Question

A block of wood floats in water with $\dfrac{2}{3}$ of its volume submerged. Its relative density is?

Answer 1

Hint: This question utilizes the Archimedes principle. We know that when an object is immersed in a fluid, it experiences a buoyant force. Using the mass density relation and buoyant force equation, the answer can be easily found.

Formulae used :
Archimedes principle $weight{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} {\kern 1pt} body = weight{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} liquid{\kern 1pt} {\kern 1pt} displaced$
$\rho = \dfrac{m}{V}$ where $\rho $ is the density of the body, $m$ is the mass of the body and $V$ is the volume occupied by the body.

Complete answer:
According to the given question,
Weight of the wooden block is equal to the weight of the volume of water displaced by $\dfrac{2}{3}$ of the wooden block .
Thus, we get
$weight{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} {\kern 1pt} block = Buoyant{\kern 1pt} {\kern 1pt} force$
$ \Rightarrow V{\rho _b}g = \dfrac{2}{3}V{\rho _{water}}g$ -----------------(i)
Here, $V$ is the total volume of the block, ${\rho _b}$ is the density of the block, ${\rho _{water}}$ is the density of the water and $g$ is the acceleration due to gravity.
From eq (i), $g$ and $V$ is cancelled from both sides, thus we get
$ \Rightarrow {\rho _b} = \dfrac{2}{3}{\rho _{water}}$ ---------------------(ii)
Now, we know that the value of ${\rho _{water}} = 1gc{m^{ - 3}}$ . Substituting this value in eq (ii) , we get
$ \Rightarrow {\rho _b} = \dfrac{2}{3} \times 1gc{m^{ - 3}} = 0.67gc{m^{ - 3}}$
Therefore, the relative density of the wooden block is $0.67gc{m^{ - 3}}$.

Note: We actually calculate the weight of the displaced water using the formula $F = mg$ . Here, we replace the $m$ with $V\rho $ using the formula $\rho = \dfrac{m}{V}$ . Using this replacement, we get $F = V\rho g$ which we use here to solve the question.