Point masses $1,2,3$ and 4 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 $X$-axis will be

• Correct answer
  A

  $43 \mathrm{~kg}-\mathrm{m}^2$
• B

  $34 \mathrm{~kg}-\mathrm{m}^2$
• C

  $27 \mathrm{~kg}-\mathrm{m}^2$
• D

  $72 \mathrm{~kg}-\mathrm{m}^2$

Solution

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles. $I=I_1+I_2+I_3+I_4$ $\therefore I=m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+m_4 r_4^2$ $I=(1 \times 0)+(2 \times 0)+\left(3 \times 3^2\right)+4(-2)^2$ $I=0+0+27+16=43 \mathrm{~kg}-\mathrm{m}^2$

Hence, the correct answer is option: 'A'.