Question

A glass tube of length 133cm and uniform cross-section is to be filled with mercury so that the volume of the tube unoccupied by mercury remains the same at all temperatures. If cubical coefficient of glass and mercury are $26 \times 10^{-6}/^{\circ}C$ and $182 \times 10^{-6}/^{\circ}C$ respectively, calculate the length of the mercury column.
A. 10 cm
B. 15 cm
C. 19 cm
D. 25 cm

Answer 1

Recall the relation between the coefficient of cubical expansion and the volume of tube or fluid column, where we define this coefficient to be an intrinsic property of the material of the tube or the fluid and is defined as the change in volume of one unit of volume of glass or mercury when heated by one unit of temperature.
Using the fact that the change in volume of the mercury in the tube remains the same throughout the expansion of both the glass and mercury, equate the expressions for the change in volume of the two, and by expanding the volume in terms of area and length, obtain the final result for the length of the mercury column by substituting the values given to us.

Formula used:
Coefficient of cubical expansion $\gamma = \dfrac{\left(\dfrac{\Delta V}{V}\right)}{\Delta T}$

Complete step-by-step answer:
We know that the coefficient of cubical expansion defines the fractional change in volume of the material when the material is subjected to a change in temperature via heating. It is quantified as:
$\gamma = \dfrac{\left(\dfrac{\Delta V}{V}\right)}{\Delta T}$
In the context of our question, the same is defined as the fractional change in the volume of the hydraulic fluid due to a change in its temperature at constant pressure.
We are given that the volume of the tube unoccupied by mercury remains that same at all temperatures.
$\Rightarrow \Delta V = \gamma V T$
$\Rightarrow \Delta V_{glass} = \Delta V_{Hg}$
We know that the area occupied by the mercury is equivalent to the area accommodated by the glass tube. Expressing their volumes in terms of area occupied and length:
$\Rightarrow \gamma_{glass} (a_0 l_{glass}).T = \gamma_{Hg} ((a_0) l_{Hg}).T$
$\Rightarrow l_{Hg} = l_{glass}.\dfrac{\gamma_{glass}}{\gamma_{Hg}}$
Substituting the values, we get:
$\Rightarrow l_{Hg} = 133 \times \left(\dfrac{26 \times 10^{-6}}{182 \times 10^{-6}}\right)$
$\Rightarrow l_{Hg} = 19\;cm$
Therefore, the correct choice would be C. 19 cm

So, the correct answer is “Option C”.

Note: Remember that if the volume remains the same throughout this means that there is a balance in the relative expansion of the glass and the mercury, which is why we are able to assign the area that they occupy the same value. This compensatory expansion is brought about by the length of the glass tube and the mercury column in order to maintain and account for the same changes in volume.