A right circular cone is divided by a plane parallel to its base in two equal volumes. In what ratio will the plane divide the axis of the cone
Let VAB be a cone of height h and base radius r. Suppose it is cut by a plane parallel to the base of the cone at point O′. Let O′A′=r1 and VO′=h1
Clearly,
∴VOVO′=OAO′A′⇒hh1=rr1
It is given that
Volume of cone VA'B' = volume of the frustum ABB'A'
13πr21h1=13π(r2+r21+rr1)(h−h1)
⇒r21h1=(r2+r21+rr1)(h−h1)
⇒1={(rr1)2+1+rr1}(hh1−1)⇒1=(hh1)3−13⇒(hh1)3=2⇒hh1=21/3
Hence the ratio =h1h−h1=1(hh1−1)=121/3−1