Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electromagnetic Waves question

2019 · 10 Apr · Shift 2 · Q49
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Electromagnetic Waves
  5. /2019 · 10 Apr · Shift 2 · Q49

Electromagnetic Waves question

2019 · 10 Apr · Shift 2 · Q49

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface has an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be :
  1. A
    6.3 × 10–4 Ns
  2. B
    5.0 × 10–3 Ns
  3. C
    1.4 × 10–6 Ns
  4. D
    3.5 × 10–6 Ns
View written solutionFree

Correct answer: B

  1. Given data
  • Energy flux (intensity): I=25 W cm−2I = 25\ \text{W cm}^{-2}I=25 W cm−2
  • Area: A=25 cm2A = 25\ \text{cm}^2A=25 cm2
  • Time: t=40 min=2400 st = 40\ \text{min} = 2400\ \text{s}t=40 min=2400 s

For a completely absorbing surface, the momentum transferred by light is:

p=Ecp = \frac{E}{c}p=cE​

where EEE is the total incident energy and c=3×108 m/sc = 3 \times 10^8\ \text{m/s}c=3×108 m/s.

  1. Find total power incident on the surface

P=IA=25×25=625 WP = IA = 25 \times 25 = 625\ \text{W}P=IA=25×25=625 W

  1. Find total energy incident in 40 min

E=Pt=625×2400=1.5×106 JE = Pt = 625 \times 2400 = 1.5 \times 10^6\ \text{J}E=Pt=625×2400=1.5×106 J

  1. Find momentum transferred

Since the surface is absorbing,

p=Ec=1.5×1063×108=5.0×10−3 N sp = \frac{E}{c} = \frac{1.5 \times 10^6}{3 \times 10^8} = 5.0 \times 10^{-3}\ \text{N s}p=cE​=3×1081.5×106​=5.0×10−3 N s

  1. Match with options

5.0×10−3 N s5.0 \times 10^{-3}\ \text{N s}5.0×10−3 N s

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

This matches our derived answer.

PreviousNext

More from Electromagnetic Waves

  • If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin [2π×2×1015(t−cx​)] then the maximum electric field…2019 · MCQ
  • The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression E(x,y)=10j​cos[(6x+8z)]. The magnetic field B…2019 · MCQ
  • An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering…2019 · MCQ
  • A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by : [Given permittivity of space ∈ 0 = 9 × 10–12 SI units, Speed of light c = 3 ×…2019 · MCQ
  • An electromagnetic wave is represented by the electric field E=E0​nsin[ωt+(6y−8z)]. Taking unit vectors in x, y and z directions to be i,j​,k…2019 · MCQ
  • A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in…2019 · MCQ
  • A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave…2019 · MCQ
  • The mean intensity of radiation on the surface of the Sun is about 108 W/m2 . The rms value of the corresponding magnetic field is closet to :2019 · MCQ