Question

The front compound wall is decorated by wooden spheres of diameter 21cm, placed on small supports, eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm 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 cost 5 paise per $c{{m}^{2}}$.

Answer 1

Hint: We have to find the surface area of each sphere and surface area of each cylinder and then we will multiply it by 8 to get the total surface area of 8 spheres and 8 cylindrical support. Then we will multiply the surface area of the sphere by the cost of silver paint and multiply the surface area of the cylinder by cost of black paint, then we will add them to get the final answer.

Complete step-by-step answer:

The surface area of sphere(S) =$4\pi {{r}^{2}}$ , r = radius of sphere
The surface area of cylinder$\left( {{S}_{2}} \right)$ = $2\pi Rh$ ,R = radius of cylinder, h = height of cylinder
Now using this we get,
But some of the area will be covered by the cylinder hence $\left( {{S}_{1}} \right)$ will be, $4\pi {{r}^{2}}$- $\pi {{R}^{2}}$
 $\left( {{S}_{1}} \right)$ = $4\pi {{\left( \dfrac{21}{2} \right)}^{2}}-\pi {{\left( 1.5 \right)}^{2}}=441\pi -2.25\pi $
$\left( {{S}_{2}} \right)$ = $2\pi \times 1.5\times 7$ $=21\pi $
But some of the area will be covered by the cylinder hence
Now the cost of paint will be for 8$\left( {{S}_{1}} \right)$ and 8$\left( {{S}_{2}} \right)$ is,
$\begin{align}
  & 8\left( 441\pi -2.25\pi \right)\times 25+8\left( 21\pi \right)\times 5 \\
 & =87750\pi +840\pi \\
 & =88590\pi \\
\end{align}$
Hence the cost of paint in paise will be $88590\pi =278172.6$

Note: The formula for calculating surface area of sphere and cylinder must be kept in mind and all the units should be verified carefully while solving this type of question and the calculation should also be cross checked to avoid mistakes.