# 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?

Solution:

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

=1372×48/(3×343)

= 64

Hence the number of spheres made is 64.

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