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JEE Main 2024 (30 Jan, Shift 2) · Photons: Energy, Momentum & Radiation Pressure

Photons: Energy, Momentum & Radiation Pressure question from JEE Main 2024 (30 Jan, Shift 2)

Q15JEE Main 2024 · 30 JAN · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure

If the total energy transferred to a surface in time tt is 6.48×105 J6.48 \times 10^{5}\text{ J}, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

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Answer: (a)

  1. For complete absorption, momentum delivered p=Ecp = \dfrac{E}{c}
  2. p=6.48×1053×108=2.16×10−3 kg m/sp = \dfrac{6.48\times10^{5}}{3\times10^{8}} = 2.16\times10^{-3}\text{ kg m/s}
  3. So the answer is (a) 2.16×10−3 kg m/s2.16\times10^{-3}\text{ kg m/s}.
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