Question -
Answer -
Given,
The radius of theCylindrical tub (r) = 5 cm
Height of theCylindrical tub (H) = 9.8 cm
Height of the coneoutside the hemisphere (h) = 5 cm
Radius of the hemisphere= 3.5 cm
Now, we know that
The volume of theCylindrical tub (V1) = ╧Аr2H
V1┬а=╧А(5)2┬а9.8
V1┬а=770 cm3
And, the volume of theHemisphere (V2) =┬а2/3 ├Ч ╧А ├Ч r3
V2┬а=┬а2/3├Ч 22/7 ├Ч 3.53
V2┬а=89.79 cm3
And, the volume of theHemisphere (V3) =┬а23 ├Ч ╧А ├Ч r ├Ч 2h
V3┬а=┬а2/3├Ч 22/7 ├Ч 3.52┬а├Ч 5
V3┬а=64.14 cm3
Thus, total volume (V)= Volume of the cone + Volume of the hemisphere
=┬аV2┬а+V3
V = 89.79 + 64.14 cm3┬а
= 154 cm3
So, the total volumeof the solid = 154 cm3
In order to find thevolume of the water left in the tube, we have to subtract the volume of thehemisphere and the cone from the volume of the cylinder.
Hence, the volume ofwater left in the tube = V1┬атАУ V2
= 770 тАУ 154
= 616 cm3
Therefore, the volumeof water left in the tube is 616 cm3.