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

$\bf{\underline{\underline{Answer:}}}$

$\therefore 12\:cm$

$\bold{\underline {Given:}}$

$Here ,\: radius \:of \:cylindrical \:tub\: ( r ) = 16m \\ and\: rise\: in\: water\: lever(h) =9cm$

$\bold{\underline {To\:find:}}$

$The \:radius\: of\: the\: ball = ?$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

Volume of the water displaced by iron ball in cylindrical tub = $\pi r^2h$

=$\pi(16)^2×9cm^3$

Now , let x be the radius of the ball .

Volume of spherical iron ball = $\dfrac{4}{3}\pi x^3$

Clearly , volume of spherical ball = volume of water displaced in cylindrical tub

•°• $\dfrac{4}{3}\pi x^3 = \pi (16)^2×9$

$⇝ \dfrac{4}{3} x^3 = 16×16×9$

$⇝ x^3 = \dfrac{16×16×9×3}{4}$

$⇝ 16×4×9×3=(4×3)^3$

$⇝ x = 4×3=12cm$

$\bf{Hence ,\: radius \:of \:the\: ball\: is \:12 cm .}$