A tank is filled upto a height h with a liquid and is placed on a platform of height h from ground.

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9-1.Fluid Mechanics

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A tank is filled upto a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get maximum range $x_m$ a small hole is punched at a distance of $y$ from the free surface of the liquid. Then

A

$x_m$ = $2\ h$

B

$x_m$ = $1.5\ h$

C

$y = h$

D

$A$ and $C$ both

Solution

velocity of efflux at $y V_{x}=\sqrt{2 g y}$

At y, vertical component is zero

The distance to ground is $h=(L-y)$ if Lis the distance from ground to top of vessel

Hence, time required to reach ground $t=\sqrt{2(L-y) / g}$

Horizontal Range is $V_{x} \times t$

As we have to maximize range, we differentiate $V_{x}$ twith $y$

We get range maximum for $y=L / 2=h$

Accordingly, if we put $y=h$ in range equation, we get range $=2 h$

Standard 11

Physics

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