Physics M.C.Q (1 Marks)

$d ^{2}=( h + R )^{2}- R ^{2}$

$= h ^{2}+ R ^{2}+2 RH - R ^{2}$

$d ^{2}= h ^{2}+2 Rh$

$\text { as } R \gg \gg \text { h then }$

$d \approx \sqrt{2 Rh } . \ldots . \text { (1) }$

Now, if coverage is to be increased $3$ times

$3 d =\sqrt{2 Rh ^{\prime}} \ldots \text {. (2) }$

Divide $2$ and $1 \frac{3 d }{ d }=\sqrt{\frac{2 R h^{\prime}}{2 R h}}$

$9=\frac{ h ^{\prime}}{ h }$

$9\,h = h \text { ' }$

$\text { If } h =100 m \text { then tower of height } 900 m \text { is } \text { required }$

$=\sqrt{2 R}\left(\sqrt{h_{T}}+\sqrt{h_{R}}\right)$

$=\sqrt{2 \times 64 \times 10^{5}}(\sqrt{25}+\sqrt{49})$

$=192 \sqrt{5} \times 10^{2} m$

$K=192$

$(b)$ $EPR$ is low for longer wavelength.

$(c)$ yes we want to avoid mixing up signals transmitted by different transmitter simultaneously.

$(d)$ Low frequency signals sent to long distance by superimposing with high frequency.

$=2 \times 10^{4} \,Hz =20 \times 10^{3} \,Hz$

$=20 \,kHz$

$d _{2}=2 d _{1}$

$\sqrt{2 Rh _{2}}=2 \sqrt{2 Rh _{1}}$

$h _{2}=4 h _{1}=500\,m$

$\Delta h =500 \,m -125 \,m =375 \,m$

$=\frac{6-2}{6+2} \times 100 \%=50 \%$

$d =\sqrt{2 \times 6400 \times h \times 10^{-3}}( h$ in $m )$

Area $=\pi d ^{2}$

$=\left(\pi \times 2 \times 6400 \times h \times 10^{-3}\right) km ^{2}$

$6.03 \times 100000=100 \times \pi \times 2 \times 6400 \times 10^{-3} h$

$h =\frac{6.03 \times 10^{5}}{10 \times \pi \times 128}$

$h =150\,m$

$V _{\min }=20$

$\%$ modulation =$\left(\frac{ V _{\max }- V _{\min }}{ V _{\max }+ V _{\min }}\right) 100$

$\left(\frac{60-20}{60+20}\right) 100 \Rightarrow\left(\frac{40}{80}\right) 100$

$50 \%$

Frequency $f_{2}=10\,MHz$

$\lambda_{1}=\frac{ c }{ f _{1}}$

$\lambda_{2}=\frac{ c }{ f _{2}}$

Wavelength bandwidth $=\lambda_{2}-\lambda_{1}=20\,m$

$A_{m}=$ amp. of modulating signal

$A_{C}=$ amp. of canrier wave

FM broadcast

Modulating frequency $=20\,k Hz = f$

Deviation ratio $=\frac{\text { frequency deviation }}{\text { modulating frequency }}=\frac{\Delta f}{ f }$

$\Rightarrow$ frequency deviation $-\Delta f = f \times 10$

$=20\,kHz \times 10=200\,kHz$

$\Rightarrow$ Bandwidth $=2( f +\Delta f )$

$=2(20+200)\,kHz$

$=440\,kHz$

$\mu \leq 1$ to avoid distortion

because $\mu>1$ will result in interference between career frequency \& message frequency.

$f_{c}+f_{m}, f_{c}, f_{c}-f_{m}$

Band width $=2 f_{m}$

$=\frac{2 \omega_{ m }}{2 \pi}=\frac{6.28 \times 10^{6}}{3.14}$

$=2\,MHz$

Given $2 A _{ m }=8$

$A _{ m }+ A _{ c }=9 \Rightarrow A _{ c }=5$

$\therefore m =\frac{4}{5}=0.8$

Side band frequencies are $f_{c}-f_{m} \,\& \,f_{c}+f_{m}$. which are almost $f_{c}$

$\lambda=\frac{ c }{ f _{ c }}$

Minimum length of antenna =

$\frac{ c }{ f _{ c } 4}=\frac{\lambda}{4}=\frac{3 \times 10^{8}}{3.5 \times 10^{9} \times 4}$

$=21.4\, mm$

$x(t)=20 \sin (1000 \; \pi t)$

$\Rightarrow \omega_{ c }=10^{7} \; \pi$

$\omega_{ m }=10^{3} \; \pi$

$A _{ C }=40$

$A_{m}=20$

Equation of modulated wave $=\left( A _{ C }+ A _{ m } \sin \omega_{ m } t \right)$

$\sin \omega_{ c } t$

$= A _{c}\left(1+\frac{ A _{ m }}{ A _{ c }} \sin \omega_{ m } t \right) \sin \omega_{ c } t$

$= A _{ c }\left(1+\mu \sin \omega_{ m } t \right) \sin \omega_{ c } t , \quad \mu=\frac{ A _{ m }}{ A _{ c }}$

$=A_{c} \sin \omega_{c} t+\frac{\mu A_{c}}{2}\left[\cos \left(\omega_{c}-\omega_{m}\right) t-\cos \left(\omega_{c}+\omega_{m}\right) t\right]$

Amplitude of minimum frequency = $\frac{\mu A _{ c }}{2}=\frac{ A _{ m }}{ A _{ c }} \times \frac{ A _{ c }}{2}=\frac{ A _{ m }}{2}=10$

$\Rightarrow \lambda=100 m$

$A =\pi d ^{2}$

$A =\pi 2 Rh$

$=3.14 \times 2 \times 6400 \times \frac{30}{1000}$

$A =1205.76 km ^{2}$

$A \simeq 1206 km ^{2}$

$h =\frac{ D ^{2}}{8 R }=\frac{45^{2}}{8 \times 6400} km \cong 39.55\, m$

$\omega_{ m }=1.57 \times 10^{8}=2 \pi f _{ m }$

$BW =2 f _{ m }=\frac{10^{8}}{2}\,Hz =50 \,MHz$

Range $( i )=\sqrt{2 Rh }$

Range $(ii)$ $=\sqrt{2 Rh }+\sqrt{2 Rh ^{\prime}}$

where $h =20\, m \& h ^{\prime}=5\, m$

$Ans =\frac{\sqrt{2 Rh ^{\prime}}}{\sqrt{2 Rh }} \times 100\, \%=\frac{\sqrt{5}}{\sqrt{20}} \times 100\, \%=50\, \%$

$\%$ modulation $=\frac{20}{80} \times 100$

$\%$ modulation $=25 \%$

$f _{ c }=1 MHz =1000 kHz$

Band width $=2 f _{ m }=4 kHz$

$\therefore$ Side frequencies will be

$=f_{c} \pm f_{m}$

$=(1000 \pm 2) kHz$

$=998 kHz$ and $1002 kHz$

So statement$-I$ and statement$-II$ both are correct.

$=\frac{16-8}{16+8}=\frac{8}{24}=\frac{1}{3}=0.33$

$x \times 10^{-2}=0.33$

$x=33$

$=2 \times 5\, kHz$

$=10\, kHz$

$\therefore$ No of stations accommodate

$=\frac{90}{10}=9$

${f}_{{m}}=\frac{12560}{2 \pi}$

$=2000 {Hz}$

$=\sqrt{2 {R}}\left(\sqrt{{h}_{1}}+\sqrt{{h}_{2}}\right)$

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

$=3578(7.07+8.94)$

$=57.28 \,{Km}$

${d}_{{m}}=\left(\sqrt{2 \times 6400 \times 10^{3} \times 320}+\sqrt{2 \times 6400 \times 10^{3} \times 2000}\right) \,{m}$

${d}_{{m}}=224\,{km}$

$\lambda$ $:$ wavelength of signal

${h}\,<\,\lambda$

$\lambda\,>\,{h}$

$\lambda\,>\,400 {m}$

$\qquad=12\, {kHz}$

$\therefore {N}=\frac{6 {MHZ}}{12 {kHZ}}=\frac{6 \times 10^{6}}{12 \times 10^{3}}=500$

$\Rightarrow v_{\max }=V_{m}+V_{C}$

$200=V_{m}+160$

$V_{m}=40$

$\therefore$ Peak voltage $A_{m}=40$

$A_{\min }=A_{c}-A_{m}=250-150=100$

$\frac{A_{\min }}{A_{\max }}=\frac{100}{400}=\frac{1}{4}=\frac{50}{200}$

$x=200$

$150 \,{km}=\sqrt{2 \times\left(6.5 \times 10^{6}\, {m}\right) {H}_{{T}}}$

$\left(150 \,{km} \times 10^{3}\right)^{2}=2 \times 6.5 \times 10^{6} {H}_{{T}}$ ${H}_{{T}}=1731\, {m}$

Population covered $=\left(\pi {r}^{2}\right)\left(2000 / {km}^{2}\right)$ $=3.14 \times(150)^{2} \times 2000=1413 \times 10^{5}$

$ =2 \times 10^{5} Hz =200 KHz $

$\mu=\frac{A_{m}}{A_{c}}=\frac{20}{20}=1$

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

${d}=\sqrt{2 {R}}\left[\sqrt{{h}_{{T}}}+\sqrt{{h}_{{R}}}\right]$

${d}=\sqrt{2 {R}}[\sqrt{{x}}+\sqrt{160-{x}}]$

$\frac{{d}({d})}{{dx}}=0$

$\frac{1}{2 \sqrt{{x}}}+\frac{1(-1)}{2 \sqrt{160-{x}}}=0$

$\frac{1}{\sqrt{{x}}}=\frac{1}{\sqrt{160-{x}}}$

${x}=80\, {m}$

${d}_{\max }=\sqrt{2 \times 6400}\left[\sqrt{\frac{80}{1000}}+\sqrt{\frac{20}{1000}}\right]$

$=\frac{80 \sqrt{2} \times 2 \sqrt{80}}{10 \sqrt{10}}$

$=8 \times 2 \times \sqrt{2} \times 2 \sqrt{2}=64\, {km}$

$\Rightarrow \frac{a}{b}=1$

$A_{\max }=A_{c}+A_{m}=12$

$A_{\max }=A_{c}-A_{m}=3$

$\Rightarrow A_{c}=\frac{15}{2}\, \& \,A_{m}=\frac{9}{2}$

$\text { Modulation index }=\frac{A_{m}}{A_{c}}=\frac{9 / 2}{15 / 2}=0.6$

$\Rightarrow x=1$

$V _{ m }=[5+3 cos 6820 t ] \sin \left(2 \pi \times 10^{4} t \right)$

$V _{\max }=5+3=8$

$V _{\min }=5-3=2$

$ \therefore \quad \mathrm{n}=\frac{(0.01) \mathrm{f}}{6 \times 10^{6}}$ $=\frac{\frac{3}{8} \times 10^{13}}{6 \times 10^{6}} $

$=\frac{1}{16} \times 10^{7}=6.25 \times 10^{5} $

$=\frac{\omega_{\mathrm{c}} \pm \omega_{\mathrm{m}}}{2 \pi}$

$=\frac{(5.5 \pm 0.22) \times 10^{5}}{2 \times \frac{22}{7}}$

$=89.25,\,85.75$

$(1)$ Amplitude varies as $8-10\, \mathrm{V}$ or $9 \pm 1$

$(2)$ Two time period $100$ $\mu s$ (signal wave) and $8\, \mu s$ (carrier wave)

Hence signal is $\left[9 \pm 1 \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}_{1}}\right)\right] \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}_{2}}\right)$

$=9 \pm 1 \sin \left(2 \pi \times 10^{4} t\right) \sin 2.5 \pi \times 10^{5} t$