1

### JEE Main 2018 (Online) 16th April Morning Slot

A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of 80% will be :
A
7 V
B
28 V
C
11.2 V
D
22.4 V
2

### JEE Main 2019 (Online) 9th January Evening Slot

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 $\times$ 108m/s, h = 6.6 $\times$ 10$-$34 J-s)
A
3.75 $\times$ 106
B
3.86 $\times$ 106
C
6.25 $\times$ 105
D
4.87 $\times$ 105

## Explanation

Given,

$\lambda$ = 800 nm

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

= 3.75 $\times$ 1014 Hz

Available frequency for signal bandwith

= 1% of F

= 3.75 $\times$ 1014 $\times$ ${1 \over {100}}$

= 3.75 $\times$ 1012 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 $\times$ 105
3

### JEE Main 2019 (Online) 10th January Morning Slot

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
4

### JEE Main 2019 (Online) 10th January Evening Slot

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

NEET