Thermodynamics – Two Multiple Choice Questions

Often simple questions on thermodynamics may cause unexpected confusion. The reason for the confusion is usually the lack of your understanding of the fundamentals. If you don’t have any confusion in respect of the following questions, well and good!

(1) A sample of an ideal gas initially having internal energy U₁ and pressure P₁ expands adiabatically and performs work W. Heat energy Q is then added to the gas at constant volume so that its pressure is increased to the initial value P₁. As a result of the above processes, the internal energy of the gas

(a) decreases by Q – W

(b) increases by Q – W

(c) decreases by Q

(d) increases by Q

(e) remains unchanged

Since the gas expands adiabatically and and outputs mechanical energy, the internal energy of the gas is decreased. [Remember that during adiabatic process there is no heat transfer between the gas and the surroundings]. On adding heat energy to the gas at constant volume the internal energy of the gas is increased. There is no work involved since the volume is constant (isochoric process).

Evidently the internal energy of the gas increases by Q – W.

[You will obtain the answer from the mathematical statement of the 1^st law of thermodynamics: ∆Q = ∆U + ∆W where ∆Q is the heat energy supplied to the system by the surroundings, ∆W is the work done by the system on the surroundings and ∆U is the increase in the internal energy of the system].

(2) In the given PV diagram, I is the initial state and F is the final state. The gas goes from I to F by (i) IAF (ii) IBF (iii) ICF. The heat absorbed by the gas is

(a) the same in all three processes

(b) the same in (i) and (ii)

(c) greater in (i) than in (ii)

(d) the same in (i) and (iii)

(e) greater in (iii) than in (ii)

This question appeared in Kerala engineering entrance 2009 question paper.

We have ∆Q = ∆U + ∆W

Since the same initial point (I) and the same final point (F) are given for all the three paths, the change in the internal energy (∆U) is the same for all the three processes. During the path IAF the net work done by the gas (∆W) is positive since the work done on the gas during the compression IA (given by the area under the line IA) is less than the work done by the gas during its expansion AF. No work is involved during the process IBF since the volume is constant (isochoric process). During the path ICF the net work done (∆W) by the gas is negative since the work done by the gas during its expansion IC is less than the work done on the gas during its compression CF.

Therefore, the heat absorbed by the gas (∆Q) is greater in (i) than in (ii).

You will find useful multiple choice questions with solution at apphysicsresources as well as at physicsplus.

Kerala Medical Entrance (KEAM) 2009 Questions on Optics

Physics questions in the Kerala Medical Entrance 2009 question paper were generally simple compared to those in the Kerala Engineering Entrance 2009 question paper. Today we will discuss the questions on optics included in the Medical Entrance question paper. Here are those three questions:

(1) A ray of light suffers minimum deviation in equilateral prism P. Additional prisms Q and R of identical shape and of same material as that of P are now combined as shown in figure. The ray will now suffer

(a) greater deviation

(b) no deviation

(c) same deviation as before

(d) total internal reflection

(e) smaller deviation

On combining additional prisms Q and R with P, we obtain a portion of an equilateral prism of the same material. Since the incident ray is such that the deviation produced by the prism P is minimum, the refracted ray will be parallel to the base of the prism P and hence it will pass parallel to the base of the combined prism, suffering the same deviation (minimum deviation) as before. The correct option is (c).

(2) When light is scattered by atmospheric atoms and molecules, the amount of scattering of light of wave length 440 nm is A. The amount of scattering for light of wave length 660 nm is

(a) (4/9) A

(b) 2.25 A

(c) 1.5 A

(d) 0.66 A

(e) A/5

The answer for this question is based on Rayleigh’s scattering formula which says that the amount of light scattered is inversely proportional to the fourth power of the wave length of light. Therefore, in the two cases we have respectively,

A α 1/440⁴ and

x α 1/660⁴ where x is the amount of scattering for light of wave length 660 nm.

Dividing, A/x = 660⁴/ 440⁴ = (3/2)⁴ = 81/16 = 5 nearly.

Therefore, x = A/5.

(3) In the measurement of the angle of a prism using a spectrometer, the reading of first reflected image are Vernier I: 320° 40' Vernier II: 140° 30' and those of the second reflected image are Vernier I: 80° 38'; Vernier II: 260° 24'. Then the angle of the prism is

(a) 59° 58'

(b) 59° 56'

(c) 60° 2'

(d) 60° 4'

(e) 60° 0'

When a parallel beam of light falls symmetrically on the two faces of the prism, the angle between the rays reflected from these faces is 2A where A is the angle of the prism. If you have a clear understanding of the experimental determination of the angle of the prism using this method, you will definitely be able to find the answer to the above question since you will know that the reading of Vernier I was first inecreased from 320° 40' to 360° on rotating the telescope to view the reflected image from the second face. The 360° mark is the same as 0° mark. So after reaching the zero reading the telescope has rotated through 80° 38'.

The difference between Vernier I readings is (360° - 320° 40') + 80° 38' = 39° 20'+ 80°' = 119° 58'. 38

The difference between Vernier II readings is 260° 24' - 140° 30' = 119° 54'.

The mean of the difference between the readings is 119° 56'.

Therefore, the angle of the prism is (119° 56')/2 = 59° 58'

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions on Rotational Motion

Four questions related to rotational motion were included in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper. Here are those questions with answers:

(1) Four identical thin rods each of mass M and length L, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

(1) (²/₃) ML²

(2) (¹³/₃) ML²

(3) (¹/₈) ML²

(4) (⁴/₃) ML²

The moment of inertia of a thin rod about an axis through the mid point of the rod and perpendicular to its length is ML²/12. The axis chosen in the problem is a parallel axis at a distance L/2. The moment of inertia of the rod bout this parallel axis is ML²/12 + M (L/2)² = ML²/3

[We have used the parallel axis theorem: I = I_CM + Ma² where I_CM is the the moment of inertia about the axis through the centre of mass and I is the moment of inertia about the parallel axis at separation a].

Note that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. The answer therefore is (⁴/₃) ML²

(2) A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis perpendicular to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

(1) ωM /(M + 2m)

(2) ω(M + 2m) /M

(3) ωM /(M + m)

(4) ω(M – 2m) /(M + 2m)

This question is often seen in entrance question papers. Its answer is based on the law of conservation of angular momentum:

I₁ω₁ = I₂ω₂ where I₁ and I₂ are the initial and final moments of inertia and ω₁ and ω₂ are the initial and final angular velocities respectively.

Therefore we have

MR²ω = (MR² + 2 mR²) ω₂

This gives ω₂ = ωM /(M + 2m)

(3) If F is the force acting on a particle having position vector r and τ be the torque of this force about the origin, then

(1) r.τ > 0 and F.τ <>

(2) r.τ = 0 and F.τ = 0

(3) r.τ = 0 and F.τ ≠ 0

(4) r.τ ≠ 0 and F.τ = 0

Many students are liable to skip the above question at the first glance itself. But this is a very simple question if you remember that the torque τ is the ‘cross product’ (vector product) of r and F:

τ = r×F

Thus the torque vector is perpendicular to both r and F. Therefore the ‘dot product’ (scalar product) of r and τ as well as that of F and τ is zero. The correct option is (2).

(4) Two bodies of mass 1 kg and 3 kg have position vectors i + 2 j + k and –3 i – 2 j + k respectively. The centre of mass of this system has a position vector

(1) –2 i – j + k

(2) 2 i – j – 2 k

(3) – i + j + k

(4) –2 i + 2 k

If two point masses m₁ and m₂ have position vectors r₁ and r₂ their centre of mass has position vector R given by

R = (m₁ r₁ + m₂ r₂) /( m₁ + m₂)

Substituting appropriate values,

R = [1(i + 2 j + k) + 3(–3 i – 2 j + k)] /(1+3)

Or, R = (– 8 i – 4 j + 4 k) /4 = –2 i – j + k

Option (1) is correct.

You will find all the posts related to rotational motion on this site by clicking on the label ‘rotation’ below this post.

Additional questions with answers can be seen at physicsplus.

Kerala Medical Entrance (KEAM) 2008 Questions on Nuclear Physics

Try not to be a person of success, but rather a person of virtue.

– Albert Einstein

Here are the two questions from nuclear physics which were included in KEAM (Medical) 2008 question paper:

(1) If the mass defect of ₈O¹⁶ nucleus is 0.128 amu, then the binding energy per nucleon of oxygen is

(a) 8.2 MeV

(b) 7.45 MeV

(c) 7.3 MeV

(d) 7.1 MeV

(e) 8.15 MeV

One atomic mass unit (amu) is equivalent to 931 MeV. Therefore, the total binding energy of the ₈O¹⁶ nucleus is 0.128×931 MeV.

Since there are 16 nucleons in the ₈O¹⁶ nucleus, the binding energy per nucleon of oxygen is (0.128×931)/16 = 7.45 MeV, very nearly.

(2) Two radioactive samples have decay constants 15x and 3x. If they have the same number of nuclei initially, the ratio of number of nuclei after a time 1/6x is

(a) 1/e

(b) e/2

(c) 1/e⁴

(d) 2e/3

(e) 1/e²

The number N of nuclei at time t is given by

N = N₀e^–λt where N₀ is the initial number, e is the base of natural logarithms, and λ is the decay constant.

The required ratio is (N₀e^–15x/6x)/ (N₀e^–3x/6x) = e^–2.5/ e^–0.5 = e^–2 = 1/e²

Three questions from nuclear physics were included in the physics question paper of Kerala Engineering Entrance (KEAM) 2008 examination. You will find those questions with solution here