Answer:
Explanation:
Given
specific gravity of oil=0.79
height of oil column is h
height of water column is 5h
Gauge pressure at bottom is equivalent to [tex]26.9\ mm[/tex]
[tex]P_{gauge}=\rho_{Hg} gh_{Hg}[/tex]
[tex]P_{gauge}=13.6\times 10^3\times 9.8\times 26.9\times 10^{-3}\ Pa[/tex]
Pressure due to oil and water at bottom
[tex]P=\rho _{oil}hg+\rho _{w}5hg[/tex]
[tex]P=0.79\rho _whg+5\rho _whg=5.79\rho _{w}hg[/tex]
[tex]P_{gauge}=P[/tex]
[tex]13.6\times 10^3\times 9.8\times 26.9\times 10^{-3}=5.79\times 10^3\times 9.8\times h[/tex]
[tex]h=63.18\ mm[/tex]
