
A right cylinder, a right cone and a hemisphere have the same height and the same base area. Then the ratio of their volumes is:
A. $2:1:4$
B. $1:2:3$
C. $3:1:2$
D. $1:3:2$
Answer
568.5k+ views
Hint:In this question, we know volume of the cylinder, cone and the hemisphere formulas that is $\pi {r^2}h,\dfrac{1}{3}\pi {r^2}h,\dfrac{2}{3}\pi {r^3}$ respectively. Then we take their ratio and use the given information of the same height and the same base area and calculate the ratio of their volumes.
Complete step-by-step answer:
We are given the cylinder, cone and the hemisphere having the same base area and height. So consider one cylinder, cone and the hemisphere. The base area of the cylinder with radius $r$ is $\pi {r^2}$.
So the base area of the cone and the hemisphere is also $\pi {r^2}$. Now let the height of the cylinder be $h$
Then the height of the cone is also \[h\].
Now we know that the height of the hemisphere is also the radius of the hemisphere.
So we can say that \[h = r\]$ - - - - - \left( 1 \right)$
Now taking volume of the cylinder: volume of the cone: volume of the hemisphere
We know that the volume of the cylinder, cone and the hemisphere that is $\pi {r^2}h,\dfrac{1}{3}\pi {r^2}h,\dfrac{2}{3}\pi {r^3}$ respectively.
So $\pi {r^2}h:\dfrac{1}{3}\pi {r^2}h:\dfrac{2}{3}\pi {r^3}$
Now using (1) we get,
$1:\dfrac{1}{3}:\dfrac{2}{3}$
Multiplying 3 in above ratios we get,
$3:1:2$
So the ratio is $3:1:2$
So, the correct answer is “Option C”.
Note:The tricky part of this question is to know that height of the hemisphere is also the radius of the hemisphere. And from this point, we get all the shapes.Students should remember formulas of volume of cone , cylinder and hemisphere to solve these types of questions.
Complete step-by-step answer:
We are given the cylinder, cone and the hemisphere having the same base area and height. So consider one cylinder, cone and the hemisphere. The base area of the cylinder with radius $r$ is $\pi {r^2}$.
So the base area of the cone and the hemisphere is also $\pi {r^2}$. Now let the height of the cylinder be $h$
Then the height of the cone is also \[h\].
Now we know that the height of the hemisphere is also the radius of the hemisphere.
So we can say that \[h = r\]$ - - - - - \left( 1 \right)$
Now taking volume of the cylinder: volume of the cone: volume of the hemisphere
We know that the volume of the cylinder, cone and the hemisphere that is $\pi {r^2}h,\dfrac{1}{3}\pi {r^2}h,\dfrac{2}{3}\pi {r^3}$ respectively.
So $\pi {r^2}h:\dfrac{1}{3}\pi {r^2}h:\dfrac{2}{3}\pi {r^3}$
Now using (1) we get,
$1:\dfrac{1}{3}:\dfrac{2}{3}$
Multiplying 3 in above ratios we get,
$3:1:2$
So the ratio is $3:1:2$
So, the correct answer is “Option C”.
Note:The tricky part of this question is to know that height of the hemisphere is also the radius of the hemisphere. And from this point, we get all the shapes.Students should remember formulas of volume of cone , cylinder and hemisphere to solve these types of questions.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

