Question

The vertical limbs of a U shaped tube are filled with a liquid of density p upto a height h on each side. The
horizontal portion of the U tube having length 2h contains a liquid of density 2p. The U tube is moved
horizontally with an accelerator g/2 parallel to the horizontal arm. The difference in heights in liquid levels in
the two vertical limbs, at steady state will be​

Answer 1

Answer:

height difference of liquid in two arms will be equal to "h"

Explanation:

As we know that U shaped tube is moving horizontally with constant acceleration a = g/2

now we know that horizontal section of the liquid is accelerating with the tube

so as per Newton's 2nd Law we will have

$F_{net} = ma$

$\rho A(2h) = m$

so we have

$\rho A(2h) \times \frac{g}{2} = (P_1 - P_2) A$

$\rho (2h) \times (\frac{g}{2}) = (\rho g h_1 - \rho g h_2)$

$h = (h_1 - h_2)$

so height difference of liquid in two arms will be equal to "h"

#Learn

Topic : Fluid Statics

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