 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. 65B. 55C. 45D. 35

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 .
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$= {V_1} = \dfrac{1}{3}(\pi {(5)^2}h + \pi {(r)^2}h)$
$= {V_2} = \dfrac{1}{3}\left( {\pi {{\left( {\dfrac{{12.1}}{2}} \right)}^2}h + \pi {{(r)}^2}h} \right) \\ \Rightarrow {V_2} = \dfrac{1}{3}\left( {\pi {{\left( {6.05} \right)}^2}h + \pi {{\left( r \right)}^2}h} \right) \\$
Given $\dfrac{{{V_2}}}{{{V_1}}} = 1.21$
$\Rightarrow \dfrac{{\dfrac{1}{3}\left( {\pi {{\left( {6.05} \right)}^2}h + \pi {{\left( r \right)}^2}h} \right)}}{{\dfrac{1}{3}\left( {\pi {{\left( 5 \right)}^2}h + \pi {{\left( r \right)}^2}h} \right)}} = 1.21$
$\Rightarrow {\left( r \right)^2} = \dfrac{{6.3525}}{{21}}$
$\Rightarrow r = \dfrac{{11}}{2} = 5.5 \\ \\$
$r = 5.5cm = 55mm$