Two balls of same material and finish have their diameters in the ratio of 2 : 1 and both are heated to same temperature and allowed to cool by radiation. Rate of cooling by big ball as compared to smaller one will be in the ratio of
A. 1 : 1
B. 2 : 1
C. 1 : 2
D. 4 : 1
Answer: Option D
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Cooling rate of big sphere / Cooling rate of small sphere = d2 big / d2small
Given, sphere A,B are of some martial
Also heated to same temperature
given that d
A
=2d
B
------- (1)
(d= diameter of sphere)
we know that
Rate of cooling = σeAT
4
(Stepons law of cooling)
Since A,B are of same material ⇒σA=σB
Also given T
A
=T
B
.
⇒RateofcodingαA(Area)
⇒
Rate of cooling of B
Rate of cooling of A
=
(Area effective)B
(Area effective)A
=
r
2
r
2
=
d
2
d
2
⇒
Q
B
Q
A
=(
d
B
d
A
)
2
=(
d
B
2d
B
)
2
=
1
4
Q
A
:Q
B
=4:1
Also heated to same temperature
given that d
A
=2d
B
------- (1)
(d= diameter of sphere)
we know that
Rate of cooling = σeAT
4
(Stepons law of cooling)
Since A,B are of same material ⇒σA=σB
Also given T
A
=T
B
.
⇒RateofcodingαA(Area)
⇒
Rate of cooling of B
Rate of cooling of A
=
(Area effective)B
(Area effective)A
=
r
2
r
2
=
d
2
d
2
⇒
Q
B
Q
A
=(
d
B
d
A
)
2
=(
d
B
2d
B
)
2
=
1
4
Q
A
:Q
B
=4:1
Correct Answer is D
Rate of heat transfer by radiation = A x T4
Temperature is same for both spheres
A = Surface area
Surface area of sphere = π d2
Cooling rate of big sphere / Cooling rate of small sphere = d2 big / d2small
Hence cooling rate will takes place in the ratio of 4 : 1