Question

A glass flask is filled up to a mark with 50 cc of mercury at ${18^0}C$. If the flask and content are heated to ${38^0}C$, how much mercury will be above the mark? (α for glass is 9$\times$${10^{ - 6}}$/C and coefficient of real expansion of mercury is 180 $\times$ ${10^{ - 6}}$/C)
A. $0.85cc$
B. $0.46cc$
C. $0.153cc$
D. $0.05cc$

Answer 1

Hint: Usually Any object or fluid when heated it expands and when cooled it contracts. The same applies for glass and mercury too. But mercury is inside the glass and as both are combinedly heated both expands. We need to contract increment in flask volume from increment in mercury volume to get apparent increment.

Formula used:
$\eqalign{
  & \gamma = 3\alpha \cr
  & \dfrac{{\Delta v}}{v} = \gamma \Delta T \cr
  & \dfrac{{\Delta v}}{v} = 3\alpha \Delta T \cr} $

Complete answer:
If there exists a case where when heated only mercury gets expanded and glass won’t expand then this problem would have been easier.
But when heated glass also expands so for the observer he sees the apparent expansion of mercury not the real expansion of the mercury.

Apparent expansion of mercury = real expansion of mercury – expansion of the glass
For mercury
$\dfrac{{\Delta v}}{v} = \gamma \Delta T$
Where $\Delta v$ is changed in real volume of the mercury.
v = initial reading volume
$\gamma $ = coefficient of real expansion of the mercury.
$\Delta T$ = change in the temperature.
Hence by applying the above formula we get
$\Delta v = v\gamma \Delta T = (50cc)(180 \times {10^{ - 6}})({38^0}C - {18^0}C) = 0.180cc$ …eq 1
For glass
$\dfrac{{\Delta v}}{v} = 3\alpha \Delta T$
Where $\Delta v$ is the change in volume of the glass.
 v = initial reading volume
$\alpha $= coefficient of linear expansion of the glass
$\gamma = 3\alpha $= coefficient of volumetric expansion of the glass
$\Delta T$ = change in the temperature.
Hence by applying the above formula we get
$\Delta v = v(3\alpha )\Delta T = (50cc)(3 \times 9 \times {10^{ - 6}})({38^0}C - {18^0}C) = 0.027cc$ …eq 2
From equation 1 and 2 by subtracting equation two from equation 1 we get apparent mercury volume increase which will be
$0.180cc - 0.027cc = 0.153cc$

So the answer would be option C.

Note:
For glasses we consider isotropic expansion or contraction when temperature changes. Hence we had taken its volumetric coefficient as thrice of linear coefficient i.e in all 3 dimensions. Even though mercury expands more in real since glass also expands simultaneously mercury expansion seems to be less. If mercury is poured in grey cast iron and heated then expansion seems to be more because grey cast iron contracts on heating.