Physics M.C.Q (1 Marks)

$ = \frac{{6.63 \times {{10}^{ - 34}}J{\rm{ - }}s \times 3 \times {{10}^8}m{s^{ - 1}}}}{{0.73 \times 1.6 \times {{10}^{ - 19}}J}}$

$= 1703\, nm$

$= 9\,\left[ {1 + \frac{{0.16}}{2}} \right]$ $( m = 40\% = 0.4)$

$= 9 (1.08) = 9.72\, kW$

 $\therefore {P_c} = 1500\,\left[ {\frac{2}{{2 + 1}}} \right]$ $m = 100\% = 1$

$= 1000 \,W$

Now, ${P_{LSB}} = \frac{{{m^2}}}{4} \times {P_c} = \frac{1}{4} \times 600 = 150\,W$

$\therefore \Delta f = \frac{{200}}{2} = 100\,kHz$

Now ${m_f} = \frac{{\Delta f}}{{{f_m}}} = \frac{{100}}{{10}} = 10$

$\therefore$ New deviation $ = 2({m_f}\,{f_m}) = 2 \times 4.5 \times 6 = 54\,kHz.$

Since the number of quantisation level is $16$

$⇒ 2n = 16 ⇒ n = 4$

$\therefore$ bit rate = sampling rate $×$ no. of bits per sample

$= 8000 × 4 = 32,000\, bits/sec.$

$= 1592 \,kHz$

${m_f} = \frac{\delta }{{{\nu _m}}} = \frac{{{\rm{Frequency \,variation\, }}}}{{{\rm{Modulating \,frequency\,,}}}} = \frac{{10 \times {{10}^3}}}{{2 \times {{10}^3}}} = 5$

Now, $m = \frac{{{V_{\max }} - {V_{\min }}}}{{{V_{\max }} + {V_{\min }}}} = \frac{{12 - 4}}{{12 + 4}} = \frac{8}{{16}} = \frac{1}{2} = 0.5 = 50\% $

 Low side band frequency

$ = {f_c} - {f_m} = 1500\,kHz - 10\,kHz = 1490\,kHz$

Upper side band frequency

$ = {f_c} + {f_m} = 1500\,kHz + 10\,kHz = 1510\,kHz$

==> $1800 = {P_c}\left( {1 + \frac{{{{(1)}^2}}}{2}} \right)$

==> ${P_c} = 1200W$

$(i)$ These signals are relatively of short range.

$(ii)$ If every body started transmitting these low frequency signals directly, mutual interference will render all of them ineffective.

$(iii)$ Size of antenna required for their efficient radiation would be larger $i.e$. about $75 km.$

We know that two separate wave front originating from two whereas sources produce interference,Secondary wavelets originating from different parts of same wave front constitute diffraction. So given statement on option $(A)$ is valid.

$f=1 GHz =1 \times 10^{9} \,Hz$

and $c=3 \times 10^{8} \,m / s$

Distance travelled by laser between two detector,

$D =\frac{c}{f}=\frac{3 \times 10^{8}}{1 \times 10^{9}}=0.3 \,m =30 \,cm$

This is the farthest distance between two detector, so that they can see the same pulse simultaneously.

$=\frac{\text { Amplitude of mod ulating signal }}{\text { Amplitude of carrier wave }}$

$\mu=\frac{1}{2}$

$=0.5$

$FM$ Broadcast $\rightarrow 88-108\,MHz$

Television $\rightarrow 54-890\,MHz$

Salellite communication $\rightarrow 3.7-4.2\,GHz$

$\therefore$ $A-II, B-I, C-IV, D-III$

Signal frequency $f _{ m }=5\,kHz$

Carrier wave frequency $f_c=2\,MHz$

$f _{ c }=2000\,KHz$

The resultant signal will have band width of frequency given by

${\left[\left(f_c+f_m\right)-\left(f_c-f_m\right)\right]}$

$\Rightarrow[(2000+5)-(2000-5)]\,kHz$

$\Rightarrow 10\,kHz$

$d _{ M }=2 \times \sqrt{2 \times 6.4 \times 10^6 \times 80}=64\,km$

$A _{ c }- A _{ m }=80$

$\therefore A _{ c }=100$

$A _{ m }=20$

Modulation index $=\frac{20}{100}=\frac{1}{5}$

Amplitude of each sideband

$= A _{ c } \frac{( mod \text { ulation index) }}{2}$

$=100 \times \frac{1}{10}=10 \text { volt }$

$V _{ C }=\frac{100 \pi}{2 \pi}=500\,Hz$

Modulating wave frequency

$V _{ m }=\frac{4 \pi}{2 \pi}=2\,Hz$

$\therefore V_{ c }- V _{ m }, V _{ c }, V _{ c }+ V _{ m }$

$=498\,Hz,500\,Hz,502\,Hz$.

$\mu=\frac{A_m}{A_c}$

$\mu_1=\frac{X}{Y}$

$\mu_2=\frac{X}{2 Y}$

$\frac{\mu_1}{\mu_2}=\frac{2}{1}$

$R _1=\sqrt{2 Rh _1}$

$h _2= h _1+\left( h _1 \times \frac{21}{100}\right)=1.21 h _1$

$\therefore R _2=\sqrt{2 Rh _2}=\sqrt{2 R (1.21) h _1}=1.1 \sqrt{2 Rh _1}$

$\therefore R _2=1.1 R _1$

$\% \text { increase in range }$

$=\frac{ R _2- R _1}{ R _1} \times 100=\left(\frac{ R _2}{ R _1}-1\right) \times 100$

$=(1.1-1) \times 100=10 \%$

$d =\sqrt{2 h _{ T } \cdot R }+\sqrt{2 h _{ R } \cdot R }$

$=\sqrt{2 \times 98 \times 6400 \times 10^3}+0=\frac{112}{\sqrt{10}}\,km$

$\text { So area }=\pi d ^2$

$=3.14 \times \frac{112^2}{10}=3942\,km ^2$

$A _{ m }=3\,V$

Maximum amplitude of modulated wave

$A _{\max }= A _{ c }+ A _{ m }=15+3=18\,V$

Minimum amplitude of modulated wave

$A _{\min }= A _c- A _{ m }=15-3=12\,V$

$\therefore \frac{ A _c+ A _{ m }}{ A _{ c }- A _{ m }}=\frac{18}{12}=\frac{3}{2}$

$=\frac{(4\,km )^2}{2(6400 km )}=\left(\frac{1}{800}\right)\,km =1.25\,m$

$c ( t )=15 \sin (1000 \pi t)$

$f _{ i }=\frac{\omega_{ c }}{2 \pi}=\frac{1000 \pi}{2 \pi}=500\,Hz$

Equation of modulated wave

$m(t)=10 \sin (4 \pi t)$

$f_m=\frac{\omega_m}{2 \pi}=\frac{4 \pi}{2 \pi}=2\,Hz$

Frequencies contained in resultant Amplitude modulated wave are $(500-2)\,Hz,500\,Hz$ and $(500+2)\,Hz$

$=\sqrt{2 \times 64 \times 10^5 \times 180}+\sqrt{2 \times 64 \times 10^5 \times 245}$

$=\left\{\left(8 \times 6 \times 10^3\right)+\left(8 \times 7 \times 10^3\right)\right\} m$ $=(48+56)\,km$

$=104\,km$

$=\frac{14-6}{14+6}=0.4$

$=2 \times 3\,kHz$

$=6\,kHz$

Minimum amplitude of modulated wav $=A_c-A_w=3$

$\therefore A_c-0.6 A_c=3 \Rightarrow 0.4 A_c=3$

$A_c=\frac{3}{0.4}=\frac{15}{2}=7.5\,V$

$A_w=0.6 A_c=4.5\,V$

$\therefore$ Maximum amplitude $= A _{ c }+ A _{ m }$

$=7.5+4.5=12\,V$