Communication System – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

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Previous Years AIEEE/JEE Mains Questions

Q. This questions has Statement-1 and Statement-2. Of the four choice given after the statements, choose the one that best describes the two statements.

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.

Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.

(1) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of statement-1

(2) Statement-1 is true, Statement-2 is true, Statrment-2 is not the correct explanation of Statement-1

(3) Statement-1 is false, Statement-2 is true

(4) Statement-1 is true, Statement-2 is false

[AIEEE-2011]

Sol. (2)

Q. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance up to which it can detect object located on the surface of the earth (Radius of earth $\left.=6.4 \times 10^{6} \mathrm{m}\right)$ is :

(1) 40 km (2) 64 (3) 80 km (4) 16km

[AIEEE-2012]

Sol. (3)

$\mathrm{d}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6400 \times 0.5}=80 \mathrm{km}$

Q. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

(1) 10.62kHz (2) 5.31 MHz (3) 5.31 kHz (4) 10.62MHz

[JEE Main-2013]

Sol. (3)

$\mathrm{f}_{\mathrm{max}} \leq \frac{\sqrt{\frac{1}{\mathrm{m}^{2}}-1}}{2 \pi \mathrm{RC}}$

$\mathrm{f}_{\max } \leq \frac{8}{6 \times 2 \pi \times 100 \times 10^{3} \times 250 \times 10^{-12}}$

$\mathrm{f}_{\max } \leq \frac{8 \times 10^{6}}{12 \pi \times 25}$

$\mathrm{f}_{\max } \leq 8.4925 \mathrm{kHz}$

Q. A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are –

(1) 2005 kHz, 2000 kHz and 1995 kHz

(2) 2000 kHz and 1995 kHz

(3) 2 MHz only

(4) 2005 kHz and 1995 kHz

[JEE Main-2015]

Sol. (1)

Frequency present after modulation

$\mathrm{f}_{\mathrm{c}}, \mathrm{f}_{\mathrm{c}} \pm \mathrm{f}_{\mathrm{s}}$

$\Rightarrow 2000 \mathrm{KHz}, 2005 \mathrm{KHz}$ and $1995 \mathrm{KHz}$

Q. Choose the correct statement :

(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.

(2) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(3) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

[JEE Main-2016]

Sol. (2)

Q. In amplitude modulation, sinusoidal carrier frequency used is denoted by $\omega_{\mathrm{c}}$ and the signal frequency is denoted by $\omega_{\mathrm{m}}$. The bandwidth $\left(\Delta \omega_{\mathrm{m}}\right)$ of the signal is such that $\Delta \omega_{\mathrm{m}}<<\omega_{\mathrm{c}}$. Which of the following frequencies is not contained in the modulated wave ? [JEE Main-2017]

Sol. (3)

Refer NCERT Page No. 526 Three frequencies are contained $\omega_{\mathrm{m}}+\omega_{\mathrm{c}}, \omega_{\mathrm{c}}-\omega_{\mathrm{m}} \& \omega_{\mathrm{c}}$

Q. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ? [JEE Main-2018]

Sol. (2)

Since the carrier frequency is distributed as band width frequency, so 10% of 10 GHz = n × 5 kHz

where n = no of channels $\frac{10}{100} \times 10 \times 10^{9}=n \times 5 \times 10^{3} \mathrm{n}=2 \times 10^{5}$ telephonic channels