Question

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black Find the cost of paint required if silver paint costs 25 paise per $c{m}^2$ and black paint costs 5 paise per $c{m}^2$

Answer 1

Diameter of each spereres = 21 cm
$\therefore \left(R\right)=\frac{21}{2}cm$
Radius of each cylinder (r) = 1.5 cm
and height (h) = 7 cm

Now surface area of one sphere $=4\pi R^2$
$=4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}c{m}^2=1386c{m}^2$
and surface area of one cylinder $=2\pi rh$
$=2\times \frac{22}{7}\times 1.5\times 7c{m}^2=66c{m}^2$
$\therefore$ Surface area of 8 spheres $=8\times 1386c{m}^2=11088c{m}^2$
Surface area of 8 cylinders tops
$=8\times \pi r^2=8\times \frac{22}{7}\times 1.5\times 1.5c{m}^2=56.57c{m}^2$
and surface area of 8 cylinders
$=8\times 66=528c{m}^2$
and surface area of spheres excluding base area $=11088-56.57=11031.43$
Rate of silver painting the spheres
= 25 p. per cm
$Totalcost=\frac{11031.43\times 25}{100}=Rs.2757.86$
and rate of black painting the cylinder
= 5 p. per $c{m}^2$
$\therefore Costofpaintingit=Rs.\frac{528\times 5}{100}=\frac{2640}{100}=Rs.26.40$
Hence, total cost $=2757.86+26.40=Rs.2784.26$