Kerala Engineering Entrance 2009 Questions (MCQ) on Alternating Current Circuits

Imagination is more important than knowledge
– Albert Einstein

Alternating current circuits were discussed on this site in the post dated 26th October 2006. You will find the post here.

Today we will discuss the questions from alternating currents which appeared in Kerala Engineering Entrance 2009 question paper. Here are the questions:

(1) In an LCR series ac circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become

(a) 10 V

(b) 20/√2 V

(c) 20√2 V

(d) 10/√2 V

(e) 20 V

Since the voltages across L and C are of the same value, the circuit is at resonance and Lω = 1/Cω where ω is the angular frequency. Under resonance the voltage across the resistor is equal to the supply voltage. So the supply voltage is 10 V.

When the inductor is short circuited, the LCR circuit reduces to an RC circuit and the supply voltage is the vector sum of the voltages across R and C so that we have

10 = √(V_C² + V_R²)

The voltage drops V_C and V_R across the capacitor and resistor respectively are equal in value since Lω = 1/Cω = R as judged from the equal voltage drops across L, C and R in the series LCR circuit. Therefore, from the above equation V_C = V_R = 10/√2 V

(2) A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6 A. The impedance of the device is

(a) 60 Ω

(b) 50 Ω

(c) 80 Ω

(d) 100 Ω

(e) 120 Ω

Efficiency, η = P_o/P_i with usual notations. Therefore, the output power is given by

P_o = η P_i = 0.9 ×4000 W = 3600 W.

In the problem nothing is mentioned about the power factor cos φ. At your level the load is usually assumed to be effectively resistive so that the power factor may be assumed to be unity. Therefore we have

P_o = I_o²R where R is the load resistance (or, the load impedence in this case).

This gives R = P_o /I_o² = 3600 /6² = 100 Ω.

(3) The impedance of an R-C circuit is Z₁ for a frequency f and Z₂ for frequency 2f. Then Z₁/Z₂ is

(a) between 1 and 2

(b) 2

(c) between ½ and 1

(d) ½

(e) 4

The impedance of an R-C circuit is √(R² + 1/C²ω²).

Since the angular frequency ω = 2πf we have

Z₁/Z₂ = [√(R² + 1/C²ω²)] /[√(R² + 1/4C²ω²)]

For large value of ω the value of Z₁/Z₂ tends to 1.

For small value of ω the value of Z₁/Z₂ tends to 2.

Therefore Z₁/Z₂ lies between 1 and 2.

(4) When a circular coil of radius 1 m and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf induced is 100 V. The coil is unwound and then rewound into a circular coil of radius 2 m. If it is rotated now, with the same speed, under similar conditions, the new peak value of emf developed is

(a) 50 V

(b) 25 V

(c) 100 V

(d) 150 V

(e) 200 V

When a coil having N turns and area A rotates in a uniform magnetic field B with angular velocity ω, the peak value of the emf (V_max) induced in the coil is given by

(V_max) = NABω

The area of the coil in the second case is 4A since the radius is doubled. But the number of turns will be N/2 in the second case since the length of the wire required for a turn is doubled.

If V₁ and V₂ are the peak values of the emf in the two cases, we have

V₁/V₂ = 100 /V₂ = NABω / [(N/2)×(4A)Bω]

Or, 100 /V₂ = ½ from which V₂ = 200 volt.

AIEEE 2009 Multiple Choice Questions on Optics

I believe in standardising automobiles, not human beings

– Albert Einstein

The following three questions were included from optics in the All India Engineering/Architecture Entrance Examination (AIEEE) 2009 question paper:

(1) A transparent solid cylindrical rod has a refractive index of 2/√3. It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure. The incident angle θ for which the light ray grazes along the wall of the rod is

(1) sin^–1(1/√3)

(2) sin^–1(1/2)

(3) sin^–1(√3/2)

(4) sin^–1(2/√3)

If n is the refractive index and θ_c is the critical angle of the material of the rod, we have n = 1/sinθ_c so that

sinθ_c = 1/n = √3/2.

If r represents the angle of refracion (as shown in the figure), we have

1×sinθ = nsin r = (2/√3) ×sin(90º – θ_c)

Or, sinθ = (2/√3) ×cos θ_c

But cos θ_c = ½ since sin θ_c = √3/2

[You may imagine a right angled triangle with opposite side √3 and hypotenuse 2. The adjacent side will be 1]

Substituting for cos θ_c, we obtain

sinθ = (2/√3) × ½ = 1/√3 so that θ = sin^–1(1/√3)

(2) In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coordinates of P will be

(1) (4f, 4f)

(2) (2f, 2f)

(3) (f/2, f/2)

(4) (f, f)

This is a well known experiment which most of you might have performed. As such most of you know the answer: (2f, 2f)

We have 1/v – 1/u = 1/f

Applying the Cartesian sign convention, v is positive and u is negative so that

1/v + 1/u = 1/f

Since the straight line makes an angle of 45º, we have v = u at P and each must be equal to 2f to satisfy the equation, 1/v + 1/u = 1/f

(3) A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

(1) 776.8 nm

(2) 393.4 nm

(3) 885.0 nm

(4) 442.5 nm

The fringe width β is given by

β = λD/d where λ is the wave length of light, D is the distance between the double slit and the screen and d is the double slit separation.

Since three fringes due to the light of wave length 590 nm exactly occupy the space occupied by four fringes due to the unknown light, we can write

3×590×D/d = 4λD/d where λ is the wave length of the unknown light (in nanometer).

This gives λ = 442.5 nm

You will find additional multiple choice questions (with solution) at the following locations:

1. apphysicsresources(i)

2. aphysicsresources(ii)

Kerala Engineering Entrance (KEAM) 2008 Questions on Nuclear Physics

Science without religion is lame, religion without science is blind.

– Albert Einstein

KEAM Engineering 2009 Exam will begin on 25^th of this month. Here are the questions on nuclear physics which appeared in last year’s physics eam:

(1) The nuclear radius of a certain nucleus is 7.2 fm and it has a charge of 1.28×10^–17C.The number of neutrons inside the nucleus is

(a) 136

(b) 142

(c) 140

(d) 132

(e) 126

The nuclear radius R is related to the mass number A as

R = 1.2 A^1/3

On substituting for R in this equation we obtain A^1/3 = 6 so that A = 216.

Since the total charge of the nucleus (carried by all the protons) is given as 1.28×10^–17 coulomb and the charge on a proton is 1.6×10^–19 coulomb, the number of protons (Z) in the nucleus is (1.28×10^–17) /(1.6×10^–19) = 80.

Therefore, the number of neutrons in the nucleus, N = A – Z = 216 – 80 = 136.

(2) The energy released in the fission of 1 kg of ₉₂U²³⁵ is (Energy per fission = 200 MeV)

(a) 5.1×10²⁶ eV

(b) 5.1×10²⁶ J

(c) 8.2×10¹³ J

(e) 5.1×10²³ MeV

1 kg of ₉₂U²³⁵ contains 1/0.235 moles and each mole contains 6.02×10²³ atoms. Therefore1 kg of ₉₂U²³⁵ contains (6.02×10²³)/ 0.235 atoms.

Therefore, the energy E released in the fission of 1 kg of ₉₂U²³⁵ is given by

E = (6.02×10²³) ×200 / 0.235 MeV = 5.1×10²⁶ MeV = 5.1×10³² eV

Since 1 eV = 1.6×10^–19 joule, the above energy is (5.1×10 ³²)×(1.6×10^–19) joule. This works out to approximately 8.2×10¹³ J.

(3) Which one of the following statements is true if half life of a radioactive substance is 1 month?

(a) 7/8^th part of the substance will disintegrate in 3 months

(b) 1/8^th part of the substance will remain undecayed at the end of 4 months

(c) The substance will disintegrate completely in 4 months

(d) 1/16^th part of the substance will remain undecayed at the end of 3 months

(e) The substance will disintegrate completely in 2 months

The number of nuclei N remaining undecayed at the end of n half lives is given by

N = N₀/2ⁿ where N₀ is the initial number.

Therefore, at the end of 3 months (3 half lives in the present case) the number remaining undecayed is N₀/2³ = N₀/8.

This means that 7/8^th part of the substance will disintegrate in 3 months [Option (a)].

You will find a few more multiple choice questions (with solution) on nuclear physics here.

All India Engineering/Architecture Entrance Examination (AIEEE) 2009 Question on Transients in LR Circuit

It is not by sitting still at a grand distance and calling the human race larvae that men are to be helped.

– Albert Einstein

The following question on transients in LR circuit was asked in the All India Engineering/Architecture Entrance Examination (AIEEE) 2009:

An inductor of inductance L = 400 mH and resistors of resistances R₁ = 2 Ω and R₂ = 2 Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

(1) 12 e^–5t V

(2) 6 e^–5t V

(3) (12/t) e^–3t

(4) 6(1– e^–t/0.2) V

The exponential growth of current I in an LR circuit is given by

I = I₀ (1– e^–Rt/L) where I is the current at the instant t, I₀ is the final maximum current equal to V/R where V is the voltage applied across the series LR combination and e is the base of natural logarithms. The potential drop across L at the instant t is L dI/dt

Therefore potential drop across L at the instant t = L (I₀R/L) e^–Rt/L = I₀R e^–Rt/L

In the present problem I₀ = 6 A (since V = 12 volt and R = R₂ = 2 Ω) and L = 400 mH = 0.4 H.

Therefore, the potential drop across L at the instant t = 6×2×e^–2t/0.4 = 12 e^–5t volt [Option (1)].

Let us try the following question involving transients in a CR circuits:

Suppose the inductor in the above question is replaced by a capacitor of capacitance 1000 μF. The switch S is closed at time t = 0. The charging current flowing through the capacitor as a function of time is

(1) 12 e^–500t A

(2) 6 e^–500t A

(3) 12e^–t/1000 A

(4) 6(1– e^–t/500) A

The exponential growth of charge Q on a capacitor in a CR circuit is given by

Q = Q₀ (1– e^–t/RC) where Q is the charge at the instant t, Q₀ is the final maximum charge which is equal to CV. The charging current I flowing through the capacitor is given by

I = dQ/dt = (Q₀/RC) e^–t/RC = (CV/RC) e^–t/RC = (V/R) e^–t/RC

Here V = 12 volt, C = 1000 μF = 0.001 F and R = R₂ = 2 Ω

Therefore, the charging current I flowing through the capacitor = 6× e^{–t/(2×0.001)} = 6 e^–500t A.

You will find a few useful multiple choice questions (with solution) on transients here at apphysicsresources. You can find more questions on transients in LR and CR circuits at other locations on the same site.

IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on Nuclear Physics

Our greatest weakness lies in giving up. The most certain way to succed is always to try just one more time.

– Thomas Alva Edison

The following three questions [(i), (ii) and (iii)] from nuclear physics were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper. They are simple as you will realize.

Paragraph for Questions (i), (ii) and (iii)

Scientists are working hard to develop unclear fusion reactor. Nuclei of heavy hydrogen, ₁H², known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D–D reaction is ₁H²+₁H² → ₂He³ + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ₁H² nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t₀ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt₀ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 10¹⁴ s/cm³.

It may be helpful to use the following : Boltzmann constant k = 8.6 × 10^–5 eV/K; e²/4πε₀ = 1.44×10^–9 eVm

Question (i):

In the core of nuclear fusion reactor, the gas becomes plasma because of

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between the deuteron–electron pairs

(D) the high temperature maintained inside the reactor core.

The correct option is (D) since the plasma state is achieved at high temperatures.

Question (ii):

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10^–15 m is in the range

(A) 1.0×10⁹ K < T < 2×10⁹ K

(B) 2.0×10⁹ K < T < 3×10⁹ K

(C) 3.0×10⁹ K < T < 4×10⁹ K

(D) 4.0×10⁹ K < T < 5×10⁹ K

The total kinetic energy of the two deuterons is 2×1.5 kT = 3 kT. At the required separation r (equal to 4×10^–15 m), the entire kinetic energy gets converted into electrostatic potential energy:

3 kT = e²/4πε₀r

Or, 3×8.6×10^–5×T = 1.44×10^–9/(4×10^–15) since e²/4πε₀ = 1.44×10^–9 eVm

Therefore, T = 1.4×10⁹ K. [Option (A)].

Question (iii):

Results of calculations for four different designs of a fusion reactor using D–D reaction are given below. Which of these is most promising based on Lawson criterion ?

(A) deuteron density = 2.0×10¹² cm^–3, confinement time = 5.0×10^–3 s

(B) deuteron density = 8.0×10¹⁴ cm^–3, confinement time = 9.0×10^–1 s

(C) deuteron density = 4.0×10²³ cm^–3, confinement time = 1.0×10^–11 s

(D) deuteron density = 1.0×10²⁴ cm^–3, confinement time = 4.0×10^–12 s.

Lawson number is nt₀. Out of given options, Lawson number is greater than 5 × 10¹⁴ s/cm³ for option (B) since n = 8.0×10¹⁴ cm^–3 and t₀ = 9.0×10^–1 s.

Therefore the correct option is (B).

You will find some useful multiple choice questions on nuclear physics here

AIEEE 2008- An Imaginary Question on Bohr Model

The following question was included in the AIEEE 2008 question paper:

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr’s model to this system, the radius of the n^th orbital of the system is found to be ‘r_n’ and the kinetic energy of the electron to be ‘T_n’. Then which of the following is true?

(1) T_n α 1/n², r_n α n²

(2) T_n independent of n, r_n α n

(3) T_n α 1/n, r_n α n

(4) T_n α 1/n, r_n α n²

The force k/r supplies the centripetal force for the circular motion of the electron so that we have

k/r = mv²/r where ‘m’ is the mass and ‘v’ is the speed of the electron.

Therefore, mv² = k which is independent of the quantum number ‘n’. The kinetic energy T_n of the electron is ½ mv² which is therefore independent of the quantum number ‘n’.

Also, v = √(k/m).

The angular momentum of the electron in the n^th orbit of radius r_n is mvr_n and in the Bohr model mvr_n = nh/2π where ‘h’ is Planck’s constant. Substituting for ‘v’ we have

m×√(k/m) ×r_n = nh/2π

This gives r_n α n. So the correct option is (2).

In the above question the attractive force on the electron was imagined to be inversely proportional to the distance just for the sake of testing your problem solving skill. In a real hydrogen atom the force is certainly inversely proportional to the square of the distance. You will find questions on real Bohr model on this site by clicking on the label ‘Bohr model ‘ below this post.