Question 971866
Volume of liquid = volume of cone
={{{(1/3)*pi*r^2*h}}} (where symbols have usual meanings)
={{{(1/3)pi*(15/2)^2*12}}}
Now volume occupied by this liquid in cylindrical tin is given by
{{{pi*R^2*H}}}
={{{pi*(10/2)^2*H}}}

Equating both the volumes,
{{{(1/3)pi*(15/2)^2*12 = pi*(10/2)^2*H}}}
=> {{{H=9}}}cms.