## Wednesday, September 27, 2006

### Kerala Engineering Entrance test-2006 – Two Questions

The following two questions were omitted by a fairly bright student who appeared for Kerala Engineering Entrance test (I wonder what is so unattractive about these questions!):
(1) The momentum of a body is increased by 25%. The kinetic energy is increased by about
(a) 25% (b) 5% (c) 56% (d) 38% (e) 65%
The kinetic energy of a body is given by E = p2/2m. Therefore, when the momentum ‘p’ is increased by 25%, the kinetic energy E is increased by about 56%. (Note that 1.252 = 1.5625).
(2) If the potential energy of a gas molecule is U = M/r6 – N/r12, M and N being positive constants, then the potential energy at equilibrium must be
(a) zero (b) M2/4N (c) N2/4M (d) MN2/4 (e) NM2/4

You know that the molecule will be in equilibrium if its potential energy U is a minimum. For this, dU/dr = 0. Therefore, -6Mr-7 + 12Nr-13 = 0 from which r6 = 2N/M. Therefore, r12 = 4N2/M2. Substituting these values in the expression (given in the question) for U, the potential energy at equilibrium is M/(2N/M) – N/(4N2/M2) = M2/2N – M2/4N = M2/4N [Option (b)].

## Tuesday, September 26, 2006

If you have any genuine doubt to be cleared, on topics in Physics which will be useful to students desirous of appearing for Medical and Engineering entrance tests, you may express it through the option for ‘comments’ provided below each post. If you have a Blogger user name, you can log in using that. If you do not have a Blogger user name you can easily get one . [For obtaining a free Blogger account, click on ‘comments’ below any post and in the window that opens up click on ‘sign up here’ which appears above the word verification characters. You will be taken through the very easy steps for obtaining your free Blogger account (user name)]. Even if you do not have a Blogger user name, you can post your comment using the provision for ‘other’ or ‘anonymous’.

## Sunday, September 24, 2006

### Will this confuse the students?

I think the following question which appeared in a Medical Entrance question paper will confuse the students:
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de Broglie wave length of the particle is
(a) 25% (b) 75% (c) 60% (d) 50% (e) 30%
The answer given to this question is 75%. In multiple choice questions, the most suitable choice need be picked out and indeed the answer then is 75%. The answer will be exactly 75% if you mean that the final kinetic energy of the particle is 16 times the initial kinetic energy, as shown below:
Kinetic energy, E = p2/2m where ‘p’ is the linear momentum and ‘m’ is the mass. If E is increased to 16 times the initial value, ‘p’ is increased to 4 times the initial value. Since de Broglie wave length λ = h/p where ‘h’ is Planck’s constant, the final wave length will be λ/4. The percentage change in the wave length is therefore 75%.
If the kinetic energy is increased by 16 times as stated in the question, the final kinetic energy will be 17 times the initial kinetic energy and the final momentum will be 4.123 times the initial momentum. The final wave length will be 0.243 times the initial wave length and the percentage change in the wave length will be 75.5%.
In the present question, the answers are almost the same. But suppose the words ‘increased by 16 times’ were replaced by ‘increased by 4 times’. The difference between the answers will be significant. If you mean that the final energy is 4 times, you should modify the words as ‘increased to 4 times’ or ‘made 4 times’. I thought of posting this because I have seen this type of confusing wording in many instances.

## Friday, September 22, 2006

### Two Questions on Properties of liquids

The weight of a floating body is equal to the weight of the displaced liquid. Question setters often find this law of floatation handy while setting Medical and Engineering test papers. Consider the following question:
A piece of wood having volume ‘V’ and density ‘d’ floats at the interface of two immiscible liquids of densities ‘ρ’ and ‘σ’ respectively. If ρ > d > σ, the ratio of the volumes of the parts of the wooden piece in the rarer and denser liquids is
(a) (ρ-d)/ (d-σ) (b) (d-σ)/(ρ-d) (c) (ρ-d)/(d-σ) (d) (ρ+d)/(ρ+σ) (e) (d-σ)/(ρ+σ)

Equating the weight of the wooden piece to the weight of the displaced liquids, we have,
Vdg = vσg + (V-v)ρg where ‘v’ is the volume of the part of the wooden piece in the rarer liquid (of density ‘ρ’). Rearranging this equation, we have,
v(ρ-σ) = V(ρ-d) so that v/V = (ρ-d)/ (ρ-σ).
Therefore, v/(V-v) = (ρ-d)/(d-σ)
The correct option is (a).
Now consider the following M.C.Q. which appeared in the Kerala Medical Entrance test paper of 2006:
Blood is flowing at the rate of 200cm3/s in a capillary of cross sectional area 0.5m2. The velocity of flow in mm/s is
(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 (e) 0.5
This is a very simple question. From the equation of continuity (av = constant), we have,
0.5 v = 200×10-6 so that v = 4×10-4m/s = 0.4mm/s [option (d)].

## Tuesday, September 19, 2006

### Properties of Fluids

In any Medical and Engineering entrance test paper you will find a few questions based on the properties of fluids. Let us discuss some typical questions. The following simple question is taken from the A.F.M.C.2004 test paper:
Application of Bernoulli’s theorem can be seen in
(a) dynamic lift of aeroplane (b) hydraulic press (c) helicopter (d) none of these

The correct option is (a). The shape of the aeroplane wings is such that when it moves forward, the air molecules at the top of the wings have a greater velocity (relative to the wings) compared to the air molecules at the bottom. Therefore, in accordance with Bernoulli’s principle, the pressure at the top of the wings is less than that at the bottom. This results in a dynamic lift of the wings which balances the weight of the plane.
Let us now consider another question involving Bernoulli’s principle:

Water is flowing steadily through two horizontal pipes of radii 3cm and 6cm connected in series. The speed of water in the first pipe is 2m/s and the pressure of water in it is 2×104 pascal. The pressure of water in the second pipe will be nearly
(a) 2×104 pascal (b) 2.2×104 pascal (c) 2.4×104 pascal (d) 2.6×104 pascal (e) 3×104 pascal

The speed v2 of water in the second pipe is given (from the equation of continuity, a1v1 = a2v2 ) by v2 = 2×a1/a2 = 2×1/4 = 0.5 m/s. (The ratio of cross section areas is ¼ since the radius of the second pipe is twice that of the first).
As per Bernoulli’s principle, we have P1 + (ρv12 )/2 + ρgh = P2 + (ρv22)/2 + ρgh with usual notations. Therefore, P2 = P1 + ρ(v12- v22)/2 =2×104 + 1000(22 – 0.52)/2 = 2.1875×104 pascal. The correct option therefore is (b).
The following question involving surface tension may appear to be difficult for some of you. See how simple it is:
Two soap bubbles of radii ‘R’ and ‘r’ (R > r) are touching each other. The radius of curvature of the common surface at the region of contact is
(a) (R+r)/2 (b) R – r (c) (R – r)/2 (d) Rr/(R+r) (e) Rr/(R-r)
At the contact region, the radius of curvature (R') of the soap film is governed by the net excess of pressure 4T/r – 4T/R where T is the surface tension. Therefore, we have 4T/r – 4T/R = 4T/R' from which R' = Rr/R-r [option (e)].
Consider now the following M.C.Q.:

Two rain drops are falling through air with the same terminal velocity of 8cm/s. If they coalesce, the terminal velocity of the combined single drop will be (in cm/s)
(a) 8/√2 (b) 16 (c) 8×41/3 (d) 8×21/3 (e) 8√
2
On equating the apparent weight of a rain drop to the viscous force (air resistance) opposing the downward motion of the drop, we have, 4/3πr3(ρ-σ)g = 6πrηv, with usual notations. Therefore, the velocity, v α r2…………(1)
Since the drops have the same terminal velocity, it follows that they have the same radius. The radius (r1 ) of the single large drop (when the two drops coalesce) is given by
2× 4/3πr3 = 4/3 πr13. Therefore, r1=21/3 r.
The velocity of the combined single drop, v1 α [21/3 r]2 -------------- (2)
From equations (1)&(2), v1= 41/3v = 41/3 × 8, since v=8cm/s. The correct option therefore is (c).

## Wednesday, September 13, 2006

### Molecular speeds

The root mean square velocity ‘c’ of a gas molecule is given by the following relations:
(1) c = √(3kT/m)
(2) c = √(3RT/M)
(3) c = √(3P/ρ)
The first relation gives the r.m.s. speed in terms of Boltzman’s constant ‘k’ and the molecular mass ‘m’. The second one gives the r.m.s. speed in terms of universal gas constant ‘R’ and the molar mass ‘M’. The third one gives the r.m.s. speed in terms of the pressure and density of the gas.
Note that the r.m.s. speed is directly proportional to the square root of the absolute temperature of the gas and also that the r.m.s. speed of the molecules of a given gas is constant at a constant temperature. At constant temperature, even if you change the pressure, the r.m.s. speed is unchanged (as given by the third relation) since the density ‘ρ’ is directly proportional to the pressure ‘P’.
The gas molecules obey Maxwell’s distribution law and the most probable velocity is given by cmp = √(2kT/m) = √(2RT/M) = √(2P/ρ).
Now, consider the following M.C.Q.:
Nitrogen gas molecules have r.m.s.speed ‘v’ at -23˚C. Their r.m.s speed at 477˚C will be
(a) 4.55v (b) 20.7v (c) 9v (d) 1.732v (e) 1.414v
Note that the temperatures are given in degree Celsius. Convert them to the Kelvin scale to obtain 250K and 750K. The rise in the temperature is 3 times. Since the r.m.s. speed is directly proportional to the the square root of the absolute temperature, the final speed is √3 times the initial speed. So, the answer is 1.732v.
If you want to write this in a mathematical step, you have v/v' = √(250/750) from which you obtain the final r.m.s. speed v'.
Now, let us discuss the folowing question:
Root mean square speed of oxygen gas molecules at a certain temperature is 11.2 km/s. The gas is cooled so that the pressure is halved without any change in its density. The final value of the root mean square speed is
(a) 11.2 km/s (b) 7.9km/s (c) 6.4km/s (d) 5.6 km/s (e) 2.8km/s
As the density is unchanged, the volume of the gas is constant. Applying Charles law (P/T = P'/T'), we have P/T = (P/2)/T' from which T' = T/2.
(You can dispense with these steps and simply argue that the temperature must be halved when the pressure is halved at constant volume). Since the RMS speed is directly proportional to the square root of the absolute temperature, the final speed is 11.2×1/√2 = 7.9 km/s.
You can arrive at the answer in no time if you remember the expression for pressure ‘P’ in the form, P = ⅓ρc2. Since the density ‘ρ’ is unchanged, the RMS speed ‘c’ should become 1/√2 times the initial value when the pressure becomes half the initial value.
Let us consider another question:
At what temperature will the molecules of nitrogen have the same r.m.s. speed as the molecules of oxygen at 27˚C?
(a) -10.5˚C (b) -34.5˚C (c) -262.5˚C (d) 262.5˚C (e) 23.62˚C
Since the r.m.s. speed is given by c = √(3RT/M) we have √(3RTn/Mn) =√(3RTo/Mn) where the suffix ‘n’ and suffix ‘o’ refer to nitrogen and oxygen respectively.
Therefore, Tn/28 = 300/32. Note that the molar mass of nitrogen is 28 and that of oxygen is32 and we have substituted the temperature in Kelvin. This yields the value 262.5K as the temperature of nitrogen molecules. The value in Celsius scale is -10.5˚C.
Consider now the following simple question which is often found in test papers:
If ‘v’ and ‘c’ are respectively the velocity of sound in a gas and the r.m.s. speed of the gas molecules at a particular temperature and γ is the ratio of specific heats, then
(a) v = c (b) v = √γc (c) v = c√(γ/3) (d) v = √(γc/3) (e) v = √(3γ/c)
The correct option is (c) since we have v = √(γP/ρ) [Newton-Laplace equation] and c = √(3P/ρ). Therefore, v/c = √(γ/3) from which v = c√(γ/3).
To conclude this post, let me ask you a simple question: Do you know why the moon does not possess an atmosphere? Many of you might be knowing the reason: The escape velocity on the moon’s surface is 2.4 km/s only (compared to 11.2 km/s for the earth) because of the smaller values of acceleration due to gravity and the radius in the case of the moon, compared to the values in the case of the earth. [Remember v=√(2gR)]. Molecular speeds on the moon can attain values significant compared to 2.4 km/s and hence gas molecules can escape into the outer space.

## Tuesday, September 12, 2006

### Photons and electrons

Let us discuss some questions involving photons and electrons. The following question high lights one basic difference between photons and electrons:
The kinetic energies of an electron and a photon are in the ratio 9:4. Their momenta are in the ratio
(a) 9:4 (b) 3:2 (c) 1:3 (d) 4:3 (e) 3:4
The kinetic energy of an electron (or any such particles of matter) is directly proportional to the square of its momentum (since E = p2 /2m with usual notations) where as the kinetic energy of a photon is directly proportional to its momentum (since E = mc2 = mc.c = pc). The ratio of the momenta of the electron and the photon is therefore √9: 4 = 3:4 [Option (e)].
You will encounter many situations in which you will be required to calculate the velocity of a particle of charge ‘q’ and mass ‘m’ accelerated by a voltage ‘V’. The equation you have to use is ½ mv2 = qV from which the velocity, v = √(2qV/m).
A special case is that of the electron, which you will encounter quite often. When you substitute the mass and the charge of the electron, the velocity is given (approximately) by v = 6×105√V. Remember this relation. You can save the valuable time you will have to spend on certain calculations. For instance, consider the following M.C.Q.:

An electron at rest is accelerated by 2500 volts. Its final velocity is approximately (in m/s)
(a) 3×106 (b) 3×107 (c) 2.5×108 (d) 2.5×106 (e) ) 1.5×108

If you use the relation, v = 6×105√V , you will get the correct option (b) in no time.
[You should note that we have not considered the relativistic increase in the mass of the electron in the above approximate equation. If the accelerating voltage is higher, the relativistic increase in the mass is significant and the above equation cannot be used. Even in the above problem, the velocity of the electron is 10% of the velocity of light and the relativistic increase in the mass begins to be exhibited].
Now consider the following question:
A laser source has output power of 300 mW at a wave length of 6630Ǻ. The number of photons emitted from this laser source every minute is
(a) 6×1019 (b) 6×1018 (c) 6×1017 (d) 6×1016 (e) 6×1014
If the number of photons emitted per second is ‘n’, we have nhν = nhc/λ = 300×10-3 so that n = 0.3 λ/hc where ‘h’ is Planck’s constant, ‘c’ is the speed of light and ‘λ’ is the wave length. The number of photons emitted per minute is 60n = (60×0.3×6630×10-10)/(6.63×10-34×3×108) = 6×1019 [Option (a)].
Energy and Wave Length of a Photon
There are many situations in which you will have to find the energy of a photon in electron volt if its wave length in Angstrom Units (A.U.) is given and vice versa. For instance, suppose you are asked to find out the minimum wave length of X-rays that can be obtained from an X-ray tube operating with an anode voltage of 20kV. Since the X-ray photon of minimum wave length will have the entire energy of 20000 electron volt (of the electron striking the target in the X-ray tube), you can write,20000 e = hc/λ where ‘e’ is the electronic charge, ‘h’ is Planck’s constant, ‘c’ is the speed of light in free space and λ is the wave length in metre.Instead of using the above equation for calculating λ, you may remember that in the case of a photon,λE = 12400 where λ is in Angstrom and the energy E is in electron volt.The answer to our present problem therefore is,
λ = 12400/20000 = 0.62 A.U.
Do you know why germanium and silicon are unsuitable for making light-emitting diodes?
The band gap for producing visible light should be at least 12400/7000 electron volt (since the maximum wave length of visible light is roughly 7000 Angstrom). This works out to be nearly 1.8eV, which is greater than the band gap of Ge and Si.

## Saturday, September 09, 2006

### Multiple Choice Questions on Magnetism

The following question which appeared in the Kerala Engineering Entrance test paper of 2006 has been popular among question setters for long:
A magnetized wire of magnetic moment M and length L is bent in the form of a semicircle of radius ‘r’. The new magnetic moment is
(a) M (b) M/2π (c) M/π (d) 2M/π (e) zero
The pole strength of the magnet, p = M/L. The pole strength of the magnet is unchanged, but the moment is changed since the poles come closer on bending the wire, thereby changing the magnetic length of the magnet from L to L'
We have L' = 2r = 2L/π so that the new magnetic moment = pL' = (M/L) ×(2L/π) = 2M/π [Option (d)].
Consider now the following M.C.Q.:
In a hydrogen atom the electron is making 6.6×1015 revolutions per second around the nucleus in an orbit of radius 0.528 Å. The equivalent magnetic dipole moment is approximately ( in Am2)
(a)10-10 (b) 10-15 (c) 10-23 (d) 10-25 (e) 10-17

The orbiting electron is equivalent to a circular current loop, whose magnetic dipole moment is given by M = IA where I is the equivalent current and A is the area of the loop. Therefore, M = (q/T)×(πr2) = qf πr2 where q is the electronic charge, T is the orbital period, r is the orbital radius and f is the frequency of revolution of the electron.
Thus M = 1.6×10-19×6.6×1015×π×(5.28×10-10)2 . This yields a value nearly 10-23 [Option (c)].
Let us now consider the following question:
Two short magnets of dipole moments M and 2M are arranged on the table so that the axial line of the weaker magnet and the equatorial line of the stronger magnet are coinciding. If the separation between the magnets is 2d, what is the magnetic flux density midway between these magnets? Ignore the earth’s magnetic field.
(a) μ0M/4πd3 (b) 3μ0M/4πd3 (c) (μ0M/4πd3)
3 (d) (μ0M/4πd3)5 (e) (μ0M/4πd3
)8
At the point midway between the magnets, the axial field (μ0/4π)(2M/d3) of the magnet of moment M and the equatorial field (μ0/4π)(2M/d3) of the magnet of moment 2M are acting at right angles so that the net field there is √2×(μ0/4π)(2M/d3) = μ0M/4πd3)√8. The correct option therefore is (e).

## Wednesday, September 06, 2006

### Work Done in Lifting a Body

When you lift a body of mass ‘m’ from the earth’s surface through a small height ‘h’, the work ‘W’ done by you against gravitational force, as you know, is given by W = mgh. This expression gives you the correct value only if ‘h’ is negligibly small compared to the radius ‘R’ of the earth.
If the height ‘h’ is comparable to the radius ‘R’ of the earth, the expression gets modified as W = mgh/[1+(h/R)].
If h = nR the above expression can be written as W = mgRn/(n+1).
In all these expressions, ‘g’ is the surface value of acceleration due to gravity.
Let us consider the following multiple choice questions:
(1)The work done in lifting a body of mass ‘m’ from the earth’s surface, through a height ‘h’ is mgR/3 where ‘g’ is the acceleration due to gravity on the earth’s surface and ‘R’ is the radius of the earth. Then ‘h’ is equal to
(a) R (b) R/2 (c) R/3 (d) R/4 (e) R/6

We have W = mgh/[1+(h/R)] so that mgR/3 = mgh/[1+(h/R)]. This gives h = R/2 [option (b)].
(2) A body of mass ‘m’ is projected at an angle of 45˚ from the moon’s surface by giving it kinetic energy equal to mgR where ‘R’ is the radius of the moon and ‘g’ is the acceleration due to gravity on the moon’s surface. Then
(a)the final potential energy of the body will be zero (b) the final kinetic energy of the body will be positive (c) the body will reach a height equal R (d) the body will reach a height between R and 2R (e) the body will reach a height between 2R and 4R.
The potential energy of the body on the moon’s surface is –GMm/R = -mgR, on substituting g = GM/R2. When the body is given kinetic energy equal to mgR, its total energy becomes zero and it escapes into the outer space where its potential energy and kinetic energy are zero. The correct option therefore is (a).
You can find more multiple choice questions (with solution) of similar type here

## Friday, September 01, 2006

### Questions on Kinetic Theory of Gases

The important relations you should remember in kinetic theory of gases are the following:
(1) Pressure exerted by a gas, P = (1/3) nmc2 = (1/3)ρc2 = nkT where ‘n’ is the number of molecules per unit volume, ‘m’ is the molecular mass, ‘c’ is the r.m.s. speed of the molecule ‘ρ’ is the density of the gas, ‘k’ is Boltzman’s constant and T is the absolute temperature.
(2) R.M.S. speed of molecule, c = √(3P/ρ) = √(3RT/M) = √(3kT/m).
Remember all the three expressions for r.m.s.speed. The second one gives the r.m.s.speed in terms of molar mass M and the universal gas constant R. The third one gives the r.m.s. speed in terms of molecular mass ‘m’ and Boltzman’s constant ‘k’.
(3) Average translational kinetic energy of any type of gas molecule is (3/2)kT since translational motion along three directions only are possible in our three dimensional space.
(4) If the molecule has ‘n’ degrees of freedom, the average kinetic energy per molecule is (n/2 )kT.
The following points are worth noting in the present context:
(i) A mono atomic gas molecule has 3 degrees of freedom and has translational kinetic energy only [equal to (3/2)kT ].
(ii) A diatomic molecule has 5 degrees of freedom (three translational and two rotational). In this case, the total average kinetic energy per molecule is (5/2 )kT.
(iii) Tri-atomic and polyatomic molecules have 6 degrees of freedom (three translational and three rotational). The total average kinetic energy per molecule is ( 6/2 )kT = 3kT.
The K.E. per mole in all the above cases is N times (N is the Avogadro number). Since Nk=R, the average K.E. per mole is (3/2 )RT for mono atomic, (5/2 )RT for diatomic and 3RT for triatomic and polyatomic gas molecules.
Note that the molar heat capacity at constant volume (CV) is obtained by putting T=1(corresponding to a temperature rise of 1K) in the above expressions. The values are therefore (3/2 )R for mono atomic gas, (5/2)R for diatomic gas and 3R for triatomic and polyatomic gases.
The molar heat capacity of a gas at constant pressure (CP) is given by CP = CV + R. Therefore, the values are (5/2)R for mono atomic gas, (7/2)R for diatomic gas and 4R for triatomic and polyatomic gases.
It will be useful to remember the relation connecting the ratio of specific heats ‘γ’ and the number of degrees of freedom ‘f’:
γ = 1+ (2/f)
Note: In the above discussion, the vibrational modes of the molecules have been ignored. Even though the above values are in agreement with the values obtained from experiment in the case of several gases, there are discrepancies in the case of certain diatomic gases and several polyatomic gases. The vibrational modes also are therefore to be taken into account in more rigorous treatment.
Let us now consider the following question:
The average translational kinetic energy of a helium gas molecule (molar mass 4) at a particular temperature is 0.05electron volt. The average translational kinetic energy of an oxygen molecule (molar mass 32) at the same temperature will be
(a) 0.4eV (b) 0.08eV (c) 0.2eV (d) 0.05eV (e) 0.1eV
Don’t be concerned about the type of the gas and the molar mass. The translational kinetic energy of all types of molecules is the same at a given temperature. The correct option therefore is (d).
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
An electron tube was sealed off during manufacture at a pressure of 1.2×10-7 mm of mercury at 27˚C. Its volume is 100cm3. The number of molecules that remain in the tube is
(a) 2×1016 (b) 3×1015 (c) 3.86×1011 (d) 5×10 11 (e) 2.5×1012
You may do this problem by using one of the following relations:
(1) P = nkT (2) PV = rT where ‘r’ is the gas constant for the mass of gas in the electron tube.
If you use the first relation, you should remember the value of Boltzman’s constant ‘k’, which is 1.38×10-23 . Since k = R/N you can substitute the values of the universal gas constant R (= 8.31 J/mol/K ) and the Avogadro number N (= 6.02×1023 ), which you should definitely remember, to get ‘k’. The number of molecules per unit volume therefore is,
n = P/(kT) = hdg/(kT)
The number of molecules in the electron tube is n×V = hdgV/(kT)
= (1.2×10-10×13600×9.8×100×10-6 ) / (1.38×10-23 ×300) = 3.86×1011 [Option (c)].
If you use the second relation you will write PV = n’RT where n’ is the number of moles of gas in the tube. Therefore, n’ = PV/(RT).
The number of molecules in the tube = n’N = PVN/(RT) = hdgVN/(RT).
Note that this is the same relation as we obtained in the first method since k = R/N.
If you are not very quick in arithmetical manipulations, don’t attempt questions like this during your initial trial.
Consider now the following M.C.Q.:
1.2 mole of helium gas (mono atomic), 0.4 mole of nitrogen (diatomic) and 0.4 mole of oxygen (diatomic) are contained in a vessel of volume 10 litre at a temperature of 27˚C. The pressure of this mixture of gases is (in Nm-2)
(a) 5×105 (b) 106 (c) 2.5×105 (d) 2.5×106 (e) 5×106
Some of you may have certain doubts regarding this simple question. Your doubts are added when you see the distractions ‘mono atomic’ and ‘diatomic’. But your doubts will be cleared the moment you remember the expression for pressure in the form, P = nkT. Here ‘n’ is the number of molecules per unit volume, ‘k’ is the Boltzman constant and T is the absolute temperature. The type of the gas does not come into the picture and you require the number density ‘n’ only for substituting in the expression for P. Since the mixture contains 2 moles (1.2+0.4+0.4), the total number of molecules in the mixture is 2N where N is Avogadro number. The number of molecules per unit volume (n) therefore is 2N/(10×10-3 ) = N/(5×10-3). Therefore, P = NkT/(5×10-3). But, k = R/N so that P = RT/(5×10-3) = 8.3×300/(5×10-3) = 5×105Nm-2.
Let us consider another question:
Oxygen and hydrogen gases are at the same temperature T. The kinetic energy of an oxygen molecule will be
(a) 32 times the kinetic energy of a hydrogen molecule (b) 16 times the kinetic energy of a hydrogen molecule (c) twice the kinetic energy of a hydrogen molecule (d) 4 times the kinetic energy of a hydrogen molecule (e) the same as that of the hydrogen molecule.

The correct option is (e). The kinetic energy (total) of a gas molecule is (n/2)kT where ‘n’ is the number of degrees of freedom. Oxygen and hydrogen are diatomic and they have the same degrees of freedom (n=5) and hence the same kinetic energy at a given temperature.
Certain simple questions may confuse you if you are in a hurry and you may give wrong answers! Here is one such question:
To decrease the volume of an ideal gas by 10% at constant temperature, the pressure should be increased by
(a) 5% (b) 8.91% (c) 10% (d) 11.1% (e) 12.25%

Don’t pick out option (c) in a hurry. We have PV = P’×0.9V so that P’ = P/0.9 = 1.111P. The increase in pressure is 11.1% [option (d)].
Let us discuss one more question:
When a gas contained in a closed vessel is heated through 1˚C, the pressure of the gas increases by 0.2%. the final temperature of the gas is
(a) 200K (b) 361K (c) 500K (d) 501K (e) 601K
We have, P/T = 1.002P/(T+1) since the volume is constant. Solving this we obtain T = 500K. The final temperature is T+1 = 501K.