Question

A $\text{U}$ tube of base length $l$ filled with 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 height $h$ is given by

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

Answer 1

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


For the given situation, liquid of density $2\rho$ should be behind that of liquid of density $\rho$.

From the right limb,

$P_A=P_{atm}+\rho gh$

$P_B=P_A+\rho a\frac{l}{2}$

$\therefore P_B=P_{atm}+\rho gh+\rho a\frac{l}{2}$

$P_C=P_B+\left(2\rho \right)a\frac{l}{2}$

$\Rightarrow P_C=P_{atm}+\rho gh+\frac{3}{2}\rho al.....\left(i\right)$

But from the left limb

$P_C=P_{atm}+\left(2\rho \right)gh..........\left(ii\right)$

From Eqs $\left(i\right)$ and $\left(ii\right)$

$P_{atm}+\rho gh+\frac{3}{2}\rho al$ = $p_{atm}+2\rho gh$

$\therefore h=\frac{3al}{2g}$

Hence, option $\left(\text{B}\right)$ is correct.