Tag: physics

Questions Related to physics

$5 g$ 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

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

  1. $1.5 g$

  2. $10 g$

  3. $5 g$

  4. $15 g$

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Correct Option: B

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? 

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

  1. $1cm$

  2. $2 cm$

  3. $3 cm$

  4. $4 cm$

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Correct Option: A
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 } $

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

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

  1. $1:9$

  2. $1:3$

  3. $9:1$

  4. $3:1$

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Correct Option: D
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 } $

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?

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

  1. $1 cm$

  2. $0.1 cm$

  3. $2 cm$

  4. $4 cm$

Show answer
Correct Option: A
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 } $

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

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

  1. Increases by $2\%$

  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases

Show answer
Correct Option: A
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 $2\%$.

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:

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

  1. $2400$

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

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

  4. $2.68 \times {10^{19}} \times {10^{11}}$

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Correct Option: B

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=$10 m/s^2]$

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

  1. $ \dfrac{P _{0}}{\rho g}\ $

  2. $ \dfrac{2P _{0}}{\rho g}\ $

  3. $ \dfrac{P _{0}}{2\rho g}\ $

  4. $ \dfrac{3P _{0}}{\rho g}\ $

Show answer
Correct Option: A
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

A tank $4m$ high is half filled with water then filled to the top with a liquid of density $0.60 g/cc$ what is the pressure at the bottom of the tank due to these liquids? (take $g=10ms^{-2}$)

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

  1. $1.6 \times 10^3Nm^{-2}$

  2. $3.2\times 10^{-3}$

  3. $1.6 \times 10^4Nm^{-2}$

  4. $3.2 \times 10^4Nm^{-2}$

Show answer
Correct Option: C

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$^{2}$):

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

  1. 0.77 km

  2. 7.7 km

  3. 77 km

  4. 0.077 km

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Correct Option: B
Explanation:

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

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

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

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

  2. $hdg$

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

  4. $hg$

Show answer
Correct Option: B
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.