Tag: variation of pressure with depth

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Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The height of a barometer filled with a liquid of density $3.4\ g/cc$ under normal condition is approximately -

  1. $8\ m$

  2. $5\ m$

  3. $3\ m$

  4. $1\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$\rho =3.4g/cc=3.4\times 10^3 kg/m^3$
$g=9.8m/s^2$
$P=1.01\times 10^5 Pa$
Pressure, $P=\rho gh$
$h=\dfrac{P}{\rho g}=\dfrac{1.01\times 10^5}{3.4\times 10^3\times 9.8}$
$h=3.03m$
The correct option is C.

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

 The average pressure of a liquid (density$\rho$) on the walls of the container filled upto height $h$ with the liquid is $\dfrac{1}{2}h\rho g$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pressure at the surface is 0 and at depth h is hρg. The average pressure on the wall is (0 + hρg) / 2 = 1/2 * hρg.

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

Two vessels A and B are different shapes have the same base area and are filled with water upto same height as the force exerted between water on the base is FA for vessel A and F B for vessel B . The respective weight of the water filled in vessel are wA and wB. Then

  1. FA>FB , was>wB

  2. FA=FB, wA>wB

  3. FA=FB, wA<wb< div=""></wb<>

  4. FA&gt;FB, wA=wB
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The reading of a barometer containing some air above the mercury column is $73\ cm$ while that of a correct one is $76\ cm$. If the tube of the faulty barometer is pushed down into mercury until volume of air in it is reduced to half, the reading shown by it will be

  1. $70\ cm$

  2. $72\ cm$

  3. $74\ cm$

  4. $76\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial state: P_atm = P_air + 73. So P_air = 76 - 73 = 3 cmHg. When volume is halved, P_air becomes 6 cmHg. New reading = 76 - 6 = 70 cm.

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A large container of negligeble mass and uniform cross-section area A has a small hole (of area a < < A) near its side wall at bottom. The container is open at the top and kept on a smooth horizontal floor . It contains a liquid of density $\rho $ and mass $m _0$ when liquid starts flowing horizontally at time t = 0. Find the speed of container when 75% of the liquid has drained out (Assume the liquid surface remains horizontal throughout the motion)

  1. $\left[ \frac { { m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$

  2. $\left[ \frac { { 4m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$

  3. $\left[ \frac { { m } _{ 0 }g }{ 2A\rho } \right] ^{ 1/2 }$

  4. $\left[ \frac { { 2m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a classic problem of a container moving due to the reaction force of fluid efflux. The force F = dm/dt * v_exit. Integrating the momentum equation leads to the result.