A cylindrical tube, open at both ends, is made of metal.

Question:

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.

Solution:

Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, = 25 cm
Outer radius of the cylindrical tube, R = (5.2 + 0.8) cm = 6 cm

$\therefore$ Volume of the metal $=$ external volume $-$ internal volume

$=\pi R^{2} h-\pi r^{2} h \quad$ (where $R$ and $r$ are the outer and inner radii, respectively)

$=\frac{22}{7} \times 25 \times\left(6^{2}-5.2^{2}\right)$

$=\frac{22}{7} \times 25 \times(36-27.04)$

$=\frac{22}{7} \times 25 \times 8.96=704 \mathrm{~cm}^{3}$

 

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