Question

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

Answer 1

Answer:

Given:

A beaker contains liquid of density ρ.

The beaker is accelerated up with

a m/s².

To find:

Pressure due to liquid column at a depth "h" from the free surface of the liquid.

Concept:

In order to calculate the pressure exerted , we will consider a cylindrical column of water of height h.

Since the beaker is accelerated upwards, the column experiences pseudo-force downwards , hence total acceleration :

$\sf{a_{total} = (g + a)}$

Calculation:

Weight of the column be w , height be h and Cross-section area be s.

$\sf{w = m \times a_{total}}$

$\sf{ \implies \: w = \{m \times (g + a) \}}$

$\sf{ \implies \: w = \{(s \times h \times \rho) \times (g + a) \}}$

Now pressure exerted :

$\sf{ \implies \: p = \dfrac{w}{area} }$

$\sf{ \implies \: p = \dfrac{ \{(s \times h \times \rho) \times (g + a)}{s} }$

$\sf{ \implies \: p = \{\rho \times (g + a) \times h \}}$

So final answer :

$\boxed{ \large{ \green{ \bold{\sf{ \: p = \{\rho \times (g + a) \times h \}}}}}}$

Answer 2

$\underline{ \boxed{ \bold{ \rm{ \huge{ \purple{ \mathfrak{Answer}}}}}}} \\ \\ \star \rm \: \red{Given} \\ \\ \implies \rm \: density \: of \: liquid = \rho \\ \\ \implies \rm \: beaker \: is \: accelerated \: in \: upward \\ \rm \: direction \: with \: acc. \: a \\ \\ \implies \rm \: height \: of \: liquid \: in \: beaker = H \\ \\ \star \rm \: \red{To \: Find} \\ \\ \implies \rm \: pressure \: due \: to \: height \: at \: the \\ \rm \: bottom \: below \: the \: free \: surface \: of \\ \rm \: of \: the \: liquid \\ \\ \star \rm \: \red{Concept} \\ \\ \implies \rm \: here \: beaker \: is \: accelerated \: in \\ \rm \: upward \: direction \: with \: acc. \: a \: so \\ \rm \: pseudo - force \: acts \: in \: downward \\ \rm \: direction...$$\\ \\ \therefore \rm \: net \: force \: in \: downward \: direction \\ \rm \: is \: given \: by... \\ \\ \rm \: \dagger \: F{ \tiny{net}} = weight \: force + pseudo \: force \\ \\ \dagger \rm \: \underline{ \pink{ \bold{ \rm{F{ \tiny{net}} = m(g + a)}}}} \: \dagger \\ \\ \star \rm \: \red{Formula} \\ \\ \implies \rm \: formula \: of \: pressure \: in \: terms \: of \\ \rm \: area \: and \:perpendicular \: force \: is \: given \\ \rm \: by... \\ \\ \dagger \: \underline{ \boxed{ \bold{ \rm{ \purple{P = \frac{F{ \tiny{perpendicular}}}{A} }}}}} \: \dagger \\ \\ \star \rm \: \red{Calculation} \\ \\ \implies \rm \: P = \frac{m(g + a)}{A} \\ \\ \rm \: \dagger \: here \: perpendicular \: force \: is \: net \: force \\ \rm \: and \: A \: indicates \: area \: of \: free \: surface \\ \rm \: of \: liquid \: in \: beaker... \\ \\ \dagger \rm \: we \: know \: that \: \boxed{ \rm{\blue{m = \rho \times V}}} \\ \\ \therefore \rm \: P = \frac{ \rho \times V \times (g + a)}{A} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{P = \rho \times (g + a) \times H}}}}} \: \red{ \star} \\ \\ \star \rm \: \red{Additional \: Inf.} \\ \\ \implies \rm \: pressure \: at \: base \: point \: due \: to \\ \rm \: height \: does\: not \: depend \: on \: area \: of \\ \rm \: free \: surface... \\ \\ \implies \rm \: pressure \: due \: to \: height \: in \: non - \\ \rm \: acceleratory \: motion \: is \: given \: by... \\ \\ \dagger \: \underline{ \boxed{ \bold{ \rm{ \green{P = \rho \times g \times H}}}}} \: \blue{ \star}$