Question

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Answer 1

Radius of cylinderical tub (r) = 16 cm
Height of water in it (h) = 30 cm
$\therefore$ Volume of water in it $=\pi r^{2}h=\pi (16{)}^2\times 30c{m}^3=\pi \times 256\times 30c{m}^3=7680\pi c{m}^3$
By droppping a ball in it,
Water level raised = 9 cm
$\therefore$ Now volume of raised water
$=\pi \times (16{)}^2\times 9=256\times 9\times \pi c{m}^3=2304\pi c{m}^3$
$\therefore$ Volume of ball $=2304\pi c{m}^3$
Then radius $=3\sqrt{\frac{Volume}{\frac{4}{3}\pi }}=3\sqrt{\frac{2304\pi \times 3}{4\times \pi }}=3\sqrt{576\times 3}=3\sqrt{1728}=3\sqrt{(12{)}^3}=12cm$