Question

A solid floats in a liquid at $20{}^\circ C$ with $75\%$ of it immersed in a liquid. When the liquid is heated to $100{}^\circ C$ the same body floats with $80\%$ of it immersed in the liquid. The coefficient of real expansion of the liquid will be given as,
$\begin{aligned} & A.8\times {{10}^{-4}}\dfrac{1}{{}^\circ C} \\\ & B.8.33\times {{10}^{-4}}\dfrac{1}{{}^\circ C} \\\ & C.8.33\times {{10}^{-5}}\dfrac{1}{{}^\circ C} \\\ & D.8\times {{10}^{-5}}\dfrac{1}{{}^\circ C} \\\ \end{aligned}$

Answer 1

In order to float the body with $75\%$ inside liquid, the buoyant force at that temperature will be equivalent to its weight. In the same way for making a body float with $80\%$ inside liquid, the buoyant force at that temperature will be identical to its weight. Use these conditions in the form of equations and then take the ratio of coefficient of linear expansion by rearranging both the equations. These all may help you to solve this question.

Complete answer:
First of all let us check the first case. In order to make a body floating with $75\%$ inside liquid, the buoyant force at that temperature will be equivalent to its weight.
This can be shown in mathematical expression such that,
$Mg=0.75V{{\rho }_{20{}^\circ C}}g$
Where $M$be the mass of the solid, $g$ be the acceleration due to gravity,$V$is the volume occupied by the solid, ${{\rho }_{20{}^\circ C}}$ is the coefficient of real expansion of the liquid at $20{}^\circ C$.
Now let us look at the second case. In order to make a body float with $80\%$ inside liquid, the buoyant force at that temperature will be equivalent to its weight. That is, this equation can be expressed in the form of a mathematical expression,
$Mg=0.8V{{\rho }_{100{}^\circ C}}g$
Where $M$ be the mass of the solid, $g$ be the acceleration due to gravity, ${{\rho }_{100{}^\circ C}}$is the coefficient of real expansion of the fluid at $100{}^\circ C$and $V$ be the volume occupied.
Let us take the ratio between these values in order to get the ratio between the coefficients of expansion.
$\dfrac{Mg}{Mg}=\dfrac{0.8V{{\rho }_{100{}^\circ C}}g}{0.75V{{\rho }_{20{}^\circ C}}g}$
Cancelling the common terms and rearranging the equation as well,
$\dfrac{{{\rho }_{100{}^\circ C}}}{{{\rho }_{20{}^\circ C}}}=\dfrac{0.75}{0.8}$
This will be equal to the expression,
$\dfrac{{{\rho }_{100{}^\circ C}}}{{{\rho }_{20{}^\circ C}}}=\dfrac{0.75}{0.8}=\dfrac{1}{1+\gamma \Delta T}$
Where $\Delta T$be the temperature difference which is given as,
$\Delta T=100-20=80{}^\circ C$
Therefore after the rearrangements we can write that,
$\gamma =8.33\times {{10}^{-4}}\dfrac{1}{{}^\circ C}$

Therefore the correct answer is option B.

Note:
The ratio of the variation in the volume and its original volume in every $1{}^\circ C$ rise in temperature is described as a coefficient of real expansion. The unit is given as ${}^\circ {{C}^{-1}}$. This can be expressed also in ${{K}^{-1}}$.