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 $cm^2$ and black paint costs $5$ paise per $cm^2$.

(N/A) $1$. For the spheres:
Diameter of each sphere = $21 \, cm$,so radius $r = 10.5 \, cm$.
The surface area of one sphere is $4 \pi r^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386 \, cm^2$.
However,a small part of the sphere is covered by the support. The support is a cylinder of radius $1.5 \, cm$. The area of the circle covered on the sphere is $\pi \times (1.5)^2 = \frac{22}{7} \times 2.25 \approx 7.07 \, cm^2$.
Surface area to be painted silver per sphere = $1386 - 7.07 = 1378.93 \, cm^2$.
For $8$ spheres,total silver area = $8 \times 1378.93 = 11031.44 \, cm^2$.
Cost of silver paint = $11031.44 \times 0.25 = ₹ 2757.86$.
$2$. For the supports:
Each support is a cylinder with radius $r = 1.5 \, cm$ and height $h = 7 \, cm$.
Curved surface area of one cylinder = $2 \pi rh = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66 \, cm^2$.
Total area to be painted black for $8$ supports = $8 \times 66 = 528 \, cm^2$.
Cost of black paint = $528 \times 0.05 = ₹ 26.40$.
Total cost = $2757.86 + 26.40 = ₹ 2784.26$.