Question

A telephonic communication service is working at carrier frequency of $$10\,GHz.$$  Only $$10\% $$ of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of $$5\,kHz$$ ?

A. $$2 \times {10^3}$$
B. $$2 \times {10^4}$$
C. $$2 \times {10^5}$$  
D. $$2 \times {10^6}$$
Answer :   $$2 \times {10^5}$$
Solution :
If $$n = $$ no. of channels
$$\eqalign{ & 10\% \,{\text{of}}\,10\,GHz = n \times 5\,KHz \cr & {\text{or,}}\,\frac{{10}}{{100}} \times 10 \times {10^9} = n \times 5 \times {10^3} \Rightarrow n = 2 \times {10^5} \cr} $$

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Modern Physics >> Modern Physics Miscellaneous

Releted Question 1

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $$6\,eV$$  fall on it is $$4\,eV.$$  The stopping potential, in volt, is

A. 2
B. 4
C. 6
D. 10
Releted Question 2

Electrons with energy $$80\,keV$$  are incident on the tungsten target of an X-ray tube. $$K$$-shell electrons of tungsten have $$72.5\,keV$$  energy. X-rays emitted by the tube contain only

A. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$
B. a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
C. the characteristic X-ray spectrum of tungsten
D. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$  and the characteristic X-ray spectrum of tungsten.
Releted Question 3

The intensity of X-rays from a Coolidge tube is plotted against wavelength $$\lambda $$ as shown in the figure. The minimum wavelength found is $${\lambda _C}$$ and the wavelength of the $${K_\alpha }$$ line is $${\lambda _K}.$$  As the accelerating voltage is increased
Modern Physics Miscellaneous mcq question image

A. $${\lambda _K} - {\lambda _C}$$   increases
B. $${\lambda _K} - {\lambda _C}$$   decreases
C. $${\lambda _K}$$ increases
D. $${\lambda _K}$$ decreases
Releted Question 4

The potential difference applied to an X-ray tube is $$5k\,V$$  and the current through it is 3.2$$mA.$$  Then the number of electrons striking the target per second is

A. $$2 \times {10^{16}}$$
B. $$5 \times {10^{6}}$$
C. $$1 \times {10^{17}}$$
D. $$4 \times {10^{15}}$$

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