Filling the Beaker with Water isn't so easy

The highest stability is achieved when the center of mass is the lowest possible. Let $h$ be the height of the water when this happens.

Consider only the water (ignore the container). The mass of the water is $(\pi r^2h)\rho = 1000\pi r^2h$ . The center of mass of the water body is at height $h/2$ .

Now we combine the water and the container to find the height of the net center of mass. This is given by

$x_{net} = \frac{x_{water}m_{water}+x_{container}m_{container}}{m_{water}+m_{container}}$ $x_{net} = \frac{(1000\pi r^2h)(h/2) + (0.1)(0.1)}{1000\pi r^2h + 0.1}=\frac{500\pi r^2h^2+0.01}{1000\pi r^2h+0.1}$

For the lowest value of $x_{net}$ , we differentiate it with respect to $h$ and get $\frac{(1000\pi r^2h +0.1)(1000\pi r^2h) - (500\pi r^2h^2 +0.01)(1000\pi r^2)}{(1000\pi r^2h +0.1)^2} = 0$

Simplifying this gives us $h(1000\pi r^2h +0.1) -(500\pi r^2h^2 +0.01) =0$ $500\pi r^2h^2 +0.1h - 0.01 =0$ Substituting $r = 0.03$ and solving the quadratic equation, we get $h = 0.05587 \text{m}$ (ignoring the negative root)

Therefore $[d] = 55 \text{mm}$ and $10[d] = \fbox{550}$