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Physics Question on thermal properties of matter

The heat passing through the cross-section of a conductor, varies with time ‘t’ as Q(t)=αt–βt2+γt3Q(t) = αt – βt^2 + γt^3. (α,βα, β and γγ are positive constants.) The minimum heat current through the conductor is

A

α−β22γ\alpha-\frac{\beta^2}{2\gamma}

B

α−β23γ\alpha-\frac{\beta^2}{3\gamma}

C

α−β2γ\alpha-\frac{\beta^2}{\gamma}

D

α−3β2γ\alpha-\frac{3\beta^2}{\gamma}

Answer

α−β23γ\alpha-\frac{\beta^2}{3\gamma}

Explanation

Solution

Heat through cross section of rod
Q = αt – βt2 + γt3
so heat current =dQdt=\frac{dQ}{dt}
heat current =dQdt=α−2βt+3γt2=\frac{dQ}{dt}=\alpha-2\beta t +3\gamma t^2
for heat current to be minimum
d2Qdt2=−2β+6γt=0\frac{d^2Q}{dt^2}=-2\beta+6\gamma t=0
t=2β6γ=(β3γ)t=\frac{2\beta}{6\gamma}=(\frac{\beta}{3\gamma})
so minimum heat current
dQdt∣minimum=α−2β×β3γ×β29γ2\frac{dQ}{dt}|_{\text{minimum}}=\alpha-2\beta\times\frac{\beta}{3\gamma}\times\frac{\beta^2}{9\gamma^2}
=α−2β23γ+β22γ=\alpha-\frac{2\beta^2}{3\gamma}+\frac{\beta^2}{2\gamma}
=(α−β23γ)=(\alpha-\frac{\beta^2}{3\gamma})

So, the correct answer is (B) : α−β23γ\alpha-\frac{\beta^2}{3\gamma}