Two large, identical water tanks, 1 and 2, kept on the top of a building of height , are filled with water up to height in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. [2024]

Question

Two large, identical water tanks, 1 and 2, kept on the top of a building of height $H$ , are filled with water up to height $h$ in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are $t_{1}$ and $t_{2}$ , respectively. If $H = (\frac{16}{9}) h,$ then the ratio $\frac{t_{1}}{t_{2}}$ is ______. [2024]

Smart Achievers · Accepted Answer

(3)

$From principle of continuity a_{1} v_{1} = a_{2} v_{2}$

$a \sqrt{2 g h} = - A \frac{d h}{d t}$

$d t = - \frac{A}{a} \frac{d h}{\sqrt{2 g h}}$

$t = \int d t = - \frac{2 A}{a \sqrt{2 g}} (\sqrt{h_{f}} - \sqrt{h_{i}})$

$t = \frac{2 A}{a \sqrt{2 g}} (\sqrt{h_{i}} - \sqrt{h_{f}})$

$For tank 1: h_{i} = h, h_{f} = 0$

$∴$ $t_{1} = \frac{2 A}{a \sqrt{2 g}} (\sqrt{h})$

$For tank 2: h_{f} = \frac{16 h}{9}, h_{i} = h + H = \frac{25 h}{9}$

$t_{2} = \frac{2 A \sqrt{h}}{a \sqrt{2 g}} (\frac{5}{3} - \frac{4}{3}) = \frac{2 A \sqrt{h}}{a \sqrt{2 g}} \times \frac{1}{3}$

$∴$ $\frac{T_{1}}{T_{2}} = 3$

Solution

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Solution

Ans. (3) From principle of continuity a1v1=a2v2 a2gh=-Adhdt dt=-Aadh2gh t=∫dt=-2Aa2g(hf-hi) t=2Aa2g(hi-hf) For tank 1: hi=h, hf=0 ∴ t1=2Aa2g(h) For tank 2: hf=16h9, hi=h+H=25h9 t2=2Aha2g(53-43)=2Aha2g×13 ∴ T1T2=3

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