Mα4πR²σln(32)
Mα4πR²σln(16/3)
Mα16πR²σln(32)
Mα16πR²σln(16/3)
Answer is A. For a small change in temperature, heat loss is dQ = MSdT = MαT³dT
Differentiating w.r.t dt above, we get
dQ/dt = -4πR²σ(T⁴ - T0⁴) = MαT³dT/dt
Assuming T0 is negligible compared to T, we get
-4πR²σT⁴ = MαT³dT/dt
dT/T = -4πR²σ/Mα dt
Integrating both sides from 3T0 to 2T0 and 0 to t respectively, we get
∫₃ᵀ₀²ᵀ₀ dT/T = -4πR²σ/Mα ∫₀ᵗ dt
ln(2/3) = -4πR²σt/Mα
t = Mαln(3/2)/4πR²σ = Mαln(32)/4πR²σ