The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm.

(b) 80080 cm3
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

$R=35 \mathrm{~cm}, r=14 \mathrm{~cm}, h=40 \mathrm{~cm}$

Volume of the bucket = Volume of the frustum of the cone

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

$=\frac{1}{3} \times \frac{22}{7} \times 40 \times\left[(35)^{2}+(14)^{2}+(35 \times 14)\right] \mathrm{cm}^{3}$

$=\left(\frac{880}{21} \times 1911\right) \mathrm{cm}^{3}$

$=80080 \mathrm{~cm}^{3}$

Hence, the volume of the bucket is $80080 \mathrm{~cm}^{3}$.