Solve the following :

The perpendicular bisector passes from $50 \mathrm{~cm}$ mark. So, there will be 49 particles on LHS and 50 particles on RHS.

$\mathrm{I}=\left[49(1)^{2}+51(1)^{2}\right]+\left[48(2)^{2}+52(2)^{2}\right] \ldots \ldots \ldots+\left[1(49)^{2}+99(49)^{2}\right]+100(50)^{2}$

$=(100)\left[1^{2}+2^{2}+\ldots \ldots \ldots+50^{2}\right]$

$=(100)\left[\frac{50 \times(50+1)(2 \times 50+1)}{6}\right] \mathrm{gm}-\mathrm{cm}^{2}$

$\mid \approx 0.43 \mathrm{~kg} / \mathrm{cm}^{2}$

$[\because$ Sum of square of $n$ natural numbers,

$\left.\Delta=\frac{n(n+1)(2 n+1)}{6}\right]$