Question

A vessel containing a liquid of density \( d \) moves down with an acceleration \( a \,(a < g) \). The pressure due to the liquid at a depth \( h \) below the free surface of the liquid is

• A. \( hgd \)
• B. \( h(g-a)d \)
• C. \( h(g+a)d \)
• D. \( h\left(\frac{g}{a}\right)d \)
• E. \( h\left(\frac{a}{g}\right)d \)

Hint:

In accelerating frames, replace \( g \) with effective gravity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem deals with fluid pressure within a non-inertial (accelerating) frame of reference. When the container accelerates, the "effective" acceleration due to gravity changes, which in turn affects the pressure.
Step 2: Key Formula or Approach:
The pressure in a fluid at rest in an inertial frame is given by \(P = h\rho g\), where `g` is the acceleration due to gravity. In an accelerating frame, `g` is replaced by the effective acceleration, \(g_{eff}\). Consider a small element of liquid of mass `m` inside the vessel. The forces acting on it are gravity (`mg` downwards) and the buoyant force (`F_b` upwards). The net force causes the acceleration `a`. The apparent weight of the liquid is reduced. From the frame of reference of the accelerating vessel, there is a pseudo-force `ma` acting upwards, opposing the acceleration. The net downward force on an object of mass m inside the liquid is \(F_{net} = mg - ma = m(g-a)\). So, the effective acceleration due to gravity is \(g_{eff} = g-a\). The pressure at depth `h` is then given by \(P = h \times d \times g_{eff}\).
Step 3: Detailed Explanation:
Let's consider a column of liquid of height `h` and cross-sectional area `A`. The mass of this liquid column is \(m = \text{volume} \times \text{density} = (hA)d\). The weight of this column is \(W = mg = hAdg\). The force exerted by the pressure at the bottom of the column is \(F_{bottom}\). The net force on this column provides its downward acceleration `a`. \(F_{net} = W - F_{upward\_pressure} = ma\). However, a simpler way is to consider the concept of effective gravity. When a frame of reference accelerates downwards with acceleration `a`, the apparent weight of any object of mass `m` inside it becomes \(m(g-a)\). So, the effective acceleration due to gravity is \(g_{eff} = g-a\). The pressure at a depth `h` is calculated using the standard formula but with \(g_{eff}\) instead of `g`. \[ P = \text{depth} \times \text{density} \times g_{eff} \] \[ P = h \times d \times (g-a) \] \[ P = h(g-a)d \] Step 4: Final Answer:
The pressure is h(g-a)d.