Bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$, it flows out…

Bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$, it flows out in $t$ seconds. If water is filled to a height $4 h$, it will flow out in time
  1. $t$
  2. $4 t$
  3. $2 t$
  4. $\frac{t}{4}$

Solution

$ \text { The given situation is shown in the figure, } $
Time taken to flow out of water upto height $h$ from hole completely, $ t=\sqrt{\frac{2 h}{g}} $ When, $\quad h^{\prime}=4 h$, then $ t^{\prime}=\sqrt{\frac{2 h^{\prime}}{g}}=\sqrt{\frac{2 \times 4 h}{g}}=2 \sqrt{\frac{2 h}{g}}=2 t $ [from Eq. (i)]

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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