Question

The fraction of the volume of a glass flask that must be filled with mercury so that the volume of the empty space may be the same at all temperatures is given by $ ({{\alpha }_{glass}}=9\times {{10}^{-6}}/{}^\circ C,\,{{\gamma }_{Hg}}=18.9\times {{10}^{-5}}/{}^\circ C) $
(A) $ \dfrac{1}{2} $
(B) $ \dfrac{1}{7} $
(C) $ \dfrac{1}{4} $
(D) $ \dfrac{1}{5} $

Answer 1

Hint
The volumetric expansion formula should be used to carry out the calculation. As the volume of the empty space remains the same, thus, we have to equate the change in volume of glass flask and that of the mercury to find the fraction of the volume of the glass flask.
 $ \Delta V = V\gamma \Delta T $
Where, $ \Delta V $ is the change in volume, $ \gamma $ is a volumetric coefficient and $ \Delta T $ is the change in temperature.

Complete step by step solution
The question states that the volume of the empty space of the glass vessel will be at all temperatures when the volume of the mercury after expansion equals the volume of the glass flask after expansion.
Now we will express this statement in terms of the equations.
So, the expression used to represent the change in the volume is given as follows.
 $ \Delta V = V\gamma \Delta T $
As the temperature remains the same, thus, the equation can be reduced to the form, $ {V_0}\gamma $
The diagram representing the glass flask filled with the mercury is as follows.

Now equate the expressions of mercury and glass flask.
 $ {V_g}{\alpha _g} = {V_{Hg}}{\gamma _{Hg}} $ …… (1)
As $ {\alpha _g} $ is a linear coefficient, it should be converted to the volumetric coefficient, as the problem is all about the volumetric expansion.
So, the equation that converts linear to volumetric coefficient is as follows
 $ \gamma = 3\alpha $
Substitute this value in the equation (1), to carry out the further calculations.
So, we get,
 $
  {V_g}(3{\alpha _g}) = {V_{Hg}}{\gamma _{Hg}} \\
   \Rightarrow 3{V_g}{\alpha _g} = {V_{Hg}}{\gamma _{Hg}} \\
  $
Now substitute the given values in the above equation.
 $
 \Rightarrow 3 \times {V_g} \times 9 \times {10^{ - 6}} = {V_{Hg}} \times 18.9 \times {10^{ - 5}} \\
   \Rightarrow \dfrac{{{V_{Hg}}}}{{{V_g}}} = \dfrac{{27 \times {{10}^{ - 6}}}}{{189 \times {{10}^{ - 6}}}} \\
  $
Upon further carrying out the calculations, we get,
 $
 \Rightarrow \dfrac{{{V_{Hg}}}}{{{V_g}}} = \dfrac{{27}}{{189}} \\
   \Rightarrow \dfrac{{{V_{Hg}}}}{{{V_g}}} = \dfrac{1}{7} \\
  $
 $ \therefore $ The fraction of the volume of a glass flask that must be filled with mercury so that the volume of the empty space may be the same at all temperatures is given by $ \dfrac{1}{7} $ .

Thus, the option (B) is correct.

Note
Being an important part of the calculation, the conversion of the linear coefficient to the volumetric coefficient should be taken care of, as the whole calculation is based on finding the change in the volumetric expansion.