Question

A cylindrical tub radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub and the level of the water is raised by 6.75 cn. Find the radius of the ball.

Answer 1

We have

Radius of cylinder(r)=12 m

Initial height of the cylinder (h) = 20 cm

Final height of the cylinder (H)= (20 + 6.75) cm

=26.75 cm

So,

Volume of the sphere = Volume of cylinder at final height - Volume of cylinder at initial height

$\frac{4}{3}\pi R^3=\pi r^{2}H-\pi r^{2}h$

$\frac{4}{3}{R}^3=r^2(H-h)$

$\sqrt[3]{3\frac{6.75\times {144}^3}{4}}$

$\sqrt[3]{3729}=9$

Hence radius is 9 cm