

### Step 6: Compare with Simple Pendulum
For a simple pendulum, the time period is given by:
\[ T = 2\pi \sqrt{\frac{l_{simple}}{g}} \]
Setting the two expressions for \( T \) equal gives:
\[ 2\pi \sqrt{\frac{2}{g}} = 2\pi \sqrt{\frac{l_{simple}}{g}} \]
Squaring both sides and simplifying:
\[ \frac{2}{g} = \frac{l_{simple}}{g} \]
Thus, we find:
\[ l_{simple} = 2 \, m \]

### Step 7: Find the Equivalent Length
Since the effective length for the compound pendulum is given by:
\[ l_{effective} = \frac{3}{2} r \]
Substituting \( r = 1 \, m \):
\[ l_{effective} = \frac{3}{2} \times 1 = 1.5 \, m \]

### Final Answer
The equivalent length of the simple pendulum is:
\[ \boxed{1.5 \, m} \]