A cylindrical vessel and conical vessel is having same radius and height. The volume of cylindrical vessel is 27π litres. Radius of the conical vessel is 9 cm. a) Find the volume of the conical vessel b) What is the height both vessels?
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A cylindrical vessel and conical vessel is having same radius and height. The volume of cylindrical vessel is 27π litres. Radius of the conical vessel is 9 cm. a) Find the volume of the conical vessel b) What is the height both vessels?
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Answer:
height = 3cm.
Step-by-step explanation:
radius = 9cm
volume of cylinder = 27^
volume of cylinder = ^r²h
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that, a cylindrical vessel and conical vessel is having same radius and height.
Radius of cone and cylindrical vessel, r = 9 cm
Volume of cylindrical vessel = 27 π litres = 27000 π cm^3
Let assume that
Height of cone and cylinder be h cm.
We know, Volume of cylinder of radius r and height h is given by
[tex]\boxed{ \rm{ \:Volume_{Cylinder)} \: = \: \pi \: {r}^{2} \: h \: }} \\ \\ [/tex]
So, on substituting the values, we get
[tex]\rm \: 27000\pi \: = \: \pi \: \times \: {9}^{2} \times h\\ [/tex]
[tex]\rm \: 27000 \: = \: 81 \times h\\ [/tex]
[tex]\bf\implies \:h \: = \: \dfrac{27000}{81} = \dfrac{1000}{3} \: cm \\ [/tex]
Now, we know that Volume of cone of radius r and height h is given by
[tex]\boxed{ \rm{ \:Volume_{Cone)} \: = \:\dfrac{1}{3} \: \pi \: {r}^{2} \: h \: }} \\ \\ [/tex]
So, on substituting the values of r and h is given by
[tex]\rm \: Volume_{Cone)} = \dfrac{1}{3} \times \dfrac{22}{7} \times {9}^{2} \times \dfrac{1000}{3} \\ [/tex]
[tex]\rm \: Volume_{Cone)} = \dfrac{22}{7} \times 9 \times 1000 \\ [/tex]
[tex]\rm \: Volume_{Cone)} = \dfrac{22 \times 9000}{7} \\ [/tex]
[tex]\rm \: Volume_{Cone)} = 9000 \: \pi \: {cm}^{3} \\ [/tex]
[tex]\bf\implies \:Volume_{Cone)} \: = \: 9\pi \: litres \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
Short Cut Trick :-
As we know,
[tex]\rm \: \:Volume_{Cone)} \: = \:\dfrac{1}{3} \: \pi \: {r}^{2} \: h \: \\ \\ [/tex]
[tex]\rm \: \:Volume_{Cone)} \: = \:\dfrac{1}{3} \: Volume_{Cylinder)} \: \\ \\ [/tex]
[tex]\rm \: \:Volume_{Cone)} \: = \:\dfrac{1}{3} \: \times \: 27\pi \: \\ \\ [/tex]
[tex]\bf\implies \:Volume_{Cone)} \: = \: 9 \: \pi \: litres \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
Additional Information :-
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]