Question

# Three discs $$A, B$$ and $$C$$ having radii $$2, 4$$ and $$6\ cm$$, respectively, are coated with carbon black. Wavelengths for maximum intensity for the three discs are $$300, 400$$ and $$500\ nm$$, respectively. If $$Q_{A}, Q_{B}$$ and $$Q_{C}$$ are powers emitted by $$A, B$$ and $$C$$, respectively, then

A
QA will be maximum
B
QB will be maximum
C
QC will be maximum
D
QA=QB=QC

Solution

## The correct option is A $$Q_{B}$$ will be maximumAccording to Wein's displacement law $$\lambda T=constant$$so  the ratio of temperatures of the three discs will be $$\dfrac{1}{3}:\dfrac{1}{4}:\dfrac{1}{5}=20:15:12$$and the ratio of the area of the discs$$=1:4:9$$Now according to Stephan's law $$Q \alpha A\times T^{4}$$so ratio $$Q_a:Q_b:Q_c=1\times(20)^{4}:4\times (15)^{4}:9\times (12)^{4}$$on solving we will get that $$Q_b$$ is maximumPhysics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More