Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen: Overview, Questions, Preparation

This is a Long Answer Type Questions as classified in NCERT Exemplar

as we know that I= I_{0 $e^{-\propto x}$}

And I= 25%of I₀=

I=I₀/4

I₀/4= I_{0 $e^{-\propto x}$}

I₀ cancel from both sides

¼= $e^{-\propto x}$

Taking log on both sides log1 -log4= - $\propto x$ loge

X= log4/ $\propto$

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height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410^-3 = 3⁸

So x=606 km after solving 
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2×6400 = 565m

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Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage $*100$

= $\frac{50-10}{50+10}*100$ = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 $*$  30= 20V

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in first case crowding spectrum is visible

If we want more wave to be modulated then more crowding will occur and more mixing up of signal.

But we want to accommodate this we use higher band width and frequency career wave

Frequency of career wave is 20MHz

Bandwidth require for modulation is 3kHz/2= 1.5kHz

For demodulating we need to reciprocal it

1/f= 1/20MHz= 0.5 $*$ 10^-7s

For modulation = 1/1.5KHz= 0.7 $*$ 10³s

According to first option =RC= 1 $*$ 0.01= 10^-5s

So it will demodulated

According to 2^nd option- RC= 10^-4s

So it can also be demodulated

According to third option – RC= 10^-8s

So it cannot be modulated

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When we send a signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.

But, the frequency of TV signals are 60 MHz which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

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As frequency of B is more than A so it has more refractive index also and if a wave have higher refractive index then it has less angle of refraction. So wave B travel more in ionosphere.

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Maximum amplitude = Am+Ac =12 while minimum amplitude is Ac-Am=3

Using elimination method = 2Ac= 18

Ac=9 and Am= 6V

So modulating index m = 6/9= 2/3

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Frequency tuned amplifier is $\frac{1}{2\pi \sqrt[]{LC}}=1MHz$

$\sqrt[]{LC}$ =1/2 $\pi \times {10}^6$

$LC=2.54\times {10}^{-14}s$

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In modulating signal we add career wave or noise signal to send it to receiver end and this wave is varied from time to time that is why more noise appear.

But in frequency modulation frequency is not varied so less noise appear.

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Total distance = 5km and loss is 2 dB/km

So total loss = 5 (2)= 10 dB

Total gain in amplifier  10+20= 30dB and gain in signal is 20dB

So by the formula 20= 10log_{10 $\frac{p_o}{p_i}$}

log_{10 $\frac{p_o}{p_i}$} = 2

so po/pi= 10²

so Po= Pi (100)= 101mW

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Range = $\sqrt[]{2hR}=\sqrt[]{2\times 20\times 6.4\times {10}^6}=16km$

Area = $\pi R^2=3.14\times 16\times 16$ =803.84km²

When H= 25 m

Range = $\sqrt[]{2hR}+\sqrt[]{2HR}=\sqrt[]{2\times 20\times 6.4\times {10}^6}+\sqrt[]{2\times 25\times 6.4\times {10}^6}$ = 33.9km

Area= 3.14 $\times 33.9\times 33.9=3608.52km$ ²

percentage increase = $\frac{3608.52-803.84}{803.84}\times 100=348.9$ %

This is a  Short Answer Type Questions as classified in NCERT Exemplar

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Frequency, fmax= 9 (N_max)^1/2

For F1 layer frequency is 5MHz

So 5 $*{10}^6=9\left(Nmax\right)$ ^1/2

Nmax= (5/9 $*{10}^6$ )²= 3.086 $*$ 10¹¹/m³

For F2 layer 8MHz

8 $*{10}^6=9\left(Nmax\right)$ ^1/2

Nmax= (8/9 $*{10}^6$ )²= 7.9 $*$ 10¹¹/m³

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In amplitude modulated wave side frequency contain the information and from the question the power will be ω_c, (ω_c – ω_m) and (ω_c + ω_m)

So to minimise cost of radiation we use both (ω_c – ω_m) and (ω_c + ω_m)

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Answer- b

Ground wave propagation – 530KHz to 1710KHz

Sky wave propagation- 1710KHz- 40MHz

Space wave propagation- 54MHz to 42GHz 

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Answer- a

length of building is given by l= 500m

And $\lambda ~$ 4l= 4 $\times$  100 =400mf

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Answer-b

p = 1kW= 1000W

Attenuation of signal = -2dB/km and total path length = 5km

Gain in dB= 5 $*-2=-10dB$

Gain in dB= 10 log (p₀/pi)……. (i)

-10=10 log (p₀/pi)= -10log (p_i/p_o)

logP_i/p_o=1 = log (p_i/p_o)= log10

pi/p₀=10= 1000W= 10p₀

p₀=100W

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Answer- a

Frequency of career signal = 1MHz and frequency = 3KHz= 0.003MHz

So frequency of side bands = 1 $?$ 0.003

So 1.003MHz and 0.997MHz

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- frequency of career wave and frequency of amplitude modulated wave is same which is w_c .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- The device which follows square law is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices.

Characteristics shown by (ii) corresponds to square law device. Some part of (i) also

Follow square law.

Hence, (ii)and (iv) can be used for modulation.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- Here, in this question, the frequency of modulated signal received becomes more, which is possible with the poor bandwidth selection of amplifiers.

This happens because bandwidth in amplitude modulation is equal to twice the

Frequency of modulating signal. But, the frequency of male voice is less than that of a female.

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Answer- b

Explanation- it is represented by a diagram given below

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Explanation-equation of modulating wave is, m (t)= A_msinw_mt

Carrier wave equation is C_m (t)= (A_c+A_msinw_mt)sinw_ct

 = A_c [1+ $\frac{A_m}{A_c}sin{w}_{m}t$ ]sinw_ct

Also $\frac{A_m}{A_c}$ = M

C_mt= (A_c+A_{c $\times \propto sin{w}_{m}t$} )sinw_ct

Ac $\times \propto$ =K

Sinw_mt= V_m

equation having phase $\varnothing$

so equation is A_c+K₂V_mt}+sinw_ct+ $?$ ]

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Answer- a, b, d

Explanation- frequency is given by Vm=15KHz

So wavelength is given by $?=\frac{c}{v}=\frac{3*{10}^8}{15*{10}^3}=\frac{1}{5}*{10}^{5}m$

Also we know that l = $?/4$  = 5km

The audio signals are of low frequency waves. Thus, they cannot be transmitted through sky

Waves as they are absorbed by atmosphere.

If the size of the antenna is less than 5 km, the effective power transmission would be very

Low because of deviation from resonance wavelength of wave and antenna length.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- w_m=3KHz

And w_c= 1.5MHz= 1500KHz

By using these two 1500 $?$ 3= 1503 and 1497

Bandwidth = 2w_m= 2 $\times 3$ = 6KHz

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, c, d

Explanation- height of tower= 240m

For line of sight communication maximum distance would be d= $\sqrt[]{2Rh}$

So d= $\sqrt[]{2*6.4*{10}^6*240}$  = 55.4km

So distance under this are communicable

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- production of modulated wave is given by frequency of upper side band – frequency of lower side band so in first three cases it is clearly visible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- modulation index m= A_m/A_c

If m>1 then A_m>A_c

maximum modulation frequency m_f= frequency deviation / maximum frequency of modulating wave

here if m>1 then it means overlapping of both sides resulting loss of information.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is

transformed into electric signals.

The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes.

Thus, (a) and (b) would produce analog signal and (c) and (d) would produce digital

signals.

$\mathrm{}580$

$\mathrm{}{X}_1=-3t^2+8t+10$

${\stackrel{?}{v}}_1=(-6t+8)\hat{i}=2\hat{i}$

$Y_2=5-8t^3$

${\stackrel{?}{\mathrm{v}}}_2=-24{\mathrm{t}}^2\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\sqrt[]{v}=\left|{\stackrel{?}{v}}_2-{\stackrel{?}{v}}_1\right|=|-24\hat{j}-2\hat{i}|$

$\sqrt[]{v}=\sqrt[]{{24}^2+2^2}$

$v=580$

For elastic collision   ${\mathrm{K}\mathrm{E}}_{\mathrm{i}}={\mathrm{K}\mathrm{E}}_{\mathrm{r}}$

$\frac{1}{2}\mathrm{m}\times 25+\frac{1}{2}\times \mathrm{m}\times 9=\frac{1}{2}\mathrm{m}\times 32+\frac{1}{2}m{v}^2$

$34=32+v^2$

$\mathrm{K}\mathrm{E}=\frac{1}{2}\times 0.1\times 2=0.1\mathrm{J}=\frac{1}{10}$

$x=1$

$\mathrm{}V=\sqrt[]{\frac{B}{?}}$

$\frac{V_{\text{pipe}}}{V_{\text{air}}}=\frac{\sqrt[]{\frac{B}{2?}}}{\sqrt[]{\frac{B}{?}}}=\frac{1}{\sqrt[]{2}}$

${\mathrm{V}}_{\text{pipe}}=\frac{{\mathrm{V}}_{\text{air}}}{\sqrt[]{2}}$

${\mathrm{f}}_{\mathrm{n}}=\frac{(\mathrm{n}+1)V_{\text{pipe}}}{2\mathcal{l}}\mathrm{};\mathrm{}{\mathrm{f}}_1-{\mathrm{f}}_0=\frac{V_{\text{pipe}}}{2\mathcal{l}}=\frac{300}{2\sqrt[]{2}}$

$=105.75\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.41$ $\text{exception 25:}$ )

$=106.05\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.414$ $\text{exception 25:}$ )