Point masses 1, 2, 3 and 4 text{ kg} are lyin | vedclass

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(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of th

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$43 	ext{ kg} \cdot 	ext{m}^2$

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(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.For the $x$-axis, the perpendicular distance of a point $(x, y, z)$ is $r = \sqrt{y^2 + z^2}$.$1$. For mass $m_1 = 1 	ext{ kg}$ at $(0,0,0)$: $r_1 = \sqrt{0^2 + 0^2} = 0 	ext{ m}$. So, $I_1 = 1 	imes 0^2 = 0 	ext{ kg} \cdot 	ext{m}^2$.$2$. For mass $m_2 = 2 	ext{ kg}$ at $(2,0,0)$: $r_2 = \sqrt{0^2 + 0^2} = 0 	ext{ m}$. So, $I_2 = 2 	imes 0^2 = 0 	ext{ kg} \cdot 	ext{m}^2$.$3$. For mass $m_3 = 3 	ext{ kg}$ at $(0,3,0)$: $r_3 = \sqrt{3^2 + 0^2} = 3 	ext{ m}$. So, $I_3 = 3 	imes 3^2 = 27 	ext{ kg} \cdot 	ext{m}^2$.$4$. For mass $m_4 = 4 	ext{ kg}$ at $(-2,-2,0)$: $r_4 = \sqrt{(-2)^2 + 0^2} = 2 	ext{ m}$. So, $I_4 = 4 	imes 2^2 = 16 	ext{ kg} \cdot 	ext{m}^2$.The total moment of inertia $I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + 27 + 16 = 43 	ext{ kg} \cdot 	ext{m}^2$.

Point masses $1, 2, 3$ and $4 \text{ kg}$ are lying at the points $(0,0,0), (2,0,0), (0,3,0)$ and $(-2,-2,0)$ respectively. The moment of inertia of this system about the $x$-axis will be:

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