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For the angle of contact , radius of the tube r, surface tension t, the density of fluid  the rise in the column is given by The radii of the two limbs r1 and r2 are 3.0 mm and 1.5 mm respectively  The level in the limb of diameter 6.0 mm is The level in the limb of diameter 3.0 mm is The difference in the heights is h2 - h1 = 4.96 mm
Since the angle of contact is obtuse the Pressure will be more on the Mercury side. This pressure difference is given as The dip of mercury inside the narrow tube would be equal to this pressure difference The mercury dips down in the tube relative to the liquid surface outside by an amount of 5.34 mm. 
Speed of the wind above the upper wing surface is v1 = 234 km h-1 Speed of the wind over the lower wing is v2 = 180 km h-1 Let the pressure over the upper and lower wing be P1 and P2 Let the plane be flying at a height of h The density of air is  Applying Bernoulli's Principle at two points over the upper and lower wing we get Area of each wing is a = 25 m2 The net upward force on the plane...
The diameter of the artery is d The viscosity of blood is  The density of blood is  The average velocity is given by  Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have
The diameter of the artery is  The viscosity of blood is  The density of blood is  The average velocity is given by  Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have
The density of whole blood  Gauge Pressure  Height at which the blood container must be placed so that blood may just enter the vein
Since the height of the water level in the vessels is the same the Pressure at the bottom would be equal. As the area of the bottom is also the same the Force exerted by the water on the bottom would be the same. The difference in the reading arises due to the fact that the weight depends on the volume of the water inside the container which is more in the first vessel. The vertical component...
As we know Specific Gravity of Mercury is 13.6 therefore 13.6 cm of water column would be equal to  1 cm of Mercury column. The pressure at the Mercury Water interface in the right column = Atmospheric Pressure + 1 cm of Mercury = 77 cm of Mercury The difference in Pressure due to the level of the Mercury column = Pressure at the Mercury Water interface - Absolute Pressure of the Glass...
In figure (a) Gauge Pressure = 20 cm of Mercury Absolute Pressure = Atmospheric Pressure + Gauge Pressure Absolute Pressure = 76 + 20= 96 cm of Mercury In figure (b) Gauge Pressure = -18 cm of Mercury Absolute Pressure = Atmospheric Pressure + Gauge Pressure Absolute Pressure = 76 + (-18)= 58 cm of Mercury
Pressure in the waterside at the bottom is Pressure in the acid side at the bottom is The pressure difference across the door is Area of the door, a = 20 cm2  The force necessary to keep the door closed is Note: The dimensions of the door are small enough to neglect pressure variations near it.
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