Tag: atmospheric pressure and its consequences

Questions Related to atmospheric pressure and its consequences


when an air particle speeds up, it actually pushes            . Complete the sentence

  1. Less

  2. More

  3. Inside

  4. Outside


Correct Option: A
Explanation:

According to Bernoulli's theorem        $ P + \rho gh+ \dfrac{1}{2}\rho v^2= constant$
$\implies$  When $v$ is more, then pressure $(P)$ is less.
Thus when an air particle speeds up, it actually pushes less.

wherever the air pressure is higher, there will be a stronger force or push against an object. Is the statement True or False

  1. True

  2. False


Correct Option: B
Explanation:

Pressure      $P = \dfrac{F}{A}$                $\implies P \propto F$

Thus greater the pressure, stronger will be the force.

The ordinary pressure of air surrounding us is

  1. 15 pounds

  2. 16 pounds

  3. 14.2 pounds

  4. 14.7 pounds


Correct Option: D
Explanation:

$Answer:-$ D

The ordinary pressure of the air surrounding us is 14.7 pounds per square inch, but the pressure can change when the wind blows or an object, like a car or airplane, accelerates.

One important principle to remember is that wherever the air pressure is higher, there will be a stronger force or push against an object. It’s also helpful to know that when an air particle speeds up, it actually “pushes” less. Imagine that fast-moving air particles are in so much of a hurry that they don’t have time to apply force. This principle is used in airplane wings to make planes fly. When a plane moves along the runway, the air above the wing speeds up more, lowering the pressure, so that the air below the wing can push the plane upward.

Specify heat of mercury is

  1. $0.03 cal/gm/^oC$

  2. $0.3 cal/gm/^oC$

  3. $0.0003 cal/gm/^oC$

  4. $0.003 cal/gm/^oC$


Correct Option: A
Explanation:

Specific heat of mercury is $0.03 \ cal/ gm/^{o} C$.

Hence, option A is correct.

A boy blows a balloon of volume $1$ litre to pressure of $10$ atmospheres in $10$ sec, the power is

  1. $10.13w$

  2. $1.013w$

  3. $20.26w$

  4. $0.506w$


Correct Option: B

Ice of water at $ 100^o $ is given 540 cal of beat and the steam formed occupation 1671 cc at the atmosphere pressure. Then workdone against atmosphere pressure in this procces is nearly

  1. 540 cal

  2. 500 cal

  3. 40 cal

  4. 100 cal


Correct Option: A

In the above problem what is the pressure necessary to accomplish this task $(g=9.8\ m/s^{2})$

  1. $1.87\times 10^{5}\ N/m^{2}$

  2. $18.7\times 10^{5}\ N/m^{2}$

  3. $2.87\times 10^{5}\ N/m^{2}$

  4. $28.7\times 10^{5}\ N/m^{2}$


Correct Option: C

What is the mass of $2$ litres of nitrogen at $22.4$ atmospheric pressure and $273\ K$ 

  1. $28\ g$

  2. $14 \times 22.4 g$

  3. $56\ g$

  4. $none\ of\ these$


Correct Option: A
Explanation:

Consider STP conditions:
 
1) 1 mol = 22.4 L of gas; 1 mol of Nitrogen gas = 14 g
 
2) Therefore, 22.4 L Nitrogen gas = 14 g → 2 L Nitrogen = 1.25 g
 

3) This is at a standard pressure of 1 atm. So, 22.4 atm pressure corresponds to 22.4 × 1.25 = 28 g  

A sphere floats in water just submerged as shown in the diagram. If atmospheric pressure is neglected

  1. Force exerted by liquid on lower half of sphere is $\dfrac{4\rho\pi\ R^{2}}{3}g$

  2. Force exerted by liquid on lower half of sphere and upper half of sphere is same in magnitude

  3. Ratio of magnitude of forces exerted by liquid on lower and upper half of sphere is $5:1$

  4. Ratio of magnitude of forces exerted by liquid on lower and upper half of sphere is $4:1$


Correct Option: C

A barometer tube, containing mercury, is lowered in a vessel containing mercury until only $50 cm$ of the tube is above the level of mercury in the vessel. If the atmospheric pressure is $75 cm$ of mercury, what is the pressure at the top of the tube

  1. $33.3 kPa$

  2. $66.7kPa$

  3. $3.33 MPa$

  4. $6.67 MPa$


Correct Option: A
Explanation:

Since the mercury rises only $50cm$ it implies the rest of the excess atmospheric pressure is being countered by the pressure at the top of the tube.
So pressure at top=atmospheric pressure - $50cm$ of Hg=$(75-50)cm=25cm$ of Hg=$1/3\times 100kPa= 33.33kPa$