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Q) A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Fine the total height of the room if it Contains (\frac{1408}{21}) m3 of air. Take (\pi =\frac{22}{7})

Ans:  Let h be the height of cylindrical part and r be the radius of hemisphere.

Volume of room = \pi r^2h + \frac{2}{3} \pi r^3 = \frac{1408}{21}

Given that r = \frac{h}{2}  or h = 2r

\therefore      Volume of room =  \pi r^2 (2r) + \frac{2}{3} \pi r^3 = \frac{1408}{21}

=          \frac{8}{3} \pi r^3 = \frac{1408}{21}

=          \frac{8}{3} x \frac{22}{7} r^3 = \frac{1408}{21}

=          8 x 22 r3  = 1408

=          r3  = 8   or  r = 2 m

=          h  = 2 r = 2 x 2 = 4 m

Therefore, height of the room is 4 m

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