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.

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# 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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This answer was edited.(i)Given surface area of the solid metallic sphere = 1256 cm^{2}4πR^{2}= 12564×3.14×R^{2}= 1256R^{2}= 1256/4×3.14R^{2}= 100R = 10Hence the radius of solid sphere is 10 cm.(ii)Volume of the solid sphere = (4/3)πR^{3}= (4/3)×3.14×10^{3}= (4000/3)× 3.14cm^{3}= 12560/3 cm^{3}Radius of the cone, r = 2.5 cmHeight of the cone, h = 8 cmVolume of the cone = (1/3)πr^{2}h= (1/3)×3.14×2.5^{2}×8= 157/3 cm^{3}Number of cones made = Volume of the solid sphere/ Volume of the cone= (12560/3)÷( 157/3)= (12560/3)×( 3/157)= 12560/157= 80Hence the number of cones made is 80.