Question-165893

Question Number 165893 by Tawa11 last updated on 09/Feb/22

Commented by otchereabdullai@gmail.com last updated on 18/Feb/22

$$\mathrm{nice} \\ $$

Answered by som(math1967) last updated on 10/Feb/22

Vol. of hemispherical tank=(2/3)π×5^3 m^3 vol. of water filled in 1 sec from pipe=π×(0..2)^2 ×2 m^3 time taken to fill the tank =((2×π×5^3 )/(3×π×(0.2)^2 ×2))=((125)/(.12)) sec =17.4minute

$${Vol}.\:{of}\:{hemispherical}\:{tank}=\frac{\mathrm{2}}{\mathrm{3}}\pi×\mathrm{5}^{\mathrm{3}} {m}^{\mathrm{3}} \\ $$$${vol}.\:{of}\:{water}\:{filled}\:{in}\:\mathrm{1}\:{sec}\:{from} \\ $$$${pipe}=\pi×\left(\mathrm{0}..\mathrm{2}\right)^{\mathrm{2}} ×\mathrm{2}\:{m}^{\mathrm{3}} \\ $$$${time}\:{taken}\:{to}\:{fill}\:{the}\:{tank} \\ $$$$=\frac{\mathrm{2}×\pi×\mathrm{5}^{\mathrm{3}} }{\mathrm{3}×\pi×\left(\mathrm{0}.\mathrm{2}\right)^{\mathrm{2}} ×\mathrm{2}}=\frac{\mathrm{125}}{.\mathrm{12}}\:{sec} \\ $$$$=\mathrm{17}.\mathrm{4}{minute} \\ $$

Commented by Tawa11 last updated on 10/Feb/22