Question

A cylindrical can whose base is horizontal and is of internal radius $3.5$ cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the can exactly. The depth of water in the can before the sphere was put, is

Answer 1

Depth of water in the can before sphere is put inside it

=volume of water/area of cross section of can

$=\frac{\frac{2\pi R^3}{3}}{\pi R^2}=2\frac{R}{3}=2\times \frac{3.5}{3}$

$=\frac{7}{3}cm$