Five particles of mass = 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

The correct option is 0.1 kg m².

Step-by-step Explanation:

1. Identify the given parameters:

Number of particles,
$n=5$

Mass of each particle,
$m=2\text{kg}$

Radius of the circular disc,
$R=0.1\text{m}$

The mass of the disc is negligible, meaning it does not contribute to the moment of inertia of the system.

2. Moment of Inertia of the System:
The moment of inertia
$I$
of a system of particles about a given axis is calculated as the sum of the moment of inertia of each individual particle:

$I=\sum _{i=1}^n{m}_i{r}_i^2$

Since all 5 particles are attached to the rim of the disc, the distance
$r_i$
of each particle from the axis passing through the centre and perpendicular to the disc's plane is exactly equal to the radius
$R$
of the disc.

Therefore, the expression for the total moment of inertia simplifies to:

$I=n·m{R}^2$

3. Calculation:
Substitute the given values into the equation:

$I=5·\left(2\text{kg}\right)·(0.1\text{m}{)}^2$

Simplify the squared term:

$(0.1\text{m}{)}^2=0.01{\text{m}}^2$

Calculate the final product:

$I=5·2·0.01{\text{kg m}}^2$

$I=10·0.01{\text{kg m}}^2$

$I=0.1{\text{kg m}}^2$

Thus, the moment of inertia of the system about the given axis is 0.1 kg m².