Indian National Astronomy Olympiad – 2015 Question Paper
INAO INA O – 2015
Roll Number: rorororo - rorororo - rorororo Duration: Three Hours
Date: 31st January 2015 Date: Maximum Maxim um Marks: 100
E S C B H Please Note:
• Please write your roll number on top of this page in the space provided. • Before starting, please ensure that you have received a copy of the question paper containing total 3 pages (5 sides).
• There are total 9 questions. Maximum marks are indicated in front of each question. questions, ons, the process involv involved ed in arriving at the solution is more important • For all questi than the answer itself. Valid assumptions / approximations are perfectly acceptable. Please write your method clearly, explicitly stating all reasoning.
spaces are provided in the question question paper for the rough work. No rough work • Blank spaces should be done on the answer-sheet.
• No computational aides like calculators, log tables, slide rule etc. are allowed. • The answer-sheet must be returned to the invigilator . You can take take this question booklet back with you.
• Please be advised that tentative dates for the next stage are as follows: – Orientation Camp (Junior): 29th April to 7th May 2015. – Orientation Camp (Senior): 2nd May to 7th May 2015.
– Selection Camp (Jr. + Sr.): 26 th May to 5th June 2015.
– Participation in both parts (Orientation and Selection) is mandatory for all participants. Useful Physical Constants
Mass of the Earth Radius of the Earth Mass of Jupiter Speed of Light Astronomical Unit Solar Luminosity Gravitational Constant Avogadro constant Charge of an electron Permeability of free space Stephan’s Constant
M E E RE M J J c 1 A. U. L⊙ G N a e µ0 σ =
≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈
6.0 6.4 2.0 3.0 1.5 3.8
24
× 10 kg × 10 m × 10 kg × 10 m/s × 10 m × 10 W × 10 m kg s 6.0 × 10 mol 1.6 × 10 C 4 π × 10 N A × 10 W m K 20 3
6
27 8
11 26
−11
23
3
−1
−2
−1
−19 −7
17
−8
3
HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088
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1. In science fiction novel “The Evitable Conflict”, Isaac Asimov suggested that, in a futuristic world, all food items may be replaced with a nutritious powder made of wheat whe at flour. Wha Whatt siz sizee of world world popul populatio ation n can be sus sustain tained ed with this food stap staple? le? You may assume that the average annual yield of wheat to be about 3. 3.6 to tonn nnee per hectare (present day actual figure) and the geography of the Earth to remain similar to the present world.
E S C B H
( 10 )
Solution:
Surface Area of the Earth = 4π 4π r2 = 5.2 1014 m2 Only 30% of it is land. Thus, total land mass is about 1. 1 .5 1014 m2 . By roug rough h est estima imate, te, ign ignori oring ng moun mountain tains, s, des deserts erts,, arid regions, regions, fore forests sts,, to towns wns / cities etc., only 10% of the land is cultivable. Thus, cultivable land is only 1. 1.5 1013 m2 i.e. 1.5 billion hectare. If all this land is used to cultivate wheat then, annual total wheat production will be 1.5 109 3600 = 5.4 trillion kg. Now each person on average needs 2000-3000 calories per day, which may be obtaine obta ined d by less than half kg of wheat flour. One can est estima imate te daily intak intakee in different differe nt ways ways but the num number ber obtained would be roughly similar. similar. e.g. One can argue that a typical human eats about 250 gm of food per meal and you need 3 meals mea ls per da day y. Th Thus us total is 750 gm. Ho Howe wev ver, water conten contentt in all food ite items ms is more than half. half. Th Thus, us, actual actual weigh weightt of food is only half half of that. Or one can argue each roti needs about 25 gm of flour and if you are not eating anything else, then one may end up eating 15-20 rotis per day. Thus, each person would need about 400 gm wheat flour per day. i.e. about 145 kg per year. i.e. sustainable population would be 5. 5.4 1012 /145 = 40 billion people.
×
×
×
×
×
×
Note: As with all estima estimation tion questions, the arguments presented presented and validit validity y of assumptions is more important than the exact numbers.
2. Match the following (each entry in column A may map with zero or one or more entries in column B but each entry in column B has exactly one matching entry in column A).
Column A Satellite Galaxy St a r Star St ar cl clus uste terr (g (gro roup up)) Dwarf Planet Asteroid Constellation Planet
1
Column B Whirlpo ol Sirius Ceres Peg egas asus us Pallas Puppis Pluto P ho b o s Pleiades Pollux
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Solution: Satellite - Phobos Star - Sirius, Pollux Dwar Dw arff Pl Plan anet et - Pl Plut uto, o, Ce Cere ress Cons Co nstel tella lati tion on - Pega egasu sus, s, Pup Puppi piss
E S C B H Galaxy - Whirlp o ol Cluster - Pleiades Ast ster eroi oid d - Pal alllas Plane Pl anett -
3. Two identical satellites A and B are launched in equatorial circular orbits of period 4.8 hours. hours. Sate Satelli llite te A rotat rotates es in the sense of rotation rotation of the Eart Earth h whi while le satellite satellite B rotates in the opposite sense. The orbits are separated slightly to avoid collision. On a particular day, at 12 noon, both the satellites were seen exactly overhead by an observer on the Earth’s equator. What is the minimum duration of time after which both these satellites will again be seen exactly overhead from the same place?
(6)
Solution:
As the earth’s rotation period is 24 hours, we have to find apparent time period of these two satellites with respect to an observer on the Earth.
∴
ωA = = ω ω 0 1 1 = T A T 0 T A =
−ω − T 1
E
E E
=
1 4.8
− 241
24
= 6 hr hrss (5 1) ωB = ω 0 + ωE 24 = 4hrs ∴ T B = (5 + 1)
−
L.C.M. of the two L.C.M. two app apparen arentt periods is 12 hour hours. s. Th Thus, us, the tw twoo sat satell ellite itess wil willl be seen overhead together after 12 hours.
4. For each of the following statements, state if the statement is true or false and give a single line justification for your answer in each case. 1. All stars stars except except pole star rise in the east and set in the west. west.
2. The Earth’s axis is changing its direction slowly because its magnetic pole is not properly aligned with the geographic pole. 3. Cosmi Cosmicc rays are not part of electromagnetic electromagnetic spectrum. 4. Some sunspot sunspotss can be b e bigger than than the Earth.
5. Jup Jupite iterr can fit in the spa space ce between between the Eart Earth’s h’s surface surface and the Moon’s orbit. orbit. 6. That the Earth does not fall into the Sun is a direct consequence consequence of the Newton’s third Law of motion. 7. The time taken by Mars Orbiter Mission Mission (Mangalyaan) (Mangalyaan) from the moment it left the Earth’s orbit till it was captured by the gravity of Mars, is well determined by the Kepler’s laws. 2
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8. On a new moon day day,, the Earth’s shadow shadow covers the moon fully fully..
E S C B H 9. The stars which which appea appearr brighter brighter are als alsoo clos closer er to the Sun.
10. Po Pole le star is vis visibl iblee from all the locations locations from the Earth. Earth.
Solution:
1. False. As an example, several several stars around pole star would never never go below b elow horizon. 2. False. Althou Although gh the Earth’s axis is indeed precessing, precessing, this is not related to magnetic field of the Earth. 3. True. Cosmi Cosmicc rays is a term used for particle showers. showers.
4. True. The Sun’s radius is 100 times bigger bigger than the Earth and many times we see big sunspots on the solar disk. A sunspot which is barely of the size of the Earth will not be visible through small telescope. 5. True. Jupite Jupiterr is just 10-15 times bigger than the Earth whereas whereas lunar orbit is 60 times bigger. 6. False. Action and reaction reaction forces act on differen differentt bodies bo dies and hence reaction reaction force cannot be seen as the one balancing the gravitational force. 7. True. The path taken taken by MOM afte afterr leaving leaving the sphere sphere of influence influence of the Earth was essentially part of an elliptic orbit around the Sun. 8. True. From the time it lef leftt the Earth’s Earth’s orbit till the time it wa wass capt captured ured by Martian gravity, it’s trajectory was part of an effective orbit around the Sun. 9. Fals alse. e. On a new moon day, day, the moon is between between the Sun and the Earth. Earth. So the Earth’s shadow cannot fall on the Moon.
10. Fals alse. e. The visible visible brightne brightness ss depends on dis distanc tancee of the sta starr as well well as its intrinsic brightness. 11. False. Pol Polaris aris is not visi visible ble from southern hemisphere. hemisphere.
5. A big hall is 15 mete meterr long, 10 meter wide and 5 met meter er high. Tw Twoo ants are seated seated inside this hall at one of the upper corners. On the edge, diagonally opposite to the ants, there is a sugar cube at a height k height k meters above the ground (see diagram). One of the ants decides to go down the vertical edge, across the floor diagonal, and then up the oppos opposite ite vertic vertical al edge to reac reach h the sugar cube. The second second ant tries to tak takee shortestt route along the walls shortes walls without touching touching either the floor or the ceiling. ceiling. Both the ants move with the same uniform speed and you can assume that no time is spent in changing changing the dir directi ection on at an any y poin point. t. Fin Find d con conditi dition on on k such that the first ant reaches the sugar cube first. 3
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E S C B H Sugar k
15
10
Ants
5
Solution:
√
The path length for the first ant = h = h + l2 + w2 + k The path length for the second ant = (l + w)2 + (h (h Our condition is,
√ h+ l √
2
+ w2 + k <
− k)
2
(l + w)2 + (h (h
2
− k) (15 + 10) + (5 − k) ∴ 5 + 15 + 10 + k < √ (5 + k) + 325 325 < < 625 + (5 − k) √ squaring, (5 + k) + 325 + 2 325(5 + k) < < 625 625 + (5 − k) √ ✚ + 10k ✚ − 10 25 10k + k + 325 + 10 13(5 + k) < < 625 625 + ✚ 25 10k k + k ✚ √ 10 13(5 + k) + 20k √ √ 20k < 300 5 13 + (2 + 13) 13)k k < 30 √ 5 × (6 − 13) √ k< 2
2
2
2
2
2
2
2
2
(2 +
13)
approximately, k < 2
6. A ray of light enters an assembly of plane mirrors (from left) as shown in the figure below belo w and undergoes undergoes reflection reflection at all the four mir mirrors rors.. Aft After er the last reflection reflection,, the ray travels in vertical direction and enters in the detecting instrument. In the figure, the dotted lines are normals for respective mirrors and dashed lines are exactly vertical tic al or hori horizon zontal tal.. The third mirror mirror is exactly exactly horizont horizontal. al. It is kno known wn that angl anglee θ ◦ is 50 . Angle φ is unknown. Find angle p. φ
mirrror2
mirror4
mirror3
p
mirror1
θ
4
φ
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Solution:
E S C B H
Let angle of incidences at the four mirrors be i1 , i2 , i3 and i4 respectively. At mirror 1,
(90◦
◦
+ p = = 90 − i − θ) + p = θ θ − p ∴ i =
1
1
For ray from mirror 1 to mirror 2, 90◦ + θ + i1 + 90◦
◦
◦
− i + 180 − φ = 360 2
∴ θ θ + +
i1 = = i i 2 + φ ∴ i 2 = 2θ p
− −φ
For ray from mirror 2 to mirror 3, 90◦
◦
− i + φ + i = 180 ∴ φ − i + i = 90 φ − (2 (2θθ − p − φ) + i = 90 2
3
2
3
◦ ◦
3
∴ i 3 =
90◦ + 2θ 2θ
− p − 2φ
For ray from mirror 3 to mirror 4,
i3 = 2i4 ∴ 2 2ii4 = 90◦ + 2θ 2θ
− 2φ − p
At mirror 4,
90◦ + 2θ 2θ
90◦ + i4 + φ = 180◦ ◦ ∴ i 4 + φ = 90 2i4 + 2φ 2 φ = 180◦ 2φ p + 2φ 2φ = 180◦ ∴ p p = = 2θ 90◦ = 2 50◦ 90◦ = 10◦ ∴ p p =
− −
− × −
7. A spaceship spaceship sends sends a proton beam with cross-se cross-secti ctional onal area of 100 m2 at a speed of 0.01 c in the plane of the Earth’s magnetic equator. The closest distance of the beam to the center of the earth is 1. 1.28 105 km. The magnetic field at the midpoint of the distance separating the proton beam and the center of the earth is zero.
×
(a) Find the number of protons emitted by the spaceship per second. 5
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(b) As seen by the alien controlling the proton beam from the spaceship, would the Earth be on the left side or the right side?
(3)
(c) If mass of the spaceship is 1667 tonne, how much will be its recoil velocity due to this proton beam?
(3)
E S C B H
Note: The strength of the Earth’s magnetic field as measured on the Earth’s surface at magnetic equator is about 5 10−5 Tesla.
×
Solution:
Let 2r 2r0 the shortest distance between the proton beam and the Earth. Let B Let B E be strength of the magnetic field at the Earth’s surface (with effective pole strength M E E ). Let BOE be the magnetic field at r0 due to earth’s magnetic field and BOP is the magnetic field due to I p . As close to the Earth’s surface, the Earth’s magnetic field acts like a dipole, BE =
⇒ µ 4M π 0
E E
µ0 M E E 4 π Re3
3 = RE BE
3 µ0 M E RE BE E BOE = = 4 π r03 r03 r0 = 10 RE ∴ B OE = 0.001 BE µ0 I P P BOP = 2 π r0 BOE = BOP µ0 I P P 0.001 BE = 2 π r0 0.001 BE r0 ∴ I P P = 2 4µπ 0.001 5 10−5 6.4 = 2 10−7 ∴ I P 107 A P = 1.6
|
| |
|
×
0
× ×
×
×
× × 10
7
Charge on a proton is same as that on an electron I P P = eN 1.6 107 N = 1.6 10−19 = 1026 protons per second
× ×
(b) The field lines from the Earth’s magnetic field go from (geographic) southern hemisphere to northern hemisphere. To cancel this field, by virtue of right hand
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rule, the spaceship should be placed such that for an observer on the ship, the Earth would be seen on the right side.
E S C B H
(c) As protons are same as Hydrogen ions, one can say N A protons would weigh about 1 g. Thus, M ship = M M beam ship vship = beam vbeam 1 10−3 N 0.01 c vship = N A M ship ship −3 1 10 1 1026 = 6 1023 1660 vship = 0.3m/s
× ×
× ×
× × × ×
× 3 × 10 × 10
6
3
8. India plans to launch N launch N communication communication satellites, all of which are to be in a geostationary orbit around the equator. All satellites are arranged equi-spaced in the same orbit. Find all the possible values of N such N such that each satellite can see all the other satellites of the series at any instant of time.
Solution:
For geostationary orbits,
2 GM E E T E a = 4π 2 20 10−11 3 = 3
×
2
24
2
× 6 × 10 × (86400) 40
21
= 8.64 10 4.2 107 m ∴ a (R + h) 42000km
≈ ≈
×
×
First of all, if the number of satellites are even then for every satellite, there will always be another satellite at diametrically opposite position. Clearly any satellite cannot see the diametrically opposite satellite. So n cannot be even. If we draw tangents to the Earth’s surface from any satellite (as shown in the figure), it becomes clear that minimum angle between the two consecutive satellites as seen from the Earth would be ∡BC D . B
R
A
R+h
C
D
∡BC D =
2
× ∡BAD
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=2
× 2 × ∡ABC
E S C B H
R R + h 4 6400 42000 4 rad 7
= 4 si sin n−1
≈ × ≈
As a circle has 2π 2π radians, N N can can take following values: N 3, 5, 7, 9, 11 Note: Exact calculations show that N N should should be les lesss than 11. Th Thus, us, 11 will be taken as marginal case. Solutions with or without 11 will get full credit.
∈ { ∈
}
9. The given given graph is a scatter-plot scatter-plot of all exoplanets exoplanets from the Open Exoplanet Catalogue , including those discovered by the Kepler miss mission ion.. Thi Thiss Kepler mission is a special space based instrument, which has been continuously observing same small part of the sky from the start of its operation to measure any small changes in the brightness of the ind indivi ividua duall star starss in that particular particular direction direction.. The scales scales on the graph are as follows: Top - log of orbital period of the planet in years Right - log of mass of the planet as a multiple of mass of the Earth Bottom - log of orbital radius of the planet Left - log of mass of the planet as a multiple of mass of Jupiter For example, on the left side, 0 would correspond to log10 (m/M J = M M J J ) = 0 i.e. m = J . As a way to simplify the problem, let us assume that all the parent stars of these exoplanets have exactly same mass as that of the Sun. (a) Complete the scale for the log10 P P ((years years)) and log10 m(M E E ) axes by providing values on the respective axes.
(4)
(b) Mercury’s distance from the Sun is about 40% that of the Earth and its mass is about 20 tim times es smaller. smaller. Nep Neptune tune’s ’s orbital radius radius is 30 times larger than the Earth’s orbital radius and Neptune is 17 times more massive than the Earth. Mark the approximate positions of Mercury and Neptune on the graph.
(4)
We further assume that all the parent stars emit same total energy per second (called Luminosity) as the Sun and all exoplanet orbits are circular. (c) If a planet can sustain liquid water water on its surface, it is said to be in the “habitable zone” around the parent star. Mark approximate edges of the habitable zone on the graph. State clearly the assumptions made in the process.
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One way of de One dete tect ctin ingg exoplanets is called the Tra ransit nsit metho method d (green dots do ts in the the graph graph). ). If a pl plane anett cro cross sses es (tr (tranansits) in front of its parent star’s disk as seen from the earth, then the observed visual brightness of the star drops by a small amount because the planet blocks some of the light. Most of the exoplanets detected by this method were discovered by the Kepler mission, mission, which will complete it’s sixth year of operat operation ion in a few months months time. To con confirm firm a dete detecti ction, on, it is necessary necessary that at least three transits of the given exoplanet should have been observed by the Kepler mission.
E S C B H
(d) Assume that the average density of the exoplanet is almost the same as that of the host star. It is known that the Kepler mission mission can detect about 0.01% variation in the ligh lightt coming from the star. Using this information, information, draw two two straight lines indicating indicating detection limit limit for the transit method by Kepler spacecraft. Giv Givee brief justification of your answer.
Another way is called the Radial Velocity method (blue dots). Here, we note that the host star and the exoplanet are both orbiting around the common centre of mass (C.M.) and hence the host star shows a periodic variation in its velocity toward ardss or away away from the observ observer er (us (us). ). As the planet is not emitting any light, we just measure the velocity of the host star using the Doppler effect. This orbital velocity of the host star around the common centre of mass can help us determi determine ne mass of the exoplanet. (e) Suppose the smallest change in radial velocity possible to detect by a certain spectrograph is 8 m/s. If we draw a straight straight line line on the graph show showing ing this limit, what would be the slope of this line? Draw this line on the graph. Looking at the exoplanets plotted on the graph, can we conclude that there exist some other instruments that could detect smaller changes in the radial velocity? Justify.
√
Note: Although these two methods account for most of the exoplanets discovered,
few of the exoplanets (also shown in this plot) are discovered by other methods like direct imaging, microlensing and timing analysis.
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(5)
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E S C B H Solution:
(a) The orbital radius of the Earth is 1 AU and its orbital period is 1 year. Thus, for the Earth, log10 a(AU ) = log10 p p((year year)) = 0
Using this one can put marks on the top scale. Note that the markings are closer than bottom scale as p2 a3 . mJ For right side scale, we note that = 300. Th Thus, us, the mar mark k appr approx oxima imatel tely y at mE the same level level as 0 on the left scale correspon corresponds ds to rati ratioo of 300. Usi Using ng this we make marks on the right scale.
∝
(b) Use the orbital radius and mass ratio values to put these planets at appropriate places. (c) Let us assume assume the planet planet is in ther thermal mal equilib equilibriu rium. m. Th Thus, us, amount amount of heat absorbed would be same as the amount heat emitted. L⊙ πr 2 = 4πr 2 σT 4 2 4πa 1 L⊙ ∴ a a = = 4T 2 πσ
We noti notice ce that thi thiss equ equatio ation n is independen independentt of planet paramete parameters. rs. Th Thus, us, the boundaries will be vert vertical ical lines. At inner boundary, boundary, the equil equilibrium ibrium temperature
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should be 273 K and at outer boundary it should be 373 K.
E S C B H aout = =
ain
4
×
1 (270)2
22 7
26
4
× 10 × × 10 17 3
−8
1 1017 4 7.3 104 1.5 1011 17 100 1.1 A. U. 4 7.3 1.5 100 0.6 A. U. 4 13 13..7 1.5
×
4
× × × × = 2× × × ≈ = 2× × × ≈
Only approximate estimation is expected here.
(d) As the observer (us) is very very far from the star, the physical separation between planet and the star is negligible and the drop in intensity of star light during dur ing a plan planet et tran transit sit depends only on the ratio of cros cross-s s-secti ectional onal area. Th Thus, us, one should draw a horizontal line. A pl = Ast =
V pl V st st
2
R pl Rst
2 3
2 3
M pl 0.0001 = M st st M pl 0.0001 = M st st M pl = 1 10−6M st st = 0.001 M J J 3
2
×
Thus, one sho Thus, should uld draw hori horizon zontal tal lin linee whe where re log10 m(M J 3. Furt urthe her, r, as J ) = mentioned in the beginning, Kepler has been in operation for nearly 6 years and it requires at least 3 transits to flag a detection. Thus, typical orbital period for exoplanets discovered by Kepler should be 2 years or smaller. Draw corresponding vertical line.
−
(e) As the orbit is circular, vst = = a a st ω and moments about the centre of mass are balanced. mst ast = m pl a pl
vst = a st ω = m pl a pl mst 2 2 m pl a pl 2 vst = mst =
m pl a pl ω mst 2π 2 π T 4π 2 T 2 mst
× ×
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by Kepler’s third law,
E S C B H 2
vst
≈
2 2 m pl a pl mst
G = mst
×
× aG 3
pl
2 m pl a pl
Thus, for any given velocity detection limit,
G 2 m pl 2 mst vst 2 GM E 2 2 a pl (in A. U.) = m pl (in M E ) 2 (1 A. U.)m U.)mst vst 2 1 1 GM E log 10 ∴ [log10 m pl ] = [log 10 a pl ] 2 2 2 (1 A. U.)m U.)mst vst a pl =
×
1 = [log10 a pl ] 2
− × × × − × × × × − 1 log 10 2
3
−11
20
10
1.5
1011
(6
2
24 2
10 )
1030
1 1 = [log10 a pl ] log 10 10−3 2 2 1 = [log10 a pl ] + 1. 1 .5 2
× (√ 8)
2
This is equation of a straight line in the form y = mx + c Thus, us, on the given given mx + c.. Th plot, we have to draw a straight line with slope 0. 0.5. Further, points ( 2, 0.5) and (3,, 3) are on the line. Once we draw the line, we realise that a lot of blue points (3 are below this line. Thus, present detection limit should be lower than this value.
−
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E S C B H 13
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Notes for Junior Group
E S C B H
• Do not solve question 7(a). Assume that the answer for question 7(a) is 10
26
photons per second.
• Question 4 will have 20 marks. • Question 7(b) and 7(c) will have 4 marks each. • Question 9(c) will have 6 marks.
Notes for Junior Group
• Do not solve question 7(a). Assume that the answer for question 7(a) is 10
26
photons per second.
• Question 4 will have 20 marks. • Question 7(b) and 7(c) will have 4 marks each. • Question 9(c) will have 6 marks.
Notes for Junior Group
• Do not solve question 7(a). Assume that the answer for question 7(a) is 10
26
photons per second.
• Question 4 will have 20 marks. • Question 7(b) and 7(c) will have 4 marks each. • Question 9(c) will have 6 marks.
Notes for Junior Group
• Do not solve question 7(a). Assume that the answer for question 7(a) is 10 photons per second.
• Question 4 will have 20 marks. • Question 7(b) and 7(c) will have 4 marks each. • Question 9(c) will have 6 marks.
prepared using LATEX2ǫ
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