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 cmThickness of the tube = ½ cm = 0.5 cmExternal radius of tube = 3+0.5 = 3.5 cmHeight of the tube, h = 21 cmVolume of the tube =(R^{2}-r^{2})h= (3.5

^{2}-3^{2})×21= (12.25-9)×21

= (3.25)×21

= 68.25 cm

^{3}Height of the cone, h = 7 cm

Let r be radius of cone.

Volume of cone = (1/3)r

^{2}h= (1/3)r

^{2}×7= (7/3)r

^{2}Since tube is melted and changed into a cone, their volumes remain same.

(7/3)r

^{2 }= 68.25r

^{2}= 68.25×3/7 = 29.25Taking square root on both sides

r = 5.4 cm

Hence the radius of the cone is 5.4 cm.