Question

A cork of density $0.5gc{m^{ - 3}}$ floats on a calm swimming pool. The fraction of the cork’s volume which is under water is
(A) $0\% $
(B) $25\% $
(C) $10\% $
(D) $50\% $

Answer 1

Hint: Calculate the weight of cork and the weight of water displaced because of it. Then use the principle of flotation to solve the equation.

Formula used:
${W_c} = {W_w}$
Where,
${W_c}$is the weight of cork and
${W_w}$ is the weight of water displaced.

Complete step by step answer:Let the density of cork be ${\rho _c}$
Let the density of water be ${\rho _w}$.
We know that the density of water is 1.
$\therefore {\rho _w} = 1$
Let the volume of the cork be $V.$
Let the volume of water displaced when the cork was put into the swimming pool be $V'$
Let the fraction of cork’s volume which is under water be $k$.
Then,
$V' = kV$
Let the mass of the cork be ${m_1}$
Then ${\rho _c} = \dfrac{{{m_1}}}{V}$
$ \Rightarrow {m_1} = V{\rho _c}$
Since it is given that ${\rho _c} = 0.5$
$ \Rightarrow {m_1} = V \times 0.5$
Let the mass of water displaced be ${m_2}$
Then,
${\rho _w} = \dfrac{{{m_2}}}{{V'}}$
$ \Rightarrow {m_2} = {V^1}{\rho _w}$
Since , the density of water is ${\rho _w} = 1,$we get
${m_2} = V' = kV$ $(\because V' = kV)$
Now, according to the principle of floating
Weight of cork is equal to the weight of water displaced
$ \Rightarrow {m_1}g = {m_2}g$
$ \Rightarrow {m_1} = {m_2}$
$ \Rightarrow V \times 0.5 = V' \times 1$
$ \Rightarrow V \times 0.5 = kV$
$ \Rightarrow k = 0.5 = \dfrac{1}{2}$
Therefore, half of the cork will be submerged under the water.
Which can be said as, $50\% $ of the cork is submerged under the water.

Note:Density of water is not given in the question. So you can solve the question if you know that the density of water is equal to 1. You need to know the formula of mass in terms of density so that you could compare weight in terms of density.