Moment of inertia of a straight wire about an axis perpendicular to the wire passing through one of its end is I.
Now the same wire is bent into a ring of two turns , then the moment of inertia would be

Options

A. `((pi^(2))/(3))l`
B. `((pi^(2))/(12))l`
C. `((3)/(16pi^(2)))l`
D. `((3)/(4pi^(2)))l`

Answer

To find the moment of inertia of a straight wire bent into a ring of two turns, we can follow these steps:

### Step 1: Understand the Moment of Inertia of the Straight Wire
The moment of inertia \( I \) of a straight wire about an axis perpendicular to the wire passing through one of its ends is given as:
\[ I = \frac{ML^2}{3} \]
where \( M \) is the mass of the wire and \( L \) is its length.

### Step 2: Relate Length of the Wire to the Ring
When the wire is bent into a ring of two turns, the total length of the wire remains the same. The circumference of a single turn of the ring is \( 2\pi r \), so for two turns, it becomes:
\[ 2 \times (2\pi r) = 4\pi r \]
Setting this equal to the length of the wire \( L \):
\[ 4\pi r = L \]
From this, we can solve for \( r \):
\[ r = \frac{L}{4\pi} \]

### Step 3: Calculate the Moment of Inertia of the Ring
The moment of inertia \( I' \) of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I' = MR^2 \]
Substituting \( r \) from the previous step:
\[ I' = M \left(\frac{L}{4\pi}\right)^2 \]
This simplifies to:
\[ I' = M \frac{L^2}{16\pi^2} \]

### Step 4: Substitute for \( ML^2 \)
From the expression for the moment of inertia of the straight wire, we know:
\[ ML^2 = 3I \]
Substituting this into the equation for \( I' \):
\[ I' = \frac{3I}{16\pi^2} \]

### Conclusion
Thus, the moment of inertia of the wire bent into a ring of two turns is:
\[ I' = \frac{3I}{16\pi^2} \]