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 $cm^{2}.$

(D) $1$. Radius of a wooden sphere $(R) = \frac{21}{2} = 10.5\, cm$.
$2$. Surface area of one sphere $= 4\pi R^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386\, cm^2$.
$3$. Area covered by the cylindrical support on the sphere $= \pi r^2 = \frac{22}{7} \times 1.5 \times 1.5 \approx 7.07\, cm^2$.
$4$. Surface area to be painted silver for one sphere $= 1386 - 7.07 = 1378.93\, cm^2$.
$5$. Total area for $8$ spheres $= 8 \times 1378.93 = 11031.44\, cm^2$.
$6$. Cost of silver paint $= 11031.44 \times 0.25 = ₹ 2757.86$.
$7$. Curved surface area of one cylindrical support $= 2\pi rh = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66\, cm^2$.
$8$. Total area to be painted black for $8$ supports $= 8 \times 66 = 528\, cm^2$.
$9$. Cost of black paint $= 528 \times 0.05 = ₹ 26.40$.
$10$. Total cost $= 2757.86 + 26.40 = ₹ 2784.26$.