Saturday, April 14, 2012

KEAM (Engineering) 2010 Questions (MCQ) on Electronics

"You may never know what results come of your actions, but if you do nothing, there will be no results."
–Mahatma Gandhi

Questions on electronics will be generally interesting to most of you. Today we will discuss questions in this section which appeared in Kerala engineering entrance (KEAM - Engineering) 2010 question paper. Here are the questions with their solution:
(1) A full wave rectifier with an a.c. input is shown:

The output  voltage across RL is represented as

The rectified output voltage will be a direct voltage but there will be very large amount of ripples. The capacitor C acts as a filter to remove the ripples; but there will still be a small amount of  ripples in the output. Therefore the correct option is (e).
(2) In the given circuit the current through the battery is
(a) 0.5 A
(b) 1 A
(c) 1.5 A
(d) 2 A
(e) 2.5 A
Since the diode D1 is reverse biased, no current will flow through the D1 branch. Diodes D2 and D3 are forward biased and hence the battery drives currents through the 20 Ω resistor and the series combination of the two 5 Ω resistors.
The current driven through the 20 Ω resistor is 10 V/20 Ω = 0.5 A.
The current driven through the 10 Ω resistor is 10 V/10 Ω = 1 A.
Therefore, total current through the battery is 0.5 A + 1 A = 1.5 A
(3) The collector supply voltage is 6 V and the voltage drop across a resistor of 600 Ω in the collector circuit is 0.6 V, in a transistor connected in common emitter mode. If the current gain is 20, the base current is
(a) 0.25 mA
(b) 0.05 mA
(c) 0.12 mA
(d) 0.02 mA
(e) 0.07 mA
We have ICRC = 0.6 V where IC is the collector current and RC is the resistance in the collector circuit.
Therefore,  IC×600 Ω = 0.6 V from which IC = 0.6/600 A = 10–3 A = 1 mA.
Since the current gain β is given by
            β = IC/IB where IB is the base current, we have
            IB = IC/β = 1 mA/20 = 0.05 mA.
(4) A pure semiconductor has equal electron and hole concentration of 1016 m–3. Doping by indium increases nh to 5×1022 m–3. Then the value of ne in the doped semiconductor is
(a) 106 m–3
(b) 1022 m–3
(c) 2×106 m–3
(d) 1019 m–3
(e) 2×109 m–3
According to the law of mass action we have
            ni2 = nenh where ni is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, ne is the electron concentration in the doped semiconductor and nh is the hole concentration in the doped semiconductor.
Therefore ne = ni2/nh = (1016)2/(5×1022) = 2×109 m–3

Sunday, April 01, 2012

Multiple Choice Questions on Dimensions of Physical Quantities [Including KEAM (Engineering) 2011 question]

I believe in standardizing automobiles, not human beings

– Albert Einstein

In most of the entrance examination question papers you will find at least one question on dimensions of physical quantities. Here are a few typical multiple choice questions in this section:

(1) Which one among the following quantities is a dimensional constant?

(a) Dielectric constant of water

(b) Speed of light in free space

(c) Viscosity of water

(d) Ratio of specific heats of a diatomic gas

(e) Reynolds number

Options (a) and (c) are not constants since the dielectric constant and viscosity depend on other parameters. Options (d) and (e) are dimensionless numbers.

[Reynolds number (used in assessing the turbulence of fluids) is the ratio of inertial force to force of viscosity]

The correct option is the speed of light in free space which is a fundamental constant with dimensions LT–1.

(2) If F denotes force and t time, then in the equation F = at–1 + bt2, the dimensions of a and b respectively are

(a) LT–4 and LT–1

(b) LT–1 and LT–4

(c) MLT–4 and MLT–1

(d) MLT–1 and MLT–4

(e) MLT–3 and MLT–2

The above question appeared in Kerala Engineering Entrance ((KEAM) 2011 question paper.

The dimensions of at–1 and bt2 have to be that of force which is MLT–2. Therefore the dimensions of a must be MLT–1 and the dimensions of b must be MLT–4.

The correct option is (d).

The following question appeared in Karnataka CET 2004 question paper:

(3) The physical quantity having the same dimensions as Planck’s constant h is

(1) Boltzmann constant

(2) force

(3) linear momentum

(4) angular momentum

Most of you will remember that the unit of h is joule second. Therefore h has the dimensions of the product of work and time which is ML2T–2×T = ML2T–1.

Angular momentum is the moment of linear momentum and has dimensions L×MLT–1 = ML2T–1.

The correct option is (4).

The following question appeared in EAMCET 2008 Engineering Entrance Exam question paper. You will answer it in no time if you remember that the dimensions of Planck’s constant are those of angular momentum.

(4) The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is

(a) zero

(b) – 1

(c) 5/3

(d) 1

Since the dimensions of Planck’s constant are those of angular momentum, it follows that the dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is zero.

[In terms of E, L and G which are assumed as fundamental quantities in the above question, Planck’s constant (h) has zero dimension in E, one dimension in L and zero dimension in G].