WBJEE 2019 Physics Solved Paper

According to the question,
Area of both bodies A and B=A
Temperature of body A=27∘C=27+273K
Temperature of body B=177∘C=177+273K
Now, by Stefan-Boltzmann law, thermal energy radiated per second by a body
Q=σAT4
where,
A = Area
T= temperature and
σ= Stefan-Boltzmam's constant
So, the ratio of thermal energy radiated per second by A to that by B is
‌

Q1

Q2

=‌

σA

σA

(‌

T1

T2

)4
‌ Now, ‌‌

Q1

Q2

=(‌

T1

T2

)4
‌

Q1

Q2

=(‌

273+27

273+177

)4
‌

Q1

Q2

=(‌

300

450

)4
‌

Q

Q2

=(‌

2

3

)4=(‌

16

81

)
Ratio of thermal energy radiated per secondQ1:Q2=16:81