A cylinderical tank of height 0.4 m is open at the top and has a diameter 0.16m. Water is filled in it up to height of 0.16 m. Find the time taken to empty the tank through a hole of radius `5 xx 10^(-3)m` in its bottom.
Text Solution
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The velocity of efflux through a small hole at a depth h is given by ` v = sqrt(2gh)` According to the principle of continuity we get ` - piR^(2) (dh)/(dt) = pir^(2) v` Since the rate of fall of water level in the tank is ` - (dh//dt)` `rArr - R^(2) (dh)/(dt) = r^(2) sqrt(2gh)` ` rArr - (dh)/(sqrt(h)) = (r^(2))/(R^(2)) sqrt(2g) .dt` Integrating we get ` - int _(0.16)^(0) (dh)/(sqrt(h)) = ((0.005)/(0.08))^(2) sqrt(2 xx 9.8 ) int_(0)^(t) dt ` Therefore , `t = 46.26 ` sec
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