Particles of masses 1 g, 2 g, 3 g, ……., 100 g, are kept at the marks 1 cm, 2 cm, 3 cm,……, 100 cm, respectively on a meter scale. Find the moment of inertia of the system particles about a perpendicular bisector of the meter scale.

Perpendicular bisection will be at the mark 50cm on the scale.

1,2,3,4,….49 masses are placed at one side.

50 masses (51,52…..100)

Are placed at other side and 50th mass is at middle.

Consider the two particles at the position 49 cm and 51 cm.
Moment of inertia due to these two particles will be

$\begin{array}{r}=49\times (1{)}^2+51\times (1{)}^2\\ =100\times 1=100\mathrm{gm}-{\mathrm{cm}}^2\end{array}$

Similarly, if we consider 48th and 52nd term we will get

$100\times (2{)}^2\mathrm{gm}-{\mathrm{cm}}^2$

Therefore, we will get 49 such set and one lone particle at 100 cm. Therefore, total moment of inertia is

$=100\left(1^2+2^2+3^2+\dots \dots ..+{49}^2\right)+100(50{)}^2$

Using, $\sum _{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$

$=100\left(\frac{\left(49\right)\left(50\right)\left(99\right)}{6}\right)+100(50{)}^2$

$=4292500gm-c{m}^2$

$=0.43kg-m^2$