TS EAMCET · Physics · Wave Optics
The efficiency of a bulb of power 60 W is \(16 \%\). The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of 2 m from the bulb is \(\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)\)
- A \(24 \mathrm{Vm}^{-1}\)
- B \(16 \mathrm{Vm}^{-1}\)
- C \(9 \mathrm{Vm}^{-1}\)
- D \(12 \mathrm{Vm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(12 \mathrm{Vm}^{-1}\)
Step-by-step Solution
Detailed explanation
For electric bulb, \(P=60 \mathrm{~W}, \eta=16 \%\) \(\therefore\) Intensity of bulb, \(\mathrm{I}_B=\frac{P}{4 \pi \mathrm{r}^2}\) \(\therefore \quad P=\frac{60}{4 \pi \times(2)^2}=\frac{15}{4 \pi}\) Intensity of radiation, \(I_R=\frac{1}{2} \mathbb{C} \varepsilon_0 E_0^2\)…
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