Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

About 20% of the power of a 100w bulb is converted to visible radiation. Assuming that the  radiation is emitted isotropically and neglecting reflection the average intensity of visible  radiation at a distance of 5m is  α25π w/m2. The value of α is ________

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Power of the bulb , P = 100w.  power of visible radiation p1=20%  of  
P=20100×100=20w 
Average intensity of radiation 
 I=P14πr2 I=204π(5)2=525π
 I=α25πw/m2  (given)
α = 5 
 

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon