Transmission Of Heat Question 115
Question: Three discs A, B and C having radii 2m, 4m, and 6m respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are Qa, Qb, and Qc respectively
[IIT-JEE (Screening) 2004]
Options:
A) Qa is maximum
B) Qb is maximum
C) Qc is maximum
D) Qa = Qb = Qc
Show Answer
Answer:
Correct Answer: B
Solution:
Radiated power$ P=A\varepsilon \sigma T^{4} $
Therefore P$ \propto AT^{4} $ From Wein’s law, $ {\lambda _{m}}T= $ constant
Therefore $ T\propto \frac{1}{{\lambda _{m}}} $ \ $ P\propto \frac{A}{{{({\lambda _{m}})}^{4}}}\propto \frac{r^{2}}{{{({\lambda _{m}})}^{4}}} $
Therefore $ Q _{A}:Q _{B}:Q _{C}=\frac{2^{2}}{{{(300)}^{4}}}:\frac{4^{2}}{{{(400)}^{4}}}:\frac{6^{2}}{{{(500)}^{4}}} $ \ $ Q _{B} $ will be maximum.
BETA
