Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwith. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 $$ \times $$ 10^{8}m/s, h = 6.6 $$ \times $$ 10^{$$-$$34} J-s)

A

3.75 $$ \times $$ 10^{6}

B

3.86 $$ \times $$ 10^{6}

C

6.25 $$ \times $$ 10^{5}

D

4.87 $$ \times $$ 10^{5}

Given,

$$\lambda $$ = 800 nm

$$ \therefore $$ f = $${{3 \times {{10}^8}} \over {800 \times {{10}^{ - 9}}}}$$

= 3.75 $$ \times $$ 10^{14} Hz

Available frequency for signal bandwith

= 1% of F

= 3.75 $$ \times $$ 10^{14} $$ \times $$ $${1 \over {100}}$$

= 3.75 $$ \times $$ 10^{12} Hz

One TV signal needs = 6 MHz band width

$$ \therefore $$ Total number of channel possible

= $${{3.75 \times {{10}^{12}}} \over {6 \times {{10}^6}}}$$

= 6.25 $$ \times $$ 10^{5}

$$\lambda $$ = 800 nm

$$ \therefore $$ f = $${{3 \times {{10}^8}} \over {800 \times {{10}^{ - 9}}}}$$

= 3.75 $$ \times $$ 10

Available frequency for signal bandwith

= 1% of F

= 3.75 $$ \times $$ 10

= 3.75 $$ \times $$ 10

One TV signal needs = 6 MHz band width

$$ \therefore $$ Total number of channel possible

= $${{3.75 \times {{10}^{12}}} \over {6 \times {{10}^6}}}$$

= 6.25 $$ \times $$ 10

2

A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower is LOS (Line of Sight) mode ? (Given : radius of earth = 6.4 × 10^{6} m).

A

40 km

B

65 km

C

48 km

D

80 km

Maximum distance upto which signal can be broadcasted is

d_{max} = $$\sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$

where h_{T} and h_{R} are heights of transmitter tower and height of reserver respectively. Putting all values -

d_{max} = $$\sqrt {2 \times 6.4 \times 106} \left[ {\sqrt {104} + \sqrt {40} } \right]$$

on solving, d_{max} = 65 km

d

where h

d

on solving, d

3

The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot ?

A

2900 kHz

B

2750 kHz

C

2250 kHz

D

2000 kz

f_{carrier} = $${{250} \over {0.1}}$$ = 2500 KHZ

$$ \therefore $$ Range of signal = 2250 Hz to 2750 Hz

Now check all options : for 2000 KHZ

f_{mod} = 200 Hz

$$ \therefore $$ Range = 1800 KHZ to 2200 KHZ

$$ \therefore $$ Range of signal = 2250 Hz to 2750 Hz

Now check all options : for 2000 KHZ

f

$$ \therefore $$ Range = 1800 KHZ to 2200 KHZ

4

An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 $$ \times $$ 10^{4}
t)] sin(5.5 $$ \times $$ 10^{5}
t). Here t is in
seconds. The sideband frequencies (in kHz) are, [Given $$\pi $$ = 22/7]

A

892.5 and 857.5

B

89.25 and 85.75

C

1785 and 1715

D

178.5 and 171.5

V(t) = 10 + $${3 \over 2}$$ [2cos A sinB]

= 10 + $${3 \over 2}$$ [sin(A+B) $$-$$ sin(A $$-$$ B)]

= 10+$${3 \over 2}$$[sin (57.2 $$ \times $$ 10^{4} t) $$-$$ sin(52.8 $$ \times $$ 10^{4} t)]

$$\omega $$_{1} = 57.2 $$ \times $$ 10^{4} = 2$$\pi $$f_{1}

f_{1} = $${{57.2 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}} = 9.1 \times {10^4}$$

$$ \simeq $$ 91KHz

f_{2} = $${{52.8 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}}$$

$$ \simeq $$ 84 KHz

Side band frequency are

f_{1} = f_{c} $$-$$ f_{w} = $${{52.8 \times {{10}^4}} \over {2\pi }}$$ $$ \simeq $$ 85.00 kHz

f_{2} = f_{c} + f_{w} = $${{57.2 \times {{10}^4}} \over {2\pi }}$$ $$ \simeq $$ 90.00 kHz

= 10 + $${3 \over 2}$$ [sin(A+B) $$-$$ sin(A $$-$$ B)]

= 10+$${3 \over 2}$$[sin (57.2 $$ \times $$ 10

$$\omega $$

f

$$ \simeq $$ 91KHz

f

$$ \simeq $$ 84 KHz

Side band frequency are

f

f

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