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

Question 18 Mensuration Exercise 17.5

Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.

Solution:

The capacity of a sphere is its volume. It is the area that the sphere occupies. The sphere has a round, three-dimensional shape. Its shape is determined by three axes: the x, y, and z axes.

Given radius of each sphere, r = 2 cm

Volume of a sphere = (4/3)π r3

= (4/3)π ×23

= (4/3)π ×8

= (32/3) π cm3

Volume of 8 spheres = 8×(32/3)π

= (256/3)π cm3

Let R be radius of new sphere.

Volume of the new sphere = (4/3)π R3

Since 8 spheres are melted and casted into a single sphere, volume remains same.

(4/3)π R3 = (256/3)π

4R3 = 256

R3 = 256/4 = 64

Taking cube root

R = 4 cm

Hence the radius of the new sphere is 4 cm.

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