Question

A U-tube of base length ${}^{\prime }{l}^{\prime }$ filled with the same volume of two liquids of densities $\rho$ and $2\rho$ is moving with an acceleration ${}^{\prime }{a}^{\prime }$ 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,

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

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

$P_C=P_B+\left(2\rho \right)a\frac{l}{2}=P_{atm}+\rho gh+\frac{3}{2}\rho al$

But from the left limb,

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

From both the equations of $P_C$

$P_{\text{atm}}+\rho gh+\frac{3}{2}\rho al=P_{\text{atm}}+2\rho gh$
$\Rightarrow h=\frac{3al}{2g}$