Question

Q14. A black metal foil is warmed by radiation from a small sphere at temperature 'T' and at a distance 'd'. It is found that the power received by the foil is P. If both the temperature and distance are doubled, the power received by the foil will be :

(A) 16 P (B) 4 P (C) 2 P (D) P

Answer 1

Hey mate,

● Answer - 4P

● Explanation-
# Given-
T1 = T
d1 = d
T2 = 2T
d2 = 2d
P1 = P
P2 = ?

# Solution-
Power received by the foil due to radiation is -
P ∝ (T^4-T°^2)
When T°<<<T
P ∝ T^4
Also,
P ∝ 1/d^2
Hence,
P ∝ (T^4/d^2)

Here,
P1 ∝ (T1^4/d1^2)
P2 ∝ (T2^4/d2^2)

Therefore,
P2/P1 = (T2^4/d2^2) / (T1^4/d1^2)
P2/P1 = (T2^4×d1^2) / (T1^4×d2^2)
P2/P = (16T^4×d^2) / (T^4×4d^2)
P2/P = 4
P2 = 4P

Power received by the foil due to radiation is 4P.

Hope that is useful...