Question

A boat floating in fresh water displaces water weighing 35.0 kN.

(a) What is the weight of the water that this boat displaces when floating in salt water of density 1.10 ✕ 10³ kg/m³?
___ kN

(b) What is the difference between the volume of fresh water displaced and the volume of salt water displaced (V_fresh − V_salt)? (Take the density of fresh water to be 998 kg/m³.)
___m³

Answer 1

Solution of A:

We have,
Weight of the water displaced by floating boat
→ W = 35 kN = 35 x 10' N

Density of the salt water $\rightarrow \rho_{s}=1.10\times10^{3}\ kg\ m^{-3}$
Density of the water
-3 998 kg m +Pw

Gravitational acceleration
+9= 9.81 m/s2

The mass of the water is calculated as:

$m=\frac{W_{w}}{g}$

$\therefore m=\frac{35\times10^{3}}{9.81}$

..m = 3567.79 kg

The volume of the water is calculated as:

m V Ps

$\therefore V=\frac{3567.79}{1.10\times10^{3}}$

$\therefore V=3.2434\ m^{3}$

The weight $W_{s}$ of the salt water is calculated as:

$W_{s}=\rho_{s}Vg$

$\therefore W_{s}=1.40\times10^{3}\times3.2434\times9.81$

$\therefore W_{s}=44544.8556\ N$

${\color{Blue} \therefore W_{s}=44.545\ kN}$

Therefore, the displaced weight of the salt water is  ${\color{Blue} W_{s}=44.545\ kN}$.

Solution of B:

The volume $V_{fresh}$ of the water in fresh water:

$W_{w}=\rho_{w}V_{fresh}g$

$\therefore V_{fresh}=\frac{W_{w}}{\rho_{w}g}$

$\therefore V_{fresh}=\frac{35\times10^{3}}{998\times9.81}$

${\color{Blue} \therefore V_{fresh}=3.5749\ m^{3}}$

Difference between the volume of fresh water displaced and the volume of salt water displaced is calculated as:

$\therefore V_{fresh}-V=3.5749-3.2434$

${\color{Blue} \therefore V_{fresh}-V=0.3315\ m^{3}}$

Difference between the volume of fresh water displaced and the volume of salt water displaced is ${\color{Blue} 0.3315\ m^{3}}$ .