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Multiple choice physics pressure in fluids and atmospheric pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

$5 g$ of water rises in the bore of capillary tube when it is dipped in water. If the radius of bore capillary tube is doubled, the mass of water that rises in the capillary tube above the outside water level is

  1. $1.5 g$

  2. $10 g$

  3. $5 g$

  4. $15 g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The height of water in a capillary tube is inversely proportional to the radius (h = 2T / rρg). If the radius is doubled, the height is halved. Since mass m = πr^2 h ρ, and h is proportional to 1/r, then m is proportional to r^2 * (1/r) = r. Doubling the radius doubles the mass.

Multiple choice physics pressure in fluids and atmospheric pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube? 

  1. $1cm$

  2. $2 cm$

  3. $3 cm$

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

Since we know that height of capillary rise in inversely proportional to radius of capillary.

i.e.    height $\alpha $ $\dfrac { 1 }{ radius } $
         $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $
         $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

Multiple choice physics pressure in fluids and atmospheric pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $2.2 cm$ and $6.6 cm$. The ratio of radii of the tubes will be

  1. $1:9$

  2. $1:3$

  3. $9:1$

  4. $3:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,

height $\propto \dfrac { 1 }{ radius } $

$\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \Rightarrow \dfrac { 2.2cm }{ 6.6cm } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $

So, $\boxed { \dfrac { { r } _{ 1 } }{ { r } _{ 2 } } =3 } $

Multiple choice physics pressure in fluids and atmospheric pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube?

  1. $1 cm$

  2. $0.1 cm$

  3. $2 cm$

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

It is to be remembered that

height of a capillary rise $\propto \dfrac { 1 }{ radius\quad of\quad capillary } $
hence,
          $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{  2} }{ { r } _{ 1 } } $
          $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

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 value of $g$ at a place is decreased by $2\%$. The barometric height of the mercury 

  1. Increases by $2\%$

  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that,

$P=h\rho g$

Then,

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

So, $h\propto \dfrac 1g$

If the value of $g$ decreased by $2\%$ then $h$ will increase by $2\%$.
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The residual pressure of a vessel at ${27^0}C$ is  $1 \times {10^{ - 11}}N/{m^2}$. The number of molecules in this vessel is nearly:

  1. $2400$

  2. $2.4 \times {10^9}$

  3. ${10^{ - 11}} \times 6 \times {10^{23}}$

  4. $2.68 \times {10^{19}} \times {10^{11}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law PV = NkT, where P = 10^-11, T = 300K, and k = 1.38 * 10^-23. For a unit volume (1 m^3), N = P / kT = 10^-11 / (1.38 * 10^-23 * 300) approx 2.4 * 10^9.

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 the half depth of a lake is equal to $\dfrac{3}{4}$ times the pressure at its bottom, then find the depth of the lake . [Take g=$10 m/s^2]$

  1. <span>$ \dfrac{P _{0}}{\rho g}\ $</span>

  2. <span>$ \dfrac{2P _{0}}{\rho g}\ $</span>

  3. <span>$ \dfrac{P _{0}}{2\rho g}\ $</span>

  4. <span>$ \dfrac{3P _{0}}{\rho g}\ $</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let depth of the lake be $h$ and pressure at bottom $= P$
Then $P=P _{0}+\rho gh\rightarrow (1)$    $(P _{0}=$ atmospheric pressure, $\rho $ = density of water)
At half depth $(h/2)$ pressure is $\dfrac{3P}{4}$ then :
$\dfrac{3P}{4}=P _{0}+\rho g\dfrac{h}{2}\rightarrow (2)$
On subtracting equation 2 from 1 we get :
$\dfrac{P}{4}=\rho g\dfrac{h}{2}$
$\Rightarrow P=2\rho gh$, substituting this value of $P$ in equation 1:
$2\rho gh=P _{0}+\rho gh$
$\Rightarrow h=\dfrac{P _{0}}{\rho g}\rightarrow $ Depth of the lake
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 air density were uniform, then the height of the atmosphere above the sea level to produce a normal atmospheric pressure of 1.0 x 10$^{5}$ Pa is(density of air is 1.3 kg/m$^{3}$ , g $=$ 10m/s$^{2}$):

  1. 0.77 km

  2. 7.7 km

  3. 77 km

  4. 0.077 km
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\rho g h = 1 \times 10^5$
$\Rightarrow h = \dfrac {10^5}{\rho g} = \dfrac {10^5}{1.3 \times 10} = 7.7 km$

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 exerted by a liquid at depth $h$ is given by:

  1. $\displaystyle \dfrac{h}{dg}$

  2. $hdg$

  3. $\displaystyle \dfrac{h}{d}$

  4. $hg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.

The pressure exerted by a liquid at a point depends on its vertical depth and density of the liquid only. It is independent of the shape of the container. The pressure at the bottom of the three vessels of different shapes containing the same liquid acts equally in all directions.
Thus, the pressure exerted by the liquid of height $h$ is given as P=hdg, where $h$ is the height, $d$ is the density and $g$ is the acceleration due to gravity.