The moment of inertia of meter scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position on the scale in `kg-m^(2)` is (Breadth of the scale is negligible)

Options

A. 0.078
B. 0.104
C. 0.148
D. 0.208

Answer

(a)

Explanation

`m=0.6kg`
Moment of inertia, `I=m_(1)((PQ)/(2))^(2)+m_(2)((QR)/(2))^(2)`
`=(0.6)/(100)xx20xx((20xx10^(-2))/(2))^(2)+((0.6xx4)/(5))xx((80)/(2)xx10^(-2))^(2)`
`=(0.65)/(5)xx10^(-2)+(2.4)/(5)xx(4xx10^(-1))^(2)=0.078kg-m^(2)`