Question

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

Answer 1

The radius of the ball is 10.5 cm³.

Step-by-step explanation:

Consider the provided information.

Radius of water tank is 14 cm and water level is 3 cm below the top.

When iron ball is dropped it water 3003 cm³ of  water flows out.

Volume of empty tank = $\pi r^2 h$

The volume of ball = Volume of empty tank + Volume of water came out

The volume of ball = $\pi r^2 h$ + Volume of water came out

The volume of ball = $\frac{22}{7}\times14^2\times3$ + 3003 cm³

The volume of ball = 1848 cm³ + 3003 cm³

The volume of ball = 4851 cm³

The volume of sphere is: $V=\frac{4}{3}\pi r^3$

Substitute the respective values in the above formula.

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

$r^3= \dfrac{4851\times3\times7}{22\times4}$

$r^3=1157.625$

$r=10.5$

Hence, the radius of the ball is 10.5 cm³.

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