Question

The diameter of one of the bases of a 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

The correct option is B $55$

Initial diameter of base $=100mm=10cm$

Final diameter of base $10+\frac{21}{100}\times 10=12.1cm$

Let the radius of top be $r$ and height be $h$

Initial volume$=V_1=\frac{1}{3}(\pi {\left(5\right)}^{2}h+\pi {\left(r\right)}^{2}h)$

Final volume$=V_2=\frac{1}{3}(\pi {\left(\frac{12.1}{2}\right)}^{2}h+\pi {\left(r\right)}^{2}h)=\frac{1}{3}(\pi {\left(6.05\right)}^{2}h+\pi {\left(r\right)}^{2}h)$

Given $\frac{V_2}{V_1}=1.21$

$\frac{\frac{1}{3}(\pi {\left(6.05\right)}^{2}h+\pi {\left(r\right)}^{2}h)}{\frac{1}{3}(\pi {\left(5\right)}^{2}h+\pi {\left(r\right)}^{2}h)}=1.21\Rightarrow {\left(r\right)}^2=\frac{6.3525}{21}\Rightarrow r=\frac{11}{2}=5.5$

$r=5.5cm=55mm$

Hence, option $B$ is correct.