INAO Problems 2008-2014

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INAO Question Papers

(2008-2014)

Compiled By

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Indian National Astronomy Olympiad – 2008

Junior Category

Roll Number:

Roll Number

Model Solutions

INAO – 2008 Date: 2nd _{February 2008}

Duration: Three Hours Maximum Marks: 100 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.

• In part A and part C, there are 20 multiple choice questions each. For each question, only one of the four alternatives is correct. Mark the correct answer on the answer sheet provided separately. Each correct answer adds +3 marks to your score. In part A, every wrong answer carries penalty of -1 marks. There is no negative marking in part C. • In part B, there are 4 analytical questions of 10 marks each. The answer to each question

must be written in the blank space provided below each question. • For the rough work, use the page(s) marked as rough sheet. • Only non-programmable calculators are allowed.

• Return BOTH the question paper booklets and the answer sheet to the invigilator. DO NOT TAKE THIS BOOKLET BACK WITH YOU.

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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• 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 Junior Science Camp by putting a cross mark (’X’) in the corresponding box. Mark only one box indicating your 1st _{preference. You will be automatically considered for the}

second preference if the 1st _{choice is not available.}

• Use only black or blue pen to put ’X’ marks on the answersheet. Do not use any other ink or pencil.

Useful Physical Constants

Mass of Earth ME ≈ 5.97 × 1024 kg Radius of Earth RE ≈ 6.4 × 106 m Mass of Sun M⊙ ≈ 1.99 × 10 30 kg Radius of Sun R⊙ ≈ 7 × 108 m Speed of Light c ≈ 3 × 108 m/s Astronomical Unit 1 A. U. ≈ 1.5 × 1011 m Gravitational Constant G ≈ 6.67 × 10−11 m3_{/(Kg s}2₎ Gravitational Acceleration g ≈ 9.8 m/s2 Speed of Sound (at room temperature in air) cs ≈ 340 m/s

Space for Rough Work

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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? (a) 500 km (b) 5 km (c) 15 km (d) 50 km

4. A regular barometer is thrown from the top of a building.If the barometer is freely falling, what will be the height of the mercury column?

(a) 100 cm (b) 76 cm (c) 50 cm (d) 0 cm

5. P. Eclipses are not distributed evenly throughout the year, but happen 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. 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?

String Heavy Ball A B D C Simple Pendulum

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(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.

(c) The drum on Earth will vibrate for a longer time than the one in space. (d) There will be no difference in terms of the vibrations or sound.

9. Which of the following Venn diagrams would be BEST suited to represent the three categories of animals?

- Animals that give us MEAT - Animals that give us EGGS - Animals that give us MILK

(a) (b) (c) (d) The answer is (a). 10. 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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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. As a result the said question was removed from evaluation.

15. A nutty professor discovers a way to shrink objects in size using lasers and mistakenly shrinks his three teenage kids by a factor of 100. These kids then stray away into the garden where they see their pet dog. How many of the kids can climb onto the dog to get a ride home if the dog can bear a weight of 20 kg?

(a) None (b) One (c) Two (d) All 16. 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).

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

Battery Lighted Bulb 4 1 3 2

(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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(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 likely to feel colder and why?

(a) The foot on the granite as it will absorb heat away from the foot more quickly.

(b) The foot on the carpet as it will absorb heat away from the foot more quickly. (c) The foot on the granite because the granite is at a lower temperature. (d) The foot on the carpet because the carpet is at a lower temperature.

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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. Thus, one has to finde L.C.M. of 11 and 30.

L.C.M. is 330. i.e. after 330 intercalary days are introduced, both calenders will match.

i.e. they will match after 30 years.

Thus, his birthday in 2038 will again mark start of the lunar calander. Note: Brute force method should not be given more than 7 marks.

22. When R1 and R2 are connected in series, the current in the circuit is 2A. When R1

and R2 are connected in parallel, the current in the circuit is 4A. Find the values of

R1 and R2.

Solution: For a given V,

V = IA(R1+ R2) = IB R1R2 (R1 + R2) 2(R1+ R2) = 4 R1R2 (R1+ R2) (R1+ R2)2 = 2R1R2 R21+ R 2 2 = 0

The said conditions cannot be true.

Note: Problem is solvable if one assumes non-zero internal resistance for the bat-tery. 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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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’ be-tween the two lenses in terms of f1 and f2.

Solution: 1 f = 1 f1 + 1 f2 − d f1f2

d is the distance between the two lenses.

If the incident beam as well as the emerging beam are parallel beams, then 1 f = 0

which gives the condition, d = f1+ f2

If one lens is concave and other convex, it will become, d = f1− f2

Thus, it is only possible if focal length of the convex lens is more than that of the concave lens.

If both lenses are convex, d is always positive. Hence it is always possible.

Note: The ray diagrams should be such that above relations could be inferred by measuring respective lengths on the ray diagrams.

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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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 r2

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.

Hence the thickness would be roughly 10nm. (Acceptable 1-100 nm)

Typical width is half of refill diameter. Thus, length, l = V t d = π r2 h 2t r = π × 0.05 × 12 2 × 10 × 10−7cm l ≈ 9.5 km.

Note: Answer is not important for this order of magnitude estimation question. Approach to the problem should be judged for marks.

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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

(b) Change in ionic constitution across the membrane (c) Flow of neuro-transmitters across the axon

(d) Change in pressure across the axon

27. Bitter substances can be tasted in minute amounts but larger amounts are needed to taste sweet substances. Based on this observation which of the following reasons is more appropriate?

(a) The bitter receptors are more sensitive than the sweet receptors (b) The bitter receptors are less sensitive than the sweet receptors

(c) The bitter substances dissolve more easily than the sweet substances (d) There are more bitter receptors in mouth than the sweet receptors 28. Gustatory cells are stimulated by

(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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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 (b) Sensory division of PNS

(c) Sympathetic division (d) Parasympathetic division

33. The rate of a simple chemical reaction normally decreases as the reaction approaches completion. This is because

(a) The reactant molecules individually become less active

(b) With the progress of the reaction, the temperature goes down and hence the reaction slows down

(c) The products inhibit the reaction

(d) The concentration of the reactants decreases 34. Froth flotation is generally used for the ore dressing of

(a) Oxide ores (b) Carbonate ores (c) Phosphate ores (d) Sulfide ores 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) H2O (c) C2H5OH (d) N2O5

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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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) N2O3 (c) N2O5 (d) N2O4

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

40. A certain current when passed through a CuSO4 solution for 100 seconds, deposits

0.3175 g of copper. The current passed (in A) is (a) 4.83 (b) 9.65 (c) 0.965 (d) 0.483

41. For the redox reaction, the correct coefficients of the reactants for the balanced reac-tion are MnO4− + C2O42−+ H + −→ M n2++ CO2+ H2O (a) 2 5 16 (b) 16 5 2 (c) 5 16 2 (d) 2 16 5

Note: In the original paper, there were typographical errors in the chemical equation. Thus, the question was removed from evaluation.

42. The pH of 0.1M CH3COOH (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

INAO – 2008 Date: 2nd

February 2008 Duration: Two and half Hours Maximum Marks: 100 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 correct answer on the answer sheet provided separately. Mark a cross (X) in the corresponding box on the answer sheet.

• In part B, there are 4 analytical questions of 10 marks each. The answer to each question must be written in the blank space provided below each question.

• For the rough work, use the page(s) marked as rough sheet. • Only non-programmable calculators are allowed.

• Return the ENTIRE question paper booklet and the answersheet back to the invig-ilator. DO NOT TAKE THIS BOOKLET BACK WITH YOU.

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: 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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• 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 ink or pencil.

Useful Physical Constants

Mass of Earth ME ≈ 5.97 × 1024 kg Radius of Earth RE ≈ 6.4 × 106 m Mass of Sun M⊙ ≈ 1.99 × 10 30 kg Radius of Sun R⊙ ≈ 7 × 10 8 m Speed of Light c ≈ 3 × 108 m/s Astronomical Unit 1 A. U. ≈ 1.5 × 1011 m/s Gravitational Constant G ≈ 6.67 × 10−11 m3 /(Kg s2 ) Gravitational Acceleration g ≈ 9.8 m/s2

Speed of Sound (at room temperature in air) cs ≈ 340 m/s

Specific Heat of water Cw ≈ 4.186 × 103 J/kgoC

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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.

(d) Both the statements are correct and ‘Q’ is not the reason of ‘P’. 3. When you stand on the ground, what is the distance of the horizon from you?

(a) 500 km (b) 5 km (c) 15 km (d) 50 km

4. A regular barometer is thrown from the top of a building.If the barometer is freely falling, what will be the height of the mercury column?

(a) 100 cm (b) 76 cm (c) 50 cm (d) 0 cm

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).

(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

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(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?

String Heavy Ball A B D C Simple Pendulum (a) A (b) B (c) C (d) D The answer is (c).

8. 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 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.

(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

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(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.

(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’. 13. P. Temperature is not constant in an adiabatic process.

Q. Adiabatic processes do not obey ideal gas equation. Which of the following options 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’. 14. A star is seen rising from Calcutta (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

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?

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(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.

(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. 18. Three balls are thrown from the top of cliff along paths P, Q and R. If their initial

speeds are the same and there is no air resistance, under what conditions will the balls strike the ground below with the same speed?

R Q

P

(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?

Battery Lighted Bulb 4 1 3 2

(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.

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(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).

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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.

Intercalary days are compensated by a month whenever they exceed 30 days. Thus, one has to find L.C.M. of 45 and 30.

L.C.M. is 90. i.e. after 90 intercalary days are introduced, both calenders will match.

i.e. they will match after 8 years.

Thus, his birthday in 2016 will again mark start of the lunar calander. Important: Brute force method should not be given more than 7 marks. 22. Find R1, R2 and R3 in the circuit diagram, given the conditions below:

P Q B C R 1 3 R R2 A

(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.

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Solution: For a given V, V = 2 (R1+ R3) R2 (R1+ R2+ R3) (1) = 4 (R2+ R3) R1 (R1+ R2+ R3) (2) = 3 (R1+ R2) R3 (R1+ R2+ R3) (3) Thus, (4) 4(R2+ R3) R1 = 2(R1+ R3) R2 (5) 2R1R2 = 2R2R3− 4R1R3 R1R2 = R2R3− 2R1R3 (6) 4(R2+ R3) R1 = 3(R1+ R2) R3 (7) R1R3 = 3R2R3− 4R1R2 (8) R1R3 = 3R2R3− 4R2R3+ 8R1R3 R2R3 = 7R1R3 R2 = 7R1 (9) 2(R1+ R3) R2 = 3(R1+ R2) R3 (10) R2R3 = 2R1R2− 3R1R3 (11) 7R1R3 = 14R 2 1− 3R1R3 10R1R3 = 14R 2 1 R3 = 7 5R1 (12)

The resistances are in ratio R1 : R2 : R3 = 1 : 7 :

7 5 (13) (R1+ R2+ R3) = 47 5 R1 (14) R3 = V (R1+ R2+ R3) 3(R1+ R2) = 47 5V R1 3 × 8R1 R3 = 47 120V (15) R1 = 5 7× 47 120V R1 = 47 168V (16) R2 = 7 × 47 168V R2 = 47 24V (17)

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

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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’ be-tween the two lenses in terms of f1 and f2.

(d) For what combinations of lenses, the said arrangement is not possible? Solution: 1 f = 1 f1 + 1 f2 − d f1f2

d is the distance between the two lenses.

If the incident beam as well as the emerging beam are parallel beams, then 1 f = 0

which gives the condition, d = f1+ f2

If one lens is concave and other convex, it will become, d = f1− f2

Thus, it is only possible if focal length of the convex lens is more than that of the concave lens.

If both lenses are convex, d is always positive. Hence it is always possible. If both lenses are concave, it is not possible at all.

Note: The ray diagrams should be such that above relations could be inferred by measuring repective lengths on the ray diagrams.

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 infer the d = f1+ f2 relation by purely observing ray diagrams.

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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 r2

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.

Hence the thickness would be roughly 10nm. (Acceptable 0.5-50 nm)

Typical width is half of refill diameter. Thus, length, l = V t d = π r2 h 2t r = π × 0.05 × 12 2 × 10 × 10−7cm l ≈ 9.5 km.

Note: Answer is not important in this order of magnitude estimation question. Approach to the problem should be judged for marks.

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Junior Category

Roll Number:

Roll Number

Model Solutions

INAO – 2009 Date: 31st _{January 2009}

• 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 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

⊠ ⊠

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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Mass of the Earth ME ≈ 5.97 × 1024 kg

Radius of the Earth RE ≈ 6.4 × 106 m

Mass of the Sun M⊙ ≈ 1.99 × 1030 kg

Radius of the Sun R⊙ ≈ 7 × 108 m

Radius of the Moon Rm ≈ 1.7 × 106 m

Speed of Light c _{≈ 3 × 10}8 _m/s

Astronomical Unit _{1 A. U. ≈ 1.5 × 10}11 _m

Gravitational Constant G _{≈ 6.67 × 10}−11 _m3_{/(Kg s}2₎

Gravitational Acceleration on the Earth g _{≈ 9.8 m/s}2

Gravitational Acceleration on the Moon gm ≈ 1.6 m/s2

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Section A: (10 questions × 3 marks each)

1. If ax _{= b}y _{= c}z _{and b}2 _{= ac, then y = ?}

(a) 2xz x+z (b) xz x+z (c) √ 2xz (d) √xz Solution: a = byx c = byz b2 _{= ac = b}y xb y z = byx+ y z 2 = y x + y z = y(x + z) xz y = 2xz z + x Ans = (a)

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

(a) (b) (c) (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) r4 _{(b) r}2 _(c) 1 r2 (d) 1 r3

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Solution: The distance between two spheres = 2r and masses are the same ⇒ F = GM 2 (2r)2 = G( 4πr3 3 ∗ ρ) 2 4r2 ⇒ F ∝ r 6 r2 F ∝ r4 Thus ans = (a)

4. A copper cube and a wooden cube of same size are initially at room temperature. Then they are kept in an enclosure at 50◦

c. What can we conclude about the tem-peratures 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 ca-pacity and thermal conductivity of the materials.

(d) Both temperatures will be practically the same, as they are in the enclosure for 5 hours.

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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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. A repairman on the T. V. tower finds his water bottle leaking at the rate of 5ml per second. He drops the bottle and it reaches the ground straight. If he was at a height of 125m at that time and there was 200ml of water left in the bottle, the amount of water left in the bottle (neglecting air resistance) just before it hit the ground was (a) 175ml (b) 50ml (c) 100ml (d) 200ml

Solution: Ans = (d).

The force acting on a bottle and water in it, is the gravitational force (constant acceleration g). Therefore because of this free fall motion, bottle and water containing in it, will have same velocity.

Hence just before hitting the ground, water bottle will contain the same amount, 200ml of water in it.

Note: In the actual question paper, there was a typographic error with height of tower specified as 125km rather than 125m. However, as the solution is inde-pendent 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 ob-stacle 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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00 11 00 11 00 11 Obstacle A B C Screen . (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) 00 11 00 11 00 00 11 11 A B C Screen Obstacle N M L P Q R S K .

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 ,} which means MN

QS = AL AK.

Similarly, △CMN & △ CP R are congruent, which means MN

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

(a) u t _(b) t u (c) t u (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. Also since speed can not assume negative values, even option (b) is incorrect. In fact, (b) represents correct velocity-time graph in case of inelastic collisions. 10. Two glass tubes filled with water are held vertical and connected by a plastic tube

as shown in the figure. Pans are mounted on top of each piston such that (weight of the piston + pan)A = (weight of the piston + pan)B

radius of the piston A = 1.0cm and radius of piston B = 1.5cm.

A 30.0gm of mass is added in pan B, what is the mass required in pan A to balance 30.0gm in pan B? 00000 11111 000 111 000 111 0000 1111 0000 1111 0000 1111 000 111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 000 111 000 111 000 111 000 111 000 111 000 111 000 1110000011111 000 111 000 111 00 11 00 11 00 11 00 11 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 00 11 00 11₀₁ 00 11 0 1 0 1 00 1101 0 1 00 11 00 11 piston pan water B A . (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.

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Now pressure in both tubes should be the same. ∴_{P =} F1 A1 = F1 A2 m1g πr2 1 = m2g πr2 2 m1 1 = 30 1.52 ∴_m₁ ₌ 30 2.25 ∴_m₁ ₌ 30₉ 4 ⇒ m1 = 13.3gm.

Hence pan A required 13.3gm of mass to balance pan B.

Section B: (4 questions × 5 marks each)

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

(a) Andromeda galaxy and Milky Way galaxy are approaching each other. (b) Two unlike, free charges move towards each other.

(c) External work is required to compress a spring.

(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

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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 hour from both the ends, from 4am to 11pm. These busses run at average speed of 45km/hr. Taxies also run on the same route at 60km/hr with regular interval of 30min from 5am to 10pm. Following statements are based upon the number of taxies or busses crossed (not overtaken) only during travelling i.e. excluding instances of arrival and departure. Select the correct statement(s).

(a) First bus crosses 6 taxis. (b) Last taxi crosses 5 buses.

(c) Bus left at 8pm crosses 10 taxis. (d) Taxi left at 12 noon crosses 6 buses. Solution: Ans = (a), (b), (c) and (d)

Bus covers the distance in 4 hours, taxi in 3 hours.

First ST bus is at 4am and it will reach its destination at 8am. 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.

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(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).

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 trian-gles △ABC and △ADC, as shown.

A B C D P1 P2 h1 h2

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)

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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 1 2d

2_.

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 solar disk covered at the time of the maximum eclipse. (Given: cos−1 1

26 ≈ 0.49π

rad).

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)

A

E

C

B

D

t

r

q

p

(1 mark)

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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). A( [ACB) = πr2 s θ1 2π p + q = r 2 sθ1 2 . A(\ADB) = πr2 m θ2 2π q + r + t = r 2 mθ2 2 . where r = A(△ACD) and t = A(△BCD)

Also from figure, r = t. ⇒ p = r 2 sθ1 2 − r2 mθ2 2 + 2A(△ACD). (2 marks) Now using Hero’s formula for area of a triangle,

A(△ACD) = ps(s − rm)(s − rs)(s − h) where, rs = 13 ′ rm = 15 ′ h = 7′ s = rs+ rm+ h 2 ∴_{s =} 13 + 15 + 7 2 = 35 2 ′ ∴_{A(△ACD) =} r 35 2 × (35 − 30) 2 × (35 − 26) 2 × (35 − 14) 2 = r 35 2 × 5 2 × 9 2 × 21 2 = s 5 × 7 × 3 4 2 × 3 = 105 4 × √ 3 ≈ (173.2 + 8.7)₄ ≈ 91₂ (2 marks)

Now to find θ1and θ2 draw a perpendicular line BE to line CD.

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Let EC = x and EB = y. ∴_{In △ ECB,} x2_{+ y}2 _{= r}2 s. and in △ EDB, (x + h)2+ y2 = rm2 2hx + h2+ rs2 = r 2 m ∴_{x =} r 2 m− r2s− h2 2h = 0.5 ′ ∴cosθ1 = x rs = 0.5 13 = 1 26 ⇒ θ1 = 0.49π (2 marks) Similarlly, cosθ2 = x + h rm = 0.5 + 7 15 = 1 2 ⇒ θ2 = π 3 (1 mark)

These are half angles only. Total angles are double these values. ⇒ p = r 2 s× 2 × θ1 2 − r2 m× 2 × θ2 2 + 2A(△ACD) ≈ 169 × 0.49π − 225 × π 3 + 2 × 91 2 ≈ π 84 − 75 +91 × 7₂₂ ≈ π 9 + 637 22 ≈ π(9 + 28.95) ≈ 38π arcmin2 _{(1.5 marks)} and A(Sun) = πr2 s = 169π arcmin2 (0.5 marks)

∴ _{The amount of solar surface covered by moon at the time of maximum eclipse} = (1 − _A(Sun)p ) × 100% ≈ (1 − 38 169) × 100% ≈ (1 −2.923₁₃ ) × 100% ≈ (1 − 0.22) × 100%. ≈ 78 %. (1 mark)

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

b + 1 c + 1

d

where a,

b, c, d are positive integers. Find a, b, c, d. Solution: 3.1416 = 3 + 1416 10000 = 3 + 1 10000 1416 = 3 + 1 9912 + 88 1416 = 3 + 1 7 + 1 1416 88 = 3 + 1 7 + 1 1408 + 8 88 = 3 + 1 7 + 1 16 + 1 88 8 = 3 + 1 7 + 1 16 + 1 11 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.

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

V A R K + | + + | |

Here, R is the unknown resistance, V the voltmeter, A the ammeter and K is the key. He obtained following readings :

V (v) 1 2 3 4 5 6 I (mA) 1.40 2.83 4.26 5.68 7.11 8.54 (a) (9 marks) Plot appropriate graph to represent the data. (b) (2 marks) Find the value of R.

(c) (1 mark) From the graph, what will be the voltage across the resistance when I = 8mA?

Solution:

(a) From the given observation table we have plotted voltage (v) versus current (mA) graph. 1 3 5 6 1 2 4 7 8 Y−axis 10 X−axis

V

I

(mA)

I

v/s

V

(V) (V) 2 3 X−axis : 1cm = 1mA Y−axis : 1cm = 1 volt (mA) 9 4 5 6 0 7 O(0,0)

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• 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. (1.5 mark) • Chart Title, Axis Titles etc. should be properly written. (1 mark) • The scales and preferably origin should be clearly specified at the top right

corner. (1 mark)

• Points used for finding slope should be well separated, well marked and preferably not from the existing dataset. (1 mark) • Line should be continuous and well balanced. (1.5 mark) • 8mA point should be marked on the graph paper (as shown). (1 mark) (b) Hence from the above graph, (2 marks)

slope = (y2− y1) (x2− x1) ⇒ R ≈ (4.2 − 0.7)V (6 − 1)mA ≈ (3.5) (5 × 10−3₎Ω ∴_{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

= 4 × π × (6.4 × 108₎2 _cm2

Now normal size of A4 paper

a = (20 × 30) cm2

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To cover the whole earth we will require (3 marks) 4 × π × 6.42 × 1016 20 × 30 papers ≈ 82 × 10 14 papers

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

We take, density of a paper ≈ density of wood = 0.5 gm/cc (1 mark) (It should be definitely less than water as even crumpled paper floats on the water) ∴_{mass of a paper ≈ 600cm}2_{× 0.01cm × 0.5gm/cm}3 ≈ 3 gm (1 to 5 gm acceptable) (1 mark) w ≈ 8.1 × 1015× 3 ≈ 2.4 × 1016gmwt ≈ 2.4 × 1013_kgwt ≈ 24 × 109 _{Tonne wt} ≈ 2.4 × 1014_N

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HBCSE

Senior Category

Roll Number:

Roll Number

Model Solutions

INAO – 2009 Date: 31st _{January 2009}

• 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 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

⊠ ⊠

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

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HBCSE

Mass of the Earth ME ≈ 5.97 × 1024 kg

Radius of the Earth RE ≈ 6.4 × 106 m

Mass of the Sun M_⊙ _{≈ 1.99 × 10}30 _kg

Radius of the Sun R_⊙ _{≈ 7 × 10}8 _m

Radius of the Moon Rm ≈ 1.7 × 106 m

Speed of Light c _{≈ 3 × 10}8 _m/s

Astronomical Unit _{1 A. U. ≈ 1.5 × 10}11 _m

Gravitational Constant G _{≈ 6.67 × 10}−11 _m3_{/(Kg s}2₎

Gravitational Acceleration on the Earth g _{≈ 9.8 m/s}2

Gravitational Acceleration on the Moon gm ≈ 1.6 m/s2

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION Tata Institute of Fundamental Research

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Section A: (10 Q × 3 marks each)

1. If ax _{= b}y _{= c}z _{and b}2 _{= ac, then y = ?}

(a) 2xz x+z (b) xz x+z (c) √ 2xz (d) √xz Solution: a = byx c = byz b2 _{= ac = b}y xb y z = byx+ y z 2 = y x + y z = y(x + z) xz y = 2xz z + x Ans = (a)

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

(a) (b) (c) (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) r4 _{(b) r}2 _(c) 1 r2 (d) 1 4r2 1

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Solution: The distance between two spheres = 2r and masses are the same ⇒ F = GM 2 (2r)2 = G( 4πr3 3 ∗ ρ) 2 4r2 ⇒ F ∝ r 6 r2 F ∝ r4 Thus ans = (a)

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 ca-pacity and thermal conductivity of the materials.

(d) Both temperatures will be practically the same, as they are in the enclosure for 5 hours.

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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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 |_|Q| is given by,

(a) cos θ (b) sec θ _{(c) − cos θ} _{(d) − sec θ}

Solution: See figure

θ

α

_α

P

R

Q

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

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,

T2 = 4π 2 GMm r3 T2 ₌ 4π 2 gR2r 3

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

T2 ₌ 4π2 1.6 × (1.7 × 106₎2(1.8 × 10 6₎3 ≈ 4 × 10 × 1.8 3 1.6 × 1.72 × 10 6 ≈ 45 × 1.8 2 1.72 × 10 6 T ≈ 6.7 × 18₁₇ × 103 ≈ 6.7 + 6.7 17 × 103 ≈ 7100sec T ≈ 118min.

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 ob-stacle and the screen. Which of these light source positions will result in the longest shadow of the obstacle on the screen?

00 00 11 11 00 00 11 11 00 11 Obstacle A B C Screen . (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.

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00 11 00 00 11 11 00 00 11 11 A B C Screen Obstacle N M L P Q R S K .

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 ,} which means MN

QS = AL AK.

Similarly, △CMN & △ CP R are congruent, which means MN

P R = AL 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.

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)

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10. Find the resultant focal length for following system where the radius of curvature is 15cm. 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 000000000000000000000000 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 111111111111111111111111 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 glass _air water (a) 40 cm (b) 60 cm (c) 120 cm _{(d) ∞}

Solution: Refractive Index : nw = 4₃ ; ng = 3₂

Refractive Index of glass relative to water:wng =

ng

nw

= 9 8 Refractive Index of air relative to water: wna=

na

nw

= 3 4 Using the lens maker formula,

1 f = (n − 1)( 1 R1 + 1 R2 )

focal length for plano-convex glass-in-water lens, where R1 = ∞; R2 = 15cm is,

1 f1 = (wng− 1)( 1 R1 + 1 R2 ) = ( 9 8 − 1) 15 = 1 120 ∴_f₁ _{= 120 cm}

Similarly, focal length for plano-concave air-in-water lens, where R1 = -15cm;

R2 = ∞ is, 1 f2 = (wna− 1)( 1 R1 + 1 R2 ) = ( 3 4 − 1) −15 = 1 60 ∴_f₂ _{= 60 cm}

Hence resultant focal length will be 1 f = 1 f1 + 1 f2 = ( 1 120 + 1 60) = 1 40 ∴_{f = 40 cm. Ans is (a).}

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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”.

(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. 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.

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

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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).

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.

(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).

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 trian-gles △ABC and △ADC, as shown.