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

Hint:
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.