Съдържание

Instructions Table of Constants Условие Решение

INPhO 2018, Теория — Задача 4. An hour glass on a weighing scale

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2018

Indian National Physics Olympiad – 2018 · 28 януари 2018 г. · 9 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

An hour glass is placed on a weighing scale. Initially all the sand of mass $m_0$ kg in the glass is held in the upper reservoir (ABC) and the mass of the glass alone is $M$ kg. At $t=0$ , the sand is released. It exits the upper reservoir at constant rate $dm/dt=\lambda$ kg/s where $m$ is the mass of the sand in the upper reservoir at time $t$ sec. Assume that the speed of the falling sand is zero at the neck of the glass and after it falls through a constant height $h$ it instantaneously comes to rest on the floor (DE) of the hour glass. Obtain the reading on the scale for all times $t>0$ . Make a detailed plot of the reading vs time.

Whole native hourglass and weighing scale, upper ABC, lower DE, constant h arrow and dial.

Summary answer field: Reading on the scale: [blank].

Complete blank Reading (kg) versus time frame, both axes, arrows and labels.

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

Решение

Velocity of grain when reaches to bottom $v=\sqrt{2gh}$ . At $t=0$ , sand is released. Time taken $t_1=\sqrt{2h/g}$ .

$0<t<t_1$

For $t<0$ , the scale was $(m+M)g$ . As $t=0\to t_1$ , more sand will enter the mid air portion and hence the reading will drop,

\lambda=\frac{dm}{dt}\Rightarrow m=\lambda t\qquad (11)

\text{Weight }(W_1)=(M+m)g-\lambda t_1g\qquad (12)

$t_1<t<t_2$

Here $t_2=m/\lambda$ is the time when all the sand has left the upper reservoir. Just after $t=t_1$ , force on the scale will partly due to weight of sand as given in above equation and partly due to impulse provided by falling sand.

\text{Impulse force}=v\frac{dm}{dt}=\lambda\sqrt{2gh}\qquad (13)

\text{Weight }(W_2)=[(M=m)g-\lambda t_1g]+\lambda\sqrt{2gh}=(M+m)g\qquad (14)

$t_2<t<t_3$

Here $t_3=m/\lambda+\sqrt{2h/g}$ . This is the time span between the sand has left upper reservoir to when all sand has reached the floor of the hour glass. The weight is

W_3=W_2+\lambda(t-t_2)g\qquad (15)

For $t>t_3$

weight is

W_4=(M+m)g

Full native scale-reading graph, three vertical levels m+M±λ√(2h/g), t1/t2/t3, discontinuities, origin, axis arrows and units.

Source note: These are literal printed key expressions. The question defines $m$ as the remaining mass in the upper reservoir but gives $dm/dt=+\lambda$ ; the key reuses $m$ both for flowing mass ( $m=\lambda t$ ) and in total-weight/time expressions instead of the initial $m_0$ . Eq. (12) prints $t_1$ although the native graph depicts a reading that decreases throughout the first interval. Eq. (14) prints an equals sign in $(M=m)$ , while its final expression uses $(M+m)$ . The graph assumes the pictured order $t_1<t_2<t_3$ ; the source does not supply a separate case for an upper reservoir that empties before the first grain arrives. No replacement formula or additional unprinted case has been inserted.

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