The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is `"4 kg m"^(2)`. The diameter of the ring is

Options

A. 2 m
B. 4 m
C. 5 m
D. 6 m

Answer

To find the diameter of the circular ring given its moment of inertia and mass, we can follow these steps:

### Step 1: Write down the formula for the moment of inertia of a circular ring.
The moment of inertia \( I \) of a circular ring about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I = m \cdot r^2 \]
where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the ring,
- \( r \) is the radius of the ring.

### Step 2: Substitute the known values into the formula.
From the question, we know:
- \( I = 4 \, \text{kg m}^2 \)
- \( m = 1 \, \text{kg} \)

Substituting these values into the equation:
\[ 4 = 1 \cdot r^2 \]

### Step 3: Solve for the radius \( r \).
Now, we can solve for \( r^2 \):
\[ r^2 = 4 \]
Taking the square root of both sides gives:
\[ r = \sqrt{4} = 2 \, \text{m} \]

### Step 4: Calculate the diameter of the ring.
The diameter \( D \) of the ring is twice the radius:
\[ D = 2r = 2 \cdot 2 = 4 \, \text{m} \]

### Final Answer:
The diameter of the ring is \( 4 \, \text{m} \).

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