Q123 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Physics
A pressure of $1 \mathrm{~mm}$ of mercury is equivalent to
- A$1 \mathrm{mPa}$
- B$13.33 \mathrm{~Pa}$
- Correct answerC$133.3 \mathrm{~Pa}$
- D$133.3 \mathrm{~Pa}$
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, $p=\rho g h$
$$
\begin{aligned}
& =13.6 \times 10^3 \times 9.8 \times 10^{-3} \\
& =133.28 \mathrm{~Pa} \simeq 133.3 \mathrm{~Pa}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.