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.**In the given detector circuit, the suitable value of carrier frequency is

Solution

Using 1/f

Using 1/f

_{carrier }<<RC We get time constant, RC=1000×10^{-12}=10^{-9}s Now v=1/T=1/10^{-9}=10^{9}Hz Thus the value of carrier frequency should be much less than 10^{9}Hz, say 100 kHz**Q2.**In earth's atmosphere, for F

_{2}-layer, the virtual height and critical frequency in night time are

Solution

Virtual height of F

Virtual height of F

_{2}-layer is 350 km and critical frequency is 6 MHz**Q3.**The bit rate for a signal, which has a sampling rate of 8 kHz and where 16 quantisation levels have been

used is

Solution

If n is the number of bits per sample, then number of quantisation levels =2

If n is the number of bits per sample, then number of quantisation levels =2

^{n}Since the number of quantisation levels is 16 ⇒2^{n}=16⇒n=4 ∴ bit rate = sampling rate × no. of bits per sample =8000×4=32,000 bits/s

**Q4.**A broken ligament is being ‘welded’ back in place using 20 ms pulses from a 0.5 W laser operating at a wavelength of 632 nm. The number of photons in 5 pulses of laser are

Solution

The power of laser is 0.5 W, let n photon/s are incident by laser pulse on the broken ligament, then n ×hv=0.5 W ⇒nhc/Î»=0.5 ⇒n=0.5×Î»/hc =0.5×632×10

The power of laser is 0.5 W, let n photon/s are incident by laser pulse on the broken ligament, then n ×hv=0.5 W ⇒nhc/Î»=0.5 ⇒n=0.5×Î»/hc =0.5×632×10

^{-9}/6.626×10^{-34}×3×10^{8}=1.59×10^{18}photon/s So, number of photons contained in 5 pulses are, n×5×(20×10^{3})=1.59×10^{23}**Q5.**Small vales of numerical aperture (NA) decrease the pulse dispersion but increase losses due to

Solution

Pulse dispersion ∝ numerical aperture of fiber. When numerical aperture of an optical fibre is small, then the energy losses will increase due to micro-bending

Pulse dispersion ∝ numerical aperture of fiber. When numerical aperture of an optical fibre is small, then the energy losses will increase due to micro-bending

**Q6.**A modem is a

Solution

Modulating and demodulating device

Modulating and demodulating device

**Q7.**Ozone layer blocks the radiations of wavelength

Solution

Ozone layer extends from 30 km to nearly 50 km above the earth’s surface in ozone sphere. This layer absorbs the major part of ultraviolet radiations coming from the sun and does not allow them to reach the earth’s surface. The range of ultraviolet radiations is 100 â„« to 4000 â„«. Thus, it blocks the radiations of wavelength less than 3×10

Ozone layer extends from 30 km to nearly 50 km above the earth’s surface in ozone sphere. This layer absorbs the major part of ultraviolet radiations coming from the sun and does not allow them to reach the earth’s surface. The range of ultraviolet radiations is 100 â„« to 4000 â„«. Thus, it blocks the radiations of wavelength less than 3×10

^{-7}m (or 3000 â„«).**Q8.**Optical fibre works on the principle of

Solution

Total internal reflection of light

Total internal reflection of light

**Q9.**The maximum range, d

_{max}of radar is

Solution

Maximum Range of the radar is given by R

Maximum Range of the radar is given by R

_{max}=( P_{t}A^{2}S/4Ï€Î»^{2}P_{min})^{1/4}Where P_{t}: peak value of transmitted power A: capture area of the receiving antenna S: radar cross-sectional area Î»: wavelength of RADAR wave P_{min}: minimum receivable power of the receiver**Q10.**Large band width for higher data rate is achieved by using

Solution

High frequency carrier wave provides a larger band width.

High frequency carrier wave provides a larger band width.