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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)

best_answer

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 $1\ cm^2$ silver = $25\ paise = Rs.\ 0.25$

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

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

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 $66\ cm^2$ silver = $Rs. (0.05\times66) = Rs.\ 3.3$

$\therefore$ The 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 = $Rs.\ (8\times348.0325) = Rs.\ 2784.26$

Posted by

HARSH KANKARIA

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