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 R_L 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 I_CR_C = 0.6 V where I_C is the collector current and R_C is the resistance in the collector circuit.

Therefore,  I_C×600 Ω = 0.6 V from which I_C = 0.6/600 A = 10^–3 A = 1 mA.

Since the current gain β is given by

            β = I_C/I_B where I_B is the base current, we have

            I_B = I_C/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 n_h to 5×10²² m^–3. Then the value of n_e 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

            n_i² = n_en_h where n_i is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, n_e is the electron concentration in the doped semiconductor and n_h is the hole concentration in the doped semiconductor.

Therefore n_e = n_i²/n_h = (10¹⁶)²/(5×10²²) = 2×10⁹ m^–3

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].

Electrostatics - Multiple Choice Questions on Electric Field and Potential

Today we will discuss a few simple, but interesting, multiple choice questions on electrostatics. Here are the questions:

(1) ABC is an equilateral triangle. When positive charges Q and 2Q are placed at points A and B, the electric potential at the mid point (O) of AB is found to be 180 V. What is the electric potential at the vertex C of the triangle under this condition?

(a) 60 V

(b) 90 V

(c) 120 V

(d) 180 V

(e) 240 V

The distance of the vertex C of the triangle from the charges is twice the distance from the point O. Since the electric potential due to a point charge is inversely proportional to the distance, the potential at the vertex C must be half the value at C. So the answer is 90 V.

[Mathematically, you will write (with usual notations)

(1/4πε₀)(3Q/r) = 180 and

(1/4πε₀)(3Q/2r) = x

This gives x = 90 V]

(2) Equal and opposite point charges + Q and – Q are fixed at the corners A and B of an equilateral triangle (Fig.). If a negative point charge (– q) is placed at the vertex C of the triangle, the net force on this charge is directed

(a) towards right

(b) towards left

(c) towards A

(d) towards B

(e) towards the mid point of side AB

The force on the charge – q due to the charge + Q is attractive and is directed along CA. The force due to the charge – Q is repulsive and is directed along BC. Since these forces have the same magnitude, their resultant acts parallel to BA. The net force on the charge – q is thus directed leftwards [Option (b)].

(3) A cube of side 4 cm has a constant electric potential of 2 volt on its surface. If there are no charges inside the cube, the potential at a distance of 1 cm from the centre of the cube is

(a) zero

(b) 0.25 V

(c) 0.5 V

(d) 1 V

(e) 2 V

Since the surface of the cube is an equipotential surface and there are no charges inside the cube, the electric field inside the cube must be zero. This means that the electric potential everywhere inside the cube is 2 V itself [Option (e)].

(4) A spherical conductor of radius R has a central cavity (Fig.) with a negative point charge (–q) located at the centre of the cavity (without touching the surface of the cavity). The electric fiel at a point P distant r from the centre of the sphere is

(a) q/4πε₀r² directed towards the sphere

(b) q/4πε₀r² directed away from the sphere

(c) q/4πε₀r² directed upwards

(d) q/4πε₀r² directed downwards

(e) zero

The negative charge at the centre of the cavity will induce positive charge q on the surface of the cavity and negative charge –q on the outer surface of the sphere. The situation is similar to that of a spherical conductor carrying a charge –q so that the magnitude E of the electric field at the point P is given by

E = q/4πε₀r²

Since the field is due to negative charge, the direction of the field is towards the centre of the sphere.

IIT-JEE Multiple Correct Answer Type Questions on Thermodynamics

"Hate the sin, love the sinner."
–Mahatma Gandhi

Today we will discuss two multiple correct answer type questions on thermodynamics. These were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(C) work done during the path A → B → C is zero

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since a curve representing an isothermal in a PV diagram must be a hyperbola and not an arc of a circle.

We have PV = nRT with usual notations. The value of V is smaller at D compared to that at B where as the value of P is the same. The product PV = nRT is therefore smaller at D. This means that the temperature at D is less than that at B and hence statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operations ABCDA is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

So the correct options are (B) and (D).

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(C) C_P/C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas.

In the case of ideal gases C_P – C_V = R, the universal gas constant. Therefore statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas compared to a mono-atonic ideal gas. Therefore, C_P + C_V is larger for a diatomic ideal gas. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For mono-atomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Obviously statement (C) is wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a mono-atomic ideal gas.

Therefore, options (B) and (D) are correct.

You will find a useful post on thermodynamics here.

Kerala Engineering Entrance 2007 Questions on Communication Systems

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Questions on communication systems at the level expected from you are generally simple and interesting. Today we will discuss three questions (MCQ) on communication systems which appeared in Kerala Engineering Entrance 2007 question paper.

(1) The time variations of signals are given as in (A), (B) and (C).

Point out true statement from the following

(a) A, B and C are analog signals

(b) A and B are analog, but C is digital signal

(c) A and C are digital, but B is analog signal

(d) A and C are analog, but B is digital signal

(e) A, B and C are digital signals (e)

Digital signals will have two discrete levels only, corresponding to the zero level and one level as shown in graph (B). Analog signals have various instantaneous values, as shown in graphs (A) and (C). The correct option therefore is (d).

(2) A photo detector used to detect the wave length of 1700 nm, has energy gap of about

(a) 0.073 eV

(b) 1.2 eV

(c) 7.3 eV

(d) 1.16 eV

(e) 0.73 eV

The energy gap of a photo detector should be equal to or less than the energy of the photon which it is intended to detect. The product of the energy of a photon in electron volt and the wave length in Angstrom is 12400. Therefore, the energy of the 1700 nm photon is 12400/ 17000 electron volt = 0.73 eV. So, the correct option is (e). Option (a) is not suitable since a semiconductor with a gap as low as 0.073 eV (if at all available) will be unreliable due to the breaking of bonds by thermal excitation.

[You can use the expression hc/λ for calculating the energy of the photon in joule and then convert it into electron volts by dividing it by the electronic charge of 1.6×10^–19 joule. But it will be more difficult and time consuming].

(3) The optical fibres have an inner core of refractive index n₁ and a cladding of refractive index n₂ such that

(a) n₁ = n₂ (b) n₁ ≤ n₂ (c) n₁ < n₂ (d) n₁ > n₂ (e) n₁ ≥ n₂

The cladding has to be rarer than the core so as to totally reflect the beam of light entering the optical fibre. So, the correct option is (d).

Apply for Kerala Engineering Agricultural Medical Entrance Examinations 2012 (KEAM 2012)

"Strength does not come from physical capacity. It comes from an indomitable will."

– Mahatma Gandhi

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 2012-13.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BSc. Hons. (Forestry)

(c) Veterinary BVSc. & AH

(d) Fisheries BFSc.

(e) Engineering B.Tech. [including B.Tech. (Agri. Engg.), B.Tech. (Food Engg.) Courses

under the Kerala Agricultural University and B.Tech. (Dairy Science & Tech.) under the Kerala Veterinary & Animal Sciences University]

(f) Architecture B.Arch.

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

23.04.2012 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

24.04.2012 Tuesday10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture, Veterinary and Fisheries courses)

25.04.2012 Wenesday10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

26.04.2012 Thursday10.00 A.M. to 12.30 P.M. Paper-II: Biology.

All candidates are required to apply online through the website www.cee.kerala.gov.in for the Entrance Examination for admission to Medical, Agriculture, Veterinary, Fsheries, Engineering and Architecture Courses.

All details regarding KEAM 2012 can be obtained from the website http://www.cee-kerala.org

Hurry up! The Print out of your Application and other relevant documents to be submitted with the Application, are to be sent to the Commissioner for Entrance Examinations so as to reach him before 5 p.m. on 15.02.2012 (Wednesday).

* * * * * * * * * * * * * * *

To access earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the label ‘Kerala’ below this post.

Optics: Karnataka CET Questions (MCQ) on Lenses

Today we will discuss two questions (MCQ) involving lenses. The first one appeared in Karnataka CET 2011 question paper and the second one appeared in Karnataka CET 2010 question paper.

(1) Two thin lenses have a combined power of + 9 D. When they are separated by a distance of 20 cm, their equivalent power becomes + 27/5 D. Their individual powers (in dioptre) are ……..

(1) 4, 5

(2) 3, 6

(3) 2, 7

(4) 1, 8

We have P₁ + P₂ = 9 where P₁ and P₂ are the powers of the two lenses.

[Note that combined power means 1/f₁ + 1/f ₂ which is the effective power when the lenses are in contact. Power is the reciprocal of focal length and when the lenses are in contact, we have 1/F = 1/f₁ + 1/f ₂ where f₁ and f₂ are the focal lengths of the lenses and F is the combined focal length].

When the lenses are separate by a distance d, the power becomes P₁ + P₂ – d P₁P₂

[This follows from the expression, 1/F = 1/f₁ + 1/f ₂ – d/f₁f₂].

Therefore, we have

P₁ + P₂ – d P₁P₂ = 27/5

Or, 9 – (0.2× P₁P₂) = 27/5

[You have to substitute the separation in metre since the power is the reciprocal of focal length in metre].

From the above equation, 18/5 = 0.2× P₁P₂

Or, P₁P₂ = 18

Evidently the values given in option (2) satisfy the above condition so that we have P₁ = 3 and P₂ = 6

(2) The focal length of a plano-convex lens is ‘f’ and its refractive index is 1.5. It is kept over a glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes 2f. Then the refractive index of the liquid is ……

(1) 1.5

(2) 2

(3) 1.25

(4) 1.33

In the case of a lens placed in air or vacuum, lens maker’s equation is

1/f = (n – 1)(1/R₁ – 1/R₂) where n is the refractive index of the lens.

[General form of lens maker’s equation is 1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are the radii of curvature of its faces, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed].

In the case of the plano-convex lens, the equation becomes

1/f = (n – 1)(1/R₁ – 1/∞) = (n – 1)(1/R₁)

Since n = 1.5, R₁ = 0.5f ……..(i)

The focal length of the combination (F) of the plano convex lens (of focal length f) and the plano-concave liquid lens (of focal length f_ℓ)is given by

1/F = 1/f + 1/f_ℓ

Therefore, 1/2f = 1/f + 1/f_ℓ

This gives f_ℓ = –2f………(ii)

But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (n_ℓ) of the material (liquid) as

1/f_ℓ = (n_ℓ – 1)(–1/R₁ – 1/∞) = (n_ℓ – 1)(– 1/R₁)

[Note that the radius of the concave surface, which the liquid lens presents to the incident light, has to be negative in accordance with the Cartesian sign convention. Click on the label ‘Cartesian sign convention’ below this post to learn more about this convention].

Therefore, 1/(–2f) = – (n_ℓ – 1)/0.5f , on substituting for R₁ and f_ℓ from equations (i) and (ii).

This gives 2 n_ℓ = 2.5 so that n_ℓ = 1.25