Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoRadiationMedium2 minQB

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The temperature of a well stirred liquid kept open to a cold surrounding is plotted against time. The value of $sec^{2} θ_{1}$ is :-

Options:

Answer:

Solution:

According to Newton's law of cooling $\frac{dT}{dt} \propto Δ T$

$\Rightarrow \frac{tan θ _{1}}{tan θ _{2}} = \frac{35 - 20}{25 - 20} = \frac{15}{5} = \frac{3}{1} \Rightarrow tan^{2} θ_{1} = 9 tan^{2} θ_{2} ∵ sec^{2} θ_{1} = 1 + tan^{2} θ_{1} \Rightarrow sec^{2} θ_{1} = 1 + 9 tan^{2} θ_{2}$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation

⏱ 2mℹ️ Source: QB

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