Съдържание

Instructions Table of Constants Условие Решение

INPhO 2018, Теория — Задача 7. Water level in an inverted cylinder

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2018

Indian National Physics Olympiad – 2018 · 28 януари 2018 г. · 15 т.

Внимание: Бележка към темата

All seven matching archived keys are retained, including alternative ladder methods, unequal-resistor generalization, accepted ranges and the full handwritten graph. Printed refraction, hourglass, Fibonacci and notation inconsistencies remain literal with separate source notes. The full original includes nineteen blank detailed-answer sheets and two graph papers.

Native cover and instructions

Indian National Physics Olympiad – 2018

Date: 28 January 2018. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,8\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ .

I permit/do not permit (strike out one) HBCSE to reveal my academic performance and personal details to a third party.

Besides the International Physics Olympiad (IPhO) 2018, do you also want to be considered for the Asian Physics Olympiad (APhO) 2018? For APhO 2018 and its pre-departure training, your presence will be required in Delhi and Vietnam from April 28 to May 15, 2018. In principle, you can participate in both Olympiads. Yes/No.

Full Name (BLOCK letters) Ms./Mr.: [blank].

Extra sheets attached: [blank box]. Date: [blank]. Centre (e.g. Kochi): [blank]. Signature: [blank].

(Do not write below this line)

Instructions

This booklet consists of 27 pages (excluding this sheet) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. 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.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Rest mass of proton	$m_p$	$1.67\times10^{-27}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
Molar mass of water		$18.02\ \mathrm{g\cdot mol^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	4
2	8
3	12
4	9
5	10
6	17
7	15
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover and marking policy

Solutions of Indian National Physics Olympiad – 2018

The key cover repeats the date, roll-number form, APhO participation question, nine instructions, thirteen constant rows, seven-question score table and HBCSE address of the question cover, but prints “16 pages (excluding this sheet)”. The retained key has eighteen PDF pages: the cover, sixteen numbered pages and one appended handwritten graph sheet.

On its next page the key states: ANY ALTERNATIVE METHOD OF SOLUTION TO ANY QUESTION THAT IS SCIENTIFICALLY AND MATHEMATICALLY CORRECT, AND LEADS TO THE SAME ANSWER WILL BE ACCEPTED WITH FULL CREDIT. PARTIALLY CORRECT ANSWERS WILL GAIN PARTIAL CREDIT.

Условие

A small circular hole of diameter $d$ is punched on the side and the near the bottom of a transparent cylinder of diameter $D$ . The hole is initially sealed and the cylinder is filled with water of density $\rho_w$ . It is then inverted onto a bucket filled to the brim with water. The seal is removed, air rushes in and height $h(t)$ of the water level (as measured from the surface level of the water in the bucket) is recorded at different times ( $t$ ). The figure below and the table in part (c) illustrates this process. Assume that air is an incompressible fluid with density $\rho_a$ and its motion into the cylinder is a streamline flow. Thus its speed $v$ is related to the pressure difference $\Delta P$ by the Bernoulli relation. Take the outside pressure $P_0$ to be atmospheric pressure $=1.00\times10^5\ \mathrm{Pa}$ .

Whole native inverted cylinder/bucket, D/d/h dimension arrows, hole, levels and base.

a) Obtain the dependence of the instantaneous speed $v_w$ of the water level in the cylinder on $h$ .

Summary answer field: $v_w=$ [blank]. [3 т.]

b) Obtain the dependence of $h$ on time.

Summary answer field: $h=$ [blank]. [3 т.]

c) The table gives the height $h$ as function of time $t$ . Draw a suitable linear graph ( $t$ on $x$ axis) from this data on the graph paper provided. Two graph papers are provided with this booklet in case you make a mistake.

$t$ (sec)	$h$ (cm)
0.57	21.54
1.20	20.10
1.81	18.67
2.47	17.23
3.07	15.80
3.86	14.36
4.55	12.92
5.34	11.49

Full native green blank graph sheet, printed page25, question-number field, header and every grid boundary.

Full native green blank graph sheet, printed page26, question-number field, header and every grid boundary.

[4 т.]

d) From the graph and the following data: $D=6.66\ \mathrm{cm}$ , $\rho_a=1.142\ \mathrm{kg/m^3}$ , $\rho_w=1.000\times10^3\ \mathrm{kg/m^3}$ obtain

i. The height $h_0$ at $t=0$ .

Summary answer field: $h_0=$ [blank].

ii. The value of $d$ .

Summary answer field: $d=$ [blank].

iii. The initial speed ( $v_w$ ) of the water level.

Summary answer field: $v_w(t=0)=$ [blank].

Detailed answers can be found on page numbers: [blank]. [5 т.]

Решение

(a)

at time $t$ from pressure balance equation

P_0=p+h\rho_wg

\Delta P=P_0-p=h\rho_wg

From Bernoulli’s principle

\frac12v^2\rho_a=h\rho_wg=\Delta P

v=\sqrt{\frac{2\rho_wg}{\rho_a}}\sqrt h

The speed of air gushing in through the small hole is related to speed $v_w$ of fall of water level by

\frac{\pi d^2}4v=\frac{\pi D^2}4v_w\qquad (28)

v_w=\frac{d^2}{D^2}\sqrt{\frac{2\rho_wg}{\rho_a}}\sqrt h\qquad (29)

(b)

v_w=-\frac{dh}{dt}=\frac{d^2}{D^2}\sqrt{\frac{2\rho_wg}{\rho_a}}\sqrt h

\sqrt h=\sqrt{h_0}-\frac{d^2}{D^2}\sqrt{\frac{\rho_wg}{2\rho_a}}\,t

(c)

The appended original handwritten graph sheet is retained in full.

Complete original scanned handwritten graph sheet: native Page26 header, eight data points, t/√cm axes, scales, intercept4.78, slope construction and h0=22.85cm, all grid boundaries.

Native handwritten annotations: The horizontal axis is $t$ (in sec), the vertical values are in $\sqrt{\mathrm{cm}}$ , the marked origin is $(0,3.25)$ , and the intercept is $4.78$ . The vertical handwritten label uses $\sqrt{h_0}$ even though the plotted values vary with $t$ .

Scale: On Xaxis $\to1\ \mathrm{cm}=0.33\ \mathrm{sec}$ ; Yaxis $\to1\ \mathrm{cm}=0.062\sqrt{\mathrm{cm}}$ .

Marked coordinates: $(1.48,4.40)$ and $(4.42,3.63)$ .

\text{Slope}=\frac{4.40-3.63}{1.48-4.42}=-2.61\times10^{-1}\ \frac{\sqrt{\mathrm{cm}}}{\mathrm{sec}}

\Rightarrow\sqrt{h_0}=4.78

h_0=22.85\ \mathrm{cm}

(d) i.

22.0\le h_0\ (\mathrm{cm})\le24.0

(d) ii.

-0.27\le\text{slope}\ (\sqrt{\mathrm{cm}}/\mathrm s)\le-0.25

0.13\le d\ (\mathrm{cm})\le0.14

(d) iii.

2.35\le v_w\ (\mathrm{cm/s})\le2.65

Source note: The key’s final PDF page is a scanned completed version of the native graph sheet marked “INPhO 2018 Page 26 Detailed Answers”. It contains the eight plotted data points, fit, construction lines, marked coordinates, units, scales, intercept and height calculation. The whole sheet remains available, including its printed header and the original handwritten $\sqrt{h_0}$ vertical label. The preceding printed key gives the accepted ranges; no unprinted point table or additional fitted answer has been supplied.

Оригинал в Архива: INPhO2018-Question.pdf · решение от архива: INPhO2018-Solution-20180213.pdf