Question

The diameter of one of the bases of the truncated cone is $100mm$. If the diameter of this base is increased by $21\% $such that it still remains a truncated cone with the height and the other base unchanged , the volume also increases by $21\% $. The radius of the other base (in $mm$) is
A. 65
B. 55
C. 45
D. 35

Answer 1

Hint: In this question find the initial and final volume of truncated cone by Volume of Frustum of Cone $ = \dfrac{1}{3}\pi \left( {{r_1}^2 + {r_2}^2 + {r_1}{r_2}} \right)h$ and also find $\dfrac{{{V_2}}}{{{V_1}}} = 1.21$. Use this to find the radius of the other base .
Complete step-by-step answer:
According to the question it is given , that initial diameter of base$ = 100mm = 10cm$
Final diameter of base $10 + \dfrac{{21}}{{100}} \times 10 = 12.1cm$
Let the radius of top be $r$ and height be $h$
Initial volume\[ = {V_1} = \dfrac{1}{3}(\pi {(5)^2}h + \pi {(r)^2}h)\]
Final volume
\[
   = {V_2} = \dfrac{1}{3}\left( {\pi {{\left( {\dfrac{{12.1}}{2}} \right)}^2}h + \pi {{(r)}^2}h} \right) \\
   \Rightarrow {V_2} = \dfrac{1}{3}\left( {\pi {{\left( {6.05} \right)}^2}h + \pi {{\left( r \right)}^2}h} \right) \\
 \]
Given $\dfrac{{{V_2}}}{{{V_1}}} = 1.21$
\[ \Rightarrow \dfrac{{\dfrac{1}{3}\left( {\pi {{\left( {6.05} \right)}^2}h + \pi {{\left( r \right)}^2}h} \right)}}{{\dfrac{1}{3}\left( {\pi {{\left( 5 \right)}^2}h + \pi {{\left( r \right)}^2}h} \right)}} = 1.21\]
$ \Rightarrow {\left( r \right)^2} = \dfrac{{6.3525}}{{21}}$
$
   \Rightarrow r = \dfrac{{11}}{2} = 5.5 \\
    \\
$
$r = 5.5cm = 55mm$

Note: Whenever we come up with such types of questions, ask to find out the radius of the base of a truncated cone which is a cone with the tip straight cutoff . The base is the larger circle and the top surface is the smaller circle. So, to get the radius, find the volume of frustum of cone and for solving such types of questions it is always advisable to remember the concepts and formulas .