The volume of air space in the passenger compartment of a1800 kgcar is5.00 m³. The volume of the motor and front wheels is 0.750 m, and the volume of the rear wheels, gas tank, and trunk is 0.800 m³; water cannot enter these two regions. The car rolls into a lake.(a) At first, no water enters the passenger compartment. How much of the car, in cubic meters, is below the water surface with the car floating (Figure)?(b) As water slowly enters, the car sinks. How many cubic meters of water are in the car as it disappears below the water surface? (The car, with a heavy load in the trunk, remains horizontal.)

• The mass of the car,$m_{car}=1800\mathrm{kg}$
• The volume of air space in the passenger compartment, $v_a=5.00{\mathrm{m}}^3$
• The volume of the motor and front wheels,$V_m=0.750{\mathrm{m}}^3$
• The volume of the rear wheels, gas tank, and trunk, $V_r=0.800{\mathrm{m}}^3$

We can use the concept of Archimedes’ principle and expression of density. When the car is fully submerged in the water, then a buoyant force from the surrounding fluid acts on it. This force has a magnitude equal to the weight of the water displaced by the car.

Formulae:

Force applied on body (or weight),$F_b=m_{f}g$ (i)

Density of a substance, role="math" localid="1661156475036" $\mathrm{\rho }=\frac{m}{V}$ (ii)

The volume of the car below the water is$V_{car}$. When the car is floating in the water,the buoyant force acts on it.Using equation (i) & (ii) and given values, we get

$$\begin{gathered} F_b=W_{car} \\ {\rho }_w{V}_{car}g=W_{car} \\ {V}_{car}=\frac{W_{car}}{{\rho }_{w}g} \\ {V}_{car}=\frac{m_{car}g}{{\rho }_{w}g} \\ =\frac{m_{car}}{{\rho }_w} \\ =\frac{1800kg}{1000kg/m^3} \\ =1.80m^3 \end{gathered}$$

Hence, the volume of the car below water surface is$1.80m^3$

The total volume of the car is $V$ and the volume of the water in the car is$V_w$ . According to Archimedes’ principle, net force can be given using equation (i) as:

$$\begin{gathered} F_g-F_{car}=m_{f}g \\ {m}_{w}g-W_{car}=m_{f}g \\ {m}_{w}g-m_{car}g=m_{f}g\left(\because \mathrm{From}\mathrm{equation}\left(\mathrm{ii}\right)\right) \\ {\rho }_{w}gV-m_{car}g={\rho }_{w}g{V}_w \\ {\rho }_{w}V-m_{car}={\rho }_w{V}_w \\ {V}_w=\frac{{\rho }_{w}V-m_{car}}{{\rho }_w} \\ {V}_w=V-\frac{m_{car}}{{\rho }_w} \\ =\left(V_a+V_m+V_r\right)-\frac{m_{car}}{{\rho }_w}\left(\because \mathrm{Total}\mathrm{volume},V=V_a+V_m+V_r\right) \\ =\left(5.00{\mathrm{m}}^3+0.750{\mathrm{m}}^3+0.800{\mathrm{m}}^3\right)-\frac{1800\mathrm{kg}}{1000\mathrm{kg}/{\mathrm{m}}^3} \\ =4.75{\mathrm{m}}^3 \end{gathered}$$

Hence, the volume of water in the car is$4.75{\mathrm{m}}^3$