Tag: fluid pressure

Questions Related to fluid pressure

By sucking through a straw, a student can reduce the pressure in his lungs to $750 mm$ of Hg (density = $\displaystyle 13.6{ gm }/{ { cm }^{ 3 } }$). Using the straw, he can drink water from a glass upto a maximum depth of: 

  1. $10 cm$

  2. $75 cm$

  3. $13.6 cm$

  4. $1.36 cm$


Correct Option: C
Explanation:

Pressure difference created $= 10 mm$ of Hg 
This must be equal to the pressure of water column being created in the straw. If height of water column be $h$

$h \rho g = \dfrac{10}{10} \times 13.6 g$
$hg \times 1 = 13.6 g$
$h = 13.6 cm$

$1$ millibar is equal to a pressure of

  1. ${10}^{5}pa$

  2. $100pa$

  3. $1000pa$

  4. ${10}^{-3}pa$


Correct Option: B
Explanation:

We know that $1$ bar $= 10^5 Pa$

So, $1$ millibar $=10^{-3}$ bar $=10^{-3}\times 10^5 Pa=100 Pa$

One gram mole of oxygen is enclosed in a vessel at a temperature of $27$ and at one atmospheric pressure. The vessel is thermally insulated and is moved with a constant speed $u _0$. Calculate $u _0$ if the rise in temperature is $1K$ when the vessel is suddenly stopped?

  1. $35.6 m/s$

  2. $45 m/s$

  3. $90 m/s$

  4. $60 m/s$


Correct Option: B
Explanation:

since internal energy depends only on the temperature ,

as the temperature changes,
the change in internal energy is
$\Delta U= \dfrac{nfR \Delta T}{2}$

vessel contains a gas of mass M,
change in kinetic energy $= \dfrac{nMv^2}{2}$

$\dfrac{nfR \Delta T}{2}$= $\dfrac{nMv^2}{2}$

$\dfrac{2}{\dfrac{c _p}{c _v} - 1}$

$f = \dfrac{2}{\dfrac{2}{5}}$ = 5

putting value

$v = 45 m/s$

What will be the pressure in pascal due to a mercury column of $76$cm. Density of mercury is $13.6$ g$cm^2$, $g=980$ cm$s^2$?

  1. $1.0129\times 10^5$Pa

  2. $1.0129\times 10^4$Pa

  3. $1.0129\times 10^3$Pa

  4. $1.0129\times 10^2$Pa


Correct Option: A
Explanation:

 h$=76cm$

Pressure=?

Formula to be used : $P=hgd$

Where h = height of mercury column and

 d= density of mercury$=13.6 gm/cc$

acceleration due to gravity$=980 cm/s²$

$P=hgd=13.6\times 980\times 76=1012928dyne.cm²$

The height of the mercury column in a simple barometer is $h$. As the tube is inclined with the vertical at an angle $\alpha$, the length of the mercury column along the length of the tube will become:

  1. $h\cos{\alpha}$

  2. $\dfrac{h}{\cos{\alpha}}$

  3. $h\sin{\alpha}$

  4. $\dfrac{h}{\sin{\alpha}}$


Correct Option: B