IIT-JEE 2008 Linked Comprehension Type Multiple Choice Questions Inviolving Thermal Physics and Properties of Fluids

Today we will discuss three linked comprehension type multiple choice questions which appeared in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

A small spherical monoatomic ideal gas bubble (γ = 5/3) is trapped inside a liquid of density ρ_ℓ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T₀, the height of the liquid is H and the atmospheric pressure is P₀ (Neglect surface tension).

1. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it:

(A) Only the force of gravity

(B) The force due to gravity and the force due to the pressure of the liquid

(C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid

(D) The force due to gravity and the force due to viscosity of the liquid

The force due to the pressure of the liquid is the buoyancy force. Therefore, besides the buoyancy force the forces acting on the bubble are the force due to gravity and the force due to viscosity of the liquid [Option (D)].

2. When the gas bubble is at height y from the bottom, its temperature is

(A) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^2/5

(B) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^2/5

(C) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^3/5

(D) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^3/5

Since there is no heat exchange, the process is adiabatic for which

T P^{[(1- γ)/ γ]} = constant where γ is the ratio of specific heats.

[Many of you might be remembering this as T ^γ P^{(1- γ)} = constant].

At the bottom pressure = P₀ + ρ_ℓ gH and temperature = T₀

At height y from the bottom pressure = P₀ + ρ_ℓ g(H –y) and temperature = T (let us say)

Therefore, T₀(P₀ + ρ_ℓ gH)^–2/5 = T[P₀ + ρ_ℓ g(H –y)]^–2/5 so that

T = T₀[(P₀ + ρ_ℓ gH) /{P₀ + ρ_ℓ g(H –y)}]^–2/5

Or, T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

So option (B) is correct.

3. The buoyancy force on the gas bubble is (Assume R is the universal gas constant)

(A) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^2/5/(P₀ + ρ_ℓ gy) ^7/5]

(B) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

(C) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^3/5/(P₀ + ρ_ℓ gy) ^8/5]

(D) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^3/5{P₀ + ρ_ℓ g(H –y)}^2/5]

Force of buoyancy, F= Vρ_ℓ g

But V = nRT/P so that F = nRTρ_ℓ g /P

Here P = P₀ + ρ_ℓ g(H –y) and T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

Substituting, F = nRρ_ℓ g T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5/ [P₀ + ρ_ℓ g(H –y)]

Or, F = ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

So option (B) is correct.

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Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2009 (KEAM 2009), Kerala

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 2009-10.
(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS
(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)
(c) Veterinary BVSc. & AH
(d) Engineering B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]
(e) Architecture B.Arch.

Dates of Exam:
Engineering Entrance Examination (For Engineering courses except Architecture):
20.04.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.
21.04.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses):
22.04.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.
23.04.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Sale of Application will commence on : 27-01-2009
Last Date for Submission of Application: 26-02-2009
You will find complete details at http://www.cee-kerala.org/

You will find many multiple choice questions of the type expected to appear in KEAM 2009 on this site. If you want to go through earlier KEAM questions only, type in ‘Kerala’ in the search box at the top left and hit the enter key.

Kerala Engineering Entrance 2008 Questions on Communication Systems

Four questions on communication systems appeared in the KEAM (Engineering) 2008 test paper. Generally you will be able to answer all the questions in this section within the stipulated time. It will be a good idea to answer questions from simple sections like this before proceeding to attempt difficult and time consuming ones. Here are the questions with their solutions:

(1) A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively

(a) 2512 kHz and 2508 kHz

(b) 2522 kHz and 2488 kHz

(c) 2502 kHz and 2498 kHz

(d) 2522 kHz and 2498 kHz

(e) 2512 kHz and 2488 kHz

This question should have been stated like this:

A carrier wave wave of frequency 2.51 Mz is amplitude modulated with a signal wave of frequency 12 kHz. The upper and lower side band frequencies are respectively

(a) 2512 kHz and 2508 kHz

(b) 2522 kHz and 2488 kHz

(c) 2502 kHz and 2498 kHz

(d) 2522 kHz and 2498 kHz

(e) 2512 kHz and 2488 kHz

Since there is just one frequency (12 kHz) in the modulating signal, there is just one frequency in the upper side band. The upper side frequency is 2.51 MHz + 12 kHz = 2510 kHz + 12 kHz = 2522 kHz.

Similarly the lower side frequency is 2.51 MHz – 12 kHz = 2510 kHz – 12 kHz = 2498 kHz [Option (d)].

(2) Which of the following statements is wrong?

(a) Ground wave propagation can be sustained at frequencies 500 kHz to 1500 kHz

(b) Satellite communication is useful for the frequencies above 30 MHz

(c) Sky wave propagation is useful in the range of 30 to 40 MHz

(d) Space wave propagation takes place through tropospheric space

(e) The phenomenon involved in sky wave propagation is total internal reflection

The frequency range 500 kHz to 1500 kHz is the medium wave band for AM sound broadcast which you know is based on ground wave propagation. So option (a) is correct. Option (b) is correct since frequencies above 30 MHz will not be reflected by the ionosphere. For the same reason option (c) is wrong.

Since the questions you get in this test are single answer type your answer option is (c).

(3) The principle used in the transmission of signals through an optical fibre is

(a) total internal reflection

(b) reflection

(c) refraction

(d) dispersion

(e) interference

In an optical fibre the refractive index is decreased as the ray proceeds radially outwards from the centre and hence it undergoes total internal reflection[Option (a)].

(4) In satellite communication

1. the frequency used lies between 5 MHz and 10 MHz

2. the uplink and downlink frequencies are different

3. the orbit of geostationary satellite lies in the equatorial plane at an inclination of 0º.

On the above statements

(a) only 2 and 3 are true

(b) all are true

(c) only 2 is true

(d) only 1 and 2 are true

(e) only 1 and 3 are true

Statement 1 is false since 5 MHz and 10 MHz will be reflected by the ionosphere. Statements 2 and 3 are true in the case of geostationary satellite. So the correct option is (a).

IIT-JEE 2008 Straight Objective Type (Single Answer Multiple Choice) Questions on Conservation of Energy

The following questions involving the law of conservation of energy are simple even though they may appear to be not so at the first reading:

(1) A block (B) is attached to two unstretched springs S₁ and S₂ with spring constants k and 4 k respectively (see fig.1). The other ends are attached to identical supports M₁ and M₂ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (fig. 2) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is

(A) 4

(B) 2

(C) ½

(d) ¼

On displacing the block B towards wall 1, spring S₁ gets compressed through x and acquires potential energy ½ kx². When the spring S₁ springs back to its original unstretched condition, it pushes the block B towards wall 2 and compresses the spring S₂ through y. In this compressed condition of spring S₂ the entire kinetic energy of the block B is transferred to spring S₂. Since the spring S₁ is free to move with the block B, it is unstretched and hence we have

½ kx² = ½ (4k) y²

This gives y/x = ½

(2) A bob of mass m is suspended by a massless string of lengh L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

(A) θ = π/4

(B) π/4 < θ < π/2

(C) π/2 < θ < 3π/4

(D) 3π/4 < θ < π

The sum of the kinetic energy and potential energy of the of the bob in the displaced position must be equal to the kinetic energy at the position A. Therefore we have

½ M (V/2)² + MgL(1 – cos θ) = ½ MV²

The critical velocity (for just tracing the vertical circle) V = √(5gR) = √(5gL)

Substituting this value we obtain

gL(1 – cos θ) = 15gL/8 so that cos θ = – 7/8

Therefore 3π/4 < θ < π.

IIT-JEE 2008 Reasoning Type Question on Rotation

The following question appeared in IIT-JEE 2008 question paper under Reasoning Type Questions.

STATEMENT -1

Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.

and

STATEMENT -2

By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT -2 is a correct explanation for Statement-1

(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for STATEMENT -1

(C) STATEMENT -1 is True, STATEMENT -2 is False

(D) STATEMENT -1 is False, STATEMENT -2 is True

The acceleration (a) of a body of mass M rolling down an inclined plane is given by

a = g sinθ/[1 + (I/MR²)]

where θ is the angle of the plane (with respect to the horizontal), I is the moment of inertia (about the axis of rolling) and R is the radius of the body.

Since the moments of inertia of solid cylinder and hollow cylinder are respectively MR²/2 and MR² the acceleration a is greater for the solid cylinder. Therefore, the solid cylinder will reach the bottom of the inclined plane first.

Since the cylinders have the same mass and are at the same height they have the same initial gravitational potential energy Mgh. This potential energy gets converted into translational and rotational kinetic energies (obeying the law of conservation of energy) when the cylinders roll down the incline and the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

The correct option is (D).

Let us consider an ordinary type of multiple choice question now:

A, B and C are three equal point masses (m each) rigidly connected by massless rods of length L forming an equilateral triangle as shown in the adjoining figure. The system is first rotated with constant angular velocity ω about an axis perpendicular to the plane of the triangle and passing through A. Next it is rotated with the same constant angular velocity ω about the side AB of the triangle. If K₁ and K₂ are the kinetic energies of the system in the first and the second cases respectively, the ratio K₁/K₂ is

(a) 2/3

(b) 4/3

(c) 8/3

(d) 10/3

(e) 16/3

The rotational kinetic energy (K) is given by

K = ½ Iω²

Since the angular velocity is the same in the two cases, the ratio of kinetic energies must be equal to the ratio of moments of inertia.

Thereore, K₁/K₂ = I₁/I₂ = 2mL²/m(Lsin60º)²

[Note that in the second case the moment of inertia of the system is due to the single mass at C which is at distance Lsin60º from the axis AB].

Thus K₁/K₂ = 2/(√3/2)² = 8/3.

AIPMT 2008 Question on Feed back Amplifier

The following question on negative feed back amplifier is simple even though Plus Two students will normally be unprepared to answer it:

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be

(1) 10

(2) 1.25

(3) 100

(4) 90

If the voltage gain without feed back is A_v and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (A_fb) with feed back is given by

A_fb = A_v/(1– βA_v)

In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by

A_fb = A_v/(1+ βA_v)

Since A_fb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,

10 = A_v/(1+ 0.09A_v)

This gives A_v = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:

When negative feed back is applied in a transistor amplifier

(1) its voltage gain is decreased

(2) its band width is decreased

(3) distortion produced by the amplifier is decreased

(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.

All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) to be conducted on 26-4-2009 are being distributed from 5.12.2008 and will continue till 5.1.2009. Candidates can apply for AIEEE 2009 either on the prescribed Application Form or make application ‘Online’. Visit the site http://aieee.nic.in immediately for details. Apply for the exam without delay.

You will find many old AIEEE questions (with solution) on this site. You can access all of them by typing ‘AIEEE’ in the search box at the top left of this page and clicking on the adjacent ‘search blog’ box.

Electromagnetic Induction- Two Multiple Choice Questions

You should be able to find the answers for the following questions without difficulty. For each question you should take about one minute at the most (if you have understood basic principles in this section thoroughly).

(1) A long straight conductor carries a current that increases at a constant rate. The current induced in the circular conducting loop placed near the straight conductor (fig.) is

(a) zero

(b) constant and in the clockwise direction

(c) constant and in the anticlockwise direction

(d) increasing and in the clockwise direction

(e) increasing and in the anticlockwise direction

A current flows in the circular loop because of the increasing magnetic flux linked with it. This increase in the magnetic flux is to be opposed in accordance with Lenz’s law. The magnetic field lines produced by the straight conductor are directed normally into the plane of the circular coil (away from the reader). Therefore, the direction of the magnetic field lines produced by the induced current in the circular coil must be directed normal to the plane of the circular coil and towards the reader. The induced current should therefore flow in the anticlockwise sense.

The magnitude of the induced current is constant since the rate of increase of flux is constant. The correct option is (c).

(2) In the circuit shown a coil of inductance 0.5 H and negligible resistance is connected to a 12 V battery of negligible internal resistance using a resistive network. If I₁ is the current delivered by the battery immediately after closing the switch and I₂ is the current (delivered by the battery) long time after closing the switch,

(a) I₁ = 0.5 A and I₂ = 0.6 A

(b) I₁ = 0.6 A and I₂ = 0.6 A

(c) I₁ = 0.5 A and I₂ = 0.5 A

(d) I₁ = 0 A and I₂ = 0.5 A

(e) I₁ = 0.6 A and I₂ = 0.5 A

The current through the inductance cannot rise abruptly on closing the switch (because of the time constant L/R of the LR circuit). At the moment of switching the circuit on, the circuit behaves as though the inductance branch is absent. The initial current is therefore given by

I₁ = 12 V/ (16+8) Ω = 0.5 A

The current through the inductance rises exponentially with time t from zero to the final steady value I [in accordance with the equation I = I₀(1– e^–Rt/L) with usual notations]. After a long time, the current through the inductance becomes steady and there is no voltage drop across the inductance. The inductance is in effect short circuited and the final steady current is given by

I₂ = 12 v/(16+4) Ω = 0.6 A

[Note that the parallel combined value of 8 Ω and 8 Ω is 4 Ω]

The correct option is (a).

You will find many useful posts on electromagnetic induction on this site. You can access them by clicking on the label ‘electromagnetic induction’ below this post.

Joint Entrance Examination for Admission to IITs and other Institutions- (IIT-JEE 2009)

The Joint Entrance (2009) Examination for Admission to IITs and other Institutions (IIT-JEE 2009) will be held on April 12, 2009 (Sunday) as per the following schedule:

09:00 – 12:00 hrs Paper – 1
14:00 – 17:00 hrs Paper – 2
Application materials of the Joint Entrance Examination 2009 will be issued with effect from 19th November, 2008. On-line submission of Application will also commence on the same day.
Application materials can be purchased from designated branches of Banks and all IITs between 19.11.2008 and 24.12.2008 by paying Rs. 500/- in case of SC/ST/PD/Female candidates, Rs.1000/- in case of male general/OBC candidates and US$ 100 in the case of candidates appearing in Dubai centre.
The last date for postal request of application materials is 16-12-2008.
The last date for receipt of the completed application at the IITs is 24th December, 2008.
Online Submission of Application: This facility will be available between 19.11.2008 and 24.12.2008 between 8:00 hrs and 17:00 hrs. through the JEE websites of the different IITs.
The JEE websites of the different IITs are given below:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Delhi: http://jee.iitd.ac.in
IIT Guwahati: http://www.iitg.ac.in/jee
IIT Kanpur: http://www.iitk.ac.in/jee
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Madras: http://jee.iitm.ac.in
IIT Roorkee: http://www.iitr.ac.in/jee
Visit one of these web sites (http://jee.iitm.ac.in) for more details.