Question

If area of cross-section of both the limbs are equal, find the difference in heights of the liquid in the two limbs when the system is given an acceleration of $2{\text{m/s}}^2$ in horizontal direction as shown in figure. Neglect the width of the limbs.

• A. $15\text{cm}$
• B. $18\text{cm}$
• C. $20\text{cm}$
• D. $10\text{cm}$

Answer 1

The correct option is D $10\text{cm}$

Let the difference in heights of the liquid in the two limbs be $x$ and $P_0$ be the atmospheric pressure.


Since fluid is accelerating horizontally in rightward direction, hence $P_A>P_B$.
For pressure at $A$,
$\Rightarrow P_A=P_B+\rho aL$
$\Rightarrow P_A=P_0+(\rho \times 2\times 0.5)$
[$\because P_B=P_0$, as the limb is open to atmosphere]
$\Rightarrow P_A=P_0+\rho ...\left(1\right)$
Due to vertical height, the pressure difference can also be written as,
$P_A=P_0+\rho gx$
$\Rightarrow P_A=P_0+10\rho x...\left(2\right)$
From Eq.$\left(1\right)\&\left(2\right)$,
$P_0+\rho =P_0+10\rho x$
$\Rightarrow x=\frac{1}{10}$
$\therefore x=0.1\text{m}=10\text{cm}$
Hence, difference in heights of the liquid in the two limbs is $10\text{cm}$.