Question

In a cylindrical vessel of radius 10 cm, containing some water, 9000 small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius 0.5 cm, then find the rise in the level of water in the vessel.

Answer 1

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Answer 2

The rise in the level of water in the vessel = 15 cm.

Given:

• Cylindrical vessel of radius 10 cm.
• 9000 small spherical balls are dropped which are completely immersed in water.
• If each spherical ball is of radius 0.5 cm.

To find:

The rise in the level of water in the vessel.

Formula used:

• Volume of sphere = $\frac{4}{3}$ × $\pi$ × (Radius)³
• Volume of cylinder =  $\pi$ × (Radius)² × (rise in the level of water in the vessel.)

Explanation:

• Volume of 9000 spherical ball = $\frac{4}{3}$ × $\pi$ × (Radius)³×9000
• Volume of 9000 spherical ball = $\frac{4}{3}$ × $\pi$ × (0.5)³ × 9000
• Volume of cylinder = Volume of 9000 spherical ball

         $\pi$ × (10)² × (rise in the level) = $\frac{4}{3}$ × $\pi$ × (0.5)³ × 9000

        "$\pi$" get cancel each other, we get

               100 × (rise in the level) = $\frac{4}{3} \times (0.5)^3 \times 9000$

                          Rise in the level = $\frac{1}{100} \times \frac{4}{3} \times 0.5 \times 0.5 \times 0.5 \times 9000$

                          Rise in the level = 15 cm.

Therefore, the rise in the level is 15 cm.

To learn more...

1. The volume of a cube is 0.125 cm.find the edge. if the length of the edge is increased three times then how much time will the volume be increased

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2. A cuboid has volume four times the volume of a cube whose each edge is 7 m.Find the volume of the cuboid.

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