JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of \(24\,W\) The radius of curvature of hemisphere is \(10\,cm\) and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is \(..........\times 10^{-8}\,N\).
- A \(3\)
- B \(2\)
- C \(1\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(\text { Force }=\int PdA \cos \theta\) \(=\frac{2 I }{ C } \int dA \cos \theta=\frac{2 I }{ C } \pi R ^2=2 \frac{ p _0}{4 \pi R ^2} \cdot \frac{\pi R ^2}{ C }\) \(=\frac{ p _0}{2 C }=\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} N \text { (Ans) }\)
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