Kerala Engineering Entrance (KEAM -Engineering) 2011 Questions (MCQ) on Transistor Amplifier and Oscillator

Today we will discuss two multiple choice questions involving transistors which appeared in Kerala Engineering Entrance (KEAM -Engineering) 2011 question paper. These questions are simple as is the case with questions in this section at the level expected in the case of 12^th grade (Plus Two) students.

(1) In a common emitter transistor amplifier, the output resistance is 500 KΩ and the current gain β = 49. If the power gain of the amplifier is 5×10⁶, the input resistance is

(A) 325 Ω

(B) 165 Ω

(C) 198 Ω

(D) 225 Ω

(E) 240 Ω

Power gain A_p is given by

A_p = β_ac A_v where β_ac is the current gain and A_v is the voltage gain (for small a.c. signals).

Since A_v = β_acR_o/R_i where R_o is the output resistance and R_i is the input resistance, we have

A_p = β_ac²R_o/R_i

Therefore R_i = β_ac²R_o/A_p = 49²×(500×10³) /(5×10⁶) = 240 Ω, very nearly.

(2) A transistor oscillator is (i) an amplifier with positive feedback (ii) an amplifier with reduced gain (iii) the one in which dc supply energy is converted into ac output energy. Then

(A) all (i), (ii) and (iii) are correct

(B) only (i) and (ii) are correct

(C) only (i) and (iii) are correct

(D) only (ii) and (iii) are correct

(E) only (ii) is correct

Statements (i) and (iii) are correct where as statement (ii) is incorect. So the correct option is (c).

Multiple Choice Questions on Magnetic Force on Moving Charges

Questions involving magnetic force on moving charges are included in most of the medical, engineering and other degree entrance examinations. Here are some simple questions which may easily tempt you to commit mistakes:

(1) The magnetic Lorentz force equation is F = q v×B. In this equation

(a) F, v and B must be mutually perpendicular.

(b) F must be perpendicular to v but not necessarily to B.

(c) F must be perpendicular to B but not necessarily to v.

(d) v must be perpendicular to both F and B

(e) F must be perpendicular to both B and v.

The equation F = q v×B gives the magnetic force F on a charge ‘q’ when it moves with velocity v in a magnetic field B. The angle between the velocity v and the field B can be any value, but the magnetic force F is at right angles to both v and B. So the correct option is (e).

[The vector product v×B which gives the force F indeed demands that F is at right angles to both v and B].

(2) A charged particle moving in the north east direction at right angles to a magnetic field experiences a force vertically upwards. This charged particle will not experience any force if it moves towards

(a) south west direction

(b) north

(c) east

(d) west

(e) north west

Since the magnetic force is vertical, the magnetic field must be horizontal.

Since the charged particle is moving in the north east direction at right angles to the magnetic field, it follows that the magnetic field must be directed either north west or south east (Fig.). So it will not experience any force if it moves towards north west or south east. The correct option is (e).

[You can use Fleming’s left hand rule to obtain the directions easily. See yourself that if the magnetic field in the above question is along the north west direction, the charge on the particle must be positive to obtain a vertically upward magnetic force. If the magnetic field in the above question is along the south east direction, the charge on the particle must be negative].

(3) A proton traveling vertically downwards experiences a southward force due to a magnetic field directed at right angles to its path.. An electron traveling northward in the same magnetic field will experience a magnetic force directed

(a) downwards

(b) upwards

(c) towards east

(d) towards west

(e) towards south east

Since the proton (which is positively charged) experiences a southward force while traveling vertically downwards, the perpendicular magnetic field must be acting towards the east.

[As required by Fleming’s left hand rule, hold the fore-finger, middle finger and thumb of your left hand in mutually perpendicular directions, with the middle finger pointing downwards (in the present case) and the thumb pointing southwards. The fore-finger then points towards the east].

If the proton were to move northward in this magnetic field, it would experience a downward magnetic force. Since the electron is negatively charged, it will experience an upward magnetic force [Option (b)]..

Karnataka CET Multiple Choice Questions on Combination of Capacitors

The following question appeared in Karnataka Common Entrance Test (CET) 2010 question paper. Even though at the first reading it may appear somewhat difficult for you on seeing the circuit, a careful look at the circuit will assure you that it’s simple.

All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the points A and B is

(a) 1.5 C

(b) 6 C

(c) C

(d) 3 C

The first three capacitors are in parallel, giving an effective capacitance 3 C. The last three capacitors also are in parallel, giving an effective capacitance 3 C.. These two parallel combinations are connected in series. Therefore, the effective capacitance between the points A and B is 3 C/2 = 1.5 C.

The following question appeared in the Karnataka CET 2008 question paper:

How many 6 μF, 200 V condensers are needed to make a condenser of 18 μF, 600 V?

(1) 9

(2) 18

(3) 3

(4) 27

Since the voltage rating of each 6 μF capacitor is 200 V, you should connect 3 capacitors in series to obtain the required voltage rating of 600 V. But then the effective capacitance of the series combination of these three capacitors will be 2 μF (6/3 = 2). To obtain the required capacitance of 18 μF, you need to connect nine such series combinations in parallel. So the total number of capacitors required is 9×3 = 27.

The following question also appeared in the Karnataka CET 2008 question paper:

The total energy stored in the condenser system shown in the figure will be

(1) 2 μJ

(2) 4 μJ

(3) 8 μJ

(4) 16 μJ

The series combination of 6 μF and 3 μF gives an effective capacitance of (6×3)/(6+3) = 2 μF. Since this is in parallel with a 2 μF capacitor, the effective capacitance C across the 2 volt battery is 4 μF. The energy stored in the system of capacitors is ½ CV² = (½)×(4×10^–6) ×2² = 8×10^–6 J = 8 μJ.

You will find a few more multiple choice questions (with solution) in this section here.

Kerala Engineering Entrance (KEAM) 2011 Questions on Kinematics in One Dimension

“Common sense is the collection of prejudices acquired by age eighteen.”

– Albert Einstein

Today we will discuss three questions (MCQ) on one dimensional kinematics. These questions were included in the KEAM (Engineering) question paper. They are of the type often seen in similar entrance tests. Here are the questions with their solution:

(1) A bus begins to move with an acceleration of 1 ms^–2. A man who is 48 m behind the bus starts running at 10 ms^–1 to catch the bus. The man will be able to catch the bus after

(a) 6 s

(b) 5 s

(c) 3 s

(d) 7 s

(e) 8 s

If the man can catch the bus after t seconds we have

10 t = 48 + ½ at² where a is the acceleration of the bus (a = 1 ms^–2).

[10 t is the distance covered by the man in time t and ½ at² is the distance covered by the bus in the same time. Since the man is 48 m behind the bus, he has to cover an additional distance of 48 m].

Since a = 1 ms^–2 the above equation reduces to

t² – 20 t + 96 = 0

Therefore t = [–(–20) ±√{20² – (4×1×96}] /(2×1)

Or, t = [20 ±√16] /2 = 12 s or 8 s

The smaller time 8 s is the answer.

(2) A car moves a distance of 200m. It covers first half of the distance at speed 60 km h^–1 and the second half at speed v. If the average speed is 40 km h^–1, the value of v is

(a) 30 km h^–1

(b) 13 km h^–1

(c) 60 km h^–1

(d) 40 km h^–1

(e) 20 km h^–1

The total distance moved is 200 m = 0.2 km

The total time (in hours) taken by the car is (0.1/60) + (0.1/v)

Since the average velocity is 40 km h^–1, we have

40×[(0.1/60) + (0.1/v)] = 0.2

Or, 4/v = 0.2 – (4/60) = 8/60 from which v = 240/8 = 30 km h^–1

[If v₁ and v₂ are the velocities while traversing the two halves of the path length s, the average velocity v_av is given by

v_av = s/[(s/2v₁) + (s/2v₂)]

Or v_av = 2v₁v₂/( v₁+v₂)

Since this final equation contains velocities only you can make substitutions without confusion (no need of conversion of metre into kilometre when distances are to be handled).

Therefore, 40 = (2×60×v)/(60+v) from which v = 30 km h^–1]

(3) A particle is moving with constant acceleration from A to B in a straight line AB. If u and v are the velocities at A and B respectively then its velocity at the mid point C will be

(a) [(u² + v²)/2u]²

(b) (u + v)/2

(c) (v – u)/2

(d) √[(u² + v²)/2]

(e) √(v² – u²)/2

If ‘s’ is the length of the path AB, the velocity of the particle changes from u to v when it moves through the distance ‘s’. Therefore we have,

v² – u² = 2as from which the acceleration, a = (v² – u²)/2s

If’ ’v_c ’ is the velocity of the particle at the mid point C of the path AB, we have v_c² = u² + 2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v_c = √[(u² + v²)/2].

AIEEE 2011 Question (MCQ) on Magnetic Field due to Current Carrying Conductors

"Seven Deadly Sins:

Wealth without work

Pleasure without conscience

Science without humanity

Knowledge without character

Politics without principle

Commerce without morality

Worship without sacrifice."

– Mahatma Gandhi

The following multiple choice question included in the All India Engineering/Architecture Entrance Examination (AIEEE) 2011 will be interesting and useful to you:

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is

(1) μ₀ I/4πR

(2) μ₀ I/π²R

(3) μ₀ I/2π²R

(4) μ₀ I/2πR

It would have been better if the magnitude of the magnetic induction at the axis was asked for since the words along the axis usually means directed along the axis. [The component of field directed along the axis is zero].

Forget about it. The question setter requires you to calculate the magnetic flux density at points on the axis (of the semicircular ring shaped cross section of the infinitely long wire). In the adjoining figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width dℓ.

With reference to the figure, we have dℓ = Rdθ.

The magnetic flux density due to the above strip is shown as dB₁ in the figure. It has an X-component dB₁ sinθ and a Y-component dB₁cosθ. When we consider a similar strip of the same with dℓ located symmetrically with respect to the Y-axis, we obtain a contribution dB₂ to the flux density. The flux density dB₂ has the same magnitude as dB₁. It has X-component dB₂ sinθ and Y-component dB₂cosθ. The X-components of dB₁ and dB₁ are of the same magnitude and direction and they add up. But the Y-components of dB₁ and dB₁ are in opposite directions and have the same magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.

The wire strip of width dℓ cn be imagined to be an ordinary thin straight infinitely long wire carrying current Idℓ/πR since the total current I flows through the semicircular cross section of perimeter πR. Putting dB₁ = dB₂ = dB we have

dB = μ₀ (Idℓ/πR)/2πR = μ₀ (IRdθ/πR)/2πR = μ₀Idθ/2π²R

The X-component of the above field is (μ₀I/2π²R) sinθ dθ

The field due to the entire conductor is B = ₀∫^π [(μ₀I/2π²R) sinθ]dθ

Or, B = μ₀I/π²R since ₀∫^π sinθ dθ = 2

Kerala Engineering Entrance (KEAM) 2011 Question (MCQ) on Zener Diode

“Everything should be made as simple as possible, but not simpler”

– Albert Einstein

The following multiple choice question on zener diode was included in Kerala Engineering Entrance (KEAM) 2011 question paper. If you identify that R₁ is the current limiting resistor and R₂ is the load resistor in the simple shunt voltage regulator making use of a zener diode, you won’t have any confusion. Here is the question:

In the circuit given the current through the zener diode is

(a) 10 mA

(b) 6.67 mA

(c) 5 mA

(d) 3.33 mA

(e) 0 mA

Since the zener breakdown voltage (across the zener diode) is 10 V, the voltage across the resistor R₂ is 10 V. The current through R₂ is 10 V/1500 Ω = 0.00667 A = 6.67 mA.

The voltage drop across R₁ is 15 V – 10 V = 5 V.

Therefore, the current through R₁ is 5 V/500 Ω = 0.01 A = 10 mA

The above current gets divided between the resistor R₂ and the zener diode. Therefore, the current through the zener diode is 10 mA – 6.67 mA = 3.33 mA.

By clicking on the label ‘zener diode’ below this post you can access similar questions on zener diodes posted on this site.

IIT-JEE 2011 Questions (MCQ) on Simple Harmonic Motion

“Do not worry about your problems with mathematics, I assure you mine are far greater.”

– Albert Einstein

The following two questions on simple harmonic motion were included in the IIT-JEE 2011 question paper. Both questions are simple for those who have mastered the fundamental principles:

(1) A wooden block performs SHM on a frictionless surface with frequency, ν₀. The block carries a charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(C) of changed frequency and with shifted mean position

(D) of changed frequency and with the same mean position

When the uniform electric field is switched on, the positively charged block experiences a constant force in the direction of the electric field and so the mean position of the block is shifted. But the frequency (ν) of oscillation of the block is unchanged since it depends only on the mass m of the block and the force constant k of the spring, in accordance with the relation

ν = (1/2π) [√(k/m)]

The correct option is (A).

(2) A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x₁(t) = A sin ωt and x₂(t) = A sin(ωt + 2π/3). Adding a third sinusoidal displacement x₃(t) = B sin(ωt + φ) brings the mass to a complete rest. The value of B and φ are

(A) √2 A, 3π/4

(B) A, 4π/3

(C) √3 A, 5π/6

(D) A, π/3

The particle in the above question is subjected to two simultaneous simple harmonic motions (SHM) of the same frequency and amplitude in the same direction. the resultant motion of the particle is simple harmonic as is evident by adding x₁(t) and x₂(t):

x₁(t) + x₂(t) = A sin ωt + A sin(ωt + 2π/3) = A sin(ωt +π/3)

The amplitude of the resultant simple harmonic motion is A itself but the initial phase of the motion is now π/3.

As the particle remains at rest on adding the third simple harmonic motion, x₃(t) = B sin(ωt + φ) the amplitude (B) of the third SHM must be A itself, but it must be 180º (or, π radian) out of phase. In other words, the initial phase (φ) of the third simple harmonic motion must be π/3 + π = 4π/3

The correct option is (B).

You can find a useful post on simple harmonic motion here.