Answer
Verified
468.9k+ views
Hint: Here first we assume radius and height for all the three shapes. Then put radius and height in the formula of volume of all the three shapes. Then find the ratio. Use formula for the volume of the cylinder,$V=\pi {{r}^{2}}h$ , Volume of the cone,$V=\frac{1}{3}\pi {{r}^{2}}h$ and Volume of the hemisphere,$V=\frac{2}{3}\pi {{r}^{2}}h$
Complete answer:
Let the radius of the base be r and height be h in cylinder, cone and hemisphere.
As we know that height of the hemisphere = radius of the hemisphere$\Rightarrow h=r$.
Volume of the cylinder = $\pi {{r}^{2}}h=\pi {{r}^{2}}r=\pi {{r}^{3}}$.
Volume of the cone = $\frac{1}{3}\pi {{r}^{2}}h=\frac{1}{3}\pi {{r}^{2}}r=\frac{1}{3}\pi {{r}^{3}}$ [taken r=h]
Volume of hemisphere = $\frac{2}{3}\pi {{r}^{3}}$.
So, now the ratio of the volumes of the cylinder, cone and hemisphere will be -
\[\Rightarrow \pi {{r}^{3}}:\frac{1}{3}\pi {{r}^{3}}:\frac{2}{3}\pi {{r}^{3}}\][Common factor $\pi {{r}^{3}}$ is removed from each terms]
\[\Rightarrow 1:\frac{1}{3}:\frac{2}{3}\][Multiplying all the terms with $''3''$ ]
\[=3:1:2\].
Therefore the required answer is - The ratio of volume of a cylinder: volume of cone: volume of hemisphere of same radius and same height is $3:1:2$
Hence, from the given multiple choices, the option C is the correct answer.
Note:: In such types of problems where every term is unknown take the help of the standard general formula, and accordingly follow the given conditions. Also, keep in mind which terms are taken in ratio as $\frac{1}{2}\text{ and }\frac{2}{1}$ make the major difference. So wisely take the ratio by converting the word statement in the proper mathematical form.
Complete answer:
Let the radius of the base be r and height be h in cylinder, cone and hemisphere.
As we know that height of the hemisphere = radius of the hemisphere$\Rightarrow h=r$.
Volume of the cylinder = $\pi {{r}^{2}}h=\pi {{r}^{2}}r=\pi {{r}^{3}}$.
Volume of the cone = $\frac{1}{3}\pi {{r}^{2}}h=\frac{1}{3}\pi {{r}^{2}}r=\frac{1}{3}\pi {{r}^{3}}$ [taken r=h]
Volume of hemisphere = $\frac{2}{3}\pi {{r}^{3}}$.
So, now the ratio of the volumes of the cylinder, cone and hemisphere will be -
\[\Rightarrow \pi {{r}^{3}}:\frac{1}{3}\pi {{r}^{3}}:\frac{2}{3}\pi {{r}^{3}}\][Common factor $\pi {{r}^{3}}$ is removed from each terms]
\[\Rightarrow 1:\frac{1}{3}:\frac{2}{3}\][Multiplying all the terms with $''3''$ ]
\[=3:1:2\].
Therefore the required answer is - The ratio of volume of a cylinder: volume of cone: volume of hemisphere of same radius and same height is $3:1:2$
Hence, from the given multiple choices, the option C is the correct answer.
Note:: In such types of problems where every term is unknown take the help of the standard general formula, and accordingly follow the given conditions. Also, keep in mind which terms are taken in ratio as $\frac{1}{2}\text{ and }\frac{2}{1}$ make the major difference. So wisely take the ratio by converting the word statement in the proper mathematical form.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If x be real then the maximum value of 5 + 4x 4x2 will class 10 maths JEE_Main
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE