Tag: physics

Questions Related to physics

We feel most comfortable at relative humidity at:

  1. < 50%

  2. 50%

  3. > 50%

  4. 100%


Correct Option: B
Explanation:

Relative humidity is the ratio of the current absolute humidity to the highest possible absolute humidity which depends on the air temperature. A reading of 100 percent relative humidity means that the air is totally saturated with water vapour and cannot hold any more, creating the possibility of rain. We feel most comfortable at relative humidity 50%.

On a day during the Monsoon season, the relative humidity at a place is 90% and the temperature is $15^\circ$ (C) The saturation vapour pressure of water at this temperature is $0.0169\, \times\, 10^5$ Pa. The partial pressure of water vapour in the air on that day is

  1. $0.9\, \times\, 10^5$ Pa

  2. $0.0169\, \times\, 10^5$ Pa

  3. $0.0152\, \times\, 10^5$ Pa

  4. $1.0\, \times\, 10$^5$ Pa


Correct Option: C
Explanation:
Relative humidity=$\dfrac { \text{Partial pressure of }{ H } _{ 2 }0 }{ \text{Vapour pressure of } { H } _{ 2 }0 }\times100\%$
Let the partial pressure of $H _20$ be $x\rho a$
$90=\dfrac { x }{ 0.169\times { 10 }^{ 5 } } \times 100\\ $
$\dfrac { 90 }{ 100 } \times \dfrac { 169 }{ { 10 }^{ 4 } } \times { 10 }^{ 5 }=x$
$x=169\times9$
$x=1521\rho a$
$=0.0152\times10^5\rho a$

When air is saturated, it cannot hold ____________.

  1. more water vapour

  2. more air

  3. more carbon dioxide

  4. more oxygen


Correct Option: A
Explanation:

When a volume of air at a given temperature holds the maximum amount of water vapour, the air is said to be saturated.

At dew point. RH is _______.

  1. 10%

  2. 30%

  3. 50%

  4. 100%


Correct Option: D
Explanation:
The dew point is the temperature at which air is saturated with water vapor, which is the gaseous state of water. The relative humidity is 100 percent when the dew point and the temperature are the same.

The relative humidity on a day when partial pressure of water vapour is $0.012 \times 10^5 Pa$ at $12^oC$ is (Take vapour pressure of water at this temperature as $0.016 \times 10^5$ Pa)

  1. 70%

  2. 40%

  3. 75%

  4. 25%


Correct Option: C
Explanation:

Given,

Partial pressure of water vapour $={ 0.125\times 10 }^{ 5 }$ Pa
Vapour pressure of water$={ 0.016\times 10 }^{ 5 }$ Pa
Relative humidity at a given temperature (R) 
$=\dfrac { Partial\quad pressure\quad of\quad water\quad vapour }{ vapour\quad pressure\quad of\quad water } \ =\dfrac { { 0.125\times 10 }^{ 5 } }{ { 0.016\times 10 }^{ 5 } } \ =0.75\ =75%$

Formation of dew is an example for ......

  1. Boiling

  2. Melting

  3. Condensation

  4. Evaporation


Correct Option: C
Explanation:

It is a example of condensation.Condensation is the process in which gas changes into a liquid when it touches a cooler surface.


Dew drops are formed due to condensation of water vapors. Air around us contains water vapors which we call moisture or humidity. Hot air contains more moisture as compared to cold air. During the night when the hot air comes into contact with some cold surface, water vapor present in it condenses on the cold surface in the form of droplets. These tiny drops of water are called dew drops

When it is raining, the dew point is:

  1. 0$^o$C

  2. 60$^o$C

  3. 100$^o$C

  4. room temperature


Correct Option: D
Explanation:
The dew point is the temperature at which the air can exactly hold the amount of moisture present. When raining, at any given temperature, the atmosphere can hold so much water vapour. So the dew point is equal to the room temperature.

Communication satellites move in orbits of radius 44,400 km around the earth find the acceleration of such a satelite, assuming that the only force acting on, it is that  due to the earth.Mass of the earth $ = 6 \times {10^{24}}Kg.$

  1. $0.2\;m{s^{ - 2}}$

  2. $2.1\;m{s^{ - 2}}$

  3. $4.2\;m{s^{ - 2}}$

  4. $0.3\;m{s^{ - 2}}$


Correct Option: A

A particle is projected vertically upward and is at a height h after $\displaystyle { t } _{ 1 }$ seconds and again after $\displaystyle { t } _{ 2 }$ seconds, then:

  1. $\displaystyle h={ gt } _{ 1 }{ t } _{ 2 }$

  2. $\displaystyle h=\frac { 1 }{ 2 } { gt } _{ 1 }{ t } _{ 2 }$

  3. $\displaystyle h=\frac { 2 }{ g } { t } _{ 1 }{ t } _{ 2 }$

  4. $\displaystyle h=\sqrt { { gt } _{ 1 }{ t } _{ 2 } } $


Correct Option: B
Explanation:

Let u be the velocity of the projection and O be the point of projection. Let P be the point in the path of the particle such that OP=h. Then,
$\displaystyle h=ut-\frac { 1 }{ 2 } { gt }^{ 2 }\Rightarrow { gt }^{ 2 }-2ut+2h=0$...(i)
Clearly, $\displaystyle { t } _{ 1 }{ t } _{ 2 }$, are two roots of this equation.
$\displaystyle \therefore \quad { t } _{ 1 }{ t } _{ 2 }=\frac { 2h }{ g } $
$\displaystyle \Rightarrow \quad h=\frac { 1 }{ 2 } g{ t } _{ 1 }{ t } _{ 2 }$

In case of an orbiting satellite, if the radius of orbit is decreased.

  1. Its KE decreases

  2. Its PE decreases

  3. Its ME decreases

  4. Its speed decreases


Correct Option: B,C
Explanation:

P.E of orbiting satellite at distance $r$ from centre of earth

$=\cfrac{-GMm}{r}\\K.E=\cfrac{GMm}{2R}$ and $M.E=\cfrac{-GMm}{2R}$

If $R$ decreased K.E increases.

P.E and M.E become more negative, thus decrease.