Liquid is filled in a container upto a height of `H`. A small hole is made at the bottom of the tank. Time taken to empty from `H` to `(H)/(3)` is `t_(0)`. Find the time taken to empty tank from `(H)/(3)` to zero.
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`t=(A)/(a) sqrt((2H)/(g))` (to empty the complete tank) Now, `t_(Hrarro)=t_(Hrarr(H)/(3))+tunderset(3)overset(H)rarr0` Given,`(A)/(a) sqrt((2H)/(g))=t_(0)+(A)/(a) sqrt((2H//3)/(g))` from here find, `(A)/(a) sqrt((2H//3)/(g))`.
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