Kerala Engineering Entrance 2008 and other Multiple Choice Questions (MCQ) on Work and Energy

Setting an example is not the main means of influencing others; it is the only means.

– Albert Einstein

Let us discuss a few multiple choice questions on work and energy.

(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms^–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?

(a) 10 J

(b) –10 J

(c) – 1 J

(d) 1 J

(e) 9.8 J

When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv²) of the sphere but its sign is negative.

Therefore, the answer is – ½ mv² = – ½ ×0.020×10² = –1 J.

(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

(a) 2 × 10⁵ J to 5 × 10⁵ J

(b) 2 × 10⁴ J to 5 × 10⁴ J

(c) 2 × 10³ J to 5 × 10³ J

(d) 200 J to 500 J

(e) 20 J to 50 J

This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.

In the 100 m dash the velocity v is almost constant so that we have

v = 100 m/10 s = 10 ms^–1

We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.

His kinetic energy will be ½ mv² = ½ ×80×10² = 4000 J so that the correct option is (c).

Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.

(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t_A and t_B, then the ratio t_A/ t_B is

(a) ½

(b) 2

(c) 2/5

(d) 5/6

(e) 1/5

The concept of impulse will be very useful here. The impulse received by A and B in the times t_A and t_B are respectively F t_A and F t_B where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F t_A and F t_B.

Since the kinetic energy is p²/2m where p is the momentum, we have

(F t_A)²/2m_A = (F t_B)²/2m_B

Therefore, t_A/ t_B = √( m_A/ m_B) = √(20/5) = 2.

(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is

(a) 20

(b) 40

(c) 50

(d) 30

(e) 35

The resutant force (F) on the particle is the sum of the forces given by

F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.

The displacement (s) of the particle is given by

s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)

The work done (W) is given by

W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.

You will find similar useful multiple choice questions (with solution) at AP Physics Resources.

KEAM 2009 Rescheduled

The Commissioner for Entrance Examinations, Govt. of Kerala, has notified that the dates of the Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering Degree Courses 2009 (KEAM 2009), Kerala have been changed (for the second time) as follows:

Modified Dates of Exam:

Engineering Stream:

25.05.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

26.05.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Stream:

27.05.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

28.05.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

For more details visit the site http://www.cee-kerala.org/ where you will find information regarding NATA (National Aptitude Test in Architecture) required for those who apply for Architecture

Angular Momentum- Two Multiple Choice Questions

How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though some times sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our happiness depends.

– Albert Einstein

(1) A thin uniform wooden rod AB of length L and mass M is hinged without friction at the end B. A small block of clay of mass m moving horizontally with velocity v srikes the end B of the rod and gets stuck to it. The angular velocity of the system about A just after the collision is

(a) mv/(mL + ML)

(b) 3mv/(mL + ML)

(c) (mv+ ML)/ 3mL

(d) mv/(3mL + ML)

(e) 3mv/(3mL + ML)

The angular momentum of the system about the point A just before collision is the angular momentum of the block of clay which is equal to mvL. (Note that the perpendicular distance of the line of action of the linear momentum (mv) of the block of clay from the point A is L).

The angular momentum of the system about the point A just after the collision is Iω where I is the total moment of inertia of the rod and clay and ω is the angular velocity of the system immediately after the collision.

We have I = ML²/3 + mL²

Equating the angular momenta before and after collision, we have

mvL = (ML²/3 + mL²) ω

Therefore ω = 3mv/(3mL + ML).

(2) A particle is projected at an angle of 60º with the horizontal with linear momentum of magnitude p. The horizontal range of this projectile is R. Just before the projectile strikes the ground at A, what is the magnitude of its angular momentum about an axis perpendicular the plane of motion and passing through the point of projection? Neglect air resistance.

(a) pR

(b) pR/2

(c) (√3) pR /2

(d) (√3) pR

(e) Zero

This is a very simple question. The magnitude of the linear momentum of the projectile just before it strikes the ground will be equal to p. The magnitude of the angular momentum at the moment will be p×Lever arm = p×ON = p×R sin 60º = (√3) pR /2.

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).