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Question

The thickness of a hollow metallic cylinder is 2 cm. It is 70 cm long with outer radius of 14 cm. Find the volume of the metel used in making the cylinder assuming that it is open at both the ends Also find its weight if the metal weighdds 8 g per cm3.

Solution
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Let $$t$$ is thickness of the cylinder
$${ r }_{ 1 }$$ is outer radius of the cylinder
$${ r }_{ 2 }$$ is inner radius of the cylinder
$$h$$ is height of the cylinder

$$\therefore t=2 cm$$
$${ r }_{ 1 }=14 cm$$
$$h=70 cm$$

Now, inner radius is calculated as,
$${ r }_{ 2 }={ r }_{ 1 }-t$$
$$\therefore { r }_{ 2 }=14-2=12 cm$$

1) Volume of the hollow cylinder is given by,
$$V=c/s\quad area\times height\quad of\quad cylinder$$
$$\therefore V=\left( \pi { { r }_{ 1 } }^{ 2 }-\pi { { r }_{ 2 } }^{ 2 } \right) \times h$$

$$\therefore V=\pi \left( 14^{ 2 }-12^{ 2 } \right) \times 70$$

$$\therefore V=\frac { 22 }{ 7 } \left( 196-144 \right) \times 70$$

$$\therefore V=22\times 52\times 10$$

$$\therefore V=11440\quad { cm }^{ 3 }$$

2) Given wright of the metal, $$\rho =8\quad \frac { g }{ { cm }^{ 3 } } $$

Let $$m$$ = mass of the cylinder
$$\therefore \rho =\frac { m }{ V } $$

$$\therefore 8=\frac { m }{ 11440 } $$

$$\therefore m=8\times 11440$$

$$\therefore m=91520 gm$$

$$\therefore m=91.520 kg$$

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