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. `(11)/(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 a 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 \(\frac{AB}{BC} = 2\), we can let:
- \(BC = x\)
- \(AB = 2x\)

Since \(ABCD\) is a rectangle, the opposite sides are equal:
- \(AD = BC = x\)
- \(CD = AB = 2x\)

### Step 2: Calculate the total length of the wire
The total length \(l\) of the wire frame is the sum of the lengths of all sides:
\[ l = AB + BC + CD + AD = 2x + x + 2x + x = 6x \]
From this, we can express \(x\) in terms of \(l\):
\[ x = \frac{l}{6} \]

### Step 3: Determine the mass distribution
The mass of the wire is uniformly distributed. Since the length of each side is proportional to its mass, we can find the mass of each side:
- Length of \(AB = CD = 2x = \frac{2l}{6} = \frac{l}{3}\)
- Length of \(BC = AD = x = \frac{l}{6}\)

Using the ratio of lengths to find mass:
- Mass of \(AB = CD = \frac{m}{3}\) (since they are each \(\frac{l}{3}\) of the total length)
- Mass of \(BC = AD = \frac{m}{6}\) (since they are each \(\frac{l}{6}\) of the total length)

### Step 4: Calculate the moment of inertia for each side about \(BC\)
1. **For sides \(AB\) and \(CD\)**:
The moment of inertia of a rod about an axis through one end is given by:
\[ I = \frac{1}{3} m L^2 \]
For \(AB\) and \(CD\):
- Length \(L = \frac{l}{3}\)
- Mass \(m = \frac{m}{3}\)

Thus, the moment of inertia for each side is:
\[ I_{AB} = I_{CD} = \frac{1}{3} \left(\frac{m}{3}\right) \left(\frac{l}{3}\right)^2 = \frac{1}{3} \cdot \frac{m}{3} \cdot \frac{l^2}{9} = \frac{ml^2}{81} \]

2. **For side \(AD\)**:
The moment of inertia for side \(AD\) about \(BC\):
- Length \(L = \frac{l}{6}\)
- Mass \(m = \frac{m}{6}\)

Thus, the moment of inertia is:
\[ I_{AD} = \frac{1}{3} \left(\frac{m}{6}\right) \left(\frac{l}{6}\right)^2 = \frac{1}{3} \cdot \frac{m}{6} \cdot \frac{l^2}{36} = \frac{ml^2}{648} \]

### Step 5: Calculate the total moment of inertia about \(BC\)
Now, we sum the moments of inertia of all sides:
\[ I_{total} = I_{AB} + I_{CD} + I_{AD} = \frac{ml^2}{81} + \frac{ml^2}{81} + \frac{ml^2}{54} \]

To add these fractions, we need a common denominator. The least common multiple of \(81\) and \(54\) 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 Answer
Thus, the moment of inertia of the wire frame about the side \(BC\) is:
\[ \boxed{\frac{7ml^2}{162}} \]