Question

The densities of wood and benzene at ${0^ \circ }C$ are $880\,kg/{m^3}$ and $900\,kg/{m^3}$ respectively. The coefficient of volume expansion is $1.2 \times {10^{ - 3}}{/^ \circ }C$ for wood and $1.5 \times {10^{ - 3}}{/^ \circ }C$ for benzene. What is the temperature at which a piece of wood will just sink in benzene?
A. ${73^ \circ }C$
B. ${83^ \circ }C$
C. ${42^ \circ }C$
D. ${120^ \circ }C$

Answer 1

Hint: The condition at which an object just sinks in a liquid is that density of both object and liquid should be equal. So, here the wood will just sink in benzene when the density of wood becomes equal to the density of benzene. The relation between density and temperature is given by the following equation,
$\rho = \dfrac{{{\rho _0}}}{{1 + \gamma \theta }}$, Where ${\rho _0}$ is the initial density, $\gamma $ is a coefficient of volume expansion and $\theta $ is the temperature change. By using this relation we can find the temperature at which density of wood becomes equal to density of water.

Complete step by step answer:
Given,
density of wood, ${\rho _w} = 880\,kg/{m^3}$
density of benzene, ${\rho _b} = 900\,kg/{m^3}$
coefficient of volume expansion for wood, ${\gamma _w} = 1.2 \times {10^{ - 3}}{/^ \circ }C$
coefficient of volume expansion for benzene, ${\gamma _b} = 1.5 \times {10^{ - 3}}{/^ \circ }C$
The condition at which an object just sinks in a liquid is that density of both object and liquid should be equal.
Therefore, in this case the wood will just sink in benzene when the density of wood becomes equal to the density of benzene. So, we need to find the temperature at which density of wood becomes equal to the density of water.
The relation between density and temperature is given by the following equation
$\rho = \dfrac{{{\rho _0}}}{{1 + \gamma \theta }}$
Where ${\rho _0}$ is the initial density, $\gamma $ is a coefficient of volume expansion and $\theta $ is the temperature change.
Let us assume that the change in temperature required for obtaining equal density is $\theta $ and $\rho $ be the final density of both wood and benzene.
For wood new density is given as
 $\rho = \dfrac{{{\rho _w}}}{{1 + {\gamma _w}\theta }}$
For benzene new density is given as
 $\rho = \dfrac{{{\rho _b}}}{{1 + {\gamma _b}\theta }}$
Now equate both equations.
$\dfrac{{{\rho _b}}}{{1 + {\gamma _b}\theta }} = \dfrac{{{\rho _w}}}{{1 + {\gamma _w}\theta }}$
On substituting the given values. We get,
$\dfrac{{900\,kg/{m^3}}}{{1 + 1.5 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }} = \dfrac{{880\,kg/{m^3}}}{{1 + 1.2 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }}$
$ \Rightarrow \dfrac{{1 + 1.2 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }}{{1 + 1.5 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }} = \dfrac{{880\,kg/{m^3}}}{{900\,kg/{m^3}}}$
$ \Rightarrow \dfrac{{1 + 1.2 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }}{{1 + 1.5 \times {{10}^{ - 3}}{/^ \circ }C \times \theta }} = 0.977$
$ \Rightarrow 1 + 1.2 \times {10^{ - 3}}{/^ \circ }C \times \theta = 0.977 \times (1 + 1.5 \times {10^{ - 3}}{/^ \circ }C \times \theta )$
$ \Rightarrow 1 + 1.2 \times {10^{ - 3}}{/^ \circ }C \times \theta = 0.977 + 1.4655 \times {10^{ - 3}}{/^ \circ }C \times \theta $
$\therefore \theta = {83^ \circ }C$
This is the temperature at which wood will just sink in benzene.
So, the correct answer is option B.

Note: The value of temperature that we get is the temperature change Since the initial temperature was given as ${0^ \circ }C$ final temperature is the same as ${83^ \circ }C$. When any other initial temperature is mentioned make sure you add the value obtained as temperature change to the initial temperature to get the final temperature.