Three particles, each having a mass of 200 g , are placed at the...

Let the three particles be A, B, and C, each of mass $m = 200 \text{ g} = 0.2 \text{ kg}$.

Side length of the equilateral triangle, $a = 10 \text{ cm} = 0.1 \text{ m}$.

Let the axis of rotation pass through particles A and B. The perpendicular distances of particles A and B from the axis are zero ($r_A = 0$, $r_B = 0$).

The perpendicular distance of particle C from the axis AB is the altitude of the equilateral triangle:

r_C = a \sin 60^{}^{{}^{\circ}} = \dfrac{\sqrt{3}}{2}a

The moment of inertia of the system about the axis AB is given by:

$I = m_A r_A^2 + m_B r_B^2 + m_C r_C^2$

$I = m(0)^2 + m(0)^2 + m \left( \dfrac{\sqrt{3}}{2}a \right)^2$

$I = \dfrac{3}{4} m a^2$

Substituting the given values:

$I = \dfrac{3}{4} \times 0.2 \times (0.1)^2$

$I = \dfrac{3}{4} \times 0.2 \times 0.01$

$I = 0.15 \times 0.01 = 1.5 \times 10^{-3} \text{ kg-m}^2$