As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. The antenna current of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is sinusoidal modulated. The percentage modulation is nearly
Solution
√(1+m2/2)=It/Ic =8.93/3=1.11625 m2/2=(1.11625)2-1 m=0.70=70%
√(1+m2/2)=It/Ic =8.93/3=1.11625 m2/2=(1.11625)2-1 m=0.70=70%
Q2.What is the value of frequency at which electromagnetic wave must be propagated for the D-region of atmosphere to have a refractive index of 0.5. Electron density for D-region is 400 electrons/cc
Solution
Here, N=400 electronic/cc=400×106 electrons m-3 μ=0.5=√(1-(81.56 N)/v2 ) =√(1-(81.56×400×106)/v2 ) On solving, v=208.4 kHz.
Here, N=400 electronic/cc=400×106 electrons m-3 μ=0.5=√(1-(81.56 N)/v2 ) =√(1-(81.56×400×106)/v2 ) On solving, v=208.4 kHz.
Q3. If a number of sine waves with modulation indices n1,n2,n3….. modulate is carrier wave, then total modulation index (n) of the wave is
Solution
√(n1-n2+n3……)
√(n1-n2+n3……)
Q4. The highest frequency of radiowaves which when sent at some angle towards the ionosphere, gets reflected from that and returns to the earth is called
Solution
The highest frequency of radiowaves that can be reflected by the ionosphere is called maximum usable frequency
The highest frequency of radiowaves that can be reflected by the ionosphere is called maximum usable frequency
Q5.The main objective of an optical source is
Solution
The main objective of optical source is to convert the electrical energy into the optical energy
The main objective of optical source is to convert the electrical energy into the optical energy
Q6. The normalised fiber frequency is expressed by (a=fiber core radius, λ0=free space wavelength, μ1,μ2= refractive index of core and cladding)
Solution
The normalised fiber frequency, v=2Ï€a/λ0 (μ12-μ22 )1/2 v=2Ï€a/λ0 ×100=1%
The normalised fiber frequency, v=2Ï€a/λ0 (μ12-μ22 )1/2 v=2Ï€a/λ0 ×100=1%
Q7.The velocity of electromagnetic waves in a medium is2×108 ms-1. The dielectric constant of the medium is
Solution
√k=vc/v=(3×108)/(2×109 )=3/2;k=9/4=2.25
√k=vc/v=(3×108)/(2×109 )=3/2;k=9/4=2.25
Q8.The range of frequencies allotted for FM radio is
Solution
88 to 108 MHz
88 to 108 MHz
Q9.Which frequency range is used for optical communication?
Solution
Optical communication is made of communication by which we can transfer the information from one place to another through optical carrier waves. Light frequencies used in optical communication system lie between 1014 Hz to 4×1014 Hz (ie, 100,000 to 400,000 GHz.)
Optical communication is made of communication by which we can transfer the information from one place to another through optical carrier waves. Light frequencies used in optical communication system lie between 1014 Hz to 4×1014 Hz (ie, 100,000 to 400,000 GHz.)
Q10. A TV tower has a height of 75 m. What is the maximum area upto which this TV communication can be possible?
Solution
Area covered=Ï€×2Rh =22/7×2×6.4×106×75 =3017.1 km2
Area covered=Ï€×2Rh =22/7×2×6.4×106×75 =3017.1 km2