Question

A cone of height $16cm$ and volume $2948c{m}^3$ floats in a liquid of density $3.2gc{m}^{-3}$ such that its base lies below the surface of the liquid. If the area of cross section of conical region above the liquid surface is $6c{m}^2$, find the density of the cone.

Answer 1

$H=16cm$ , volume $V=2948c{m}^3$

$V=2948=\frac{\pi R^{2}H}{3}$

$R^2=\frac{3\times 2948}{\pi H}$....(1)

The area of cross section of conical region above the liquid surface is $6c{m}^2$.

The the radius of that section be r , $6=\pi r^2$

$r^2=\frac{6}{\pi }$ ....(2)

Let the height of that cone be h , Now the triangle $ABC$ and $AD{A}_2$ are similar so $\frac{R}{r}=\frac{H}{h}$

Using the equation 1 and 2,

we get $h=1.67cm$

So the volume of cylinder immersed into the liquid $V_1=2948-\frac{\pi r^{2}h}{3}=2944.66c{m}^3$

Weight of liquid =buoyancy force

$V\times {\rho }_{cone}\times g=V_1\times {\rho }_l\times g$

$2948\times {\rho }_{cone}=2944.66\times 3.2$

${\rho }_{cone}=3.196c{m}^{-3}$