Monday, January 25, 2010

Karnataka CET Questions (MCQ) on Digital Circuits

I give below 4 questions which appeared in Karnataka Common Entrance Test (CET) question papers of the years 2009, 2008, 2007 and 2006 respectively:

(1) In the following combination of logic gates, the outputs A, B and C are respectively

(1) 0, 1, 0

(2) 1, 1, 0

(3) 1, 0, 1

(4) 0, 1, 1

In circuit (A), the output of the input NAND gate is 0 since both inputs are 1. After inversion it becomes 1 and the final OR gate produces output 1.

In circuit (B), the output NAND gate gets 1 and 0 as its inputs and hence the final output is 1.

In the circuit (C), the NOR gate at the input side supplies 0 as one input to the AND gate at the output side. Hence the final output is 0.

Therefore, the correct option is (2).

(2) To get an output Y = 1 from the circuit shown, the inputs A, B and C must be respectively

(1) 0, 1, 0

(2) 1, 0, 0

(3) 1, 0, 1

(4) 1, 1, 0

In all cases the OR gate will provide a high (level 1) input to the AND gate at the output side. The AND gate should have both inputs at level 1 in order to have its output at 1. This is possible only if input C is 1. So the correct option is (3).

(3) The truth table given below is for

(A and B are the inputs, y is the output)

(1) NAND

(2) XOR

(3) AND

(4) NOR

The output y is the complement of AND operation and hence the answer is NAND [Option (1)].

(4) Identify the logic operation performrd by the cicuit given below

(1) NOT

(2) AND

(3) OR

(4) NAND

If you know De Morgan’s theorem, you will immediately obtain the answer as AND. The NOR gate at the output side gets inputs which are complement of A and complement of B. According to De Morgan’s theorem, (NOT A OR NOT B) is the same as [NOT (A AND B)]. Since the output gate is not an OR gate, but a NOR gate, the output is (A AND B).

Therefore, given circuit performs AND operation [Option (2)].

[If you do not know De Morgan’s theorem, you may try substituting values 0 and 1 for A and B to obtain the answer].

Monday, January 18, 2010

Two Questions (MCQ) from Wave Optics

Here are two questions from wave optics. The first question is meant for testing your understanding of coherent sources and interference while the second one is meant for testing your knowledge of the interference pattern produced by Young’s double slit.

(1) Two coherent sources emitting light of wave length λ and amplitude A produce an interference pattern on a screen. The maximum intensity of the interference pattern is I. If the light sources are not coherent but the wave length and amplitude are λ and A respectively, the maximum intensity on the screen will be

(a) I/4

(b) I/2

(c) I

(d) √I

(e) 2√I

At the interference maximum the amplitudes of the light waves get added. Therefore, the resultant amplitude at the interference maximum is A + A = 2A.

The intensity is directly proportional to the square of the amplitude so that we have

I α 4A2

Or, I = k×4A2……………(i)

where k is the constant of proportionality

Since the amplitude of each light wave is A, the intensity (I0) of light produced by each source is given by

I0 = k×A2

This gives I0 = I/4 from (i).

Each source will produce uniform intensity I0 everywhere on the screen and the resultant intensity everywhere will be 2I0 (since there are two sources) and we have

2I0 = 2×I/4 = I/2

The correct option is (b).

(2) The angular width of the interference fringes obtained in a double slit experiment using light of wave length λ is found to be θ. If the entire experimental arrangement is immersed in water having refractive index 4/3, the angular fringe width will be

(a) θ/4

(b) θ/3

(c) θ

(d) 4θ/3

(e) 3θ/4

The angular fringe width is λ/d where d is the separation between the slits. Therefore we have

θ = λ/d.

When the arrangement is submerged in water of refractive index n, the wave length of light is reduced to λ' given by

λ'= λ/n = 3λ/4 since n = 4/3

The angular fringe width now becomes θ'= λ'/d = 3λ/4d = 3θ/4.

You will find some useful multiple choice questions in this section at AP Physics Resources: Multiple Choice Questions on Interference and Diffraction .

Friday, January 01, 2010

“New Year’s day is every man’s birthday.”

– Charles Lamb

Happy New Year…