Question

A pycnometer weighs 40 gm when empty and 1040 gm when filled with mercury at $0^\circ {\text{C}}$. On heating to $100^\circ {\text{C}}$, 10 gm of mercury overflows. If the coefficient of real expansion of mercury is $0.0002/^\circ {\text{C}}$, the coefficient of cubical expansion of glass is:
A. $0.00001/^\circ {\text{C}}$
B. $0.0003/^\circ {\text{C}}$
C. $0.0002/^\circ {\text{C}}$
D. $0.0001/^\circ {\text{C}}$

Answer 1

Calculate the apparent expansion coefficient of the mercury using the expression for the volume expansion with respect to the temperature. The real expansion coefficient of the mercury is the sum of cubical expansion of the glass and the apparent expansion of the mercury. Then solve for the coefficient of cubical expansion of the glass.

Formula used:
${\gamma _{real}} = {\gamma _{glass}} + {\gamma _{app}}$
Here, ${\gamma _{real}}$ is the real expansion coefficient, ${\gamma _{glass}}$ is the cubical expansion coefficient of glass and ${\gamma _{app}}$ is the apparent coefficient of expansion.

Complete step by step answer:
We have given that the initial mass of the mercury is 1000 gm and when it is heated to $100^\circ {\text{C}}$, 10 gm of mass overflows. We have the relation,
${\gamma _{app}} = \dfrac{{{\text{Mass overflows}}}}{{{\text{Initial mass}} \times \Delta T}}$
Here,${\gamma _{app}}$ is the apparent coefficient of expansion of mercury and $\Delta T$ is the change in the temperature.

Substituting the values, we get,
${\gamma _{app}} = \dfrac{{{\text{10}}}}{{1000 \times 100}}$
$\Rightarrow {\gamma _{app}} = 0.0001/^\circ {\text{C}}$
We know that the real expansion of the mercury is the sum of cubical expansion of the glass and the apparent expansion of the mercury. Therefore,
${\gamma _{real}} = {\gamma _{glass}} + {\gamma _{app}}$
$\Rightarrow {\gamma _{glass}} = {\gamma _{real}} - {\gamma _{app}}$

Substituting $0.0002/^\circ {\text{C}}$ for ${\gamma _{real}}$ and $0.0001/^\circ {\text{C}}$ for ${\gamma _{app}}$ in the above equation, we get,
${\gamma _{glass}} = 0.0002 - 0.0001$
$\therefore {\gamma _{glass}} = 0.0001/^\circ {\text{C}}$
Therefore, the cubical expansion of the glass is $0.0001/^\circ {\text{C}}$.

So, the correct answer is option D.

Note: Always remember when we heat the substance placed in the glass; both glass and the substance undergo volume expansion. To determine the apparent expansion coefficient of the mercury, we have used the expression for the volume expansion, $\Delta V = \gamma {V_i}\Delta T$, where, $\Delta V$ is the change in the volume which can be treated as change in the weight, $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in the temperature.