UPSEE 2017 Solved Paper

Given,
temperature of black body in first condition
T1=50°C Emitted power in first condition, P1=P
Temperature of black body in second condition
T2=100°C
We know that,
P∝TA
Hence,

P2

P1

=

T24

T14

⇒

P2

P

=(

273+100

273+50

)4
⇒

P2

P

=(

373

323

)4

P2

P

=(1.15)4

P2

P

=1.77≈2
P2=2P