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Damped Oscillations

A U-tube stan...

Question

A U-tube stands vertically such that its arm A is towards left and arm B is towards right and the distance AB is L. When it is at rest, a liquid stands at the same height in the two arms i.e, $H_{A} = H_{B}$ . If the tube is given an acceleration a towards right, then

A

H_{B} = H_{A} + \frac{g L}{a}

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B

H_{B} = H_{A} + \frac{a L}{g}

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C

H_{A} = H_{B} + \frac{g L}{A}

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D

H_{A} = H_{B} + \frac{a L}{g}

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Solution

The correct option is D $H_{A} = H_{B} + \frac{a L}{g}$
Mass of liquid in horizontal position is $A L ρ$ .
The pseudo force acting is $A L ρ a$ (a is the acceleration)
Force due to the pressure difference is: $(H_{A} ρ g - H_{B} ρ g) A$
$(H_{A} - H_{B}) ρ g A = A L ρ a$
$H_{A} - H_{B} = \frac{A L ρ a}{ρ g A} = \frac{a L}{g}$

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