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1

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
An amplitude modulated signal is plotted below :



Which one of the following best describes the above signal ?
A
(1 + 9sin (2$$\pi $$ $$ \times $$ 104 t)) sin(2.5$$\pi $$ $$ \times $$ 105t) V
B
(9 + sin (4$$\pi $$ $$ \times $$ 104 t)) sin(5$$\pi $$ $$ \times $$ 105t) V
C
(9 + sin (2$$\pi $$ $$ \times $$ 104 t)) sin(2.5$$\pi $$ $$ \times $$ 105t) V
D
(9 + sin (2.5$$\pi $$ $$ \times $$ 104 t)) sin(2$$\pi $$ $$ \times $$ 104t) V

Explanation

Analysis of graph says

(1)   Amplitude varies as 8 $$-$$ 10 V or 9 $$ \pm $$ 1

(2)  Two time period as

100 $$\mu $$s (signal wave) & 8 $$\mu $$s (carrier wave)

Hence signal is $$\left[ {9 \pm 1\sin \left( {{{2\pi t} \over {{T_1}}}} \right)} \right]\sin \left( {{{2\pi t} \over {{T_2}}}} \right)$$

= 9 $$ \pm $$ 1sin (2$$\pi $$ $$ \times $$ 104t) sin 2.5$$\pi $$ $$ \times $$ 105 t
2

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 $$ \times $$ 104 t)] sin(5.5 $$ \times $$ 105 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

Explanation

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 $$ \times $$ 104 t) $$-$$ sin(52.8 $$ \times $$ 104 t)]

$$\omega $$1 = 57.2 $$ \times $$ 104 = 2$$\pi $$f1

f1 = $${{57.2 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}} = 9.1 \times {10^4}$$

$$ \simeq $$ 91KHz

f2 = $${{52.8 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}}$$

$$ \simeq $$ 84 KHz



Side band frequency are

f1 = fc $$-$$ fw = $${{52.8 \times {{10}^4}} \over {2\pi }}$$ $$ \simeq $$ 85.00 kHz

f2 = fc + fw = $${{57.2 \times {{10}^4}} \over {2\pi }}$$ $$ \simeq $$ 90.00 kHz
3

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
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

Explanation

fcarrier = $${{250} \over {0.1}}$$ = 2500 KHZ

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

Now check all options : for 2000 KHZ

fmod = 200 Hz

$$ \therefore $$   Range = 1800 KHZ to 2200 KHZ
4

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)
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 × 106 m).
A
40 km
B
65 km
C
48 km
D
80 km

Explanation

Maximum distance upto which signal can be broadcasted is

dmax = $$\sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$

where hT and hR are heights of transmitter tower and height of reserver respectively. Putting all values -

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

on solving, dmax = 65 km

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