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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. Find the total height of the room if it Contains (\frac{1408}{21}) m3 of air. Take (\pi =\frac{22}{7})

Ans: 

Step1: Let h be the height of cylindrical part and r be the radius of hemisphere.  A room is in the form of cylinder surmounted by a hemi-spherical dome.

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

Step 2: Given that r = \frac{h}{2}  or h = 2r

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

=   \frac{8}{3} \pi r^3

=   \frac{8}{3} x \frac{22}{7} r^3

=   \frac{176}{21} r^3

Step 3: Given that the volume of air in the room is  \frac{1408}{21}

∴   \frac{176}{21} r^3 = \frac{1408}{21}

∴  176 r3  = 1408

∴   r3  = 8   

∴  r = 2 m

Step 4:  Since,  h  = 2 r A room is in the form of cylinder surmounted by a hemi-spherical dome.

∴ h =  2 x 2 = 4 m

Step 5: Total Height of the room is:

h + r = 4 +2 = 6m

Therefore, height of the room is 6 m.

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