An Important Question of M.L Aggarwal book of class 10 Based on Mensuration Chapter for ICSE BOARD.
Here a hollow metallic cylindrical tube has an internal radius and height . The thickness of the metal of the tube is 1/2 cm. The tube is melted and cast into a right circular cone of given height.
Find the radius of the cone.
This is the Question Number 12, Exercise 17.5 of M.L Aggarwal.
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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 1/2 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.
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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 = (R2-r2)h
= (3.52-32)×21
= (12.25-9)×21
= (3.25)×21
= 68.25 cm3
Height of the cone, h = 7 cm
Let r be radius of cone.
Volume of cone = (1/3)r2h
= (1/3)r2×7
= (7/3)r2
Since tube is melted and changed into a cone, their volumes remain same.
(7/3)r2 = 68.25
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.