Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • About of the power of a <img alt="100 \mathrm{~W}" src="https://entrancecorner.oncodecogs.com/gif.latex?100%20%5C

About 6 \% of the power of a 100 \mathrm{~W} light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 8 \mathrm{~m} is (Assume that the radiation is emitted isotropically and neglect reflection.)
 

Option: 1

3.5 \times 10^{-3} \mathrm{Wm}^{-2}



 


Option: 2

5.1 \times 10^{-3} \mathrm{Wm}^{-2}


Option: 3

7.2 \times 10^{-3} \mathrm{Wm}^{-2}


Option: 4

2.3 \times 10^{-3} \mathrm{Wm}^{-2}


Answers (1)

best_answer

Here, power of bulb =100 \mathrm{~W}

As intensity, \mathrm{\mathrm{I}=\frac{\text { Power of visible light }}{\text { area }}=\frac{100 \times 6 / 100}{4 \pi(8)^2}=7.2 \times 10^{-3} \mathrm{Wm}^{-2}}

Hence option 3 is correct.

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question