Q : 2     The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig \small 13.32. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius \small 1.5\hspace{1mm}cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per \small cm^2 and black paint costs 5 paise per \small cm^2.
 

            

Answers (1)

Given,

The radius of the wooden spheres = r_1 = \frac{21}{2}\ cm

\therefore The surface area of a single sphere = 4\pi r_1^2

\\ = 4\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2} \\ = 22\times3\times21

= 1386\ cm^2

Again, the Radius of the cylinder support = \small r_2 = 1.5\hspace{1mm}cm

Height of the support = h = 7\ cm

\therefore The base area of the cylinder = \pi r_2^2 = \frac{22}{7}\times1.5\times1.5 = 7.07\ cm^2

Now, Cost of painting 1\ cm^2silver = 25\ paise = Rs.\ 0.25

\therefore Cost of painting 1 wooden sphere = Cost of painting (1386-7.07)\ cm^2silver

Rs. (0.25\times1378.93) = Rs.\ 344.7325

Now, Curved surface area of the cylindrical support = 2\pi r_2 h

\\ = 2\times\frac{22}{7}\times1.5\times7 \\ = 22\times3 \\ = 66\ cm^2

Now, Cost of painting 1\ cm^2 black = 5\ paise = Rs.\ 0.05

\therefore Cost of painting 1 such stand = Cost of painting 66\ cm^2silver = Rs. (0.05\times66) = Rs.\ 3.3

\thereforeThe total cost of painting 1 sphere and its support = Rs.\ (344.7325+3.3) = Rs.\ 348.0325

Therefore, total cost of painting 8 such spheres and their supports = Rs.\ (8\times348.0325) = Rs.\ 2784.26

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