A cylindrical vessel of radius $4 \; cm$ contains water. A solid sphere of radius $3 \; cm$ is lowered into water until it is completely immersed. Find the rise in the water level in the vessel.
Let the initial level of water in the cylindrical vessel $= h \; cm$
Radius of the cylindrical vessel $= r = 4 \; cm$
Initial volume of water $= \pi r^2 h = \pi \times 4^2 \times h = 16 \pi h \; cm^3$
Let the radius of the sphere $= R = 3 \; cm$
Volume of sphere $= \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$
Let the new height of water $= H \; cm$
New volume of water $= \pi r^2 H = 16 \pi H \; cm^3$
$\therefore \;$ We have,
$16 \pi h + 36 \pi = 16 \pi H$
i.e. $\;$ $16 \pi \left(H - h\right) = 36 \pi$
i.e. $\;$ $H - h = \dfrac{36}{16} = 2.25 \; cm$
i.e. $\;$ Rise in water level $= 2.25 \; cm$