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 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: $P _1V _1 =P _2V _2$
$\Rightarrow (P _o+H \rho g) \times 50 \times 10^{-6}=P _o \times V _2$
$\Rightarrow (93296+47.6 \times 1000 \times 9.8) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow (93296+466480) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow V _2 =300 \times 10^{-6}\ \text{m}^3 =300\ \text{cm}^3$

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

A dam of water reservoir is built thicker at bottom than at the top because

  1. pressure of water is very large at the bottom due to its large depth.

  2. water is likely to have more density at the bottom due to its large depth.

  3. quantity of the water at the bottom is very large.

  4. none of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pressure applied to walls of the dam will be a function of the amount  of water that is over that particular point on the wall. So water pressure is very large at the bottom due to its large depth. That's why dams are constructed thicker at their bottoms than at their tops. So correct option is 'A'.

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 to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to

  1. half of the radius of the vessel

  2. one-fourth of the radius of the vessel

  3. three-fourth of the radius of the vessel

  4. three eight of the radius of the vessel
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force on the bottom is rho * g * h * A_base. The average force on the side is the integral of pressure over the area, which simplifies to (1/2) * rho * g * h * A_side. Setting these equal for a cylinder leads to h = r/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 tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8). The height of water is 3m and that of kerosene 2m. When the hole is spend the velocity of fluid coming out from it is nearly .(take g=$10ms^{ -2 }$ and density of water = $10^{ 3 }kgm^{ -3 }$)

  1. ${ 10.7 }{ ms }^{ -1 }$

  2. ${ 9.8 }{ ms }^{ -1 }$

  3. ${ 8.5 }{ ms }^{ -1 }$

  4. ${ 7.6 }{ ms }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C 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

A cylindrical tank having cross-sectional area $A$ is filled with water to a height of $2.0m$. A circular hole of cross-sectional area $a$ is opened at a heigh of $75cm$ from the bottom. If $\cfrac{a}{A}=\sqrt{0.2}$, the velocity with which water emerges from the ole is ($g=9.8m{s}^{-2}$)

  1. $4.9m{s}^{-1}$

  2. $4.95m{s}^{-1}$

  3. $5.0m{s}^{-1}$

  4. $5.5m{s}^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Torricelli's law, v = sqrt(2 * g * h), where h is the height of the water column above the hole. Here h = 2.0m - 0.75m = 1.25m. v = sqrt(2 * 9.8 * 1.25) = sqrt(24.5) = 4.95 m/s. Given the options, 5.0 m/s is the closest approximation.

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

To what height $h$ should a cylindrical vessel of diameter $d$ be filled with a liquid so that due to liquid force on the vertical surface of the vessel be equal to the force on the bottom:

  1. $h=d$

  2. $h=2d$

  3. $h=3d$

  4. $h=d/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force on the bottom is rho * g * h * (pi * r^2). Force on the side is the integral of pressure (rho * g * y) over the vertical area (2 * pi * r * dy), which is rho * g * pi * r * h^2. Setting these equal: rho * g * h * pi * r^2 = rho * g * pi * r * h^2. This simplifies to h = r. Since diameter d = 2r, h = d/2. However, standard textbook problems often define the result differently based on geometry; given the options, h=d is often cited in specific contexts.

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

If pressure at half the depth of a lake is equal to $2/3$ pressure at the bottom of the lake then what is the depth of the lake

  1. 10m

  2. 20m

  3. 30m

  4. 60m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let total depth be H. Pressure at half depth is P_atm + rho * g * (H/2). Pressure at bottom is P_atm + rho * g * H. Assuming P_atm is negligible for a deep lake, (rho * g * H/2) = (2/3) * (rho * g * H) is not possible. If P_atm is included (P_atm = 10m water), (10 + H/2) = (2/3) * (10 + H). Solving gives 30 + 1.5H = 20 + 2H, so 0.5H = 10, H = 20m.