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Question: Consider two hot bodies \(B _ { 2 }\) which have temperatures \(40 ^ { \circ } \mathrm { C }\). The ...

Consider two hot bodies B2B _ { 2 } which have temperatures 40∘C40 ^ { \circ } \mathrm { C }. The ratio of the respective rates of cooling t=0t = 0 will be

A

R1:R2=3:2R _ { 1 } : R _ { 2 } = 3 : 2

B

R1:R2=5:4R _ { 1 } : R _ { 2 } = 5 : 4

C

R1:R2=2:3R _ { 1 } : R _ { 2 } = 2 : 3

D

R1:R2=4:5R _ { 1 } : R _ { 2 } = 4 : 5

Answer

R1:R2=3:2R _ { 1 } : R _ { 2 } = 3 : 2

Explanation

Solution

Initially at t = 0

Rate of cooling (R) ∝ Fall in temperature of body (θ – θ0)

⇒ R1R2=θ1−θ0θ2−θ0=100−4080−40=32\frac { R _ { 1 } } { R _ { 2 } } = \frac { \theta _ { 1 } - \theta _ { 0 } } { \theta _ { 2 } - \theta _ { 0 } } = \frac { 100 - 40 } { 80 - 40 } = \frac { 3 } { 2 }