Question

A test tube has lower part hemispherical and upper part is cylindrical with same radius. If 5159/6 cubic cm of water is added, the test tube will be just completely filled. But if 2002/3 of water is added, 5 cm of height of the will remain empty. Find the radius and the height of the cylindrical part. ​

Answer 1

Radius and the height of the cylindrical part is 3.5 cm. and 20 cm. respectively.

Given:

• A test tube has lower part hemispherical and upper part is cylindrical with same radius.
• If 5159/6 cubic cm of water is added, the test tube will be just completely filled.
• If 2002/3 of water is added, 5 cm of height of the will remain empty.

Find:

        Radius and the height of the cylindrical part. ​

Formula used:

• Volume of hemisphere = $\frac{2}{3}$×$\pi$×(radius)³
• Volume of cylinder = $\pi$ ×(radius)² (Height)

Explanation:

• If 5159/6 cubic cm of water is added, the test tube will be just completely filled.
• If 2002/3 of water is added, 5 cm of height of the will remain empty.
• So from above data it can be concluded that Volume which fill remaining     5 cm cylinder = $\frac{5159}{6}$ - $\frac{2002}{3}$ = $\frac{1155}{6}$
• Volume that 5 cm. height cylinder = $\pi$ ×(radius)² (5)

                                                  $\frac{1155}{6}$  = $\pi$ ×(radius)² (5)

                                             (radius)² =  $\frac{1155}{6}$×$\frac{1}{5\pi }$

                                              radius = 3.5 cm.

• Given that radius of cylinder and hemisphere is same
• Therefore, Volume of hemisphere = $\frac{2}{3}$×$\pi$×(radius)³

                                                         =$\frac{2}{3}$×$\pi$×(3.5)³

                                                         = 89.797 cm.³

• Total volume of cylinder and hemisphere is equal to  $\frac{5159}{6}$  ......Given
• so total volume =  $\pi$ ×(radius)² (Height)  + $\frac{2}{3}$×$\pi$×(radius)³

                    $\frac{5159}{6}$ =  $\pi$ ×(3.5)² (Height) + 89.797 cm.³

                  Height = 20 cm.

To learn more....

1 What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

https://brainly.in/question/9846950

2 A sphere and right circular cylinder of same radius have equal volume .by what percent does diameter of cylider exceed its height???please sir reply fast its vry urgent for me.

https://brainly.in/question/301651

Answer 2

The radius and the height of the cylindrical part are 17.49 cm and 10.67 cm.

Step-by-step explanation:

Volume of water that completely fills the test tube = 5159/6 cm³

Total volume = Volume of cylinder + Volume of hemisphere

The volume of cylinder is given by the formula:

V₁ = πr²h

The volume of hemisphere is given by the formula:

V₂ = 2/3 πr³

Now,

Total volume = πr²h + 2/3 πr³

5159/6 = πr²h + 2/3 πr³

5159/6 = πr² (h + 2/3r) → (equation 1)

The volume of the water filled with 5 cm empty = 2002/3 cm³

Total volume = πr²(h - 5) + 2/3 πr³

2002/3 = πr²(h - 5) + 2/3 πr³

2002/3 = πr² (h - 5 + 2/3r) → (equation 2)

On dividing equation (1) by (2), we get,

(5159/6)/(2002/3) = (πr² (h + 2/3r))/(πr² (h - 5 + 2/3r)) → (equation 3)

On subtracting equation (1) from (2), we get,

2002/3 - 5159/6 = πr² (h - 5 + 2/3r) - πr² (h + 2/3r)

- 1155/6 = πr² (-5)

1155/6 = 22/7 × r² × (5)

r² = 192.5 × 0.3181 × (5)

∴ r = 17.49 cm

On substituting value of r in equation 3, we get,

(5159/6)/(2002/3) = (πr² (h + 2/3r))/(πr² (h - 5 + 2/3r))

(5159/6)/(2002/3) = (h + 2/3r)/(h - 5 + 2/3r)

5159/4004 = (h + 2/3r)/(h - 5 + 2/3r)

5159(h - 5 + 2/3r) = 4004(h + 2/3r)

5159h - 25795 + 3439.3r = 4004h + 2669.3r

5159h - 25795 + 3439.3(17.49) = 4004h + 2669.3(17.49)

5159h - 25795 + 60153.357 = 4004h + 46686.057

60153.357 - 25795 - 46686.057 = 4004h - 5159h

-12327.7 = -1155h

h = 12327.7/1155

∴ h = 10.67 cm