Question

the hight of cylinder is 2h and the radious of bace is half taking this measurement prepare the new cylinder . what is
the relationship with the volume of first cylinder?​

Answer 1

Answer:

$\huge\mathfrak{\red{ Answer }}$

2:1

$\implies$ The volume of the second cylinder is twice the volume of the first cylinder.

Step-by-step explanation:

$\huge\underline{\overline{\mid{\bold{\red{\mathcal{Correct \ Question:- }}\mid}}}}$

$\bold\hookrightarrow$ The height of a cylinder, whose initial height is h and radius is r, is doubled (that is, 2h ) and the radius is taken as half of the initial radius ( that is, r/2 )to make an another cylinder. Find the relationship of the volume of the second cylinder with the volume of the first cylinder?

$\huge\underline{\overline{\mid{\bold{\green{\mathcal{Hint:- }}\mid}}}}$

$\bullet{\leadsto}$ We will find the Volume of the first cylinder and then the volume of the second cylinder.

$\bullet{\leadsto}$ Then, we will Find the relationship between the two cylinders.

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Formulas:- }}\mid}}}}$

Related to a cylinder

Curved surface area :

• 2πrh

Total Surface area :

• 2πr(r+h)

Volume :

• πr^2h

$\huge\underline{\overline{\mid{\bold{\purple{\mathcal{Required \ Solution:- }}\mid}}}}$

Given, Radius of the 1st cylinder is r units

Height is h

$\therefore Volume \: :-$

$\pi {r}^{2} h \: \: \: {units}^{3}$

According to the question;

Radius of the 2nd cylinder is r/2 units

Height is 2h

$\therefore Volume \: is$

$\pi { (\frac{r}{2}) }^{2} 2h \: \: {units}^{3}$

$= > \: \pi \: \frac{ {r}^{2} }{4} 2h \: \: {units}^{3}$

Now, The required relationship is

$\fbox{\red{ Volume of 1st cylinder / Volume of 2nd cylinder}}$

$\implies$ $\huge \frac{\pi {r}^{2}h }{\pi \frac{ {r}^{2} }{4}2h }$

$\huge \frac{1}{ \frac{2}{4} } = \frac{1}{2} \times 4 = \frac{2}{1}$

OR

2:1

That is, the volume of the second cylinder is twice the volume of the first cylinder.

$\tiny\mathfrak{\red{ \ \ \ \ \ \ \ @MissTranquil}}$

Answer 2

Answer:

hope the above answer helps you