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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.
Excess Pressure inside a bubble is given by             (It's double the usual value because of the presence of 2 layers in case of soap bubble) where T is surface tension and r is the radius of the bubble Atmospheric Pressure is  The density of soap solution is  The pressure at a depth of 40 cm (h) in the soap solution is  Total Pressure inside an air bubble at that depth
Surface Tension of Mercury is  The radius of the drop of Mercury is r = 3.00 mm Excess pressure inside the Mercury drop is given by Atmospheric Pressure is  Total Pressure inside the Mercury drop is given by
As the liquid and the temperature is the same in all three the surface tension will also be the same. Since the length is also given to be equal (40 cm) in all three cases the weight being supported is also the same and equal to .
Cross-sectional area of cylindrical tube is a1 = 8.0 cm2 The total area of the 40 fine holes is a2 Speed of liquid inside the tube is v1 = 1.5 m min-1 Let the speed of ejection of fluid through the holes be v2 Using the continuity equation
By the continuity equation, the velocity of the non-viscous liquid will be large at the kink than at the rest of the tube and therefore pressure would be lesser here by Bernoulli's principle and the air column above it, therefore, should be of lesser height. Figure (a) is therefore incorrect.
The speed of air above and below the wings are given to be v1 = 70 m s-1 and v2 = 63 m s-1 respectively.  Let the pressure above and below the wings be p1 and p2 and let the model aeroplane be flying at a height h from the ground. Applying Bernoulli's Principle on two points above and below the wings we get The pressure difference between  the regions below and above the wing is 605.15 Pa The...
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