The moment of inertia of a 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)

Step-by-step solution

m=0.6 ~kg Mass per unit length = 0.6 100 ~kg / cm Mass of part A B, m₁= 0.6 100 20= 0.6 5 ~kg Mass of part B C, m₂= 0.6 100 80= 0.6 4 5 = 2.4 5 ~kg Moment of inertia, I=m₁ ( A B 2 )^2+m₂ (B C)^2 2 = 0.6 5 ( 20 2 10⁻² )^2+ 2.4 5 ( 80 2 10⁻² )^2= 0.6 5 10⁻²+ 2.4 5 (4 10⁻¹ )^2= 0.6 5 10⁻²+ 2.4 5 16 10⁻²= ( 0.6+38.4 5 ) 10⁻²=7.8 10⁻²=0.078 ~kg - m ^2