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A cylindrical tank of height 0.4m is open at the top and has a diameter 0.16m. Water is filled in it uo to height of 0.16m. Find the time taken to empty the tank through a hole of radius 5xx10^(-3)m in its bottom. |
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Answer» Solution :The VELOCITY of efflux through the hole `V=sqrt2gh` Let R, r nr the radius of cylindrical TANK and hole and .dh. is the GEIGHT of WATER in the .dt.sec useprinciple of continuity at the top and hole . then `-piR^(2)(dh)/(dt)=pir^(2)v = pir^(2) sqrt2gh` or `-(dh)/(sqrth)=(r^(2))/(R^(2))sqrt2g dt` integrating it wuthin the condition of PROBLEM, `-int_(h)^(0)(dh)/(sqrth)=((5xx10^(-3))/(0.08))^(2) sqrt2xx9.8 int_(0)^(t)dt` `2sqrth=((0.005)/(0.008))^(2)sqrt2xx9.8xxt` (or)`t=(2xxsqrt0.16)/(((0.005)/(0.08))^(2) xx sqrt2 xx 9.8) =46.26s` |
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