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)].

Clear Your Doubts

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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.

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)].

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) 8×4^1/3 (d) 8×2^1/3 (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=2^1/3 r.
The velocity of the combined single drop, v₁ α [2^1/3 r]² -------------- (2)

From equations (1)&(2), v₁= 4^1/3v = 4^1/3 × 8, since v=8cm/s. The correct option therefore is (c).

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 c_mp = √(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 √(3RT_n/M_n) =√(3RT_o/M_n) where the suffix ‘n’ and suffix ‘o’ refer to nitrogen and oxygen respectively.
Therefore, T_n/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.

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.

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).

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

ONAM GREETINGS

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 (C_V) 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 (C_P) is given by C_P = C_V + 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.

Questions on Satellites

The following questions on satellites are simple, but slightly different from the common direct questions. See whether you can solve them yourself. Go through the solution given, only after your trial.
(1)A satellite is moving in an orbit of radius ‘r’ around the earth. A second satellite is moving in an orbit of radius (1.02)r. Then the orbital period of the second satellite is greater than that of the first by approximately
(a) 0.2% (b) 2% (c) 3% (d) 10.2% (e) 6%
We have T² α r³ (Kepler’s law) from which T α r^3/2. Since the increment in the orbital radius ‘r’ is 0.02r, the fractional increment in the orbital period ‘T’ is (∆T)/T = (3/2)(∆r)/r = (3/2) × 0.02r/r = 0.03. The percentage increase is 0.03×100 = 3% [option (c)].
(2) If the orbital period of a satellite is ‘T’, its kinetic energy is directly proportional to
(a) T (b) T^-1 (c) T^3/2 (d) T^2/3 (e) T^-2/3

We have, kinetic energy, E = GMm/2r where G is the gravitational constant, M is the mass of the earth ‘m’ is the mass of the satellite and ‘r’ is the radius of the orbit. Therefore, E α 1/r. Since T² α r³ in accordance with Kepler’s law, we can write r α T^2/3. Therefore, E α T^-2/3. The correct option is (e).

Questions on Bohr Atom Model

You will usually find a couple of questions pertaining to Bohr model of hydrogen atom in most entrance test papers. Questions in this section are interesting and often simple. Consider the following M.C.Q.:
Which state of the triply ionised Be+++ has the same orbital radius (for the electron) as that of the hydrogen atom in the ground state?
(a) First excited state (b) Second excited state (c) Third excited state (d) Fourth excited state (e) Ground state
The Be atom has lost 3 electrons since it is triply ionised and it behaves as a hydrogen like atom. Therefore, the radius of the orbit of quantum number ‘n’ is r n²/z where ‘r’ is the Bohr radius (radius of the electron orbit in the hydrogen atom in the ground state) and ‘z’ is the atomic number of the hydrogen like atom. Since z = 4 for Be, we have r = r n²/4 from which n = 2, which means the first excited state. So, the correct option is (a).
Now consider the following questions which appeared in the Kerala Engineering Entrance Test of 2002:
(1) The energy of an electron in excited hydrogen atom is -3.4 eV. Then, according to Bohr’s theory, the angular momentum of the electron in Js is
(a) 2.11×10^-34 (b) 3×10^-34(c) 3×10^-34 (d) 0.5×10^-34
Since the energy of the electron is -3.4 eV, the electron is in the second orbit. This is checked using the energy (in eV) expression E = -13.6/n² = -3.4 eV, when n = 2.
Since the angular momentum is nh/2π, on substituting for h (=6.63×10^-34 Js) and n (=2) we obtain the first option as the correct answer.
(2) Consider the spectral line resulting from the transition from n = 2 to n = 1, in atoms and ions given below. The shortest wave length is produced by
(a) hydrogen atom (b) deuterium atom (c) singly ionized helium (d) doubly ionized helium (e) doubly ionized lithium

The correct option is (e). The expression for the energy of an electron in a hydrogen like atom is E = (- 13.6 z²)/n². The energy for a given value of ‘n’ is maximum for lithium in the present case (since z = 3). The energy difference also is therefore maximum (for the given transition) in the case of lithium, giving rise to the shortest wavelength.

You will find more questions with solution at AP Physics Resources: AP Physics B– Multiple Choice Questions (for practice) on Atomic Physics and Quantum Effects

Questions on Electromagnetic Waves

When an electromagnetic wave is propagating through any medium including empty space, the direction of propagation of the wave is related to the directions of the electric and magnetic field vectors (E and B) in a definite manner. Many students are found to commit mistake in marking the directions of these vectors. If you remember that the vector product E×B always points along the direction of propagation of the electromagnetic wave, you wont commit mistake in specifying the directions of these vectors. Now consider the following M.C.Q.:
In a plane electromagnetic wave, the magnetic field vector is along the negative X-direction and the direction of propagation is along the positive Y-direction. Then the direction of the electric field vector must be along
(a) positive Z-direction (b) negative X-direction (c) negative Z-direction
(d) positive Y-direction (e) negative Y-direction
Since the direction of the E has to be perpendicular to both the direction of B and the direction of propagation, it has to be in the positive or negative Z-direction. Since there are two options [(a) and (c)] you have a 50-50 chance to hit the bull’s eye. But don’t take such chances. The magnetic field vector B is along the negative X-direction and so you can get the vector product E×B along the positive Y-direction only if E is along the negative Z-direction [option(c)].
You might have noted that our eyes are most sensitive to light of wave length about 550 nm because the sun radiates maximum energy in the form of light at this wavelength. This wavelength is related to the surface temperature T of the sun in accordance with Wein’s law which states that λm T = constant where λm is the wave length at which maximum energy is radiated. If you express λm in cm and T in Kelvin the constant is 0.29 cmK. Now consider the following question:
If the temperature of the sun were twice the present value, the radiation of the sun would be mostly in
(a) microwave region (b) infra red region (c) visible region (d) X-ray region (e) ultraviolet region.
The correct option is (e) because the wave length of maximum energy radiation has to be reduced to half the present value when the temperature is doubled (in accordance with Wien’s law). The present value of λm is about 550 nm and so when the temperature is doubled, the value of λm should become 275 nm, which is in the ultra violet region.

Questions on Elasticity

Here is a question involving Young’s modulus. It’s a simple question, but there is a good chance of committing a mistake if you are not careful enough.
A wooden plank, used as a bridge over a canal sags by 10mm due to its own weight. If the thickness of the wooden plank is doubled, the sag will be
(a) 5mm (b) 2.5mm (c) 1.25mm (d) 10mm (e) 20mm
The expression for the depression (sag) ‘δ’ of a centrally loaded beam is,
δ = (WL³)/4Ybd³ where W is the load (W=mg), L is the length of the beam, Y is the Young’s modulus, ‘b’ is the breadth of the beam and ‘d’ is its thickness(depth).
In the above problem, no separate load is suspended from the centre of the beam, but the weight of the beam acts through its centre to depress it. When a beam of twice the thickness is used, its weight appearing in the numerator is doubled and the value of d³ appearing in the denominator becomes 8 times. Therefore, the depression is one-fourth, equal to 2.5 mm [option (b)].
Let us consider another question which also may baffle you slightly:
A thick rope of length L is hanging from the ceiling of a room. If Y and ρ are respectively the Young’s modulus and density of its material, its increase in length due to its own weight is
(a) (ρgL²)/2Y (b) (ρgL²)/Y (c) (ρgL)/Y (d) Y/ρgL² (e) data insufficient.
Since the Young’s modulus is given by the usual expression,Y = FL/Al where L is the original length, ‘l’ is the increase in length, F is the elongating force and A is the area of cross section, we have,
l = FL/YA. Here F = ALρg, which is the weight of the rope. You have to substitute the length L as it is, in the expression for F. But the weight of the rope acts through its centre of gravity and hence you have to substitute L/2 instead of L in the expression for the increase in length. The answer therefore is (ρgL²)/2Y [option (a)].

Friction Moves the Car and Friction Stops the Car

It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.

– Richard Feynman

 
When you think of friction in detail you will find that it is an ordinary force with extraordinary features. Its direction is always self adjusting and is opposite to the direction of the force which tries to move a body. The magnitude of frictional force is self-adjusting up to the limiting value.
Have you ever thought how exactly a car moves forward? The engine of the car does not exert any forward force. It just spins the wheels of the car. The portion of the wheel (tyre) in contact with the ground moves backwards. The frictional force between the wheel and the ground is therefore directed forwards. It is this frictional force which moves the car forward.
When you apply the brakes, the braking mechanism does not apply any backward force. It just stops the spin of the wheels. The forward frictional force ceases. But the car will continue to move forward due to inertia. The wheel will then slide forward. Immediately, a backward frictional force is called into play and the car stops after traveling some distance. Now, consider the following questions:(1) If μ is the coefficient of friction between the road and the tyre of a car, the minimum time in which the car can cover a distance ‘s’, starting from rest is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ (e) independent of μThe correct option is not (b) but is (c). We have s = 0 + ½ at². Since the driving force is the frictional force μmg, acceleration a = μg. . Therefore s = ½ μg t² and hence t α 1/√μ. The correct option is (c).(2) If μ is the coefficient of friction between the road and the tyre of a car moving with a velocity ‘v’, the minimum distance in which it can be stopped is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ
(e) independent of μThe correct option is (b) which you can obtain using the equation, 0 = v² – 2as, on substituting for a = μg. Thus , stopping distance, s = v²/2μg.
Note this equation. It says that the stopping distance of any vehicle is directly proportional to the square of the speed of the vehicle.(3) A cube of mass 2kg is kept pressed against a vertical wall by applying a horizontal force of 60N. The coefficient of friction between the cube and the wall is 0.5. What is the frictional force between the cube and the wall?
(a) 60N
(b) 30N
(c) 19.6N
(d) 9.8N
(e) zeroDon’t be tempted to chose option (b) as the answer. Option (b) gives the maximum possible frictional force μR, where R is the normal reaction. Here friction is just sufficient to balance the weight of the cube (mg). So the correct option is (c).Certain simple questions may confuse you occasionally. Consider the following question:A wooden block of mass 3kg is resting on an inclined plane of inclination 30˚. If the coefficient of static friction between the block and the plane is 0.8, the frictional force between the block and the plane is (g=10m/s²)
(a) 20.8 N
(b) 24N
(c) 30N
(d) 12.5N
(e) 15NYou may be inclined to calculate the force as μmg cosθ, which yields the value20.8N, which is not the correct answer. Remember, μmg cosθ is the maximum possible frictional force (limiting friction). Since the block is resting on the inclined plane, the frictional force required is just enough to balance the component of the weight along the plane, which is mg sinθ = 15N. The correct option therefore is (e). Since frictional force is self adjusting (up to the limiting value), the value gets automatically adjusted to 15N in the present case.
Now consider the following MCQ:A heavy uniform chain is lying on a horizontal table with a certain portion hanging over one edge. If the coefficient of static friction between the chain and the table is 0.25, what is the maximum fraction of the length of the chain that can hang over the edge of the table?
(a) 10% (b) 20% (c) 33% (d) 45% (e) 50%If ‘L’ is the full length of the chain and ‘l’ the maximum length that can overhang, we have, mlg = μm(L-l)g where m is the mass per unit length (linear density) of the chain. The weight of the hanging portion tries to displace the chain while the frictional force between the table and the remaining portion tries to restore the chain. We have equated their magnitudes in the limiting case. From this we get l = 0.25(L-l) so that l/L = 0.2. The fraction written as percentage is 20%.An insect inside a hemispherical bowlHere is an interesting problem:An insect is trapped in a hemispherical bowl of radius R kept on a horizontal table. Up to what height can it crawl if the angle of friction is θ?
(a) Rcosθ
(b) Rsinθ
(c) R(1-cosθ)
(d) R(1-sinθ)
(e) RtanθAs the insect crawls upwards, the component of its weight along the surface of the bowl increases and the opposing frictional force also increases. When the insect reaches the maximum possible height, the frictional force is maximum and is equal to mg sinθ where θ is the angle of friction. This means that the inclination (with the horizontal) of the tangent to the hemispherical surface at the limiting position of the insect is equal to the angle of friction,θ. The angle between the vertical radius (of the hemispherical surface) and the radius drawn from the limiting position of the insect also is equal to θ, as ascertained from simple geometry.
Therefore, the maximum height attained by the insect is R-Rcosθ = R(1-cosθ).

Millikan’s oil drop experiment

Millikan’s famous oil drop experiment which brought him a Nobel Prize in Physics will always shine brightly in the memory of any one who loves Physics. Let us consider the following M.C.Q. related to Millikan’s experiment:
In an oil drop experiment (Millikan’s), an oil drop carrying a charge Q is held stationary between the plates by applying a potential difference of 400V. To keep another drop of half the radius stationary, the potential difference had to be increased to 600V. The charge on the second drop is
(a) Q/24
(b) Q/12
(c) Q/6
(d) 3Q/2
(e) 2Q/3
In the case of the first drop, we have
w = (400/d)×Q ……….. (1)
where ‘w’ is the apparent weight of the first drop and ‘d’ is the separation between the plates.
In the case of the second drop, the apparent weight is w/8 since the volume (and hence the mass) of the drop is reduced to one-eighth 
[Note that the volume is directly proportional to the cube of the radius]
Therefore we have,
w/8 = (600/d)×q ………... (2)
where ‘q’ is the charge on the second drop. Dividing eq(1) by eq(2), we obtain q = Q/12.
Consider now the following simple question which is meant for high lighting the quantum nature of electric charge:
An experimenter obtained the following values for the charge (in coulomb) on five different drops in Millikan’s oil drop experiment. Which one is the most unlikely value?
(a) 3.2×10^-19
(b) 4.8×10^-18
(c) 1.6×10^-17
(d) 8×10^-18
(e) 5.6×10^-19.
Remember that the minimum electric charge in nature is the electronic charge, which is equal to 1.6×10^-19 coulomb. All charges will be integral multiples of this minimum value. So, the unlikely value is 5.6×10^-19 coulomb [option (e)].
The number of electrons passing per second through any section of a conductor or a region of space to produce a given current is a constant and is independent of the voltage. 
Consider the following M.C.Q. which appeared in the I.I.T. screening test paper of 2002:
The potential difference applied to an X-ray tube is 15kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is
(a) 2×10¹⁶
(b) 5×10⁶
(c) 1×10¹⁷
(d) 4×10¹⁵
As the current is 3.2mA, the charge reaching the target per second is 3.2 millicoulomb. Since the electronic charge is 1.6×10^-19 coulomb, the number of electrons striking the target per second is (3.2×10^-3)/(1.6×10^-19) = 2×10¹⁶.
The accelerating voltage of 15kV is just a distraction in the question.

Questions on Surface Tension

God used beautiful mathematics in creating the world.

– P. A. M. Dirac

 
You might be remembering the expression for the excess of pressure ‘p’ inside a liquid drop: p = 2T/r where ‘T’ is the surface tension of the liquid and ‘r’ is the radius of the drop. The excess pressure is due to surface tension by which the free surface of a liquid behaves like a stretched membrane. The excess of pressure inside a bubble is 4T/r, which is twice that inside a drop because a bubble has two free surfaces (inner and outer).
Consider the following question which often finds a place in Medical and Engineering Entrance test papers:
Two soap bubbles A and B are obtained at the ends of a glass tube with a stop cock at its middle. The stop cock is closed initially and the bubble B is larger in size. If the stop cock is opened what happens to the bubbles?
(a) Nothing
(b) The size of A decreases and the size of B increases
(c) The size of A increases and the size of B decreases
(d) Both A and B will decrease in size
(e) Both A and B will increase in size.The excess pressure inside the smaller bubble A is greater and hence air from it will rush to B through the tube. The size of A will decrease further while the size of B will increase. The correct option therefore is (b).
You should note that the pressure on the concave side of a liquid surface is always greater by 2T/r. Consider the following question:
A tall metallic jar has a small hole of radius 0.07mm at its bottom. What is the approximate depth up to which the jar can be lowered vertically in water before any water penetrates in to it through the hole? (Surface tension of water = 0.073N/m)
(a) 7cm
(b) 11cm
(c) 16cm
(d) 21cm
(e) 27cmWater will begin to enter the jar through the hole when the hydrostatic pressure exceeds the allowed excess pressure of 2T/r on the concave side of the water surface at the hole. Therefore, the depth ‘h’ up to which the jar can be lowered safely is given by
hdg = 2T/r so that h = 2T/dgr = 2×0.073/(1000×9.8×0.00007) = 0.21m =21cm.

Let us consider one more question involving surface tension:
If T is the surface tension of soap solution, the amount of work done in increasing the radius of a soap bubble from r to 2r is
(a) (64πr²)T
(b) (32πr²)T
(c) (24πr²)T
(d) (16πr²)T
(e) (8πr²)T

Work done = Increase in surface area ×T = 2 [4π(2r)² - 4πr² )]T = (24πr²)T. So, the correct option is (c). Note that the bubble has two surfaces.

The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:Water rises in a capillary tube to a height ‘h’. Choose FALSE statement regarding capillary rise from the following:
(a) On the surface of Jupiter, height will be less than h.
(b) In a lift moving up with constant acceleration, height is less than h.
(c) On the surface of moon, height is more than h.
(d) In a lift moving down with constant acceleration, height is less than h.
(e) At the poles, height is less than that at equatorThe expression for surface tension (T) in terms of capillary rise in a capillary tube is
T = hrdg/2cosθ where ‘h’ is the capillary rise (or fall), ‘r’ is the radius of the tube, ‘d’ is the density of the liquid, ‘g’ is the acceleration due to gravity and ‘θ’ is the angle of contact. When ‘g’ is less, the capillary rise ‘h’ is more and vice versa. The value of ‘g’ is smaller in options (c) and (d) only so that the value of ‘h’ is greater in these two cases. So, option (d) is false. [Note that when a lift moves down with acceleration ‘a’, the effective value of acceleration due to gravity is (g-a)]

Introductory Remarks

This site is intended to deal with topics in Physics for helping students aspiring to become Engineering and Medical Professionals. I will discuss typical multiple choice questions (M.C.Q.) of the type you can expect to find in Medical and Engineering Entrance test papers of various institutions. Depending on the feedback from the viewers, other types of questions also will be included in due course. As Physics is not limited by regional boundaries, you will find that the topics discussed will be useful to you irrespective of your geographical location. An important case in point is the usefulness of the posts for students preparing for Graduate Record Examinations (GRE) and AP Physics B and C Examinations
The minimum I expect from the users of this site is some genuine interest in the subject. If you remember that the scientific and technological progress you find all around today is, to a very large extent, thanks to the hard work of physicists, you will find pleasure in learning the beautiful principles of Physics with enthusiasm. Finding answers to questions in Physics will then become a pleasing task for you. Be confident. Don’t be under the impression that Physics is a tough subject. It is a very interesting subject. You can definitely score high marks in Physics if you show genuine interest in it.
Many among you might be interested more in Physics than in Medicine or Engineering, but you may be forced to take up these professional courses by the pressure of circumstances. Don’t be discouraged. You will definitely find scope for applying the principles of Physics in almost all situations where you might be destined to work.

Your comments on the posts will be most welcome. Please do not hesitate to post crtical comments and suggestions