## Monday, January 28, 2008

### Kerala Engineering, Architecture & Medical Entrance Examinations 2008

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 2008-09.

(a) Medical: (i) MBBS (ii) BDS (iii) B.Pharm. (iv) B.Sc. (Nursing) (v) B.Sc. (MLT)

(vi) BAMS (vii) B.H.M.S (viii) B.S.M.S (Siddha) (ix) B.Sc. Nursing (Ayurveda)

(x) B.Pharm. (Ayurveda) and (xi) B.P.T (Physiotherapy).

(b) Agriculture: (i) B.Sc. Hons. (Agriculture) (ii) B.F.Sc. (Fisheries) (iii) B.Sc. Hons. (Forestry)

(c) Veterinary: B.V.Sc. & 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.

The last date for the receipt of completed applications by the Commissioner for Entrance Examinations is 29-2-2008 Friday (before 5 PM).

## Monday, January 21, 2008

### Two Kerala Engineering Entrance 2007 Questions on Linear Motion

The following questions appeared in Kerala Engineering Entrance 2007 question paper:

(1) Two balls are dropped to the ground from different heights. One ball is dropped two seconds after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is (g = 10 ms–2)

(a) 20 m

(b) 80 m

(c) 170 m

(d) 40 m

(e) 160 m

The initial height of the first ball is the distance travelled by it in 5 seconds and is given by

x1 = 0×5 + (½)×10×52, using the relation, x = v0t + (½)at2

Therefore, x1 = 125 m.

The time of travel of the second ball is 3 seconds and hence its initial height is given by

x2 = 0×3 + (½)×10×32 = 45 m.

The difference in initial heights is x1x2 =125 – 45 = 80 m.

(2) A ball is thrown vertically upwards with a velocity of 25 ms–1 from the top of a tower of height 30 m. How long will it travel before it hits the ground?

(a) 6 s

(b) 5 s

(c) 4 s

(d) 12 s

(e) 10 s

Let us take all downward vectors positive. The displacement is 30 m and is positive. The acceleration due to gravity, g (equal to 10 ms–2) also is positive. But, the velocity of projection is negative (being upwards).

From the equation, x = v0t + (½)at2, we have

30 = – 25 t + (½)×10×t2

Or, t2 – 5 t – 6 = 0, from which t = [5±√(25+24)]/ 2 = + 6 s or –1 s.

Since negative time is impossible, the answer is 6 s.

[You could have taken upward quantities as positive if you wanted to, but you would get the same answer].

## Thursday, January 10, 2008

### Communication Systems: MCQ on Amplitude modulation

Here are some questions (on amplitude modulation) of the type you are most likely to encounter in your entrance exam:

(1) A carrier wave of peak value 16 V is used to transmit a note of frequency 1000 Hz. What should be the peak value of the modulating signal to have a modulation index of 60% using amplitude modulation?

(a) 12 V

(b) 11.2 V

(c) 10.4 V

(d) 9.6 V

(e) 7.2 V

This question demands from you merely a basic knowledge of the modulation process and you are expected to calculate the modulation index (μ) using the equation,

μ = Am/Ac where Am and Ac are the amplitudes of the modulating signal and the carrier respectively.

Therefore, 0.6 = Am/16, from which Am = 9.6 V. (The frequency of the modulating signal is just a distraction in this question).

[In a practical modulator, the modulating signal amplitude may not be μ times the carrier amplitude. For instance, if the carrier is applied on the base side and the modulating signal is applied on the collector side of a transistor, the carrier amplitude (at the base) will be much less than the modulating signal amplitude, even though the modulation index is less than 100%. Of course, the carrier will appear in the amplified form at the collector. The modulation index (μ) in all situations will be correctly obtained if you consider it as the ratio of the amplitude of variation of the envelope of the amplitude modulated carrier to the amplitude of the unmodulated carrier. But problems with the above type of statement are often seen].

(2) The maximum amplitude of an amplitude modulated wave is 12 V and the minimum amplitude is 4V. The modulation percentage is

(a) 20

(b) 30

(c) 50

(d) 60

(e) 80

The modulation percentage is the modulation index expressed as percentage.

The maximum amplitude of an amplitude modulated (AM) wave is the sum of the amplitude of the unmodulated carrier and the amplitude of variation of the envelope of the AM wave.

The minimum amplitude of an AM wave is the difference between the amplitude (Ac) of the unmodulated carrier and the amplitude (Am) of variation of the envelope of the AM wave.

Therefore, we have (Ac + Am) = 12 V

(Ac Am) = 4 V

From these equations, Ac = 8 V and Am = 4 V so that

μ = Am/Ac = 4/8 = 0.5 = 50%

(3) An unmodulated carrier has a peak value of V volt. When it is amplitude modulated with a sine wave, the maximum peak to peak value (of the modulated carrier) is 3 V volt. The modulation index is

(a) 0.33

(b) 0.5

(c) 0.6

(d) 0.66

(e) 0.8

The maximum peak value (maximum amplitude) of the modulated carrier is 3V/2 volt.

Therefore, (Ac + Am) = 3V/2

Since the amplitude of the unmodulated carrier (Ac) is V volt, the amplitude (Am) of the variation of the envelope of the modulated carrier is V/2. Therefore, The modulation index is given by

μ = Am/Ac = (V/2)/V = 0.5

(4) An amplitude modulated wave has modulation index 0.6. If the power carried by the carrier component in the modulated wave is Pc , what is the power carried by the upper and lower side bands (together)?

(a) 0.6 Pc

(b) 0.54 Pc

(c) 0.36 Pc

(d) 0.46 Pc

(e) 0.18 Pc

The total power (P) carried by an amplitude modulated wave is given by

P = Pc[1+ (μ2/2) where Pc is the carrier power and μ is the modulation index.

Therefore, the power carried by the two side bands together is

Pc ×μ2/2 = Pc×(0.6)2/2 = 0.18 Pc.

[This shows how much power is wasted in the carrier component which does not carry any information. Further, we need only one side band to extract the modulating signal. This points to the benefit of the suppressed carrier single side band transmission].

(5) An AM radio station using double side band transmitted carrier system employs a carrier of frequency 1.2 MHz. If the modulating signal frequency is 2 kHz, the frequencies present in the modulated carrier are

(a) 1.2 MHz and 1.4 MHz

(b) 1 MHz, 1.2 MHz and 1.4 MHz

(c) 1.198 MHz, 1.2 MHz and 1.202 MHz

(d) 1.198 MHz and 1.202 MHz

(e) 1.2 MHz and 1.202 MHz

The double side band transmitted carrier system is the one employed by ordinary broadcast stations. The amplitude modulated wave will contain the carrier frequency (fc) as well as the upper and lower side frequencies fc + fm and fcfm. Therefore, the correct option is (c).