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

As the fluid velocity increases the dissipative forces become important as turbulence rises due to which drag due to friction forces increases.

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 .

Total weight supported by the film
Since a soap film has two surfaces, the total length of the liquid film is 60 cm.
Surface Tension is T

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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