A cylinder, a cone and a hemisphere are of the same base and of same height. Find the ratio of their volumes.
Answer
543.7k+ views
Hint:- We had to only write the formula of volume of cylinder, cone and hemisphere with same height and radius and then divide them to get the ratio of their volumes.
Complete step-by-step answer:
As the height and radius of cylinder, cone and hemisphere are the same.
So, let their height be h units.
And their radius is r units.
Now as we know that the height of the hemisphere is the radius of the hemisphere.
So, r = h (because h is the height of all shapes and r is the radius of all shapes)
So, as we know that if h is the height and r is the radius of cylinder then its volume is calculated as \[\pi {r^2}h\]
Let the volume of the cylinder is \[{V_1}\].
So, \[{V_1} = \pi {r^2}h = \pi {r^3}\] (because h = r)
As we know that if h is the height and r is the radius of cone then its volume is calculated as \[\dfrac{1}{3}\pi {r^2}h\]
Let the volume of the cone is \[{V_2}\].
So, \[{V_2} = \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3}\pi {r^3}\] (because h = r)
As we know that if r is the radius of hemisphere then its volume is calculated as \[\dfrac{2}{3}\pi {r^3}\]
Let the volume of the hemisphere is \[{V_3}\].
So, \[{V_3} = \dfrac{2}{3}\pi {r^3}h\]
Now the ratio of the volumes of cylinder, cone and hemisphere is the ratio of \[{V_1}\], \[{V_2}\] and \[{V_3}\].
So, \[{V_1}:{V_2}:{V_3} = \pi {r^3}:\dfrac{1}{3}\pi {r^3}:\dfrac{2}{3}\pi {r^3} = 1:\dfrac{1}{3}:\dfrac{2}{3}\]
On multiplying ratio be 3. We get,
\[{V_1}:{V_2}:{V_3} = 3:1:2\]
Hence, the volume of the cylinder cone and hemisphere are in ratio 3 : 1 : 2.
Note:- Whenever we come up with this type of problem the first, we had to write the formula of all the given shapes using same parameters and the we check whether any parameter is missing in any formula like here in the formula of volume of hemisphere, height is not present but we know that the height of the hemisphere is same as its radius. So, we had to replace height with radius in the formula of volume of all given shapes because they all had the same height and radius. And after that we can easily divide their volume to get the required ratio of their volumes.
Complete step-by-step answer:
As the height and radius of cylinder, cone and hemisphere are the same.
So, let their height be h units.
And their radius is r units.
Now as we know that the height of the hemisphere is the radius of the hemisphere.
So, r = h (because h is the height of all shapes and r is the radius of all shapes)
So, as we know that if h is the height and r is the radius of cylinder then its volume is calculated as \[\pi {r^2}h\]
Let the volume of the cylinder is \[{V_1}\].
So, \[{V_1} = \pi {r^2}h = \pi {r^3}\] (because h = r)
As we know that if h is the height and r is the radius of cone then its volume is calculated as \[\dfrac{1}{3}\pi {r^2}h\]
Let the volume of the cone is \[{V_2}\].
So, \[{V_2} = \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3}\pi {r^3}\] (because h = r)
As we know that if r is the radius of hemisphere then its volume is calculated as \[\dfrac{2}{3}\pi {r^3}\]
Let the volume of the hemisphere is \[{V_3}\].
So, \[{V_3} = \dfrac{2}{3}\pi {r^3}h\]
Now the ratio of the volumes of cylinder, cone and hemisphere is the ratio of \[{V_1}\], \[{V_2}\] and \[{V_3}\].
So, \[{V_1}:{V_2}:{V_3} = \pi {r^3}:\dfrac{1}{3}\pi {r^3}:\dfrac{2}{3}\pi {r^3} = 1:\dfrac{1}{3}:\dfrac{2}{3}\]
On multiplying ratio be 3. We get,
\[{V_1}:{V_2}:{V_3} = 3:1:2\]
Hence, the volume of the cylinder cone and hemisphere are in ratio 3 : 1 : 2.
Note:- Whenever we come up with this type of problem the first, we had to write the formula of all the given shapes using same parameters and the we check whether any parameter is missing in any formula like here in the formula of volume of hemisphere, height is not present but we know that the height of the hemisphere is same as its radius. So, we had to replace height with radius in the formula of volume of all given shapes because they all had the same height and radius. And after that we can easily divide their volume to get the required ratio of their volumes.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the name of Japan Parliament?

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

