Question

A closed cubical vessel is given a horizontal acceleration $a$. Find the value of $a$ such that pressure at the mid point $M$ of $\text{AC}$ is equal to pressure at $N$.

• A. $2g$
• B. $3g$
• C. $g$
• D. $4g$

Answer 1

The correct option is A $2g$

For the given accelerated container, the pressure difference in the liquid along horizontal direction is given by,
$\Delta P=\rho ax...\left(i\right)$
The pressure will increase in a direction opposite to that of acceleration of the container.


From figure, applying Eq.$\left(i\right)$
$P_M=P_C+(\rho a\times 2.5)...\left(1\right)$
Similarly,
$P_A=P_C+(\rho a\times 5)...\left(2\right)$
The pressure difference in the liquid due to vertical depth is given as:
$\Delta P=\rho gh$
Since $P_A>P_N$
$\Rightarrow P_N=P_A-\rho g\times 5...\left(3\right)$
$[\because h=5\text{m}]$

From eq.$\left(1\right),\left(2\right),\left(3\right)$
$P_N=(P_C+\rho a\times 5)-\rho g\times 5...\left(4\right)$
Equating pressure at $M\&N$ as given in question,
$P_M=P_N\Rightarrow P_C+\rho a\times 2.5=P_c+\rho a\times 5-\rho g\times 5$
$\Rightarrow 5\rho g=2.5\rho a$
$\therefore a=2g$
Acceleration of the closed cubical vessel is $2g$