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A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is $10.4 cm$ and its length is $25 cm .$ The thickness of the metal is $8 mm$ everywhere. Calculate the volume of the metal.

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543

c m 3

704

c m 3

532

c m 3

786

c m 3

answer is B.

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Detailed Solution

Given,
Length of a hollow tube (h) = 25

cm

Inner diameter = 10.4 cm
Thickness of the metal = 8 mm.
Radius =

D i a m e t e r 2

So, inner radius (r) is,

r = 10.4 2 ⇒ r = 5.2 cm

Since, thickness of the metal = 8 mm

⇒

Thickness of the metal = 0.8

cm

{1cm = 10mm}
So, outer radius(R) = Thickness + Inner radius
R= 0.8+5.2

⇒

R=6

cm

The volume of the material contained in a hollow cylindrical shell,
Volume =

π R 2 − r 2 h

⇒ V o l u m e = 227 × 62 − 5.2 2 × 25

⇒ V o l u m e = 5507 36 − 27.04

⇒ V o l u m e = = 5507 × 8.96

⇒ V o l u m e = = 5507 × 896100

⇒ V o l u m e = = 704 c m 3

Therefore, the volume of the metal is

704 c m 3

.
Hence option (2) is correct.

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