Question

A U-tube of base length '$l$' filled with the same volume of two liquids of densities $\rho$ and $2\rho$ is moving with an acceleration '$a$' on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height $h$ is given by

• A. $\frac{a}{2g}l$
• B. $\frac{3a}{2g}l$
• C. $\frac{a}{g}l$
• D. $\frac{2a}{3g}l$

Answer 1

The correct option is B $\frac{3a}{2g}l$

For the given situation, liquid of density $2\rho$ should be behind that of $\rho$.
From the right limb, $P_A=P_{atm}+\rho gh$
$P_B=P_A+\rho a\frac{l}{2}=P_{atm}+\rho gh+\rho a\frac{l}{2}$
$P_C=P_B+\left(2\rho \right)a\frac{l}{2}=P_{atm}+\rho gh+\frac{3}{2}\rho al$ ....(i)
But from the left limb, $P_C=P_{atm}+\left(2\rho \right)gh$ .....(ii)
From Eqs. (i) and (ii),
$P_{atm}+\rho gh+\frac{3}{2}\rho al=P_{atm}+2\rho gh$
$\Rightarrow h=\frac{3al}{2g}$