Radiation

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Ratio of radii of two cylinders is 1 : 2 and that of their lengths is 2 : 3. The cylinders are heated to the same temperature and allowed to cool in the same surrounding. Then the ratio of their initial rate of cooling is

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Solution

$\begin{array}{l} m s (- \frac{d θ}{d t}) = e . σ A . (θ^{4} - θ_{o}^{4}) \\ \Rightarrow (π r^{2} l . ρ) s (- \frac{d θ}{d t}) = e . σ (2 π r l) (θ^{4} - θ_{o}^{4}) \end{array}$
$\begin{array}{l} ∴ - \frac{dθ}{dt} = \frac{2 e . σ (θ^{4} - θ_{o}^{4})}{sρ r} \\ ∴ \frac{R_{1}}{R_{2}} = \frac{r_{2}}{r_{1}} = 2 . \end{array}$

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Radiation

Remember concepts with our Masterclasses.

Ratio of radii of two cylinders is 1 : 2 and that of their lengths is 2 : 3. The cylinders are heated to the same temperature and allowed to cool in the same surrounding. Then the ratio of their initial rate of cooling is

ms−dθdt = e.σ A. θ4−θo4⇒ πr2l.ρs−dθdt = e.σ 2πrlθ4−θo4∴ −dθdt = 2 e.σ θ4−θo4sρ r ∴ R1R2= r2r1=2.

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