3 particles of mass 10 gm each are placed at the corners of an equilateral triangle of side 5 cm. The M.I. of the system about a perpendicular axis to the plane passing through a corner of the triangle will be : (in kg-`m^(2)`)

Options

A. `5xx10^(-2)`
B. `5xx10^(-3)`
C. `5xx10^(-4)`
D. `5xx10^(-5)`

Answer

To find the moment of inertia (M.I.) of the system of three particles placed at the corners of an equilateral triangle about a perpendicular axis passing through one corner, we can follow these steps:

### Step 1: Understand the Configuration
We have three particles, each of mass \( m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} \), placed at the corners of an equilateral triangle with a side length of \( L = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \).

### Step 2: Identify the Axis of Rotation
We need to find the moment of inertia about an axis perpendicular to the plane of the triangle and passing through one of the corners. Let's denote the corners of the triangle as A, B, and C.

### Step 3: Calculate the Moment of Inertia for Each Particle
The moment of inertia \( I \) for a point mass about an axis is given by:
\[ I = m \cdot r^2 \]
where \( r \) is the distance from the axis of rotation to the mass.

- For particle A (located at the axis), the distance \( r_A = 0 \):
\[ I_A = m \cdot r_A^2 = 10 \times 10^{-3} \cdot 0^2 = 0 \]

- For particle B (distance \( r_B = L \)):
\[ I_B = m \cdot r_B^2 = 10 \times 10^{-3} \cdot (5 \times 10^{-2})^2 = 10 \times 10^{-3} \cdot 25 \times 10^{-4} = 2.5 \times 10^{-5} \, \text{kg m}^2 \]

- For particle C (distance \( r_C = L \)):
\[ I_C = m \cdot r_C^2 = 10 \times 10^{-3} \cdot (5 \times 10^{-2})^2 = 10 \times 10^{-3} \cdot 25 \times 10^{-4} = 2.5 \times 10^{-5} \, \text{kg m}^2 \]

### Step 4: Sum the Moments of Inertia
Now, we can find the total moment of inertia \( I_{\text{total}} \) by summing the contributions from all three particles:
\[ I_{\text{total}} = I_A + I_B + I_C = 0 + 2.5 \times 10^{-5} + 2.5 \times 10^{-5} = 5 \times 10^{-5} \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia of the system about the given axis is:
\[ \boxed{5 \times 10^{-5} \, \text{kg m}^2} \]