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Questions Related to physics

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.

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 vessel with a small hole at the bottom is filled with water and kerosene. The height of the water column is 20 cm and that of the kerosene is 25 cm. the velocity with which water flows out the hole is

  1. 2 m/s

  2. 4 m/s

  3. 2.8 m/s

  4. 1 m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Torricelli's law: v = sqrt(2gh_eff). Pressure at the hole is due to water and kerosene. P = h_w * ρ_w * g + h_k * ρ_k * g. h_eff = h_w + h_k * (ρ_k/ρ_w). h_eff = 0.2 + 0.25 * (0.8) = 0.2 + 0.2 = 0.4m. v = sqrt(2 * 10 * 0.4) = sqrt(8) = 2.82 m/s.

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

If the atmospheric pressure is 76 cm of Hg at what depth of water in a lake the pressure will becomes 2 atmospheres nearly.

  1. $862 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pressure at that depth $= 2$ atmosphere $= 2 \times 76\, cm$ of $Hg$

Pressure is due to atmosphere $+$ pressure due to column of 
water $= 2 \times  76\, cm$ of $Hg$ 
$\implies 76\, cm$ of $Hg +$ depth $\times$ density of 
water $\times g = 2 \times  76 cm$ of $Hg$
Or 

$h \times d \times g = 76\, cm$ of $Hg$
Or 

$h = 76 \,cm \times  13600 \times \dfrac{g }{ 1000} \times g$  

Note: pressure due to $h$ meter of $Hg =h\times density\,of\,mercury \times g$$ 

Cancelling $g$ we have 

$h = 13.6 \times 76 = 1033.6 \,cm $ ( as cm is taken for atmosphere answer too comes in cm)

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

The pressure at the bottom of a lake, due to water is $4.9 \times 10 ^ { 6 } \mathrm { N } / \mathrm { m } ^ { 2 }$ . Whatis the depth of the lake? 

  1. 500$\mathrm { m }$

  2. 400$\mathrm { m }$

  3. 300$\mathrm { m }$

  4. 200$\mathrm { m }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P = hρg. 4.9 * 10^6 = h * 10^3 * 9.8. h = 4.9 * 10^6 / 9800 = 500m.