The moment of inertia of a meter scale of mass $0.6 \mathrm{~kg}$ about an axis perpendicular to the scale and located at the $20 \mathrm{~cm}$ position on the scale in $\mathrm{kg}-\mathrm{m}^2$ is : (Breadth of the scale is negligible)

• Correct answer
  A

  0.078
• B

  0.104
• C

  0.148
• D

  0.208

Solution

$m=0.6 \mathrm{~kg}$
Mass per unit length $=\frac{0.6}{100} \mathrm{~kg} / \mathrm{cm}$ Mass of part $A B, m_1=\frac{0.6}{100} \times 20=\frac{0.6}{5} \mathrm{~kg}$ Mass of part $B C, m_2=\frac{0.6}{100} \times 80=\frac{0.6 \times 4}{5}=\frac{2.4}{5} \mathrm{~kg}$ Moment of inertia, $I=m_1\left(\frac{A B}{2}\right)^2+m_2 \frac{(B C)^2}{2}$ $=\frac{0.6}{5} \times\left(\frac{20}{2} \times 10^{-2}\right)^2+\frac{2.4}{5} \times\left(\frac{80}{2} \times 10^{-2}\right)^2$ $=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times\left(4 \times 10^{-1}\right)^2$ $=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times 16 \times 10^{-2}$ $=\left(\frac{0.6+38.4}{5}\right) \times 10^{-2}$ $=7.8 \times 10^{-2}$ $=0.078 \mathrm{~kg}-\mathrm{m}^2$

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