A conical vessel of base radius 9 cm and height 20 cm is full of water. A part of this water is now poured into a hollow cylinder, closed at one end, till the cylinder is completely filled with water. If the base radius and the height of the cylinder are 6 cm and 10 cm respectively, find the volume of water which is left in the cone. (Taken = 3.14)
Whith solution please
Answers
Answer:
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Answer:
ANSWER:-
Given that:-
First Phase:-
Conical vessel of radius 9 cm
Height of the conical vessel is 20 cm
Second Phase:-
The water is poured into a cylinder, lidless
Base is 6 cm and height is 10 cm.
Volume of water now left in cone? What we need to do now? Suspicious!
\boxed{\sf{Volume \ of \ water - Volume \ of \ Cylinder = Required \ Answer}}
Volume of water−Volume of Cylinder=Required Answer
Confusion gone! Let's Do!
\boxed{\rm{Volume \ of \ Cone = \dfrac{1}{3} \pi \times r^2 \times h}}
Volume of Cone=
3
1
π×r
2
×h
Where r is 9 cm
Where h is 20 cm
\boxed{\bf{Volume \ of \ Cylinder = \pi \times r^2 \times h}}
Volume of Cylinder=π×r
2
×h
Where r is 6 cm
Where h is 10 cm
\rm{Volume \ of \ Cone = \dfrac{1}{3} \times 3.14 \times (9)^2 \times 20}Volume of Cone=
3
1
×3.14×(9)
2
×20
\rm{Volume \ of \ Cone = \dfrac{1}{3} \times 3.14 \times 81 \times 20}Volume of Cone=
3
1
×3.14×81×20
\boxed{\boxed{\rm{Volume \ of \ Cone = 1695.6 \ cm^3 }}}
Volume of Cone=1695.6 cm
3
Next phase:-
\bf{Volume \ of \ Cylinder= 3.14 \times (6)^2 \times 10}Volume of Cylinder=3.14×(6)
2
×10
\bf{Volume \ of \ Cylinder = 3.14 \times 36 \times 10}Volume of Cylinder=3.14×36×10
\boxed{\boxed{\bf{Volume \ of \ Cylinder = 1130.4 \ cm^3}}}
Volume of Cylinder=1130.4 cm
3
Now, let us subtract the values to get the answer.
\boxed{\sf{Volume \ of \ water - Volume \ of \ Cylinder = Required \ Answer}}
Volume of water−Volume of Cylinder=Required Answer
[Remember?]
\implies 1695.6-1130.4⟹1695.6−1130.4
\boxed{\sf{\longrightarrow 565.2 \ cm^3 is \ the \ required \ answer.}}
⟶565.2 cm
3
is the required answer.
| \longleftarrow∣⟵ Amount of water left.