Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Broadcasting antennas are generally

A. Omni directional type

B. Vertical type

C. Horizontal type

D. None of these

SOLUTION

Solution : B

Broadcasting antennas are generally vertical type.

Question 2

The attenuation in optical fiber is mainly due to

A. Absorption

B. Scattering

C. Neither absorption nor scattering

D. Both (a) and (b)

SOLUTION

Solution : D

A very small part of light energy is lost from an optical fiber due to absorption or due to light leaving the fiber as a result of scattering of light sideways by impurities in the glass fiber.

Question 3

The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to

h^{\frac{1}{2}}

B. h

h^{\frac{3}{2}}

h^{2}

SOLUTION

Solution : A

d = $\sqrt{2 h R} \Rightarrow d α h^{\frac{1}{2}}$

Question 4

Laser beams are used to measure long distances because

A. They are monochromatic

B. They are highly polarised

C. They are coherent

D. They have high degree of parallelism

SOLUTION

Solution : D

Laser beams are perfectly parallel, so that they are very narrow and can travel a long distance without spreading. This is the feature of laser while monochromatic and coherent are characteristics only.

Question 5

An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulating index

A. 0.20

B. 5.0

C. 0.67

D. 1.5

SOLUTION

Solution : B

The formula for modulation index is given by
$m_{f} = \frac{δ}{v_{m}} = \frac{F r e q u e n c y v a r i a t i o n}{M o d u l a t i n g f r e q u e n c y} = \frac{10 \times 10^{3}}{2 \times 10^{3}} = 5$

Question 6

Consider telecommunication through optical fibres. Which of the following statements is not true.

A. Optical fibres may have homogeneous core with a suitable cladding.

B. Optical fibres are subject to electromagnetic interference from outside.

C. Optical fibres can be of graded refractive index

D. Optical fibres have extremely low transmission loss

SOLUTION

Solution : B

Optical fibres are not subjected to electromagnetic interference from outside.

Question 7

For television broadcasting, the frequency employed is normally

A. 30-300 MHz

B. 30-300 GHz

C. 30-300 KHz

D. 30-300 Hz

SOLUTION

Solution : A

VHF (Very High Frequency) band having frequency range 30 MHz to 300 MHz is typically used for TV and radar transmission.

Question 8

What is the modulation index of an over modulated wave

A. 1

B. >1

C. <1

D. Zero

SOLUTION

Solution : B

When m > 1 then carrier is said to be over modulated.

Question 9

In frequency modulation

A. The amplitude of modulated wave varies as frequency of carrier wave

B. The frequency of modulated wave varies as amplitude of modulating wave

C. The amplitude of modulated wave varies as amplitude of carrier wave

D. The frequency of modulated wave varies as frequency of modulating wave

E. The frequency of modulated wave varies as frequency of carrier wave

SOLUTION

Solution : B

The process of changing the frequency of a carrier wave (modulated wave) in accordance with the audio frequency signal (modulating wave) is known as frequency modulation (FM).

Question 10

A transmitter supplies 9 kW to the aerial when unmodulated. The power radiated when modulated to 40% is

A. 5 kW

B. 9.72 kW

C. 10 kW

D. 12 kW

SOLUTION

Solution : B

$P_{t}$ = $P_{c} [1 + \frac{m^{2}}{2}]$ = 9 [1 + $\frac{{(0.4)}^{2}}{2}$ ]
= 9 [1 + $\frac{0.16}{2}$ ] ( $∵$ m = 40 % = 0.4)
= 9 (1.08) = 9.72 kW

Question 11

A photodetector is made from a semiconductor In $_{0.53} G a$ $_{0.47} A s$ with $E_{g}$ = 0.73 eV. What is the maximum wavelength, which it can detect

A. 1000 nm

B. 1703 nm

C. 500 nm

D. 173 nm

SOLUTION

Solution : B

Limiting value of hv is $E_{g}$ , such that hv = $\frac{h c}{λ}$ = $E_{g}$
or $λ = \frac{h_{c}}{E_{g}} = \frac{6.63 \times 10^{- 34} J - s \times 3 \times 10^{8} m s^{- 1}}{0.73 \times 1.6 \times 10^{- 19} J}$
= 1703 nm

Question 12

A 500 Hz modulating voltage fed into an FM generator produces a frequency deviation of 2.25 kHz. If amplitude of the voltage is kept constant but frequency is raised to 6 kHz then the new deviation will be

A. 4.5 kHz

B. 54 kHz

C. 27 kHz

D. 15 kHz

SOLUTION

Solution : B

$m_{f}$ = $\frac{δ}{f_{m}} = \frac{2250}{500}$ = 4.5
$∴$ New deviation = 2( $m_{f} f_{m}$ ) = 2 $\times$ 4.5 $\times$ 6 = 54 kHz.

Question 13

A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 $\times 10^{6}$ m.

A. 40 km

B. 44 km

C. 45.5 km

D. 48.5 km

SOLUTION

Solution : C

$\sqrt{2 \times 64 \times 10^{5} \times 32} + \sqrt{2 \times 64 \times 10^{5} \times 50}$ = 64 $\times$ 10^2 $\times$ $\sqrt{10}$ + 8 $\times$ 10^3 $\times$ $\sqrt{10}$ m = 144 $\times$ 10^2 $\times$ $\sqrt{10}$ = 45.5 km

Question 14

The audio signal used to modulate 60 sin 2 $π 10^{6}$ t is 15sin 300 $π$ t. The depth of modulation is

A. 50%

B. 40%

C. 25%

D. 15%

SOLUTION

Solution : C

$m_{a} = \frac{E_{m}}{E_{c}} = \frac{15}{60} \times 100$ = 25 %

Question 15

In AM, the centpercent modulation is achieved when

A. Carrier amplitude = signal amplitude

{C a r r i e r a m p l i t u d e}^{- 1}

= signal amplitude

C. Carrier frequency = signal frequency

{C a r r i e r f r e q u e n c y}^{- 1}

= signal frequency

SOLUTION

Solution : A

When signal amplitude is qual to the carrier amplitude, the amplitude of carrier ware varies between 2A and zero.

$m_{a} = \frac{A m p l i t u d e c h a r g e o f c a r r i e r}{A m p l i t u d e o f n o r m a l c a r r i e r} = \frac{2 A - A}{A} \times 100$ = 100%

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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