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INPhO 2022, Теория — Задача 3

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Indian Olympiad Qualifier in Physics (IOQP) 2021-2022 — Part II: Indian National Physics Olympiad (INPhO) · 13 март 2022 г. · 9 т.

Указания

This booklet consists of 4 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.
Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Условие

Consider an air filled spherical balloon comprised of elastic material of surface tension $\gamma = 500\,\mathrm{kg/s^2}$ . The pressure outside the balloon is the atmospheric pressure ( $P_{\mathrm{atm}} = 101\,\mathrm{kPa}$ ) and the density of air outside is $\rho_{\mathrm{atm}} = 1.22\,\mathrm{kg/m^3}$ .

The balloon starts deflating slowly. Assume that the average velocity of air inside the balloon is negligible, and air leaves the balloon in a streamline fashion. Consider $\gamma$ to be constant throughout, and the air to be incompressible.

(a) Write an expression for the time $t$ required to deflate the balloon through a small opening of cross-sectional area $A$ from an initial radius $R_0$ to a final radius $R$ . [8 т.]

(b) Obtain the value of this time for $A = 1 \times 10^{-5}\,\mathrm{m^2}$ , $R_0 = 0.15\,\mathrm{m}$ , and $R = 0.05\,\mathrm{m}$ . [1 т.]

Отговори

Решение

Pressure of air inside the balloon of radius $r$ and given surface tension $\gamma$ is

P = P_{\mathrm{atm}} + \frac{4\gamma}{r} \qquad (3.1)

Schematic of a spherical balloon of radius R (dashed radius) with a short nozzle of area A on its right; point 1 is just inside the balloon, point 2 at the nozzle exit, and an arrow v shows the outflowing air.

Consider the schematic diagram of deflating the balloon. Just inside the balloon, at point (1), gas can be treated stationary. Outside, at point (2), it can be treated flowing out with speed $v$ . Then by Bernoulli’s equation

P_1 = P_2 + \frac{\rho v^2}{2} \qquad (3.2)

Outside pressure and density are $P_{\mathrm{atm}}$ and $\rho_{\mathrm{atm}}$ respectively. Also, the inside pressure is given by the Eq. (3.1). Thus

P_{\mathrm{atm}} + \frac{4\gamma}{r} = P_{\mathrm{atm}} + \frac{\rho_{\mathrm{atm}} v^2}{2} \qquad (3.3)

\Rightarrow v = \sqrt{\frac{8\gamma}{\rho_{\mathrm{atm}} r}} \qquad (3.4)

Volume of air flowing through hole of area $A$ per sec at $P_{\mathrm{atm}}$ is

\frac{dV}{dt} = Av \qquad (3.5)

\frac{d}{dt}\left(\frac{4\pi r^3}{3}\right) = Av \qquad (3.6)

4\pi r^2 \frac{dr}{dt} = Av \qquad (3.7)

We use Eq. (3.4) in the above equation, which yields

r^{5/2} dr = \frac{A}{4\pi}\sqrt{\frac{8\gamma}{\rho}} dt \qquad (3.8)

Integrating from $R_0$ to $R_1$

t = \frac{4\pi}{7A}\sqrt{\frac{\rho_{\mathrm{atm}}}{2\gamma}}\left[R_0^{7/2} - R_1^{7/2}\right] \qquad (3.9)

\qquad (3.10)

In Eq. (3.1) if the extra pressure taken as $2\gamma/R$ then

t = \frac{4\pi}{7A}\sqrt{\frac{\rho_{\mathrm{atm}}}{\gamma}}\left[R_0^{7/2} - R_1^{7/2}\right] \qquad (3.11)

For the given values, Eq. (3.9) gives $t = 8.02\,\mathrm{s}$ and Eq. (3.10) yields $11.22\,\mathrm{s}$ . Both Eqs. (3.9) and (3.10) and accordingly the calculated values are considered correct.

Оригинал в Архива: IOQP2022-PartII-Questions-en.pdf · официални решения: IOQP2022-PartII-Solutions-20220503.pdf