Let us discuss the following multiple choice question which appeared in the Kerala Engineering entrance 2005 test paper:
If a radio receiver is tuned to 855kHz radio wave, the frequency of local oscillator in kHz is
(a) 1510 (b) 455 (c) 1310 (d) 1500 (e) 855
If a radio receiver is tuned to 855kHz radio wave, the frequency of local oscillator in kHz is
(a) 1510 (b) 455 (c) 1310 (d) 1500 (e) 855
Since the radio frequency mentioned in the question is 855 kHz, it must an AM (amplitude modulation) receiver. AM receivers are super heterodyne receivers with intermediate frequency (IF) equal to 455 kHz, by convention.. So the local oscillator in the receiver has a frequency higher by 455 kHz compared to the frequency of the received radio wave. Its frequency is therefore (855+455) kHz = 1310 kHz.
Consider the following MCQ which high lights the component frequencies in an AM wave:
A carrier of frequency 1.2 MHz is amplitude modulated by a microphone output of frequency 800 Hz. The frequencies present in the amplitude modulated carrier are (in kHz)
(a)1.2 MHz and 800 Hz (b) 400 kHz, 1.2 MHz and 2 MHz (c)1.04 MHz,1.2 MHz and 2 MHz (d) 1.1992 MHz, 1.2 MHz and 1.2008 MHz (e) 1.1992 MHz and 1.2008 MHz
The correct option is (d) since the AM wave contains the carrier frequency, the upper side frequency (which is the sum of the carrier frequency and the modulating signal frequency) and the lower side frequency (which is the difference between the carrier frequency and the modulating signal frequency). While finding the upper and lower side frequencies, it will be convenient if you convert the frequencies in to kilohertz or megahertz. Thus, the lower side frequency = (1200 – 0.8)kHz = 1199.2 kHz = 1.1992 MHz. Similarly, the upper side frequency = (1200 + 0.8) kHz = 1200.8 kHz = 1.2008 MHz.
Now consider the following question:
A semiconductor is to be selected for use as a photo detector to detect radiation of wavelength 620 nm. The appropriate semiconductor will have a band gap of
(a) 2eV (b) 2.2eV (c) 2.4eV (d) 2.8eV (e) both (a) and (b)
The band gap of the semiconductor should not exceed the energy of the photon. The wavelength of the photon being 620 nm which is equal to 6200 Angstrom Units (A.U), its energy is 12400/6200 = 2.0 eV (Note that the product of the wave length in A. U. and the energy in eV for a photon is 12400). The option (a) alone is therefore suitable.
The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:
If α and β are the current gains in the CB and CE configurations respectively of the transistor circuit, then (β – α)/αβ =
(a) ∞ (b) 1 (c) 2 (d) 0.5 (e) zero
Since β = α/(1- α), the given ratio, (β – α)/αβ =[α/(1- α) – α]/αβ = [1/(1- α) -1]/β =[1-(1- α)]/(1-α)β = α/(1- α)β = β/β =1. So, the correct option is (b).
Consider the following MCQ which high lights the component frequencies in an AM wave:
A carrier of frequency 1.2 MHz is amplitude modulated by a microphone output of frequency 800 Hz. The frequencies present in the amplitude modulated carrier are (in kHz)
(a)1.2 MHz and 800 Hz (b) 400 kHz, 1.2 MHz and 2 MHz (c)1.04 MHz,1.2 MHz and 2 MHz (d) 1.1992 MHz, 1.2 MHz and 1.2008 MHz (e) 1.1992 MHz and 1.2008 MHz
The correct option is (d) since the AM wave contains the carrier frequency, the upper side frequency (which is the sum of the carrier frequency and the modulating signal frequency) and the lower side frequency (which is the difference between the carrier frequency and the modulating signal frequency). While finding the upper and lower side frequencies, it will be convenient if you convert the frequencies in to kilohertz or megahertz. Thus, the lower side frequency = (1200 – 0.8)kHz = 1199.2 kHz = 1.1992 MHz. Similarly, the upper side frequency = (1200 + 0.8) kHz = 1200.8 kHz = 1.2008 MHz.
Now consider the following question:
A semiconductor is to be selected for use as a photo detector to detect radiation of wavelength 620 nm. The appropriate semiconductor will have a band gap of
(a) 2eV (b) 2.2eV (c) 2.4eV (d) 2.8eV (e) both (a) and (b)
The band gap of the semiconductor should not exceed the energy of the photon. The wavelength of the photon being 620 nm which is equal to 6200 Angstrom Units (A.U), its energy is 12400/6200 = 2.0 eV (Note that the product of the wave length in A. U. and the energy in eV for a photon is 12400). The option (a) alone is therefore suitable.
The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:
If α and β are the current gains in the CB and CE configurations respectively of the transistor circuit, then (β – α)/αβ =
(a) ∞ (b) 1 (c) 2 (d) 0.5 (e) zero
Since β = α/(1- α), the given ratio, (β – α)/αβ =[α/(1- α) – α]/αβ = [1/(1- α) -1]/β =[1-(1- α)]/(1-α)β = α/(1- α)β = β/β =1. So, the correct option is (b).
2 comments:
I'm impressed by the selfless effort of VM. Congrats.
Are there anymore sites/bloggers like this?
Thank you Ajithkumar. I found howthingswork.virginia.edu/ useful.This is a site that explains the Physics of everyday life.
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