Question

A $U-tube$ of base length ${}^{\prime }{\ell }^{\prime }$ filled with the same volume of two liquids of densities $p$ and $2p$ 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. $\frac{al}{g}$
• B. $\frac{3al}{2g}$
• C. $\frac{2al}{3g}$
• D. $\frac{al}{2g}$

Answer 1

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

The pressure at the bottoms of the left hand side and right hand side columns are,

$\begin{array}{c}{P}_1=p_0+2\rho ghand\\ P_2=p_0+\rho gh\end{array}$

Where $p_0$ is the atmospheric pressure.

The pressure at the left-hand side must be greater because a net force to the right is needed to accelerate the liquid in the horizontal part of the tube.

The masses in the left hand side and the right hand side of the liquid is $2\rho \left(\frac{L}{2}\right)Aand\rho \left(\frac{L}{2}\right)A$

where $A$ is cross sectional area of the tube.

$\begin{array}{c}\rho ghA=\left[2\rho \left(\frac{L}{2}\right)A+\rho \left(\frac{L}{2}\right)A\right]fwheref=acceleration.\\ \Rightarrow gh=\frac{3L}{2}f\\ \therefore h=\frac{3a}{2g}l\end{array}$

Option $C$ is correct .