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

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.

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 the pressure will becomes 2 atmospheres nearly.

  1. $826 cm$

  2. $932 cm$

  3. $982 cm$

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

Let required depth be $h$


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 

$h\times d\times  g = 2 \times 76 cm$ of $Hg$
Or 

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

$h = \dfrac{76\, cm \times  13\times  g}{1000 \times g}$  ( 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 depth of the dam is 240 m. The pressure of water is (Take $g=10 m/{ s }^{ 2 }$ density of liquid = $1000 kg/{ m}^{ 3})$

  1. $24\times { 10 }^{ 5 }N/{ m }^{ 2 }$

  2. $12\times { 10 }^{ 4 }N/{ m }^{ 2 }$

  3. $10\times { 10 }^{ 3 }N/{ m }^{ 2 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pressure exerted by a liquid column is calculated using the formula P = h * rho * g. Substituting the given values: 240 m * 1000 kg/m^3 * 10 m/s^2 = 24 * 10^5 N/m^2.

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 on a swimmer $20$ m below the surface of water at sea level is

  1. $1.0$ atm

  2. $2.0$ atm

  3. $2.5$ atm

  4. $3.0$ atm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$P _0=1atm=1\times 10^5 Pa$
$h=20m$
$\rho=1000kg/m^3$
$g=10m/s^2$
The pressure on a swimmer $20m$ below the surface of water at sea level is
$P=P _0+\rho gh$
$P=1\times 10^5+1000\times 10\times 20$
$P=3\times 10^5$
$P=3atm$
The correct option is D.

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 $4.9 \times 10^{6} N/m^{2}$. What is the depth of the lake?

  1. 500m

  2. 400m

  3. 300m

  4. 200m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$P=4\times 10^6\,N/m^2$

$\rho=1000kg/m^2$

We have,

$P=\rho g h$

Then,

$h=\dfrac{P}{\rho g}$

$=\dfrac{4\times 1066}{1000\times 9.8}=\dfrac{1000}{2}=500\,m$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A ball o mass m and density p is immersed in a liquid of density 3 p ar a depth h and released. to what height will the ball jump up above the surface of liquid ?(neglect the resistance of water and air)

  1. h

  2. 2h

  3. 3h

  4. 4h
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Archimedes' principle and conservation of energy, the buoyant force is greater than the weight of the ball. The ball gains kinetic energy while submerged, and the height it jumps above the surface is determined by the work done by the buoyant force minus the potential energy lost, resulting in h.

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

Water is being poured into a vessel at a constant rate $ qm^2/s $. There is small aperture of cross-section area 'a' at the bottom of the vessel.The maximum level of water level of water in the vessel is proportional to

  1. q

  2. $ q^2 $

  3. $ \frac {1}{a} $

  4. $ \frac {1}{a^2} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At steady state, the rate of inflow q equals the rate of outflow, which is given by a * v = a * sqrt(2 * g * h). Thus, q = a * sqrt(2 * g * h). Solving for h gives h = q^2 / (2 * g * a^2), meaning h is proportional to 1/a^2.