Question
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984kg m^{–3}$. Determine the height of the wine column for normal atmospheric pressure.
✓
Answer
Density of mercury, $\rho_1= 13.6\times10\text{kg/m}$
Height of the mercury column, h = 0.76m.
Density of French wine, $\rho_2=984\text{kg/ m}^3$
Height of the French wine column = $h_2$
Acceleration due to gravity, $g = 9.8m/s^2$
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column.
= Pressure in the French wine column.
$\rho_1\text{h}_1\text{g}=\rho_2\text{h}_2\text{g}$
$\text{h}_2=\frac{\rho_1\text{h}_1}{\rho_2}$
$=\frac{13.6\times10^3\times0.76}{986}$
$10.5\text{m}$
Hence, the height of the French wine column for normal atmospheric pressure is 10.5m.
Height of the mercury column, h = 0.76m.
Density of French wine, $\rho_2=984\text{kg/ m}^3$
Height of the French wine column = $h_2$
Acceleration due to gravity, $g = 9.8m/s^2$
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column.
= Pressure in the French wine column.
$\rho_1\text{h}_1\text{g}=\rho_2\text{h}_2\text{g}$
$\text{h}_2=\frac{\rho_1\text{h}_1}{\rho_2}$
$=\frac{13.6\times10^3\times0.76}{986}$
$10.5\text{m}$
Hence, the height of the French wine column for normal atmospheric pressure is 10.5m.
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