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Question
A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in this tube. Find its thickness.
Solution
Given that the volume of the iron in the tube 192 cm3
Let the thickness of the tube = x cm
Side of the external square = ( 5 + 2x ) cm
∵ Ext. vol. of the tube its internal vol.= volume of iron in the tube, we have,
( 5 + 2x )( 5 + 2x ) x 8 - 5 x 5 x 8 = 192
( 25 + 4x2 + 20x ) x 8 - 200 = 192
200 + 32x2 + 160x - 200 = 192
32x2 + 160x - 192 = 0
x2 + 5x - 6 = 0
x2 + 6x - x - 6 = 0
x( x + 6 ) - ( x + 6 ) = 0
( x + 6 )( x - 1 ) = 0
( x - 1 ) = 0
x = 1
Therefore, thickness is 1 cm.
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