ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.5 in Chapter 17 - Mensuration

Question 24 Mensuration Exercise 17.5

The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm.

Calculate the number of cones recast. (Use π = 3.14).



No matter how different the new shape is, when we change one solid shape into another, its volume stays the same. In fact, the volume of the large sphere is equal to the sum of the volumes of the many little cones that result from melting a single large sphere.

Volume of the solid sphere = (4/3)π R3

= (4/3)π x 103

= (4000/3) π cm3

= 12560/3 cm3

Radius of the cone, r = 2.5 cm

Height of the cone, h = 8 cm

Volume of the cone = (1/3)π r2h

= (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.

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