A pressure of $1 \mathrm{~mm}$ of mercury is equivalent to
A pressure of $1 \mathrm{~mm}$ of mercury is equivalent to
- $1 \mathrm{mPa}$
- $13.33 \mathrm{~Pa}$
- $133.3 \mathrm{~Pa}$
- $133.3 \mathrm{~Pa}$
Solution
Height of mercury column
$
h=1 \mathrm{~mm}=10^{-3} \mathrm{~m}
$
Density of mercury, $\rho=13.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
$
g=9.8 \mathrm{~ms}^{-2}
$
$\therefore$ Pressure, $p=\rho g h$
$
\begin{aligned}
& =13.6 \times 10^3 \times 9.8 \times 10^{-3} \\
& =133.28 \mathrm{~Pa} \simeq 133.3 \mathrm{~Pa}
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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