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

Question 13 Mensuration Exercise 17.5

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is ½ cm. The tube is melted and cast into a right circular cone of height 7 cm.

Find the radius of the cone correct to one decimal place.



When we convert one solid shape to another, its volume remains the same, no matter how different the new shape is. So here, metallic cylindrical tube got converted into right circular cone.

Given internal radius of the tube, r = 3 cm

Thickness of the tube = ½ cm = 0.5 cm

External radius of tube = 3+0.5 = 3.5 cm

Height of the tube, h = 21 cm

Volume of the tube = π (R2-r2)h

=π (3.52-32)×21

= π (12.25-9)×21

= π (3.25)×21

= 68.25 π cm3

Height of the cone, h = 7 cm

Let r be radius of cone.

Volume of cone = (1/3)π r2h

= (1/3)π r2×7

= (7/3)π r2

Since tube is melted and changed into a cone, their volumes remain same.

(7/3)π r2 = 68.25π

r2 = 68.25×3/7 = 29.25

Taking square root on both sides

r = 5.4 cm

Hence the radius of the cone is 5.4 cm.

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