Question

If a sphere of the maximum volume is placed inside a hollow right circular cone with radius 'r' and slant height 'l' such that the base of the cone touches the sphere, then the volume of the sphere is

• A. $\frac{4}{3}\pi {\left(\frac{l+r}{l-r}\right)}^3$
• B. $\frac{4}{3}\pi r^3{\left(\frac{l-r}{l+r}\right)}^{\frac{3}{2}}$
• C. $\frac{4}{3}\pi {\left(\frac{l-r}{l+r}\right)}^3$
• D. $\frac{4}{3}\pi r^3{\left(\frac{l+r}{l-r}\right)}^{\frac{3}{2}}$

Answer 1

The correct option is B

$\frac{4}{3}\pi r^3{\left(\frac{l-r}{l+r}\right)}^{\frac{3}{2}}$

Consider the biggest cross-section of the cone as an isosceles triangle, therefore, the circle inscribed in the triangle will be the biggest cross-section of the sphere.


We know that radius $\times$ semi perimeter = Area of the triangle
A little calculation will lead to the answer, i.e.,
$\frac{4}{3}\pi r^3{\left(\frac{l-r}{l+r}\right)}^{\frac{3}{2}}$