The thickness of a hollow metal cylinder is $2 c m .$ It is $70 c m$ long with outer radius of $14 c m .$ Find the volume of metal required to make the cylinder, assuming it is open at both ends. Also, find its weight if the metal weighs $8 g$ per $c m^{3} .$

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Solution

$T h i c k n e s s o f c y l i n d e r = 2 c m$

$L e n g t h o f c y l i n d e r (h) = 70 c m$

$O u t e r R a d i u s (R) = 14 c m$

$T h e n$ $I n n e r R a d i u s (r) = o u t e r r a d i u s - t h i c k n e s s$

$= 14 - 2$ $= 12 c m$

$V o l u m e o f h o l l o w c y l i n d e r = π h (R^{2} - r^{2})$

$= \frac{22}{7} \times 70 \times (14^{2} - 12^{2})$

$= 22 \times 10 \times (196 - 144)$ $= 22 \times 10 \times 52$

$= 11440 c m^{3}$

$W e i g h t o f 1 c m^{3} = 8 g m$

$S O, w e i g h t o f 11440 c m^{3} = 11440 \times 8$

$= 91520 g m$ $= \frac{91520}{1000} k g$

$= 91.52 k g$

$H e n c e,$ $W e i g h t o f t h e m e t a l i s 91.52 k g .$

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