Question

A conical flask is full of water. The flask has base radius a and height 2a. The water is poured into a cylindrical flask of base-radius $ \dfrac{{2a}}{3} $ . Find the height of water in the cylindrical flask.
A. $ \dfrac{5}{2}a $
B. $ 2a $
C. $ \dfrac{3}{2}a $
D. $ 1a $

Answer 1

Hint: Complete water from conical flask is poured into a cylindrical flask. So the volume occupied by water in the cylindrical flask is equal to the volume of conical flask. Find the volume of conical and cylindrical flasks by substituting the given measurements in their formulas. Equate the volume of conical flask with the volume of cylindrical flask and find the height of cylindrical flask.

Complete step-by-step answer:
Volume of a cone is $ \dfrac{1}{3}\pi {r^2}h $ , where r is the base radius of the cone, h is the height of the cone and $ \pi = \dfrac{{22}}{7} $
Volume of a cylinder is $ \pi {r^2}h $ , where r is the base radius of the cylinder, h is the height of the cylinder and $ \pi = \dfrac{{22}}{7} $
Step-by-step Solution:
We are given that the conical flask has base radius a and height 2a and the cylindrical flask has base-radius $ \dfrac{{2a}}{3} $ .
We have to find the height of water in the cylindrical flask.

Volume of the conical flask is $ \dfrac{1}{3}\pi {r^2}h $ , where r is the base radius of the flask, h is the height of the flask and $ \pi = \dfrac{{22}}{7} $
 $
\Rightarrow Volum{e_{cone}} = \dfrac{1}{3}\pi {r^2}h \\
  r = a,h = 2a,\pi = \dfrac{{22}}{7} \\
\Rightarrow Volum{e_{cone}} = \dfrac{1}{3} \times \dfrac{{22}}{7} \times {a^2} \times 2a = \dfrac{{44}}{{21}}{a^3}cube.units \\
  $
Volume of the cylindrical flask is $ \pi {r^2}h $ , where r is the base radius of the flask, h is the height of the water in flask and
$\Rightarrow \pi = \dfrac{{22}}{7} $
 $
\Rightarrow Volum{e_{cylinder}} = \pi {r^2}h \\
  r = \dfrac{{2a}}{3},h = ?,\pi = \dfrac{{22}}{7} \\
  \Rightarrow Volum{e_{cylinder}} = \dfrac{{22}}{7} \times {\left( {\dfrac{{2a}}{3}} \right)^2} \times h \\
 \Rightarrow Volum{e_{cylinder}} = \dfrac{{22}}{7} \times \dfrac{{4{a^2}}}{9}h = \dfrac{{88{a^2}h}}{{63}}cube.units \\
  $
Volume of the conical flask = Volume of the cylindrical flask
 $
 \Rightarrow Volum{e_{cone}} = Volum{e_{cylinder}} \\
 \Rightarrow \dfrac{{44}}{{21}}{a^3} = \dfrac{{88{a^2}h}}{{63}} \\
 \Rightarrow h = \dfrac{{44}}{{21}}{a^3} \times \dfrac{{63}}{{88{a^2}}} \\
 \Rightarrow h = \dfrac{{{a^3}}}{{{a^2}}} \times \dfrac{{44}}{{21}} \times \dfrac{{63}}{{88}} \\
  \therefore h = \dfrac{3}{2}a \\
  $
Therefore the height of the water in the flask is $ \dfrac{3}{2}a $ units.

So, the correct answer is “Option C”.

Note: The volume of cone is one third of volume of a cylinder i.e. $\dfrac{1}{3}\pi {r^2}h$, because a cone which has same base radius and height as a cylinder will have lesser volume than the cylinder and 3 cones can be fitted into that cylinder, hence the volume of cone is one third of the volume of the cylinder.