four bodies of mass 1kg, 2kg,3kg, and 4 kg are situated at four corners of a square of side 2m, take the origin at the center of square. find m.i. if the system passing through the centre and perpendicular to the plane of square

You probably mean to ask:

"Find the moment of inertia of the system about an axis passing through the center.."
Each mass is R = sqrt2 meters from the center.

I = (1 + 2 + 3 + 4)*R^2 = 20 kg*m^2

To find the moment of inertia (I) of this system, we need to calculate the individual moment of inertia for each body and then add them up.

1. Let's start with the body of mass 1 kg located at one corner of the square. The distance (r) of this body from the origin is the half-diagonal of the square, which is √2 m. The moment of inertia of this body (I₁) about the origin is given by I₁ = m₁ * r₁², where m₁ is the mass of the body and r₁ is the distance from the origin.

I₁ = 1 kg * (√2 m)² = 2 kg⋅m²

2. Next, the body of mass 2 kg is located at another corner of the square. The distance (r) of this body from the origin is also √2 m (since it is the same half-diagonal). The moment of inertia of this body (I₂) can be calculated similarly:

I₂ = 2 kg * (√2 m)² = 4 kg⋅m²

3. The body of mass 3 kg is situated at another corner of the square. The distance (r) of this body from the origin is the side length of the square, which is 2 m. Applying the same formula, the moment of inertia of this body (I₃) is:

I₃ = 3 kg * (2 m)² = 12 kg⋅m²

4. Lastly, the body of mass 4 kg is located at the remaining corner of the square. Its distance (r) from the origin is also 2 m. Therefore, the moment of inertia of this body (I₄) is given by:

I₄ = 4 kg * (2 m)² = 32 kg⋅m²

Now, to find the moment of inertia (I) of the system, we simply add up the individual moment of inertia moments:

I = I₁ + I₂ + I₃ + I₄
= 2 kg⋅m² + 4 kg⋅m² + 12 kg⋅m² + 32 kg⋅m²
= 50 kg⋅m²

Therefore, the moment of inertia of the system, passing through the center of the square and perpendicular to the plane of the square, is 50 kg⋅m².