The curve between lo g e T 2 J and T 1 is

We know that J = A T 2 e − b / T T 2 J = A e − b / T Taking log both side, we get lo g e T 2 J = lo g e ( A ⋅ e − b / T ) lo g e T 2 J = lo g e A − T b Comparing with y = m x + C we get correct option (a).

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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicMedium2 minPYQ_2008

PhysicsMediumsingle choice

The curve between $lo g_{e} \frac{J}{T ^{2}}$ and $\frac{1}{T}$ is

Options:

Answer:

Solution:

We know that $J = A T^{2} e^{- b / T}$ $\frac{J}{T ^{2}} = A e^{- b / T}$ Taking log both side, we get $lo g_{e} \frac{J}{T ^{2}} = lo g_{e} (A \cdot e^{- b / T})$ $lo g_{e} \frac{J}{T ^{2}} = lo g_{e} A - \frac{b}{T}$ Comparing with $y = m x + C$ we get correct option (a).

Stream:AIIMSSubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: PYQ_2008

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