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 \(Nm^{-2}\)? \((g =10ms^{-2})\)
1.17 x 107 Nm-2
6.80 x 105Nm-2
1.17x105Nm-2
1.03 x 105Nm-2
1.36 x104 Nm-2
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Discussions (5)
Total pressure on object=P+P1, where P and P1 are the atmospheric pressure and pressure due to the depth 10cm respectively.
P=0.76×10×13600=103360 N/m^2
P1=0.1×10×13600=13600 N/m^2
P+P1=103360+13600=1.17×10^5 N/m^2
Alternatively; 10cm+76cm= 86cm =0.86m
0.86×10×13600= 116960=1.17×10^5 N/m^2
Total pressure on object=P+P1, where P and P1 are the atmospheric pressure and pressure due to the depth 10cm respectively.
P=0.76×10×13600=103360 N/m^2
P1=0.1×10×13600=13600 N/m^2
P+P1=103360+13600=1.17×10^5 N/m^2
The selected answer is wrong:
P=hpg,for Christ's sake h is the height which is 10cm not 76cm! Therefore
P=10*10*13600=1.03*10^5
