An Important Question of M.L Aggarwal book of class 10 Based on Mensuration Chapter for ICSE BOARD.
The surface area of a solid metallic sphere is given. It is melted and recast into solid right circular cones of given radius and height.
Calculate the radius of the solid sphere and the number of cones recast.
This is the Question Number 27, Exercise 17.5 of M.L Aggarwal.
Deepak BoraNewbie
The surface area of a solid metallic sphere is 1256 cm square. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate (i) the radius of the solid sphere. (ii) the number of cones recast. (Use pi (π) = 3.14).
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(i)Given surface area of the solid metallic sphere = 1256 cm2
4πR2 = 1256
4×3.14×R2 = 1256
R2 = 1256/4×3.14
R2 = 100
R = 10
Hence the radius of solid sphere is 10 cm.
(ii)Volume of the solid sphere = (4/3)πR3
= (4/3)×3.14×103
= (4000/3)× 3.14cm3
= 12560/3 cm3
Radius of the cone, r = 2.5 cm
Height of the cone, h = 8 cm
Volume of the cone = (1/3)πr2h
= (1/3)×3.14×2.52×8
= 157/3 cm3
Number of cones made = Volume of the solid sphere/ Volume of the cone
= (12560/3)÷( 157/3)
= (12560/3)×( 3/157)
= 12560/157
= 80
Hence the number of cones made is 80.