If it takes 5,minutes to fill a 15,litre buck | vedclass

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(D) Given: Diameter of tap $D = \frac{2}{\sqrt{\pi}}\,cm = \frac{2}{\sqrt{\pi}} 	imes 10^{-2}\,m$.Radius $r = \frac{D}{2} = \frac{1}{\sqrt{\pi}} 	im

Vedclass · Accepted Answer

$5500$

Vedclass · Answer

(D) Given: Diameter of tap $D = \frac{2}{\sqrt{\pi}}\,cm = \frac{2}{\sqrt{\pi}} 	imes 10^{-2}\,m$.Radius $r = \frac{D}{2} = \frac{1}{\sqrt{\pi}} 	imes 10^{-2}\,m$.Volume flow rate $Q = \frac{15\,litres}{5\,minutes} = \frac{15 	imes 10^{-3}\,m^3}{300\,s} = 5 	imes 10^{-5}\,m^3/s$.Area of cross-section $A = \pi r^2 = \pi \left( \frac{1}{\sqrt{\pi}} 	imes 10^{-2} ight)^2 = 10^{-4}\,m^2$.Velocity $v = \frac{Q}{A} = \frac{5 	imes 10^{-5}}{10^{-4}} = 0.5\,m/s$.Reynolds number $R_e = \frac{ho v D}{\eta}$,where $ho = 10^3\,kg/m^3$,$v = 0.5\,m/s$,$D = \frac{2}{\sqrt{\pi}} 	imes 10^{-2}\,m$,and $\eta = 10^{-3}\,Pa\cdot s$.$R_e = \frac{10^3 	imes 0.5 	imes (2/\sqrt{\pi}) 	imes 10^{-2}}{10^{-3}} = \frac{5 	imes (2/\sqrt{\pi})}{10^{-3}} = \frac{10}{\sqrt{\pi}} 	imes 10^3 \approx \frac{10}{1.772} 	imes 10^3 \approx 5642$.The closest value is $5500$.

If it takes $5\,minutes$ to fill a $15\,litre$ bucket from a water tap of diameter $\frac{2}{\sqrt{\pi}}\,cm$,then the Reynolds number for the flow is (density of water $= 10^3\,kg/m^3$ and viscosity of water $= 10^{-3}\,Pa\cdot s$) close to

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