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Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 12
1.
8416 - 2A16 is equal to
5A16
4A16
7A16
15A16
Answer: Option
Explanation:

(2A)16 = (00101010)2

2's complement = 11010110 = (D6)16

(84)16 + (D6)16 ⇒ 1(5A)16 carry discarded ⇒ (5A)16 .

2.
Calculate the electrostatic force of repulsion between two mica particles at a distance of 10-13 m from each other. Charge on a particle is 3.2 x 10-19 C. (relative permittivity for mica is 5)
1.843 x 10-15 N
9.2 x 10-2 N
14.41 x 10-9 N
none of these
Answer: Option
Explanation:

By Coulomb's law:

Electrostatic force = =

= 18.43 x 10- 3 N.

3.
An electromechanical closed loop control system has the following characteristics equation :s3 + 6Ks2 + (K + 2)s = 8
where K is the forward gain of the system. The condition for closed loop stability is
K = 0.528
K = 2
K = 0
K = -2528
Answer: Option
Explanation:

s3 + 6ks2 + (k + 2)s + 8 = 0

a0 = 1, a1 = 6k, a2 = k + 2, a3 = 8

Apply R.H. criteria.

4.
What is the propagation constant for air filled wave guide with dimensions a = 1.59" and b = 0.795" at 4.95 GHz. The wave impedance is
400.94 Ω
562.94 Ω
502.80 Ω
492.80 Ω
Answer: Option
Explanation:

.

5.
An antenna array comprises of two dipoles that are separated by wavelength. The dipoles are fed by currents of the same magnitude and 90° in phase. The array factor is:
2 cos - 2 sin
2 cos + 2 sin
2 cos
2 sin
Answer: Option
Explanation:

Array factor

φ = 90°.

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