Question

A rock weighs 1400N in air and when it is submerged in freshwater of density $998{\text{kg/}}{{\text{m}}^3}$, its apparent weight is 900N. Find the volume of the rock in ${{\text{m}}^3}$.
A) $1.4 \times {10^2}$
B) $6 \times {10^{ - 2}}$
C) $9 \times {10^{ - 2}}$
D) $5.1 \times {10^{ - 2}}$

Answer 1

Hint: According to Archimedes principle, when a rock is submerged in freshwater (or any liquid) it displaces some amount of water and the weight of this displaced water will be equal to the loss of weight of the body when it is in submerged in water.

Formula used:
The weight of the liquid displaced by an object when it is submerged in it is given by, ${W_{dis}} = \rho gV$ where $\rho $ is the density of the liquid, $g$ is the acceleration due to gravity and $V$ is the volume of the liquid displaced.

Complete step by step answer:
Step 1: List the data given in the question.
It is given that the weight of rock in the air is ${W_a} = 1400{\text{N}}$ and when it is submerged in freshwater it becomes ${W_{fw}} = 900{\text{N}}$ .
The density of water is given as $\rho = 998{\text{ kg/}}{{\text{m}}^3}$
Step 2: Find the change in weight of the rock when it is submerged in freshwater.
The loss of weight of the rock, when submerged in freshwater is given by $\Delta W = {W_a} - {W_{fw}}$
Substituting the values of ${W_a} = 1400{\text{N}}$ and ${W_{fw}} = 900{\text{N}}$ in the above equation we get, $\Delta W = 1400 - 900 = 500{\text{N}}$
Thus the loss of weight is $\Delta W = 500{\text{N}}$ --------- (1).
Step 3: Find the weight of water displaced when the rock was submerged in freshwater.
Given, the density of freshwater $\rho = 998{\text{ kg/}}{{\text{m}}^3}$ .
Now the weight of water displaced is given by, ${W_{dis}} = \rho gV$ where $\rho $ is the density of the liquid in which it is submerged, $g$ is the acceleration due to gravity and $V$ is the volume of the liquid displaced.
Let $V$ be the volume of the rock and it is equal to the volume of the water displaced.
Substituting the values for $\rho = 998{\text{ kg/}}{{\text{m}}^3}$ in the above equation we get, ${W_{dis}} = 998Vg$ ------ (2)
Step 4: Using equation (1) and (2) obtain the volume of the rock.
 From equation (1) we have the loss of weight is $\Delta W = 500{\text{N}}$ and from equation (2) we have the weight of water displaced as ${W_{dis}} = 998Vg$ .
By the Archimedes principle, the loss of weight is equal to the weight of water displaced.
i.e., $\Delta W = {W_{dis}}$
Substituting the values for $\Delta W = 500{\text{N}}$ and ${W_{dis}} = 998Vg$ in the above equation we get, $500 = 998Vg$ or, on rearranging we get, $V = \dfrac{{500}}{{998 \times 9.8}} = 0.051{{\text{m}}^3}$

Therefore, the volume of the rock is $5.1 \times {10^{ - 2}}$. Hence option (D) is correct.

Note:
Here, the rock is said to be submerged (fully submerged) in freshwater. This is why the volume of the rock becomes equal to the volume of the fluid displaced. If it was partially submerged, the two volumes would be different.