A solid is in the form of a right circular cone mounted on a hemisphere
The radius of the hemisphere is 2.1 cmn and the height of the cone is
4 cm. The solid is placed in a cylindrical tub full of water in such a
way that the whole solid is submerged in water. If the radius of the
cylinder is 5 cm and its height is 9.8 cm, find the volume of the water
left in the tub
(CBSE 1996C, 2000)
Answers
Answer:
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Step-by-step explanation:
Given, radius of the hemisphere,r=3.5cm
Now, since the solid is in the form of a right circular cone mounted on a hemisphere,
then radius of base of the cone=radius of the hemisphere
⇒ radius of the base of the cone=r=3.5cm
Height of the cone,h=4cm
So,
volume of the solid=volume of the cone+ volume of the hemisphere
⇒ volume of the solid=
3
1
πr
2
h+
3
2
πr
3
⇒ volume of the solid=
3
1
πr
2
(h+2r)
⇒ volume of the solid=
3
1
×
7
22
×(4+7)=141.16cm
3
Now, radius of the base of the cylindrical vessel,r
1
=5cm
Height of the cylindrical vessel,h
1
=10.5cm
∴ Volume of the water in the cylindrical vessel==πr
1
2
h
1
=
7
22
×25×10.5=825cm
3
Now, when the solid is completely submerged in the cylindrical vessel full of water, then
volume of the water displaced by the solid= volume of solid
Hence, volume of the water left in the vessel= volume of the water in the vessel- volume of solid
=(825−141.16)cm
3
=683.84cm
3
.
Answer:
683.84 is your answer friend