A diatomic molecule is formed by two atoms which may be treated as mass points `m_1` and `m_2` joined by a massless rod of length `r`. Then the moment of inertia of molecule about an axis passing through centre of mass and perpendicular to the rod is:

Options

A. Zero
B. `(m_(1)+m_(2))r^(2)`
C. `((m_(1)m_(2))/(m_(1)+m_(2)))r^(2)`
D. `((m_(1)+m_(2))/(m_(1)m_(2)))r^(2)`

Answer

To find the moment of inertia of a diatomic molecule formed by two atoms treated as mass points \( m_1 \) and \( m_2 \) joined by a massless rod of length \( r \), we will follow these steps:

### Step 1: Understand the System
We have two masses \( m_1 \) and \( m_2 \) connected by a massless rod of length \( r \). We need to find the moment of inertia about an axis passing through the center of mass and perpendicular to the rod.

### Step 2: Find the Center of Mass
The position of the center of mass (CM) for the two masses can be found using the formula:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Assuming \( m_1 \) is at \( x = 0 \) and \( m_2 \) is at \( x = r \), we have:
\[ x_{CM} = \frac{m_1 \cdot 0 + m_2 \cdot r}{m_1 + m_2} = \frac{m_2 r}{m_1 + m_2} \]

### Step 3: Determine Distances from the Center of Mass
The distance of \( m_1 \) from the center of mass is:
\[ d_1 = x_{CM} = \frac{m_2 r}{m_1 + m_2} \]
The distance of \( m_2 \) from the center of mass is:
\[ d_2 = r - x_{CM} = r - \frac{m_2 r}{m_1 + m_2} = \frac{m_1 r}{m_1 + m_2} \]

### Step 4: Calculate the Moment of Inertia
The moment of inertia \( I \) about the center of mass is given by:
\[ I = m_1 d_1^2 + m_2 d_2^2 \]
Substituting \( d_1 \) and \( d_2 \):
\[ I = m_1 \left(\frac{m_2 r}{m_1 + m_2}\right)^2 + m_2 \left(\frac{m_1 r}{m_1 + m_2}\right)^2 \]

### Step 5: Simplify the Expression
Calculating each term:
\[ I = m_1 \frac{m_2^2 r^2}{(m_1 + m_2)^2} + m_2 \frac{m_1^2 r^2}{(m_1 + m_2)^2} \]
Factoring out \( \frac{r^2}{(m_1 + m_2)^2} \):
\[ I = \frac{r^2}{(m_1 + m_2)^2} \left(m_1 m_2^2 + m_2 m_1^2\right) \]
This simplifies to:
\[ I = \frac{m_1 m_2}{m_1 + m_2} r^2 \]

### Final Answer
Thus, the moment of inertia of the diatomic molecule about an axis passing through the center of mass and perpendicular to the rod is:
\[ I = \frac{m_1 m_2}{m_1 + m_2} r^2 \]