
A right cylindrical vessel is full of water. How many right cones having the same diameter and height as that of the right cylinder will be needed to store that water?
(A) \[2\]
(B) \[4\]
(C) \[3\]
(D) \[5\]
Answer
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Hint: We will assume the radius of the cylindrical vessel be \[r\] and the height of the cylindrical vessel be \[h\] and \[n\] right cones are needed to store that water. The right cones have the same diameter and height as that of the right cylinder. So, we get the radius of the cone will be \[r\] and the height of the cone will be \[h\]. Volume of water is constant so we will write, Volume of cylindrical vessel \[ = \] \[n \times \] volume of cone. On simplifying this we get the result.
Complete step by step answer:
Let the radius of the cylindrical vessel be \[r\] and the height of the cylindrical vessel be \[h\].
As given in the question, the right cones have the same diameter and height as that of the right cylinder. So, we get the radius of the cone as \[r\] and the height of the cone as \[h\].
Let \[n\] right cones are needed to store that water.
Volume of a cylinder is equal to \[\pi {R^2}H\], where \[R\] is the radius of the cylinder and \[H\] is the height of the cylinder.
Also, volume of a cone is equal to \[\dfrac{1}{3}\pi {R^2}H\], where \[R\] is the radius of the cone and \[H\] is the height of the cone.
Now, as we know in this question, the volume of water is constant.
So, we can write,
\[ \Rightarrow \] Volume of cylindrical vessel \[ = \] \[n \times \] volume of cone
Putting the values, we get
\[ \Rightarrow \pi {R^2}H = n \times \dfrac{1}{3}\pi {R^2}H\]
On cross multiplication, we get
\[ \Rightarrow n = \dfrac{{\pi {R^2}H}}{{\dfrac{1}{3}\pi {R^2}H}}\]
On simplification, we get
\[ \Rightarrow n = 3\]
Therefore, the number of right cones needed to store the water is \[3\]. Hence, option (C) is correct.
Note:
A right cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base and a right cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base.
Complete step by step answer:
Let the radius of the cylindrical vessel be \[r\] and the height of the cylindrical vessel be \[h\].
As given in the question, the right cones have the same diameter and height as that of the right cylinder. So, we get the radius of the cone as \[r\] and the height of the cone as \[h\].
Let \[n\] right cones are needed to store that water.
Volume of a cylinder is equal to \[\pi {R^2}H\], where \[R\] is the radius of the cylinder and \[H\] is the height of the cylinder.
Also, volume of a cone is equal to \[\dfrac{1}{3}\pi {R^2}H\], where \[R\] is the radius of the cone and \[H\] is the height of the cone.
Now, as we know in this question, the volume of water is constant.
So, we can write,
\[ \Rightarrow \] Volume of cylindrical vessel \[ = \] \[n \times \] volume of cone
Putting the values, we get
\[ \Rightarrow \pi {R^2}H = n \times \dfrac{1}{3}\pi {R^2}H\]
On cross multiplication, we get
\[ \Rightarrow n = \dfrac{{\pi {R^2}H}}{{\dfrac{1}{3}\pi {R^2}H}}\]
On simplification, we get
\[ \Rightarrow n = 3\]
Therefore, the number of right cones needed to store the water is \[3\]. Hence, option (C) is correct.
Note:
A right cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base and a right cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base.
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