A small circular membrane is 10cm below the surface of a pool of mercury when the barometric height is 76 cm of mercury. If the density of mercury is 13600\(kgm^{-3}\), what is the pressure on the membrane in \(Nm^{-2}\)? \((g =10ms^{-2})\)
Pressure at a depth for fluid with constant density such as mercury is given as
\(p = p_{0} + \rho hg\)
where \(p_{0}\) = atmospheric pressure.
\(p_{0} = \rho hg =13600 \times \frac{76}{100} \times 10\)
= \(103,360 Nm^{-2}\)
\(p = 103,360 + (13600 \times \frac{10}{100} \times 10)\)
\(p = 103,360 + 13,600\)
\(p = 116,960 Nm^{-2} \approxeq 1.17 \times 10^{5} Nm^{-2}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}