Question

Find the lifting force of a $4\;{\text{kg}}$ cork lifebelt in sea water if the densities of cork and seawater are $0.2 \times {10^3}\;{\text{kg/}}{{\text{m}}^3}$ and $1.03 \times {10^3}\;{\text{kg/}}{{\text{m}}^3}$ respectively.
(A) $163\;{\text{N}}$
(B) $273\;{\text{N}}$.
(C) $119\;{\text{N}}$.
(D) $289\;{\text{N}}$.

Answer 1

In this question, calculate the volume of the cork life belt and also calculate the mass of the displaced water and the difference in buoyant force. Multiply the buoyant force by the gravitational constant to calculate the value of lift force.

Complete step by step answer:
The lifted force is defined as force acting on a body in upward direction. The lift force is always worked in the opposite upward direction. The SI unit of the lifted force is Newton, and which is denoted by ${\text{N}}$.
Let us consider the mass of the cork is ${\text{m}}$ and consider the density as rho$\rho$. And consider the ${V_{cork}}$ as the volume of cork.
And we are given the mass of the cork is $4\;{\text{kg}}$ and the density of water is $0.2 \times {10^3}\;{\text{kg/}}{{\text{m}}^3}$ and the density of the sea water is given $1.03 \times {10^3}\;{\text{kg/}}{{\text{m}}^3}$.
First we have to calculate the volume of the cork of the life belt,
${V_{cork}} = \dfrac{m}{\rho }$
Now we substitute the values in the formula for the volume of cork.

$${V_{{\text{cork}}}} = \dfrac{4}{{0.2 \times {{10}^3}}} \\\ {V_{{\text{cork}}}} = 2 \times {10^{ - 2}}\;{{\text{m}}^3} \\\$$

Now, the mass of the water displaced is equal to the product of the density of the sea water and the volume of cork life belt.
${m_{{\text{sea}}}} = {\rho _{{\text{sea}}}} \times {V_{{\text{cork}}}}$
Substitute the values in the above equation

$${m_{{\text{sea}}}} = \left( {1.03 \times {{10}^3}} \right)\left( {2 \times {{10}^{ - 2}}} \right) \\\ {m_{{\text{sea}}}} = 20.6\;{\text{kg}} \\\$$

The buoyant force is the force that occurred on the objects that are submersed in fluids and which exerts an upward buoyant force. And the difference in buoyant force is equal to the water displaced minus the weight of the body.
Calculate the difference in buoyant force,

$${F_{\text{B}}} = {m_{sea}} - {m_{cork}} \\\ {F_{\text{B}}} = 20.6\;{\text{kg}} - 4\;{\text{kg}} \\\ {F_{\text{B}}} = 16.6\;{\text{kg}} \\\$$

So, the lifted force is the product of the difference in the buoyant force and the gravitational force. And where $\left( {\text{g}} \right)$ is the gravitational constant and take the value of $g$ as $9.8\;{\text{m/se}}{{\text{c}}^2}$
${F_{{\text{lift}}}} = {F_{\text{B}}} \times {\text{g}}$
Now, Substitute the values

$${F_{{\text{lift}}}} = 16.6 \times 9.8 \\\ {F_{{\text{lift}}}} = 163\;{\text{N}} \\\$$

Therefore, the lifting force is $163\;{\text{N}}$. So, the option ${\text{A}}$ is correct.

Note: Do not forget to convert all the values into the standard units And here we have to take the value of gravitational force $\left( {\text{g}} \right)$ as $9.8\;{\text{m/se}}{{\text{c}}^{\text{2}}}$ instead of $10\;{\text{m/se}}{{\text{c}}^{\text{2}}}$. Calculation should be done carefully.