Particles having masses of 1 g , 2 g , 3 g , ..., 100 g are placed...

The perpendicular bisector of a metre scale passes through its midpoint at $50 \text{ cm}$.

The moment of inertia of the system about this axis is given by:

$I = \sum_{i=1}^{100} m_i r_i^2$

where $m_i = i \text{ g}$ and the distance from the axis is $r_i = |i - 50| \text{ cm}$.

$I = \sum_{i=1}^{100} i (i - 50)^2$

Let $j = i - 50$. As $i$ varies from $1$ to $100$, $j$ varies from $-49$ to $50$.

$I = \sum_{j=-49}^{50} (j + 50) j^2 = \sum_{j=-49}^{50} j^3 + 50 \sum_{j=-49}^{50} j^2$

Since $j^3$ is an odd power, the sum of symmetric terms is zero, giving $\sum_{j=-49}^{49} j^3 = 0$.

$\sum_{j=-49}^{50} j^3 = 50^3 = 125000$

For the second term, using the sum of squares formula $\sum_{j=1}^{n} j^2 = \dfrac{n(n+1)(2n+1)}{6}$:

$\sum_{j=-49}^{50} j^2 = 50^2 + 2 \sum_{j=1}^{49} j^2$

$\sum_{j=1}^{49} j^2 = \dfrac{49 \times 50 \times 99}{6} = 40425$

$\sum_{j=-49}^{50} j^2 = 2500 + 2(40425) = 83350$

Substituting these values back into the equation for $I$:

$I = 125000 + 50(83350) = 125000 + 4167500 = 4292500 \text{ g cm}^2$

Converting the units to $\text{kg m}^2$:

$I = 4292500 \times 10^{-3} \text{ kg} \times (10^{-2} \text{ m})^2$

$I = 4292500 \times 10^{-7} \text{ kg m}^2 = 0.42925 \text{ kg m}^2$

Rounding to two decimal places, we get $0.43 \text{ kg m}^2$.

Answer: $0.43 \text{ kg m}^2$