Question

A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

Answer 1

Hint:In the above equation we will use the concept that the volume of sphere dropped in the right circular cylindrical vessel is equal to the volume increased in the cylindrical vessel. Also, we will use the formulae of volume of sphere and volume of right circular cylindrical vessel which is given as below:Volume of sphere = \[\dfrac{4}{3}\pi {{R}^{3}}\] where R is the radius of the sphere.Volume of the right circular cylinder = \[\pi {{R}^{2}}h\] where R is radius and h is the height of the vessel.

Complete step-by-step answer:
Let us first draw the figure and mark the details.

We have been given,
Diameter of the sphere = 6 cm
Radius (R) of the sphere = \[\dfrac{6}{2}=3\] cm
Diameter of cylindrical vessel = 12 cm
Radius (R) of the cylindrical vessel = \[\dfrac{12}{2}=6\] cm
Volume of sphere = \[\dfrac{4}{3}\pi {{R}^{3}}\] = \[\dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}c{{m}^{3}}\]
Volume of cylindrical vessel = \[\pi {{r}^{2}}h=\pi \times {{6}^{2}}\times hc{{m}^{3}}\]
Here, the volume of the sphere = Volume increased in the cylindrical vessel.
\[\begin{align}
  & \therefore \dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}=\pi \times {{\left( 6 \right)}^{2}}\times h \\
 & \dfrac{4}{3}\times 3\times 3\times 3=6\times 6\times h \\
\end{align}\]
\[h=\dfrac{4\times 3\times 3}{6\times 6}=1\] cm
Therefore, there is a 1 cm level of water rise in the cylindrical vessel.

Note: Just be careful while doing calculation as there is a chance that you might make a mistake. Also, remember that we have been given the diameter of the sphere as well as the right circular cylindrical vessel so we will have to convert them into radius.Students should remember the formulas of volume of sphere and cylindrical body for solving these questions.