Question

Water is filled in a right cylindrical tank with base radius 14cm , such that water level is 3 cm below the top. when an iron ball is dropped in the tank,3003 cm cube of water flows out. Find the radius of the ball.​

Answer 1

The radius of the ball = 10.5 cm

Step-by-step explanation:

radius = 14 cm

let h be the height of the cylinder

volume of tank = $\pi r^2h$

= $\pi \times (14)^2\times h$

now the water level is 3 cm below the top

height of the water level = (h-3) cm

volume of the water in the tank = $\pi \times(14)^2\times(h-3)$

when an  iron ball is dropped in the tank,3003 cm cube of water flows out.

volume of iron ball = (volume of cylindrical tank - volume of water in tank) + 3003

= $(\pi (14)^2h - \pi (14)^2 (h-3)) +3003$

= $(\pi (14)^2h - \pi (14)^2 h + \pi (14)^2\times 3) +3003$

= $588\pi +3003$

= $588\times\frac{22}{7} +3003$

= 4851 cm³

let the radius of the ball = r

volume of iron ball

$\frac{1}{3}\pi  r^3 = 4851$

$\frac{4}{3}\times \frac{22}{7}\times r^3 = 4851$

on solving

r³ = 1157.625

r = $\sqrt[3]{1157.625}$

r = 10.5 cm

hence,

The radius of the ball = 10.5 cm

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