IIT JEE , AIEEE Forum - Physics , Mechanics

pinnaka

Hmm,Let us think about this Question
You have water initially poured into the U tube so therefore the levels in both arms must be equal say x from the base of the u tube
Upon pouring oil through one of the sides such that the water in the other branch increase by 25cm, so this means that the water in the side in which there is oil goes down by 25cm. Now the oiul floats on the water. For the system to be equilibrium the pressure at all depths must be equal. Now This trick probelm
we will start the analysis from the bottom of the cube
at the bottom of the tube
we have the pressures at the bottom due to the both arms to be equal

P_a + r_oil * h *g + r_water *(x-25) g = P_a + r_water *(x+25)g
now here h is the height of the column containing oil

now the above equation goes like the below

h = r_water (50)/(r_oil)
h = 62.5cm
so therefore the height of the oil column is equal to 62.5cm
but
Now analyzing the pressure in the both the arms at the end of the oil column
r_water * (y) = r_oil * 62.5)
y = 50cm
y is the height of the water column above the oil water interface (in the other side)

now the difference in heights
62.5cm ? 50 cm = 12.5cm

So the answer is 12.5cm

----Hope this Helps
----Stuart Anderson
---stuartanderson_cal@yahoo.com

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