Tag: floating bodies

Questions Related to floating bodies

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

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

  1. $826 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$

Show answer
Correct Option: D
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 $cm$).

The depth of the dam is 240 m. The pressure of water is (Take $g=10 m/{ s }^{ 2 }$ density of liquid = $1000 kg/{ m}^{ 3})$

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

  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

Show answer
Correct Option: A

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

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

  1. $1.0$ atm

  2. $2.0$ atm

  3. $2.5$ atm

  4. $3.0$ atm

Show answer
Correct Option: D
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.

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?

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

  1. 500m

  2. 400m

  3. 300m

  4. 200m

Show answer
Correct Option: A
Explanation:
Given,

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

$\rho=1000kg/m^2$

We have,

$P=\rho g h$

Then,

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

$=\dfrac{4\times 1066}{1000\times 9.8}=\dfrac{1000}{2}=500\,m$

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)

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

  1. h

  2. 2h

  3. 3h

  4. 4h

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Correct Option: A

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

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

  1. q

  2. $ q^2 $

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

  4. $ \frac {1}{a^2} $

Show answer
Correct Option: D

A column of mercure of lenath $h = 10 \mathrm { cm }$ is contained in the middle of a narrow horizontal tube of length $1 \mathrm { m } ,$ closed at both ends. The air in both halves of the tube is under a pressure of $P _ { 0 } = 76 \mathrm { cm }$ of mercury. The tube is now slowly made vertical. The distance moved by mercury will be approximately

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

  1. $4.5$ $\mathrm { cm }$

  2. $3.0$ $\mathrm { cm }$

  3. $2.5$ $\mathrm { cm }$

  4. $1.2$$ $\mathrm { cm }$

Show answer
Correct Option: B

The volume of an air bubble increases by $ \mathrm{x} \%  $ as it rises from the bottom of a lake to its surface. If the height of the water barometer is H, the depth of the lake is

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

  1. $

    \left(\dfrac{H+x}{100}\right)^{2}

    $

  2. $

    \dfrac{H x}{(100+x)}

    $

  3. $

    \dfrac{H x}{100}

    $

  4. $

    \dfrac{100 H}{x}

    $

Show answer
Correct Option: C
Explanation:
We have,

$P _1V _1=P _2V _2$

$V _2=V _1+\dfrac{x}{100}V _1$

$P _1V _1=P _2(V _2+\dfrac{x}{100}V _1)$

$P _1=P _2(1+\dfrac{x}{100})$

But,

$P _2=1\,atm$

Then,

$P _1=P _2+\dfrac hH$

$1+\dfrac hH=1+\dfrac{x}{100}$

$h=\dfrac{xH}{100}$

A water tank is 20$\mathrm { m }$ deep. If the waterbarometer reads $10 \mathrm { m } ,$ the pressure at thebottom of the tank is

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

  1. 2 atmosphere

  2. 1 atmosphere

  3. 3 atmosphere

  4. 4 atmosphere

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Correct Option: C

A cylindrical can open at the bottom end lying at the bottom of a lake $47.6\ \text{m}$ deep has $50\ \text{cm}^3$ of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become $($atmospheric pressure $= 70\ \text{cm}$ of Hg and density of Hg $= 13.6\ \text{g/cc)}$:

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

  1. $350\ \text{cm}^3$

  2. $300\ \text{cm}^3$

  3. $250\ \text{cm}^3$

  4. $22\ \text{cm}^3$

Show answer
Correct Option: B
Explanation:

$P _o= 70\ \text{cm}$ of Hg $=70 \times 10^{-2} \times 13600 \times 9.8 =93296\ \text{Pa}$
Using Boyle's law: $P _1V _1 =P _2V _2$
$\Rightarrow (P _o+H \rho g) \times 50 \times 10^{-6}=P _o \times V _2$
$\Rightarrow (93296+47.6 \times 1000 \times 9.8) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow (93296+466480) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow V _2 =300 \times 10^{-6}\ \text{m}^3 =300\ \text{cm}^3$