INAO Problems 2008-2014

INAO Question Papers (2008-2014)

Compiled By Science Olympiad Blog

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Indian National Astronomy Olympiad – 2008 Junior Category

Roll Number:

Roll Number

Model Solutions Date: 2nd February 2008 Maximum Marks: 100

INAO – 2008 Duration: Three Hours Please Note:

• The examination consists of three parts. This question booklet containing parts A and B should be returned to invigilators at the end of 2.5 hours. At that time, second question booklet containing part C of the paper will be given to you. You will get 30 minutes to solve part C. m •

• • • •

oom In part A and part C, there are 20 multiple choice .c.c questions each. For each question, only one of the four alternatives is correct. Mark the correct answer on the answer sheet a amarks l +3 l provided separately. Each correct answer adds to your score. In part A, every ee wrong answer carries penalty of -1 marks. There is no negative marking in part C. m m d d In part B, there are 4 analytical questions aa of 10 marks each. The answer to each question must be written in the blank space oprovided below each question. lo l For the rough work, use the n page(s) n marked as rough sheet. ww oo Only non-programmable calculators are allowed. d d . . Return BOTH the question paper booklets and the answer sheet to the invigilator. DO ww NOT TAKE THIS BOOKLET BACK WITH YOU. ww ww Please fill in all the data below correctly. The contact details provided here would be used for all further correspondence.

Full Name (BLOCK letters) Ms. / Mr.:

Male / Female

Date of Birth (dd/mm/yyyy):

Name of the school / junior college:

Class: VIII / IX / X / XI

Board: ICSE / CBSE / State Board / Other

Full Residential address (include city and PIN code):

Telephone (with area code): Email address:

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Instructions for using the answersheet: • Write the name at the top of the answer sheet. • On the left side there is space provided for roll number. Write your INAO roll number in the squares with exactly one digit per square. • Below each of the digits of roll number, mark corresponding digit by a cross mark (’X’). i.e. if your roll number is 40001, then you will put X on 4, 0, 0, 0 and 1 in the corresponding columns. • Below the roll number, you should mark your preference for Either Astronomy camp or m mcorresponding box. Mark only Junior Science Camp by putting a cross mark (’X’) in the oo st one box indicating your 1 preference. You c will be . .c automatically considered for the st second preference if the 1 choice is not available. a

l la e • Use only black or blue pen to put ’X’ marks e on the answersheet. m ink or pencil. ddm aa o l o Useful nnl Physical Constants w Mass of Earth ME oow Radius of Earth RE dd . . Mass of Sun M⊙ w w Radius of Sun R⊙ w Speed of Light w c w Astronomical Unit 1 A. U. w Gravitational Constant Gravitational Acceleration Speed of Sound (at room temperature in air)

Do not use any other

≈ ≈ ≈ ≈ ≈ ≈ G ≈ g ≈ cs ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 9.8 m/s2 340 m/s

Space for Rough Work

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INAO – Jr – 2008

Part A: Multiple Choice Questions 1. The unit on a graph paper is changed in scale from 1 cm to 1 inch. What will be the change in area of the unit cell? (a) 84%

(b) 254%

(c) 545%

(d) 645%

2. Of the Galilean Moons, which is the farthest from Jupiter? (a) Io

(b) Europa

(c) Ganymede

(d) Callisto

3. When you stand on the ground, what is the distance of the horizon from you?

m m o .cc A regular barometer is thrown from the topaof .a building.If the barometer is freely l a falling, what will be the height of the mercury eel column? (a) 100 cm (b) 76 cm (c) 50 cmm (d) 0 cm ddm aathroughout the year, but happen only in certain P. Eclipses are not distributed evenly o l lo months of a given year. n Q. Orbit of the Moon (aroundnthe Earth) makes an angle of roughly 5 degrees to the ww orbit of Earth (around o Sun). o Which of the following options is correct? d .d . (a) Statement ww‘P’ is correct but ‘Q’ is incorrect. ww ‘P’ is incorrect but ‘Q’ is correct. (b) Statement ww the statements are correct and ‘Q’ is the correct reason of (c) Both

(a) 500 km 4.

5.

(b) 5 km

(c) 15 km

(d) 50 kmo

‘P’. (d) Both the statements are correct and ‘Q’ is not the reason of ‘P’. 6. When the ball at the end of the string swings to its lowest point, the string is cut by a sharp knife as shown. Assuming no air resistance, what will be the path of the ball? Simple Pendulum

String

A B Heavy Ball C D

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INAO – Jr – 2008

(a) A

(b) B

(c) C

(d) D

The answer is (c). 7. If we ever make contact with aliens, which of our fundamental units is likely to match theirs? (In other words, which of these units is universally fundamental?) (a) Kelvin

(b) Light year

(c) a.m.u. (Atomic Mass Unit)

(d) None of these

8. If the person beats drum on the Earth and an astronaut beats an identical drum in space, what will be the differences in the effects? (a) There will be no vibration in the drum in space. (b) There will be vibration in space but no sound.

m (c) The drum on Earth will vibrate for a longer time than the one in space. m o

o (d) There will be no difference in terms of c vibrations or sound. .cthe

9.

10.

aa. l Which of the following Venn diagrams would eelbe BEST suited to represent the three categories of animals? m - Animals that give us MEAT ddm a - Animals that give us EGGSo a l o - Animals that give us MILK nnl w oow .d .d ww ww (b) (c) (d) The answer is (a). (a) ww P: Gravitational force exerted by Saturn on a human being is approximately same as that exerted by another human being standing a few cm away. Q. Saturn has very low density. (Additional data: Mass of Saturn = 5 × 1026 kg, Distance of Saturn = 1.4 × 109 km) (a) Statement ‘P’ is correct but ‘Q’ is incorrect. (b) Statement ‘P’ is incorrect but ‘Q’ is correct. (c) Both the statements are correct and ‘Q’ is the correct reason of ‘P’. (d) Both the statements are correct and ‘Q’ is not the reason of ‘P’.

11. For the Earth, if the perihelion were 147 million km, approximately what will be the aphelion for the Earth? Aphelion: Point farthest from the Sun in the orbit of a body about the Sun. Perihelion: Point nearest from the Sun in the orbit of a body about the Sun. (a) About 2 times the Perihelion, 300 million km (b) About 3 times the Perihelion, 450 million km (c) Slightly more than the perihelion, about 155 million km (d) Exactly the same as the perihelion, 147 million km

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INAO – Jr – 2008

12. A star is seen rising from Kolkata (23.5 ◦N 92 ◦ E) at 7:00 pm IST, at what time IST will it be seen to rise from Mumbai (19 ◦N 72 ◦E)? (a) 5:40 pm

(b) 7:00 pm

(c) 7:20 pm

(d) 8:20 pm

13. To separate various components of air, the principle that different gases have different . . . . . . . . . can be used. (a) Boiling Points

(b) Density

(c) Color

(d) Molecular weight

14. How many zeros will be as the ending digits of 120! ? (120! = 1 × 2 × ..... × 119 × 120) (a) 25

(b) 26

(c) 27

(d) 28

Note: In the original paper, the choices given had typographical errors. Thus, the correct answer 28 was not listed amogst the options. m As a result the said question was removed from evaluation. oom

.cc

. 15. A nutty professor discovers a way to shrink objects lala in size using lasers and mistakenly shrinks his three teenage kids by a factor e of 100. These kids then stray away into the e garden where they see their pet dog. How many of the kids can climb onto the dog to m m get a ride home if the dog can bear d ad weight of 20 kg? (a) None 16.

a

a (c) Two loo(d) All

(b) One

nnl Which of the following is true? ww oo (a) cos 80◦ = − sin 10◦ dd . cos 240◦ (b) cos 120◦ =.− w ◦ w ◦ (c) sin 135 ww= − sin 270 (d) sin 330◦ = sin 210◦ ww

The answer is (d).

17. A battery is connected by wires to a bulb as shown below and the bulb glows. Through which points does the charge flow?

1 Battery 4

2 Lighted Bulb 3

(a) 1-2-3-4-1. Charge flows through the battery also. (b) 1-2-3-4. Charge flows through the wires and bulb only. (c) 2-3.Charge flows only through the bulb. (d) There is no flow of the charge in the circuit. 18. Every object exerts gravitational force on every other object - The force exerted by an object is higher if its mass is higher. Consider 2 magnets - a bigger magnet P and a smaller one Q. Which of the following will be true?

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INAO – Jr – 2008

(a) Magnet P will exert a greater magnetic force than Q. (b) The magnetic forces exerted by P and Q will be the same. (c) Magnet Q will exert a greater magnetic force than P. (d) We cannot tell from the sizes, as gravity and magnetism are unrelated. 19. What is the value of ‘F’ in the following equation if A, B, C, D, E and F are non-zero numbers? ABCDEF × 6 = DEFABC (a) 1

(b) 3

(c) 5

(d) 7

20. On a cold winter day, if I stand on the edge of a carpet with one foot on the carpet and one on the smooth granite floor surface, which foot is mlikely to feel colder and why?

m

oo heat away from the foot (a) The foot on the granite as it willcabsorb . c more quickly. a . l a e

(b) The foot on the carpet as it will absorb e l heat away from the foot more quickly.

m (c) The foot on the granite because mthe granite is at a lower temperature. d

aad the carpet is at a lower temperature. (d) The foot on the carpet because o lo nnl w oow .d .d ww ww ww

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INAO – Jr – 2008

Part B: Analytical Questions 21. A year in Solar calendar consists of 365 days and the same in Lunar calendar consists of 354 days. The additional days in Solar calendar are kept as balance every year. Whenever the number of balance days exceeds 30, an additional month of 30 days is added to the lunar year to offset the difference. The cycle goes on. Anwesh, whose birthday falls on 1st January, noticed that in the year 2008, his birthday coincided with the start of the lunar year. In which earliest future year, his birthday will again coincide with the start of the lunar year? (Ignore leap days.) Solution: Every year the Solar year lags by 11 days. Intercalary days are compensated by a month whenever they exceed 30 days. m Thus, one has to finde L.C.M. of 11 and 30. oom L.C.M. is 330. i.e. after 330 intercalary days.c are introduced, both calenders will .c a match. l a eel i.e. they will match after 30 years. Thus, his birthday in 2038 will again mmark start of the lunar calander.

22.

ddm aa o Note: Brute force method should l lonot be given more than 7 marks. nn w oow in series, the current in the circuit is 2A. When R1 When R1 and R2 are connected and R2 are connected in parallel, the current in the circuit is 4A. Find the values of .d .d R1 and R2 . ww ww w Solution:w For a given V, V

= IA (R1 + R2 ) R1 R2 = IB (R1 + R2 ) R1 R2 2(R1 + R2 ) = 4 (R1 + R2 ) 2 (R1 + R2 ) = 2R1 R2 R12 + R22 = 0 The said conditions cannot be true. Note: Problem is solvable if one assumes non-zero internal resistance for the battery. Those who reach till previous step, get 9 out of 10. Last point is reserved for those who take internal resistance into account.

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INAO – Jr – 2008

23. You are given two lenses of focal lengths f1 and f2 respectively. (a) Is it possible to arrange them in such a way that both incident beam and emergent beam of light will be parallel beams for the following cases? 1. One concave and one convex lens 2. Both convex lenses Draw the ray diagrams. (b) Are the incident and emergent beams parallel to each other? (c) By observing the ray diagrams, state the condition on the distance ‘d’ between the two lenses in terms of f1 and f2 .

m oom 1 cc d 1 1 + −. . = f f1 fl2aaf1 f2 l d is the distance between the two lenses.ee m If the incident beam as well as the emerging beam are parallel beams, then f1 = 0 m dfd2 which gives the condition, d = f1a+ a If one lens is concave and other convex, it will become, d = f1 − f2 loo Thus, it is only possible if n focall length of the convex lens is more than that of the n concave lens. ww oo d is always positive. Hence it is always possible. If both lenses are convex, d . .d Note: The ray w diagrams should be such that above relations could be inferred by w w measuring respective lengths on the ray diagrams. w ww Solution:

The two beams will be parallel to each other if both lenses are parallel to each other. Knowledge of the first equation is not expected. Students should inferr the d = f1 + f2 relation by purely observing ray diagrams.

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INAO – Jr – 2008

24. If Aniket starts drawing a straight line with a brand new typical ball-point pen, how long line can be drawn before he finishes his refill? Explicitly state all the assumptions you make. Solution: A typical ball pen refill has length of 12 cm. (Acceptable 10-15 cm) Typical refill diameter is 1 mm. (Acceptable 0.5 - 2 mm) Hence Total volume of ink is π r 2 h = π(0.05)2 × 12 cc V ≈ 0.1 cc Typical thickness of writing is of the size of finite number of molecules. Size of one ink molecule can be taken to be 0.5 - 1nm. m m Hence the thickness would be roughly 10nm. oo .c.c (Acceptable 1-100 nm) a Typical width is half of refill diameter. l la ee Thus, length, 2m πr h π × 0.05 × 12 V = d m = cm l= t d a2td r 2 × 10 × 10−7 a l ≈ 9.5 km. loo

nnl ww Note: Answer is not o important for this order of magnitude o Approach to the problem should be judged for marks. .d .d ww ww ww

estimation question.

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INAO – Jr – 2008

Part C: Multiple Choice Questions

Roll Number: Roll

Number

25. Resting membrane potential difference is a negative value which means: (a) No charge outside and negative charge inside of the membrane (b) Positive charge outside and more negative charge inside of the membrane (c) Zero charge inside and negative charge outside of the membrane (d) Positive charge inside and more negative charge outside of the membrane 26. Nerve impulse is... (a) Flow of electrons across the axon

m o cc (c) Flow of neuro-transmitters across the. axon . a a l (d) Change in pressure across the axon eel m Bitter substances can be tasted in minute ddmamounts but larger amounts are needed to taste sweet substances. Based onathis a observation which of the following reasons is o more appropriate? l lo n n are more sensitive than the sweet receptors (a) The bitter receptors ww oo are less sensitive than the sweet receptors (b) The bitter receptors d (c) The bitter. substances dissolve more easily than the sweet substances .d ww (d) There are more bitter receptors in mouth than the sweet receptors ww Gustatoryw cells w are stimulated by (b) Change in ionic constitution acrossothem membrane

27.

28.

(a) Dissolved chemicals (b) Pressure (c) Temperature (d) Texture 29. The cornea of one person can be transplanted from one person to another with little or no possibility of rejection as it is beyond the reach of immune system because... (a) The cornea has no blood vessels (b) The cornea is a dead tissue (c) The cornea has no nerve endings (d) The cornea kills the cells of immune system 30. Find out the mismatched pair in the following combinations. (a) Nearsightedness — longer than normal eyeball (b) Farsightedness — shorter than normal eyeball (c) Myopia — loosened-up extrinsic eye muscles (d) Astigmatism — cylindrical lens

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INAO – Jr – 2008

31. Which sequence of events occur when a person looks at a star at night (a) pupils constrict → suspensory ligaments relax → lenses become less convex (b) pupils dilate → suspensory ligaments become taut → lenses become less convex (c) pupils dilate → suspensory ligaments become taut → lenses become more convex (d) pupils constrict → suspensory ligaments relax → lenses become more convex 32. Which nervous system conducts impulses from CNS to voluntary muscles? (a) Motor division of PNS

m oom .c.c (d) Parasympathetic division a l a eel The rate of a simple chemical reaction normally decreases as the reaction approaches m completion. This is because ddm aa (a) The reactant molecules individually become less active o l lo (b) With the progress n of the reaction, the temperature goes down and hence the n reaction slows down ww oinhibit o the reaction (c) The products d . .d (d) The concentration of the reactants decreases ww wwis generally used for the ore dressing of Froth flotation ww (a) Oxide ores (b) Carbonate ores (c) Phosphate ores (d) Sulfide ores (b) Sensory division of PNS (c) Sympathetic division

33.

34.

35. The atomic property which is not periodic is (a) Atomic radius (b) Mass number (c) Electronegativity (d) Ionization energy 36. The largest number of molecules is present in 1 g of (a) CO2

(b) H2 O

(c) C2 H5 OH

(d) N2 O5

The answer is (c). 37. An isotope of the parent element is produced with the emission of (a) one α- and one β- particle (b) one α- and two β- particles (c) two α- and one β- particles (d) two α- and two β- particles

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INAO – Jr – 2008

38. A compound was found to contain nitrogen and oxygen in the ratio nitrogen 28 g and oxygen 80 g. The formula of the compound is (a) NO

(b) N2 O3

(c) N2 O5

(d) N2 O4

The answer is (c). 39. Acetic acid is a weak electrolyte because (a) Its molecular mass is high (b) It is a covalent compound (c) It is highly unstable (d) Its ionization is very small

m m for 100 seconds, deposits 40. A certain current when passed through a CuSO4osolution o 0.3175 g of copper. The current passed (in A) is .cc (a) 4.83

41.

(b) 9.65

(c) 0.965

.

(d) 0.483laa

eel For the redox reaction, the correct coefficients of the reactants for the balanced reacm ddm tion are aa + 2− 2+ MnO 4− + C2 O + CO2 + H2 O 4 + H −→ Mn o l lo nn (a) 2 5 16 w (b) 16 5 2 oow (c) 5 16 2.dd . (d) 2 16w 5 w ww paper, there were typographical errors in the chemical equation. Note: In the original w was removed from evaluation. Thus, the w question

42. The pH of 0.1M CH3 COOH (dissociation constant of acetic acid is 1.80 × 105 at 25 ◦ C) will be (a) 1.0

(b) 2.9

(c) 1.8

(d) 0.2

43. Which kind of number pyramid will fit for the following example? Grass — Deer — Flea — Leptomonas (parasite of flea). (a) (b) (c) (d) The answer is (d).

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Indian National Astronomy Olympiad – 2008 Senior Category

Roll Number:

Roll Number

Model Solutions Date: 2nd February 2008 Maximum Marks: 100

INAO – 2008 Duration: Two and half Hours Please Note:

• The question paper consists of two parts. In part A, there are 20 multiple choice questions with 3 marks for each correct answer and -1 mark for each wrong answer. In each question, only one of the four alternatives is correct. Mark the m correct answer on the answer sheet provided separately. oom Mark a cross (X) in the corresponding box on the answer sheet. .cc • • • •

aa. l In part B, there are 4 analytical questions eelof 10 marks each. The answer to each question must be written in the blank space provided below each question. m m d For the rough work, use the page(s) marked as rough sheet. aad o l o are allowed. Only non-programmable calculators nnl ww paper booklet and the answersheet back to the invigReturn the ENTIRE o question oTHIS BOOKLET BACK WITH YOU. ilator. DO NOT TAKE d .d . w w Please fill in all the data below correctly. The contact details provided here wwwould be used for all further correspondence. ww

Full Name (BLOCK letters) Ms. / Mr.:

Male / Female

Date of Birth (dd/mm/yyyy):

Name of the school / junior college:

Class: IX / X / XI / XII

Board: ICSE / CBSE / State Board / Other

Full Residential address (include city and PIN code):

Telephone (with area code): Email address:

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Instructions for using the answersheet: • Write the name at the top of the answer sheet. • On the left side there is space provided for roll number. Write your INAO roll number in the squares with exactly one digit per square. • Below each of the digits of roll number, mark corresponding digit by a cross mark (’X’). i.e. if your roll number is 40001, then you will put X on 4, 0, 0, 0 and 1 in the corresponding columns. • Use only black or blue pen to put ’X’ marks on the answersheet. Do not use any other m ink or pencil. o m

o .c.c aa Useful Physical lConstants eel m Mass of Earth ME m d d Radius of Earth RE aa o Mass of Sun M ⊙ l o Radius of Sun R⊙ nnl Speed of Light c ww o Astronomical Unit 1 A. U. o d d . Gravitational Constant . G ww Gravitational Acceleration g Speed of Sound w (at cs wroom temperature in air) Specific Heatw of water Cw w

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 3 × 108 m/s 1.5 × 1011 m/s 6.67 × 10−11 m3 /(Kg s2 ) 9.8 m/s2 340 m/s 4.186 × 103 J/kg oC

Space for Rough Work

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INAO – Sr – 2008

Part A: Multiple Choice Questions 1. The dimensions of Boltzmann constant (k) are (a) M1 L2 T1

(b) M1 L2 T−1

(c) M2 L1 T−1

(d) M1 L2 T−2

Note: Objections are raised over this question, because temperature dimensions are not specified. Thus, Question was removed from evaluation. 2. P. One can see absorption lines in the Solar spectrum. Q. The core of the Sun has temperature of more than 1 million degree Celsius and the Solar surface has temperature of about 6000 degree Celsius. m Which of the following options is correct? o m

c o a lisalcorrect. (b) Statement ‘P’ is incorrect but ‘Q’ ee (c) Both the statements arem correct and ‘Q’ is the correct reason of m d ‘P’. aad oocorrect and ‘Q’ is not the reason of ‘P’. (d) Both the statements lare nnl When you stand on the ground, ww what is the distance of the horizon from you? oo (c) 15 km (d) 50 km (a) 500 km (b) 5 km d . .d A regular barometer ww is thrown from the top of a building.If the barometer is freely falling, what w will be the height of the mercury column? w ww (b) 76 cm (c) 50 cm (d) 0 cm (a) 100 cm . .c (a) Statement ‘P’ is correct but ‘Q’ is incorrect.

3.

4.

5. P. Eclipses are not distributed evenly throughout the year, but happens only in certain months of a given year. Q. Orbit of the Moon (around the Earth) makes an angle of roughly 5 degrees to the orbit of Earth (around Sun). Which of the following options is correct? (a) Statement ‘P’ is correct but ‘Q’ is incorrect. (b) Statement ‘P’ is incorrect but ‘Q’ is correct. (c) Both the statements are correct and ‘Q’ is the correct reason of ‘P’. (d) Both the statements are correct and ‘Q’ is not the reason of ‘P’. 6. For the Earth, if the perihelion were 147 million km, approximately what will be the aphelion for the Earth? Aphelion: Point farthest from the Sun in the orbit of a body about the Sun. Perihelion: Point nearest from the Sun in the orbit of a body about the Sun. (a) About 2 times the Perihelion, 300 million km

1

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INAO – Sr – 2008

(b) About 3 times the Perihelion, 450 million km (c) Slightly more than the perihelion, about 155 million km (d) Exactly the same as the perihelion, 147 million km 7. When the ball at the end of the string swings to its lowest point, the string is cut by a sharp knife as shown. Assuming no air resistance, what will be the path of the ball? Simple Pendulum

String

8.

m oomA .c.c a B l la e e Heavy Ball m C ddm D aa lo lo (a) A (b) B (c) C (d)nn D w The answer is (c). oow .d If we ever make contact .dwith aliens, which of our fundamental units is likely to match theirs? (In otherw words, w which of these units is universally fundamental?) w wLight year (c) a.m.u. (Atomic Mass Unit) (d) None of these (a) Kelvin (b) ww

9. P. In a dynamo, it is necessary that coil should be moving and magnet should be stationary and not vice versa. Q. Magnetic force is perpendicular to the direction of motion of charges. Which of the following options is correct? (a) Statement ‘P’ is correct but ‘Q’ is incorrect. (b) Statement ‘P’ is incorrect but ‘Q’ is correct. (c) Both the statements are correct and ‘Q’ is the correct reason of ‘P’. (d) Both the statements are correct and ‘Q’ is not the reason of ‘P’. 10. If a person beats drum on the Earth and an astronaut beats an identical drum in space, what will be the differences in the effects? (a) There will be no vibration in the drum in space. (b) There will be vibration in space but no sound. (c) The drum on Earth will vibrate for a longer time then the one in space. (d) There will be no difference in terms of the vibrations or sound. 11. An Olympiad student fails to get a medal. The angry team leader launches him into a rocket far into space & then throws him out. The student just grazes the upper edge of the atmosphere during his trajectory. He will... 2

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INAO – Sr – 2008

(a) Go around the Earth (b) Crash on the Moon (c) Get lost into deep space (d) Reach the Olympiad venue again 12. P: Gravitational force exerted by Saturn on a human being is approximately same as that exerted by another human being standing a few cm away. Q. Saturn has very low density. (Additional data: Mass of Saturn = 5 × 1026 kg, Distance of Saturn = 1.4 × 109 km) (a) Statement ‘P’ is correct but ‘Q’ is incorrect. (b) Statement ‘P’ is incorrect but ‘Q’ is correct.

m o cc ‘Q’ is not the reason of ‘P’. (d) Both the statements are correct .and . a l la ee process. P. Temperature is not constant in an adiabatic Q. Adiabatic processes do not obey ideal gas equation. m m d Which of the following options is correct? aad (a) Statement ‘P’ is correct but ‘Q’ is incorrect. lo lo n n but ‘Q’ is correct. (b) Statement ‘P’ is incorrect ww (c) Both the statements oo are correct and ‘Q’ is the correct reason of ‘P’. d . .d (d) Both the statements are correct and ‘Q’ is not the reason of ‘P’. ww A star is seenw rising from Calcutta (23.5 ◦N 92 ◦ E) at 7:00 pm IST, at what time IST w ◦ ◦ will it be seen wwto rise from Mumbai (19 N 72 E)? mcorrect reason of ‘P’. (c) Both the statements are correct and ‘Q’o is the

13.

14.

(a) 5:40 pm

(b) 7:00 pm

(c) 7:20 pm

(d) 8:20 pm

15. A broom with a long handle balances at its centre of gravity as shown in the figure. If you cut the broom into two parts through the centre of gravity and then weigh each part, which part will weigh more?

(a) The part with the broom will weigh more. (b) The part without the broom will weigh more. (c) Both the parts will weigh the same. (d) It wold depend on the weight of the broom. 16. Star A has temperature 4000 ◦K and star B has temperature 40, 000 ◦K. If the two stars have roughly same radii, which of the following statements is not true? (a) B is more luminous than A. 3

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INAO – Sr – 2008

(b) A emits more in IR than in UV. (c) B emits more in UV than in IR. (d) A emits more in IR than B. Note: In the actual question the word “NOT” was missing and the fact that both stars are assumed to have same radii was not explicitly stated. Thus, the question was removed from evaluation. 17. Every object exerts gravitational force on every other object - The force exerted by an object is higher if its mass is higher. Consider 2 magnets - a bigger magnet P and a smaller one Q. Which of the following will be true? (a) Magnet P will exert a greater magnetic force than Q.

m (b) The magnetic forces exerted by P and Q will be the same. m oo

(c) Magnet Q will exert a greater magnetic .ccforce than P.

a . ee Three balls are thrown from the top of cliff along paths P, Q and R. If their initial m m d speeds are the same and there is noaairdresistance, under what conditions will the balls a strike the ground below with theosame l lo speed? nn P ww oo d . .d ww Q ww R ww l la and magnetism are unrelated. (d) We cannot tell from the sizes, as gravity

18.

(a) This will happen if the mass of each ball is the same. (b) This will happen if the distance traveled by each ball is the same. (c) This cannot happen unless the paths of the balls are identical. (d) This will always happen - No additional condition is required. 19. A battery is connected by wires to a bulb as shown below and the bulb glows. Through which points does the charge flow?

1

2

Battery

Lighted Bulb 3

4

(a) 1-2-3-4-1. Charge flows through the battery also. (b) 1-2-3-4. Charge flows through the wires and bulb only. (c) 2-3.Charge flows only through the bulb. 4

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INAO – Sr – 2008

(d) There is no flow of the charge in the circuit. 20. Which of the following is true? (a) cos 80◦ = − sin 10◦ (b) cos 120◦ = − cos 240◦ (c) sin 135◦ = − sin 270◦ (d) sin 330◦ = sin 210◦ The answer is (d).

m oom .c.c a l a eel m ddm aa o l o nnl w oow .d .d ww ww ww

5

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INAO – Sr – 2008

Part B: Analytical Questions 21. A year in Solar calendar consist of 365.25 days and the same in Lunar calendar consist of 354 days. The additional days in Solar calendar are kept as balance every year. Whenever the number of balance days exceeds 30, an additional month of 30 days is added to the lunar year to offset the difference. The cycle goes on. Anwesh, whose birthday falls on 1st January, noticed that in the year 2008, his birthday coincided with the start of the lunar year. In which earliest future year, his birthday will again coincide with the start of the lunar year? Solution: Every year the Solar year lags by 11.25 days. After 4 years, number of intercalary day will be integer, i.e. 45. m Intercalary days are compensated by a month whenever oom they exceed 30 days. Thus, one has to find L.C.M. of 45 and 30. .cc . L.C.M. is 90. i.e. after 90 intercalary days lalaare introduced, both calenders will ee match. i.e. they will match after 8 years. m ddm Thus, his birthday in 2016 will again mark start of the lunar calander.

22.

aa o l o Important: Brute force method nnl should not be given more than 7 marks. w oow ddthe circuit diagram, given the conditions below: Find R1 , R2 and R3.in . ww A ww ww R R 2

1

B

C

P

R3

Q

(a) If P is connected to A & Q is connected to C then the current in the circuit is 2A. (b) If P is connected to A & Q is connected to B then the current in the circuit is 4A. (c) If P is connected to C & Q is connected to B then the current in the circuit is 3A.

6

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INAO – Sr – 2008

Solution: For a given V, V

(R1 + R3 ) R2 (R1 + R2 + R3 ) (R2 + R3 ) R1 = 4 (R1 + R2 + R3 ) (R1 + R2 ) R3 = 3 (R1 + R2 + R3 ) = 2

(1) (2) (3)

Thus, 4(R2 + R3 ) R1 = 2(R1 + R3 ) R2 2R1 R2 = 2R2 R3 − 4R1 R3 m o3o−m R1 R2 = R2 R 2R1 R3

(4) (5)

.c.c a a 1 + R2 ) R3 4(R2 + R3 ) R1 e=l l3(R e R1 m R3 = 3R2 R3 − 4R1 R2 m d1d R R aa 3 = 3R2 R3 − 4R2 R3 + 8R1 R3 oR2 R3 = 7R1 R3 lo l nn R2 = 7R1 w oow 2(R1 + R3 ) R2 = 3(R1 + R2 ) R3 .d .d R2 R3 = 2R1 R2 − 3R1 R3 ww 7R1 R3 = 14R12 − 3R1 R3 ww ww 10R1 R3 = 14R12 R3 = The resistances are in ratio (R1 + R2 + R3 ) = R3 = R3 = R1 = R1 = R2 = R2 =

7 R1 5

R1 : R 2 : R 3 = 1 : 7 :

(6) (7) (8)

(9) (10) (11)

(12) 7 5

47 R1 5 47 V R1 V (R1 + R2 + R3 ) = 5 3(R1 + R2 ) 3 × 8R1 47 V 120 5 47 × V 7 120 47 V 168 47 7× V 168 47 V 24

(13) (14)

(15)

(16)

(17)

Note: If one finds ratio of resistance correctly, 8.5 marks out of 10 should be awarded.

7

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INAO – Sr – 2008

23. You are given two lenses of focal lengths f1 and f2 respectively. (a) Is it possible to arrange them in such a way that both incident beam and emergent beam of light will be parallel beams? Show all (at least 3)possible cases (for different lens combinations) with ray diagrams. (b) Are the incident and emergent beams parallel to each other? (c) By observing the ray diagrams, state the condition on the distance ‘d’ between the two lenses in terms of f1 and f2 . (d) For what combinations of lenses, the said arrangement is not possible? Solution:

1 1 d 1 m + − om = f f1 f2 f1 fo 2

.cc

d is the distance between the two lenses. a . l a 1 If the incident beam as well as the emerging eel beam are parallel beams, then f = 0 which gives the condition, d = f1 + fm 2 If one lens is concave and other convex, ddmit will become, d = f1 − f2 Thus, it is only possible if focal a length a of the convex lens is more than that of the o o l concave lens. l If both lenses are convex, dnis nalways positive. Hence it is always possible. wwit is not possible at all. If both lenses are concave,

oo d Note: The ray diagrams . .d should be such that above relations could be inferred by measuring repective ww lengths on the ray diagrams. ww The twow beams w will be parallel to each other if both lenses are parallel to each other. Knowledge of the first equation is not expected. Students should infer the d = f1 + f2 relation by purely observing ray diagrams.

8

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INAO – Sr – 2008

24. If Aniket starts drawing a straight line with a brand new typical ball-point pen, how long line can be drawn before he finishes his refill? Explicitly state all the assumptions you make. Solution: A typical ball pen refill has length of 12 cm. (Acceptable 10-15 cm) Typical refill diameter is 1 mm. (Acceptable 0.5 - 2 mm) Hence Total volume of ink is π r 2 h = π(0.05)2 × 12 cc V ≈ 0.1 cc Typical thickness of writing is of the size of finite number of molecules. Size of one ink molecule can be taken to be 0.5 - 1nm. m m Hence the thickness would be roughly 10nm. oo .c.c (Acceptable 0.5-50 nm) a Typical width is half of refill diameter. l la ee Thus, length, 2m πr h π × 0.05 × 12 V = d m = cm l= t d a2td r 2 × 10 × 10−7 a l ≈ 9.5 km. loo

nnl ww Note: Answer is notoimportant in this order of magnitude o Approach to the problem should be judged for marks. .d .d ww ww ww

estimation question.

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Download ebooks at http://www.downloadmela.com/ebooks/ebookdisplay.html Download Video tutorials at http://www.downloadmela.com/videotutorials/videotutorials.html Download placement papers,interview questions,certification dumps and many more at http://www.downloadmela.com/companypapers/itpapers.html m

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Indian National Astronomy Olympiad – 2009

A-PDF Watermark DEMO: Purchase from www.A-PDF.com to remove the watermark

Junior Category

Roll Number:

Roll Number

Model Solutions INAO – 2009 Duration: Three Hours

Date: 31st January 2009 Maximum Marks: 100

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 (6 sides). • In Section A, there are 10 multiple choice questionsm with 4 alternatives out of which m and -1 mark for each ooanswer only 1 is correct. You get 3 marks for each correct .c.c wrong answer. •

•

lala In Section B, there are 4 multiple choiceequestions with 4 alternatives each, out of e which any number of alternatives may be correct. You get 5 marks for each correct m m answer. No marks are deducted fordd any wrong answers. You will get credit for aa mark all correct choices and no wrong the question if and only ifoyou l lo choices. There is no partial nncredit. For both these sections,w you have to indicate the answers on the page 2 of the oow answersheet by putting a dd × in the appropriate box against the relevant question . number, like this: . w Q.NO. (a) w Q.NO. (a) (b) (c) (d) ww(b) (c) (d) 22 w  w ⊠   OR 35 ⊠ ⊠   Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

• In Section C, there are 5 analytical questions totalling 50 marks. • Blank spaces are provided in the question paper for the rough work. No rough work should be done on the answer-sheet. • No calculators are allowed. • The answer-sheet must be returned to the invigilator. You can take this question booklet back with you.

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Useful Physical Constants Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Speed of Light Astronomical Unit Gravitational Constant Gravitational Acceleration on the Earth Gravitational Acceleration on the Moon

ME RE M⊙ R⊙ Rm c 1 A. U. G g gm

m oom .c.c Space for RoughaWork l a eel m ddm aa o l o nnl w oow .d .d ww ww ww

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 1.7 × 106 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 9.8 m/s2 1.6 m/s2

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INAO – Jr – 2009

Section A: (10 questions × 3 marks each)

1. If ax = by = cz and b2 = ac, then y = ? √ √ 2xz xz (a) x+z (b) x+z (c) 2xz (d) xz Solution:

y

a = bx y c = bz y y b2 = ac = b x b z y y = bx+z m y y om 2 = + o x z.cc y(xla +a z). = eexzl 2xz y d=m m d z+x

2.

aa o Ans = (a) l o nnl ww o Each of the figures below, odepict a constellation. d .d . ww ww ww (a)

(b)

(c)

Find the odd one out.

(d)

Solution: The constellations are (a) Leo (b) Taurus (c) Scorpio (d) Canis Major. First three are zodiac signs whereas the fourth one is not. Ans = (d) 3. Gravitational force between two identical uniform solid gold spheres of radius r each in contact is proportional to (a) r 4

(b) r 2

(c)

1 r2

(d)

1 r3

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INAO – Jr – 2009

Solution: The distance between two spheres = 2r and masses are the same ⇒F =

GM 2 (2r)2 3

G( 4πr ∗ ρ) 3 = 4r 2 6 r ⇒F ∝ 2 r F ∝ r4

2

Thus ans = (a)

4.

m oom .c.c A copper cube and a wooden cube of samela size a are initially at room temperature. ◦ l Then they are kept in an enclosure at 50 e c.eWhat can we conclude about the temperatures attained by both cubes afterm 5 hours? m d d (a) Tcopper > Twood as thermal conductivity of copper is greater than that of aa wood. lo lo n (b) Twood > Tcopper as specific heat capacity of wood is greater than that of n w copper. oow (c) The temperatures d will depend on the interplay between specific heat ca.d . pacity w and thermal conductivity of the materials. w (d) Both temperatures will be practically the same, as they are in the enclosure ww for 5 hours. ww Solution: Both copper and wooden cube will have same temperature, as that of the enclosure, because the time is sufficiently long to bring them in thermal equilibrium with their surroundings. Ans = (d)

5. If the product of all the numbers from 1 to 100 is divisible by 2n , then what is the maximum possible value for n? (a) 128

(b) 97

(c) 64

(d) 87

Solution: There are 50 numbers between 1 to 100, which are divisible by at least first power of 2. There are 25 numbers, which are divisible by at least second power of 2, i.e. 4. However, as they are already counted once in previous step, we count them again for only single power of 2 in this step. Continuing in the same fashion, there are 12 numbers, which are divisible by at least third power of 2, i.e. 8.

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INAO – Jr – 2009

There are 6 numbers, which are divisible by at least forth power of 2, i.e. 16. There are 3 numbers, which are divisible by at least fifth power of 2, i.e. 32. There is only 1 number divisible by sixth power of 2, i.e. 64. Summing, (50 + 25 + 12 + 6 + 3 + 1) = 97 is the number of times factor 2 appears in the product. So 297 is the highest power of 2, which will be a factor of the product of all the numbers from 1to 100. Ans = (b)

mleaking at the rate of 5ml per 6. A repairman on the T. V. tower finds his water bottle oom second. He drops the bottle and it reaches the ground cc straight. If he was at a height . of 125m at that time and there was 200ml ofawater . left in the bottle, the amount of a l water left in the bottle (neglecting air resistance) e l just before it hit the ground was e m m (a) 175ml (b) 50ml (c) 100ml dd (d) 200ml aa o l lo n n Solution: Ans = (d). w w and water in it, is the gravitational force (constant The force acting on aobottle o acceleration g). Therefore because of this free fall motion, bottle and water .dd containing in it, will. have same velocity. w hitting the ground, water bottle will contain the same amount, Hence just beforew wwin it. 200ml of water w Note: In w the actual question paper, there was a typographic error with height of tower specified as 125km rather than 125m. However, as the solution is independent of height, answer doesn’t change. 7. In which of the following cities, your shadow will be the shortest, on the 15th of June? (a) Delhi

(b) Bhopal

(c) Bangalore

(d) Thiruvanantpuram

Solution: The latitude of Bhopal is closest to the Tropic of Cancer. On the 15th June (summer solstice – 21st June), the Sun will be almost at zenith and therefore there we can see our shortest Shadow. Ans = (b). 8. In the following figure, A, B, C are three light source positions with respect to the obstacle and the screen. Which of these light source positions will result in the longest shadow of the obstacle on the screen?

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INAO – Jr – 2009 11 00 A

11 B00

Screen

11 00 C

Obstacle

. (a) A (b) C

m

(c) A and C form shadows of same length, while B forms a smaller shadow. oom

c l a eel m Solution: Ans = (d) ddm aa o A l o nnl w B oow C Obstacle .d .d ww ww ww

. .c of same length. (d) All the three light sources will formashadows

11 00 00 11

L

K P

M

11 00 11 00

Screen Q

N

R

S

. From the above figure, P R is the shadow because of light source C and QS is the shadow because of the light source A. Now line AK is perpendicular to screen from A, which meets line MN at L. Since MN k KS, ∴ △ AMN and △ AQS are congruent , MN AL which means = . QS AK Similarly, △CMN & △ CP R are congruent, MN AL which means = . PR AK =⇒ QS=PR. Therefore, shadows of the obstacle formed by either light source A or C are the same. In the same manner, we can prove that the shadow formed by the light source B is also of the same length.

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INAO – Jr – 2009

9. Which of the following represents the correct speed-time graph, for a ball bouncing frequently from a fixed surface? u

u

u t

t

(a)

(b)

t

(c)

(d) None.

Solution: Ans = (d). From the kinematical equation of motion v = (u + at) , we can say that the speed-time graph should be a linear one. So options (a) and (c) are not correct. m m o Also since speed can not assume negative values, even cco option (b) is incorrect. In . fact, (b) represents correct velocity-time graph.in case of inelastic collisions.

10.

lala ee Two glass tubes filled with water are m held mvertical and connected by a plastic tube d d as shown in the figure. Pans are mounted on top of each piston such that aa (weight of the piston + pan)A l=o(weight of the piston + pan)B o l radius of the piston A = 1.0cm nnand radius of piston B = 1.5cm. A 30.0gm of mass is added in ww pan B, what is the mass required in pan A to balance o 30.0gm in pan B? o d .d . ww pan ww piston ww 0000 1111 111 000 1111 0000 111 000 1111 0000 111 000 111 000 111 000 A B 1111 0000 111 000 1111 0000 111 000 111 000 1111 0000 1111 0000 0 11111 11111 00000 0000 111 000 water 1111 0000 111 000 11 00 11 00 1 0 1 0 11 00 11 00 11 00 1 0 1 0 11 00 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 1 00 0 11 11 00 1 0

(a) 67.5gm

(b) 30.0gm

(c) 13.3gm

.

(d) 24.0gm

Solution: Ans = (c) Let, mass in pan A = m1 , mass in pan B = m2 = 30gm. radius of the piston A = r1 = 1.0cm and radius of piston B = r2 = 1.5cm.

5

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INAO – Jr – 2009

Now pressure in both tubes should be the same. F1 F1 = A1 A2 m1 g m2 g = πr12 πr22 m1 30 = 1 1.52 30 ∴ m1 = 2.25 30 ∴ m1 = 9

∴P =

m m

oo ⇒ m1 = 13.3gm. 4

11.

.c.c a Hence pan A required 13.3gm of mass to balance l a pan B. eel m ddm aa o l lo Section B: (4nquestions × 5 marks each) n ww Which of the following observations support the statement that “Every system tends o to adjust by itself to d haveominimum Potential Energy”. . .d (a) Andromeda ww galaxy and Milky Way galaxy are approaching each other. (b) Twow unlike, free charges move towards each other. w ww work is required to compress a spring. (c) External (d) A powerful magnet can deflect a compass needle from equilibrium position. Solution: Ans = (b), (c) and (d). Andromeda Galaxy and the Milky Way are moving around the centre of our local group of galaxies. In the course of this motion, they just happen to be coming closer to each other. Their mutual gravitational attraction does not play any significant role in this motion.

12. In one of the truly revolutionary finds of the 20th century, Howard Carter discovered discovered tomb of Egyptian Pharaoh (emperor) Tutankhamen in 1922. Along with the mummy following items were also removed from the tomb. Which of these items could have been carbon dated to fix the period of the Pharaoh? (a) Fragments of glass (b) Bronze Razor (c) Dried Fruits (d) Leather Shoe

6

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INAO – Jr – 2009

Solution: Carbon dating relies on the fact that all living objects have 12 C and 14 C in a fixed ratio to each other. The radioactive 14 C keeps decaying to 12 C, however, it is replaced by food intake consumed by living organisms. After death, the decay process continues, but there is no replacement for decayed 14 C. Thus the ratio starts changing. In the list above, the first two are not made from any organic / living substances. Thus, they cannot be used for carbon dating. Ans = (c) and (d). 13. There is a regular bus service between Pune and Mumbai (180km apart) at every m hour from both the ends, from 4am to 11pm. These busses run at average speed m o o of 45km/hr. Taxies also run on the same route c at 60km/hr with regular interval of . are .c based upon the number of taxies 30min from 5am to 10pm. Following statements a a l or busses crossed (not overtaken) only during eel travelling i.e. excluding instances of arrival and departure. Select the correct statement(s).

m ddm aa o (b) Last taxi crosses 5 buses. l lo n (c) Bus left at 8pm crosses n 10 taxis. ww (d) Taxi left at 12 noon crosses 6 buses. oo d . .d ww Solution: Ans = ww (a), (b), (c) and (d) Bus covers the distance in 4 hours, taxi in 3 hours. ww is at 4am and it will reach its destination at 8am. First ST bus (a) First bus crosses 6 taxis.

Thus, during the journey, it will meet all taxis to have started from the other city from 5 am till 7:30 am. i.e. 5:00, 5:30, 6:00, 6:30, 7:00, 7:30 = 6 taxis Last taxi starts at 10:00 pm, by which 6:00 pm bus would have arrived at the bus station. It reaches well past 11:00 pm (time of last bus). Thus, it will meet all the buses after the 6:00 pm bus. i.e. 7:00, 8:00, 9:00, 10:00, 11:00 = 5 busses. The taxi leaving at 5:00 pm would have arrived by 8:00 pm. Thus bus will meet all taxis from 5:30 pm till the last taxi which leaves at 10:00pm i.e. 5:30, 6:00, 6:30, 7:00, 7:30, 8:00, 8:30, 9:00, 9:30, 10:00= 10 taxis The 8:00 am bus would have arrived by noon. Thus the taxi meets 9:00am, 10:00am, 11:00am, 12:00noon, 1:00pm, 2:00pm buses on the way = 6 buses

14. Which of the following statement(s) is(are) useful, in estimating distances in the Universe? (a) Some time Venus can be seen transiting over the solar disc.

7

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INAO – Jr – 2009

(b) Stars with no proper motion appear to change their position in the sky when viewed six months apart. (c) Stars exhibit Doppler shift. (d) All supernovae of Type Ia, have same absolute brightness. Solution: With Doppler shift we can estimate the velocity of stars but not the distance. The Earth-Sun distance was successfully estimated for the first time using Venus transit method. Option (b) talks of parallax method. The absolute magnitudes of Supernovae is useful standard candle for cosmological distances. Hence, ans= (a), (b) and (d). m

α.

oom .c.c a l a eel m Section C: Analytical Questions ddm aa (8 marks) What will be area of o the largest cyclic quadrilateral that can be inscribed o l in a given circle? Justify your answer qualitatively (formal proof not necessary). nnl ww o Solution: Divide any quadrilateral ABCD, inscribed in a circle, into two trianddo . . gles △ABC and △ADC, as shown. ww ww B ww h1

P1

A

C P2

h2

D

The area of the triangle is given by the formula, 1 2

× base × height.

The common base of the two triangles △ABC and △ADC is given by AC. The total area, of the quadrilateral will be sum of the areas of the two triangles i.e., 1 2

× l(AC) × (h1 + h2 )

8

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INAO – Jr – 2009

To maximize the total area, l(AC) as well as (h1 + h2 ) should be maximized. Now the maximum possible length that can be fitted inside a circle has to be its diameter, d. Consequently, the total area of the quadrilateral ABCD would be maximum if, (h1 + h2 ) = d and l(AC) = d Thus, the maximum area can be 12 d2 . Noting that the the base and height are perpendicular to each other, it is clear that the said quadrilateral is a square. • If correct area with no justification: 2 marks.

m • If correct area with incorrect justification: o 3m marks.

β.

cco . • If justification only considers cyclic rectangles: 4 marks. . lala ee resulting in wrong area: 7 marks. • If correct justification, but minor error m ddm • Any correct method will receive aa full consideration. lo loshe saw a solar eclipse when the size of the solar (12 marks) Jayshree claimednthat n ww disk was 26′ and that of the lunar disk was 30′ . She also claimed that at the time of the oo between the centres of the two disks was 7′. Qualitatively maximum eclipse, distance d show that she could. not .d have observed a total eclipse. Find the percentage
of the −1 1 w solar disk covered w at the time of the maximum eclipse. (Given: cos ≈ 0.49π 26 ww rad). ww Solution: At the time of the maximum eclipse, the centres were 7′ away from each other. However, the radius of the solar disk is smaller than that of the lunar disk by just 2′ . Thus, she must not have viewed the total solar eclipse. Let us find percentage of the maximum partial eclipse. (1 mark)

p q E

A

B C t

r

D

(1 mark)

9

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INAO – Jr – 2009

In the figure above, points A and B make an angle of θ1 and θ2 with centre of the Solar and the Lunar disks (i.e. C and D) respectively. rs denotes the solar radius, rm the lunar radius and h the separation between the two centres at the closest approach (all in arcminutes). [ = πrs2 A(ACB) p+q =

θ1 2π

rs2 θ1 . 2

2 \ = πrm A(ADB)

θ2 2πm

m

2 o rm θ2 q + r + t = .cc.o 2.

lala where r = A(△ACD) and t = A(△BCD) ee Also from figure, r = t. m m d 2a d 2 r θ1 r θ2 (2 marks) ⇒ p = s − om a+ 2A(△ACD). 2 l lo 2 nn Now using Hero’s formula for area of a triangle, ww oo p A(△ACD) d . .d = s(s − rm )(s − rs )(s − h) where, w w ww rs = 13′ ww rm = 15′

h = 7′ rs + rm + h s = 2   13 + 15 + 7 35 = ∴s = ′ 2 2 r 35 (35 − 30) (35 − 26) (35 − 14) ∴ A(△ACD) = × × × 2 2 2 r2 35 5 9 21 × × × = 2 2 2 2 s 2 5×7×3 = ×3 4 105 √ = × 3 4 (173.2 + 8.7) ≈ 4 91 ≈ (2 marks) 2

Now to find θ1 and θ2 draw a perpendicular line BE to line CD.

10

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INAO – Jr – 2009

Let EC = x and EB = y. ∴ In △ ECB, x2 + y 2 = rs2 . and in △ EDB, 2 (x + h)2 + y 2 = rm 2 2hx + h2 + rs2 = rm r 2 − rs2 − h2 ∴x = m = 0.5′ 2h 0.5 1 x = = ∴ cosθ1 = rs 13 m 26 m ⇒ θ1 = 0.49πcoo (2 marks) . .c Similarlly, a a l x +l h 0.5 + 7 1 = cosθ2 = ee = rm 15 2 m m π ⇒a θ2dd = (1 mark) 3 a

loo n

l angles are double these values. These are half angles only.n Total ww r2 × 2 × θ 1 s

⇒o po=

.d .d ww ww ww

≈ ≈ ≈ ≈ ≈

2 rm × 2 × θ2 + 2A(△ACD) 2 2 π 91 169 × 0.49π − 225 × + 2 × 2  3 91 × 7 π 84 − 75 + 22   637 π 9+ 22 π(9 + 28.95) 38π arcmin2 (1.5 marks)

−

and A(Sun) = πrs2 = 169π arcmin2

(0.5 marks)

∴ The amount of solar surface covered by moon at the time of maximum eclipse p ) × 100% A(Sun) 38 (1 − ) × 100% 169 2.923 (1 − ) × 100% 13 (1 − 0.22) × 100%. 78 %. (1 mark)

= (1 − ≈ ≈ ≈ ≈

11

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INAO – Jr – 2009

γ. (8 marks) The famous Indian astronomer, Aryabhata, expressed the value of π in 1 what we now know as continuing fractions i.e..π = 3.1416 = a + where a, 1 b+ 1 c+ d b, c, d are positive integers. Find a, b, c, d. Solution: 1416 10000 1 m = 3+ m 10000 oo .c.c 1416 a l la1 e = 3+e 9912 + 88 m m d 1416 aad 1 o= 3 + lo l 1 nn 7+ 1416 w oow 88 d .d 1 . = 3+ ww 1 ww 7+ 1408 + 8 ww 88 1 = 3+ 1 7+ 1 16 + 88 8 1 = 3+ 1 7+ 1 16 + 11 Therefore a = 3, b = 7, c = 16, d = 11. 3.1416 = 3 +

• If attempted to solve polynomial: 1 mark. • If approximate reciprocals found correctly: 6 marks. • If approximate reciprocals found correctly and final answer is tallied back: full marks. • For each wrong value out of a, b, c, d: –1 mark each.

12

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INAO – Jr – 2009

δ. Mehul performed an experiment to verify Ohm’s law. He connected following circuit to measure voltage and current.

|

+

K

| A +

R

V |

+

m m

oo A the ammeter and K is the Here, R is the unknown resistance, V the voltmeter, c . .c key. He obtained following readings : lala e3e 4 V (v) 1 2 5 6 I (mA) 1.40 2.83m 4.26 5.68 7.11 8.54 ddm aa to represent the data. (a) (9 marks) Plot appropriate graph o (b) (2 marks) Find the value loflo nn R. (c) (1 mark) From the graph, ww what will be the voltage across the resistance when o I = 8mA? o d .d . w w Solution: ww (a) Fromw the given observation table we have plotted voltage (v) versus current w (mA) graph.

V (V)

v/s

I (mA) X−axis : 1cm = 1mA Y−axis : 1cm = 1 volt O(0,0)

Y−axis

7 6 5

V(V)

4 3 2 1 X−axis 0

1

2

3

4

5

6

I

7

8

9

10

(mA)

13

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INAO – Jr – 2009

• Optimum amount of graph paper should be utilized.

(1 mark)

• Choice of scale should be convenient.

(1 mark)

• Optimum scale for voltage on larger axis: y-axis – 2 big sq. = 1V, x-axis – 1 big sq. = 1mA • Optimum scale for current on larger axis: y-axis – 2 big sq. = 1V, x-axis – 2 big sq. = 1mA • Points should be clearly marked. • Chart Title, Axis Titles etc. should be properly m written.

(1.5 mark) (1 mark)

oom • The scales and preferably origin should .be ccclearly specified at the top right . corner. (1 mark) lala ee • Points used for finding slope should be well separated, well marked and m m preferably not from the existing (1 mark) d dataset. aad • Line should be continuous (1.5 mark) oand well balanced. lo l nn • 8mA point should be marked on the graph paper (as shown). (1 mark) ww oo d (b) Hence from the above (2 marks) . .d graph, ww (y2 − y1 ) slope = ww (x2 − x1 ) ww (4.2 − 0.7)V ⇒R ≈

(6 − 1)mA (3.5) Ω ≈ (5 × 10−3 ) ∴ R ≈ 700 Ω (2 mark)

(c) From the graph, for I = 8mA; V= 5.6V

(1 mark)

ǫ. If the entire surface of the earth is covered using A4 size (size of your answer sheet) sheets of paper, what will be the total weight of paper used? Solution: The total surface area of the Earth

(2 marks)

A = 4πR2 2 = 4 × π × (6.4 × 108 ) cm2 Now normal size of A4 paper a = (20 × 30) cm2

14

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INAO – Jr – 2009

To cover the whole earth we will require

(3 marks)

4 × π × 6.42 × 1016 papers ≈ 82 × 1014 papers 20 × 30 Now thickness of a typical 100 page notebook is 1cm (or thickness of question paper + answer sheet was about 2mm). So thickness of a single paper = 0.01cm = 0.1mm. (1 mark) ∴ mass of a single A4 paper : m = volume × density of a paper

m m

oo We take, density of a paper ≈ density of wood = 0.5 gm/cc (1 mark) c . .c crumpled paper floats on the (It should be definitely less than water as even a a water) ll ee ∴ mass of a paper ≈ 600cmm × 0.01cm × 0.5gm/cm3 m d ≈ 3a gm d(1 to 5 gm acceptable) a w ≈ lo 8.1 o× 1015 × 3 l n ≈n2.4 × 1016 gmwt w 13 oow≈ 2.4 × 10 kgwt 9 .d .d ≈ 24 × 10 14Tonne wt ww ≈ 2.4 × 10 N ww ww 2

∴ Total weight of paper used is about 24 Trillion Kgwt.

prepared using LATEX2ǫ

(1 mark)

(2 marks)

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Indian National Astronomy Olympiad – 2009 Senior Category

Roll Number:

Roll Number

Model Solutions

Please Note:

Date: 31st January 2009 Maximum Marks: 100

E

INAO – 2009 Duration: Three Hours

• 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 (6 sides).

CS

• In Section A, there are 10 multiple choice questions with 4 alternatives out of which only 1 is correct. You get 3 marks for each correct answer and -1 mark for each wrong answer. • In Section B, there are 4 multiple choice questions with 4 alternatives each, out of which any number of alternatives may be correct. You get 5 marks for each correct answer. No marks are deducted for any wrong answers. You will get credit for the question if and only if you mark all correct choices and no wrong choices. There is no partial credit. • For both these sections, you have to indicate the answers on the page 2 of the answersheet by putting a × in the appropriate box against the relevant question number, like this: Q.NO. (a) (b) (c) (d) Q.NO. (a) (b) (c) (d) 22

 ⊠  

OR

35

⊠ ⊠  

HB

Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

• In Section C, there are 5 analytical questions totalling 50 marks. • Blank spaces are provided in the question paper for the rough work. No rough work should be done on the answer-sheet. • No calculators are allowed.

• The answer-sheet must be returned to the invigilator. You can take this question booklet back with you.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Useful Physical Constants ME RE M⊙ R⊙ Rm c 1 A. U. G g gm

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 1.7 × 106 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 9.8 m/s2 1.6 m/s2

E

Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Speed of Light Astronomical Unit Gravitational Constant Gravitational Acceleration on the Earth Gravitational Acceleration on the Moon

HB

CS

Space for Rough Work

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – Sr – 2009

Section A: (10 Q × 3 marks each)

1. If ax = by = cz and b2 = ac, then y = ? √ √ 2xz xz (a) x+z (b) x+z (c) 2xz (d) xz

E

Solution: y

Ans = (a)

CS

a = bx y c = bz y y b2 = ac = b x b z y y = bx+z y y + 2 = x z y(x + z) = xz 2xz y = z+x

2. Each of the figures below depict a constellation. Find the odd one out.

(b)

(c)

HB

(a)

(d)

Solution: The constellations are (a) Leo (b) Taurus (c) Scorpio (d) Canis Major. First three are zodiac signs whereas the fourth one is not. Ans = (d)

3. Gravitational force between two identical uniform solid gold spheres of radius r each in contact is proportional to (a) r 4

(b) r 2

(c)

1 r2

(d)

1 4r 2

1

INAO – Sr – 2009

Solution: The distance between two spheres = 2r and masses are the same GM 2 (2r)2

⇒F =

3

2

E

∗ ρ) G( 4πr 3 = 4r 2 6 r ⇒F ∝ 2 r F ∝ r4 Thus ans = (a)

CS

4. A copper cube and a wooden cube of volume 10−3 m3 each are initially at room temperature. They are then moved to an enclosure of ambient temperature 50◦ C. What can we conclude about the temperatures attained by both cubes after 5 hours? (a) Tcopper > Twood as thermal conductivity of copper is greater than that of wood. (b) Twood > Tcopper as specific heat capacity of wood is greater than that of copper. (c) The temperatures will depend on the interplay between specific heat capacity and thermal conductivity of the materials. (d) Both temperatures will be practically the same, as they are in the enclosure for 5 hours.

HB

Solution: Both copper and wooden cube will have same temperature, as that of the enclosure, because the time is sufficiently long to bring them in thermal equilibrium with their surroundings. Ans = (d)

5. If the product of all the numbers from 1 to 100 is divisible by 2n , then what is the maximum possible value for n? (a) 128

(b) 97

(c) 64

(d) 87

Solution: There are 50 numbers between 1 to 100, which are divisible by at least first power of 2. There are 25 numbers, which are divisible by at least second power of 2, i.e. 4. However, as they are already counted once in previous step, we count them again for only single power of 2 in this step. Continuing in the same fashion, there are 12 numbers, which are divisible by at least third power of 2, i.e. 8.

2

INAO – Sr – 2009

E

There are 6 numbers, which are divisible by at least forth power of 2, i.e. 16. There are 3 numbers, which are divisible by at least fifth power of 2, i.e. 32. There is only 1 number divisible by sixth power of 2, i.e. 64. Summing, (50 + 25 + 12 + 6 + 3 + 1) = 97 is the number of times factor 2 appears in the product. So 297 is the highest power of 2, which will be a factor of the product of all the numbers from 1to 100. Ans = (b) − → − → 6. Two vectors P and Q are acting at a point such that their resultant is perpendicular − → − → − → | to Q . If θ is the angle between P and Q then |P is given by, |Q| (b) sec θ

(c) − cos θ

(d) − sec θ

CS

(a) cos θ

P

R

α

θ Q α

Solution: See figure

HB

α = 180 − θ Q = cos α P = cos(180 − θ) = − cos θ P ∴ = − sec θ Q

7. What will be the approximate period of Chandrayaan moving in an orbit 100 km above the moon’s surface? (a) 57 min

(b) 30 min

(c) 118 min

(d) 79 min

Solution: Let R = radius of the Moon, r = radius of the orbit Using Kepler’s Law, 4π 2 3 r GMm 4π 2 3 r = gR2

T2 = T2

3

INAO – Sr – 2009

Radius of orbit of Chandrayaan from center of the moon is, r = 1.7×106 m+100km

≈ ≈ T ≈ ≈

CS

≈ T ≈

4π 2 (1.8 × 106 )3 1.6 × (1.7 × 106 )2 4 × 10 × 1.83 × 106 1.6 × 1.72 45 × 1.82 × 106 1.72 6.7 × 18 × 103 17   6.7 6.7 + × 103 17 7100sec 118min.

E

T2 =

is the period of Chandrayaan orbit. The calculation above is a back of the envelope calculation to estimate the period. Ans = (c) 8. In the following figure, A, B, C are three light source positions with respect to the obstacle and the screen. Which of these light source positions will result in the longest shadow of the obstacle on the screen? 11 00 A

00 11

00 B11

HB

Screen

00 11 00 11 C

Obstacle

.

(a) A

(b) C

(c) A and C form shadows of same length, while B forms a smaller shadow.

(d) All the three light sources will form shadows of same length.

Solution: Ans = (d)

4

INAO – Sr – 2009

00 11

00 11 A

L

K P

11 00 00 B11

M

11 00 C

Obstacle

Screen Q

E

N R

S

.

CS

From the above figure, P R is the shadow because of light source C and QS is the shadow because of the light source A. Now line AK is perpendicular to screen from A, which meets line MN at L. Since MN k KS, ∴ △ AMN and △ AQS are congruent , MN AL which means = . QS AK Similarly, △CMN & △ CP R are congruent, MN AL which means = . PR AK =⇒ QS=PR.

Therefore, shadows of the obstacle formed by either light source A or C are the same. In the same manner, we can prove that the shadow formed by the light source B is also of the same length.

HB

9. The following figure shows skeleton chart of the Orion constellation. Approximate direction “North” is marked with the letter ....

(a) A

(b) B

(c) C

(d) D

Solution: The Orion’s head is northwards. Ans = (a)

5

INAO – Sr – 2009

10. Find the resultant focal length for following system where the radius of curvature is 15cm.

111111111111111111111111 000000000000000000000000 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111

E

11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111

glass

(b) 60 cm

(c) 120 cm

air

(d) ∞

CS

(a) 40 cm

water

Solution: Refractive Index : nw =

4 3

; ng =

3 2

9 ng = nw 8 na 3 Refractive Index of air relative to water: w na = = nw 4 Refractive Index of glass relative to water:w ng =

Using the lens maker formula, 1 1 1 = (n − 1)( + ) f R1 R2

focal length for plano-convex glass-in-water lens, where R1 = ∞; R2 = 15cm is, ( 98 − 1) 1 1 1 1 = (w ng − 1)( + )= = f1 R1 R2 15 120

HB

∴ f1 = 120 cm

Similarly, focal length for plano-concave air-in-water lens, where R1 = -15cm; R2 = ∞ is, ( 3 − 1) 1 1 1 1 = (w na − 1)( + )= 4 = f2 R1 R2 −15 60

∴ f2 = 60 cm

Hence resultant focal length will be 1 1 1 1 1 1 = + =( + )= f f1 f2 120 60 40

∴ f = 40 cm. Ans is (a).

6

INAO – Sr – 2009

Section B: (4 Q × 5 marks each)

11. Which of the following observations support the statement that “Every system tends to configure itself to have minimum Potential Energy”.

E

(a) Andromeda galaxy and Milky Way are approaching each other. (b) Two unlike, free charges move towards each other. (c) External work is required to compress a spring.

(d) Light iron dust moves towards powerful magnet in close vicinity.

CS

Solution: Ans = (b), (c) and (d). Andromeda Galaxy and the Milky Way are moving around the centre of our local group of galaxies. In the course of this motion, they just happen to be coming closer to each other. Their mutual gravitational attraction does not play any significant role in this motion.

12. Consider a sealed frictionless piston cylinder assembly where the piston mass and atmospheric pressure above the piston remain constant. A gas in the cylinder is heated and hence it expands. Which of the following is / are true? (a) The density of the gas will increase.

(b) The pressure of the gas will decrease.

(c) The internal energy of the system will remain the same. (d) In this process work is done by the gas.

HB

Solution: As gas in the cylinder is heated and it expands so work is done by the gas. At the new equilibrium, the density of the gas would have decreased (same mass in larger volume), internal energy would have increased (proportional to change in temperature) and the pressure would have remained the same (as that of atmospheric pressure + pressure due to piston mass). Ans = (d).

13. In one of the truly revolutionary finds of the 20th century, Howard Carter discovered the tomb of the Egyptian Pharaoh (emperor) Tutankhamun in 1922. Following items were removed from the tomb, along with the mummy of the Pharaoh. Which of these items could have been carbon dated to fix the period of the Pharaoh? (a) Fragments of glass

(b) Golden Bracelets (c) Dried Fruits

(d) Leather Shoe

7

INAO – Sr – 2009

Solution: Carbon dating relies on the fact that all living objects have 12 C and C in a fixed ratio to each other. The radioactive 14 C keeps decaying to 12 C, however, it is replaced by food intake consumed by living organisms. After death, the decay process continues, but there is no replacement for decayed 14 C. Thus the ratio starts changing. In the list above, the first two are not made from any organic / living substances. Thus, they cannot be used for carbon dating. Ans = (c) and (d).

E

14

14. Which of the following phenomena is / are useful, in estimating distances in the Universe? (a) Some time Venus can be seen transiting over the solar disc.

CS

(b) Stars with no proper motion appear to change their position in the sky when viewed six months apart. (c) Stars exhibit Doppler shift.

(d) All supernovae of Type Ia have same absolute brightness.

HB

Solution: With Doppler shift we can estimate the velocity of stars but not the distance. The Earth-Sun distance was successfully estimated for the first time using Venus transit method. Option (b) talks of parallax method. The absolute magnitudes of Supernovae is useful standard candle for cosmological distances. Hence, ans= (a), (b) and (d).

Section C: Analytical Questions

α. (8 marks) What will be area of the largest cyclic quadrilateral that can be inscribed in a given circle? Justify your answer qualitatively (formal proof not necessary). Solution: Divide any quadrilateral ABCD, inscribed in a circle, into two triangles △ABC and △ADC, as shown.

8

INAO – Sr – 2009

B

E

h1

P1

A

C

P2

h2

D

CS

The area of the triangle is given by the formula, 1 2

× base × height.

The common base of the two triangles △ABC and △ADC is given by AC. The total area, of the quadrilateral will be sum of the areas of the two triangles i.e., 1 2

× l(AC) × (h1 + h2 )

To maximize the total area, l(AC) as well as (h1 + h2 ) should be maximized. Now the maximum possible length that can be fitted inside a circle has to be its diameter, d. Consequently, the total area of the quadrilateral ABCD would be maximum if, (h1 + h2 ) = d and l(AC) = d

HB

Thus, the maximum area can be 21 d2 . Noting that the the base and height are perpendicular to each other, it is clear that the said quadrilateral is a square. • If correct area with no justification: 2 marks. • If correct area with incorrect justification: 3 marks. • If justification only considers cyclic rectangles: 4 marks. • If correct justification, but minor error resulting in wrong area: 7 marks.

• Any correct method will receive full consideration.

β. (12 marks) Jayshree claimed that she saw a solar eclipse when the size of the solar disk was 26′ and that of the lunar disk was 30′ . She also claimed that at the time of the maximum eclipse, distance between the centres of the two disks was 7′ . Qualitatively show that she could not have observed a total eclipse. Find the percentage
of the −1 1 ≈ 0.49π solar disk covered at the time of the maximum eclipse. (Given: cos 26 9

INAO – Sr – 2009

rad).

E

Solution: At the time of the maximum eclipse, the centres were 7′ away from each other. However, the radius of the solar disk is smaller than that of the lunar disk by just 2′ . Thus, she must not have viewed the total solar eclipse. Let us find percentage of the maximum partial eclipse. (1 mark)

p

q

A

E

B

C t

CS

r

D

(1 mark)

In the figure above, points A and B make an angle of θ1 and θ2 with centre of the Solar and the Lunar disks (i.e. C and D) respectively. rs denotes the solar radius, rm the lunar radius and h the separation between the two centres at the closest approach (all in arcminutes).

HB

[ = πrs2 A(ACB) p+q =

θ1 2π

rs2 θ1 . 2

2 \ = πrm A(ADB)

θ2 2π

2 rm θ2 . q+r+t = 2

where r = A(△ACD) and t = A(△BCD) Also from figure, r = t. ⇒p=

2 rs2 θ1 rm θ2 − + 2A(△ACD). 2 2

10

(2 marks)

INAO – Sr – 2009

Now using Hero’s formula for area of a triangle, A(△ACD) where, rs rm h

=

p

s(s − rm )(s − rs )(s − h)

CS

E

= 13′ = 15′ = 7′ rs + rm + h s = 2   13 + 15 + 7 35 ∴s = ′ = 2 2 r 35 (35 − 30) (35 − 26) (35 − 14) ∴ A(△ACD) = × × × 2 2 2 2 r 35 5 9 21 = × × × 2 2 2 2 s 2 5×7×3 ×3 = 4 105 √ = × 3 4 (173.2 + 8.7) ≈ 4 91 (2 marks) ≈ 2

HB

Now to find θ1 and θ2 draw a perpendicular line BE to line CD. Let EC = x and EB = y. ∴ In △ ECB, x2 + y 2 = rs2 . and in △ EDB, 2 (x + h)2 + y 2 = rm 2 2hx + h2 + rs2 = rm r 2 − rs2 − h2 ∴x = m = 0.5′ 2h 0.5 1 x = = ∴ cosθ1 = rs 13 26 ⇒ θ1 = 0.49π (2 marks) Similarlly, x+h 0.5 + 7 1 cosθ2 = = = rm 15 2 π (1 mark) ⇒ θ2 = 3

11

INAO – Sr – 2009

These are half angles only. Total angles are double these values.

≈ ≈ ≈ ≈ ≈

2 × 2 × θ2 rs2 × 2 × θ1 rm − + 2A(△ACD) 2 2 π 91 169 × 0.49π − 225 × + 2 × 2 3  91 × 7 π 84 − 75 + 22   637 π 9+ 22 π(9 + 28.95) 38π arcmin2 (1.5 marks)

E

⇒p =

CS

and A(Sun) = πrs2 = 169π arcmin2

(0.5 marks)

∴ The amount of solar surface covered by moon at the time of maximum eclipse p ) × 100% A(Sun) 38 ) × 100% (1 − 169 2.923 ) × 100% (1 − 13 (1 − 0.22) × 100%. 78 %. (1 mark)

= (1 −

≈

≈

≈ ≈

HB

γ. (8 marks) The famous Indian astronomer, Aryabhata, expressed the value of π in 1 what we now know as continuing fractions i.e..π = 3.1416 = a + where a, 1 b+ 1 c+ d b, c, d are positive integers. Find a, b, c, d.

12

INAO – Sr – 2009

Solution: 3.1416 = 3 +

E

= 3+

1416 10000 1 10000 1416 1 9912 + 88 1416 1 1 7+ 1416 88 1 1 7+ 1408 + 8 88 1 1 7+ 1 16 + 88 8 1 1 7+ 1 16 + 11

= 3+

CS

= 3+

= 3+

= 3+

HB

= 3+

Therefore a = 3, b = 7, c = 16, d = 11.

• If attempted to solve polynomial: 1 mark. • If approximate reciprocals found correctly: 6 marks. • If approximate reciprocals found correctly and final answer is tallied back: full marks. • For each wrong value out of a, b, c, d: –1 mark each.

δ. (10 marks) Kedar sent a container of marbles by road from Mumbai to Parag in Pune. The container was 2m × 2m × 2.5m in size, with height being the larger dimension. Marbles of 2 cm diameter were arranged to fill the entire base and then additional layers of marbles were arranged with each upper marble exactly on top of corresponding marble in the previous layer (see figure). The marbles were thus 13

INAO – Sr – 2009

CS

E

placed upto 2m height to complete the cubical structure. However, on reaching Pune, when Parag opened the container, he found the height of structure was not 2m as promised by Kedar, but something else. Kedar defended saying that marbles may have readjusted due to jiggling. Can you find new height of this marble pile?

Solution: Total number of marbles is 100×100×100 = 106 arranged in total 100 layers. The marbles will rearrange themselves so as to occupy vacant spots in the structure. We assume that there is enough jiggling in the trip from Mumbai to Pune so that uniformity is attained in the arrangement of the marbles throughout the box. There are three ways to solve this problem and all of them lead to nearly the same answer. (All three methods earn full credit).

HB

Method 1: The marbles will be rearranged in a “face-centred cubic” structure. In this structure, the bottom layer has 100×100 marbles, arranged such that each 4 marbles create a gap between them, into which a marble can be placed in the next layer. Since there are 99 × 99 such gaps, the next layer will consist of 99 × 99 marbles. The layer above that will again have 100 marbles, and so on. Thus, we will have 50 layers of 100 × 100 marbles and 50 layers of 99 × 99 marbles. But in addition, we shall have extra marbles displaced from the 99 × 99 layers. Number of marbles displaced in each such layer is (1002 − 992 ) = 199. There are total 50 such layers. Thus, total number of marbles remaining after filling 100 layers is n = 199 × 50 = 200 × 50 − 50 = (100 × 100) − 50

The 101st layer can be of 100 × 100 marbles. The number of marbles remaining are in fact smaller than this number. Thus, there will be one additional incomplete layer. In this formation each set of 4 marbles on one layer and the 1 marble on the next layer sitting in the gap between them form a pyramid. Let the height of this pyramid calculated from the plane containing the centres of the 4 marbles in

14

INAO – Sr – 2009

h s

d

2

b

1

CS

1

2

E

the first layer to the centre of the top marble be h cm. Since the radius of each marble is 1 cm, according to the left panel in the figure below, √ 2 2 − s2 h = p 4 − (12 + 12 ) = √ 2 =

1

2

Now, Distance between floor and centre of marbles in the 1st √ layer is 1 cm, st nd Vertical separation between 1 and 2 layer centres is √2 cm, Vertical separation between 2nd and 3rd layer centres is 2 cm, and so on. For 101st layer, distance between the top and marble centres will again be 1 cm. Thus, total height of the structure will be,

HB

√ H = 1 + 100 × 2 + 1 cm ≈ 1 + 141.4 + 1 cm ≈ 143.4 cm

Method 2: The marbles will be rearranged in a “hexagonal close-packed” structure. In this structure, the bottom layer has 100 marbles in the first row next to a wall, but the adjacent row has 99 marbles, fitting into the gaps of the first row. The next row will have again 100 marbles, and so on. The distance between the lines connecting the centres of one row of marbles to the next row is (see right panel of figure above) √ √ 22 − 12 cm = 3 cm. The distance from the centre of the first row to the wall is 1 cm, and the distance from the centre of the last row to the opposite wall is 1 cm. Therefore if the number of rows between opposite walls is r, then √ 1 + (r − 1) 3 + 1 = 200

15

INAO – Sr – 2009

Solving, we get r = 198/1.73 + 1 ≈ 115 (r must be rounded off to the nearest lower integer, since the walls are fixed). Thus one can put 58 rows of 100 marbles and 57 rows of 99 marbles in the first layer. That is a total of 58 × 100 + 57 × 99 = 100 + 57 × 100 × 2 − 57 = 11443 marbles in the first layer.

E

The second layer can be arranged in exactly the same manner, but now each row will fit in the gaps created by the adjacent rows in the first layer. Since the pattern of gaps is reverse of the pattern of marbles in the first layer, one can put 58 rows of 99 marbles and 57 rows of 100 marbles in the second layer. That is a total of 58 × 99 + 57 × 100 = 58 × 100 × 2 − 58 − 100 = 11442 marbles in the second layer.

CS

Thus the first two layers will consist of 11443 + 11442 = 22885 marbles. From the third layer, the pattern will repeat. Therefore, we shall have 106 /22885 ≈ 44 (rounding off to the nearest larger integer) pairs of layers, i.e., 88 layers in all (the last layer being incomplete).

HB

In this formation each set of 3 marbles on one layer and the 1 marble on the next layer sitting in the gap between them form a tetrahedron. Let the height of this tetrahedron calculated from the plane containing the centres of the 3 marbles in the first layer to the centre of the top marble be d cm. Since the radius of each marble is 1 cm, according to the right panel in the figure below, √ d = 22 − b2 r 2√ 2 = 4−( 2 − 1 2 )2 3 r 2√ 2 4−( 3) = 3 r 4 4− = 3 √ √ = 2 2/ 3 Now, Distance between floor and centre of marbles in the 1st layer is 1 cm, √ √ Vertical separation between 1st and 2nd layer centres is 2 √2/ √3 cm, Vertical separation between 2nd and 3rd layer centres is 2 2/ 3 cm, and so on. For 88th layer, distance between the top and marble centres will again be 1 cm. Thus, total height of the structure will be

H =

≈

≈ ≈

√ 87 × 2 2 √ 1+ + 1 cm 3 174 × 1.414 1+ + 1 cm 1.732 2 + 100 × 1.414 cm 144 cm

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INAO – Sr – 2009

E

Method 3: The labour involved in the above two solutions can be avoided to a great extent by recognising that the two arrangements described above are, in fact, the “face-centred cubic” and “hexagonal close packing” arrangements. In each of these cases, the “packing fraction” is 0.74, i.e., the actual volume occupied by the marbles is 74% of the apparent volume of the stack. Therefore, if V is the volume occupied in each of these arrangements, then V is given by the equation 0.74V = 106 ×

4π (1cm)3 3

(1)

Now, V = A × H, where A is the area of the stack. Here, A = 200 cm2 . Thus,

CS

H = V /A 106 × 4π 3 cm = 0.74 × 4 × 104 100 × 3.1416 ≈ cm 3 × 0.74 100 × 104.72 cm ≈ 74 ≈ 141.4 cm

We obtain an answer close to those of the first two methods. The small discrepancy among the three methods arises due to the difference in packing of the balls near the edges, i.e., close to the walls (which manifests itself in the approximations that we have to make in each case). The answer obtained in the third method is the ideal case scenario, where we did not consider any “edge effect”.

HB

(In the last method, recognising that 0.74 is actually lation.)

π √ 3 2

will simplify the calcu-

ǫ. Chiraag performed an experiment using a simple pendulum to find value of g. He measured time taken for 30 oscillations of the pendulum for various values of length (repeated thrice for each value of length), following readings were obtained. L Readings cm t1 (Sec) t2 (Sec) t3 (Sec) 20.0 26.9 26.9 27.0 25.0 30.1 30.1 30.1 30.0 32.9 32.8 32.7 35.0 35.6 35.8 35.7 40.0 38.0 38.1 38.1 45.0 40.4 40.4 40.5

Mass of the bob of the pendulum was known to be 50 gm. (a) (9 marks) Plot appropriate graph to represent the data.

(b) (2 marks) Find the value of g.

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INAO – Sr – 2009

(c) (1 mark) Using the graph, find out required length of the pendulum to get a time period of 1.24 sec.

E

Solution: The equation for period of simple pendulum is given as, s L T = 2π g ∴ T2 =

4π 2 L g

In the experiment above, time is measured for 30 oscillation, So we have to average out three readings for each length and then find time for single oscillation. Readings t1 (Sec) t2 (Sec) t3 (Sec) 26.9 26.9 27.0 30.1 30.1 30.1 32.9 32.8 32.7 35.6 35.8 35.7 38.0 38.1 38.1 40.4 40.4 40.5

tavg

T= 30t

(cm)

20.0 25.0 30.0 35.0 40.0 45.0

T2

(Sec)

(Sec)

Sec2

26.9 30.1 32.8 35.6 38.1 40.5

0.897 1.03 1.09 1.19 1.27 1.35

0.81 1.06 1.19 1.41 1.61 1.82

CS

L

• Data should be properly averaged converted to time period for one oscillation. (1.5 marks) • Proper choice of variable on x-axis.

(1 mark)

• Optimum amount of graph paper should be utilized.

HB

• T 2 values√should be computed and linear graph should be plotted. (Graph of T vs. L is also acceptable). (2 marks)

• Choice of scale should be convenient.

(0.5 mark)

• Points should be clearly marked.

• Chart Title, Axis Titles etc. should be properly written.

(1 mark) (0.5 marks)

• The scales and preferably origin should be clearly specified at the top right corner. (0.5 marks) • Points used for finding slope should be well separated, well marked and preferably not from the existing dataset. (0.5 marks) • Line should visually appear to be well balanced. • 38cm point should be marked on the graph paper (as shown). marks)

18

(1 mark) (0.5

INAO – Sr – 2009

The T 2 Vs. L graph should be plotted to find slope. 2

T Vs. L 2.0

Scale x−axis: 1unit = 5cm y−axi: 1unit = 0.2 sec Origin (0,0)

1.8

E

1.6

1.52 1.4

1.2

T2 (Sec 2 )

2

1.0

0.8

0.6

CS

0.4

0.2

38

(0,0)

5

10

15

20

25

30

35

40

45

L (cm)

From the graph,

(1.61 − .79) (40 − 20) (.82) = 4.1 × 10−2 ≈ (20)

slope ≈

4π 2 ≈ 4.1 × 10−2 g 40 ∴g ≈ 4.1 × 10−2 g ≈ 9.8 m/s2

HB

∴

(2 mark)

Using the graph, when T = 1.24 sec, T 2 = 1.54 sec2 . Thus, L ≈ 38 cm. mark)

prepared using LATEX2ǫ

(1

Indian National Astronomy Olympiad – 2010 Junior Category

Roll Number:

Roll Number

Question Paper Date: 30th January 2010 Maximum Marks: 100

INAO – 2010 Duration: Three Hours

E

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 (6 sides).

CS

• In Section A, there are 10 multiple choice questions with 4 alternatives out of which only 1 is correct. You get 3 marks for each correct answer and -1 mark for each wrong answer. • In Section B, there are 4 multiple choice questions with 4 alternatives each, out of which any number of alternatives may be correct. You get 5 marks for each correct answer. No marks are deducted for any wrong answers. You will get credit for the question if and only if you mark all correct choices and no wrong choices. There is no partial credit. • For both these sections, you have to indicate the answers on the page 2 of the answer sheet by putting a × in the appropriate box against the relevant question number, like this: Q.NO. (a) (b) (c) (d) Q.NO. (a) (b) (c) (d) 22

 ⊠  

OR

35

⊠ ⊠  

Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

HB

• In Section C, there are 5 analytical questions totaling 50 marks. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Useful Physical Constants ME RE M⊙ R⊙ Rm c 1 A. U. G cs ǫ

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 1.7 × 106 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 330 m/s 23.5◦

E

Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Speed of Light Astronomical Unit Gravitational Constant Speed of Sound in Air Inclination of the Earth’s Axis

HB

CS

Space for Rough Work

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – Jr – 2010

Section 1:Multiple Choice Questions Part A: (10 Q × 3 marks each)

1. As seen from the Earth, the stars appear to twinkle, while the planets do not because, (a) Light coming from the stars gets absorbed by interstellar dust.

E

(b) the stars are self luminous while the planets merely reflect the light. (c) Angular sizes of the stars are much smaller than the planets. (d) All the above.

CS

A wooden cube of length 5 units is painted on all faces and then cut in 85 smaller cubes of varying sizes. The sides of each of the smaller cubes are some integer number of units. All bigger cubes are carved out of the corners of the original cube. Answer the following three questions: 2. How many cubes have no side painted? (a) 15

(b) 17

(c) 20

(d) 27

HB

Solution: If you cut the bigger cube in smaller cubes with all side of 1 unit, there will be 125 total cubes. Each cube of size 2 will fuse 8 such smaller cubes together. i.e. the total number of cubes will reduce by 7. Similarly, each cube of size 3 and 4 will reduce total number of cubes by 26 and 63 respectively. Now, we have 85 total cubes, which is 40 less than 125. This means there is 1 cube of size 3 and 2 cubes of size 2 at three of the corners. Rest all are size 1. If there were 125 smaller cubes of size 1, one can say the volume which is not touched by paint on any face is equivalent to a cube of size 3 i.e. 27 smaller cubes. However, the the bigger cubes carved out of corners include some of these inner cubes as well. 8 such inner smaller cubes are included in the size 3 cube and 1 each is included in the 2 size 2 cubes. Thus, number of unpainted cubes = 27 − 8 − 1 − 1 = 17

3. How many cubes have exactly 2 sides painted? (a) 24

(b) 27

(c) 30

(d) 36

Solution: To have exactly 2 sides painted, the smaller cube should be along any of the edges but not at the corner. There are total 12 edges to the original cube. For each edge, there will be 3 cubes satisfying the condition if all cubes were of size 1. How ever, there is a size 3 cube, which includes 2 such smaller cubes each

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INAO – Jr – 2010

along 3 of the edges. Similarly, the 2 size 2 cubes will each include 1 such smaller cube each along 3 of the edges. Thus, total number of cubes = 12 × 3 − 2 × 3 − 1 × 3 − 1 × 3 = 24

(a) 69

(b) 64

(c) 62

(d) 50

E

4. If one was to cut the original cube (of size 5 units each side) with a condition that all corners would be occupied by the cubes bigger than 1 unit size, what will be the smallest total of cube pieces possible?

CS

Solution: There are two possibilities which will satisfy this condition. Either all corners will have cubes of size 2 or else there will be 1 cube of size 3 and rest all corners will have cubes of size 2. There are total 8 corners. In first case total number of cubes will be = 125 − 8 × 7 = 69 In second case, total number of cubes will be = 125 − 7 × 7 − 26 = 50 5. In a given watch, the minute and the hour hand come together successively exactly after 65 minutes. Does the watch gain or lose time and how much per hour? (a) Gains about 27 seconds. (b) Loses about 27 seconds. (c) Neither gains nor loses.

(d) Information insufficient.

HB

Solution: The angular rate of the two hands of the clock should be compared to find the interval at which the two hands coincide. 2π = ωm t − ωh t   2π 2π = t − 1 12 12 hr t = 11 60 min = 1 hr 11 300 = 65 min sec 11 3 = 65 min 27 sec 11

As the clock’s hands are coinciding faster than they ideally should, the clock is gaining time.

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INAO – Jr – 2010

6. Two parallel wires carrying current in opposite direction will (a) exert a force twisting the wires. (b) attract each other. (c) repel each other.

E

(d) not exert any force on each other. 7. If you hold a magnifying glass of focal length 10 cm in the sunlight and place a piece of paper at its focus, you can burn a hole in the paper. What could be the size of this hole? (a) 10 mm

(b) 5 mm

(c) 0.5 mm

(d) 0.1 mm

CS

Solution: If the magnification is given by M, Object distance, Object size, Image Distance, Image size by u, O, v&I respectively, M =

I ≈ 7 × 108

I ≈

≈

≈

v I = u O f 500c 0.1 × 7 × 108 500 × 3 × 108 7 × 10−3 mt. 15 0.5 mm

HB

8. During an earthquake, an earthquake monitoring Centre observed that transverse waves traveling with speed 4.5 km/s arrived at the centre 3 minutes after the longitudinal waves traveling at 8.2 km/s. Deduce the approximate distance to the epicenter. (a) 60 km

(b) 220 km

(c) 660 km

(d) 1800 km

Solution:

8.2t = 4.5 × (t + 180) 3.7t = 810 810 D = 8.2t = 8.2 × 3.7 = 2.22 × 810 D ≈ 1800km

9. A certain person nicknamed “Enthu”, encountered an automatic staircase (i.e. escalator) at a shopping complex, which was moving upward at a constant rate. Just for

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INAO – Jr – 2010

the fun of it he decided to walk up this escalator at the rate of one step a second. Twenty steps brings him to the top. Next day he goes up at two steps a second and reaching the top in 32 steps. How many steps are there in the escalator? (a) 40

(b) 60

(c) 64

(d) 80

E

Solution: Let the escalator rise with the speed of ‘n’ steps per second. On first day, Enthu reaches the top after 20 manual steps i.e. in 20 seconds. On second day, Enthu reaches the top after 32 / 2 = 16 seconds. Thus, total number of steps, = = = = =

16 × (n + 2) 32 − 20 = 12 3 20 × (3 + 1) 80

CS

20 × (n + 1) 4n ∴n ∴ Steps

10. Two stars of masses M and 3M respectively are going around each other, inq near circular orbits, with period T . The separation between them is given by D = The value of k is,

(a) 0.5

(b) 1

(c) 1.5

3

kGM T 2 . π2

(d) 3

Solution: The Centre of mass of the system will be located at distance D4 from the heavier mass. Centripetal force is provided by the mutual gravitation.

HB

G M (3M) D = (3M) ω 2 2 D 4 D 4π 2 GM = D2 4 T2 GMT 2 D3 = 2 rπ 2 3 GMT D = π2

Section B: (4 questions × 5 marks each)

11. If we throw a ball in a shallow water tank, propagation velocity of ripples on surface of the water will depend on (a) surface tension of the water

(b) depth of the water tank (c) density of the water

(d) height from which the ball was dropped 4

INAO – Jr – 2010

E

Solution: Note: The propagation velocity will vary as per the depth of the water, provided the water is not too deep. However, the derivation for this effect includes concepts beyond the level of participants. Thus, the accepted solution for this question would be ’a’ and ’c’ must be selected, ’d’ must not be selected and ’b’ would be acceptable both ways. 12. Amit decided to experiment with cannon ball by making it hollow and filling water inside it. He then punched few holes in it. After the ball was fired horizontally, he was expecting to see water jets coming out from some of the holes. Which of the following locations of holes will allow water jet to come out? (b) Back, top and sides

(c) back and bottom

CS

(a) Front, bottom and sides (d) None of the holes

Solution: Horizontally, all particles (i.e. cannonball and water particles) are flying with the same velocity. Hence water cannot come out of front / back /side holes. Vertically, all particles are experiencing free fall. Thus, gravity cannot exert any additional pressure on water and neither can water be slower than the cannonball. Hence no water can come out from top or bottom. 13. In the progress of Astronomy over the ages, we find several instances of startling new observations changing our ideas about the Universe and lead to new theories. Listed below are some milestones in the history of Astronomy and observations which necessitated them. Pick the correct statement(s). (a) Ptolemy was aware of retrograde motion of planets when he gave his model of the solar system.

HB

(b) Newton was aware of Kepler’s Laws when he gave his Law of Gravitation. (c) Einstein felt need to modify Newton’s theory of Gravitation to explain the expansion of the Universe.

(d) Existence of Cosmic Microwave Background Radiation (CMBR) led to the creation of the Big Bang Theory.

Solution: Ptolemy needed epicyclic model of the solar system as that was the only way to fit observed motions of the planets including the retrograde motion. Kepler’s laws enabled Newton to deduce r12 nature of the law of gravitation. Einstein gave General theory of relativity in 1916 and assumed the Universe to be non-expanding as per then norm. Hubble discovered the expansion of the Universe in late 20’s forcing Einstein to revisit his own theory. Existence of CMBR was one of the key predictions of the big bang theory. The theory was formulated before CMBR was discovered.

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INAO – Jr – 2010

14. Two simple pendula are hung close to each other on a thin, rigid support and are allowed to oscillate independently in planes parallel to each other. They have periods 3 seconds and 7 seconds respectively with same amplitudes and are initially released from the opposite extreme positions of each other. At which of the following times the threads of the two pendula will be coplanar again? (b) 7.88 seconds

(c) 10.50 seconds

(d) 23.10 seconds

E

(a) 1.05 seconds

Section C: Analytical Questions

α. An alien civilisation on a star far far away came to know about the Astronomy Olympiad Examination and wanted to test smartness of the students. They sent following two coded secret messages. Decode them.

CS

(a) (5 marks) First is a pictorial message in black and white colour. They sent it on radio waves in the form of ones and zeros. Find out what the picture says. 00000 00000 00000 00000 00000 00000 11111 01100 00100 00100 00011 10000 01000 10100 01000 10100 01101 10000 10001 00100 10001 11000 10001 00001 00010 00101 00100 01001 10110 01111 10100 00110 10000 01001 11000 00000 00000 00000 00000 00000 000 (b) (5 marks) Surprisingly, the aliens are also proficient in English and the following coded message is actually a sentence in English language. In the answer sheet, write down the coded sentence and also the true meaning of each alphabet in the code. “Up tpmwf uif qsfwjpvt tvcrvftujpo uijol pg uif nfttbhf tfou cz uif bsfdjcp ufmftdpqf up bmjfot.”

HB

Solution: In 1974, a pictographic message was sent to outer space by radio waves using Arecibo telescope. The total number ’digit’s in the message was product of two prime numbers. It was argued that any intelligent being can rearrange the grid in a rectangle with sides measuring these two prime numbers and then use blank square for ‘0’ and filled square for ‘1’ to reveal the picture.

In the present message same idea is used. Total number of digits is 203 which is a semiprime number; i.e. it can only be divided as 29 × 7. If the grid is rearranged in this way, following pattern is seen:

 ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ 

Marking Scheme:

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INAO – Jr – 2010

• Realising total number of digits are 203, which is a semi-prime number. (2 marks) • Attempt to draw either 29 × 7 or 7 × 29 grid.

(1 mark) (2 marks)

E

• picking the message “INAO”.

Part (b) is a substitution cipher. The word “uif” appears thrice in the sentence. Obviously, it stands for “the”. Looking at the two words, the substitution rule is inferred as “every letter is substituted by the next letter” to create the coded message. Deciphering, “To solve the previous subquestion think of the message sent by the arecibo telescope to aliens.” Marking Scheme:

CS

• Mapping the alphabets and deciphered Sentence.

(5 marks)

• Mapping the alphabets and Deciphered Sentence, but with small syntactic errors. (4.5 marks) • Mapping the alphabets only.

(4 marks)

• Deciphered Sentence but Alphabet mapping is not given.

(3 marks)

• Only partial decipherment with more than 10 letters.

(2 marks)

• Only partial decipherment with 7-10 letters.

• Only partial decipherment with “the” deciphered correctly.

(1.5 marks) (1 mark)

HB

β. (8 marks) In the following table, the first column gives various optical phenomena / instruments and the top row gives various optical effects which may help in explaining them. In the answers sheet, tick the correct effect(s) involved in each phenomena in appropriate rows. Appreciable Dispersion

Internal Reflection Reflection

Blue Sky Mirage Rainbow Smooth Convex Mirror Thick Concave Lens Solution:

7

Refraction Scattering

INAO – Jr – 2010

Int. Refle.

X

X X

Refle.

Refra.

Scat. X

X X X

X

X

E

Blue Sky Mirage Rainbow Sm. Conv. Mirror Th. Conc. Lens

Appri. Disp.

• Blue Sky (1 mark): One mark for scattering. -0.5 for each additional tick mark. • Mirage (2 marks): One each for internal reflection and refraction. 0 marks if internal reflection is not ticked.

CS

• Rainbow (2 marks): One for internal reflection; 0.5 each for dispersion and refraction. 0 marks if internal reflection is not ticked. • Mirror (1 mark): One mark for reflection. 0 if any additional / other tick marks. • Thick Lens (2 marks): One each for dispersion and refraction. Any options not mentioned above, do not carry either positive or negative credit.

HB

γ. Study the following image of the night sky (bigger version is printed in your answersheet) for mid-night on a certain day at a certain place and answer the questions below it. All answers must be marked / written on the answer-sheet.

8

CS

E

INAO – Jr – 2010

(a) (2 marks) Mark all the four directions on the map.

(b) (1 mark) Is this place in northern part of India or the southern part? (c) (3 marks) Mark and name 2 constellations each from the following two lists:

HB

• 0.5 marks each; any 2: Orion, Ursa Major, Taurus, Leo, Cygnus, Scorpio • 1 mark each; any 2: Hydra, Corvus, Aquarius, Cancer, Canis Major, Aries

(d) (2 marks) Sketch rough band of ecliptic i.e. apparent path of the Sun, the Moon and all the planets in the sky. (e) (2 marks) In which month the sky will appear like this at this time? Give reason in one line. Solution: (a) From Top clockwise: East, North, West, South. Note the position of Ursa Minor Constellation (i.e. Pole star) to the right of the map. Further, Taurus is at the bottom and Leo is at the top. (0.5 marks per direction) If the direction pointers are off by small angle (less than 10◦ ), overall -0.5 marks. (b) Ursa Minor is very close to Horizon. Hence the latitude of the place is not very high. Thus, the place is in Southern India. (-0.5 if wrong reason is mentioned.) (c) Cygnus, Scorpio, Aquarius are not on the map. Rest are clearly visible.

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INAO – Jr – 2010

E

(d) The band of ecliptic should pass through all the zodiac signs (i.e. Aries to Virgo). Better precision than this is not expected. For every zodiac sign significantly away from the ecliptic, -0.5 marks (minimum zero). (e) At mid-night, Cancer is nearly at Zenith. Hence the Sun is roughly 6 zodiac signs away i.e. in Capricorn. Thus, it will reach Vernal Equinox (in Pisces) in roughly two months. Hence, current date is about two months before 21st March. Thus, the current month is January. 0.5 marks for getting month (December February accepted). Reason 1.5 marks. δ. Mayank visited a place located at latitude φ and longitude 82.5◦ E on 21st June. He observed that at local noon, shadow of a one meter stick standing vertically on the ground was 26.8 cm long due south.

CS

(a) (5 marks) Find latitude of the place.

(b) (5 marks) Find the day on which the shadow of this stick at the local noon will be longest and find length and direction of that shadow. Note: √ sin−1 (0.268) = 15.5◦, cos−1 (0.268) = 74.5◦, tan−1 (0.268) = 15.0◦ , tan(2◦ ) ≈ 1 and 3 = 1.73 30 Solution: Shadow is 0.268 m due South on the day of the Summer solstice. This means the Sun is to the North of Zenith, i.e. the place is south of tropic of cancer i.e. in southren India. (1 mark) On that day, the Sun is directly above the tropic of cancer, i.e. 90◦ −23.5◦ = 66.5◦ away from the North Celestial Pole. (1 mark) If the altitude of the Sun at the local noon is θ, 1 1 = 0.268 tan(15◦) ◦ = tan(75 ) ∴ φ = 75◦ − 66.5◦ ∴ φ = 8.5◦ (3 marks)

HB

tan θJun =

In part (b) For a northern hemisphere place, the longest shadow of the stick will be cast on the Winter Solstice day. The Sun at local noon will be 47◦ further south than its position on the Summer Solstice day. Hence the Sun will be 180◦ − 75◦ − 47◦ = 58◦ high on the South of the Zenith. (2 marks)

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INAO – Jr – 2010

1 tan 58◦ = tan 32◦ = tan(30◦ + 2◦ ) tan 30◦ + tan 2◦ = 1 − tan 30◦ tan 2◦ √1 + 1 30 3 ≈ 1 1 − √13 × 30 √ 30 + 3 √ ≈ 30 3 − 1 31.73 31.73 ≈ ≈ 51.9 − 1 51 ≈ 0.62m

CS

E

l =

The Shadow will be 0.62 m long due North.

(3 marks)

ǫ. Sketch approximate graphs for the following situations:

(a) (4 marks) See the figure below. A tank of water (height of water column b) is kept on a electronic weighing scale. A metal cube (side a and density ρ) is hung from a spring balance and the spring balance is slowly lowered into the tank till the cube reaches the bottom of the tank. The distance of separation between the bottom of the tank and bottom of the cube is denoted by h with initial value h0 .

HB

spring balance

cube

111111111 000000000 water 000000000 111111111 000000000 tank111111111 b 000000000 111111111 000000000 111111111 a

h

00.00

weighing scale

Sketch the graph of reading on the spring balance as a function of h.

(b) (3 marks) For the situation above, sketch the graph of reading on the electronic scale as a function of h. (c) (2 marks) For the situation above, sketch the graph of sum of the reading on the electronic scale and the reading on the spring balance as a function of h.

(d) (3 marks) For a typical primary mirror used in Newtonian telescope, sketch a

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INAO – Jr – 2010

graph of object distance, u versus image distance, v. All physically measurable lengths should be taken as positive. In all cases, mark the significant points on the graph and give their coordinates.

E

Solution:

w

w a 3ρ

a 3ρ

a 3 (ρ−1)

CS

(0,0)

a3

(b−a)

b

h0

(0,0)

h

(b−a)

b

h0 h

V

w a 3ρ

2f f

HB

f 2f

(0,0)

h0

u

h

• Partial marks given on basis of shape of the curve, neatness of figure, proper nomencleture (axis titles etc.), proper markingof points and writing their co-ordinates etc. • In (a), (b) and (c), it is acceptable to take mass of water as m instead of zero. No points cut.

• The y-coordinates can be either in mass units or in force units (multiplication by g). No points cut. • Density of water can be written as ρw or its CGS values (i.e. 1) or its MKS value (i.e. 1000). No points cut.

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INAO – Jr – 2010

HB

CS

E

• In (d), the curve between u = 0 and u = f may be plotted in both quadrant 1 and quadrant 2. No points cut.

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INAO – Jr – 2010

HB

CS

E

Space for Rough Work

prepared using LATEX2ǫ

Indian National Astronomy Olympiad – 2010 Senior Category

Roll Number:

Roll Number

Question Paper Date: 30th January 2010 Maximum Marks: 100

INAO – 2010 Duration: Three Hours

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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 (6 sides).

CS

• In Section A, there are 10 multiple choice questions with 4 alternatives out of which only 1 is correct. You get 3 marks for each correct answer and -1 mark for each wrong answer. • In Section B, there are 4 multiple choice questions with 4 alternatives each, out of which any number of alternatives may be correct. You get 5 marks for each correct answer. No marks are deducted for any wrong answers. You will get credit for the question if and only if you mark all correct choices and no wrong choices. There is no partial credit. • For both these sections, you have to indicate the answers on the page 2 of the answer sheet by putting a × in the appropriate box against the relevant question number, like this: Q.NO. (a) (b) (c) (d) Q.NO. (a) (b) (c) (d) 22

 ⊠  

OR

35

⊠ ⊠  

Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

HB

• In Section C, there are 5 analytical questions totaling 50 marks. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Useful Physical Constants ME RE M⊙ R⊙ Rm c 1 A. U. G cs ǫ

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

5.97 × 1024 kg 6.4 × 106 m 1.99 × 1030 kg 7 × 108 m 1.7 × 106 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 330 m/s 23.5◦

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Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Speed of Light Astronomical Unit Gravitational Constant Speed of Sound in Air Inclination of the Earth’s Axis

HB

CS

Space for Rough Work

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – Sr – 2010

Section 1:Multiple Choice Questions Part A: (10 Q × 3 marks each)

1. As seen from the Earth, the stars appear to twinkle, while the planets do not because, A. Light coming from the stars gets absorbed by interstellar dust.

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B. the stars are self luminous while the planets merely reflect the light. C. Angular sizes of the stars are much smaller than the planets. D. All the above.

CS

A wooden cube of length 5 units is painted on all faces and then cut in 85 smaller cubes of varying sizes. The sides of each of the smaller cubes are some integer number of units. All bigger cubes are carved out of the corners of the original cube. Answer the following three questions: 2. How many cubes have no side painted? A. 15

B. 17

C. 20

D. 27

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Solution: If you cut the bigger cube in smaller cubes with all side of 1 unit, there will be 125 total cubes. Each cube of size 2 will fuse 8 such smaller cubes together. i.e. the total number of cubes will reduce by 7. Similarly, each cube of size 3 and 4 will reduce total number of cubes by 26 and 63 respectively. Now, we have 85 total cubes, which is 40 less than 125. This means there is 1 cube of size 3 and 2 cubes of size 2 at three of the corners. Rest all are size 1. If there were 125 smaller cubes of size 1, one can say the volume which is not touched by paint on any face is equivalent to a cube of size 3 i.e. 27 smaller cubes. However, the the bigger cubes carved out of corners include some of these inner cubes as well. 8 such inner smaller cubes are included in the size 3 cube and 1 each is included in the 2 size 2 cubes. Thus, number of unpainted cubes = 27 − 8 − 1 − 1 = 17

3. How many cubes have exactly 2 sides painted? A. 24

B. 27

C. 30

D. 36

Solution: To have exactly 2 sides painted, the smaller cube should be along any of the edges but not at the corner. There are total 12 edges to the original cube. For each edge, there will be 3 cubes satisfying the condition if all cubes were of size 1. How ever, there is a size 3 cube, which includes 2 such smaller cubes each

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along 3 of the edges. Similarly, the 2 size 2 cubes will each include 1 such smaller cube each along 3 of the edges. Thus, total number of cubes = 12 × 3 − 2 × 3 − 1 × 3 − 1 × 3 = 24

A. 69

B. 64

C. 62

D. 50

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4. If one was to cut the original cube (of size 5 units each side) with a condition that all corners would be occupied by the cubes bigger than 1 unit size, what will be the smallest total of cube pieces possible?

CS

Solution: There are two possibilities which will satisfy this condition. Either all corners will have cubes of size 2 or else there will be 1 cube of size 3 and rest all corners will have cubes of size 2. There are total 8 corners. In first case total number of cubes will be = 125 − 8 × 7 = 69 In second case, total number of cubes will be = 125 − 7 × 7 − 26 = 50 5. It is easier to balance on a faster bicycle moving along a straight line than a slower one because of A. conservation of linear momentum

B. conservation of angular momentum

C. conservation of linear momentum and angular momentum D. conservation of energy

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6. On a nice sunny day in Chennai, Akshay saw a supersonic fighter plane flying parallel to the ground in the sky. As a student of Aeronautical Engineering, he knew off-hand the speed of the plane to be 1.25 Mach (i.e. 1.25 times the speed of sound in air). He could hear its sonic boom, 12 seconds after the plane flew directly overhead. What is the altitude of the plane? A. 3.7 Km

B. 4.0 Km

C. 5.3 Km

D. 6.6 Km

Solution: Half angle of the cone of the sonic boom is given by sin θ = vp is velocity of the plane and vs is the speed of the sound in air.

2

vs , vp

where

INAO – Sr – 2010

vs = 0.8 1.25vs 0.6 sin θ h = cos θ vp t h 1.25 × 330 × 12 4 × 15 × 330 3 6600 mt. = 6.6 km

sin θ = ∴ cos θ =

0.8 = 0.6 h = =

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tan θ =

A. 10 mm

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7. If you hold a magnifying glass of focal length 10 cm in the sunlight and place a piece of paper at its focus, you can burn a hole in the paper. What could be the size of this hole? B. 5 mm

C. 0.5 mm

D. 0.1 mm

Solution: If the magnification is given by M, Object distance, Object size, Image Distance, Image size by u, O, v&I respectively, M =

I ≈ 7 × 108

I ≈

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≈

v I = u O f 500c 0.1 × 7 × 108 500 × 3 × 108 7 × 10−3 mt. 15 0.5 mm

≈

8. During an earthquake, an earthquake monitoring Centre observed that transverse waves traveling with speed 4.5 km/s arrived at the centre 3 minutes after the longitudinal waves traveling at 8.2 km/s. Deduce the approximate distance to the epicenter. A. 60 km

B. 220 km

C. 660 km

D. 1800 km

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Solution:

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8.2t = 4.5 × (t + 180) 3.7t = 810 810 D = 8.2t = 8.2 × 3.7 = 2.22 × 810 D ≈ 1800km

A. 40

B. 60

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9. A certain person nicknamed “Enthu”, encountered an automatic staircase (i.e. escalator) at a shopping complex, which was moving upward at a constant rate. Just for the fun of it he decided to walk up this escalator at the rate of one step a second. Twenty steps brings him to the top. Next day he goes up at two steps a second and reaching the top in 32 steps. How many steps are there in the escalator? C. 64

D. 80

Solution: Let the escalator rise with the speed of ‘n’ steps per second. On first day, Enthu reaches the top after 20 manual steps i.e. in 20 seconds. On second day, Enthu reaches the top after 32 / 2 = 16 seconds. Thus, total number of steps, = = = = =

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20 × (n + 1) 4n ∴n ∴ Steps

16 × (n + 2) 32 − 20 = 12 3 20 × (3 + 1) 80

10. Two stars of masses M and 3M respectively are going around each other, inq near circular orbits, with period T . The separation between them is given by D = The value of k is,

A. 0.5

B. 1

C. 1.5

3

kGM T 2 . π2

D. 3

Solution: The Centre of mass of the system will be located at distance

4

D 4

from

INAO – Sr – 2010

the heavier mass. Centripetal force is provided by the mutual gravitation.

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D 2 G M (3M) = (3M) ω D2 4 D 4π 2 GM = D2 4 T2 GMT 2 D3 = 2 rπ 2 3 GMT D = π2

Section B: (4 questions × 5 marks each)

A. B. C. D.

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11. If we throw a ball in a shallow water tank, propagation velocity of ripples on surface of the water will depend on surface tension of the water depth of the water tank density of the water height from which the ball was dropped

Solution: Note: The propagation velocity will vary as per the depth of the water, provided the water is not too deep. However, the derivation for this effect includes concepts beyond the level of participants. Thus, the accepted solution for this question would be ’a’ and ’c’ must be selected, ’d’ must not be selected and ’b’ would be acceptable both ways.

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12. Amit decided to experiment with cannon ball by making it hollow and filling water inside it. He then punched few holes in it. After the ball was fired horizontally, he was expecting to see water jets coming out from some of the holes. Which of the following locations of holes will allow water jet to come out? A. Front, bottom and sides D. None of the holes

B. Back, top and sides

C. back and bottom

Solution: Horizontally, all particles (i.e. cannonball and water particles) are flying with the same velocity. Hence water cannot come out of front / back /side holes. Vertically, all particles are experiencing free fall. Thus, gravity cannot exert any additional pressure on water and neither can water be slower than the cannonball. Hence no water can come out from top or bottom.

13. In the progress of Astronomy over the ages, we find several instances of startling new observations changing our ideas about the Universe and lead to new theories. Listed below are some milestones in the history of Astronomy and observations which necessitated them. Pick the correct statement(s). 5

INAO – Sr – 2010

A. Ptolemy was aware of retrograde motion of planets when he gave his model of the solar system. B. Newton was aware of Kepler’s Laws when he gave his Law of Gravitation.

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C. Einstein felt need to modify Newton’s theory of Gravitation to explain the expansion of the Universe. D. Existence of Cosmic Microwave Background Radiation (CMBR) led to the creation of the Big Bang Theory.

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Solution: Ptolemy needed epicyclic model of the solar system as that was the only way to fit observed motions of the planets including the retrograde motion. Kepler’s laws enabled Newton to deduce r12 nature of the law of gravitation. Einstein gave General theory of relativity in 1916 and assumed the Universe to be non-expanding as per then norm. Hubble discovered the expansion of the Universe in late 20’s forcing Einstein to revisit his own theory. Existence of CMBR was one of the key predictions of the big bang theory. The theory was formulated before CMBR was discovered. 14. Two simple pendula are hung close to each other on a thin, rigid support and are allowed to oscillate independently in planes parallel to each other. They have periods 3 seconds and 7 seconds respectively with same amplitudes and are initially released from the opposite extreme positions of each other. At which of the following times the threads of the two pendula will be coplanar again? A. 1.05 seconds

B. 7.88 seconds

C. 10.50 seconds

D. 23.10 seconds

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Section C: Analytical Questions

α. An alien civilisation on a star far far away came to know about the Astronomy Olympiad Examination and wanted to test smartness of the students. They sent following two coded secret messages. Decode them. (a) (5 marks) First is a pictorial message in black and white colour. They sent it on radio waves in the form of ones and zeros. Find out what the picture says. 00000 00000 00000 00000 00000 00000 11111 01100 00100 00100 00011 10000 01000 10100 01000 10100 01101 10000 10001 00100 10001 11000 10001 00001 00010 00101 00100 01001 10110 01111 10100 00110 10000 01001 11000 00000 00000 00000 00000 00000 000

(b) (5 marks) Surprisingly, the aliens are also proficient in English and the following coded message is actually a sentence in English language. In the answer sheet, write down the coded sentence and also the true meaning of each alphabet in the code. “Up tpmwf uif qsfwjpvt tvcrvftujpo uijol pg uif nfttbhf tfou cz uif bsfdjcp ufmftdpqf up bmjfot.”

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Solution: In 1974, a pictographic message was sent to outer space by radio waves using Arecibo telescope. The total number ’digit’s in the message was product of two prime numbers. It was argued that any intelligent being can rearrange the grid in a rectangle with sides measuring these two prime numbers and then use blank square for ‘0’ and filled square for ‘1’ to reveal the picture.

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In the present message same idea is used. Total number of digits is 203 which is a semiprime number; i.e. it can only be divided as 29 × 7. If the grid is rearranged in this way, following pattern is seen:

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 ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠ ⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠⊠  Marking Scheme:

• Realising total number of digits are 203, which is a semi-prime number. (2 marks) • Attempt to draw either 29 × 7 or 7 × 29 grid. • picking the message “INAO”.

(1 mark) (2 marks)

HB

Part (b) is a substitution cipher. The word “uif” appears thrice in the sentence. Obviously, it stands for “the”. Looking at the two words, the substitution rule is inferred as “every letter is substituted by the next letter” to create the coded message. Deciphering, “To solve the previous subquestion think of the message sent by the arecibo telescope to aliens.” Marking Scheme: • Mapping the alphabets and deciphered Sentence.

(5 marks)

• Mapping the alphabets and Deciphered Sentence, but with small syntactic errors. (4.5 marks) • Mapping the alphabets only.

(4 marks)

• Deciphered Sentence but Alphabet mapping is not given.

(3 marks)

• Only partial decipherment with more than 10 letters.

(2 marks)

• Only partial decipherment with 7-10 letters. • Only partial decipherment with “the” deciphered correctly.

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(1.5 marks) (1 mark)

INAO – Sr – 2010

β. (8 marks) In the following table, the first column gives various optical phenomena / instruments and the top row gives various optical effects which may help in explaining them. In the answers sheet, tick the correct effect(s) involved in each phenomena in appropriate rows. Appreciable Dispersion

Internal Reflection Reflection

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Solution:

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Blue Sky Mirage Rainbow Smooth Convex Mirror Thick Concave Lens

Refraction Scattering

Blue Sky Mirage Rainbow Sm. Conv. Mirror Th. Conc. Lens

Appri. Disp.

Int. Refle.

X

X X

Refle.

Refra.

Scat. X

X X

X

X

X

• Blue Sky (1 mark): One mark for scattering. -0.5 for each additional tick mark.

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• Mirage (2 marks): One each for internal reflection and refraction. 0 marks if internal reflection is not ticked. • Rainbow (2 marks): One for internal reflection; 0.5 each for dispersion and refraction. 0 marks if internal reflection is not ticked. • Mirror (1 mark): One mark for reflection. 0 if any additional / other tick marks.

• Thick Lens (2 marks): One each for dispersion and refraction.

Any options not mentioned above, do not carry either positive or negative credit.

γ. Study the following image of the night sky (bigger version is printed in your answersheet) for mid-night on a certain day at a certain place and answer the questions below it. All answers must be marked / written on the answer-sheet.

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INAO – Sr – 2010

(a) (2 marks) Mark all the four directions on the map.

(b) (1 mark) Is this place in northern part of India or the southern part? (c) (3 marks) Mark and name 2 constellations each from the following two lists:

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• 0.5 marks each; any 2: Orion, Ursa Major, Taurus, Leo, Cygnus, Scorpio • 1 mark each; any 2: Hydra, Corvus, Aquarius, Cancer, Canis Major, Aries

(d) (2 marks) Sketch rough band of ecliptic i.e. apparent path of the Sun, the Moon and all the planets in the sky. (e) (2 marks) In which month the sky will appear like this at this time? Give reason in one line. Solution: (a) From Top clockwise: East, North, West, South. Note the position of Ursa Minor Constellation (i.e. Pole star) to the right of the map. Further, Taurus is at the bottom and Leo is at the top. (0.5 marks per direction) If the direction pointers are off by small angle (less than 10◦ ), overall -0.5 marks. (b) Ursa Minor is very close to Horizon. Hence the latitude of the place is not very high. Thus, the place is in Southern India. (-0.5 if wrong reason is mentioned.) (c) Cygnus, Scorpio, Aquarius are not on the map. Rest are clearly visible.

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INAO – Sr – 2010

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(d) The band of ecliptic should pass through all the zodiac signs (i.e. Aries to Virgo). Better precision than this is not expected. For every zodiac sign significantly away from the ecliptic, -0.5 marks (minimum zero). (e) At mid-night, Cancer is nearly at Zenith. Hence the Sun is roughly 6 zodiac signs away i.e. in Capricorn. Thus, it will reach Vernal Equinox (in Pisces) in roughly two months. Hence, current date is about two months before 21st March. Thus, the current month is January. 0.5 marks for getting month (December February accepted). Reason 1.5 marks. δ. Mayank visited a place located at latitude φ and longitude 82.5◦ E on 21st June. He observed that at local noon, shadow of a one meter stick standing vertically on the ground was 26.8 cm long due south.

CS

(a) (5 marks) Find latitude of the place.

(b) (5 marks) Find the day on which the shadow of this stick at the local noon will be longest and find length and direction of that shadow. Note: √ sin−1 (0.268) = 15.5◦, cos−1 (0.268) = 74.5◦, tan−1 (0.268) = 15.0◦ , tan(2◦ ) ≈ 1 and 3 = 1.73 30 Solution: Shadow is 0.268 m due South on the day of the Summer solstice. This means the Sun is to the North of Zenith, i.e. the place is south of tropic of cancer i.e. in southren India. (1 mark) On that day, the Sun is directly above the tropic of cancer, i.e. 90◦ −23.5◦ = 66.5◦ away from the North Celestial Pole. (1 mark) If the altitude of the Sun at the local noon is θ, 1 1 = 0.268 tan(15◦) ◦ = tan(75 ) ∴ φ = 75◦ − 66.5◦ ∴ φ = 8.5◦ (3 marks)

HB

tan θJun =

In part (b) For a northern hemisphere place, the longest shadow of the stick will be cast on the Winter Solstice day. The Sun at local noon will be 47◦ further south than its position on the Summer Solstice day. Hence the Sun will be 180◦ − 75◦ − 47◦ = 58◦ high on the South of the Zenith. (2 marks)

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1 tan 58◦ = tan 32◦ = tan(30◦ + 2◦ ) tan 30◦ + tan 2◦ = 1 − tan 30◦ tan 2◦ √1 + 1 30 3 ≈ 1 1 − √13 × 30 √ 30 + 3 √ ≈ 30 3 − 1 31.73 31.73 ≈ ≈ 51.9 − 1 51 ≈ 0.62m

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l =

The Shadow will be 0.62 m long due North.

(3 marks)

ǫ. Sketch approximate graphs for the following situations:

(a) (4 marks) See the figure below. A tank of water (height of water column b) is kept on a electronic weighing scale. A metal cube (side a and density ρ) is hung from a spring balance and the spring balance is slowly lowered into the tank till the cube reaches the bottom of the tank. The distance of separation between the bottom of the tank and bottom of the cube is denoted by h with initial value h0 .

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spring balance

cube

111111111 000000000 water 000000000 111111111 000000000 tank111111111 b 000000000 111111111 000000000 111111111 a

h

00.00

weighing scale

Sketch the graph of reading on the spring balance as a function of h.

(b) (3 marks) For the situation above, sketch the graph of reading on the electronic scale as a function of h. (c) (2 marks) For the situation above, sketch the graph of sum of the reading on the electronic scale and the reading on the spring balance as a function of h.

(d) (3 marks) For a typical primary mirror used in Newtonian telescope, sketch a

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graph of object distance, u versus image distance, v. All physically measurable lengths should be taken as positive. In all cases, mark the significant points on the graph and give their coordinates.

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Solution:

w

w a 3ρ

a 3ρ

a 3 (ρ−1)

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(0,0)

a3

(b−a)

b

h0

(0,0)

h

(b−a)

b

h0 h

V

w a 3ρ

2f f

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f 2f

(0,0)

h0

u

h

• Partial marks given on basis of shape of the curve, neatness of figure, proper nomencleture (axis titles etc.), proper markingof points and writing their co-ordinates etc. • In (a), (b) and (c), it is acceptable to take mass of water as m instead of zero. No points cut.

• The y-coordinates can be either in mass units or in force units (multiplication by g). No points cut. • Density of water can be written as ρw or its CGS values (i.e. 1) or its MKS value (i.e. 1000). No points cut.

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HB

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• In (d), the curve between u = 0 and u = f may be plotted in both quadrant 1 and quadrant 2. No points cut.

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HB

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Space for Rough Work

prepared using LATEX2ǫ

Indian National Astronomy Olympiad – 2011 Question Paper

Roll Number:

Roll Number

Date: 29th January 2011 Maximum Marks: 100

INAO – 2011 Duration: Three Hours

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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 4 pages (8 sides).

CS

• In Section A, there are 10 multiple choice questions with 4 alternatives out of which only 1 is correct. You get 3 marks for each correct answer and -1 mark for each wrong answer. • In Section B, there are 4 multiple choice questions with 4 alternatives each, out of which any number of alternatives may be correct. You get 5 marks for each correct answer. No marks are deducted for any wrong answers. You will get credit for the question if and only if you mark all correct choices and no wrong choices. There is no partial credit. • For both these sections, you have to indicate the answers on the page 2 of the answer sheet by putting a × in the appropriate box against the relevant question number, like this: Q.NO. (a) (b) (c) (d) Q.NO. (a) (b) (c) (d) 22

 ⊠  

OR

35

⊠ ⊠  

Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

HB

• In Section C, there are 6 analytical questions totaling 50 marks. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Useful Physical Constants RE M⊙ R⊙ c G ǫ g h ℏ = 2π Na MH

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

6.4 × 106 m 2 × 1030 kg 7 × 108 m 3 × 108 m/s 6.67 × 10−11 m3 /(Kg s2 ) 23.5◦ 10 m/s2 10−34 J.s 6.023 × 1023 mol−1 1.008 a.m.u.

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Radius of the Earth Mass of the Sun Radius of the Sun Speed of Light Gravitational Constant Inclination of the Earth’s Axis Gravitational acceleration Reduced Planck constant Avogadro constant Atomic mass of Hydrogen

HB

CS

Space for Rough Work

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – 2011

Section 1:Multiple Choice Questions Part A: (10 Q × 3 marks each)

A. 63 min

C. 127 min

D. 171 min

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Solution:

B. 109 min

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1. On one starry evening, Nidhi was trying to spot an artificial polar satellite from her backyard. Typical altitude of any polar satellite is about 800km above surface of the earth. What is the typical duration after sunset for which Nidhi should try her luck?

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Height of the orbit of polar satellite is 800 Km. After Sunset, the Sunlight will reach satellite for a time (θ/ω), where ω is the angular velocity of the earth and θ is as shown in the diagram. ω = 15 deg/hr   R −1 θ = cos R+h   6.4 × 106 −1 ∴ θ = cos 6.4 × 106 + 8 × 105   8 ∴ θ = cos−1 = cos−1 (0.889) 9 √ ! 3 = 30◦ θ . cos−1 (0.866) = cos−1 2 Thus the angle is slightly less than 30◦ . Thus for this angle, time will be slightly less than 2 hours.

2. What will be the difference in potential energy (∆U) of an object of mass ‘M’, if it is lifted from the ground to a height of 2R, where R is the radius of the earth? 2GM 2GM GM GM A. B. C. D. R 3R 2R 3R 1

INAO – 2011

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Solution: Let U1 be the potential energy of the object on the ground, U2 be the potential energy at the height 2R and ∆U be the change in the potential energy   GM U1 = − R   GM U2 = − 3R ∴ ∆U = U2 − U1     GM GM + ∴ ∆U = − 3R R   2GM ∴ ∆U = 3R

CS

Note: Implicit assumption is that the mass is scaled in earth mass units. Since the assumption was not explicitly stated, the question was deemed ambiguious and was dropped from evaluation. 3. Pole star appears stationary because.....

A. Earth is not moving with respect to the pole star. B. Earth is on the axis of rotation of the pole star.

C. Both Earth and the pole star have same velocity in the Milky Way galaxy. D. None of the above.

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Solution: Pole star appears stationary because it is almost along the axis of rotation of the earth. 4. Consider a system of two converging lenses, one with focal length of 20cm and the other with focal length of 5cm, kept 50cm apart. An object is kept at 40cm from the first lens. What can be said about the image formed on the other side of the second lens? A. Erect and Real

B. Inverted and Real

C. Erect and Virtual

D. Inverted and Virtual

Solution: If the object is kept at a distance of 2f, image will also form at the distance of 2f and that will be real and inverted. The distance between two lenses is arranged in such a way that the image from the first lens forms at 2f of the second lens.

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Thus again a real image of this image will be formed at distance 2f on the other side of second lens and it will invert the inverted image. Thus, final image will be an erect one.

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5. Three rings of same dimensions, are dropped at the same time over identical cylindrical magnets as shown below. The inner diameter of each ring is greater than the diameter of the magnet.

Which of the following correctly describes the order in which the rings P, Q and R reach the bottom of the respective magnets? A. They arrive in the order P, Q, R

B. They arrive in the order P, R, Q C. Rings P and R arrive simultaneously, followed by Q.

D. Rings Q and R arrive simultaneously, followed by P.

Solution: Plastic is not a conducting material so its motion will not be affected. When ring Q will drop over the magnet, due to mutual induction eddy currents will form which oppose the downward motion of copper ring so it will take longer time to reach at bottom of magnet. Since R is not complete circular ring, circuit can not be completed but small local loops of eddy current still form in the ring so it will reach to bottom of magnet after ring P but before ring Q.

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D. None of the above

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→ − 6. A charged particle with initial velocity V enters a region with a uniform magnetic → − field B = Bbi. If it starts moving along the positive X-axis in a helical path such that → − the separation between successive loops is constant, what can be inferred about V ?. → − A. V =V b j → − B. V = −V b k → − C. V =Vyb j + Vz b k

CS

→ − → − − → Solution: Lorentz’s force is given by F =q( V × B ) → − If initial velocity vector is restricted to only the plane perpendicular to B , particle would have only circular motion since particle is also progressing along X-axis it must have parallel component along this direction. So initial velocity should be → − V = Vxbi + Vyb j + Vz b k, where Vx 6= 0 and at least one out of Vy and Vz 6= 0 7. If P QRS × 4 = SRQP , where P, Q, R and S are distinct non-zero digits. what is value of R? A. 1

B. 3

C. 5

D. 7

HB

Solution: P should be even and 4P < 10, hence P = 2. ⇒ S is 3 or 8 and 4P ≤ S ⇒S=8 ⇒ Q and R are odd and 4Q < 10 ⇒Q=1 ⇒R=7

8. From the given P-V diagram, find out the total work done by the gas, while going from state A to state C.

A. Wtot = WAC

B. Wtot = WBC − WAB

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C. Wtot = WAB + WBC D. Wtot = WAB − WBC

E

Solution: From P-V diagram Work done is = Area under the curve ∴ Total work done by the gas = Wtot = WAB + WBC The work done would be differnce in the areas under the curve. However, in case of work done, the negative sign is implicit in WBC .

A. 10Ω

CS

9. Find out the equivalent resistance at AB from given circuit, if R=10Ω

B. 100Ω

C. 5Ω

D. 6Ω

HB

Solution: We can simplify given circuit as follows

∴ Rres =6Ω

10. How many 3 digit prime numbers can be formed, using digits 5, 6 and 7? Repetition of digits is allowed. A. 20

B. 4

C. 7

D. 6

Solution: since we want it to be a prime number, last digit must be 7. Also one can repeat same digits. Hence, there are 9 possibilities: 557, 567, 577, 657, 667, 677, 757, 767, 777 Since 5 + 6 + 7 = 18 , 567 and 657 both numbers are divisible by 3. Clearly 777 is divisible by 7, so we have to check remaining 6 numbers out of which 667 is divisible by 23 and 767 is divisible by 13. Remaining 4 are prime numbers.

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Section B: (4 questions × 5 marks each)

CS

E

11. Four conducting plates A,B,C,D are arranged as shown in the figure. Plates A and C are connected to the positive terminal of a DC source and the Plates B and D are connected to the negative terminal of a DC source. A proton is kept, at the centre of this assembly. If we disturb the proton slightly from its equilibrium position, which of the following statements will describe the path followed by the proton.

A. If the proton is displaced slightly towards plate A, it will keep moving towards plate A. B. If the proton is displaced slightly towards plate D, it will keep moving towards plate D. C. If the proton is displaced slightly along diagonal of the assembly between plates B and C, it will move directly towards plate B.

HB

D. If the proton is displaced slightly towards plate C, it will come back to its original position.

Solution: The proton plate A and plate C are positively charged, hence if the proton is displaced towards plate A, then it will get repelled and return to its equilibrium position i.e. the centre of the assembly. Conversely, plate B and plate D are negatively charged, so if the proton is displaced towards them it will get attracted. Now if the proton is displaced diagonally, positive plates will repel it and negative plates will attract it. Thus it will move towards the closer negative plate. However, the motion will also have a harmonic oscillator component in direction of positive plates hence the overall motion may not be termed as “direct”.

12. A solid copper sphere is kept on an insulating stand. A charge given to it gets distributed uniformly on its surface only. Which of the following factors is/are relevant to this observation? 6

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A. Copper is a conducting material. B. Shape of the conductor is a sphere. C. Like charges repel each other.

E

D. Potential energy of the system is minimum in this configuration.

CS

Solution: In a perfect conductor, like charges are free to take up equilibrium positions in response to the Coulomb repulsion between them. Sphere being symmetric, there will not be any accumulation of charges at any point and hence there will be uniform distribution over the surface of the sphere, This is the energetically most favorable distribution of the charge. Hence the potential energy will be minimum in this case. Since the material of the sphere is a good conductor, all charges will only reside on the outer surface, whether the interior is hollow or solid. 13. A block of mass 5 kg is initially at rest on a rough horizontal surface having coefficient of static friction µs = 0.5 and coefficient of kinetic friction µk = 0.3. A gradually increasing horizontal force is applied for dragging it. Assuming g = 10 m/s2 , acceleration of the block and dragging force acting on the block could be respectively given by, A. 7 m/s2 , 50 N B. 2 m/s2 , 25 N C. 0 m/s2 , 20 N D. 3 m/s2 ,40 N Solution: Initially the mass is stationary therefore, the frictional force acting on it is Fs = µs mg i.e. 25 N. ∴ if the external horizontal force applied is smaller than Fs , then the acceleration produced is zero. Hence, choice 3 i.e. 0, 20 is correct.

HB

Similarly, if the applied horizontal force is greater than the frictional force then the acceleration produced can be calculated as follows: ma = Fh − Fk For Fh = 50N, a = 7m/s2 For Fh = 40N, a = 5m/s2

exactly Fh = 25N, we have to consider µs and not µk . 2 ∴Now a = at 0m/s

14. Two stars are seen close to each other in the sky. Star A appears brighter than Star B. Which of the following statements satisfactorily explain the difference in their observed brightness? Assume both the stars to be perfect black bodies. A. Both the stars are identical except for the fact that star A is closer to us than star B. B. Both the stars are at same distance, but star A appears yellow, where as star B appears orange. 7

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C. Both the stars are identical except for the fact that star A has smaller radius than star B. D. Both the stars are identical except for the fact that star A is less massive than star B.

CS

E

Solution: If both the stars are identical in mass and age, and if star A is closer to us than star B, then its apparent brightness will be greater than star B. Now if both the stars are at same distance and have same mass, then star A will appear brighter if it is intrinsically bright i.e. it is hotter (blue is hotter than yellow) than star B. the size and the mass of the star does not explicitly explain the brightness of a star. A star B having smaller radius than star A does not specify that star B will be brighter than star A. Similarly a more massive star need not always be brighter than a low mass star.

Section C: Analytical Questions α. (8 marks) In the following table, the first column gives the names of various bright stars in the sky and the top row gives the names of some zodiacal constellations. In the answers sheet, tick mark the constellation to which they may belong. Wrong tick marks carry negative points. Aries

Taurus

Gemini

HB

Star Name Aldebaran Antares Castor Denebola Hamal Pollux Regulus Sirius Spica

Leo

Virgo Scorpio

Solution:

Star Name Aldebaran Antares Castor Denebola Hamal Regulus Sirius Spica Pollux

Aries

Taurus X

Gemini

Leo

Virgo Scorpio X

X X

X

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X X X

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Marking scheme: • One mark for each correct answer. • -0.5 for each wrong answer.

E

• Sirius should be left blank. No marks for leaving it blank. Negative marks for putting a tick mark in that row. • If more than one constellations are ticked for same star, it is counted as wrong answer. • total 8 marks.

CS

β. (8 marks) Prof. Subramanium Chandrasekhar was first to suggest that the white dwarf stars will havean upper 2 limit on their mass, which is given by
√ 1 ℏc b Mlimit = k aπ G µe mH where ‘a’, ’µe ’ and ‘k’ are dimensionless integers, with k ≈ 1 and µe ≈ 2 is called mean number of nucleons per electron. mH is the mass of one hydrogen atom. This is famously known as ‘Chandrasekhar Mass Limit’ for which he won Nobel Prize in 1983. Find ‘a’ and ‘b’. Solution: ’b’ can be found by dimensional analysis as follows (3 marks for getting correct b)

2 1 1.44M⊙ = k aπ µe mH !b J · s × m/s 1 kg = N m2 (kg)2 kg 2 

HB

√

ℏc G

b 

kg · m2 kg · m and J = s2 s2  b m2 ·kg · s × m/s 1 s2   ∴ kg = kg·m 2 m (kg 2 )2 s2

now, N =

kg 2

⇒ b = 3/2

Similarly ’a’ can be found by substituting the values of all the constants given

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and equating it to 1.44M⊙ , which is the famous Chandrasekhar mass limit. = = √ √ √ √

aπ ≈ aπ ≈ aπ ≈ ≈ ≈ ≈ ≈

CS

aπ aπ a ⇒a

2  10−34 × 3 × 108 3/2 6.023 × 1026 ) aπ( 6.67 × 10−11 2 × 1.008 √ 36 × 1052 1 3/2 aπ( ) × 2.22 × 1015 4 30 1.44 × 2 × 10 × (2.22 × 1015 )3/2 9 × 1052 0.16 × 2 × (2.22)3/2 × 10−22 × 1022.5 q (0.32)2 × (2.22)3 × 10 √ 0.1024 × 10.7 × 10 10.9 3.47 3 √

(1mark)

E

1.44 × 2 × 10

30

Marking scheme:

• All 4 marks for getting correct answer.

• 3 marks for close enough integer answers.

• 2 marks for getting only correct order of magnitude (i.e. 0). • 0.5 marks deducted if final answer is not an integer. • +1 mark for overall clarity of solution.

HB

γ. (8 marks) Hot solar plasma is emitted from surface of a circular sunspot whose diameter is 10,000 km. When the plasma reaches the height of 16,000 km above the surface of the sun its horizontal cross section is measured to have diameter of 90,000km. Assuming that the edge of the plasma cone is parabolic, find the depth inside the sun from which the plasma started. Assume that the viscosity and magnetic permeability remains same inside and outside the solar surface.

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Solution: r2 (3marks) (r + h)2 r2 = (r + h)2 r = (r + 16000) = 8000km (3marks)

E

=

CS

d1 d2 1 ∴ 9 1 ∴ 3 ∴r

(2 marks) Marking scheme:

• Wrong parabola (x2 = 4ay) considered. Deduct 4 marks.

HB

• Linear terms in y included in the equation, grading as per merit of justification.

δ. (8 marks) Vinita studied a star for 55 days in succession. She noted down the temperature of the star everyday, which was varying in a nice symmetric manner. The data of her observations is given below. Help Vinita to find the mean temperature and the period of temperature variation of this star by any suitable method. Give proper justification for the method used.

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Temperature 5312 5264 5250 5273 5328 5400 5472 5527 5550 5536 5488 5419 5345 5284 5253 5257 5297 5363

HB

Solution:

Days 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

Days 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

Temperature 5437 5503 5543 5547 5516 5455 5381 5312 5264 5250 5273 5328 5400 5472 5527 5550 5536 5488

E

Temperature 5472 5527 5550 5536 5488 5419 5345 5284 5253 5257 5297 5363 5437 5503 5543 5547 5516 5455 5381

CS

Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

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E

Maximum values Minimum values Days Temperature Days Temperature 3 5550 9 5253 16 5547 22 5250 28 5550 34 5253 41 5547 47 5250 53 5550

CS

after exactly 25 days you get same maximum temperature and there is one peak in between. so the period is 25/2 = 12.5 days. from the data maximum ≈ 5550 minimum ≈ 5250 mean = 5400o C since data contains some incomplete period, mean of all 55 readings will give incorrect answer. Since data is symmetric, mean by merely finding peaks and taking average is reasonably correct. Marking Scheme:

• Period value: 2 marks

• Period justification: 1 mark • Mean value: 2 marks

• Correct method for finding mean: 3 marks

• If graphical method is used, upto 2 marks for correct drawing of graph.

HB

• No credit for averaging over all 55 readings.

ǫ. (9 marks) On one fine day, Akshay was watching DTH television from Madurai in Tamil Nadu (78◦ 07′ E; 9◦ 48′ N). He got a call from his IIT friend, Sujeet, who was watching DTH television of the same company from Salem in Tamil Nadu (78◦ 07′ E; 11◦ 39′ N). Both were getting their DTH signals from the same satellite located at 36000 km directly above a point at the same longitude but at latitude of 10◦ 43.5′ N. Find the angle difference in the antenna pointing for Akshay and Sujeet.

Solution:

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CS

(3 marks) Let x be the angle of the antenna with respect to horizon. Radius of the earth (r) = 6400 km. l ≈ rθ ≈ dφ x ≈ 90 − φ rθ ≈ 90 − d 1.85 6400 × ≈ 90 − 36000 2 64 1.85 ≈ 90 − × 200 36 ≈ 90 − 0.32 × 0.52 1 ≈ 90 − 6 ◦

HB

The antenna at Salem will be pointing 16 south of local zenith and antenna at ◦ ◦ Madurai 61 north of local zenith. The secular angle difference is 13 or 20′ (4 marks). The total angle difference (i.e. difference in angles measured w.r.t. local horizons) would be 20′ + 1◦ 51′ = 2◦ 11′ (2 marks)

ζ. Sketch the graphs of following functions in the space provided on the answersheet (Plotting on a graphsheet is not expected): (a) (3 marks) |x + 1| + |x − 1|

(b) (3 marks) x + sin(x). (c) (3 marks) xlog(x).

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HB

CS

E

Solution:

Criterian Shape Marking scheme: Values Slope Concept

Graph 1 Graph 2 1 2 1 – 1 – – 1

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Graph 3 1 2 – –

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HB

CS

E

Space for Rough Work

prepared using LATEX2ǫ

Indian National Astronomy Olympiad – 2012 Question Paper

Roll Number:

Roll Number

Date: 28th January 2012 Maximum Marks: 100

INAO – 2012 Duration: Three Hours

E

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 4 pages (8 sides).

CS

• In Section A, there are 10 multiple choice questions with 4 alternatives out of which only 1 is correct. You get 3 marks for each correct answer and -1 mark for each wrong answer. • In Section B, there are 2 multiple choice questions with 4 alternatives each, out of which any number of alternatives may be correct. You get 5 marks for each correct answer. No marks are deducted for any wrong answers. You will get credit for the question if and only if you mark all correct choices and no wrong choices. There is no partial credit. • For both these sections, you have to indicate the answers on the page 2 of the answer sheet by putting a × in the appropriate box against the relevant question number, like this: Q.NO. (a) (b) (c) (d) Q.NO. (a) (b) (c) (d) 22

 ⊠  

OR

35

⊠ ⊠  

Marking a cross (×) means affirmative response (selecting the particular choice). Do not use ticks or any other signs to mark the correct answers.

HB

• In Section C, there are 5 analytical questions totaling 60 marks. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Useful Physical Constants ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

6 × 1024 kg 6.4 × 106 m 2 × 1030 kg 7 × 108 m 1.7 × 106 m 3.84 × 108 m 3 × 108 m/s 1.5 × 1011 m 6.67 × 10−11 m3 /(Kg s2 ) 23.5◦ 10 m/s2

E

ME RE M⊙ R⊙ Rm dm c 1 A. U. G ǫ g

CS

Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Distance to the Moon Speed of Light Astronomical Unit Gravitational Constant Inclination of the Earth’s Axis Gravitational acceleration

HB

Space for Rough Work

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – 2012

Section 1:Multiple Choice Questions Part A: (10 Q × 3 marks each)

1. If the square of your age in seconds gives the age of the Universe in seconds, then B. you are in primary school C. you are a young adult

E

A. you haven’t started to walk yet

D. you were born before British left India

CS

Solution: Age of the Universe is about 14 billion years. If your age in years is Y , then, (Y × 365.25 × 86400)2 = 14 × 109 × 365.25 × 86400 14 × 109 ∴Y2 = 365.25 × 86400 14 ≈ × 103 ≈ 500 28 √ 500 ≈ 22 years ∴Y ≈ Thus, your age is in early 20s.

2. If f (x + y) = f (x)f (y) and f (2) = 5, then the value of f (−2) is A. 5

B. 1

Solution:

C. 0.25

D. 0.2

HB

Solution 1: f (4) = f (2)f (2) = 25 and f (2) = f (4 − 2) = f (4)f (−2) = 25f (−2) f (2) 5 Thus: f (−2) = = = 0.2 25 25

Solution 2: f (2) = f (2 + 0) = f (2)f (0) implying f (0) = 1 Thus, f (0) = f (2 − 2) = f (2)f (−2) 1 f (0) = = 0.2 Giving, f (−2) = f (2) 5

3. A container with uniform rectangular cross-section and weight M kg, falls from a cargo ship into the sea and is floating with x part of its height under water (x is a fraction less than 1). Two persons from a nearby boat, each weighing m kg board on

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the container resulting it to go down such that the top surface of the container levels in terms of x. exactly with the water level. Find the ratio M m A.

(1 − x) 2x

B.

x (1 − x)

C.

(1 − x) x

D.

2x (1 − x)

E

Solution: If V is volume of the container,

CS

M = xV & M + 2m = V M ∴ M + 2m = x 2mx = M(1 − x) 2x M = ∴ m (1 − x)

4. An electron moving uniformly in space is neither deflected nor accelerated over a long distance. Which of the following statements may describe the local conditions? A. B. C. D.

Both electric and magnetic fields are necessarily zero simultaneously. The electric field is necessarily zero. The magnetic field is necessarily zero. Neither of them are necessarily zero.

Solution: The Lorentz force equation is given by, ~ + ~v × B) ~ F~ = q(E

HB

We require F~ to be zero. The electric and magnetic fields can be so adjusted that their net effect cancels out, and the electron continues its path.

5. Pradip once observed image of the Sun on the platform of CST station (Mumbai’s main railway station). He realised that the sunrays forming the image are entering through a tiny triangular shaped window in the high ceiling. He measured the diameter of the image to be 0.175 m. Find the height of the ceiling. A. 12.5 m

B. 18.75 m

C. 22.5 m

D. 37.5 m

Solution: h = height of the ceiling; Dse = 1.5 × 1011 m ;w = diameter of image; d = 14 × 108 = diameter of sun. Dse h ≈ w d

1.5 × 1011 14 × 108 h ≈ 18.75m h ≈ .175 ×

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6. The most energy efficient direction of projection of rockets from the earth surface is A. Eastwards

B. Westwards

C. Northwards

D. Vertically upwards

E

Solution: The earth rotates from west to east. If the rocket is projected in a eastward direction, the earth’s motion acts as a boost for the projection. 7. A car travels the first 20 km eastwards with 30 km/hr, next 20 km northwards with 40 km/hr and then 20 km westwards with 50 km/hr. The average velocity of the car for the journey is 600 400 10 km/hr C. km/hr D. km/hr A. 10 km/hr B. 9 47 47

CS

Solution: Total displacement is 20 km only.

total displacement total time 20 20 20 totaltime = + + 30 40 50 47 hr = 30 20 Avg.Velocity = 47 Avg.Velocity =

30

600 = 47

HB

8. The Sun is found to be setting exactly at 6.00 pm on a given day. If the Earth’s atmosphere was only half dense as it is then the sunset would have occurred A. Slightly later than 6.00 pm

B. Slightly earlier than 6.00 pm C. Exactly at 6.00 pm

D. It depends on the latitude of the place.

Solution: We “see” the sunset a few minutes after geometric sunset as the solar disk remains visible even after going below horizon due to refraction. If the atmosphere is rarer, the sunrays will get refracted less. Thus, this additional period of visibility of solar disk is reduced. Hence, we will “see” sunset slightly earlier.

9. A knock-out tennis tournament begins with a total of 193 players. In each round, if the number of players is odd, then one player gets direct bye to the next round. No

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player can get byes in two consecutive rounds. How many matches must be played in the tournament before the winner is decided? A. 386

B. 192

C. 169

D. 97

E

Solution: The only way a player can go out of the tournament is by losing. Winner is a single player; thus, 193 − 1 = 192 players must lose. Implying 192 matches.

A. 45◦

B. 60◦

CS

10. Four metal rods all of identical dimensions and made of same material are welded together at a single point. The configuration is such that any two rods are oriented at 120◦ . The far ends of three of the rods are maintained at 90◦ temperature and that of the fourth rod is maintained at 30◦ temperature. The temperature of the junction point is C. 75◦

D. None of these.

Solution: For equilibrium condition, heat flow in and out of junction should be same. Let k represent some constant representing heat content, which may depend on dimensions and material of the rods. k(T − 30) = 3k(90 − T ) 4T = 300 T = 75

Section B: (2 questions × 5 marks each)

HB

11. Consider the following statements:

• A central eclipse is the one where central point of lunar disk exactly passes over the central point of solar disk.

• An eclipse is called a total Eclipse if it is seen as total from at least some point on the earth. • An eclipse is called a partial Eclipse if it is not seen as total / annular from any point on the earth.

Now choose the incorrect statement/s from below: A. All central eclipses are total. B. All total eclipses are central. C. All partial eclipses are non-central

D. All non-central eclipses are partial

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E

Solution: First one is incorrect as some central eclipses can be annular only. Second one is incorrect as one can get a total eclipse even when the two disks are slightly off-center w.r.t. to each other. Third one is correct. Fourth one is incorrect for the reason explained above regarding the second option. 12. Consider a pendulum with inextensible string with a lightweight magnetic bob. Underneath this arrangement iron dust is spread out. Now the bob of the pendulum starts swinging. Which of the following might be a susequent observation(s)? A. The maximum height reached by the bob will start increasing. B. The tension in the string will start increasing.

CS

C. The period of oscillation of the bob will start decreasing. D. The angular momentum of the bob remains constant.

Solution: As the oscillation starts the bob will slowly pick up iron dust and its mass will increase. As the velocity of the bob at the hightest point is zero, by conservation of energy, height of the highest point must decrease. As the mass increases, the tension in the string will increase too. Since the string is inextensible the time period will remain the same. But angular momentum, L = Iω will change, as I changes.(I is moment of inertia).

Section C: Analytical Questions

HB

α. (10 marks) In the night sky, the constellations depict several shapes of animate objects (species like birds, animals, etc) or man made instruments. List 20 such figures / instruments. The list should contain at least 4 instruments. Write both the object and the corresponding constellation. Exact name of the constellation is not necessary. The constellations must be from International Astronomical Union’s standard list of 88 constellations (i.e. Indian / Chinese / Mayan constellations not accepted). Solution:

Animals (easy): bull (Taurus), ram (Aries), eagle (Aquila), bear (Ursa Major / Ursa Minor), lion (Leo), crab (Cancer), scorpion (Scorpio), dog (Canis Major / Canis Minor), fishes (Pisces), swan (Cygnus), dolphin (Delphinus), snake (Serpens), crow (Corvus), man (Orion / Bootes etc.) Animals (difficult) - : fly (Musca), hare (Lepus), Wolf (Lupus), fox (Vulpecula), peacock (Pavo), dove (Columba), Emu (camelopardalis), small horse (Equuleus), swordfish (Dorado), crane (Grus), lizard (Lacerta), Chamaeleon, hornbill (Tucana)

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Animals (mythical): water goat (Capricornus), centaur (Centaurus), dragon (Draco), winged horse (Pegasus), a mythical snake (Hydra), a sea monster (Cetus), a bird (Apus), Phoenix

E

instruments: Telescopium, scales (Libra), Microscopium, Sextans, directional compass (Pyxis), Table (Mensa), cross (Crux), Crown (Corona Australis / Corona Borealis), harp (Lyra), arrow (Sagitta), shield (Scutum), easel (Pictor), air pump (Antila), compasses (Circinus) Each correct creature / instrument: 0.5 marks. Maximum 16 creatures allowed. β. From the graph below, find:

(a) (7 marks) The cubic polynomial describing the curve. [Note: f (7) 6= 200.]

CS

(b) (5 marks) If this curve describes orbit of a comet with the Sun at point (1,0) and the comet crosses the x-axis exactly two months apart, find the approximate position of the comet 1 month since first crossing of the x-axis.

HB

880 840 800 760 720 680 640 600 560 520 480 440 400 360 320 280 240 200 160 120 80 40 0 -40 -80 -120

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

Solution: (a) From the graph, we note that f (−4) = 0, f (8) = 0, f (0) = 800

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and f (1) = 840. Using these to find the cubic polynomial,

ax3 + bx2 + cx + d a(−4)3 + b(−4)2 + c(−4) + d = −64a + 16b − 4c + d 83 a + 82 b + 8c + d = 512a + 64b + 8c + d d a+b+c+d a+b+c 576a + 48b + 12c 48a + 4b = 4(12a + b) 64a + 8b + c + 100 24b − 3c + 900 8b + 300

E

= = = = = = = = = = =

CS

f (x) 0 0 800 840 ∴ 40 &0 ∴ −c &0 ∴0 ∴c

(2.5 marks)

Solving, we get,

f (x) = x3 − 29x2 + 68x + 800

(1)

(b) We have to divide the area enclosed by the curve and the x-axis in 2 equal halfs as divided by the line from point (1,0), to get the position for the comet after one month. Non-calculus solution: A rough counting of squares shows that the correct position would be approximately between x = 2 and x = 2.5. Calculus based solution: If the x-coordinate of the comet after one month was x0 , Z x0  Z 8 1 f (x)dx − (x0 − 1)f (x0 ) f (x)dx = 2 (2) 2 −4 −4

HB

Solving, we get,

88 x40 32x30 − + 29x20 − 868x0 + 64 × =0 2 3 3

(3)

x0 = 2 yeilds slightly positive value and x0 = 3 yeilds large negative value. Thus, the root the the equation lies very close to 2 but is higher than 2. Note: The approximate solution turns out to be x0 = 2.2071. Marking Principle: By either way, student should state (with proper justification) that the x-coordinate is close to 2 but higher than 2.

γ. (a) (6 marks) Assume that human civilisation has carved out all the mantle and the central core of the earth, and started living on the inside surface of the earth’s crust which is only about 7 km in thickness. As a consequence, people are finding that their weight has altered drastically. To feel normal weight again, one scientist suggested an idea involving tampering one of the motions of the earth. Can you describe this idea qualitatively and quantitatively?

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E

Solution: For a person standing on the inside surface of the hollow shell, there is no effect of gravity due to the shell itself (shell theorem). He / she would “feel” the weight purely due to the centrifugal force due to the rotation of earth. (2 marks) Thus, to get the normal weight back, the earth must start rotating faster. (0.5 mark) Shell thickness should be considered as negligible as compared to the radius of the earth. (0.5 mark)

CS

g = Rω 2 r g 2π = ∴ω = T R s R ∴ T = 2π g r 6.4 × 106 ≈ 2π 10 ≈ 2π × 800 T ≈ 5000 sec

Thus, the earth’s rotation period should be decreased to about 84 minutes. (2 marks) This calculation is valid only for the equator and apparent weight at other latitudes would be different. (1 mark)

HB

(b) (8 marks) An observer is sitting in a stationary spacecraft outside earth’s orbit such that at the closest point, distance between the observer and the earth is 1 AU. Sketch a rough plot (exact functional form / shape not expected) showing how the orbital angular velocity of the earth varies with time (in an year), as seen by this observer. Mark every point of maximum or minimum clearly and write its coordinates. Similarly, mark every point with zero angular velocity and write its coordinates. Assume the orbit of earth around the Sun to be circular. B

Solution:

observer

60

8

Earth

Sun

A

C R

D

CS

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INAO – 2012

The exact expression for the function above is

sin(φ) cos(φ + θ) sin(θ) R sin(θ) where tan φ = k + R(1 − cos(θ)) ω(θ) = ω0

In this, R is the orbital radius of the earth (1 AU), k is the shortest separation between the observer and the earth orbit (also 1 AU), θ is the observer - sun earth angle, φ is the earth - observer - sun angle. The coordinates for points of interest would be (0, ω0 ), (60,0), (180, ω0 /3) and (300,0), where x-axis represents θ. This θ can be recalibrated in terms of time with 30◦ interval corresponding to one month period.

HB

The sign of ω as well as start of the year point are matters of convention. Hence if this graph is flipped on x-axis or is shifted along the x-axis, the answer is equally valid. Marking scheme: 1.5 mark for each the four coordinates. 2 marks for the overall shape of the curve and proper markings of axes etc.

δ. In the 1957 science fiction novel “The black cloud” by Sir Fred Hoyle, an interstellar dark cloud is discovered appraching the solar system. We will try to retrace the steps explained in the novel to calculate speed and direction of the cloud. Two photographs given here show a small region inside the constellation of Orion. The photographs are taken exactly 2 months apart. The grid in the image has the size of 1◦ × 1◦ . By spectroscopy, it was realised that the emission line of neutral hydrogen (Rest wavelength 21.1000 cm) emitted by this cloud was showing up at 21.0789 cm wavelength.

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INAO – 2012

(a) (4 marks) In how many months (after the second image) the cloud will entirely cover the belt of Orion (seen on the left)? (b) (1 mark) Is the cloud headed directly for the solar system? Why? (c) (2 marks) If yes, then in how many months (after the second image) the cloud will arrive at the earth? If no, what will be the closest separation between the cloud and the earth? Assume the cloud to have uniform velocity throughout the journey. (d) (5 marks) If this spherically symmetric cloud is placed at the exact centre of the solar system, which planets will be engulfed by the cloud? r c−v Note: The expression of relativistic Doppler effect is given by λobserved = λemitted c+v

HB

Solution: (a) The size of the cloud is about 1 cm in the first image and 2 cm in the second image. Now, Mintaka and Alnitak are almost equidistant from the cloud’s centre (about 4.6 cm) so they would be get covered at almost same time. If the physical diameter of cloud is x and its distace from us is d, x = θd = 2θ(d − vt2 ) = 9.2θ(d − vt3 ) ∴ d = 2vt2 & 9.2vt3 = 7.2d + 2vt2 = 16.4vt2 16.4 t3 = × 2 months 9.2 = 3.565 months

i.e. In just over six weeks from the second image (about 47 days), the belt of Orion will be covered by the cloud. (b) The cloud is headed directly for the solar system as the position of its centre has not shifted from the first image to the second image.

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INAO – 2012

(c)

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d = 2vt2 = vt4 ∴ t4 = 2t2 t4 = 4 months Thus, the cloud will arrive at the solar system in exactly 2 months after the second image is taken. (d) Velocity of the cloud is given by, 2  2 λobserved 21.1 × (1 − 0.001 = = 0.998 λemitted 21.1 2c ≈ 300km/s ∴v = 1998 

CS

c−v = c+v

As per the grid, about 1.95 cm correspond to 1◦ separation in the sky. Thus, θ ≈ 0.5◦ in the first image. x = dθ = 2vt2 θ 2 × 3 × 105 × 2 × 30 × 86400 π ∴x = × 0.5 × 11 1.5 × 10 180 24 × 8.64π ≈ 360 π ≈ 1.728 × 3 x ≈ 1.8 A.U.

HB

Thus, the cloud radius would be about 0.9 A. U. Meaning, the cloud will engulf Mercury and Venus. ǫ. (12 marks) Let us assume a situation where an asteroid passes close to the earth. Some times the shortest distance of such an asteroid can be even lesser than the earthmoon distance. In such cases, the asteroid will cross the lunar orbit at two points. A particular asteroid makes a close-earth pass such that at its closest proximity to the earth, its distance from the earth is 256000 km. For simplicity, let us assume that the gravitational force of the earth acts on this asteriod only for a short duration of 1000 seconds, when the asteroid is exactly at the closest proximity point. Let us assume that the asteroid follows straight line paths with uniform velocities before and after this gravitational interaction and the velocity of the asteroid before the interaction was 6.25 km/s. Find out how much would be the deviation measured (in km) along the lunar orbit. Assume the Moon to be on the other side of the Earth in its orbit during the asteroid’s transit, hence not having any impact on the motion. Take the orbit of the Moon around the Earth to be circular. Solution:

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C

Path if undeviated B

s

Path after deviation r θ Moon

Earth d (Shortest distance)

D

E

E

A

C

R

B

Approach

~v = ~u + ~at GMt ~v = uˆj + 2 ˆi d

(3 marks)

6.67 × 10−11 × 6 × 1024 × 103 ˆ i (2.56 × 108 )2

CS

= 6.25 × 103ˆj +

F

≈ 6250ˆj + 6.1ˆi

(3marks)

As the first term on the R.H.S. is much larger as compared to the second term, the angle of deviation of the asteroid would be very small. In the figure, E denotes centre of the Earth, A closest proximity position of asteroid, BC is arc measured along the lunar orbit. AB is denoted by r and both EB and EC would equal to R. As angle of deviation is very small, Arc length BC can be approximated as length BD. Let angle BAC be θ and angle AEB be φ. Thus, angle FBC and hence approximately angle FBD will also be 90◦ − φ. Angle BFD can be approximated to 90◦ . tan(θ) =

HB

∴ tan(θ) ≈

at ≈ 10−3 u ≈ 10−3 r R BF cos(90◦ − φ) r tan(θ) sin(φ) R tan(θ) 3.84 × 108 × 10−3 384km (6marks)

sin(φ) = BD = =

= ≈ s ≈

Note: If BC is approximated to BF, only 3 marks would be awarded. Secondly, AB = r 6= R. That approach does not carry any points.

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HB

CS

E

Space for Rough Work

prepared using LATEX2ǫ

Indian National Astronomy Olympiad – 2013 Question Paper

Roll Number:

Roll Number

Date: 2nd February 2013 Maximum Marks: 100

INAO – 2013 Duration: Three Hours Please Note:

E

• 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 4 pages (8 sides). • There are total 10 questions. Maximum marks are indicated in front of each question.

CS

• For all questions, the process involved in arriving at the solution is more important than the answer itself. Valid assumptions / approximations are perfectly acceptable. Please write your method clearly, explicitly stating all reasoning. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you. • Please be advised that tentative schedule for OCSC in astronomy is from 2nd to 19th May 2013.

Useful Physical Constants

HB

Mass of the Earth Radius of the Earth Mass of the Sun Radius of the Sun Radius of the Moon Distance to the Moon Mass of Jupiter Astronomical Unit Light year

Solar Luminosity Gravitational Constant Gravitational acceleration Wien’s constant

ME RE M⊙ R⊙ Rm dm MJ 1 A. U. 1 Ly 1 Ly L⊙ G g K

≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈

6 × 1024 kg 6.4 × 106 m 2 × 1030 kg 7 × 108 m 1.7 × 106 m 3.84 × 108 m 2 × 1027 kg 1.5 × 1011 m 9.461 × 1015 m 63240 A.U. 3.826 × 1026 W 6.67 × 10−11 m3 /(Kg s2 ) 9.8 m/s2 2.898 × 10−3 m.K

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – 2013

E

1. (10 marks) Astrophysicist Freeman Dyson proposed around 50 years back that an advanced civilisation would make optimal use of the energy of the parent star by constructing a full shell around the parent star with radius equal to the orbital radius of their planet, trapping all the radiation inside. The civilisation can live on the surface of this shell. We would like to construct a “Dyson sphere” with radius equal to the orbital radius of the Earth. Let us assume that we have access to all the material within the solar system (except the Sun itself, which we would like to retain as the energy source) and we have necessary technology to solidify the material available to us. What will be rough thickness of this shell?

CS

Solution: Total mass available to us for constructing the Dyson Sphere would primarily consist of mass of all planets and satellites. As an approximation, we will say Jupiter and Saturn are of similar size. Saturn is known to have very low density so its mass will not exceed half Jupiter mass. Further, radius of Uranus and Neptune is less than half of Jupiter radius. Thus, they will be at least 10 times lighter than Jupiter. Similarly, Earth and Venus are of similar size and Mars is about half of that. There are a handful of objects with radii about a quarter of that of Earth or even smaller (Mercury, Moon, Ganymede, Callisto, Titan etc.). All these are rocky bodies and we can take typical density to be ρs = 5 gm/cc. Combined mass of all other combined bodies will not be greater than mass of Mars. From real planetary data, total mass estimate of about 1.5MJ is closer to reality.

HB

As we want shell to be solid like rock, we will assume similar density for the shell too. So if the shell has thickness l,   MJ M⊕ M⊕ 2 4πD lρs ≈ 2MJ + 2 + 2M⊕ + 2 +6 10 10 100 2MJ + 0.2MJ + 2M⊕ + 0.2M⊕ + 0.06M⊕ ∴l ≈ 4πD 2 ρs 2.2(MJ + M⊕ ) ≈ 4π(1.5 × 1011 )2 × 5 × 103 2.2(2 × 1027 + 6 × 1024 ) ≈ 4π × 2.25 × 1022 × 5 × 103 2006 10 ≈ ≈ 200π π ≈ 3 mt. Thus, the thickness will be approximately of the order of a few meters.

As is evident from the numbers, mass estimates are just rough indicators of total mass. and density can be lowered to about 2.5 gm/cc. Thus, the thickness can go up to 5 metres. Marching Scheme: Reasonable estimate of mass (≈ 1.5 − 3.0MJ ) and density (1 - 10 gm/cc) with

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reasons: 5 marks Rest of the Calculation: 5 marks

Solution:

CONSTELLATION — — Orion Cygnus Taurus — — — Virgo Scorpio

CS

STAR Polaris Rigil Kentaurus — — — Sirius Vega Regulus — —

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2. (10 marks) Following table contains names of a few constellations and names of some stars. Fill in the blanks. For missing star names, you can name any star from that constellation.

CONSTELLATION Ursa Minor Centaurus Orion Cygnus Taurus Canis Major Lyra Leo Virgo Scorpius

HB

STAR Polaris Rigil Kentaurus Betelgeuse Deneb Aldebaran Sirius Vega Regulus Spica Antares

3. (10 marks) An Indian festival called ‘Kojagiri’ is always celebrated on a Full Moon night in Autumn. As a part of festivities, many people keep a bowl of milk on the ground or a terrace under the Moon light for a few minutes, before drinking it. On this year’s ‘Kojagiri’ night, Sheetal had placed roughly 100 ml milk in a dish of size 25cm diameter for about 10 minutes exactly when the moon was overhead. Assume that density of milk is similar to the water, its specific heat capacity is 0.9 times that of water and heat loss from the milk to surroundings is negligible. Assume that full Moon reflects about 14% of the sunlight incident on its surface. What will be the change of temperature caused by the light received from the moon?

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Solution: Moon receives light energy from the Sun which is partly reflected back towards the Earth. L⊙ (1 mark) 4πD 2 L⊙ 2 Total Solar Energy incident on Moon = × πRm (1 mark) 2 4πD L⊙ 2 × πRm × α (1 mark) Total Solar Energy reflected by Moon = 2 4πD 2 πL⊙ Rm α Reflected Energy received at Earth per unit area = = Er (2 marks) 2 4πD × 2πd2m

E

Solar Energy per unit area incident on Moon =

Let us assume that this entire energy is used to heat the milk without any losses.

CS

tEr A = ms∆T (1 mark) 2 tπr 2 πL⊙ Rm α (1 mark) ∆T = × 2 V ρs 4πD × 2πd2m

2

600 × 0.1252 × 3.826 × 1026 × (1.7 × 106 ) × 0.14 = (1 mark) 10−4 × 103 × 0.9 × 4.18 × 103 × 8 × (1.5 × 1011 )2 × (3.84 × 108 )2 1 ≈ 1.4 × 10−4 (2 marks) ∆T ≈ 7000 Thus, temperature will rise by only about 10−4 Celsius.

4. (7 marks) Let A and B be two objects in the solar system orbiting each other and A is much heavier than B. Theoretically, what is the maximum possible revolution period of B?

HB

Solution: To have longest possible period, the orbital radius should also be longest. If A was any planet, orbital radius of B cannot be very high as far from the planet B will come under influence of the Sun. So for longest possible radii, we take A as the Sun and try to place B as far as possible. (1.5 marks) The other limiting factor will be the influence of other stars. As B is still part of the solar system, Sun exerts more influence on it than any other star. (1.5 marks)

Star closest to us is the Alpha-Centauri triplet. It has two stars similar to the Sun and Proxima Centauri, which is much smaller than the Sun. Thus, we can approximate the system as about 2M⊙ system about 4.3 light years away. Hence, rd at about 13 of the distance, Alpha-Centauri system will start dominating on B. 4.3 This means that maximum possible radius is about = 1.4 light years. Thus, 3 longest period will be, (1.5 marks) We can utilise the fact that orbital radius of the Earth is 1 A.U. and orbital

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period is 1 year.

T12 = T22 T1 =

= ≈

4π 2 3 r (0.5 marks) GM R13 R23 s R13 T2 R23 s (1.4 × 63240)3 1× 13 q (90000)3

E

T2 =

CS

= 3003 = 2.7 × 107 years (2 marks)

Thus, approximate period would be 27 million years.

It is acceptable if Alpha-Centauri system is assumed to have mass of just a M⊙ . 5. (8 marks) Prove that 2013 × 20124 + 1 is a composite number. Solution:

2013 × 20124 + 1 (2012 + 1) × 20124 + 1 (a + 1) × a4 + 1 where a = 2012 a5 + a4 + 1 (2 marks) a5 + a4 + a3 − a3 + 1 a3 (a2 + a + 1) − (a3 − 1) a3 (a2 + a + 1) − (a − 1)(a2 + a + 1) (a2 + a + 1)(a3 − a + 1) (20122 + 2013)(20123 − 2011) (6 marks)

HB

X = = = = = = = = =

6. (10 marks) Ayush was doing an experiment with a lens in his Physics lab. He kept a reference object at several positions and noted corresponding image sizes and positions. The same are shown in the diagram below. Unfortunately, he forgot to label the object and image positions and to mention details about the lens, because he thought them to be too trivial. Help his science teacher to make sense of his drawing by finding type of lens (convex / concave), its position and its focal length. In the diagram, O stands for Object and I for image. All lengths and heights are to scale.

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I I I

O

O

E

I O

O

O

CS

Solution: The type of lens is convex lens. In the picture, Object-Image pairs have been given same subscript. O5 is placed at exact focus so it has no corresponding image. Focal length of lens is about 2.25 cm (From the figure, 4f = 9.1 cm).

I4

I 3

I2

O

5

I 1

O4

O

O

HB

1

3

O2

Marking scheme: Lens type - 1 mark, Reasoning - 2 marks, position - 3 marks, focal length - 3 marks, cross checking - 1 mark

7. (5 marks) In a competition for the most useless inventions, Sharad presented a novel technique which makes it possible to fold an ordinary paper 50 times on itself, without any gaps between layers of folds. Calculate thickness of the folded paper after it has been folded 50 times. Note that the paper is folded further and further without opening the previous folds i.e. it is folded in half and then the folded paper is folded on the middle to reduce the area to a quarter and so on. Solution: A 100 page notebook is roughly 1 cm thick. (1 mark) −2 −4 −4 Thus, thickness of a single paper is 10 cm or 10 mt (2 × 10 − 2 × 10−5 mt accepted). (1 mark)

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If you fold the paper 50 times, its thickness will become,

E

l = 250 × 10−4 (2 marks)
5 = 210 × 10−4 = (1024)5 × 10−4 ≈ 1015 × 10−4 ≈ 1011 mt. (1 mark)

(1) (2) (3) (4)

Thus, the thickness will be of the order of the Sun-Earth Distance.

CS

8. (10 marks) Alankar was jogging downhill and he noted distance traveled by him at various intervals. Plot appropriate graph to find his acceleration and initial velocity. In the table, time (t) is in minutes and distance (d) is in meters. Both the quantities are measured from the start of the journey. t d 2 21 4 60

t d 6 117 8 192

t d 10 285 12 396

t 14 16

d 525 672

t d 18 837 20 1020

Solution: d = ut + 21 at2 Thus, if one plots a graph of distance versus time, it will be a parabolic curve. d a To linearise the graph, we change the equation as = u + t t 2 a d Now if we plot versus t, it will give a linear graph with as slope and u as t 2 y-intercept. t d/t 6 19.5 8 24

t d/t 10 28.5 12 33

HB

t d/t 2 10.5 4 15

t d/t 14 37.5 16 42

t d/t 18 46.5 20 51

Thus, u = 6 m/min and a = 4.5 m/min2 . Marking scheme: Linearisation of equation: 2 marks, graph plotting skills: 4 marks, slope: 2 marks, final answers: 2 marks. Solutions with non-linear graphs: maximum 5 marks.

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9. (10 marks) Sandesh fabricated a magic sphere of radius R, which is hollow on the inside and has perfectly reflecting inner surface. This sphere had a small hole in it. Sandesh sent a ray of light radially through this hole. It hit a plane mirror, kept at an angle of 45◦ with the incident ray, at some point beyond the centre of the sphere, but before reaching the opposite end. After undergoing one more reflection at the inner surface of the sphere, the ray came out from the hole. Find the distance from the centre of the sphere to the point where it struck the plane mirror.

CS

Solution: In the figure below, D is the point of second reflection. OD will be normal to the surface at D.

HB

∡OBD = 90◦ (1 mark) ∡BDO = ∡ADO = ∡DAO = x ∵ AO = OD (6 marks) ∴ ∡AOD = 180 − (∡ADO + ∡DAO) = 180 − 2x but ∡AOD = ∡OBD + ∡BDO 180 − 2x = 90 + x ∴ x = 30 (2 marks) R l(OB) = l(OD) sin 30 = (1 mark) 2

10. (20 marks) Answer following questions in 3 to 4 lines each. (a) The core of earth is like a perfectly conducting fluid sphere. If this core suddenly shrinks to half of its present radius, then what will happen to its magnetic field? Why? 7

INAO – 2013

(b) A satellite is revolving around the Earth in a polar orbit with 90 minute period. On 23rd September, Prasad saw it exactly overhead at the time of sunset. Within next twenty four hours how many sunrises will be seen by the satellite? Why? (c) Anand has taken a high zoom photograph of a sun-spot. The photograph does not include any part of the disk outside sun-spot. What will be the predominant colour seen in the photograph? Why?

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(d) What will happen to the Moon if the Earth vanishes suddenly? Why? (e) On a full Moon day, with respect to an observer on the Earth, does the Moon move faster (angle covered per hour) during the day time or the during night time? Why?

CS

1 Solution: (a) Magnetic flux must be conserved. Thus, B ∝ 2 . As the radius R shrinks to half, the magnetic field will increase four-fold. (b) As the satellite makes angle of 90◦ with the Sun, its orbit is exactly along the line separating day and night regions on the Earth’s surface. Hence, its one side will continuously face the Sun and hence, there will be NO sunrises seen by the satellite. (c) The sunspots have temperature of around 4500K. By Wien’s law, this corresponds to roughly red colour. Thus, the sunspots will appear red. (d) Moon will retain all its velocity at the instance of disappearance of the Earth. Thus, it will have significant non-radial velocity around the Sun and it will continue to orbit the Sun in near identical path.

HB

(e) We note that Moon revolves in the same direction as the rotation of the Earth i.e. from West to East. On the full Moon day, Moon is on the night side of the Earth. Thus, with respect to an observer on the Earth, the tangential velocities of the Moon and the Earth during the night will be in the same direction and hence relative velocity is slower. During the day tangential velocities are in opposite directions hence the relative velocity is faster. Thus, Moon will be moving faster during the day.

Space for Rough Work

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Notes for Junior Group • Question 4: Take body A to be the Sun. Kepler’s third law states that the square of orbital period will be proportional to the cube of orbital radius. • Question 5: a3 − 1 = (a − 1)(a2 + a + 1).

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• Question 8: Modify appropriate equation of motion such that all points will lie along a straight line. • Question 10 (a): If the conducting sphere suddenly changes the size, the field lines will stay trapped on the surface.

CS

• Question 10 (c): At any given temperature, the peak of thermal radiation will be Wien′ s constant at wavelength λmax = . T

Notes for Junior Group

• Question 4: Take body A to be the Sun. Kepler’s third law states that the square of orbital period will be proportional to the cube of orbital radius. • Question 5: a3 − 1 = (a − 1)(a2 + a + 1).

• Question 8: Modify appropriate equation of motion such that all points will lie along a straight line. • Question 10 (a): If the conducting sphere suddenly changes the size, the field lines will stay trapped on the surface.

HB

• Question 10 (c): At any given temperature, the peak of thermal radiation will be Wien′ s constant at wavelength λmax = . T

Notes for Junior Group

• Question 4: Take body A to be the Sun. Kepler’s third law states that the square of orbital period will be proportional to the cube of orbital radius.

• Question 5: a3 − 1 = (a − 1)(a2 + a + 1). • Question 8: Modify appropriate equation of motion such that all points will lie along a straight line. • Question 10 (a): If the conducting sphere suddenly changes the size, the field lines will stay trapped on the surface. • Question 10 (c): At any given temperature, the peak of thermal radiation will be Wien′ s constant at wavelength λmax = . T

prepared using LATEX2ǫ

Indian National Astronomy Olympiad – 2014 Question Paper

Roll Number:

Roll Number

Date: 1st February 2014 Maximum Marks: 100

INAO – 2014 Duration: Three Hours Please Note:

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• 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 (6 sides). • There are total 8 questions. Maximum marks are indicated in front of each question.

CS

• For all questions, the process involved in arriving at the solution is more important than the answer itself. Valid assumptions / approximations are perfectly acceptable. Please write your method clearly, explicitly stating all reasoning. • Blank spaces are provided in the question paper for the rough work. No rough work 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 this question booklet back with you. • Please be advised that tentative dates for the next stage are as follows: – Orientation Camp (Junior): 28th April to 6th May 2013. – Orientation Camp (Senior): 1st May to 6th May 2013.

– Selection Camp (Jr. + Sr.): 27th May to 6th June 2013.

HB

– 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 Astronomical Unit Gravitational acceleration Avogadro constant Solar Constant

ME RE MJ 1 A. U. g Na S

≈ ≈ ≈ ≈ ≈ ≈ ≈

6 × 1024 kg 6.4 × 106 m 2 × 1027 kg 1.5 × 1011 m 9.8 m/s2 6.023 × 1023 mol−1 1400 W/m2

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INAO – 2014

1. (12 marks) Read the following passage and point out scientific inaccuracies. Give very brief argument for each mistake you point out.

E

Arrival of a spacecraft on the surface of the Moon is called a Moon landing. This includes both manned and unmanned (robotic) missions. In order to get to the Moon, a spacecraft must first cross the point, beyond which the Earth’s gravitational force is zero. The only practical way of accomplishing this currently is with a rocket. At every instant, the rocket has to produce thrust to propel itself to a velocity equal to critical velocity at that height. Jet engines can be used to propel the spacecraft to the Moon, but rockets provide greater advantage in terms of power required for given mass of the spacecraft.

CS

Unlike the Earth, the Moon does not have thick atmosphere to absorb most of the solar radiation and the magnetic field of the moon is too weak to deflect the UV rays. Thus, astronauts traveling to the Moon are exposed to harmful electromagnetic radiation. Spacesuits for astronauts are specially designed by keeping this in mind. On arrival near the Moon, the spacecraft is captured by the lunar gravity in an orbit around the Moon. As the Moon is much smaller than the Earth, typically these orbits are low altitude orbits (close to the lunar surface) as compared to the polar satellites around the Earth. The spacecraft can stay in this orbit forever, however, if one needs it to land on the Moon, it has to fire its engines to change course.

HB

Landing at the Moon can be of two types. A hard landing is equivalent to crash landing on the Moon. A soft landing is a controlled landing in which priority is to maintain all equipment and astronauts (if any) inside the spacecraft safe. Needless to say, all Apollo missions had a soft landing on the Moon. This was achieved by firing reversing rockets very close the lunar surface to slow down the spacecraft. Charring of the lunar soil, due to burning of these rockets, is one of the permanent impressions left by humankind on the Moon.

Solution:

• Earth’s gravitational force goes to zero only at infinity. Close to the Moon, the force is small but not zero. • Velocity needed is bigger than the critical velocity. • For jet engines, the oxygen is taken from the air. It is not stored on-board. So they cannot be used to take spacecraft to the Moon. • UV rays are electromagnetic waves and are not affected by the magnetic field. • Low altitude orbits are not stable, as imperfections on the lunar surface result in change in gravitational force from point to point.

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• The lunar soil won’t get charred due to firing of rockets as there is no oxygen on the lunar surface. • One mark for each correct indentification of the mistake. One mark each for corresponding reasoning.

E

• If a blatantly wrong argument is resulting in incorrect identification of an inaccuracy, one mark will be subtracted. • No negative marking for subtle mistakes in reasoning.

CS

2. (5 marks) A comet is in elliptical orbit around the Sun with period T and semimajor axis a. A second comet around the Sun has same period T , but a different orbit with same eccentricity e. The two orbits are not necessarily in the same plane. Further, we also know that when the first comet is closest to the Sun in its orbit (at perihelion), the other comet is farthest from the Sun in its orbit (at aphelion). Given this configuration, what is the minimum possible distance between the two comets, when one of the comets is at its perihelion? Justify your answer by brief arguments / sketches. Solution: By visualising the configuration correctly, it will be realised that to achieve minimum possible separation, the two orbits should lie in the same plane with their semi-major axes along the same line, such that sun - perihelion-1 aphelion-2 are collinear in that sequence. (3 marks) Thus, the minimum separation will be

HB

d = daphelion,1 − dperihelion,2 = a(1 + e) − a(1 − e) = 2ae (2 marks)

3. On a flight between the cities of Oslo (61◦ N, 8◦ E) and Helsinki (60.2◦ N, 25◦ E), Mayank saw the Sun just at the horizon on the west. After 20 minutes, when he checked again, he saw the Sun was now 2◦ above the horizon. (a) (2 marks) The flight was traveling from which city to which city?

(b) (8 marks) What was the speed of the aircraft with respect to the ground (in Km/hr)? (c) (4 marks) How long will this aircraft take to traverse the distance between these two cities? Assume uniform speed during the entire flight.

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Solution: The Sun rising from the west can be observed only if we are traveling in the direction opposite to the Earth’s rotation. Thus, the flight was from Helsinki to Oslo. (2 marks)

E

In 20 minutes, the Earth would rotate by 5◦ eastwards. Mayank saw the Sun rising by 2◦ in that time. This means, the plane moved (5◦ + 2◦ ) westwards in 20 minutes. (2 marks) We can approximate latitudes of both places to φ = 60◦ .

(2 marks)

CS

θ = 5◦ + 2◦ = 7◦ π 7 π θ × = × rad s−1 ω= 20 × 60 180 1200 180 v = rω = R⊕ ω cos φ 7π = 6.4 × 106 × × cos 60◦ 1200 × 180 7π 1 28 22 = 64 × × ≈ × 1.2 × 1.8 2 0.27 7 8800 18 ≈ 326 m/s ≈ × 27 5 ≈ 1170 km/hr (4 marks)

Difference between longitudes of the two cities is 16◦ . Thus, 16 × 20 7 ≈ 46min

t=

The flight will take about 46 minutes.

(4 marks)

HB

Note: The solution above is a simplification as θ is not estimated correctly. When Maynak will see the Sun 2◦ above horizon, it would actually mean that at 60◦ latitude, the Sun has rolled back by 4◦ along its apparent path. Thus, solution using θ = 9◦ is more correct. However, students are not expected to know this at INO level and hence solution given above is accepted for full credit.

4. The skymap below corresponds to sky above Nagpur (21◦ N, 79◦ E at 09:00 am on 1st February 2014. If you are not used to using sky maps, it is important to note that sky map is usually seen lying down on the ground (feet to the South), facing the sky with map in your hand. Thus, East is on the left of the map and West is on the right. Answer the following questions: (a) (1 mark) Mark Polaris with letter ‘P’.

(b) (2 marks) Circumpolar stars for a given place are the stars, which will never go below the horizon. Draw boundary of this region and mark it by the letter ‘C’. (c) (2 marks) The celestial equator is just a projection of the Earth’s equator in the sky. It will be locus of points which are equidistant from the north and the 3

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south pole. Draw the equator on the map approximately and mark it with ‘Q’. (d) (2 marks) The ecliptic is the imaginary yearly path of the Sun in the sky. Mark this approximately on the map and mark it with ‘E’. (e) (2 marks) Mark approximate position of the Sun on the map as ‘S’.

E

(f) (2 marks) Yesterday was a new moon day. Mark the current position of the moon on the map as ‘M’. (g) (2 marks) Which star was very close to the Zenith at 06:00 am today? Mark it on the map as ‘N’.

HB

CS

(h) (2 marks) Draw a line across sky showing horizon line as at 07:00 am today as ‘H’.

Solution: Notes:

• Pole star should be exact.

• Circumpolar region should be a circle around ‘P’ with radius equal to Polaris-Northern horizon distance. • Equator should be a smooth curve passing through East and West points and below zenith (centre) in the middle. • Ecliptic should be a smooth curve roughly passing through the zodiacal signs (i.e. mostly below equator) and cutting equator in Virgo.

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• The Sun should be in Capricornus. • The Moon should be in Aquarius. • Any star reasonably close to Archturus is accepted as answer for ‘N’.

CS

E

• Horizon at 7:00am should not cut current northern horizon, roughly pass close to the Sun and cut bottom edge on the west of south.

HB

5. (12 marks) Gliese-876 is a star in the constellation of Aquarius which has four confirmed planets going around it. We know following facts about this system: • the four planets have names ‘b’, ‘c’, ‘d’ and ‘e’.

• the four planets have masses (in Earth masses) of 7M⊕ , 15M⊕ , 225M⊕ and 720M⊕ . (not necessarily in the same order)

• the orbital radii of the four planets (in A.U.) are 1.93, 30, 61, 124. (not necessarily in the same order)

• the orbital eccentricities of the planets are 0.03, 0.06, 0.21, 0.26. (not necessarily in the same order) • planet ‘e’ is farthest from the star.

• the heaviest planet has the lowest eccentricity.

• the lightest planet is also the closest to the star.

• the two giant planets are neither the closest nor the farthest from the star.

• the planet with a mass of 7M⊕ has eccentricity of 0.21 and is closer to the star than planet ‘c’. 5

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• eccentricity of planet ‘d’ is more than planet ‘e’, but less than ‘c’. • planet ‘c’ is closer to the star as compared to planet ‘b’.

E

Use the information above to correctly identify mass, orbital radius and orbital eccentricity of each of the planet. For your help, you may use the grid given in the answer sheet. The grid has six sub-grids of 4 × 4 size. You can tick mark in the boxes corresponding to matching pairs and cross mark the boxes where there is no match. Note that this grid is only for your help. It is your choice if you wish to use this grid or some other method. Final answer must be written in the second table below the grid. Only the values written in the table will be considered for giving marks.

CS

Solution:

b

mass 720

eccentricity 0.03

orbital radius 61

c

225

0.26

30

d

7

0.21

1.93

e

15

0.06

124

One mark per value.

HB

6. (a) (9 marks) A thin convex lens, of focal length of f , is placed in y-z plane. The principal axis of the lens is along the x-axis. A luminous square sheet is placed in the x-y plane with its sides parallel to x and y axes. The centre of the sheet is exactly at a distance of 2f from the centre of the lens and length of each side of the sheet is f . The sheet is slightly rotated about the y-axis to make the entire sheet visible from the other side of lens. Sketch the shape of the image of the sheet and mark all relevant lengths in terms of f . (b) (6 marks) A pin P is placed in front of a concave mirror such that its mid-point lies on the principal axis at a distance d from the pole of the mirror. An observer on the principal axis at a distance D, where D ≫ d, finds the pin and its real image to overlap. When the observer moves slightly to the left of the principal axis, the image is viewed to the right of the pin. From this observation, what can be concluded about value of d? Justify your answer.

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E

Solution: (a) The image will look as follows:

CS

• Correct drawing of object: 1 mark

• Correct extremities of image on the axis: 3 marks • Top and bottom edges curved: 2 marks

• left and right edges curved on ‘correct’ side: 3 marks

(b) As you move you observing position laterally, the closer objects appear to move faster than the farther objects. Thus, when you move your eye to the left, the one closer out of the object and the image will appear to shift to the right of other. (3 marks) It means that image distance is more than object distance. (1 mark) Hence, the object is between f and 2f . (2 marks)

HB

7. (8 marks) Chandrayaan-1, the lunar mission launched by India in 2008, had all gold wiring. A particular instrument on board the mission, was operating at the same temperature as the space surrounding the Chandrayaan. Instrument specifications demanded that no wire in the instrument should offer a resistance more than 7 mΩ. All the instruments on Chandrayaan were fabricated in ISRO labs at room temperature. If the radius of all wires in the instrument was 0.1mm (as measured in ISRO lab), what should be the maximum allowable length (as measured in ISRO lab) of any wire? Physical properties of gold: • Resistivity (at room temperature), ρ = 2.214 × 10−8 Ω m

• Temperature coefficient of resistivity, β = 0.0032 K−1

• Linear expansion coefficient, α = 1.5 × 10−5 K−1

• Density, ̺ = 2 × 104 kg/m3

• Molar heat capacity, Cp = 25 J mol−1 K−1

• Molecular Weight, W ≈ 197 g

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Solution: Mass of the wire is, M = πr12 l1 ̺ = πr22 l2 ̺ ∴ r12 l1 = r22 l2

E

Resistance of wire is given by, ρl ρl = 2 A πr ρ2 = ρ1 (1 − β∆T ) l2 = l1 (1 − α∆T ) ρ2 l2 R2 = (2 marks) πr22 ρ1 (1 − β∆T )l2 = l r2 π 1l21

CS

R=

ρ1 (1 − β∆T )l22 πl1 r12 ρ1 (1 − β∆T )(l1 (1 − α∆T ))2 = πl1 r12 ρ1 (1 − β∆T − 2α∆T + α2 ∆T 2 ) l1 = πr12 =

(2 marks)

Typical room temperature can be assumed to be 30◦ Celcius. Temperature in space is typically 3 Kelvin. Thus, temperature difference will be about 300 Kelvin. (1 mark) Now, as α2 term is too small. Further, as α is 100 times smaller than beta, it can be ignored too. (1 mark)

HB

ρ1 (1 − β∆T )l1 πr12 πr12 R2 ∴ l1 ≈ ρ1 (1 − β∆T ) (10−4 )2 × 7 × 10−3 22 ≈ × −8 2.214 × 10 × (1 − 0.0032 × 300) 7 −2 1 10 = ≈ (1 − 0.96) 4 = 25 cm (2 marks) R2 ≈

Thus, the wire can be at max 25 cm long.

8. (18 marks) We are in the year 2020 and the spacecraft Voyager-1 is at a distance of around 200 A. U. from the Sun. It is still continuously sending information about its speed with respect to inertial frame of the surrounding interstellar medium every 10 days. Following are the readings reported by the Voyager over course of 100 days. 8

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Time (days) Velocity (m/s) Time (days) Velocity (m/s)

0 1000 60 1394.4

10 1061.5 70 1467.4

20 1124.4 80 1543.1

30 1189.0 90 1621.8

40 1255.4 100 1703.8

50 1323.8

E

Swapnil claimed that this indicates existence of an unknown body directly ahead in the path of Voyager. Those of you who agree with him, find mass and distance of this unknown body from Voyager. Those who disagree with him should give alternate explanation for the data observed and justify their arguments. Note:

• For purpose of any calculations, you may ignore the influence of the Sun on Voyager. At the end of calculations, you should give reason why this is a valid assumption. Two points are reserved for this justification.

CS

• As an approximation, assume the acceleration of the Voyager to be constant during each 10 day period between the readings. • You may note that 8.52 ≈ 72.

• For calculations, you may assume one day approximately contains 90000 seconds.

Solution: We note that Voyager’s velocity is changing. As we are ignoring influence of the Sun, this can only be because of presence of some unknown object close by. By noting difference between first two readings and last two readings, we realise that the change is not the same, i.e. acceleration is changing. We try to estimate average acceleration and velocities for each period by, v−u t 1 s = = u + at t 2

a=

vav

HB

In the table below, time is in days, velocities in m/sec and acceleration in µ m/s2 . Average velocity of the Voyager is given by vav in m/s and last column is cumulative sum of vav . (8 marks) t 0 10 20 30 40 50 60 70 80 90 100

v 1000 1061.5 1124.4 1189.0 1255.4 1323.8 1394.4 1467.4 1543.1 1621.8 1703.8

v−u — 61.5 62.9 64.6 66.4 68.4 70.6 73.0 75.7 78.7 82.0

9

a — 68.3 69.9 71.8 73.8 76.0 78.4 81.1 84.1 87.4 91.2

vav — 1003 1065 1128 1192 1259 1327 1398 — — —

P

vav — 1003 2068 3195 4388 5647 6974 8372 — — —

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Now note that,

∴

X

s = vav t X s= vav × 10 × 90000

GM r2 GM P = (r0 − s)2 X X ∴ a1 (r0 − ( s) )2 = a2 (r0 − ( s) )2 1 X X2 √ √ a1 (r0 − ( s) ) = a2 (r0 − ( s) ) 1 2 √ √ 71.8(r0 − 3195 × 900000) = 81.1(r0 − 8372 × 900000) r0 r0 8.5( − 8.4) 8 − 3.2) ≈ 9( 9 × 10 9 × 108 ∴ r0 ≈ 9 × 108 × (8.4 × 18 − 3.2 × 17) ≈ 9 × 108 × 96.8 ≈ 8.7 × 1010 m ≈ 0.6 A U (5 marks)

CS

E

a=

HB

Now we use this distance to find mass of the object. P a(r0 − s)2 ∴M = G 68.3 × 10−6 × (8.7 × 1010 − 1003 × 9 × 105 )2 × 3 ≈ 20 × 10−11 68.3 × (8.6)2 × 1024 × 3 ≈ 2 ≈ 68.3 × 74.0 × 1.5 × 1024 ≈ 5 × 103 × 1.5 × 1024 ≈ 7.5 × 1027 kg ≈ 4MJ (3 marks) Now the object is about 300 times lighter than the Sun and about 400 times closer to the Voyager as compared to the Sun. Fobj Mobj 4002 r2 = 2 × Sun = FSun robj MSun 300

Thus, gravitational force by the Sun on the Voyager will be about 500 times smaller and hence can be safely ignored. (2 marks)

prepared using LATEX2ǫ