A wire of length `l` and mass `m` is bent in the form of a rectangle `ABCD` with `(AB)/(BC)=2`. The moment of inertia of this wife frame about the side `BC` is

Options

A. `22/252 ml^(2)`
B. `8/203 ml^(2)`
C. `5/136 ml^(2)`
D. `7/162 ml^(2)`

Answer

To find the moment of inertia of the wire frame bent in the form of a rectangle \(ABCD\) about the side \(BC\), we can follow these steps:

### Step 1: Define the dimensions of the rectangle
Given that the ratio of the lengths of sides \(AB\) and \(BC\) is \(2:1\), we can let:
- \(AB = 2x\)
- \(BC = x\)

### Step 2: Calculate the perimeter and relate it to the length of the wire
The perimeter \(P\) of the rectangle is given by:
\[ P = 2(AB + BC) = 2(2x + x) = 6x \]
Since the total length of the wire is \(l\), we have:
\[ 6x = l \implies x = \frac{l}{6} \]

### Step 3: Determine the lengths of the sides
Now substituting \(x\) back, we find:
- \(AB = 2x = 2 \cdot \frac{l}{6} = \frac{l}{3}\)
- \(BC = x = \frac{l}{6}\)

### Step 4: Calculate the mass distribution
The mass \(m\) of the wire is uniformly distributed along its length. The mass per unit length \(\mu\) is:
\[ \mu = \frac{m}{l} \]
Now, we can find the mass of each side:
- Mass of side \(AB\) (length \(\frac{l}{3}\)):
\[ m_{AB} = \mu \cdot \frac{l}{3} = \frac{m}{l} \cdot \frac{l}{3} = \frac{m}{3} \]
- Mass of side \(BC\) (length \(\frac{l}{6}\)):
\[ m_{BC} = \mu \cdot \frac{l}{6} = \frac{m}{l} \cdot \frac{l}{6} = \frac{m}{6} \]
- Mass of side \(CD\) is the same as \(AB\):
\[ m_{CD} = m_{AB} = \frac{m}{3} \]

### Step 5: Calculate the moment of inertia about side \(BC\)
The moment of inertia \(I\) about an axis through one end of a rod is given by:
\[ I = \frac{1}{3} m L^2 \]
1. For sides \(AB\) and \(CD\) (both have length \(\frac{l}{3}\)):
- Distance from \(BC\) to \(AB\) and \(CD\) is \(\frac{l}{6}\):
\[ I_{AB} = \frac{1}{3} \left(\frac{m}{3}\right) \left(\frac{l}{3}\right)^2 = \frac{1}{3} \cdot \frac{m}{3} \cdot \frac{l^2}{9} = \frac{ml^2}{81} \]
Since \(I_{CD} = I_{AB}\):
\[ I_{CD} = \frac{ml^2}{81} \]

2. For side \(AD\) (length \(\frac{l}{6}\)):
- The moment of inertia about \(BC\) is:
\[ I_{AD} = m_{AD} \cdot d^2 = \left(\frac{m}{6}\right) \left(\frac{l}{3}\right)^2 = \frac{m}{6} \cdot \frac{l^2}{9} = \frac{ml^2}{54} \]

### Step 6: Sum the moments of inertia
The total moment of inertia \(I_{total}\) about side \(BC\) is:
\[ I_{total} = I_{AB} + I_{CD} + I_{AD} = 2 \cdot \frac{ml^2}{81} + \frac{ml^2}{54} \]
Finding a common denominator (which is 162):
\[ I_{total} = \frac{2ml^2}{81} + \frac{3ml^2}{162} = \frac{4ml^2}{162} + \frac{3ml^2}{162} = \frac{7ml^2}{162} \]

### Final Result
Thus, the moment of inertia of the wire frame about the side \(BC\) is:
\[ \boxed{\frac{7ml^2}{162}} \]