A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use $\pi =\dfrac{22}{7}$)

Answer Verified Verified
Hint: Find the total volume of water using the formula for volume of a cone, given by $V=\dfrac{1}{3}\pi {{r}^{2}}h$, where base radius, r is 5 cm and height, h is 24 cm. Use the fact that the volume of water remains the same when it is emptied into a cylindrical vessel. For the volume of a cylindrical vessel use the formula $V=\pi {{r}^{2}}h$. Equate both the volumes to find the value of height of the cylinder.

Complete step-by-step answer:
We know that the volume of water in a vessel is the same as the volume of the vessel it is kept in. Thus, the total volume of water in the conical vessel can be calculated as the volume of the cone, given by $V=\dfrac{1}{3}\pi {{r}^{2}}h$. Using $r=5cm$ and $h=24cm$ in this formula, we get


  & V=\dfrac{1}{3}\pi {{r}^{2}}h \\

 & \Rightarrow V=\dfrac{1}{3}\pi {{\left( 5cm \right)}^{2}}\left( 24cm \right) \\

 & \Rightarrow V=\pi \left( 25c{{m}^{2}} \right)\left( 8cm \right) \\

 & \Rightarrow V=200\pi c{{m}^{3}} \\


Now, since this entire volume is transferred to a cylindrical vessel, the volume of water would be the same as the volume of the cylinder, which can be given by $V=\pi {{r}^{2}}h$. This volume would be equal to the volume of the cube and the base radius is given as 10 cm. Equating the two volumes thus gives us

$\pi {{r}^{2}}h=200\pi c{{m}^{3}}$

Substituting the value of $r=10cm$ in this equation, we get


  & \pi {{\left( 10cm \right)}^{2}}h=200\pi c{{m}^{3}} \\

 & \Rightarrow \pi \left( 100c{{m}^{2}} \right)h=200\pi c{{m}^{3}} \\

 & \Rightarrow 100h=200cm \\

 & \Rightarrow h=2cm \\


Thus the height upto which water is filled in the cylindrical vessel is 2 cm.

Note: To make calculations easier, the value of $\pi $ has not been substituted, even though it is given in the question, because $\pi $ occurs in the expression for both these volumes and hence, cancels out when the volumes are equated, thus reducing the calculations to a great extent.
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