Question

The radius of the base of a right circular cone is r. If it is cut by a plane parallel to the base at a height h from the base and the slant height of the frustum so obtained is $\sqrt{h^2+\frac{9r^2}{16}},$ then the volume of the frustum is equal to

• A. $\frac{3\pi r^{2}h}{8}$
  
• B. $\frac{5\pi r^{2}h}{16}$
  
• C. $\frac{7\pi r^{2}h}{16}$
  
• D. $\frac{9\pi r^{2}h}{16}$
  

Answer 1

The correct option is C $\frac{7\pi r^{2}h}{16}$


Lower radius = r

Let upper radius be r'(r' < r)


$\therefore l=\sqrt{h^2+(r-r^{\prime }{)}^2}$

$\Rightarrow h^2+\frac{9r^2}{16}=h^2+(r-r^{\prime }{)}^2$

$\Rightarrow \frac{9r^2}{16}=(r-r^{\prime }{)}^2$

$\Rightarrow \frac{3r}{4}=r-r^{\prime }$

$\Rightarrow r^{\prime }=r-\frac{3r}{4}=\frac{r}{4}$

$\therefore$ Volume of frustum $=\frac{1}{3}\pi h(r^2+r{r}^{\prime }+(r^{\prime }{)}^2)$

$=\frac{1}{3}\pi \times h\times (r^2+r\times \frac{r}{4}+{\left(\frac{r}{4}\right)}^2)$

$=\frac{1}{3}\pi r^{2}h(1+\frac{1}{4}+\frac{1}{16})=\frac{7}{16}\pi r^{2}h$

Hence, the correct answer is option (c).