A classic way to solve any problem of this type.
\(\displaystyle \frac{h}{H} = \frac{d}{D}\)
\(\displaystyle h:\) height of the small cylinder.
\(\displaystyle H:\) height of the big cylinder.
\(\displaystyle d:\) diameter of the small cylinder
\(\displaystyle D:\) diameter of the big cylinder
We are given the height and the volume of the big cylinder, so we can find its diameter \(\displaystyle D\).
\(\displaystyle V = \pi R^2 H = \pi \left(\frac{D}{2}\right)^2H\)
\(\displaystyle D = \sqrt{\frac{4V}{\pi H}}\)
Back to our classic formula.
\(\displaystyle \frac{h}{H} = \frac{d}{D} = \frac{d}{\sqrt{\frac{4V}{\pi H}}}\)
Now, we can find the diameter of the small cylinder.
\(\displaystyle d = \frac{h\sqrt{\frac{4V}{\pi H}}}{H} = \frac{0.30\sqrt{\frac{4 \times 10.8}{\pi \times 0.45}}}{0.45} \approx 3.68527\) m
Finally, we can find the volume of the small cylinder.
\(\displaystyle V_S = \pi r^2 h = \pi \left(\frac{d}{2}\right)^2 h = \pi \left(\frac{3.68527}{2}\right)^2 \times 0.30 = 3.2\) litres
