Home
Class 12
PHYSICS
Find the hydrostatic force (in kN) actin...

Find the hydrostatic force (in kN) acting on the circular patch of radius ` R. (R = 1m, g = 10 m // s ^ 2 ` , atmospheric pressure `= 10 ^(5 ) Pa )

Text Solution

Verified by Experts

The correct Answer is:
C
` F= (P_0 + rho_g Y_C) A `
where ` Y_C = 2R and A pi R ^ 2 `
or , ` F = [10 ^5 + (1000) (10) (2) ] pi = 120 pi kN = 377 kN `
Promotional Banner

Similar Questions

Explore conceptually related problems

What is the pressure on a swimmer 10m below the surface of lake? g=10ms^(-2) , atmospheric pressure = 1.01 xx 10^(5)Pa

At a depth of 500m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? Density of sea water is 1.03 xx 10^(3) kg//m^(3), g=10ms^(-2) . Atmospheric pressure = 1.01 xx 10^(5)Pa .

Find the thrust acting on the human body due to atmospheric pressure. Take the surface area of a man of middle size to be 1.5m^2 and atmospheric pressure (1 atm) =1.013xx10^5 Pa.

Water is filled in a flask upto a height of 20cm . The bottom of the flask is circular with radius 10 cm . If the atmospheric pressure is 1.013xx10^(5) Pa, find the force exerted by the water on the bottom . Take g=10 "ms"^(-2) and density of water =1000 "kgm"^(-3).

The pressure acting on a submarine is 3 xx10^(5) at a certain depth. If the depth is doubled the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is 1 xx 10 ^(5) Pa density of water is 10^(3) kgm ^(-3), g - 10 ms ^(-2))

The guage pressure exerted below a column of water, open to the earth's atmosphere at depth of 10 m is (density of water = 1000 kg/ m^3 , g = 10 m/ s^2 and 1 atm pressure = 10^5 Pa)

At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) what is the auge pressure? (c ) Find the force acting on the windown of area 20 cm xx 20cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. The density of sea water is 1.03 xx 10^(3) kg m^(-3) , g=10ms^(-2) . Atmospheric pressure = 1.01 xx 10^(5) Pa .

The centripetal force acting on a body of mass m moving with speed v along a circular path of radius r is given by F = (mv^(2))/(r) If the values of m , v and r are measured as 0.5 kg , 10 ms^(-1) and 0.4 m respectively to the accuracies of 0.005 kg , 0.01 ms^(-1) and 0.01 m respectively , calculate the percentage error in the force acting on the body.

There is a small mercury drop of radius 4.0mm . A surface P of area 1.0 mm^(2) is placed at the top of the drop. Atmospheric pressure = 10^(5) Pa . Surface tension of mercury = 0.465 N//m . Gravity effect is negligible. The force exerted by air on surface P is