Physics - Fluid Question with Solution | TestHub

To find the depression, first calculate the initial submerged depth. Then, calculate the additional submerged depth when the 10 kg mass is added. The difference is the depression.

Given:

Height of cylinder $H=3$ m

Mass of cylinder $M=80$ kg

Specific gravity of wood $SG=0.80$

Mass of additional body $m=10$ kg

Density of water ${\rho }_w=1000$ kg/m³

1. Calculate the density of the wood:

${\rho }_w=SG\times {\rho }_w=0.80\times 1000$ kg/m³ $=800$ kg/m³

2. Calculate the volume of the cylinder:

$V=\frac{M}{{\rho }_w}=\frac{80kg}{800kg/m^3}=0.1m^3$

3. Let $A$ be the cross-sectional area of the cylinder. Then $V=A\times H$.

$A=\frac{V}{H}=\frac{0.1m^3}{3m}=\frac{1}{30}{m}^2$

4. Initial submerged depth ($h_1$):

By Archimedes' principle, the buoyant force equals the weight of the cylinder.

${\rho }_{w}A{h}_{1}g=Mg$

$h_1=\frac{M}{{\rho }_{w}A}=\frac{80kg}{1000kg/m^3\times \frac{1}{30}{m}^2}=\frac{80\times 30}{1000}=\frac{2400}{1000}=2.4m$

5. Final submerged depth ($h_2$) when the 10 kg mass is added:

The total mass is $M_{total}=M+m=80kg+10kg=90kg$.

${\rho }_{w}A{h}_{2}g=M_{total}g$

$h_2=\frac{M_{total}}{{\rho }_{w}A}=\frac{90kg}{1000kg/m^3\times \frac{1}{30}{m}^2}=\frac{90\times 30}{1000}=\frac{2700}{1000}=2.7m$

6. Amount of depression:

Depression $=h_2-h_1=2.7m-2.4m=0.3m$

The final answer is $\boxed{0.30m}$.