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

2.3 cm

Let the initial depth of water in the can be x cm

Then, volume of initial water column = $πr^2h=π(3.5)^2×xcm^3$

Volume of the sphere = $\frac{4}{3}πr^3=\frac{4}{3}(3.5)^2x\,cm^3$

Now, volume of initial water column + Volume of the sphere = Volume of the cylinder up to the height of
water level

$∴π×(3.5)^2×x\,cm^3+\frac{4}{3}π(3.5)^3$

$=π(\frac{7}{2})^2\,7cm^3$

$∴x+\frac{4}{3}×3.5=7⇒x+\frac{14}{3}=7⇒x=7-\frac{14}{3}=\frac{7}{3}$

∴ The depth of water in the can be before the sphere was put into it = 2.3 cm