Question

Solid spherical balls each of diameter 1.4 cm are dropped into a cylindrical beaker containing water upto a height of 28 cm and are fully submerged. The diameter of the beaker is 5.6 cm. If the water in the beaker rises by 17.5 cm, then find the number of balls dropped in it.


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Answer 1

Solution :-

Let total n number of balls dropped in the cylindrical beaker.

given that,

→ Diameter of each Solid spherical ball = 1.4cm.

So,

→ radius = diameter / 2 = 1.4/2 = 0.7cm.

Now,

→ volume of one Solid spherical ball = (4/3) * π * (radius)³ = (4/3) * π * (0.7)³ cm³.

So,

→ Volume of n such spherical balls = n * {(4/3) * π * (0.7)³} cm³. ---------- Eqn.(1)

Now, we have given that,

→ The diameter of the beaker = 5.6 cm.

→ radius = (5.6/2) = 2.8cm.

and, the water in the beaker rises by 17.5 cm, because of those balls .

So,

→ Volume of water raised = π * (radius)² * height = {π * (2.8)² * 17.5} cm³ . ------------ Eqn.(2)

Now, we can conclude that,

→ Volume of all spherical balls = Volume of water raised

→ Eqn.(1) = Eqn.(2)

Putting values we get :-

→ n * {(4/3) * π * (0.7)³} = {π * (2.8)² * 17.5}

π will be cancel from both sides ,

→ n * (4/3) * (7/10)³ = (28/10)² * 17.5

→ n * 4 * 343 * 100 = 784 * 17.5 * 1000 * 3

→ 1372n = 784 * 525

→ n = (784 * 525) / 1372

→ n = 300 (Ans.)

Hence, Total 300 balls were dropped into cylindrical beaker.