The radiation pressure $1 \mathrm{~m}$ away from a $330 \mathrm{~W}$ electric bulb is

The radiation pressure $1 \mathrm{~m}$ away from a $330 \mathrm{~W}$ electric bulb is
  1. $1.25 \times 10^{-7} \mathrm{P}$
  2. $8.75 \times 10^{-8} \mathrm{P}_{\mathrm{a}}$
  3. $5.45 \times 10^{-8} \mathrm{P}_{\mathrm{a}}^{\mathrm{a}}$
  4. $8.50 \times 10^{-7} \mathrm{P}_{\text {a }}^{\text {a }}$

Solution

$ \mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{330}{4 \pi \times 1^2}=26.26 \mathrm{watt} / \mathrm{m}^2 $ So, Radiation Pressure $=\frac{\mathrm{I}}{\mathrm{C}}=\frac{26.26}{3 \times 10^8}$ $ =8.75 \times 10^{-8} \mathrm{~Pa} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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