A beaker containing a liquid of density `rho` moves up with an acceleration `a`. The pressure due to the liquid at a depth h below the free surface of the liquid is.

Options

A. `h rho g`
B. `h rho(g + a)`
C. `h rho(g - a)`
D. `2 h rho g((g - a)/(g + a))`

Answer

B

Explanation

Consider an element of the liquid of height `dy` and area of cross-section `a'` at a depth `y` below the surface of the liquid. Let `p` and `p + dp` be the liquid pressures at the upper and lower surfaces of the element.
Mass of element `dm = p a' dy`
=`[(p + dp)a' - pa'] -dm g`
`a' dp - dm g = dm a`
`a' dp = dm (g + a)`
=`rho a'(g + a) dy`
`int_0^p dp = rho (g + a) int_0^h dy`
`p = rho(g + a) h`.

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