IIT JEE , AIEEE Forum - Physics , Mechanics

kasirajan.1990

A U-tube containing a liquid is accelerated horizontally with a constant horizontally a_o .If the seperation b/w the vertical limbs is l thn find the difference in the heights of the liquid in the two arms.

aditya_arora04

Hi,

The appratus goes right

Look, let l₁ be the final height of left column, and l₂ be of right.

So, Al₁g = Al₂g + Aka₀ [ k is distance between limbs, A is area of cross-section ]

So, we get :

Difference in heights => (l₁- l₂) = ka₀/g

pannaguma

i think its tan@=a0/g = H/L
gives H=La0/g

kasirajan.1990

@pannaguma ur answer is corect..cld u explain how u arrived at this ans.!

pannaguma

just correlate the concept where you find angle of top liquid surface, contained in a body, which is accelerated horizontally........

its the same thing, here. so the pressure difference causes a height change. except that here only the extreme points are under consideration, but pressure difference remains the same.

plz rate if helpful.......

aditya_arora04

Look,

When the appratus is moving, the water which is in the horizontal part of the u-tube, will experience a force Alda₀ [ here, A is area of cross-section, l is length of the horizontal column, when we multiply we get the volume of water in that part, d is density, a0 is acc with which the appratus moves ]

If it is moving in the right direction, the final picture is as shown in pic.

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