Tag: physics

Questions Related to physics

Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level. Given $\sigma = 10 ^{-26} cm^2$

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $10^4 cm$

  2. $10^{-4} cm$

  3. $10^6 cm$

  4. $10^{-6} cm$

Show answer
Correct Option: C
Explanation:

The mean free path is given by, $\lambda=\dfrac{1}{\pi \sigma n}$

Here the proton is scattered by the molecules of the atmosphere. 
The density of the molecules of the atmosphere is $n=\dfrac{N}{V}=\dfrac{P}{kT}=\dfrac{10^6}{(1.38\times 10^{-16})(300)}=2.4\times 10^{19}$  (all units are taken in CGS unit)
So, $\lambda=\dfrac{1}{\pi(10^{-26})(2.4\times 10^{19})}=1.3\times 10^6\sim 10^6 $  $cm$

Estimate the average number of collisions per second that each $N _2$ molecule undergoes in air at room temperature and at atmospheric pressure. The diameter of a $N _2$ molecule is $0.3\ mm$.

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $0.2\ \mu m$

  2. $0.1\ \mu m$

  3. $0.1\ mm$

  4. None of these

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Correct Option: B

A container is divided into two equal parts I and II by a partition with a small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperatures $T _I=150$K and $T _{II}=300$K by connecting to heat reservoirs. Let $\lambda _I$ and $\lambda _{II}$ be the mean free paths of the gas particles in the two parts such that $d > > \lambda _I$ and $d > > \lambda _{II}$. Then $\lambda _I/\lambda _{II}$ is close to.

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $0.25$

  2. $0.5$

  3. $0.7$

  4. $1.0$

Show answer
Correct Option: C
Explanation:

Given,

Partition has hole of diameter $d$, Mean pressure between both sections is equal.

Boltzmann constant ${{K} _{B}}$

At constant pressure, Mean Free path $\lambda \ \alpha \ \sqrt{{{K} _{B}}T}$

Mean free path in ${{1}^{st}}$ section ${{\lambda } _{I}}=\sqrt{{{K} _{B}}\times 150}$

Mean free path in ${{2}^{nd}}$ section ${{\lambda } _{II}}=\sqrt{{{K} _{B}}\times 300}$

$\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}=\dfrac{\sqrt{{{K} _{B}}\times 150}}{\sqrt{{{K} _{B}}\times 300}}=0.707$

Hence, $\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}\cong 0.7$ 

Calculate the means free path of nitrogen molecule at $27^o$C when pressure is $1.0$ atm. Given, diameter of nitrogen molecule $=1.5\overset{o}{A}$, $k _B=1.38\times 10^{-23}$J $K^{-1}$. If the average speed of nitrogen molecule is $675$ $ms^{-1}$. The time taken by the molecule between two successive collisions is?

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $0.6$ns

  2. $0.4$ns

  3. $0.8$ns

  4. $0.3$ns

Show answer
Correct Option: A
Explanation:
Here, $T=27^oC=27+273=300$K.
$P=1$atm $=1.01\times 10^5$N $m^{-2}$, d$=1.5\overset{o}{A}=1.5\times 10^{-10}$m,
$k _B=1.38\times 10^{-23}$J $K^{-1}, \lambda =?$
From $\lambda =\dfrac{k _BT}{\sqrt{2}\pi d^2p}=\dfrac{1.38\times 10^{-23}\times 300}{1.414\times 3.14(1.5\times 10^{-10})^2\times 1.01\times 10^5}$
$=4.1\times 10^{-7}$m
Time interval between two successive collisions
$t=\dfrac{distance}{speed}=\dfrac{\lambda}{v} = \dfrac{4.1\times 10^{-7}}{675}=0.6\times 10^{-9}s$

Ten small planes are flying at a speed of $150$km $h^{-1}$ in total darkness in an air space that is $20\times 20\times 1.5km^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius $10$m.

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $125$h

  2. $220$h

  3. $432$h

  4. $225$h

Show answer
Correct Option: D
Explanation:
Here, $v=150$km $h^{-1}$
$N=10$
$V=20\times 20\times 1.5$ $km^3$
Diameter of plane, $d=2R=2\times 10$
$=20m=20\times 10^{-3}$km
$n=\dfrac{N}{V}=\dfrac{10}{20\times 20\times 1.5}=0.0167km^{-3}$
Mean free path of a plane
$\lambda =\dfrac{1}{\sqrt{2}\pi d^2n}$
Time elapse before collision of two planes randomly,
$t=\dfrac{\lambda}{v}=\dfrac{1}{\sqrt{2}\pi d^2nv}$
$=\dfrac{1}{1.414\times 3.14\times (20)^2\times 10^{-6}\times (0.0167)\times (150)}$
$=\dfrac{10^6}{4449.5}=224.74$h $=225$h

Estimate the mean free path for a water molecule in water vapor at $373K$,the diameter of the molecule is $2\ \times 10^{-10}\ m$ and at $STP$ number of molecular per unit volume is $2.7\ \times 10^{25}\ m^{-3}$ :

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $2.81 \times 10^{-7}\ m$

  2. $3 \times 10^{-7}\ m$

  3. $4 \times 10^{-7}\ m$

  4. $5 \times 10^{-7}\ m$

Show answer
Correct Option: A

There are two vessels of same consisting same no of moles of two different gases at same temperature . One of the gas is $CH _{4}$ & the other is unknown X. Assuming that all the molecules of X are under random motion whereas in $CH _{4}$ except one all are stationary. Calculate $Z _{1}$ for X in terms of $Z _{1}$ of $CH _{4}$. Given that the collision diameter for both gases are same & $\displaystyle (U _{rms}) _{x}=\frac{1}{\sqrt{6}}(Uav) _{CH _{4}}$.

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $\displaystyle \frac{2\sqrt{2}}{3\sqrt{\pi }}Z _{1}$

  2. $\displaystyle \frac{3\sqrt{2}}{2\sqrt{\pi }}Z _{1}$

  3. $\displaystyle \frac{2\sqrt{3}}{2\sqrt{\pi }}Z _{1}$

  4. $\displaystyle \frac{4\sqrt{2}}{3\sqrt{\pi }}Z _{1}$

Show answer
Correct Option: A
Explanation:

V, n, T $\rightarrow  same$(25) so $P\rightarrow $ also same ( P  5  25)
$\displaystyle \sigma \rightarrow same (25)$
given

$\displaystyle (v {rms})\times

x=\dfrac{1}{\sqrt{6}}(v _{avg.}) _{CH _{4}}$ &

$v _{rms}=\sqrt{\dfrac{3\pi }{8}}(v _{avg.})$ so
$\displaystyle \sqrt{\dfrac{3\pi }{8}}(v _{avg.}) _{CH _{4}}$
$\displaystyle \dfrac{(v _{avg.})x}{(v _{avg.}) _CH _{4}}=\sqrt{\dfrac{8}{3\pi }}.\frac{1}{\sqrt{6}}=\dfrac{2}{3\sqrt{\pi }}$
For X (9< ) : $\displaystyle Z _{1}=\sqrt{2}\pi \sigma ^{2}(v _{avg.}) _{x}N^{\ast }$
For CH{4} (9< ) : $\displaystyle Z _{1}=\pi \sigma ^{2}(v _{avg.}) _{CH _{4}}N^{\ast }$
Since T, P, v, n are same, $N\ast $ will also be same.
$\displaystyle



\frac{Z _{1}X}{Z _{1}(CH _{4})}=\sqrt{2}\frac{(v _{avg.}) _{x}}{(v _{avg.}) _{CH _{4}}}=\sqrt{2}.\frac{2}{3\sqrt{\pi

}}$
$\displaystyle Z _{1}(X)=Z _{1}(CH _{4}).\frac{2\sqrt{2}}{3\sqrt{\pi }}$

Which is the wrong statement out of the following?

physics reflection of light at curved surfaces uses of curved mirrors uses of spherical mirror uses of spherical mirrors

  1. A concave mirror can give a virtual image.

  2. A convex mirror can give a virtual image.

  3. A concave mirror can give a diminished virtual image.

  4. A convex mirror cannot give a real image.

Show answer
Correct Option: C
Explanation:

The properties of image formed for different object distance are listed below:
Concave mirror:
1)u<f----virtual image , enlarged.
2)f,u,2f---Real, inverted, magnified.
3)u>2f----Real, inverted, diminished.
Convex mirror:
for any object distance virtual,errect and enlarged image.

State whether true or false.
A concave mirror is always used as a street light reflector.

physics reflection of light at curved surfaces uses of curved mirrors uses of spherical mirror uses of spherical mirrors

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Convex mirrors are used as reflectors in street lights because they are able to spread the light over a wide area.

The temperature in a spherical reflector type solar cooker is raised to more than $500^o C$ by :

physics reflection of light at curved surfaces uses of curved mirrors uses of spherical mirror uses of spherical mirrors

  1. concentrating energy at one point by using a concave reflector

  2. proper insulation of the cooker

  3. concentrating energy at one point by using a covex reflector

  4. both (A) and (B)

Show answer
Correct Option: A
Explanation:

The temperature in a spherical reflector type solar cooler is raided to more than $500^\circ C $ by concentrating energy at one point by using a concave reflector.