Question:
A metallic sphere cools from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $300 \mathrm{~s}$. If atmospheric temperature around is $20^{\circ} \mathrm{C}$, then the sphere's temperature after the next 5 minutes will be close to:
Correct Option: 1
Solution:
$\frac{50-40}{300}=\beta\left(\frac{50+40}{2}-20\right)$
$\frac{40-\mathrm{T}}{300}=\beta\left(\frac{40+\mathrm{T}}{2}-20\right)$
$\therefore \mathrm{T}=\frac{100}{3}$
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