Question

When an object is immersed in water, it displaces $20{\text{kg}}$ of water. Find the buoyant force acting on the object in Newtons.
A) 100
B) 200
C) 0
D) 400

Answer 1

Hint: A Greek scientist named Archimedes discovered that when a body was immersed in a fluid partially or fully submerged, the fluid will exert a force upwards. This upward thrust acting on the body is referred to as the buoyant force. Archimedes found that the upward buoyant force acting on the body is equal to the weight of the fluid it displaces as it gets immersed in it. Here the fluid is water.

Formulae used:
The weight of a body is given by, $W = mg$ where $m$ is the mass of the body and $g$ is the acceleration due to gravity.
The buoyant force acting on a body is given by, ${F_b} = \rho gV$ where $\rho $ is the density of the fluid, $g$ is the acceleration due to gravity and $V$ is the volume of the water displaced.
The volume of a fluid is given by, $V = \dfrac{m}{\rho }$ where $m$ is the mass of the fluid and $\rho $ is the density of the fluid.

Complete step by step solution:
List the parameters known from the question.
The mass of the displaced water is given to be $m = 20{\text{kg}}$ .
Here, we take the acceleration due to gravity to be $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ .
Express the weight of the water displaced.
The weight of the water displaced will be given by, $W = mg$ -------- (1)
Substituting for $m = 20{\text{kg}}$ and $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ in equation (1) we get, $W = 20 \times 10 = 200{\text{N}}$
Thus the weight of the water displaced when the object was immersed in it is obtained to be $W = 200{\text{N}}$ .
Based on the Archimedes’ principle, obtain the buoyant force acting on the object.
Let ${F_b}$ be the buoyant force acting on the object when it is immersed in water.
According to Archimedes’ principle, this buoyant force is equal to the weight of the water that got displaced by the immersion of the object in it.
$ \Rightarrow {F_b} = W = 200{\text{N}}$
$\therefore $ the buoyant force acting on the object is obtained to be ${F_b} = 200{\text{N}}$ .

So the correct option is B.

Note: Alternate method:
The density of water is known to be $\rho = 1000{\text{kg}}{{\text{m}}^{ - 3}}$ .
Then the volume of the water displaced will be $V = \dfrac{m}{\rho } = \dfrac{{20}}{{1000}} = 0.02{{\text{m}}^3}$ .
The relation for the upward buoyant force acting on the object is expressed as
${F_b} = \rho gV$ --------- (A)
Substituting for $\rho = 1000{\text{kg}}{{\text{m}}^{ - 3}}$ , $V = \dfrac{{20}}{{1000}} = 0.02{{\text{m}}^3}$ and $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ in equation (A) we get, ${F_b} = 1000 \times 10 \times 0.02 = 200{\text{N}}$
$\therefore $ the buoyant force acting on the object is obtained to be ${F_b} = 200{\text{N}}$.