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#### If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is : Option: 1 NOT Option: 2 AND Option: 3 OR Option: 4 NAND \

The output of OR gate is 0 when all inputs are 0 and output is 1 when at least one of the input is 1.

Observing output x it is 0 when all inputs are 0 and it is 1 when at least one of the inputs is 1
therefore, the gate is OR

#### The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency , peak voltage is amplitude modulated by a sine wave of amplitude are respectively. Then the value of is______

$\begin{gathered} f_{c}=11.21 \times 10^{6} \mathrm{~Hz} \\ A_{c}=15 \\ f_{m}=7.7 \times 10^{3} \mathrm{~Hz} \\ A_{m}=5 \mathrm{~V} \end{gathered}$

\begin{aligned} \frac{a}{10}=\frac{b}{10} &=\frac{\mu A_{c}}{2} \\ \frac{a}{b} &=1 \end{aligned}

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#### The maximum amplitude for an amplitude modulated wave is found to be  while the minimum amplitude is found to be . The modulation index is where  is ______.

\begin{aligned} &\text { Maximum \: amplitude}=V_{\text {max }}=12\\ \end{aligned}

\begin{aligned} &\text { Minimum amplitude }=V_{\text {min }}=3\\ \end{aligned}

\begin{aligned} &\mu =\frac{V_{\max }-V_{\min }}{V_{\max }+V_{\min }}=\frac{9}{15}\\ &\mu =0.6\\ &\therefore x=1 \end{aligned}

#### An amplitude modulated wave is represented by  volts. The modulating frequency in kHz will be -----------------------.

$C_{m}\left ( t \right )= 10\left ( 1+0\cdot 2\cos 12560t \right )\sin \left ( 111\times 10^{4}t \right )$
$\omega _{m}= 12560= 2\pi f$
$12560= 6\cdot 28\times f$
$f= \frac{12560}{6\cdot 28}$
$= 2\cdot 0\, kHz$
$f= 2\, kHz$

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#### 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 the range of communication for Line of Sight (LoS) mode? [use radius of earth = 6400 km]Option: 1 80.2 kmOption: 2 144.1 kmOption: 3 57.28 kmOption: 4 45.5 km

$Range\; of\: communication = \sqrt{2h_{T}R}+\sqrt{2h_{R}R}$
$= \sqrt{2\times 50\times 6400\times 10^{3}}+\sqrt{2\times 80\times 6400\times 10^{3}}$
$=10\times 80\times 10\sqrt{10}+40\times 80\times 10$
$=800\left ( 10\sqrt{10}+40 \right )$
$=800\left ( 31\cdot 62+40 \right )$
$Range=57\cdot 28\times 10^{3}m= 57\cdot 28\, km$
The correct option is (3)

#### A transmitting antenna has a height of  and that of receiving antenna is . The maximum distance between them for satisfactory communication in line of sight mode is . The value of ________

$Range\, of\, communication= \sqrt{2h_{T}R}+\sqrt{2h_{R}R}$
$= \sqrt{2\times 320\times 6400\times 10^{3}}+\sqrt{2\times 2000\times 6400\times 10^{3}}$
$=80\times 80\times 10+200\times 80\times 10$
$=64000+160000$
$=224000 \, m$
$=224 \, km$

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#### An antenna is mounted on a tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of ?Option: 1Option: 2Option: 3Option: 4

Range of Transmission $= \sqrt{2hR}$

Height of transmission tower $= h= h_{building }+h_{antenna}$

$44000= \sqrt{2\times h\times 6400\times10^{3}}$

$44000\times 44000= 2h\times 6400\times10^{3}$

$h_{min}= 151.25$

$\lambda = 4h= 605m$

#### If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at , then the maximum range of LOS communication is___________

Range of LOS communication $= \sqrt{2h_{T}R}+\sqrt{2h_{R}R}$
$h_{T}+h_{R}=160\, m$
For maximum range ,
$h_{T}=h_{R}$
$\therefore h_{T}=h_{R}= 80 m$
$= 2\sqrt{2h_{T}R}$
$= 2\times \sqrt{2\times 80\times 6400\times 10^{3}}$
$= 2\times \sqrt{160\times 6400\times 10^{3}}$
$= 2\times80\times 40\times 10$
$\Rightarrow Range= 64000 m= 64\, km$

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#### A bandwidth of is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed . The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be_________. Given 27

Signal Bandwidth $=2f_{m}$

$\because$ $f_{m}=6KHz$

$\therefore$ Signal Bandwidth $=12KHz$

No of stations that can be broadcast is let say $N$

$N=\frac{6MHz}{12KHz}=\frac{6000}{12}=500$

#### A carrier wave with amplitude of is amplitude modulated by a sinusoidal base band signal of amplitude . The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is , then value of is____________.

$A_{max}= A_{c}+A_{m}= 250+150= 400$
$A_{min}= A_{c}-A_{m}= 250-150= 100$
$\frac{A_{min}}{A_{max}}= \frac{100}{400}= \frac{1}{4}= \frac{50}{200}$
$x= 200$

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