If the radii of the circular ends of a bucket 24 cm

The height of the conical bucket is h = 24 cm. The radii of the bottom and top circles are r1 = 15cm and r2 = 5cm respectively.

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(15-5)^{2}+24^{2}}$

$=\sqrt{676}$

$=26 \mathrm{~cm}$

The curved surface area of the bucket is

$=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^{2}$

$=\frac{22}{7} \times(15+5) \times 26+\pi \times 5^{2}$

$=\pi \times 20 \times 26+25 \pi$

$=545 \pi \mathrm{cm}^{2}$

Hence the curved surface area of the bucket is $545 \pi \mathrm{cm}^{2}$