Question

A mercury-glass thermometer consists of a glass bulb filled with mercury and a 

narrow glass tube attached to the glass bulb. At 0℃, the bulb is completely filled with the 

mercury and its radius is 5 mm. As the temperature increases the mercury column rise 

through the narrow glass tube whose diameter is 0.3 mm. 

What should be the minimum length of the glass tube so that this thermometer can be 

used to measure the temperatures up to 100 ℃ ?

(Hint: ignore the thermal expansion of the glass tube)

liner expansion coefficient of glass = 3.2 × 10−6 ℃−1

volume expansion coefficient of mercury = 1.82 × 10−4 ℃−1

Answer 1

Solution;

Given;

$\alpha_m=3.2×10^{-6}°C^{-1}$

$\gamma_g=1.82×10^{-4}°C^{-1}$

$r_b=5mm$

$d_t=0.3mm$

Let $A_o$be the cross sectional area of the glass tube;

$A_o=πr^2=π×(0.15×10^{-3})^2=7.069×10^{-8}m^2$

Let $V_o$ be the volume of the mercury in the bulb at 0°C;

(The bulb is in spherical shape)

$V_o=\frac43πr^3=\frac43×π×(5×10^{-3})^3=5.236×10^{-7}m^3$

Now;

Let L be the length of mercury in glass tube;

$L=\frac{\Delta V}{A_o}$

But;

$\Delta V=(\gamma_m-3\alpha_g)V_o$

Hence;

$L=\frac{V_0(\gamma_m-3\alpha_g)}{A_0}$

By substitution;

$L=\frac{5.236×10^{-7}(1.82×10^{-4}-3(3.2×10^{-4})}{7.069×10^{-8}}$

$L=1.276×10^{-3}m$

L=1.276mm

The length should be at least 1.277mm.