Find the pressure exerted below a column of water, open to the atmosphere, at depth
(i) 10 m " " (ii) 30 m
(Given, density of water `=1xx10^(3)kgm^(-3), g =10 ms^(-2))`

Find the pressure exerted below a column of water, open to the atmosphere, at depth (i) 5 m (ii) 20 m (Given, density of water = 1 xx 10^(3)"kg m"^(-3), g = 10 m s^(-2) )

Find the pressure exerted below a column of water, open to the atmosphere, at depth (i) 5 m (ii) 20 m (Given, density of water = 1 xx 10^(3)"kg m"^(-3), g = 10 m s^(-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)

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)

The pressure at a point in water is 10 N//m^(2) . The depth below this point where the pressure becomes double is (Given density of water = 10^(3) kg m^(-3) , g = 10 m s^(-2) )

The pressure at a point in water is 10 N//m^(2) . The depth below this point where the pressure becomes double is (Given density of water = 10^(3) kg m^(-3) , g = 10 m s^(-2) )

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))

Calculate the total pressure at the bottom of a lake of depth 5.1 m (atm pressure =10^(5) P), density of water =10^(3)kg//m^(3),g=10m//s^(2)

A glass full of water has a bottom of area 20 cm^(2) , top of area 20 cm , height 20 cm and volume half a litre. a. Find the force exerted by the water on the bottom. b. Considering the equilibrium of the v , after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure = 1.0 xx 10^(5) N//m^(2) . Density of water = 1000 kg//m^(-3) and g = 10 m//s^(2)

A glass full of water has a bottom of area 20 cm^(2) , top of area 20 cm , height 20 cm and volume half a litre. a. Find the force exerted by the water on the bottom. b. Considering the equilibrium of the v , after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure = 1.0 xx 10^(5) N//m^(2) . Density of water = 1000 kg//m^(-3) and g = 10 m//s^(2)