Question

The volume of the bulb of a mercury thermometer at 0°C, is V0 and cross section of the capillary is A0. The coefficient of linear expansion of glass is ag per °C and the cubical expansion of mercury lamdam per °C. If the mercury just fills the bulb at 0°C what is the length of mercury column in capillary at T°C

Answer 1

L = \frac{V_0 T (\lambda_m - 3\alpha_g)}{A_0}

Answer 2

The length of the mercury column in the capillary at T°C can be determined by considering the differential expansion of mercury and glass.

1. Initial State (0°C):

   • Volume of the glass bulb = $V_0$
   • Volume of mercury = $V_0$ (since it just fills the bulb)
   • Cross-section of the capillary = $A_0$

2. Final State (T°C):

   • Volume of the glass bulb at T°C ($V_{bulb}(T)$):
     The coefficient of linear expansion of glass is $\alpha_g$. The coefficient of cubical expansion of glass ($\gamma_g$) is approximately $3\alpha_g$.
     $V_{bulb}(T) = V_0 (1 + \gamma_g T) = V_0 (1 + 3\alpha_g T)$

   • Volume of mercury at T°C ($V_{mercury}(T)$):
     The coefficient of cubical expansion of mercury is $\lambda_m$.
     $V_{mercury}(T) = V_0 (1 + \lambda_m T)$

3. Volume of mercury in the capillary:
   The mercury expands more than the glass bulb. The excess volume of mercury that rises into the capillary is the difference between the expanded volume of mercury and the expanded volume of the bulb.
   $V_{excess} = V_{mercury}(T) - V_{bulb}(T)$
   $V_{excess} = V_0 (1 + \lambda_m T) - V_0 (1 + 3\alpha_g T)$
   $V_{excess} = V_0 [(1 + \lambda_m T) - (1 + 3\alpha_g T)]$
   $V_{excess} = V_0 [1 + \lambda_m T - 1 - 3\alpha_g T]$
   $V_{excess} = V_0 (\lambda_m T - 3\alpha_g T)$
   $V_{excess} = V_0 T (\lambda_m - 3\alpha_g)$

4. Length of mercury column:
   This excess volume ($V_{excess}$) occupies a length $L$ in the capillary with cross-section $A_0$.
   $V_{excess} = A_0 L$
   Therefore, $A_0 L = V_0 T (\lambda_m - 3\alpha_g)$
   $L = \frac{V_0 T (\lambda_m - 3\alpha_g)}{A_0}$

The length of the mercury column in the capillary at T°C is $\frac{V_0 T (\lambda_m - 3\alpha_g)}{A_0}$.