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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems

If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are

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Payal Gupta

Contributor-Level 10

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n = \frac{90 k H_{z}}{2 * 5 k H z} = 9$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems

The maximum and minimum voltage of an amplitude modulated signal are 60V and 20V respectively. The percentage modulation index will be:

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alok kumar singh

Contributor-Level 10

Percentage modulation = $\frac{(V_{m a x} - V_{m i n})}{(V_{m a x} + V_{m i n})} * 100 %$

Percentage modulation = $(\frac{60 - 20}{80 + 20}) * 100$

= 50%

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems

Draw the output signal Y in the given combination of gates.

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Payal Gupta

Contributor-Level 10

$y = \bar{\bar{A} + B} = A . \bar{B}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems

A radio can tune to any station in 6 MHz to 10 MHz band. The value of corresponding of wavelength bandwidth will be:

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Vishal Baghel

Contributor-Level 10

$λ_{1} = 6 M H z = 6 * 1 0^{6} H z$

$λ_{1} = \frac{C}{υ_{1}} = \frac{3 * 1 0^{8}}{6 * 1 0^{6}} = 50 m$

$λ_{2} = \frac{C}{υ_{2}} = 10 * 1 0^{6} = \frac{3 * 1 0^{8}}{10 * 1 0^{6}} = 30 m$

Wavelength bandwidth

$= | λ_{1} - λ_{2} | = 20 m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems

A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is:

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Vishal Baghel

Contributor-Level 10

f = 20 kHz

Deviation ratio = 10

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Waves

For a transverse wave traveling along a straight line, the distance between two peaks (crests) is 5
, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in $A \to I V; B \to I$ ) of the waves are:

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alok kumar singh

Contributor-Level 10

(n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Waves

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be

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Payal Gupta

Contributor-Level 10

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow$ $V$ $=$ $\frac{10}{3}$ $*$ $\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}}$ $=$ $\frac{10}{3}$ $*$ $\frac{4.08 * 4.16}{. 08}$ $=$ $\frac{10}{3}$ $*$ $\frac{408 * 4.16}{8}$ $=$ $7$ $0$ $7$ $.$ $2$ $m$ $/$