Kerala Engineering Entrance Questions (MCQ) on Communication Systems

Questions on communication systems are generally simple and interesting and so you can easily score marks by attempting them. The following three multiple choice questions were included in Kerala Engineering Entrance 2005 question paper and will be able to answer them in less than three minutes:

(1) If a radio receiver is tuned to 855 kHz radio wave, the frequency of local oscillator in kHz is

(a) 1510

(b) 455

(c) 1310

(d) 1500

(e) 855

Since the tuned frequency is 855 kHz, the receiver is an AM (amplitude modulation) receiver. Modern AM receivers are super heterodyne receivers employing a higher frequency local oscillator. On mixing the local oscillator output with the incoming amplitude modulated carrier, an amplitude modulated wave at intermediate frequency (IF) of 455 kHz (by convention) is produced. Since the intermediate frequency is 455 kHz, it follows that the frequency of local oscillator is 855 kHz + 455 kHz = 1310 kHz.

(2) If n₁ and n₂ are the refractive indices of the core and the cladding respectively of an optical fibre,

(a) n₁ = n₂

(b) n₁ < n₂

(c) n₂ < n₁

(d) n₂ = 2n₁

(e) n₂ = √(2n₁)

Since the optical fibre confines the light signal within the fibre by total internal reflection, the refractive index of the cladding should be less than that of the core. Therefore, n₂ < n₁ [Option (c)]

(3) A TV tower has a height of 100 m. What is the maximum distance up to which TV transmission can be received? (Radius of the earth, R = 8×10⁶ m)

(a) 34.77 km

(b) 32.7 km

(c) 40 km

(d) 40.7 km

(e) 42.75 km

We have maximum distance, d ≈ √(2Rh) where h is the height of the antenna.

Substituting given values, d ≈ √(2×8×10⁶ ×100) = 40×10³ m = 40 km.

[The mean radius of the earth is nearly 6400 km. The value is given as 8000 km in the problem to make your calculation simple.

You should remember that the height of the transmitting antenna (or receiving antenna) is the height with respect to the ground level. If an antenna is mounted on a mast of height h₁ and the mast is erected on a hill or building of height h₂, the height of the antenna will be h = h₁ + h₂]

Multiple Choice Questions (MCQ) on Alternating Currents

Most of you might have noted that electric power stations generate alternating current rather than direct current even though the majority of electrical and electronic appliances require direct current. At the power generation stage, alternating current is preferred since the current can be controlled, without power loss, using reactive elements (inductors and capacitors). Transmission of electric power over long distances without appreciable losses is possible by using transformers. The idea, as most of you should know, is to transmit electrical energy at low current and high voltage so that the Joule heating (which is proportional to the square of the current) in the transmission lines is minimized.

You will find some earlier posts on alternating currents on this site. You can access all the posts by trying a search for ‘alternating current’ using the search box provided on this page. Or, you may click on the label ‘alternating current’ below this post.

Let us discuss a few more questions on alternating currents:

(1) A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were quadrupled and the wire radius halved, the electric power dissipated would be

(a) halved

(b) same

(c) doubled

(d) quadrupled

The above question appeared in IIT 2002 screening test paper.

Electric power dissipation is given by P = V²/R. where V is the voltage induced in the coil and R is the resistance of the coil. When the number of turns is quadrupled, the induced voltage V is quadrupled so that V² becomes 16 times. But the resistance of the coil also becomes 16 times since resistance R = ρL /A where ρ is the resistivity (specific resistance), L is the length and A is the area of cross section of the wire. The length becomes 4 times when the number of turns is quadrupled and the area of cross section becomes one-fourth when the radius is halved. The resistance therefore becomes 16 times.
The power dissipated is therefore unchanged [Option (b)].

(2) An electric heater consumes 1000 watts power when connected across a 100 volt D.C. supply. If this heater is to be used with 200V, 50 Hz A.C.supply, the value of the inductance to be connected in series with it is

(a) 5.5 H

(b) 0.55 H

(c) 0.055 H

(d)1.1 H

(e)11 H

The current drawn by the heater is 1000 W/100V = 10 A . When the heater is used with A.C. supply, it will draw 10 A itself. (Note that the current and the voltage values are R.M.S. values when you deal with electric power). If ‘L’ is the inductance required, the expression for the current I is

I = V/√(R² + L²ω²)] where V, R and ω are respectively the alternating voltage, the resistance of the heater and the angular frequency of the A.C.

Substituting, 10 = 200/√[10² + L² (100π)²], since ω = 2πf where f’ is the frequency of the A.C.

Squaring and rearranging, L² = 300/100π)² from which L = √3/10π = 0.055 H.
(3) An alternating emf V = 6 cos1000t is applied across a series LR circuit of 3 mH inductance and 4 Ω resistance. The amplitude of the current is

(a) 0.6 A

(b) 1.2 A

(c) 1.4 A

(d) 1.8 A

Amplitude (maximum value or peak value) of current (I_max) is given by

I_max = V_max/√(R²+L² ω²) = 6/√[4²+(3×10^-3)²×1000²] = 1.2 A

[Note that the values of V_max and ω are obtained from the expression for the emf V which is in the form, V = V_max cos ωt].

Here is a very simple question which you should answer carefully:

(4) The voltage V applied across an A.C. circuit and the current I flowing in it are given by

V = 12 cos ωt volt and I = 20 sin ωt milliampere respectively.

The power dissipated in the circuit is

(a) 120 watt

(b) 120 milliwatt

(c) 240 watt

(d) 249 milliwatt

(e) zero

In alternating current circuits the power is given by P = V_rms I_rms cosΦ where Φ is the phase difference between the applied voltage and the resulting current. Since the voltage is a cosine function and the current is a sine function, the phase difference Φ is π/2. [Note that the voltage can be written as V = 12 sin(ωt + π/2) volt]. Therefore cosΦ (which is called the power factor) is zero. The correct option is (e).

All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) Postponed

The Central Board of Secondary Education has rescheduled the All India Engineering/Architecture Entrance Examination (online/pen & paper) 2011 (AIEEE 2011) from 24-4-2011 to 1-5-2011 in consideration of the Easter festival. The sale of Information Bulletins from all selling centres will start from 22.12.2010 instead of 15.12.2010. Rest of the things/schedule will remain unchanged.

Visit the web sites www.cbse.nic.in and www.aieee.nic.in for details.

Electronics - Multiple Choice Questions on Transistor Amplifiers

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

– Albert Einstein

Questions on transistor amplifiers at the class 12 (plus two or higher secondary) level are generally simple and interesting even though some among you may have unclear ideas. Today we will discuss a few questions on common emitter transistor amplifiers:

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(1) The adjoining figure shows a common emitter transistor amplifier which uses a silicon transistor. If the quiescent emitter current is 1 mA what is the base biasing voltage?

(a) 4.7 V

(b) 3.7 V

(c) 2.7 V

(d) 1.7 V

(e) 0 V

Because of the emitter current the voltage drop across the 1 KΩ resistor connected to the emitter is 1 V.

[1 mA×1 KΩ = (1/1000) A×1000 Ω = 1 V].

The voltage drop across the base-emitter junction of the silicon transistor is 0.7 V. Therefore, the base voltage under no signal (quiescent) condition is 1 V + 0.7 V = 1.7 V.

(2) In the amplifier circuit shown in Question No.1 what is the function of the capacitor C₁ connected across the 1 KΩ emitter resistor?

(a) To produce positive feed back.

(b) To produce negative feed back.

(c) To pass the excess signal to the ground.

(d) To act as filter capacitor for the transistor supply voltage.

(e) To bypass the signal current so that it will not flow through the emitter resistor.

The capacitor C₁ provides an easy path (bypass) for the signal component of the emitter current. If C₁ is absent the signal component of the emitter current will produce signal voltage drop across the emitter resistor, thereby reducing the signal output at the collector.

The correct option is (e).

(3) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, the quiescent base current of the transistor is very nearly equal to

(a) 1 mA

(b) 1 μA

(c) 2 μA

(d) 4 μA

(e) 5 μA

In the common emitter mode, the current amplification factor (current gain) under no signal condition (β_dc) is given by

β_dc = I_C/I_B where I_C is the collector current and I_B is the base current (both under no signal conditions).

Since the collector current is almost equal to the emitter current I_E (because of large value of β_dc), we have

β_dc ≈ I_E/I_B

Therefore I_B ≈ I_E/β_dc = 1 mA/200 = 0.005 mA = 5 μA.

(4) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, what is the voltage drop across the base biasing resistor R under quiescent conditions?

(a) 12 V

(b) 11 V

(c) 10.3 V

(d) 5.4 V

(e) 4.7 V

The quiescent base current is 5 μA as shown in answering question no.3 above. The base biasing voltage is 1.7 V as shown in answering question no.1. The power supply voltage is 12 V. Therefore, the voltage drop across the base biasing resistor R under quiescent conditions is 12 V – 1.7 V = 10.3 V.

(5) The base biasing resistor in the circuit shown in Question No.1 is

(a) 1 KΩ

(b) 4.7 KΩ

(c) 1.03 MΩ

(d) 1.87 MΩ

(e) 2.06 MΩ

The quiescent base current of the transistor is very nearly equal to 5 μA as shown in answering Question No.3. The voltage drop across the base biasing resistor R under quiescent conditions is 10.3 V as shown in answering Question No.4. Therefore, the base biasing resistor is given by

R = (10.3)V/(5 μA) = 2.06 MΩ

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You can access all questions (with solution) on electronics posted on this site by clicking on the label ‘electronics’ below this post or by trying a search for ‘electronics’ using the search box provided on this page

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Apply for All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) to be conducted on 24-4-2011 will be distributed from 15.12.2010 and will continue till 14.1.2011. Candidates can apply for AIEEE 2011 either on the prescribed Application Form or make application ‘Online’.

Online submission of the application is possible from 23-11-2010 to 14-1-2011 at the website http://aieee.nic.in

Visit the site http://aieee.nic.in for all details and information updates.

You will find many old AIEEE questions (with solution) on this site. You may search for ‘AIEEE’, using the search box at the top of this page, to access all posts in this context.

All India Pre-Medical / Pre-Dental Entrance Examination -2011 (AIPMT 2011)

Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination -2011 as per the following schedule for admission to 15% of the merit positions for the Medical/Dental Courses of India

1. Preliminary Examination …. 3^rd April, 2011 (Sunday) 10 AM to 1 PM

2. Final Examination ………… 15^th May, 2011 (Sunday) 10 AM to 1 PM

The Preliminary Examination as well as the Final Examination would be objective type.

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination either offline or online. Here are the important dates in this regard:

Sale of bulletin : 13-12-2010 to 31-12-2010

Online submission : 15-11-2010 to 31-12-2010

Last date of receipt of

(a) Offline application : 07-1-2011

(b) Online application : 07-1-2011

(Confirmation page)

Visit the site www.aipmt.nic.in for detailed information.

You will find some old AIPMT questions with solution on this blog. You can access these questions by searching for ‘AIPMT’, making use of the search box provided on this page.

IIT JEE 2010 Question (MCQ) on Magnetic Force on Current Carrying Conductors

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

The following question which appeared in IIT JEE 2010 question paper is worth noting:

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure.

When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

(a) iBL

(b) iBL /π

(c) iBL /2π

(d) iBL /4π

Considering an elemental length dL of the wire, the magnetic force F acting on the element is given by

F = idLB = iRdθB, as is clear from the figure.

The tension T developed in the wire is resolved into rectangular components T sin(dθ/2) and T cos(dθ/2) as indicated in the figure.

The magnetic force F is related to the tension T as

F = iRdθB = 2T sin(dθ/2)

Since dθ is a very small angle, sin(dθ/2) ≈ dθ/2

Therefore, the above equation gives T = iRB

But R = L /2π

Therefore, T = iBL /2π

Now, consider the following MCQ which is quite simple contrary to the impression of some of you:

Two straight, infinitely long parallel wires P and Q carry equal currents (I) in opposite directions (fig.). A square loop of side a carrying current i is arranged symmetrically in between the straight wires so that the plane of the loop is in the plane containing the wires P and Q. What is the torque acting on the current loop?

(a) Zero

(b) μ₀Iia²/2πd

(c) (μ₀Iia²/2π)[1/(d+a)]

(d) (μ₀Iia²/2π)[1/d + 1/(d+a)]

(e) (μ₀Iia²/2π)[1/d – 1/(d+a)]

The resultant magnetic forces acting on the four sides of the current loop act at their centres. The forces on the opposite sides of the loop are oppositely directed as you can easily verify using Fleming’s left hand rule. Since the lines of action of the resultant forces on opposite sides of the loop coincide, there is no lever arm to produce a torque and hence the torque is zero.

Apply for Joint Entrance Examination 2011 (JEE 2011) for Admission to IITs

Notification regarding the Joint Entrance Examination 2011 (JEE 2011) for Admission to IITs has been issued. This Joint Entrance Examination is for admission to the under graduate programmes in the IITs at Bhubaneswar, Bombay, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Kanpur, Kharagpur, Madras, Mandi, Patna, Rajasthan, Roorkee and Ropar as well as at IT-BHU Varanasi and ISM Dhanbad.

Important Dates for IIT JEE 2011:

November 01 to December 15, 2010: Online application process
November 12 to December 15, 2010: Sale of Off-line application forms.

Last date for receiving application forms (Off line or printed copy of Online) at IITs: 17:00 hrs, Monday, December 20, 2010

Date of Exam of IIT JEE 2011:

Sunday, April 10, 2011 Paper 1: 09:00 to 12:00 hrs;

Paper 2 : 14:00 to 17:00 hrs

All information regarding online application and offline application and other important details about IIT JEE 2011 can be obtained from the following websites:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Guwahati: http://www.iitg.ernet.in/jee/
IIT Delhi: http://jee.iitd.ac.in
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Kanpur: http://www.jee.iitk.ac.in
IIT Roorkee:http://jee.iitr.ernet.in/

IIT Madras: http://jee.iitm.ac.in

Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p’²/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p²/2m.

[Note that the initial kinetic energy of particle A is p²/2m and its final kinetic energy is (1/9) p²/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p’²/2M = (8/9) p²/2m

Substituting for p’ = 4p/3, we have

(16/9) (p²/2M) = (8/9) p²/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(c) 2v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2