Question

A container is having a liquid of density $\rho$ up to height $h$ . If the vessel is given an upward acceleration equal to $g$ , the pressure at the bottom of the vessel will be: ( ${P_0}$ : atmospheric pressure)
(A) ${P_0} + 2\rho gh$
(B) ${P_0} + \rho gh$
(C) ${P_0}$
(D) Zero

Answer 1

An upward acceleration of a body increases the perceived weight of the body. The total pressure in a liquid is the sum of the atmospheric pressure and that the pressure due to the weight of the liquid.
Formula used: $P = {P_0} + \rho gh$ , where ${P_0}$ is atmospheric pressure, $\rho$ is density, $g$ is acceleration due to gravity and $h$ is height of the water.

Complete step by step answer:

Formula for pressure in a liquid of density $\rho$ open to the atmosphere in a planet of acceleration due to gravity $g$ is given as
$P = {P_0} + \rho gh$ where ${P_0}$ is atmospheric pressure and $\rho gh$ is the pressure due to the weight of the liquid above the point being considered.
Now the weight of a liquid in a container of cross sectional area $A$ and height $h$ is $W = mg = \rho Vg = \rho Ahg$
We know that mass is density time volume, and volume is cross sectional area times the height.
When a body is accelerating upward or downward, there’s an additional term due to the non-inertial nature of the body. Depending on whether it accelerates upward or downward there’s a pseudo weight gain or loss. In this cases, the equation of the weight becomes
$W' = m(g + a) = \rho Ah(g + a)$ for upward acceleration or
$W' = m(g - a) = \rho Ah(g - a)$ for downward acceleration, where $W'$ is pseudo-weight, to be separated from the proper weight, and $a$ is the acceleration of the body.
In our case, the vessel is accelerating upwards with an acceleration equal to $g$ i.e. $a = g$ .
Therefore, the corrected weight of the liquid is
${W_c} = \rho Ah(g + g) = 2\rho gAh$ (replacing $a$ with $g$ )
Recall that
$P = \dfrac{F}{A} = \dfrac{W}{A}$
$F$ is force and $W$ is weight which is our force in this case.
Therefore,
Pressure due to weight ${P_w}$ is
${P_w} = \dfrac{{{W_c}}}{A} = \dfrac{{2\rho gAh}}{A}$
$A$ cancels out, which gives
${P_w} = 2\rho gh$
Now, the total pressure is the atmospheric pressure plus the pressure due to the corrected weight
$P = {P_0} + {P_w} = {P_0} + 2\rho gh$
$\therefore P = {P_0} + 2\rho gh$
Hence, the correct option is A.

Note:
This question is similar to the very common problem where the reading of a scale in a lift is supposed to be calculated for a person, when the lift is accelerating upwards or/and downwards. You only go one step further by calculating pressure. In both these cases, the recurrent problem encountered is whether to use $W' = m(g + a)$ or $W' = m(g - a)$ . You can remember easily by tying it to the concept of weightlessness. When a body is weightless (weight equals to zero), it means the body is falling at acceleration due to gravity i.e. $a = g$ (but $a$ is downward). If you insert $a = g$ into $W' = m(g - a)$ you get zero which is weightlessness. Thus, downward has the negative sign, then upward must have the positive sign.