First slide

Fluid dynamics

Question

A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes $T_{1}$ time to decrease the height of water $\frac{H}{η} (η > 1)$ ; and it takes $T_{2}$ time to take out the rest of water. If $T_{1} = T_{2}$ , then the value of $η$ is

Moderate

A

2
B

3
C

4
D

$2 \sqrt{2}$

Solution

$t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$
Now, $T_{1} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{\frac{H}{η}}]$
and $T_{2} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{\frac{H}{η}} - \sqrt{0}]$
According to problem $T_{1} = T_{2}$
$∴ \sqrt{H} - \sqrt{\frac{H}{η}} = \sqrt{\frac{H}{η}} - 0 \Rightarrow \sqrt{H} = 2 \sqrt{\frac{H}{η}} \Rightarrow η = 4$

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