Top 50 Most Repeated ELECTROMAGNETIC WAVES PYQs | JEE MAINS
A curated collection of the most important questions from ELECTROMAGNETIC WAVES, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from ELECTROMAGNETIC WAVES, fully solved with step-by-step concepts to prepare for JEE MAINS.
Use formula $f_c = 9\sqrt{N_{max}}$ and substitute value....
Read Full Step-by-Step Solution →Shannon's formula for channel capacity is $ C = B \log_2(1 + \frac{S}{N_0 B}) $. When bandwidth $ B $ is doubled, the signal-to-noise ratio per hertz ...
Read Full Step-by-Step Solution →In AM, sideband frequencies are given by $f_c \pm f_m$, where $f_c = 1000\,\text{kHz}$ and $f_m = 10\,\text{kHz}$. Thus, upper sideband =
Total bandwidth available is $5 \text{ MHz} = 5000 \text{ kHz}$. Each voice channel requires $4 \text{ kHz}$. Number of channels $= \frac{5000}{4} = 1...
Read Full Step-by-Step Solution →Bandwidth = 2 × maximum modulating frequency (fm)....
Read Full Step-by-Step Solution →Modulation index $mu = (V_{max} - V_{min}) / (V_{max} + V_{min}) = 8/12 = 2/3$....
Read Full Step-by-Step Solution →In a vacuum EM wave, (i) the fields are orthogonal, (ii) momentum $p = \frac{U}{c}$, and (iii) the speed is $c$. Statement (iv) is false because $E = ...
Read Full Step-by-Step Solution →The linear SNR is $P_{signal}/P_{noise}=10/0.1=100$. Converting to decibels: $\text{SNR}_{dB}=10\log_{10}(100)=10\times2=20\ \text{dB}$....
Read Full Step-by-Step Solution →Sky wave propagation uses HF band (3–30 MHz) reflected by ionosphere....
Read Full Step-by-Step Solution →First compute $\log_2(1+\text{SNR}) = \log_2(16) = 4$. Then $C = 3\ \text{MHz} \times 4 = 12\ \text{Mbps}$. The answer is an integer value of 12....
Read Full Step-by-Step Solution →Use formula $f_c = 9\sqrt{N_{max}}$ and substitute value....
Read Full Step-by-Step Solution →Shannon's formula for channel capacity is $ C = B \log_2(1 + \frac{S}{N_0 B}) $. When bandwidth $ B $ is doubled, the signal-to-noise ratio per hertz ...
Read Full Step-by-Step Solution →In AM, sideband frequencies are given by $f_c \pm f_m$, where $f_c = 1000\,\text{kHz}$ and $f_m = 10\,\text{kHz}$. Thus, upper sideband =
Total bandwidth available is $5 \text{ MHz} = 5000 \text{ kHz}$. Each voice channel requires $4 \text{ kHz}$. Number of channels $= \frac{5000}{4} = 1...
Read Full Step-by-Step Solution →Bandwidth = 2 × maximum modulating frequency (fm)....
Read Full Step-by-Step Solution →Modulation index $mu = (V_{max} - V_{min}) / (V_{max} + V_{min}) = 8/12 = 2/3$....
Read Full Step-by-Step Solution →In a vacuum EM wave, (i) the fields are orthogonal, (ii) momentum $p = \frac{U}{c}$, and (iii) the speed is $c$. Statement (iv) is false because $E = ...
Read Full Step-by-Step Solution →The linear SNR is $P_{signal}/P_{noise}=10/0.1=100$. Converting to decibels: $\text{SNR}_{dB}=10\log_{10}(100)=10\times2=20\ \text{dB}$....
Read Full Step-by-Step Solution →Sky wave propagation uses HF band (3–30 MHz) reflected by ionosphere....
Read Full Step-by-Step Solution →First compute $\log_2(1+\text{SNR}) = \log_2(16) = 4$. Then $C = 3\ \text{MHz} \times 4 = 12\ \text{Mbps}$. The answer is an integer value of 12....
Read Full Step-by-Step Solution →Use formula $f_c = 9\sqrt{N_{max}}$ and substitute value....
Read Full Step-by-Step Solution →Shannon's formula for channel capacity is $ C = B \log_2(1 + \frac{S}{N_0 B}) $. When bandwidth $ B $ is doubled, the signal-to-noise ratio per hertz ...
Read Full Step-by-Step Solution →In AM, sideband frequencies are given by $f_c \pm f_m$, where $f_c = 1000\,\text{kHz}$ and $f_m = 10\,\text{kHz}$. Thus, upper sideband =
Total bandwidth available is $5 \text{ MHz} = 5000 \text{ kHz}$. Each voice channel requires $4 \text{ kHz}$. Number of channels $= \frac{5000}{4} = 1...
Read Full Step-by-Step Solution →Bandwidth = 2 × maximum modulating frequency (fm)....
Read Full Step-by-Step Solution →Modulation index $mu = (V_{max} - V_{min}) / (V_{max} + V_{min}) = 8/12 = 2/3$....
Read Full Step-by-Step Solution →In a vacuum EM wave, (i) the fields are orthogonal, (ii) momentum $p = \frac{U}{c}$, and (iii) the speed is $c$. Statement (iv) is false because $E = ...
Read Full Step-by-Step Solution →The linear SNR is $P_{signal}/P_{noise}=10/0.1=100$. Converting to decibels: $\text{SNR}_{dB}=10\log_{10}(100)=10\times2=20\ \text{dB}$....
Read Full Step-by-Step Solution →Sky wave propagation uses HF band (3–30 MHz) reflected by ionosphere....
Read Full Step-by-Step Solution →First compute $\log_2(1+\text{SNR}) = \log_2(16) = 4$. Then $C = 3\ \text{MHz} \times 4 = 12\ \text{Mbps}$. The answer is an integer value of 12....
Read Full Step-by-Step Solution →Use formula $f_c = 9\sqrt{N_{max}}$ and substitute value....
Read Full Step-by-Step Solution →Shannon's formula for channel capacity is $ C = B \log_2(1 + \frac{S}{N_0 B}) $. When bandwidth $ B $ is doubled, the signal-to-noise ratio per hertz ...
Read Full Step-by-Step Solution →In AM, sideband frequencies are given by $f_c \pm f_m$, where $f_c = 1000\,\text{kHz}$ and $f_m = 10\,\text{kHz}$. Thus, upper sideband =
Total bandwidth available is $5 \text{ MHz} = 5000 \text{ kHz}$. Each voice channel requires $4 \text{ kHz}$. Number of channels $= \frac{5000}{4} = 1...
Read Full Step-by-Step Solution →Bandwidth = 2 × maximum modulating frequency (fm)....
Read Full Step-by-Step Solution →Modulation index $mu = (V_{max} - V_{min}) / (V_{max} + V_{min}) = 8/12 = 2/3$....
Read Full Step-by-Step Solution →In a vacuum EM wave, (i) the fields are orthogonal, (ii) momentum $p = \frac{U}{c}$, and (iii) the speed is $c$. Statement (iv) is false because $E = ...
Read Full Step-by-Step Solution →The linear SNR is $P_{signal}/P_{noise}=10/0.1=100$. Converting to decibels: $\text{SNR}_{dB}=10\log_{10}(100)=10\times2=20\ \text{dB}$....
Read Full Step-by-Step Solution →Sky wave propagation uses HF band (3–30 MHz) reflected by ionosphere....
Read Full Step-by-Step Solution →First compute $\log_2(1+\text{SNR}) = \log_2(16) = 4$. Then $C = 3\ \text{MHz} \times 4 = 12\ \text{Mbps}$. The answer is an integer value of 12....
Read Full Step-by-Step Solution →Use formula $f_c = 9\sqrt{N_{max}}$ and substitute value....
Read Full Step-by-Step Solution →Shannon's formula for channel capacity is $ C = B \log_2(1 + \frac{S}{N_0 B}) $. When bandwidth $ B $ is doubled, the signal-to-noise ratio per hertz ...
Read Full Step-by-Step Solution →In AM, sideband frequencies are given by $f_c \pm f_m$, where $f_c = 1000\,\text{kHz}$ and $f_m = 10\,\text{kHz}$. Thus, upper sideband =
Total bandwidth available is $5 \text{ MHz} = 5000 \text{ kHz}$. Each voice channel requires $4 \text{ kHz}$. Number of channels $= \frac{5000}{4} = 1...
Read Full Step-by-Step Solution →Bandwidth = 2 × maximum modulating frequency (fm)....
Read Full Step-by-Step Solution →Modulation index $mu = (V_{max} - V_{min}) / (V_{max} + V_{min}) = 8/12 = 2/3$....
Read Full Step-by-Step Solution →