Answer:
The atmospheric pressure is [tex]0.97622\times10^{5}\ Pa[/tex].
Explanation:
Given that,
Atmospheric pressure [tex]P_{atm}= 1.013\times10^{5}\ Pa[/tex]
drop height h'= 27.1 mm
Density of mercury [tex]\rho= 13.59 g/cm^3[/tex]
We need to calculate the height
Using formula of pressure
[tex]p = \rho g h[/tex]
Put the value into the formula
[tex]1.013\times10^{5}=13.59\times10^{3}\times9.8\times h[/tex]
[tex]h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}[/tex]
[tex]h=0.76\ m[/tex]
We need to calculate the new height
[tex]h''=h - h'[/tex]
[tex]h''=0.76-27.1\times10^{-3}[/tex]
[tex]h''=0.76-0.027[/tex]
[tex]h''=0.733\ m[/tex]
We need to calculate the atmospheric pressure
Using formula of atmospheric pressure
[tex]P=\rho g h[/tex]
Put the value into the formula
[tex]P= 13.59\times10^{3}\times9.8\times0.733[/tex]
[tex]P=0.97622\times10^{5}\ Pa[/tex]
Hence, The atmospheric pressure is [tex]0.97622\times10^{5}\ Pa[/tex].
