Question

An $U-$ 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, as shown in the figure. If the difference in heights of the liquid columns in the two vertical limbs (open to the atmosphere) is zero, then the height $h$ is given by

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

Answer 1

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

For the accelerated container, the pressure will increase in a direction opposite to that of acceleration.
$\Rightarrow P_C>P_A$
Hence the heavier liquid $\left(2\rho \right)$ will be in the left limb so that higher hydrostatic pressure difference also justifies that $P_C>P_A$.


For right limb:
$P_A=P_{atm}+\rho gh...\left(i\right)$
Due to accelerated container,
$P_B=P_A+\rho ax$
$\Rightarrow P_B=P_A+\rho a\times \frac{l}{2}...\left(ii\right)$

For left limb:
$P_C=P_{atm}+2\rho gh...\left(iii\right)$
Due to accelerated container,
$P_C=P_B+2\rho ax$
$\Rightarrow P_C=P_B+2\rho a\times \frac{l}{2}...\left(iv\right)$

From Eq.$\left(i\right)\left(ii\right),\left(iii\right),\left(iv\right)$, we get
$P_{atm}+\rho gh+\frac{3}{2}\rho al=P_{atm}+2\rho gh$
$\therefore h=\frac{3al}{2g}$