Question

A vessel in the shape of a hollow hemisphere surmounted by a cone is held with the axis vertical and vertex uppermost. If it be filled with a liquid so as to submerge half the axis of the cone in the liquid, and the height of the cone be double the radius of its base, find the value of $x$, where the resultant downward thrust of the liquid on the vessel is $x$ times the weight of the liquid that the hemisphere can hold.

• A. $15/8$
• B. $1/8$
• C. $5/8$
• D. $15/2$

Answer 1

Answer: (A)

Let $r$ be the radius of the base of the hemisphere or cone then the height of the base $=2r$ (given)
i.e., $VK=2r$ or $VL=LK=r$
Also from similar triangles $VLF$ and $VKB$, we have
$\frac{LF}{KB}=\frac{VL}{VK}=\frac{r}{2r}=\frac{1}{2}$ or $LF=\frac{1}{2}KB=\frac{1}{3}r$
$\therefore$ The volume of the frustum
$ABFE=\frac{\pi }{3}h\left(r_1^2+r_2^2+r_1{r}_2\right)$
$=\frac{1}{3}\pi r\left[r^2+{\left(\frac{1}{2}r\right)}^2+r\left(\frac{1}{2}r\right)\right]=\frac{7}{12}\pi r^3$
And the volume of the hemisphere $=\frac{2}{3}\pi r^3$
$\therefore$ Weight of the liquid contained in the vessel $=$ (volume of the frustum $+$ volume of the hemisphere) $g\rho$ $=\left(\frac{7}{12}\pi r^3+\frac{2}{3}\pi r^3\right)g\rho =\frac{5}{4}\pi r^{3}g\rho$, where $\rho$ is the density of the liquid.
Now the resultant vertical thrust on the vessel, which is partly pressed upwards and partly downwards
$=$ weight of the liquid contained
$=\frac{5}{4}\pi r^{3}g\rho =\frac{15}{8}\left(\frac{2}{3}\pi r^{3}g\rho \right)=\frac{15}{8}$ (the weight of the liquid that the hemisphere can hold)