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### JEE Main 2021 (Online) 27th August Evening Shift

An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km?
A
37.8 m
B
605 m
C
75.6 m
D
302 m

## Explanation

h : height of antenna

$$\lambda$$ : wavelength of signal

h < $$\lambda$$

$$\lambda$$ > h

$$\lambda$$ > 400 m
2

### JEE Main 2021 (Online) 26th August Evening Shift

A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is range of communication for Line of Sight (LoS) mode?

[use radius of earth = 6400 km]
A
45.5 km
B
80.2 km
C
144.1 km
D
57.28 km

## Explanation

$${d_t} = \sqrt {2R{h_1}} + \sqrt {2R{h_2}}$$

$$= \sqrt {2R} \left( {\sqrt {{h_1}} + \sqrt {{h_2}} } \right)$$

$$= {(2 \times 6400 \times {10^3})^{1/2}}(\sqrt {50} + \sqrt {80} )$$

$$= 3578(7.07 + 8.94)$$

$$= 57.28$$ km
3

### JEE Main 2021 (Online) 25th July Morning Shift

In amplitude modulation, the message signal

Vm(t) = 10 sin (2$$\pi$$ $$\times$$ 105t) volts and

Carrier signal

VC(t) = 20 sin(2$$\pi$$ $$\times$$ 107 t) volts

The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $$\alpha$$ kHz. The value of $$\alpha$$ is :
A
200 kHz
B
50 kHz
C
100 kHz
D
0

## Explanation

Bandwidth = 2 $$\times$$ fm

= 2 $$\times$$ 105 HZ = 200 KHZ
4

### JEE Main 2021 (Online) 22th July Evening Shift

What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of 150 km? The average population density around the tower is 2000/km2 and the value of Re = 6.5 $$\times$$ 106 m.
A
Height = 1241 m

Population covered = 7 $$\times$$ 105
B
Height = 1731 m

Population covered = 1413 $$\times$$ 105
C
Height = 1800 m

Population covered = 1413 $$\times$$ 108
D
Height = 1600 m

Population covered = 2 $$\times$$ 105

## Explanation

$$d = \sqrt {2{R_e}h}$$

Area covered = $$\pi$$d2

$$h = {{{d^2}} \over {2{R_e}}}$$ = 1731 m

Population covered = 2000 $$\times$$ $$\pi$$(d2)

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