Tag: introduction to atmospheric pressure

Questions Related to introduction to atmospheric pressure

Atmospheric pressure can be measured by which of the following instruments

  1. doctor's thermometer

  2. mercury barometer

  3. speedometer

  4. none of these


Correct Option: B
Explanation:

Speedometre is an instrument which is used to measure the speed of a vehicle.

Doctor's thermometre is used to measure the temperature of the body of the patient.
Mercury thermometre is used for measuring the atmospheric pressure.

The pressure energy per unit volume of a liquid is

  1. $ \frac {P}{\rho}$

  2. P

  3. P x $\rho$

  4. $ \frac {\rho}{P}$


Correct Option: B
Explanation:

Pressure energy per unit volume=$=\dfrac{Work done}{Volume}=\dfrac{W}{V}=\dfrac{F.d}{A.d}$

$=\dfrac{F}{A}$
$=P$

'The atmospheric pressure at a place is $76  cm$ of $Hg$? State its value in $bar$.

  1. $1.013\times{10}^{5}$ $bar$

  2. $1 bar$

  3. $1.013\times{10}^{-5}$ $bar$

  4. None


Correct Option: B
Explanation:

$1atm$ is the standard atmospheric pressure. 
In terms of length of mercury it is $76cm $ of Hg.

In SI units, the atmospheric pressure is $1atm=1.01325\times 10^{5}Nm^{-2}= 1 bar$

Unit for pressure used in weather maps is millibar. 1 millibar is equal to

    1. 1000 bar

    2. 100 kPa

    3. 100 Pa

    4. 1 atm


    Correct Option: C
    Explanation:

    Bar is a unit of pressure.

    $1bar=10^5 Pa$
    Hence dividing by 1000 expresses the milli form of the unit.
    $\implies 1millibar=\dfrac{10^5}{1000}Pa$
    $=100Pa$

    $1\ atm$ of pressure is equal to

    1. $1\ Torr$

    2. $\dfrac{1}{760}\ Torr$

    3. $760\ Torr$

    4. $1.013 \times 10^5\ Torr$


    Correct Option: C
    Explanation:

    The pressure due to $760$ mm of mercury is equal to atmospheric pressure. Pressure equivalent to $1$mm of mercury is $1torr$. Hence the atmospheric pressure $1atm=760torr$.

    $1\ Pa$ of pressure is equal to:

    1. $7.5 \times 10^{-3}\ Torr$

    2. $7.5 \times 10^{3}\ Torr$

    3. $760\ Torr$

    4. $1\ Torr$


    Correct Option: A
    Explanation:

    Atmospheric pressure $=1atm=760torr=1.01\times 10^3Pa$

    $\implies 1Pa=\dfrac{760}{1.01\times 10^5}torr=7.5\times 10^{-3}torr$

    A measurement referenced to the atmospheric pressure and it varies from the measurement of barometric reading and manometer reading is called as 

    1. gauge pressure

    2. absolute pressure

    3. atmospheric pressure

    4. none of the above


    Correct Option: A
    Explanation:

    Absolute pressure: It uses absolute zero (pressure in vacuum) as its zero point.

    Gauge pressure: It uses atmospheric pressure  as its zero point.
    Therefore, Absolute pressure = Atmospheric pressure + Gauge pressure
    A barometer reads the atmospheric pressure and a manometer reads absolute pressure (when closed from one end), therefore a measurement referenced to the atmospheric pressure varying from barometer reading and manometer is called as gauge pressure.

    A pressure equivalent to 1 mm of mercury is called 

    1. 1 torr

    2. 1 bar

    3. 1 pascal

    4. 1 atm


    Correct Option: A
    Explanation:

    The pressure at the bottom of $760 mm$ of mercury is equal to atmospheric pressure.

    Also, $1atm=760torr$
    Hence, $1torr$ pressure and $1mm$  of mercury are equivalent.

    1 pascal (Pa) is equal to 

    1. $1 \times 10^{5} $ bar

    2. $1 \times 10^{3} $ bar

    3. $1 \times 10^{-3} $ bar

    4. $1 \times 10^{-5} $ bar


    Correct Option: D
    Explanation:

    We have,

                   $1 atm=1\times10^{5}Pa$ ,
    and         $1atm=1bar$ ,
    hence,   $1bar=1\times10^{5}Pa$ 
    or            $1Pa=1\times10^{-5}bar$

    $1\ Torr$ of pressure is equal to:

    1. $133\ Pa$

    2. $13\ Pa$

    3. $1\ Pa$

    4. none of these


    Correct Option: A
    Explanation:

    Atmospheric pressure $=1atm=1.01\times 10^5Pa=760torr$

    $\implies 1torr=\dfrac{1.01\times 10^5}{760}Pa$$=133Pa$