Question

Torricelli's barometer uses mercury of density $13.6\times {10}^3{\text{kg/m}}^3$. If mercury is replaced by wine of density $984{\text{kg/m}}^3$, then determine the height of the wine column for normal atmospheric pressure.

• A. $760\text{m}$
• B. $76\text{m}$
• C. $10.5\text{m}$
• D. $105\text{m}$

Answer 1

The correct option is C $10.5\text{m}$

The pressure due to height column of mercury is
$P=\rho gh.......\left(i\right)$
where $\rho =$ density of mercury.
The normal atmospheric pressure is $76\text{cm}$ of $Hg$.
If $h^{\prime }$ is the height of the column of wine corresponding to one atmospheric pressure, then,
$P=\sigma g{h}^{\prime }.......\left(ii\right)$
where $\sigma =$ density of wine
Given, $\sigma =984{\text{kg/m}}^3$
Equating $\left(i\right)$ & $\left(ii\right)$, we get
$\rho gh=\sigma g{h}^{\prime }$
$\Rightarrow h^{\prime }=\frac{\rho h}{\sigma }=\frac{13600\times 0.76}{984}$
$\Rightarrow h^{\prime }=10.5\text{m}$