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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.