Question

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given 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 $perc{m}^2$ and black paint costs 5 paise $perc{m}^2$.

• A. Rs. 1784.26
• B. Rs. 2734.26
• C. Rs. 6784.26
• D. Rs. 2784.26

Answer 1

The correct option is D

Rs. 2784.26

Radius (r) of wooden sphere$=\left(\frac{21}{2}\right)cm=10.5cm$
Surface area of the wooden sphere $=4\pi r^2$
$=[4\times \frac{22}{7}\times (10.5{)}^2]c{m}^2=1386c{m}^2$
Radius $\left(r_1\right)$of the circular end of cylindrical support =1.5 cm
Height (h) of the cylindrical support = 7 cm
CSA of the cylindrical support $=2\pi rh$
$=[2\times \frac{22}{7}\times (1.5)\times 7]c{m}^2=66c{m}^2$
Area of the circular end of cylindrical support $=\pi r^2$
$=[2\times \frac{22}{7}\times (1.5{)}^2]c{m}^2=7.07c{m}^2$
Area to be painted silver $=[8\times (1386-7.07\left)\right]c{m}^2=(8\times 1378.93)c{m}^2=11031.44c{m}^2$
Cost fo ther painting with silver colour $=Rs.(11031.44\times 0.25)=Rs.2757.86$
Area to be painted black $=(8\times 66)c{m}^2=528c{m}^2$
Cost for the painting with black colour $=Rs.(528\times 0.05)=Rs.26.40$
Total cost in painting = Rs. (2757.86 + 26.40)
=Rs. 2784.26
Therefore, it will cost Rs. 2784.26 for painting in such a way.