TS EAMCET · Physics · Dual Nature of Matter
About \(20 \%\) of the power of a \(100 \mathrm{~W}\) bulb is converted to visible radiation. Assuming that the radiations is emitted isotropically and neglecting reflection, the average intensity of visible radiation at a distance of \(5 \mathrm{~m}\) is \(\frac{\alpha}{25 \pi} \mathrm{W} / \mathrm{m}^2\). The value of \(\alpha\) is
- A 15
- B 5
- C 37.5
- D 30
Answer & Solution
Correct Answer
(B) 5
Step-by-step Solution
Detailed explanation
Power, \(\mathrm{P}=\frac{100 \times 20}{100}=20 \mathrm{~W}\) The average intensity of visible radiation at a distance \[ \begin{aligned} & 5(I)=\frac{P}{4 \pi r^2}=\frac{20}{4 \pi \times(5)^2} \\ & (I)=\frac{5}{25 \pi} \\ & \alpha=5 \end{aligned} \]
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