Question

The closed container filled with water is moving with some acceleration as shown in the figure. Then $P_A-P_D$ is :-

• A. $-4\times {10}^3{\text{N/m}}^2$
• B. $+4\times {10}^3{\text{N/m}}^2$
• C. $+5.6\times {10}^5{\text{N/m}}^2$
• D. $-5.6\times {10}^4{\text{N/m}}^2$

Answer 1

The correct option is D $-5.6\times {10}^4{\text{N/m}}^2$

For the given accelerated container the pressure difference in liquid along the horizontal direction is given by,
$\Delta P=\rho ax...\left(i\right)$


For the accelerated container, the pressure will increase in a direction opposite to that of acceleration of the vessel.
$\Rightarrow P_C>P_D$
From eq.$\left(i\right)$,
$P_C-P_D=\rho ax$
$\Rightarrow P_C-P_D=1000\times 2\times 2$
$(\because x=2\text{m})$
$\Rightarrow P_C-P_D=4\times {10}^3{\text{N/m}}^2...\left(ii\right)$

The pressure difference for liquid due to vertical depth is given as:
$\Delta P=\rho gh$
$\because$ The pressure will increases in vertically downward direction
$P_C-P_A=\rho gh$
where $h=6\text{m}$ is the depth of liquid.
$\Rightarrow P_C-P_A=1000\times 10\times 6$
$\Rightarrow P_C-P_A=60\times {10}^3{\text{N/m}}^2...\left(iii\right)$
On subtracting $\left(iii\right)$ from $\left(ii\right)$,
$\Rightarrow (P_C-P_D)-(P_C-P_A)=(4\times {10}^3)-(60\times {10}^3)$
$\Rightarrow P_A-P_D=-56\times {10}^3$
$\therefore P_A-P_D=-5.6\times {10}^4{\text{N/m}}^2$
Thus, difference in pressure $(P_A-P_D)$ is $-5.6\times {10}^4{\text{N/m}}^2$