Question
CBSEENMA9003766

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in 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 per cm2 and black paint costs 5 paise per cm2.

Solution

For a wooden sphere
Diameter = 21 cm
therefore   Radius (r) = 21 over 2 space cm

∴ Surface area of a wooden sphere = 4
therefore space space straight r squared space equals 4 cross times 22 over 7 cross times open parentheses 21 over 2 close parentheses squared equals 1386 space cm squared 
∴ Surface area of a wooden sphere to be painted = 1386 – straight pi(1.5)2
equals 1386 minus 22 over 7 left parenthesis 1.5 right parenthesis squared equals 1378.93
∴ Surface area of eight wooden spheres = 1378.93 x 8 = 11031.44 cm2 Cost of painting silver @ 25 paise per cm2 = 11031.44 x 25 paise
equals straight र fraction numerator 11031.44 cross times 25 over denominator 100 end fraction equals straight र space 2757.86

For a cylindrical support
Radius (r) = 1.5 cm Height (h) = 7 cm
Surface area of a cylindrical support = 2straight pirh

equals 2 cross times 22 over 7 cross times 1.5 cross times 7 equals 66 space cm squared

∴ Surface area of eight cylindrical supports = 66 x 8 = 528 cm2
∴ Cost of painting black @ 5 paise per cm2 = 528 x 5 paise
equals straight र space fraction numerator 528 cross times 5 over denominator 100 end fraction equals bold र bold space bold space bold 26 bold. bold 40

∴ Cost of paint required = 2757.86 + 26.40 = र 2784.26 (approx.)

Sponsor Area