Physics - Fluid Question with Solution | TestHub

Solution:

Let $H$ be the height of the cone and $R$ be the base radius.

$H=25$ cm, $D=25$ cm $⟹R=12.5$ cm.

Volume of the cone $V=\frac{1}{3}\pi R^{2}H$.

When the cone floats with $h=20$ cm immersed, let $r$ be the radius of the immersed part's base.

By similar triangles, $\frac{r}{R}=\frac{h}{H}⟹r=R\frac{h}{H}=12.5\frac{20}{25}=10$ cm.

Volume of immersed part $V_{imm}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (10{)}^2(20)$.

By Archimedes' principle, weight of cone $W_{cone}={\rho }_{w}g{V}_{imm}$.

Density of cone ${\rho }_c=\frac{W_{cone}}{gV}=\frac{{\rho }_w{V}_{imm}}{V}={\rho }_w\frac{\frac{1}{3}\pi r^{2}h}{\frac{1}{3}\pi R^{2}H}={\rho }_w{\left(\frac{r}{R}\right)}^2\frac{h}{H}={\rho }_w{\left(\frac{h}{H}\right)}^3$.

${\rho }_c={\rho }_w{\left(\frac{20}{25}\right)}^3={\rho }_w{\left(\frac{4}{5}\right)}^3=0.512{\rho }_w$.

For complete immersion, the total weight must equal the buoyant force when fully submerged.

$W_{cone}+W_{add}={\rho }_{w}gV$.

$W_{add}={\rho }_{w}gV-W_{cone}={\rho }_{w}gV-{\rho }_{c}gV=({\rho }_w-{\rho }_c)gV$.

$W_{add}=({\rho }_w-0.512{\rho }_w)gV=0.488{\rho }_{w}gV$.

$W_{add}=0.488{\rho }_{w}g\frac{1}{3}\pi R^{2}H=0.488\times 1000\times 9.8\times \frac{1}{3}\pi (0.125{)}^2(0.25)$.

Mass $M_{add}=\frac{W_{add}}{g}=0.488{\rho }_{w}V=0.488\times 1000\times \frac{1}{3}\pi (0.125{)}^2(0.25)\approx 1.99$ kg.

This is approximately 2 kg.

The final answer is $\boxed{2kg}$.