Saturday, January 28, 2012

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).

Friday, January 20, 2012

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 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

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.

Tuesday, January 17, 2012

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)is given by
1/F = 1/f + 1/f
Therefore, 1/2f = 1/f + 1/f
This gives f = –2f………(ii)
But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (n) of the material (liquid) as
1/f = (n – 1)(–1/R1 – 1/) = (n – 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 – 1)/0.5f , on substituting for R1 and f from equations (i) and (ii).
This gives 2 n = 2.5 so that n = 1.25