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

Question 28 Mensuration Exercise 17.5

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm.

How many such spheres can be obtained?



No matter how different the new shape is, when we change one solid shape into another, its volume stays the same. In fact, if we melt a single large sphere into a number of smaller ones, the combined volumes of the smaller ones match the size of the large spherical.

Given surface area of the sphere = 616 cm2

4π R2 = 616

4×(22/7)R2 = 616

R2 = 616×7/4×22

R2 = 49

R = 7

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

= (4/3)π x 73

= (1372/3)π cm3

Diameter of smaller sphere = 3.5 cm

So radius, r = 3.5/2 = 7/4 cm

Volume of the smaller sphere = (4/3)π r3

= (4/3)π (7/4)3

= (343/48) π cm3

Number of spheres made = Volume of the solid metallic sphere/ Volume of the smaller sphere

= (1372/3)π ÷(343/48)π


= 64

Hence the number of spheres made is 64.

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