The diameters of the two circular ends

(a) Given, diameter of one end of the bucket

$2 R=44 \Rightarrow R=22 \mathrm{~cm}$ $[\because$ diameter, $r=2 \times$ radius $]$

and diameter of the other end,

$2 r=24 \Rightarrow r=12 \mathrm{~cm} \quad[\because$ diameter, $r=2 \times$ radius $]$

Height of the bucket, $h=35 \mathrm{~cm}$

Since, the shape of bucket is look like as frustum of a cone.

$\therefore$ Capacity of the bucket $=$ Volume of the frustum of the cone

$=\frac{1}{3} \pi h\left[R^{2}+r^{2}+R r\right]$

$=\frac{1}{3} \times \pi \times 35\left[(22)^{2}+(12)^{2}+22 \times 12\right]$

$=\frac{35 \pi}{3}[484+144+264]$

$=\frac{35 \pi \times 892}{3}=\frac{35 \times 22 \times 892}{3 \times 7}$

$=32706.6 \mathrm{~cm}^{3}=32.7 \mathrm{~L} \quad\left[\because 1000 \mathrm{~cm}^{3}=1 \mathrm{~L}\right]$

Hence, the capacity of bucket is 32.7 L.