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Electromagnetic Waves · Question 23

JEE Main 2025 (Online) 3rd April Morning ShiftMCQMedium📈 Topic in 65% papers+4−1
1

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is

  1. 3 10-8 Pascals
  2. 0
  3. 1.5 10-8 Pascals
  4. 6 10-8 Pascals

Answer

(4) 6 10-8 Pascals

Solution

The source radiates uniformly in all directions, so the intensity at r = 2 m is

I = P4 r2 = 4504 4 8.95 W m-2

For a perfectly reflecting surface (normal incidence): p = 2Ic

= 2 8.953 108 6 10-8 Pa

Answer: (4)

JEE Main 2025 (Online) 3rd April Morning Shift

Keep practising — Electromagnetic Waves

2

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is false but (R) is true
  4. (A) is true but (R) is false
JEE Main 2025 (Online) 29th January Morning Shift
3

Due to presence of an em-wave whose electric component is given by E = 100 ( t - kx) NC-1 a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

  1. 50 ( t - kx) NC-1
  2. 400 ( t - kx) NC-1
  3. 200 ( t - kx) NC-1
  4. 25 ( t - kx) NC-1
JEE Main 2025 (Online) 28th January Morning Shift
4

The magnetic field of an E.M. wave is given by B = ( 32 i + 12 j )30 [ (t - zc ) ] (S.I. Units). The corresponding electric field in S.I. units is:

  1. E = ( 12 i + 32 j )30c [ (t + zc ) ]
  2. E = ( 12 i - 32 j )30c [ (t - zc ) ]
  3. E = ( 32 i - 12 j )30c [ (t + zc ) ]
  4. E = ( 34 i + 14 j )30c [ (t - zc ) ]
JEE Main 2025 (Online) 28th January Evening Shift

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