Question Number 181053 by Acem last updated on 22/Nov/22
$${An}\:{open}\:{pipe}\:{of}\:{r}_{{o}} =\:\mathrm{10}\:{Cm},\:\ell=\:\mathrm{3}{m}\:{has}\:{an}\:{outer} \\ $$$$\:{layer}\:{of}\:{ice}\:{that}\:{is}\:{melting}\:{at}\:{the}\:{rate}\:{of}\:\mathrm{2}\pi\:{Cm}^{\mathrm{3}} \\ $$$$\:{per}\:{minute}\:{with}\:{thickness}\:{of}\:\mathrm{20}\:{mm}.\:{How}\:{many} \\ $$$$\:{days}\:{untill}\:{all}\:{the}\:{ice}\:{melts}?\:{and}\:{how}\:{fast}\:{is}\:{the} \\ $$$$\:{thickness}\:{of}\:{the}\:{ice}\:{decreasing}\:{per}\:{hour}? \\ $$
Answered by Acem last updated on 22/Nov/22
![V_(ice) = π ℓ [(r_o +δ)^2 −r_o ^( 2) ] ; δ: thickness of the ice layer (∂V/∂t)= 2π ℓ (r_o +δ) (∂δ/∂t)= −2π (∂δ/∂t)= − 0.16 mm/hr the rate of the ice is decreasing Hence 4mm per day Then it takes 5 days to fully melt](Q181200.png)
$$\:{V}_{{ice}} =\:\pi\:\ell\:\left[\left({r}_{{o}} +\delta\right)^{\mathrm{2}} −{r}_{{o}} ^{\:\mathrm{2}} \right]\:\:;\:\delta:\:{thickness}\:{of}\:{the}\:{ice}\:{layer} \\ $$$$\:\:\frac{\partial{V}}{\partial{t}}=\:\mathrm{2}\pi\:\ell\:\left({r}_{{o}} +\delta\right)\:\frac{\partial\delta}{\partial{t}}=\:−\mathrm{2}\pi \\ $$$$\:\frac{\partial\delta}{\partial{t}}=\:−\:\mathrm{0}.\mathrm{16}\:{mm}/{hr}\:\:{the}\:{rate}\:{of}\:{the}\:{ice}\:{is}\:{decreasing} \\ $$$$\:{Hence}\:\mathrm{4}{mm}\:{per}\:{day} \\ $$$$\:{Then}\:{it}\:{takes}\:\mathrm{5}\:{days}\:{to}\:{fully}\:{melt} \\ $$$$ \\ $$