A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

Internal diameter of cylindrical can = 21 cm.

Radius (R) = $\dfrac{21}{2}$ cm.

Diameter of sphere = 10.5 cm = $\dfrac{21}{2}$ cm.

Radius of sphere (r) = $\dfrac{\dfrac{21}{2}}{2} = \dfrac{21}{4}$ cm.

Let the rise in water level be h.

Rise in volume of water = Volume of sphere immersed.

$\Rightarrow πR^2h = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow \Big(\dfrac{21}{2}\Big)^2 \times h = \dfrac{4}{3} \times \Big(\dfrac{21}{4}\Big)^3 \\[1em] \Rightarrow h = \dfrac{4 \times 21^3 \times 2^2}{3 \times 4^3 \times 21^2} \\[1em] \Rightarrow h = \dfrac{4 \times 21 \times 4}{3 \times 64} \\[1em] \Rightarrow h = \dfrac{336}{192} \\[1em] \Rightarrow h = 1.75 \text{ cm}.$

Hence, the rise in water level is 1.75 cm.