Dear Readers,

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.  . A radio receiver antenna that is 2 π‘š long is oriented along the direction of the electromagnetic wave and receives a signal of intensity 5 × 10-16π‘Š/π‘š2. The maximum instantaneous potential difference across the two ends of the antenna is
  •   1.23 πœ‡π‘‰
  •   1.23 π‘šπ‘‰
  •   1.23 𝑉
  •   12.3 π‘šπ‘‰
𝐼 = 1/2Ξ΅0𝐢𝐸02 
 ⇒ 𝐸o = √2𝐼Ρ0𝑐 = √ (2 × 5 × 10-16 )/ (8.85 × 10-12 × 3 × 108)
 = 0.61 × 10-6 𝑉 /π‘š Also 𝐸0 = 𝑉0/ 𝑑 ⇒ 𝑉0= 𝐸0𝑑 = 0.61 × 10-6 × 2 = 1.23πœ‡π‘‰

Q2. The two slits are 1π‘šπ‘š apart from each other and illuminated with a light of wavelength 5 × 10-7π‘š. If the distance of the screen is 1 π‘š from the slits, then the distance between third dark fringe and fifth bright fringe is
  •   1.5 π‘šπ‘š
  •   0.75 π‘šπ‘š
  •   1.25 π‘šπ‘š
  •   0.625 π‘šπ‘š
Distance of 5th bright fringe from central fringe, 𝑋5D = 5πœ†π·/ 𝑑 …(i) 
Distance of 3rd dark fringe from central fringe, 𝑋3D = (2 × 3 − 1)πœ†π· /2𝑑 = 5 /2 πœ†π· /𝑑 …(ii) 
 From (i) and (ii) required distance 
 π‘‹5B − 𝑋3D = (5 − 5 /2 ) πœ†π· /𝑑 = 5 /2 × (5 × 10-7 × 1 )/(1 × 10-3) = 1.25 π‘šπ‘š

Q3.   . In Young’s double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 β„«hen angular position of first dark fringe is
  •   0.08°
  •   0.16°
  •   0.20°
  •   0.32°
𝑑 = 0.1 mm = 10-4,D = 20 cm = 1/5 m 
 πœ† = 5460β„« = 5.46 × 10-7
 Angular position of first dark fringe is
 ΞΈ = π‘₯/𝐷 = πœ† /2𝑑 = 5.46 × 10-7 / 2 × 10-4 
= 2.73 × 10-3 rad = 2.73 × 10-3 × 180°/πœ‹ = 0.156°

Q4.  . The maximum distance upto which TV transmission from a TV tower of height β„Ž can be received is proportional to
  •   β„Ž1/2
  •   h
  •   h
  •   h2
Distance covered by T.V. signals = √2β„Žπ‘… ⇒ maximum distance ∝ β„Ž1/2

Q5. In Young’s double slit experiment, the intensity on the screen at a point where path difference πœ† is 𝐾. What will be the intensity at the point where path difference is πœ†/4
  •   K/4
  •   K/2
  •   K
  •   ZERO
  By using phase difference πœ™ = 2πœ‹/πœ† (Ξ”) 
For path difference πœ†, phase difference πœ™1 = 2πœ‹ and for path difference πœ†/4, phase difference πœ™2 = πœ‹/2 
Also by using 𝐼 = 4𝐼0 cos2 πœ™/2 ⇒ 𝐼1 /𝐼2 = cos2(πœ™1/2) cos2(πœ™2/2) 
 ⇒ 𝐾/ 𝐼2 = cos2(2πœ‹/2)/ cos2 (πœ‹/2 2 ) = 1 /1/2 ⇒ 𝐼2 = 𝐾/ 2 .

Q6.  In Young’s double slit experiment, the separation between the slit is haled and the distance between the slits and screen is doubled. The fringe-width will
  •   Be halved
  •   Be doubled
  • Be quadrupled
  •   Remain unchanged
Let S be a slit illuminated by monochromatic light of wavelength(πœ†), let 𝑆1,𝑆2 be coherent sources and distance between them be d and distance between source and screen is D. then, fringe width ( W ) is given by 

 π‘Š = π·πœ†/𝑑 
When, 𝑑2 = 𝑑/2 ,𝐷2 = 2𝐷 
 ∴ π‘Š2 = (2𝐷)πœ†/(𝑑/2) = 4 π·πœ†/ 𝑑 = 4π‘Š 
 The fringe width is quadrupled.

Q7. The distance between the first dark and bright band formed in Young’s double slit experiment with band width B is
  •   B/4
  •   B
  •   B/2
  •   3B/2

Position of nth bright fringe π‘₯1 = π‘›πœ†π· /𝑑 
For first bright fringe 𝑛 = 1 ∴ π‘₯1 = πœ†π·/ 𝑑
 Position of nth dark fringe π‘₯2 = (2𝑛−1)πœ†π·/ 2𝑑
 For first dark fringe 𝑛 = 1 ∴ π‘₯2 = πœ†π· /2𝑑 
Now, π‘₯1 − π‘₯2 = πœ†π·/ 2𝑑 If B is the band width, then π‘₯1 − π‘₯2 =𝐡 /2

Q8. Which of the following statements indicates that light waves are transverse
  •   Light waves can travel in vacuum
  • Light waves show interference
  •   Light waves can be polarized
  •   Light waves can be diffracted
Transverse waves can be polarized only

Q9. In Young’s double slit experiment, the central bright fringe can be identified

  •   As it has greater intensity than the other bright fringes
  •   As it is wider than the other bright fringes
  •   As it is narrower than the other bright fringes
  •   By using white light instead of monochromatic light
When white light is used instead of monochromatic light, the central bright fringe becomes white, while others are coloured. Hence, distinction is made

Q10.  The observed wavelength of light coming from a distant galaxy is found to be increased by 0.5% as compared with that coming from a terrestrial source. The galaxy is
  •   Stationary with respect to the earth
  •   Approaching the earth with velocity of light
  •   Receding from the earth with the velocity of light
  • Receding from the earth with a velocity equal to 1.5 × 106π‘š/𝑠
Ξ”πœ†/πœ† = 𝑣 /𝑐 , Now Ξ”πœ† = 0.5/ 100 πœ† ⇒ Ξ”πœ† /πœ† = 0.5 /100
 ∴ 𝑣 = 0.5/ 100 × π‘ = 0.5/ 100 × 3 × 108 = 1.5 × 106π‘š/𝑠 
 Increase in πœ† indicates that the star is receding

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