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