Question #37156

A rectangular vessel when full of water takes 10 minute to be empties through an orifice in the bottom .how much time will it take to be emptied when half filledf with water ??????????????????????

Expert's answer

A rectangular vessel when full of water takes 10 minute to be empties through an orifice in the bottom. how much time will it take to be emptied when half filled with water?

**Solution:**

h1=Hhigh of the water when vessel is full;h_1 = H - \text{high of the water when vessel is full};

t1=10mintime after which vessel is empty (height h1);t_1 = 10\text{min} - \text{time after which vessel is empty (height } h_1);

h2=H2high of the water when vessel is half filled;h_2 = \frac{H}{2} - \text{high of the water when vessel is half filled};

t2time after which vessel is empty (height h2)t_2 - \text{time after which vessel is empty (height } h_2)

the emptying time is proportional to the height of water:

t=α2hg,hheight of water,αsome coefficient of proportionalityt = \alpha \sqrt{\frac{2h}{g}}, h - \text{height of water}, \alpha - \text{some coefficient of proportionality}

t1=α2Hgt_1 = \alpha \sqrt{\frac{2H}{g}}t2=α2h2g=α2H2g=αHgt_2 = \alpha \sqrt{\frac{2h_2}{g}} = \alpha \sqrt{\frac{2\frac{H}{2}}{g}} = \alpha \sqrt{\frac{H}{g}}


(2) ÷\div (1):


t2t1=αHg1αg2H=12\frac{t_2}{t_1} = \alpha \sqrt{\frac{H}{g}} \cdot \frac{1}{\alpha} \sqrt{\frac{g}{2H}} = \frac{1}{\sqrt{2}}t2=t12=10 minutes2=7 minutest_2 = \frac{t_1}{\sqrt{2}} = \frac{10 \text{ minutes}}{\sqrt{2}} = 7 \text{ minutes}


**Answer:** time after which vessel is empty, is equal to 7 minutes.

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