Question

The value of coefficient of volume expansion of glycerin is $5 \times {10^{ - 4}}{K^{ - 1}}$. The fractional change in the density of glycerin for a rise of ${40^0}C$ in its temperature is:
$\left( a \right)0.010$
$\left( b \right)0.015$
$\left( c \right)0.020$
$\left( d \right)0.025$

Answer 1

Hint: In this question use the property that change in density is the ratio of difference of density at particular temperature and density at room temperature to density at room temperature and the density at any particular temperature is given as $\rho = {\rho _o}\left( {1 + \alpha \Delta T} \right)$ so use these concepts to reach the solution of the question.
Formula used - $\rho = {\rho _o}\left( {1 + \alpha \Delta T} \right)$

Complete Step-by-Step solution:
Given data:
Coefficient of volume of expansion $\alpha = 5 \times {10^{ - 4}}{K^{ - 1}}$
And rise in temperature $\Delta T = {40^0}C$
Now as we know density $\left( \rho \right)$ of glycerin is given as
$\rho = {\rho _o}\left( {1 + \alpha \Delta T} \right)$ , where
${\rho _o}$ = density of glycerin at room temperature.
$\Delta T$ = temperature difference or change in temperature or rise in temperature.
$\alpha $ = Coefficient of volume of expansion
Now the above equation is written as
$ \Rightarrow \rho - {\rho _o} = {\rho _o}\alpha \Delta T$
$ \Rightarrow \dfrac{{\rho - {\rho _o}}}{{{\rho _o}}} = \alpha \Delta T$
Where, $\dfrac{{\rho - {\rho _o}}}{{{\rho _o}}}$ = change in density of glycerin.
Now substitute the value we have,
$ \Rightarrow \dfrac{{\rho - {\rho _o}}}{{{\rho _o}}} = 5 \times {10^{ - 4}}\left( {40} \right) = 200 \times {10^{ - 4}} = 0.020$
So this is the required change in density of the glycerin.
Hence option (C) is the correct answer.

Note – Whenever we face such types of question the key concept is always recall the formula of density of glycerin at any particular temperature which is stated above, then change in density is given as $\dfrac{{\rho - {\rho _o}}}{{{\rho _o}}}$ now just simply substitute the values and simplify we will get the required answer.