## Tuesday, August 18, 2009

### AIEEE 2009 Multiple Choice Questions (MCQ) with Solution on Electrostatics

The following four questions which appeared in the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) will be beneficial to most of the entrance test takers. Those who prepare for AP Physics C exam and physics GRE may take special note of question no.2. Here are the questions with solution:

(1) A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electric force on Q is zero, then Q/q equals

(1) – 1/√2

(2) – 2√2

(4) 1

The electric force FQQ (Fig.) on Q because of the charge Q at the opposite corner is repulsive and is given by

F1 = (1/4πε0)(Q2 /2a2) since the distance between the charges Q and Q is 2×a/√2 = a√2

The electric forces FQq and FQq (Fig.) on Q because of the charges q and q at the adjacent corners are perpendicular to each other. Each force has magnitude given by

FQq = (1/4πε0)(Qq/a2).

Their resultant F2 has magnitude √2 times the above value:

F2 = (1/4πε0)(√2 Qq/a2).

Since the net force on Q is zero, we have

F1 + F2 = 0

Therefore, (1/4πε0)(Q2 /2a2) + (1/4πε0)(√2 Qq/a2) = 0

This gives Q/q = – 2√2

[The charges Q and q must be of opposite sign so that the force between Q and q is attractive to ensure that F1 and F2 are opposite in directions and the net force on Q is zero as given in the question].

(2) Let P(r) = Qr/πR4 be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is

(1) Qr12 /3πε0 R4

(2) 0

(3) Q/4πε0 r12

(4) Qr12 /4πε0 R4

Charge dQ contained in the spherical shell of radius r and thickness dr is (Qr/πR4)(4πr2dr) = (4Q/R4) r3dr

Charge Q1 contained in the spherical volume of radius r1 is therefore given by

Q1 = 0r1 (4Q/R4) r3dr = (4Q/R4)(r14/4) = Qr14 /R4

If E is the electric field at distance r1 from the centre of the sphere, we have from Gauss theorem,

r12E = Qr14 /R4ε0 from which

E = Qr12 /4πε0 R4

[Remember that the charges distributed with spherical symmetry outside the point p will produce zero field at p and the field at p is indeed given correctly by the above expression].

(3) Two points P and Q are maintained at the potentials of 10 V and – 4 V, respectively. The work done in moving 100 electrons from P to Q is

(1) 2.24×10–16 J

(2) – 9.60×10–17 J

(3) 9.60×10–17 J

(4) – 2.24×10–16 J

This is a simple question. The potential difference ∆V between P and Q is 10 V – (– 4 V) = 14 V. Since the electrons are negatively charged, external work (positive work) has to be done to move them from the higher potential point P to the lower potential point Q.

The work done = q∆V = (100×1.6×10–19)×14 J = 2.24×10–16 J

(4) This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement 1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement 2 : The net work done by a conservative force on an object moving along a closed loop is zero.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statment-1.

(2) Statment-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

(3) Statement-1 is false, Statement-2 is true.

(4) Statement-1 is true, Statement-2 is false

Electrostatic field is conservative and the net work done by an electrostatic field on a charged particle is dependent only on the initial and final positions of the charged particle. (It is independent of the path connecting the initial and final positions).

The correct option is (1).

You will find some more useful multiple choice questions with solution from electrostatics at AP Physics Resources.

## Saturday, August 08, 2009

### Multiple Choice Questions (including KEAM 2009) on Transistor Amplifiers

Questions from electronics are generally simple and interesting at the level expected from you. Here are a few questions:

(1) A transistor amplifier circuit is operated with an emitter current of 2 mA. The collector current is 1.98 mA. The common emitter current gain (βdc) of the transistor used in the circuit is

(a) 45

(b) 50

(c) 100

(d) 125

(e) 200

Current gain (βdc) in the common emitter configuration is given by

βdc = IC /IB = 2 mA/(2 – 1.98) mA = 100

(2) A common emitter low frequency amplifier has an effective input resistance of 1 kΩ. The collector load resistance is 5 kΩ. On applying a signal, the base current changes by 10 μA and the collector current changes by 1 mA. What is the power gain of this amplifier?

(a) 100

(b) 500

(c) 1000

(d) 10000

(e) 50000

The power gain or power amplification (AP) is the product of the current gain βac and the voltage gain Av:

AP = βac× Av

Here βac = ∆IC /∆IB = (1×10–3)/(10×10–6) = 100 and

Av = βac×RL/Ri where RL is the load resistance and Ri is the input resistance. The negative sign just indicates that in the common emitter configuration the output signal is phase shifted by 180º.

[We use the load resistance RL instead of the output resistance Ro (of the amplifier stage) in the above expression since the load resistance is small compared to the transistor output resistance. (The output resistance of the amplifier stage is the parallel combined value of RL and the transistor output resistance)].

Ignoring the negative sign, Av = 100×5/1 = 500

Therefore, power gain AP = βac× Av = 100×500 = 50000.

(3) For a common emitter amplifier, the audio signal voltage across the collector resistance 2 kΩ is 2V. If the current amplification factor of the transistor is 200 and the base resistance is 1.5 kΩ, the input signal voltage and base current are

(a) 0.1 V and 1 μA

(b) 0.15 V and 10 μA

(c) 0.015 V and 1 A

(d) 0.0015 V and 1 mA

(e) 0.0075 V and 5 μA

This MCQ appeared in KEAM (Engineering) 2009 question paper.

The collector signal current ic is given by

ic = vc/Rc = 2 V/ 2 kΩ = 1 mA.

Since the current amplification (βac) of the transistor is 200, the base signal current ib is given by

ib = ic /βac = 1 mA /200 = 0.005 mA = 5 μA

The input signal is the voltage drop (ibRb) produced across the base resistance by the flow of the base current.

Therefore, the input signal voltage = ibRb = 5 μA×1.5 kΩ = 7.5×10–3 V = 0.0075 V.

(4) In an NPN transistor 108 electrons enter the emitter in 10–8 s. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification are respectively

(a) 0.8 and 49

(b) 0.9 and 90

(c) 0.7 and 50

(d) 0.99 and 99

(e) 0.88 and 88

This MCQ appeared in KEAM (Medical) 2009 question paper.

The number of electrons (108) entering the base and the time (10–8 s) given in this question are not required to solve the problem. They can serve as distraction while attempting this simple question!

Since 1% electrons are lost in the base, 99% reach the collector so that the fraction of current that enters the collector is 99/100 = 0.99.

The current amplification is the ratio of the collector current to the base current and is equal to 99/1 = 99.

To access all questions from electronics on this blog you may click on the label ‘electronics’ below this post.

## Monday, August 03, 2009

### Kinematics- KEAM (Engineering) 2009 Questions on One dimensional Motion

Today we will discuss some multiple choice questions on one dimensional motion which appeared in Kerala Engineering Entrance 2009 question paper:

(1) A ball is thrown up vertically with speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is

(a) u

(b) 2u

(c) u gt

(d) √(u2 gt)

(e) gt

If you know the concept of relative velocity correctly, this question will be quite simple for you.

The relative velocity of A with respect to B is given by

VAB = VA VB where VA and VB are respectively the velocities of A and B (with respect to the common frame of reference chosen).

The velocity of A and B at time t are respectively u gt and – gt, taking upward quantities positive and downward quantities negative.

The speed of A relative to B is u gt – (– gt) = u [Option (a)].

(2) A body is falling freely under gravity. The distances covered by the body in the first, second and third minutes of its motion are in the ratio

(a) 1 : 4 : 9

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 5 : 6

(e) 1 : 5 : 13

Since you require the ratio of distances, it is enough to consider the times in seconds (instead of minutes). Strictly, it must be mentioned that the body starts from rest. In the case of such a freely falling body, the distance covered is directly proportional to t2 since s = ut + ½ gt2 with usual notations where u = 0

The distances covered by the body in one second, two seconds and three seconds are in the ratio 1 : 4 : 9.

Therefore, the distances covered by the body in the first, second and third seconds are in the ratio 1 : (4 – 1) : (9 – 4) which is 1 : 3 : 5

(3) A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

(a) 22/3 cm

(b) 40/3 cm

(c) 20/3 cm

(d) 22/5 cm

(e) 26/5 cm

We have v2 = u2 + 2as where u is the initial velocity, v is the final velocity after suffering a displacement s and a is the acceleration.

Therefore we have

u2/4 = u2 + 2a × 0.4

Or, 3u2/4 = 2a × 0.4 …………..(i)

If the bullet comes to rest after penetrating a further distance of s1 we have

0 = u2/4 + 2a × s1

Or, u2/4 = 2a × s1 …………….(ii)

Dividing Eq(i) by Eq(ii) we get

3 = 0.4/s1 from which s1 = 0.4/3 m = 40/3 cm.

[The above question can be worked out using the work energy principle as well].

You will find similar useful multiple choice questions on kinematics (with solution) here.