ALL INDIA IJSO(STAGE-I) TEST SERIESOPEN TEST/MOCK TEST PAPER # 10 Time : 2 Hr. Date : 09-11-2014 Max. Marks : 240 GENERAL INSTRUCTIONS 1. In addition to this question paper, you are given a separate answer sheet. 2. Fill up all the entries carefully in the space provided on the OMR sheet ONLY IN BLOCK CAPITALS. Incomplete/incorrect/carelessly filled information may disqualify your candidature. 3. A student has to write his/her answers in the OMR sheet by darkening the appropriate bubble with the help of HB Pencil as the correct answer(s) of the question attempted. 4. Paper carries 80 questions each of 3 marks. 5. Any rough work should be done only on the blank space provided at the end of question paper. 6. For each correct answer gets 3 marks, each wrong answer gets a penalty of 1 mark. 7. Blank papers, clip boards, log tables, slide rule, calculators, mobiles or any other electronic gadgets in any form is "NOT PERMISSIBLE". Resonance Eduventures Pvt. Ltd. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Rajasthan)-324005 Tel. No. : 0744-3192222, 3012222, 3022222 | Fax. 022-39167222, 0744-2427144 PCCP Head Office J-2, Jawahar Nagar, Main Road, Kota (Rajasthan)-324005 Contact. No. : +91-0744-2434727, 8824078330 Website : www.pccp.resonance.ac.in E-mail : [email protected] Toll Free : 1800 200 2244 |SMS RESO DLP at 56677 |[email protected] www.dlpd.resonance.a c.in (D) 12 Find the sum of (2. b. If we move a distance ‘d’ towards the foot of the tower. the angle of elevation increases to ’y’. When the curves y = log10x and y = x-1 are drawn in the x-y plane many thies do they intersect for values x 1 ? (A) Never (B Once (C) Twice (D) More than twice 6. b and c are distinct real numbers such that a : b + c = b : c + a then : 3.52 + -------. c are all negative (C) a + b + c = 0 (D) ab + bc + ca + 1 = 0 L. (A) a. The number of isossceles triangles with integer sides such that no side is greater than 4 units is : (A) 8 7. (B) 9 (B) 17587 (C) 17937. M and N are mid points of sides AB. (B) 8 sq.+ (50)2 . Then the height h of the cylinder must be : (A) 100/7 cm (B) 18 mc (C) 15 cm (D) 200 / 11 cm 2.IJSO TEST PAPER 1. units. A cylinder of radius 6 cm and height h cm is filled with ice cream.5)2 + 102 + 12. If a. b. The ice cream is then distributed among 10 children in identical cones having hemispherical tops. then the area of triangle LMN will be : (A) 6 sq. (A) – 2010 2009 (B) – 1 (C) (D) 1 2009 1 x+1+ = 0. (A) 35175 8. c are all positive (B) a. BC and CA of triangle ABC. then the height of the tower is : (A) d tan x tan y tan y – tan x (B) d(tan y + tan x) (C) d(tan y – tan x) (D) d tan x tan y tan y tan x 5.units (D) 24 sq.5 (D) None of these If the difference of (1025 – 7) and (1024 + x) is divisible by 3 then x is equal to (A) 3 9.units The angle of elevation of the top of a tower as observed from a point on the horizontal ground is ‘x’.units (C) 12 sq.units 4.5)2 + 52 + (7. The radius of the base of the cone is 3 cm and its height is 12cm. If area of triangle ABC is 48 sq. is 2010 x 2009 2010 (D) 1 Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 2 . (C) 16 (B) 2 (C) 6 The sum of the reciprocals of the roots of the equation. BE and CF are the medians of ABC. + a20x20 . then (a + b + c + d) is equal to : (A) – 5 12. then the lengths of the corresponding sides are in the ratio : (A) 2 : 3 : 4 15. If BC = 7 and AC = 6 find AB. then the difference between the length and width of the rectangle is 8d2 p2 2 (A) 11. then : (A) cos x + cos2 x = 1 14.. 5.3 20 1 2 (D) 5. (B) 6 : 4 : 3 3 BC 2 (B) > 5 BC 3 (C) > 3 BC 2 (D) < 2 BC 3 Let (1 + 2x)20 = a0 + a1x + a2x2 + . (B) (B) 4 ± 2 2 (C) 6 (D) 5 In ABC .3 20 – 1 2 The values of k for which the expression kx2 + (k + 1)x + 2 will be a perfect square of linear factor are (A) 3 ± 2 2 18. Then. 8d 2 p 2 2 (B) (B) cos x – cos2 x = 1 (C) cos2 x + cos3 x =1 (D) cos2 x + cos4 x = 1 If the altitudes of a triangle are in the ratio 2 : 3 : 4.. The sum of lengths of segments BE and CF is : (A) < 16..3 20 3 2 (C) 5. (A) 15 units (B) 17 unit (C) 13 unit (D) None of these Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 3 . (C) 3 : 2 : 4 (D) 3 : 2 : 1 AD. If the perimeter of a rectangle is p and its diagonal is d.3 20 – 3 2 5.. 6d2 p2 2 (C) If a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5. the median from A is perpendicular to the median from B. 2a19 + 3a20 equals to : (A) 17. 3a0 + 2a1 + 3a2 + 2a3 + 3a4 + 2a5 + . 13.. 8d2 p2 4 (D) (B) – 10/3 (C) – 7/3 (D) 5/3 Which one of the following pairs of lines are consistent having unique solution (A) x + y = 7 and 2x + 2y = 14 (B) x – y = 5 and 2x – 2y = 15 (C) x – y – 1 = 0 and 4x – 4y – 15 = 0 (D) 2x + y – 6 = 0 and 4x – 2y – 4 = 0 If sin x + sin2 x = 1..10. (D The electrons move randomly but slowly drift in a direction opposite to E . current through the resistor 3 is 0. but slowly drift in the direction of E . 22.19.. In ABC. 30 cm 3 (B) 20 cm 3 (C) 40 cm 3 (D) 10 cm 3 In the presence of an applied electric field ( E ) in a metallic conductor. The input voltage ‘V’ is equal to : (A) 10 V (B) 20 V (C) 5 V (D) 15 V 2 Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 4 . (A) 72º 20. (A) The electrons move in the direction of E (B) The electrons move in a direction opposite to E (C) The electrons may move in any direction randomly. In the ladder network shown. Find the measure of A. The tangents at P and Q intersect at point T. (B) 60º (C) 80º (D) none of these PQ is a chord of length 8 cm of a circle of radius 5 cm. Find the length TP. (A) 21.25 A. D is a point on BC such that AD is the internal bisector of Suppose B = 2C and CD = AB. where Re is the radius of earth. (A) 26.23. then R is : (A) 5 (B) less than 5 (C) greater than 5 (D) between 4 and 5. With what velocity should a particle of mass m be projected from the mid point of their centres so that it may escape out to infinity. R1 and M2. R2 respectively. G(M1 M2 ) d (B) 2G(M1 M2 ) d (C) 4G(M1 M2 ) d (D) GM1M2 d In the circuit shown the readings of ammeter and voltmeter are 4A and 20V respectively. Two mirrors are inclined at an angle as shown in the figure. Their centres are d distance apart. Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 5 . (B) = 30° (C) = 60° (D) all three A body of mass m is situated at a distance 4Re above the earth's surface. Light ray is incident parallel to one of the mirrors. (B) 2mgRe (C) mgR e 5 (D) mgR e 16 The mass and radius of earth and moon are M1. How much minimum energy be given to the body so that it may escape (A) mgRe 25. The meters are non-ideal. Light will start retracing its path after third reflection if : (A) = 45° 24. 31. In the figure given below. If water has refractive index of 4/3. (D) 29.5°C. The resistance of 10 and 20 are arranged in series across 220 volt supply.04 kJ. 0 g (B) 0°C.27. The minimum mass of C that may be placed on A to prevent it from moving is : (A) 15 kg (B) 10 kg (C) 5 kg (D) 12 kg 28. Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown.5°C. 2 g (C) 37. 10 g of ice at –10ºC is added to 10g of water at 85ºC. What is the final temperature and amount of ice left in the system ?(System is kept inside an ideal insulator ) (A ) 0°C. The distance between L1 and L2 will be : (A) f1 (B) f2 (C) f1 + f2 (D) f1 . 0 g (D) 37.2. The coefficient of static friction of A with table is 0. (C) 30 kJ. 5 g Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 6 . An object is placed at 24 cm distance above the surface of a lake. then the heat produced by this combination in half an hour will be : (A) 2904 kJ. a fish will sight the object : (A) 32 cm above the surface of water (B) 18 cm over the surface of water (C) 6 cm over the surface of water (D) 6 cm below the surface of water 29. (B) 29 kJ. then at what distance from lake surface.f2 30. there are two convex lens L1 and L2 having focal length of f1 and f2 respectively. The velocity at the end of its fall is (in m/s) : (A) 35. If the body penetrates x cm into the sand.32. (D) 2/17 A machine. Reading of ammeter in ampere for the following circuit is : 2V. (A) 34. (C) 2/11 A block of mass ‘ m ‘ is attached to a spring in natural length of spring constant ‘ k ‘ . (B) 1/13 h (B) Mg 1 x (C) Mgh + Mgx h (D) Mg 1 – x An object moves with 5m/s towards right while the mirror moves with 1 m/s towards the left as shown. 3 (A) 2/15 33. which is 75% efficient. the average resistance offered by the sand to the body is : h (A) Mg x 36. Find the velocity of image. Find the maximum extension in the spring. 1 – + – + 3V. The other end A of the spring is moved with a constant velocity v away from the block . 1m/s 5m/s O mirror (A 7 m/s towards left (C) 5 m/s towards left (B) 7 m/s towards right (D) 5 m/s towards right Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 7 . 1 4 m v2 k (B) (B) 24 m v2 k (C) 1 2 m v2 k (D) 2 m v2 k 12 (C) 18 (D) 9 A body of mass M is dropped from a height h on a sand floor. 1 – 2 1V. uses 12J of energy in lifting up a 1kg mass through a certain distance. The mass is then allowed to fall through that distance. 2 + – A + 1V. if t1 is the time taken in going up while t2 in coming down to starting point. Calculate molarity of the diluted solution. What volume of a 0.8 M solution contains 100 millimoles of the solute? (A) 100 mL (B) 125 mL (C) 500 mL 42.37.25 M Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 8 . A circular coil A of radius r carries current I.5 M (C) 0. A body is projected vertically up from the ground. 4 7 g (D) upward. Another circular coil B of radius 2r carries current of I. [S = 32] (A) 0. g 2 (B) 39.5 m–1 is bent into a circle of radius 1m.75 M (D) 0. g 7 7 (A) g g upward. The equivalent resistances in ohm is : (A) 38.18) containing 49% H2SO4 by mass is diluted to 590 ml.7 M (B) 7. (D) 62. 4 7 g 6 (C) downward.5 mL 75 ml of H2SO4 (specific gravity = 1. (B) +1 (C) ( 2) (D) ( 4) In the system shown in the figure. The same wire is connected across a diameter AB as shown in figure. Taking air resistance into account. then: (A) t1 > t2 (B) t1 = t2 (C t1 < t2 (D) t1 can be greater or smaller depending upon the initial velocity of the body 41. the acceleration of the 1kg mass and the tension in the string connecting between A and B is : g 8g downward. A wire of resistance 0. The magnetic field at the centres of the circular coils are in the ratio of : (A) 3 : 1 (B) 4 : 1 (C) 1 : 1 (D) 2 : 1 40. (I). Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 9 . correct datas are : (A) Total concentration of cation(s) = 0. 16 8 O (C) 40 18 Ar. The ratio of the energy of a photon of wavelength 3000 Å to that of a photon of wavelength 6000Å respectively is: (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 1 : 3 48. (A) 1. proton. positron (C) -particle.1 M (D [Cl¯] = 0. The atomic weight of an element is 'a'.225 × 1022 (C) 1. proton. Bombardment by -particle leads to artificial disintegration in three ways.5 × 1022 (D) 1. Find out the number of photons emitted by a 60 watt bulb in one minute. Products X.125 × 1022 (B) 1. proton 50. For 100 ml of 0. neutron. The fraction of atom that is occupied by nucleus is : (A) 10–5 (B) 105 (C) 10–15 (D) None of these 46.14 M (B) Total concentration of cation(s) = 0. (II) and (III) as shown. Which of the following is/are isotones : (A) 2 1 H.43. Atomic radius is of the order of 10–8 cm and nuclear radius is of the order of 10–13 cm. 24 He 47. Y and Z respectively are : (ii) 9 Be 4 12 6 C+Y (i) 16 8 O+Z (A) -particle. neutron.3 M CaCl2 solution + 400 ml of 0. (iii) 13 7 N+X (B) positron.1 M HCl solution. if wavelength of an emitted photon is 620 nm. 40 20 Ca (D) 3 1 H.125 × 1012 49. then the correct formula for the number of moles of gas in its 'w' g is : (A) 3w a (B) w 3a (C) 3wa (D) a 3w 45. neutron The ratio of the mass number and atomic number of the resulting nuclide is: (A) 22/10 (B) 22/11 (C) 23/10 (D) 23/12 Hint : During the emission of positron from the nucleus atomic number changes. A positron is emitted from 23 11 Na . If this element occurs in nature as a triatomic gas.2 M 44. proton (D) positron. 3 1 H (B) 15 7 N.07 M (C) [Cl¯] = 0. (A) – 372. (C) – 322. If an ideal gas at 1 atmosphperic pressure. (D) – 362.5 atm (C) 5 atm (D) None of these.0 kCal.3 mL (C) 43. 54. (B) (C) (D) The standard molar enthalpies of formation of ethane.1.0 kCal. A flask is of capacity one litre.3 mL (B) 23. For a fixed amount of ideal gas at constant temperature. (B) – 352.4 atm (B) 2. which of the following plots is/are not correct : (A) 58. if it is heated from 27°C to 37°C ? Assume pressure to be constant. 5 L of a sample of a gas at 27°C and 1 bar pressure is compressed to a volume of 1000 mL keeping the temperature constant. For a fixed amount of ideal gas at constant temperature. Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 10 .1 and 68.3 mL 53. then find the final pressure : (A) 0. What volume of air will escape from the open flask. The percentage increase in pressure is : (A) 100 % (B) 400 % (C) 500% (D) 80% 55. 94. which of the following plots is correct : (A) (B) (C) (D) 56.3 mL (D) 13.0 kCal. the partial pressure of hydrogen at equilibrium would be : (A) P/2 (B) P/3 (C) P/4 (D) P/6 57.51. (A) 33. CO2 & liquid water are 21. Starting with one mole of nitrogen and 3 moles of hydrogen. Calculate the standard molar enthalpy of combustion of ethane.3 kCal respectively. is spreading from 20 cm3 to 50 cm3 at constant temperature. at equilibrium 50% of each had reacted according to the reaction : N2 (g) + 3H2 (g) 2NH3 (g) If the equilibrium pressure is P. The correct representation of Charles law is given by : (A) (B) (C ) (D) 52.0 kCal. (D) maintainance of water cycle between biotic components of ecosystem. The plasma membrane is (A) permeable. Cell Division A B C D (A) 4 2 1 3 (B) 1 4 3 2 (C) 3 1 2 4 (D) 2 3 4 1 65. Mechanical strength D. (C) taenia.004 % (C) 0. (B) ascaris. (D) hydra. Transport of food C. Match List I with List II and select the correct answer using the codes given below the lists. each having 10 chromosomes 64. Plants having seeds but lacking flowers are (A) Pteridophytes. Phloem 3. Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 11 . On mitotic division of a cell with 20 chromosomes (A) two daughter cells will be created. Conduction of water B. (D) Bryophytes. Synctial epidermis is found in (A) pheretima. (A 20 ºC (B) 30 ºC (C) 40 ºC (D) 50 ºC 60. (C) selectively permeable (D) impervious 62. (C) Angiosperms.The density of formic acid is 1. 67. each having 40 chromosomes. (B) transferring of matter and energy between the abiotic components of the biosphere. Meristem 4. (B) two daughter cells will be created. List – List – (Plant tissue) (Function) A. each having 10 chromosomes. Biogeochemical cycles are (A) dynamic but stable systems of constant interaction between the biotic and abiotic components of the biosphere. The percentage of formic acid converted to formate ion are (A) 0. The self ionization constant for pure formic acid . (B) impermeable. Phylloclade is a feature of (A) vallisneria (an aquatic plant) (C) cactus (a xerophytic plant) (B) maize (a terrestrial plant) (D) mango (a tall tree) 66.008 % 61. Xylem 2.002 % (B) 0. For which temperature the pOH of pure water can be greater than 7. (B) Gymnosperms.59.006 % (D) 0. Sclerenchyma 1. (D) two daughter cells will be created. each having 20 chromosomes. K = [ HCOOH2+ ] [HCOO] has been estimated as 10 6 at room temperature .15 g/cm3. (C) four daughter cells will be created. (C) essential in the maintenance of balance between the biotic components of the biosphere. 63. (D) unhygenic food. Process of accumulation of pesticides in the human body is called (A) biomagnification. (B) bioaccumulation. At which stage of meiosis I. DNA differs from RNA : (A) In the nature of sugar alone (C) In the nature of sugar and pyrimidines (B) In the nature of purines alone (D) None of the above The first organism was (A) aerobic (B) anaerobic (C) catabolic 76. (D) biodegradation One of the factor required for the maturation of erythrocytes is (A) Vit – D (B) Vit – A (C) Vit– B12 (D) Vit – C 72. (C) bioconcentration. Ptyalin enzyme present in which fluid of the body ? (A) Bile (B) Saliva (C) Pancreatic Juice (D) None of the above 70. (B) sarcomere. False statment amonst the following is (A) ozone is a non-poisonous gas. 78. (C) oxygen molecule combines with oxygen atom to form ozone. (B) sneezing. the F2 generation has both tall and dwarf plants. 73. 80. (C) I-band. This proves the principle of : (A) Dominance (B) Segregation (C) Independent assortment (D) Incomplete Dominence 75. the homologous chromosomes separate but are held by chiasmata ? (A) Diakinesis (B) Diplotene (C) Pachytene (D) Zygotene 69.68. Syphilis is caused by (A) Treponema pallidum (C) Neisseria (B) HIV (D) Trichomonas 74. (D) ozone is found in the stratosphere layer of the atmosphere 71. 77. 79. (D) both (A) and (B) Total number of canines in permanent dental set of human is (A) 4 (B) 6 (C) 2 (D) 12 The exchange of gases in alveoli of lungs takes place by(A) passive transport (B) simple diffusion (C) osmosis (D) active transport The functional unit of contractile system in striated muscle is (A) myosin. Space For Rough Work IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 12 . If a plant is heterozygous for tallness. (B) UV radiations split oxygen molecule into atomic oxygen. (C) sexual contact. (D) actin Microbial diseases like Syphilis and AIDS are transmitted through (A) contaminated water. 4 4 4. 3. [a. unit. Volume of icecream in cylinder = volume of icecream in 12 cones ABC ~ MNL 2 Area ABC BC = (2)2 = 4 = Area MNL LN 2 1 2 3 (6)2 × h = 10 3 (3) 12 3 (3) 36 h = 10 [36 + 18] area LMN = h = 15 cm. a b = bc ca [By AA similarity] area ABC 48 = = 12 sq. c distinct & real and b + c c + a [ a b] a (c + a) = b (b + c) ac + a2 = b2 + bc ac – bc = b2 – a2 c(a – b) = (b – a)(b + a) [ a b] –c=a+b Hence. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 B C A B C B B A C C C A A B A A D A A B Ques. LN = 1 BC 2 [by mid-point theorem] A L B In ABC tan y = h b h b = tan y N M Let the height of the tower be h.ALL INDIA IJSO(STAGE-I) TEST SERIES OPEN TEST/MOCK TEST PAPER # 10 DATE : 09-11-2014 HINTS & SOLUTIONS ANSWER KEY Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 C A A C B B C B B B A C A B C B A B B C Ans. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 D B B C C C A A C A A B B C B A D C D C Ques. a + b + c = 0. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. In ABD C tan x = h bd SOL. C C C A B D C B B D B D D B C C A B A B Ques. 1. Ans. Ans. b. 2. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 1 . 7. 4. 2 3. 10. 2 2. 1. 1 4. –2010 1 2009 1 + = 2010 = – 1 2009 Thus.8. tan y tan x 2009 x 1 +1+ =0 2010 x d = h tan x tan y 2009 x 2 2010 x 2010 =0 2010 x d tan x tan y h = tan y tan x . 4.25 × = = 20 ( 20 1) ( 40 1) 6 = 20 21 41 6 625 20 21 41 100 6 625 21 41 56 125 7 41 35875 = = = 17937. 1 or d2 + 2lb = or 2lb = p2 4 p2 – d2 4 12 such combinations are possible.+ 202] = (2. (l – b)2 = l2 + b2 – 2lb = d2 – (2. 3.5)2 [12 + 22 + 32 + 42 + -----. Isosceles triangles Since perimeter of a rectangle p = 2(l + b) and diagonal of the rectangle d = Maximum side = 4 units l2 b 2 A c B a (l + b) = b p p2 or (l + b)2 = 2 4 C a<b+c l2 + b2 + 2lb = b<a+c p2 4 c<a+b Possible combinations : 4. 3 4. log10x = 6. p2 + d2 4 8d2 p2 4 a+1=b+2=c+3=d+4=a+b+c+d+ 5=k (a + 1) + (b + 2) + (c+ 3) + (d + 4) = 4k a + b + c + d + 10 = 4K SOL. log10 x = x-1 1 or xx = 10 x This is possible for only one value of x (2 < x < 3). 4. 5. 3. 2. 3. 3 1. 2. 1 2. 4 2. h tan x = 1025 – 7 – 1024 – x h d tan y h h = tan y + d tan x 1 Given (1025 – 7 ) – (1024 + x) is divisible by 3 1024 (10 – 1) – 7 – x 9 1024 – 7 – x If divisible by 3 value of x = 2 1 d = h tan x tan y 9. 1 3. 3. 4 3.5 2 4 8d2 p2 4 or (l – b) = 11. 3 3. 4. 2 4. 2.5)2 × = 6. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 2 . 2009x2 + 2010x + 2010 = 0 –2010 2010 += and = 2009 2009 For the curves to intersect. C D 5 –10 –5= 3 3 In BGC BG + GC > BC ..e.3 20 1 = 3a0 + 2a1 + 3a2 . IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 3 ..... a(2k) = b(3k) = c(4k) 2a = 3b = 4c 2x y 18.. (i) Option (D) is correct because In GFE 2 1 i........ + 4a19 + 6a20 5. D=0 (k + 1)2 – 8k = 0 a = 3 ± 2 2 ah1= bh2= ch3 A Given : h1 : h2 : h3 = 2 : 3 : 4 3 h1 = 2k..... Let x = 1 320 = a0 + a1 + a2 . h2 = 3k and h3 = 4k So. a1 b1 4 – 2 a2 b2 13.. FG + GE > FE ... 2 Let AD = h1 BE = h2 CF = h3 Area of ABC = 1 1 1 ah1 = bh2 = ch3 2 2 2 17.... – a19 + a20 .A a + b + c + d + 5 + 5 = 4K K + 5 = 4K E F 3K = 5 G 15.. 6 4 3 3 2y B Corresponding sides are in the ratio 6 : 4 : 3.... 20 1 = a0 – a1 + a2.. (ii) Add (i) and (ii) we get BE + CF > BC + FE sin x + sin2 x = 1 BE + CF > BC + sin x = cos2 x Now sin2 x + cos2 x = 1 BC 2 BC FE 2 (cos2 x)2 + cos2 x = 1 cos4 x + cos2 x = 1 3BC 2 (1 + 2x)20 = a0 + a1x + a2x2 + . + 2a19 + 3a20... Divide by 12 a b c = = ....(i) Let x = – 1 14. + a20x20 BE + CF > 16.... B 5 K= 3 By mid point theorem a+b+c+d+5=K 1 BC 2 EF = a+b+c+d=K–5= 12......320 + 1 = 6a0 + 4a1 + 6a2 + ..(ii) Multiply equation (i) by 5 and add with equation (ii) 5.. + a20 .. 4( E x D 7/2 7/2 C 49 = 4y2 + x2 ) 4 9 = 4x2 + y2 49 = 4x2 + 16y2 - * - 40 = 15 y2 SOL.... we have OP2 = OR2 + PR2 OR2 = OP2 – PR2 = 52 – 42 = 9 OR = 3 cm. (4 – ) R = RV = 20 (4 – ) R = 20 4 – is less than 4 So that.Let D be the point of intersection of AD and BE. DEI = ABI = DBI BDE is isosceles and BD = DE = EA ED || AB . In right triangles PRT and OPT. Now AIB ~ DIE (as = y = IDE. So 5C = 180º) C = 30º.y2 = 40 8 15 3 9 = 4x2 + 2 8 3 9- 8 = 4x2 3 x 2= 19 12 16 256 TP = + 42 = + 16 3 9 2 = 400 9 TP= 20 cm. 17 Req = 7 + 4 + 9 = 20 V = IReq = 1 × 20 = 20 V Draw the angle bisector BE of ABC to meet AC in E. Since OT is perpendicular bisector of PQ. BE = CE EBA = ECD in BEA and CED BEA CED So. A = 2C. ADE = y. BC = AC (as BD = AE . . IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 4 . EA = ED If DAC = y.2 = 15 kg SOL. R is greater than 5 PR = QR = 4cm In right triangle ORP. we have TP2 = TR2 + PR2 and. EBC = ECB Now BA = CD. Apply Newton’s law for system along the string mB g = (mA + mC) × g mC = = mB – mA 9 – 10 0 .Join ED. From energy conservation PEi + KEi = PEf + KEf GM1 GM 2 1 m + – mv2 = 0 + 0 d / 2 d / 2 2 V2 = 4G (m1 +m2) d V = 4G(m1 m 2 ) d Let TR = y. OT2 = TP2 + OP2 OT2 = (TR2 + PR2) + OP2 (y + 3)2 = y2 + 16 + 25 6 y = 32 16 y= 3 27. 3 4x2 + 4y2 = AB2 4 19 4 8 = AB2 3 4 3 22. 26. 25. Also CED = EAD + EDA = 2y = A = 72º 20. DIE ) So.Since B = 2C. 19 32 2 3 3 = AB AB2 = 15/3 AB = 19. (Here m = 1kg) 18 m/s At point A body has only PE. 29. Efficiency of machine. v m A //////////////////////////////////////////////////////// During maximum extension.energy theorem between point A & C. 24 1 x= /4 1 3 observer v v m A //////////////////////////////////////////////////////// = 24 24 4 = = 32 cm 3 3 4 In the frame of observer. 1 1 2 mv2 = k x max 2 2 Distance between lens is = f1 + f2 . For series combination : total resistance = R1\ + R2 Total resistance = 10 + 20 + 30 Accordintg to ohm’s law : V = IR mv 2 k x max = 34. 33. = 75% Energy used = 12 J So. SOL. tw = 85ºC Heat lost by water Qw = 10 × (85º – 0º)× 1 = 850 cal Heat required for ice to come at 0ºC and to change into water Qi = 10×[0–(–10)] × 5 + 10× 80 cal 10× 10 × 5 + 800 = 850 cal Qw = Qi So net temperature = 0ºC So ice left =0g. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 5 . Now.) = 2904 KJ 75 =9J 100 1 mv2 = 9 2 v2 = 18 v = 31. mi = 10g. PE = mg (h + x) KE = 0 A h At point C B x C KE = 0 By applying work .mw = 10g ti = –10ºC. the block will come to rest with respect to the observer. energy used to lift the mass V 220 22 I= = = A R 30 3 Heat produced = I2RT = 12 × 22 22 30 30 60 = 3 3 ( t = 30 min.28. 30. by energy conservation. block will have initial velocity v towards left. Consider an observer moving with speed v with point A in the same direction. 35. Resistance of each samicircle. Use 1 1 1 R AB = 0.1 400 500 0 . of atom = 4 (10 8 )3 3 = 10–15.6 0. E1 2 6000 E2 1 = 3000 = 2.18 10 75 = M2 × 590 98 43. 47. no. nhc 53. 50. P1 V1 = P2 V2 P2 = 1 20 atm = 0.1 400 = = = 0.6 0 .0333 – 1 = 0. 10 Suppose at T = 27°C = 300 K T1 = 37°C = 310 K V = 1 litre V1 = ? V1 V = T T 1 V1 1 = 300 310 M1V1 M2 V2 V1 V2 V1 = 310 = 1. of solution 42.5 = 1 46. 4 = 0.0333 litre 300 Since.2)100 0. 48.5 + 1 R 4 E= 60 × 60 = 41. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 6 . of solution = 125 ml (Here nsolute = mole of solute. of nucleus Fraction = vol.64 10 34 3 10 8 nsolute M= V solution 100 10 3 0 .] 44. h F = Mg 1 x Molecular wt. n 6. 620 10 9 n = 1.3 mL Ans.8 = 1000 vol. R1 = R2 = r × 0.5 Resistance of diameter (AB1. 50 SOL. 3(0. So ratio of 23 . M1V1 = M2V2 Molarity of cation = = 23 10 Ne + 0 1 e .5= 0.5 + 0. R3 = 2r × 0. capacity of flask is 1 litre. of gas = 3a w .2 M 5 at constant pressure 0 .2 100 0.125 × 1022 vol.12 M 5 500 Molarity of Cl— = 23 11 Na atomic mass and atomic number = 49 1. Isotones have same number of neutrons. of moles of gas = 45. of solution).0333 litre = 33.4 atm Ans. 37. 52. Vsolution = vol. 3a 4 (10 13 )3 3 vol. Volume of air escaped out = 1.Work done by gravity + work done by resistance = KE at pt A – KE at pt C Mg (h + x) – Fx = 0 – 0 Fx = Mg (h + x) [Here F is the resistance offered. For H2O [H+] T.54. 59. % increase = 56.5 N2 1 0. 1 . [HCOO–] = 10–3 mol/L.004%. 60.5 + 3H2 3 1.5 PH2 1. out of 25 mol HCOOH 10–3 mol are ionised into HCOO– ions.5 P = P/2 = 3 5 1 100 = 400% 1 2NH3 0 1 K = [HCOOH2+] [HCOO–] = 10–6. 1 2. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 7 . moles of (HCOOH) in 1 litre solution = 1150 46 = 25 mol. T SOL. pH Kw T. % dissociation = 10 –3 × 100 25 = 0. 1 liter solution of HCOOH has = 1150 g mass.5 P2 = = 5 atm 0 .