Question

The front compound wall of the house is decorated by using wooden spheres of diameter 21cm placed on the small support 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 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if the silver paint costs 25 paise per $c{{m}^{2}}$ and black paint costs 5 paise per $c{{m}^{2}}$.

Answer 1

Hint: To calculate the total cost of the paint we are first required to find the surface area of the sphere and the surface area of the cylinder and then we will multiply it with their respective paints costs to get the total cost of that paints required and then add all the costs to get the required total costs of that paint. The surface area of the sphere is given by $4\pi {{r}^{2}}$ and the Surface area of the cylinder is given by $2\pi rh$.

Complete step-by-step solution:
We know from the question that the diameter of the wooden sphere is equal to 21cm and the height of each cylindrical support is 7cm and the radius is 1.5cm.
Since we have to find the total cost of paint which is required to paint 8 spheres and 8 cylinders which are supporting each of them.
Since we know that the surface area of each sphere is $4\pi {{r}^{2}}$, where r is the radius of the sphere.
So, the surface area of each sphere of diameter 21cm (i.e. radius = $\dfrac{21}{2}cm$ ) = $4\times \pi \times {{\left( \dfrac{21}{2} \right)}^{2}}c{{m}^{2}}$
So, the surface area of 8 such sphere is $8\times 4\times \pi \times {{\left( \dfrac{21}{2} \right)}^{2}}c{{m}^{2}}$ = $2\times 4\times \pi \times {{\left( 21 \right)}^{2}}c{{m}^{2}}$ =\[11083.54c{{m}^{2}}\] .
We know from the question that we are painting the sphere with silver color and the cost of painting the sphere with silver color is 25 paise per $c{{m}^{2}}$.
So, cost of paint required to paint the area of \[11083.54c{{m}^{2}}\] which is total surface area of all sphere = $25\times 11083.54paise$ = 277088.5 paise.
We also know that the surface area of each cylinder is $2\pi rh$, where r is the radius and h is the height of the cylinder.
So, surface area of each supporting cylinder whose height is 7cm and radius is 1.5 cm is $2\times \pi \times \left( 1.5 \right)\times 7c{{m}^{2}}$
So, the surface area of 8 such cylinder = $8\times 2\times \pi \times \left( 1.5 \right)\times 7c{{m}^{2}}$= \[527.787c{{m}^{2}}\]
We know from the question that we are painting the cylinder with black color and the cost of painting the cylinder with black color is 5 paise per $c{{m}^{2}}$.
So, the cost of paint required to paint the area of \[527.787c{{m}^{2}}\] which is the total surface area of all cylinders = $5\times 527.787paise$ = 2638.935 paise.
Hence, the total cost of paint required = Cost of silver paint required + Cost of black paint required
So, the total cost of paint = 277088.5 paise + 2638.935 paise = 279727.435 paisa = Rs. 2797.27435
This is our required solution.

Note: Students are required to memorize the formula and should avoid calculation mistakes, as there are more chances of making calculation mistakes. Also, students should note that 1Rs = 100 paise.