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Two holes 1 and 2 are made at depths h and 16h, respectively. Both the holes are circular but radius of hole 1 is two times the radius of hole 2.

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Initially equal volumes of liquid will flow from both the holes in unit time

Initially more volume of liquid will flow from hole 2 per unit time

After some time more volume of liquid will flow from hole 1 per unit time

After some time more volume of liquid will flow from hole 2 per unit time

answer is A, D.

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Detailed Solution

Initially

$\frac{d V_{1}}{d t} = v_{1} a_{1} = (\sqrt{2 g h}) (π) {(2 R)}^{2} . . . (i) \frac{{dV}_{2}}{dt} = v_{2} a_{2} = (\sqrt{2 g (16 h)}) (π) {(R)}^{2} . . . (ii)$

From Eqs. (i) and (ii). we can see that $\frac{d V_{1}}{d t} = \frac{d V_{2}}{d t}$

After some time v₁ and v₂ both will decrease, but decrease in the value of v₁ is more dominating. So,

$a_{1} v_{1} o r \frac{d V_{1}}{d t} < a_{2} v_{2} o r \frac{d V_{2}}{d t}$

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