Question

A cylindrical vessel of radius 8 cm and height 15 cm has some water in it. a sphere with radius 3 cm and a cone of height 9 cm are dropped into the vessel and thus the water level raised by 1 5/16 cm. what is the radius of the base of the cone

Answer 1

Answer:

Radius of the cone is 4 cm.

Step-by-step explanation:

Radius of a cylindrical vessel is 8 cm and height 15 cm.

In this vessel a sphere having radius = 3cm and cone with height 9 cm were dropped in this vessel.

Let the radius of the cone = r cm

By immersing the sphere and the cone water level raised by $1\frac{5}{16}$ cm

Since volume of the water raised in cylinder = Volume of sphere + volume of the cone

Volume of water raised in cylinder = $\pi r^{2}h$

= $\pi (8^{2})(\frac{21}{16})$

=  $84\pi$ cm³

Volume of the sphere = $\frac{4}{3}\pi r^{3}$

= $\frac{4}{3}\pi (3)^{3}$

= 36π cm³

Volume of cone = $\frac{1}{3}\pi (r)^{2}(9)$

= 3π(r)²

84π = 36π + 3π(r)²

84 - 36 = 3r²

3r² = 48

r² = 16

r = 4 cm

Therefore, radius of the cone is 4 cm.

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Answer 2

Step-by-step explanation:

Radius of the cone is 4 cm

Radius of a cylindrical vessel is 8 cm and height 15 cm.

In this vessel a sphere having radius = 3cm and cone with height 9 cm were dropped in this vessel.

Let the radius of the cone = r cm

By immersing the sphere and the cone water level raised by 1\frac{5}{16}1

16

5

cm

Since volume of the water raised in cylinder = Volume of sphere + volume of the cone

Volume of water raised in cylinder = \pi r^{2}hπr

2

h

= \pi (8^{2})(\frac{21}{16})π(8

2

)(

16

21

)

= 84\pi84π cm³

Volume of the sphere = \frac{4}{3}\pi r^{3}

3

4

πr

3

= \frac{4}{3}\pi (3)^{3}

3

4

π(3)

3

= 36π cm³

Volume of cone = \frac{1}{3}\pi (r)^{2}(9)

3

1

π(r)

2

(9)

= 3π(r)²

84π = 36π + 3π(r)²

84 - 36 = 3r²

3r² = 48

r² = 16

r = 4 cm

Therefore, radius of the cone is 4 cm.

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