AP EAMCET · PHYSICS · Dual Nature of Matter
The radiation pressure \(1 \mathrm{~m}\) away from a \(330 \mathrm{~W}\) electric bulb is
- A \(1.25 \times 10^{-7} \mathrm{P}\)
- B \(8.75 \times 10^{-8} \mathrm{P}_{\mathrm{a}}\)
- C \(5.45 \times 10^{-8} \mathrm{P}_{\mathrm{a}}^{\mathrm{a}}\)
- D \(8.50 \times 10^{-7} \mathrm{P}_{\text {a }}^{\text {a }}\)
Answer & Solution
Correct Answer
(B) \(8.75 \times 10^{-8} \mathrm{P}_{\mathrm{a}}\)
Step-by-step Solution
Detailed explanation
\[ \mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{330}{4 \pi \times 1^2}=26.26 \mathrm{watt} / \mathrm{m}^2 \] So, Radiation Pressure \(=\frac{\mathrm{I}}{\mathrm{C}}=\frac{26.26}{3 \times 10^8}\) \[ =8.75 \times 10^{-8} \mathrm{~Pa} \]
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