IIT-JEE Multiple Correct Answer Type Questions on Thermodynamics

"Hate the sin, love the sinner."
–Mahatma Gandhi

Today we will discuss two multiple correct answer type questions on thermodynamics. These were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(C) work done during the path A → B → C is zero

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since a curve representing an isothermal in a PV diagram must be a hyperbola and not an arc of a circle.

We have PV = nRT with usual notations. The value of V is smaller at D compared to that at B where as the value of P is the same. The product PV = nRT is therefore smaller at D. This means that the temperature at D is less than that at B and hence statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operations ABCDA is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

So the correct options are (B) and (D).

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(C) C_P/C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas.

In the case of ideal gases C_P – C_V = R, the universal gas constant. Therefore statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas compared to a mono-atonic ideal gas. Therefore, C_P + C_V is larger for a diatomic ideal gas. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For mono-atomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Obviously statement (C) is wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a mono-atomic ideal gas.

Therefore, options (B) and (D) are correct.

You will find a useful post on thermodynamics here.

Kerala Engineering Entrance 2007 Questions on Communication Systems

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Questions on communication systems at the level expected from you are generally simple and interesting. Today we will discuss three questions (MCQ) on communication systems which appeared in Kerala Engineering Entrance 2007 question paper.

(1) The time variations of signals are given as in (A), (B) and (C).

Point out true statement from the following

(a) A, B and C are analog signals

(b) A and B are analog, but C is digital signal

(c) A and C are digital, but B is analog signal

(d) A and C are analog, but B is digital signal

(e) A, B and C are digital signals (e)

Digital signals will have two discrete levels only, corresponding to the zero level and one level as shown in graph (B). Analog signals have various instantaneous values, as shown in graphs (A) and (C). The correct option therefore is (d).

(2) A photo detector used to detect the wave length of 1700 nm, has energy gap of about

(a) 0.073 eV

(b) 1.2 eV

(c) 7.3 eV

(d) 1.16 eV

(e) 0.73 eV

The energy gap of a photo detector should be equal to or less than the energy of the photon which it is intended to detect. The product of the energy of a photon in electron volt and the wave length in Angstrom is 12400. Therefore, the energy of the 1700 nm photon is 12400/ 17000 electron volt = 0.73 eV. So, the correct option is (e). Option (a) is not suitable since a semiconductor with a gap as low as 0.073 eV (if at all available) will be unreliable due to the breaking of bonds by thermal excitation.

[You can use the expression hc/λ for calculating the energy of the photon in joule and then convert it into electron volts by dividing it by the electronic charge of 1.6×10^–19 joule. But it will be more difficult and time consuming].

(3) The optical fibres have an inner core of refractive index n₁ and a cladding of refractive index n₂ such that

(a) n₁ = n₂ (b) n₁ ≤ n₂ (c) n₁ < n₂ (d) n₁ > n₂ (e) n₁ ≥ n₂

The cladding has to be rarer than the core so as to totally reflect the beam of light entering the optical fibre. So, the correct option is (d).

Apply for Kerala Engineering Agricultural Medical Entrance Examinations 2012 (KEAM 2012)

"Strength does not come from physical capacity. It comes from an indomitable will."

– Mahatma Gandhi

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2012-13.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BSc. Hons. (Forestry)

(c) Veterinary BVSc. & AH

(d) Fisheries BFSc.

(e) Engineering B.Tech. [including B.Tech. (Agri. Engg.), B.Tech. (Food Engg.) Courses

under the Kerala Agricultural University and B.Tech. (Dairy Science & Tech.) under the Kerala Veterinary & Animal Sciences University]

(f) Architecture B.Arch.

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

23.04.2012 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

24.04.2012 Tuesday10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture, Veterinary and Fisheries courses)

25.04.2012 Wenesday10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

26.04.2012 Thursday10.00 A.M. to 12.30 P.M. Paper-II: Biology.

All candidates are required to apply online through the website www.cee.kerala.gov.in for the Entrance Examination for admission to Medical, Agriculture, Veterinary, Fsheries, Engineering and Architecture Courses.

All details regarding KEAM 2012 can be obtained from the website http://www.cee-kerala.org

Hurry up! The Print out of your Application and other relevant documents to be submitted with the Application, are to be sent to the Commissioner for Entrance Examinations so as to reach him before 5 p.m. on 15.02.2012 (Wednesday).

* * * * * * * * * * * * * * *

To access earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the label ‘Kerala’ below this post.

Optics: Karnataka CET Questions (MCQ) on Lenses

Today we will discuss two questions (MCQ) involving lenses. The first one appeared in Karnataka CET 2011 question paper and the second one appeared in Karnataka CET 2010 question paper.

(1) Two thin lenses have a combined power of + 9 D. When they are separated by a distance of 20 cm, their equivalent power becomes + 27/5 D. Their individual powers (in dioptre) are ……..

(1) 4, 5

(2) 3, 6

(3) 2, 7

(4) 1, 8

We have P₁ + P₂ = 9 where P₁ and P₂ are the powers of the two lenses.

[Note that combined power means 1/f₁ + 1/f ₂ which is the effective power when the lenses are in contact. Power is the reciprocal of focal length and when the lenses are in contact, we have 1/F = 1/f₁ + 1/f ₂ where f₁ and f₂ are the focal lengths of the lenses and F is the combined focal length].

When the lenses are separate by a distance d, the power becomes P₁ + P₂ – d P₁P₂

[This follows from the expression, 1/F = 1/f₁ + 1/f ₂ – d/f₁f₂].

Therefore, we have

P₁ + P₂ – d P₁P₂ = 27/5

Or, 9 – (0.2× P₁P₂) = 27/5

[You have to substitute the separation in metre since the power is the reciprocal of focal length in metre].

From the above equation, 18/5 = 0.2× P₁P₂

Or, P₁P₂ = 18

Evidently the values given in option (2) satisfy the above condition so that we have P₁ = 3 and P₂ = 6

(2) The focal length of a plano-convex lens is ‘f’ and its refractive index is 1.5. It is kept over a glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes 2f. Then the refractive index of the liquid is ……

(1) 1.5

(2) 2

(3) 1.25

(4) 1.33

In the case of a lens placed in air or vacuum, lens maker’s equation is

1/f = (n – 1)(1/R₁ – 1/R₂) where n is the refractive index of the lens.

[General form of lens maker’s equation is 1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are the radii of curvature of its faces, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed].

In the case of the plano-convex lens, the equation becomes

1/f = (n – 1)(1/R₁ – 1/∞) = (n – 1)(1/R₁)

Since n = 1.5, R₁ = 0.5f ……..(i)

The focal length of the combination (F) of the plano convex lens (of focal length f) and the plano-concave liquid lens (of focal length f_ℓ)is given by

1/F = 1/f + 1/f_ℓ

Therefore, 1/2f = 1/f + 1/f_ℓ

This gives f_ℓ = –2f………(ii)

But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (n_ℓ) of the material (liquid) as

1/f_ℓ = (n_ℓ – 1)(–1/R₁ – 1/∞) = (n_ℓ – 1)(– 1/R₁)

[Note that the radius of the concave surface, which the liquid lens presents to the incident light, has to be negative in accordance with the Cartesian sign convention. Click on the label ‘Cartesian sign convention’ below this post to learn more about this convention].

Therefore, 1/(–2f) = – (n_ℓ – 1)/0.5f , on substituting for R₁ and f_ℓ from equations (i) and (ii).

This gives 2 n_ℓ = 2.5 so that n_ℓ = 1.25

Apply for All India Pre-Medical / Pre-Dental Entrance Examination-2012 (AIPMT 2012)

Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination-2012 for admission to 15% of the merit positions for the Medical/Dental Courses of India.

The exams will be conducted as per the following schedule:

1. Preliminary Examination …. 1^st April, 2012 (Sunday) 10 AM to 1 PM

2. Final Examination ………… 13^th May, 2012 (Sunday) 10 AM to 1 PM

Both examinations will be objective type.

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination 2012 (AIPMT 2012) only online.

All details can be obtained from the official site www.aipmt.nic.in

You can find many AIPMT previous questions with solution on this site. Click on the label ‘AIPMT’ below this post or try a search for ‘AIPMT’ using the search box provided on this page, to access all relevant posts.

Two Questions (MCQ) on Rotation of Rigid Bodies

“Learn from yesterday, live for today, hope for tomorrow. The important thing is to not stop questioning.”

– Albert Einstein

Today we will discuss a couple of questions involving rigid body rotation. The first question may appear to be familiar to you as it has appeared in various entrance exam question papers. The second one is not so common but you must certainly work it out yourself before going through the solution given here.

(1) A thin straight uniform rod AB (Fig.) of length L and mass M, held vertically with the end A on horizontal floor, is released from rest and is allowed to fall. Assuming that the end A (of the rod) on the floor does not slip, what will be the linear velocity of the end B when it strikes the floor?

(a) √(3L/g)

(b) √(2L/g)

(c) √(3gL)

(d) √(2g/L)

(e) √(3g/L)

When the rod falls, its gravitational potential energy gets converted into rotational kinetic energy. Therefore we have

MgL/2 = ½ I ω² …………….. (i)

where I is the moment of inertia of the rod about a normal axis passing through its end and ω is the angular velocity of the rod when it strikes the floor.

[Note that initially the centre of gravity of the rod is at a height L/2 and that’s why the initial potential energy is MgL/2]

The moment of inertia of the rod about the normal axis through its end is given by

I = ML²/3

[The moment of inertia of the rod about an axis throgh its centre and perpenicular to its length is ML²/12. On applying parallel axes theorem, the moment of inertia about a parallel axis through the end is ML²/12 + M (L/2)² = ML²/3]

Substituting for I in Eq.(i), we have

MgL/2 = ½ (ML²/3) ω²

Therefore ω = √(3g/L)

The linear velocity v of the end B of the rod is given by

v = ωL = √(3gL)

(2) A thin straight uniform rod AB (Fig.) of length L and mass M is pivoted at point O, distant L/4 from the end A. The friction at the hinge is negligible and the rod can rotate freely (about the hinge) in a vertical plane. Initially the rod is held horizontally and is released from rest. The linear velocity of the end B of the rod when it momentarily attains vertical position A₁B₁ is

(a) √(3L/4g)

(b) √(27gL/14)

(c) √(18gL/7)

(d) √(3g/4L)

(e) √(2gL)

The gravitational potential energy of the rod gets converted into rotational kinetic energy when the rod is release from its horizontal position. The centre of gravity of the rod is lowered through a distance L/4 and hence the decrease in the gravitational potential energy is MgL/4. The gain in rotational kinetic energy is ½ I ω² where I is the moment of inertia of the rod about the axis of rotation passing through the point O and ω is the angular velocity of the rod when it attains the vertical position. Therefore, from the law of conservation of energy we have

MgL/4 = ½ I ω² …………….. (i)

Here I = ML²/12 + M (L/4)² = M [(L²/12) + (L²/16)]

Substituting in Eq.(i) we have

g = [(L/6) + (L/8)] ω² = (7L/24)ω²

Or, ω = √(24g/7L)

Since the end B of the rod is at distance 3L/4 from the axis of rotation, the linear velocity v of the end B of the rod is given by

v = ω×(3L/4) = [√(24g/7L)] (3L/4)

Or, v = √[(24g×9L²) /(7L×16)] = √(27gL/14)

Apply for All India Engineering Entrance Examination 2012 (AIEEE 2012)

It’s time to apply for AIEEE 2012. Applications for the examination are to be submitted online only, from 15-11-2011 to 31-12-2011.

There are two modes of examination viz., offline examination (Pen & Paper Mode) and online examination (Computer Based Test) with dates as given below:

Offline Examination - 29th April, 2012

Online Examination- 7th May, 2012 to 26th May, 2012

You may visit the site http://aieee.nic.in for details and information updates.

You will find many old AIEEE questions (with solution) on this blog. You can access all of them by trying a search for ‘AIEEE’ making use of the search box on this page or by clicking on the label ‘AIEEE’ below this post. Make use of the ‘older posts’ button to navigate through older posts related to AIEEE.

IIT-JEE 2011 and IIT-JEE 2010 Questions on Doppler Effect

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."
– Sir Isaac Newton

Questions on Doppler Effect were discussed on this site earlier. You can access them by clicking on the label ‘Doppler effect’ below this post or by trying a search for ‘Doppler effect’ using the search box provided on this page. Questions on Doppler effect appear frequently in Entrance examination question papers. Today we will discuss two questions in this section Here is the first question which appeared in IIT-JEE 2011 question paper as a single answer type multiple choice question:

(1) A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

(A) 8.50 kHz

(B) 8.25 kHz

(C) 7.75 kHz

(D) 7.50 kH

The general expression for the apparent frequency (n’) produced due to Doppler effect is

n’ = n(v+w–v_L)/(v+w–v_S) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘v_L’ is the velocity of the listener, ‘v_S’ is the velocity of the source of sound and ‘n’ is the actual frequency of the sound emitted by the source. Note that all the velocities in this expression are in the same direction and the source is behind the listener. In other words, the listener is moving away from the source and the source is moving towards the listener.

[It will be helpful to remember that if the source moves towards the listener or the listener moves towards the source, the apparent frequency increases. If they move away, the apparent frequency decreases].

In the present case the source of sound which produces the Doppler effect is the reflected image of the siren. (The car driver will not detect any change in the frequency of the direct sound from the siren since he is moving along with the siren). The reflected image (source) of the siren moves towards the car driver with a speed of 36 km/hr and the car driver (listener) moves towards the reflected image (source) with the same speed.

Thus in the expression for apparent frequency we have ro substitute

n = 8 kHz, v_L = – 36 km/hr = 36×(5/18) ms^–1 = –10 ms^–1, w = 0, v_S = + 10 ms^–1.

Therefore, n’ = 8×(320 + 10)/(320 – 10) = 8×(33/31) kHz = 8.5 kHz.

The following question appeared in IIT-JEE 2010 question paper as an integer type question (in which the answer is a single-digit integer, ranging from 0 to 9):

(2) A stationary source is emitting sound at a fixed frequency f₀, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f₀. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms^–1.

This is similar to question no.1 above. The difference between the frquencies of the sound reflected from the cars is given by

n₁’ – n₂’ = [f₀(v+v₁)/(v–v₁)] – [f₀(v+v₂)/(v–v₂)] where v₁ and v₂ are the speeds of the two cars.

Therefore (n₁’ – n₂’)/f₀ = [(v+v₁)/(v–v₁)] – [(v+v₂)/(v–v₂)]

The quantity on the left hand side of the above equation is 1.2/100 as given in the question.

Therefore, 0.012 = [(v+v₁) (v–v₂) – (v+v₂) (v–v₁)] / [(v–v₁)(v–v₂)]

Or, 0.012 = (2 vv₁ – 2 vv₂)/ [(v–v₁)(v–v₂)] = 2v(v₁ –v₂)/v² since (v–v₁) ≈ (v–v₂) ≈ v

[The cars are moving at speeds much smaller than the speed of sound].

Therefore we have

0.012 = 2(v₁ –v₂)/v from which (v₁ –v₂) = 0.006×330 ms^–1

The difference in the speeds of the cars in km per hour is 0.006×330×(18/5) = 7.128

The answer, which is the nearest integer, is 7.