QUESTION 25 ,JEE MAINS LEVEL
Attachments:
Answers
Answered by
7
From Bernoulli's principle and Torricelli's law we have (the total energy being constant) following for a liquid of uniform density in a tank or pipe.
v²/2 + g h + P / ρ = constant
From this it is derived that for liquid (in a container) coming out of a small hole at a depth of d from surface:
velocity of liquid at the hole = v = √(2gd)
(as the pressure at the top surface and at the aperture is P_atm.)
In the vessel the liquid of density ρ is on the top. Velocity of liquid coming out from that : v1 = √(2gh/2) = √(gh)
For finding velocity v2 of the heavier liquid coming out of the hole at a depth of 3h/2 from surface, we replace the lighter (top) liquid with half its volume of the heavier liquid of density 2ρ.
So now the lower hole will be at a depth of h from the new surface.
So v2 = √(2gh)
v1/ v2 = 1/√2
v²/2 + g h + P / ρ = constant
From this it is derived that for liquid (in a container) coming out of a small hole at a depth of d from surface:
velocity of liquid at the hole = v = √(2gd)
(as the pressure at the top surface and at the aperture is P_atm.)
In the vessel the liquid of density ρ is on the top. Velocity of liquid coming out from that : v1 = √(2gh/2) = √(gh)
For finding velocity v2 of the heavier liquid coming out of the hole at a depth of 3h/2 from surface, we replace the lighter (top) liquid with half its volume of the heavier liquid of density 2ρ.
So now the lower hole will be at a depth of h from the new surface.
So v2 = √(2gh)
v1/ v2 = 1/√2
Dhruv00:
ohhh really thanx
but why u have replace the lighter liquid by the heavier one with half volume
because we know how to calculate using formula, for a liquid of uniform density.. and we dont have a formula for a mixture of different densities..
but how u replace the lighter liquid by heavier one with "HALF VOLUME"
Similar questions