Question

A beaker containing a liquid of density $\rho$ moves up with an acceleration ${}^{\prime }{a}^{\prime }$. The pressure in the liquid at a depth $h$ below the free surface of the liquid is (consider atmospheric pressure to be $P_0$)

• A. $P_0+h\rho g$
• B. $P_0-h\rho (g-a)$
• C. $P_0+h\rho (g+a)$
• D. $P_0+2h\rho g\left(\frac{g+a}{g-a}\right)$

Answer 1

The correct option is C $P_0+h\rho (g+a)$

Due to upward acceleration, pseudo force will act downwards in the frame of the beaker. So, the effective value of acceleration due to gravity will increase by ${}^{\prime }{a}^{\prime }$.
$\Rightarrow g^{\prime }=(g+a)$
Now pressure at depth $h$ is,
$P=P_0+\rho g^{\prime }h$
$\therefore P=P_0+\rho (g+a)h$