A. 100s
B. 10s
C. 1min
D. 10min
I got this question in an online interview.
The question is from Fluids Mechanical Properties in section Mechanical Properties of Fluids of Physics – Class 11
Explanation: The speed of efflux when height of water is ’h’ = \(\sqrt{2gh}\). The volume of water flowing out in time dt is Avdt = 0.005*\(\sqrt{2gh}\)*dt. Therefore the decrease in height of water level (dh) in time dt is vol flowing out / area of tank = 0.005\(\sqrt{2gh}\)dt/5.
dh = \(\sqrt{2gh}\) (0.005)dt/5
\(\int_{0}^{0.05}dh/\sqrt{h} = \int_{0}^{t}\sqrt{2g}\)0.001 dt
∴ 2\(\sqrt{0.05}\) = \(\sqrt{20}\)0.001 t
∴ t = \(\sqrt{0.2}\)*1000/\(\sqrt{20}\) = 100s.