$\frac{d T}{d t}=$ Rate of fall temperature
given $\left(\frac{d T}{d t}\right)_{P}=x\left(\frac{d T}{d t}\right)_{Q}\left[m=\text { density } \times \text { volume }=\rho \times \frac{4}{3} \pi r^{3}\right]$
$\frac{d Q}{d t}=\sigma e A T^{4}=m s \frac{d T}{d t}$
$\left(\frac{d T}{d t}\right)_{P}=\frac{\sigma e\left(4 \pi r p^{2}\right)^{4}}{\rho_{3}^{4} \pi r p^{2}}=\frac{3 \sigma e T^{4}}{\rho r p}$
$\frac{\left(\frac{d T}{d t}\right)_{P}}{\left(\frac{d T}{d t}\right)_{Q}}=\frac{r_{q}}{r_{p}}=3$
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$\frac{{dv}}{{dt}} = - 2.5\sqrt v $ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be........$s$
