Съдържание

Universal constants Astronomical data Calculus related formulas Условие (T11.1) Part 1: Inside a star (T11.2) Part 2: Evolving stars (T11.3) Part 3: Mass distribution inside a star Решение General Marking Principles

IOAA 2025, Теория — Задача 11. Stars through graphs

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2025

IOAA 2025 — Theory Examination · 50 т.

Theory Examination Instructions

You are given an envelope marked B. Please verify the Contestant Code printed on it. Inside you will find:

This Instruction Sheet and/or This Instruction Sheet in your native language (if applicable)
Data Sheet and/or Data Sheet in your native language (if applicable)
A set of Working Sheets
One envelope marked S (please verify the Contestant Code marked on it).

You are allowed to open the envelope S only when the whistle is blown to mark the start of the examination.

Inside the envelope S you will find the following:

Question Paper and/or Question Paper in your native language (if applicable)
A set of Summary Answersheets.

Total duration of this examination is 5 hours.

This examination consists of the following questions:

Questions	Marks each	Subtotal
T01 and T02	10	$2\times10=20$ marks
T03 and T04	15	$2\times15=30$ marks
T05–T08	20	$4\times20=80$ marks
T09 and T10	35	$2\times35=70$ marks
T11 and T12	50	$2\times50=100$ marks
Total		300 marks

Writing your answers:

Summary Answersheet:

For each question a separate Summary Answersheet has been provided. Write your contestant code on each page of the Summary Answersheet.
Final answer(s) for each question / each part of a question must be written in the corresponding box in the Summary Answersheet.

Working Sheets:

Your detailed work must be shown on the Working Sheet.
Write your Contestant Code on both sides of each Working Sheet.
Start a new question on a fresh Working Sheet. Write the question number in the box provided at the top.
Both sides of the Working Sheet should be used. Write only inside the boxed area.
For rough work that you do not want to be evaluated, clearly strike that part out.
For each question write the page number in the box provided at the top of the Working Sheet, starting with 1 (each sheet has 2 pages). For example, if you use only 1 Working Sheet for T01, the question number and page number should be “T01, 1” and “T01, 2” on the FRONT and BACK of that Working Sheet, respectively. Next, if you use 2 Working Sheets for T02, the question number and page number should be “T02, 1” and “T02, 2” for the first Working Sheet, and “T02, 3” and “T02, 4” on the second Working Sheet.
You may ask for extra Working Sheets, if required.
Note that failure to mark the Working Sheets, as instructed above, may result in your answers not being evaluated.

At the end of the examination:

Check that you have correctly written your Contestant Code and page numbers on all used sheets.
Arrange your sheets in the following way: Summary Answersheet of each question (starting with T01) followed by Working Sheet(s) used for that question.
Similarly stack the Summary Answersheet and Working Sheet(s) used for each question, in sequence.
Place the bunch of all your used sheets arranged in the above manner inside the smaller envelope S. All Summary Answersheets must be placed inside the envelope S irrespective of whether you have used them or not.
Only the contents of the smaller envelope S will be evaluated. Any sheet that is not placed inside this envelope will not be evaluated.
Place the smaller envelope S inside the bigger envelope B.
All other sheets, namely, Instruction Sheet(s), Question Paper(s), Data Sheet(s) and unused Working Sheet(s) should be placed directly inside the bigger envelope B.
Tie the thread seal on the envelope B.
Hand over the envelope B to the examiner. Make sure that no sheet is left on the table, except the green and the red card.

General Instructions:

If you need a visit to the toilet during the examination, please raise your hand holding the green card kept on your table. You will be escorted by a volunteer.
If you need any other help during the examination, please raise your hand holding the red card kept on your table.
Use only black or blue pen for writing. For drawing, you may use a pencil.
Some marks may be deducted if the final answer is given without detailed solution.
Some marks may be deducted if the final answers have improper number of significant digits, no units or wrong units.
The necessary values of fundamental and astronomical constants should be taken from the Data Sheet provided to you.

Original IOAA 2025 Data Sheet

Universal constants

Quantity	Symbol	Value
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Charge of electron	$e$	$1.602\times10^{-19}\ \mathrm C$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\,s}$
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m\,s^{-1}}$
Universal gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Wien’s displacement constant	$b$	$2.898\times10^{-3}\ \mathrm{m\,K}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Permeability of free space	$\mu_0$	$1.257\times10^{-6}\ \mathrm{N\,A^{-2}}$
Mass of electron	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
Mass of proton	$m_p$	$1.673\times10^{-27}\ \mathrm{kg}=938.272\ \mathrm{MeV}/c^2$
Mass of neutron	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}=939.565\ \mathrm{MeV}/c^2$
Mass of deuteron	$m_D$	$3.344\times10^{-27}\ \mathrm{kg}=1875.613\ \mathrm{MeV}/c^2$
Mass of He nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\ \mathrm{kg}=3727.181\ \mathrm{MeV}/c^2$

Astronomical data

Quantity	Symbol	Value
Mass of Sun	$M_\odot$	$1.988\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_\odot$	$6.957\times10^8\ \mathrm m$
Luminosity of Sun	$L_\odot$	$3.828\times10^{26}\ \mathrm W$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5772\ \mathrm K$
Apparent magnitude of Sun (in V-band)	$m_{V,\odot}$	$-26.74$
Absolute magnitude of Sun (in V-band)	$M_{V,\odot}$	$+4.82$
Apparent bolometric magnitude of Sun	$m_{\mathrm{bol},\odot}$	$-26.83$
Absolute bolometric magnitude of Sun	$M_{\mathrm{bol},\odot}$	$+4.74$
Solar constant (above atmosphere of Earth)	$S_\odot$	$1361\ \mathrm{W\,m^{-2}}$
Apparent angular diameter of Sun (from Earth)	$\theta_\odot$	$\approx32'$
Mass of Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_\oplus$	$6.378\times10^6\ \mathrm m$
Axial tilt of Earth	$\epsilon$	$23^\circ26'$
Inclination of the lunar orbit w.r.t. the ecliptic		$5^\circ8'43''$
Mass of Jupiter	$M_J$	$1.898\times10^{27}\ \mathrm{kg}$
Radius of Jupiter	$R_J$	$6.991\times10^7\ \mathrm m$
1 Astronomical Unit	1 au	$1.496\times10^{11}\ \mathrm m$
1 parsec	1 pc	$3.086\times10^{16}\ \mathrm m$
1 light-year	1 ly	$9.461\times10^{15}\ \mathrm m$
1 jansky	1 Jy	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 tropical year		$365.2422$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 5\ \mathrm h\ 48\ \mathrm{min}\ 46\ \mathrm s$
1 sidereal year		$365.2564$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 6\ \mathrm h\ 9\ \mathrm{min}\ 13\ \mathrm s$
Rate of precession of Vernal Equinox		$1^\circ$ per $71.6$ years

Calculus related formulas

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ .
$\frac{d}{dx}x^n=nx^{n-1}$ .
$\frac{d}{dx}\sin kx=k\cos kx$ .
$\frac{d}{dx}\cos kx=-k\sin kx$ .
$\frac{d}{dx}\tan kx=k\sec^2kx$ .
$\int x^n\,dx=\frac{x^{n+1}}{n+1}+\text{constant}$ ; for $n\ne-1$ .
$f(x)\simeq f(x_0)+\left.\frac{df}{dx}\right|_{x=x_0}(x-x_0)$ ; for $x\approx x_0$ .

Условие

Stars can be well approximated as spherically symmetric objects, and hence the radial distance $r$ from the centre can be chosen as the only independent variable in modelling stellar interiors. The mass contained within a sphere of radius $r$ is denoted by $m(r)$ . The luminosity $l(r)$ is defined as the net energy flowing outward through a spherical surface of radius $r$ per unit time. Other quantities of interest, for example, the density $\rho(r)$ , temperature $T(r)$ , hydrogen mass fraction $X(r)$ , helium mass fraction $Y(r)$ , and the nuclear energy generated per unit mass per unit time $\epsilon_\mathrm{nuc}(r)$ , are taken to be functions of $r$ . Throughout this problem we shall neglect the effects of diffusion and gravitational settling of elements inside the star.

The symbol “log” refers to logarithm to the base 10. The problem consists of three independent parts.

(T11.1) Part 1: Inside a star

The graph below shows the variation of three structural quantities, A, B, and C, as functions of the fractional radius $r/R$ in a stellar model of mass $1M_\odot$ and age 4 Gyr, where $R$ is the photospheric radius of the star. The values of the helium mass fraction at the (photospheric) surface, $Y_s$ , and the metallicity (mass fraction of all elements heavier than helium) at the (photospheric) surface, $Z_s$ , of the star are given by $(Y_s,Z_s)=(0.28,0.02)$ . All quantities shown in the plots are normalised by their respective maximum values.

Original normalized stellar-interior curves A/B/C, full axes, grid, colours and labels.

(T11.1a) Identify the three quantities A, B, and C uniquely from among the five possibilities:

T(r),\quad l(r),\quad\epsilon_\mathrm{nuc}(r),\quad X(r),\quad Y(r).

(Write A/B/C in the boxes beside the appropriate quantities in the Summary Answersheet. No justification is needed for your answer.) [6 т.]

(T11.1b) What is the mass fraction of helium at the centre, $Y_c$ , of the star? [3 т.]

(T11.1c) Sketch the remaining two quantities from the list of five (which were not identified as curves A, B, or C) given in (T11.1a), as functions of $r/R$ on the same graph in the Summary Answersheet, and label by their respective quantities. [5 т.]

(T11.2) Part 2: Evolving stars

Consider the evolution of a $1M_\odot$ star whose initial uniform composition is given by the mass fractions of helium, $Y_0=0.28$ , and metals, $Z_0=0.02$ . The figures below show the variation of different global quantities of this star as it evolves from ZAMS (Zero Age Main Sequence) till the end of helium burning in its core.

The graph below shows the evolutionary track of the star on the HR diagram (plot of $\log L/L_\odot$ vs $\log T_\mathrm{eff}$ , where $L$ is the surface luminosity and $T_\mathrm{eff}$ is the effective temperature).

Complete original Hertzsprung–Russell track with both P/Q markers, reversed temperature axis and luminosity ticks.

The figure below has four graphs which show the variation of $T_\mathrm{eff}$ (in K), $L$ (plotted as $\log L/L_\odot$ ), $R$ (plotted as $\log R/R_\odot$ ), and $Y_c$ with age (in $10^9\,\mathrm{yr}$ ) of the same star. In each of these four graphs, the insets show the variations of the respective quantities in detail between the ages of $11.86\times10^9\,\mathrm{yr}$ to $12.00\times10^9\,\mathrm{yr}$ , for greater clarity.

All four original stellar-evolution panels, full age scales and every native late-age inset.

Use these graphs to answer the questions below.

(T11.2a) What is the approximate main sequence lifetime, $t_\mathrm{MS}$ (in years), of the star? [1 т.]

(T11.2b) What is the approximate duration, $\Delta t_\mathrm{He}$ (in years), for which the star burns helium in its core? [1 т.]

(T11.2c) What fraction, $f_H$ , of the initial amount of hydrogen at its centre has been burnt when the luminosity of the star is $1L_\odot$ ? [3 т.]

(T11.2d) What is the radius of the star, $R_1$ (in units of $R_\odot$ ) when 60% of the initial amount of hydrogen at its centre has been burnt? [3 т.]

(T11.2e) What are the radii of the star, $R_P$ and $R_Q$ (in units of $R_\odot$ ), corresponding to its positions P and Q, respectively, as marked on the HR-diagram? [4 т.]

(T11.3) Part 3: Mass distribution inside a star

The equation that governs the distribution of mass inside a star is given by

\frac{dm(r)}{dr}=4\pi r^2\rho(r).

It would be convenient to express this equation in terms of three dimensionless variables, namely, the fractional mass, $q$ , the fractional radius, $x$ , and the relative density, $\sigma$ , that we define as

q=\frac mM,\qquad x=\frac rR,\qquad\sigma=\frac\rho{\bar\rho}

where $M$ and $R$ are the total mass and radius of the star, respectively, and $\bar\rho\equiv M/(\frac43\pi R^3)$ is the average density of the star. For the particular star that we shall be considering in this part, the following information is given:

The central density $\rho(x=0)=80\bar\rho$ .
Half of the star’s mass is contained within the inner 25% of its total radius, and 70% of its mass is contained within the inner 35% of its total radius.

In all subsequent parts of this question, it will be sufficient to round off all derived numerical coefficients to within 0.005.

(T11.3a) Express the above equation describing the dependence of mass and radius in terms of $x$ , $dq(x)/dx$ and $\sigma(x)$ . [2 т.]

To obtain the distribution of mass with radius, we need to know the density profile inside the star. For the purpose of this problem, we shall describe the variation of density with radius by approximate forms in two domains of $x$ :

the inner part of the star: $0\le x\le0.32$ ;
the middle part of the star: $0.32<x<0.80$ .

We do not make any approximation for the outermost part, i.e., $0.80\le x\le1.00$ .

(T11.3b) Approximation for the middle part: The variation of $\log\sigma$ , as a function of $\log x$ in the middle part of the star is shown (by the black curve) in the graph below. We shall make a linear approximation (shown as a dashed red line in the graph) for $\log\sigma$ as a function of $\log x$ in the domain $-0.5<\log x<-0.1$ , i.e., $0.32\lesssim x\lesssim0.80$ (shown by the green shaded domain). Further, we shall approximate the slope of this line by the nearest integer.

Use this approximation to write an expression for $\sigma(x)$ as a function of $x$ in the domain $0.32<x<0.80$ . [4 т.]

(T11.3c) Use the result of (T11.3b) above to derive an expression for $q(x)$ in the domain $0.32<x<0.80$ . [6 т.]

(T11.3d) Approximation for the inner part: In the inner part of the star ( $0\le x\le0.32$ ), the density may be approximated as a linear function of radius, i.e., $\sigma(x)=Ax+B$ , where $A,B$ are constants. Determine $A$ and $B$ , and hence obtain an expression for $q(x)$ in the domain $0\le x\le0.32$ . Note that the approximations adopted in the previous part and this part may lead to small discontinuities in density or mass at $x=0.32$ . [8 т.]

(T11.3e) The expressions for $q(x)$ obtained in parts (T11.3c) and (T11.3d) are approximations that describe the variation of mass with radius quite well, but only in specific regions of the star. For the domain $0.80\le x\le1$ (for which we have not derived any expression), it is possible to use appropriate extrapolation from the neighbouring region. Use these approximate expressions and given data to sketch a smooth curve (without any discontinuities either in $q(x)$ or in its derivative) for $q(x)$ vs $x$ for the entire star ( $0\le x\le1$ ) that represents the variation of mass with radius. [4 т.]

Complete original Summary Answersheet T11, SA-21, FRONT: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Complete original Summary Answersheet T11, SA-22, BACK: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Решение

Part 1: Inside a star

(T11.1a)

$X$ and $l$ increase with $r$ while $T$ , $\varepsilon_{\mathrm{nuc}}$ , $Y$ decrease with $r$ . So A and B must be $X$ and $l$ .
For a $1\,M_\odot$ star at age 4 Gyr, $X\ne0$ at the centre. So B cannot be $X$ .
Also, $l=0$ at the centre. This matches with B.
Both $X$ and $l$ must saturate to their maximum values just outside the H-burning core, and this happens at $r/R=0.3$ here.
Each of $T$ , $\varepsilon_{\mathrm{nuc}}$ , $Y$ is maximum at the centre and decreases outward. So C must be one of these.
But $Y$ cannot be (close to) zero at the surface ( $r/R=1$ ) compared to its central value.
$\varepsilon_{\mathrm{nuc}}$ becomes extremely small just outside the core ( $r/R\lesssim0.3$ ), and not at the surface.
Therefore, C must be $T(r)$ (the surface value is $\sim10^3$ K, compared to the maximum value at the centre $\sim10^6$ K).

Therefore,

Curve	Quantity	Marks
A	$X(r)$	2.0
B	$l(r)$	2.0
C	$T(r)$	2.0

No negative marking. Credit only for unique identification. No credit even for correct answer if the same letter is repeated for another quantity.

(T11.1b)

Maximum $X$ is outside the core. Therefore,

X_s=1-Y_s-Z_s=1-0.28-0.02=0.70.\qquad[1.0]

Note that $Z$ is constant throughout in absence of gravitational settling and diffusion (given).

From the graph, the central value $X_c\approx0.55X_s=0.55\times0.70=0.39$ . [1.5]

Credit 1.0 for realising that $X_c\ne0.55$ , but this is the normalised value. Credit 0.5 for correctly reading 0.53–0.55 from the graph.

Then

Y_c=1-X_c-Z_c=1-0.39-0.02=\boxed{0.59}.\qquad[0.5]

Half credit lower limit	Full credit range	Half credit upper limit
	0.59 to 0.61

(T11.1c)

The curve for $\varepsilon_{\mathrm{nuc}}$ must mirror that of $l$ . [2.5]

Any reasonable smooth curve between $(0,1)$ and $(0.3,0)$ is acceptable. Credit points:

Aspect	Marks
Curve starts at $(0,1)$	0.5
Curve goes to zero between $(0.25,0)$ and $(0.35,0)$	1.0
Curve goes smoothly to zero	0.5
Curve is overall smooth	0.5

The curve for $Y$ must mirror that of $X$ , but flatten outside the core at the normalised value $Y_s/Y_c=0.28/0.59\approx0.47$ . [2.5]

Aspect	Marks
Curve starts at $(0,1)$	0.5
Shape mirrors $X$	0.5
Curve flattens between $r/R=0.25$ and $0.35$	0.5
Flattened value between 0.45 and 0.50	1.0

Original stellar-interior answer: full A/B/C plots and handwritten helium/nuclear-generation curves, all grid and labels.

For wrong identification of A, B, C in (T11.1a), student may draw curves of $X$ , $l$ or $T$ in this part. Credit should be given for those as:

Curve	Aspect	Marks
$X$	Starts between $(0,0.5)$ and $(0,0.6)$	1.0
$X$	Shape mirrors $Y$	0.5
$X$	Goes to 1 between $r/R=0.25$ and $0.35$	1.0
$l$	Starts at $(0,0)$	1.0
$l$	Shape mirrors $\varepsilon_{\mathrm{nuc}}$	0.5
$l$	Goes to 1 between $r/R=0.25$ and $0.35$	1.0
$T$	Starts at $(0,1)$	1.0
$T$	Decreases smoothly but not linearly	0.5
$T$	Almost zero at $r/R=1.0$	1.0

Part 2: Evolving stars

(T11.2a)

During the main sequence, the central helium mass fraction increases steadily up to $Y_{c,\max}=1-Z_c$ . Here the maximum is reached at age around 8.5 Gyr from the plot of $Y_c$ vs age.

\boxed{t_{\mathrm{MS}}=8.5\times10^9\ \mathrm{yr}}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$8.0\times10^9$ yr to $9.0\times10^9$ yr

(T11.2b)

During He-burning the central $Y$ decreases steadily to zero. The inset of $Y_c$ shows ages $11.896\times10^9$ to $11.960\times10^9$ yr, corresponding to 64 Myr.

\boxed{\Delta t_{\mathrm{He}}=64\times10^6\ \mathrm{yr}}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$62\times10^6$ yr to $66\times10^6$ yr

(T11.2c)

Read the $\log L$ vs age graph to get age $\approx4\times10^9$ yr. [1.0]

At this age $Y_c\approx0.60$ , so $X_c\approx1-0.60-0.02=0.38$ . [1.0]

The fraction of original H burnt is

\boxed{f_H=1-\frac{X_c}{X_0}=1-\frac{X_c}{1-Y_0-Z_0}=0.46}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
0.39	0.45 to 0.46	0.53

Expression in % or fraction is ok.

(T11.2d)

X_0=1-Y_0-Z_0=1-0.28-0.02=0.70.

When 60% of H has burnt, $X_c=0.40X_0=0.28$ .

Y_c=1-X_c-Z_c=1-0.28-0.02=0.70.\qquad[2.0]

From the $Y_c$ graph, the age is $\sim5\times10^9$ yr. From the graph of $\log R$ , the radius is $\sim1\,R_\odot$ ( $\log(R/R_\odot)\approx0$ ).

\boxed{R_1=1\,R_\odot}.\qquad[1.0]

Any other answer is unlikely since it would involve eye approximation of reading of graphs, that is not to be credited.

(T11.2e)

For P: $\log T_{\mathrm{eff}}=3.74$ , so $T_{\mathrm{eff}}\sim5500$ K ( $L$ being flat does not help here). [1.0]

The $T_{\mathrm{eff}}$ graph gives age $\sim11\times10^9$ yr, then $\log R\sim0.25$ , so

\boxed{R_P=1.78\,R_\odot}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$1.78\,R_\odot$ to $1.80\,R_\odot$

Only credit of 0.5 in final step if left as $\log R/R_\odot=0.25$ .

For Q: $\log L=2.5$ (One can also use $\log T_{\mathrm{eff}}=3.60\implies T_{\mathrm{eff}}\approx3981$ K which would require rounding off to 3950 K at the next step). [1.0]

The inset of $\log L$ gives age $\approx11.884\times10^9$ yr. This value occurs twice; take the first, ascending RGB. $\log R\sim1.6$ , so

\boxed{R_Q=39.81\,R_\odot}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$39.8\,R_\odot$ to $40.0\,R_\odot$

Only credit of 0.5 in final step if left as $\log R/R_\odot=1.6$ . Answer should be same if using $T_{\mathrm{eff}}$ in the first step.

Part 3: Mass distribution inside a star

(T11.3a)

\frac{dq}{dx}=\frac{dq}{dm}\frac{dm}{dr}\frac{dr}{dx}\qquad[0.5]

=\frac1M\left[4\pi(Rx)^2\rho\right]R =3\left(\frac{4\pi R^3}{3M}\right)x^2\rho =3\left(\frac1{\bar\rho}\right)x^2\rho.\qquad[1.0]

\boxed{\frac{dq}{dx}=3x^2\sigma}.\qquad[0.5]

(T11.3b)

Let us denote the line as $\log\sigma=\alpha\log x+\beta$ .

From the graph, use two endpoints $(-0.5,1.0)$ and $(-0.1,-1.0)$ . Therefore,

1.0=-0.5\alpha+\beta,\qquad -1.0=-0.1\alpha+\beta.

Solving, we get $\alpha=-5$ and $\beta=-1.5$ . Therefore,

\log\sigma=-5\log x-1.5\quad\text{for }-0.5<\log x<-0.1.\qquad[2.0]

\sigma(x)=10^{-1.5}x^{-5}\quad\text{for }0.32<x<0.80.

\boxed{\sigma(x)=0.030x^{-5}}\quad\text{for }0.32<x<0.80.\qquad[2.0]

Not rounding off $10^{-1.5}$ is acceptable with full credit.

(T11.3c)

\frac{dq}{dx}=3x^2\sigma=3x^2(0.030x^{-5})=0.090x^{-3}.\qquad[1.0]

Integrating, we get

q(x)=\int0.090x^{-3}\,dx\qquad[1.0]

=-\frac{0.045}{x^2}+C\quad(C\text{ is a constant of integration}).\qquad[1.0]

Without the constant of integration, only 0.5 marks is given in the last step. Now, given that $q(x=0.35)=0.7$ . Using this, we get $C=1.067$ . [2.0]

Therefore (rounding off to within 0.005),

\boxed{q(x)=1.065-\frac{0.045}{x^2}}\quad\text{for }0.32<x<0.80.\qquad[1.0]

If consistent with rounding off at earlier steps, values between 1.065 and 1.091 for the constant term are acceptable with full credit.

(T11.3d)

\frac{dq}{dx}=3x^2\sigma=3x^2(Ax+B)=3Ax^3+3Bx^2.\qquad[1.0]

Integrating, we get

q(x)=\int(3Ax^3+3Bx^2)\,dx\qquad[1.0]

=\frac{3A}{4}x^4+Bx^3+D\quad(D\text{ is a constant of integration}).\qquad[1.0]

Without the constant of integration, only 0.5 marks is given in the last step.

Three boundary conditions applied sequentially to find the constants:

Mass at the centre is zero, i.e., $q(0)=0\implies D=0$ . [0.5]
$\rho(0)=80\bar\rho$ (given), i.e., $\sigma(0)=80\implies\boxed{B=80}$ . [1.5]
Given $q(0.25)=0.50$ . Therefore,

0.5=\frac{3A}{4}(0.25)^4+80

\implies A=\frac43\frac{0.50-80(0.25)^3}{(0.25)^4}

\implies\boxed{A=-256}.\qquad[2.0]

Therefore, putting the values of $A$ and $B$ ,

\boxed{q(x)=-192x^4+80x^3}\quad\text{for }0\le x\le0.32.\qquad[1.0]

Using continuity at $x=0.32$ instead of either of the above conditions will not be credited, since discontinuity has been indicated in the question.

(T11.3e)

Credit will be given on the following aspects:

Aspect	Marks
Curve begins at $(0,0)$ and ends at $(1,1)$	0.5
Slope is zero at both ends	1.0
Curve passes through the two given points	1.0
Initial part of the curve looks like a cubic-quartic	0.5
Latter part of the curve looks negative inverse-square with a dc shift	0.5
Curve is overall smooth	0.5

General Marking Principles

STRICT adherence to given marking scheme.
No double penalty – deduction only at the original place of error; no further penalty in consequent steps, if they are otherwise correct – if final answer is physically unviable or dimensionally incorrect, deduct for final answer also.
Significant figures: one more or one less than appropriate no. of significant figures allowed with full credit.
Units / dimensions:
- In cases where units are not mentioned in the Summary Answersheet, 0.5 deduction for missing units.
- For missing units on the same kind of quantities in a given part, deduct only once.
- For wrong dimensions in final answer because of some error in earlier steps, no credit for final answer, irrespective of no-double penalty principle.
If expression is asked in terms of x, y, …: deduction of final part credit if expression is not in terms of appropriate quantities.
Cumulative part marking: all previous part credits to be added if final expression / number is correct (unless otherwise specified).
Final answer not in Summary Answersheet, but present inside: if final answer is present on Working Sheet, but not transferred on Summary Answersheet, it is to be credited.

Бележки за изданието

The printed question, key and blank answer sheets contain differing expressions and units. The Balmer key uses different rounded colour excesses; the quasar scale changes its angular unit; the neutron-star key prints differing mass-rate units/values. The black-hole-shadow question prints an unsquared radial speed in its comparison equation, while the repeated key uses a square. Atmospheric Seeing asks for a diameter but its key describes a radius. The nucleosynthesis key changes strict inequalities, has a differently grouped decay exponent, and includes an extra time-arrow field also present on its answer sheet. The stellar-profile key omits a factor in one line that appears in the next. The Hawking key prints a separation step differing from its integral and a horizon volume differing from its prose radius. Its answer sheet labels a temperature field with “s”. These native forms are preserved, including every original blank answer page.

Оригинал в Архива: Theory-Questions.pdf · официални решения: Theory-Solution.pdf · допълнителни данни: Theory-Summary-Answersheet.pdf · допълнителни данни: Theory-Instructions.pdf · допълнителни данни: Data_Sheet.pdf · допълнителни данни: IOAA2025_General_Marking_Scheme.pdf