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JEE Main 2019 (10 Apr, Shift 2) · Momentum & Radiation Pressure

Momentum & Radiation Pressure question from JEE Main 2019 (10 Apr, Shift 2)

Q10JEE Main 2019 · 10 APR · SHIFT 2Medium

Momentum & Radiation Pressure

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:

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Answer: (b)

  1. Power on the surface P=25 W/cm2×25 cm2=625 WP = 25\ \text{W/cm}^2 \times 25\ \text{cm}^2 = 625\ \text{W}
  2. Energy in 40 min40\ \text{min}: U=625×2400=1.5×106 JU = 625 \times 2400 = 1.5 \times 10^6\ \text{J}
  3. Complete absorption: p=Uc=1.5×1063×108=5.0×10−3 N sp = \dfrac{U}{c} = \dfrac{1.5 \times 10^6}{3 \times 10^8} = 5.0 \times 10^{-3}\ \text{N s}
  4. So the answer is (b).
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