a. In a conical vessel, the water level rises at a constant rate v_0. How does the rate of water entering a vessel through a tube of section s depend on time? At time zero, the vessel is empty.b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness h. How does the speed of movement of the spot boundary depend on time, if the volume of oil q enters per unit of time? At the initial time, the spot radius is zero.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty.

        <p>

b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero.

    <h3>Solution</h3>
    <p>
        
          <center>
          <figure>
            <img src="drawing1.png"
              loading="lazy" width="200" />
            <figcaption>
              Cone vessel
            </figcaption>
          </figure>
          </center>

$a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to:
$$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$
Where $dx$ is the change in water level:

$$\frac{dV}{dt} = \pi r(t)^2 v_0$$

From Geometry,
$$r(t) = v_0 t \cdot tg(\alpha)$$

By definition, the velocity of incoming water is equal to
$$v = \frac{dV}{sdt}$$

Substituting the previous expressions:
$$v = {\pi v_0^3 t^2 \cdot tg^2(\alpha)}/s$$

$b)$ For a small time interval $dt$, the volume changes by
$$dV = q dt$$

Also the volume increment can be written as
$$dV = 2\pi r dr \cdot h$$

Thus:

$$q dt = 2\pi r dr \cdot h$$

Considering $v = \frac{dr}{dt}$,
$$\fbox{$v = \frac{q}{2\pi r h}$}$$

    <h4>Answer</h4>
    <p>
        $$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$

$$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$

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