A laminar stream is flowing vertically down f | vedclass

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(B) Let $A_1 = 1 \ cm^2$ and $A_2 = 0.5 \ cm^2$ be the cross-sectional areas at the tap and at a distance $h = 10 \ cm$ below it,respectively.From the

Vedclass · Accepted Answer

$4.9$

Vedclass · Answer

(B) Let $A_1 = 1 \ cm^2$ and $A_2 = 0.5 \ cm^2$ be the cross-sectional areas at the tap and at a distance $h = 10 \ cm$ below it,respectively.From the equation of continuity,$A_1 v_1 = A_2 v_2$,we have $1 \cdot v_1 = 0.5 \cdot v_2$,which implies $v_2 = 2v_1$.Using Bernoulli's equation between the two points: $P_0 + \frac{1}{2}ho v_1^2 + ho gh = P_0 + \frac{1}{2}ho v_2^2$.Substituting $v_2 = 2v_1$: $gh + \frac{1}{2}v_1^2 = \frac{1}{2}(2v_1)^2 = 2v_1^2$.Thus,$gh = \frac{3}{2}v_1^2$,so $v_1 = \sqrt{\frac{2gh}{3}}$.Given $g = 980 \ cm/s^2$ and $h = 10 \ cm$,$v_1 = \sqrt{\frac{2 \cdot 980 \cdot 10}{3}} = \sqrt{\frac{19600}{3}} \approx 80.83 \ cm/s$.Volumetric flow rate $Q = A_1 v_1 = 1 \ cm^2 \cdot 80.83 \ cm/s = 80.83 \ cm^3/s$.Converting to $litre/min$: $Q = 80.83 	imes 10^{-3} \ litre/s = 80.83 	imes 10^{-3} 	imes 60 \ litre/min \approx 4.85 \ litre/min$.Rounding to the nearest option,we get $4.9 \ litre/min$.

$A$ laminar stream is flowing vertically down from a tap of cross-section area $1 \ cm^2$. At a distance $10 \ cm$ below the tap,the cross-section area of the stream has reduced to $0.5 \ cm^2$. The volumetric flow rate of water from the tap is about ........ $litre/min$.

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