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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

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.

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

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

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

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

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

  4. $1.2$$ $\mathrm { cm }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the tube is vertical, the pressure in the trapped air changes according to Boyle's law (P1V1 = P2V2). The mercury column shifts to balance the pressure difference between the top and bottom air columns. Calculation based on the pressure change leads to a shift of approximately 3 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 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

  1. $

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

    $

  2. $

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

    $

  3. $

    \dfrac{H x}{100}

    $

  4. $

    \dfrac{100 H}{x}

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
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,

$P _1=P _2+\dfrac hH$

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

$h=\dfrac{xH}{100}$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

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

  1. 2 atmosphere

  2. 1 atmosphere

  3. 3 atmosphere

  4. 4 atmosphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The pressure at the bottom is the sum of atmospheric pressure and the hydrostatic pressure of the water column. Since 10 m of water equals 1 atmosphere, 20 m of water equals 2 atmospheres. Total pressure = 1 atm (atmospheric) + 2 atm (water) = 3 atmospheres.

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

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 $= 13.6\ \text{g/cc)}$:

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

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

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

  4. $22\ \text{cm}^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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