FULL SYLLABUS TEST-1 KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) | STREAM (SA)_XI ® Time: 3 Hours Max. Marks : 100. Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. INSTRUCTIONS 1. Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited. 2. The question paper consists of two parts (both contain only multiple choice questions) for 100 marks. There will be four sections in Part-A (each containing 15 questions) and four sections in Part-B (each containing 5 questions). 3. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. MARKING SCHEME : PART-A : MATHEMATICS Question No. 1 to 15 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PHYSICS Question No. 16 to 30 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. CHEMISTRY Question No. 31 to 45 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. BIOLOGY Question No. 46 to 60 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PART-B : MATHEMATICS Question No. 61 to 65 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. PHYSICS Question No. 66 to 70 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. CHEMISTRY Question No. 71 to 75 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. BIOLOGY Question No. 76 to 80 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. 4. No deduction from the total score will be made if no response is indicated for an item in the Answer sheet. 5. No candidate is allowed to carry any textual material, printed or written, bits of papers, paper, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room. 6. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page. 7. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. 8. Do not fold or make any stray marks on the Answer Sheet. Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | 73400 10333 Website : www.resonance.ac.in | E-mail : [email protected]| CIN: U80302RJ2007PLC024029 Name of the Candidate : I have read all the instructions and shall abide by them ....................................... Signature of the Candidate Roll Number : I have verified all the information filled by the candidate. ........................................ Signature of the Invigilator
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FULL SYLLABUS TEST-1
KISHORE VAIGYANIK PROTSAHAN YOJANA
(KVPY) | STREAM (SA)_XI
®
Time: 3 Hours Max. Marks : 100.
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
INSTRUCTIONS
1. Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.
2. The question paper consists of two parts (both contain only multiple choice questions) for 100 marks. There will be four sections in Part-A (each containing 15 questions) and four sections in Part-B (each containing 5 questions).
3. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.
MARKING SCHEME :
PART-A :
MATHEMATICS
Question No. 1 to 15 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PHYSICS Question No. 16 to 30 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. CHEMISTRY Question No. 31 to 45 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. BIOLOGY
Question No. 46 to 60 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PART-B :
MATHEMATICS
Question No. 61 to 65 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response.PHYSICS Question No. 66 to 70 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. CHEMISTRY Question No. 71 to 75 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. BIOLOGY
Question No. 76 to 80 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response.
4. No deduction from the total score will be made if no response is indicated for an item in the Answer sheet. 5. No candidate is allowed to carry any textual material, printed or written, bits of papers, paper, mobile phone,
any electronic device, etc., except the Admit Card inside the examination hall/room. 6. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the
bottom of each page. 7. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the
Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. 8. Do not fold or make any stray marks on the Answer Sheet.
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
1. The number of real roots of the polynomial equation x4 – x2 + 2x – 1 = 0 is (A) 0 (B) 2 (C) 3 (D) 4 2. Let P be an interior point of a convex quadrilateral ABCD and K, L, M, N be the midpoints of AB, BC,
CD, DA respectively. If area (PKAN) = 25, area(PLBK) = 36, and area(PMDN) = 41 then area(PLCM) is : (A) 20 (B) 29 (C) 52 (D) 54
3. The number of polynomials p(x) with integer coefficients such that the curve y = p(x) passes through (2, 2) and (4, 5) is
(A) 0 (B) 1 (C) more than 1 but finite (D) infinite
4. Let N1 = 255 + 1 and N2 = 165. Then
(A) N1 and N2 are coprime (B) the HCF (Highest Common Factor) of N1 & N2 is 55 (C) the HCF of N1 and N2 is 11 (D) the HCF of N1 and N2 is 33
5. Let s be the sum of the digits of the number 152 × 518 in base 10. Then (A) s < 6 (B) 6 s < 140 (C) 140 s < 148 (D) s 148 6. Let t be real number such that t2 = at + b for some positive integers a and b. Then for any choice of
positive integers a and b. t3 is never equal to (A) 4t + 3 (B) 8t + 5 (C) 10t + 3 (D) 6t + 5 7. Let p(x) = x2 + ax + b have two distinct real roots, where a, b are real numbers. Define g(x) = p(x3) for all
real number x. Then which of the following statements are true ? I. g has exactly two distinct real roots II. g can have more than two distinct real roots III. There exists a real number a such that g(x) for all real x
(A) Only I (B) Only I and III (C) Only II (D) Only II and III 8. Let x0, y0 be fixed real numbers such that 2 2
0 0x y > 1. If x, y are arbitrary real numbers such that
x2 + y2 1, then the minimum value of (x – x0)2 + (y – y0)2 is
9. The number of non-negative integer solutions of the equations 6x + 4y + z = 200 and x + y + z = 100 is :
(A) 3 (B) 5 (C) 7 (D) Infinite 10. The number of pairs (a, b) of positive real numbers satisfying a4 + b4 < 1 and a2 + b2 > 1 is (A) 0 (B) 1 (C) 2 (D) more than 2
11. Consider a triangle PQR in which the relation QR2 + PR2 = 5PQ2 holds. Let G be the point of
intersection of medians PM and QN. Then QGM is always (A) less than 45º (B) obtuse
(C) a right angle (D) acute and larger than 45º 12. Let x1, x2, …., x11 be 11 distinct positive integers. If we replace the largest of these integers by the
median of the other 10 integers, then (A) the median remains the same (B) the mean increases (C) the median decreases (D) the mean remains the same
13. All the vertices of a rectangle are of the form (a, b) with a, b integers satisfying the equation (a – 8)2 – (b – 7)2 = 5. Then the perimeter of the rectangle is (A) 20 (B) 22 (C) 24 (D) 26 14. Let N be the least positive integer such that whenever a non-zero digit c is written after the last digit of
N, the resulting number is divisible by c. The sum of the digits of N is : (A) 9 (B) 18 (C) 27 (D) 36
15. Let f be a function defined on the set of all positive integers such that f(xy) = f(x) + f(y) for all positive integers x,y. If f(12) = 24 and f(8) = 15, the value of f(48) is
(A) 31 (B) 32 (C) 33 (D) 34
PHYSICS
16. In an experiment on simple pendulum to determine the acceleration due to gravity, a student measures the length of the thread as 63.2 cm and diameter of the pendulum bob as 2.256 cm. The student should take the length of the pendulum to be
(A) 64.328 cm (B) 64.3 cm (C) 65.456 cm (D) 65.5 cm 17. A cylindrical vessel of base radius R and height H has a narrow neck of height h and radius r at one
end (see figure).The vessel is filled with water (density w) and its neck is filled with immiscible oil (density 0) Then the pressure at
H
M N
h
2r
2r
(A) M is g (h0 + Hw) (B) N is g (h0 + Hw) 2
2
R
r
(C) M is gHw (D) N is 22
20
2w
rR
hrHRg
18. A car around uniform circular track of radius R at a uniform speed v once in every T second. The
magnitude of the centripetal acceleration is ac. If the car now goes uniformly around a larger circular track of radius 2R and experiences a centripetal acceleration of magnitude 8ac, then its time period is.
19. A rigid ball rolls without slipping on a surface shown below.
Which on the following is the most likely representation of the distance travelled by the ball vs time
graph ?
(A)
time A.
distance
(B)
time B.
distance
(C)
time C.
distance
(D)
time D.
distance
20. A box filled water has a small hole on its side near the bottom. It is dropped from the top of a tower. As
it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show (A) the water coming down forming a parabolic stream. (B) the water going up forming a parabolic stream. (C) the water coming out in a straight line. (D) no water coming out.
21. If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity
profile as function of height ? Assume g = 10 m/s2.
22. A particle at a distance of 1m from the origin starts moving such that dr/d = r, where (r, ) are polar coordinates. Then the angle between resultant velocity and tangential velocity component is :
(A) 30 degrees (B) 45 degrees (C) 60 degrees (D) dependent on where the particle is 23. A student performs an experiment to determine the acceleration due to gravity g. The student throws a
steel ball up with initial velocity u and measures the height h traveled by it at different times t. The graph the student should plot on a graph paper to readily obtain the value of g is.
(A) h versus t (B) h versus t2 (C) h versus t (D) h/t versus t
24. Consider a bowl filled with water on which some black pepper powder have been sprinkled uniformly. Now a drop of liquid soap is added at the center of the surface of water. The picture of the surface immediately after this will look like
(A)
(B)
(C)
(D)
25. A solid cube of wood of side 2a and mass M is resting on a horizontal surface as shown in the figure.
The cube is free to rotate about a fixed axis AB. A bullet of mass m (<< M) and speed v is shot horizontally at the face opposite to ABCD at a height of 4a/3 from the surface to impart the cube and angular speed . It strikes the face and embeds in the cube. Then is close to (note : the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is
22Ma3
A
B
C
D
(A) Mvma
(B) Mv2ma
(C) mvMa
(D) mv2Ma
26. In Guericke's experiment to show the effect of atmospheric pressure, two copper hemispheres were
tightly fitted to each other to form a hollow sphere and the air from the sphere was pumped out to create vacuum inside. If the radius of each hemisphere is R and the atmospheric pressure is P, then the minimum force required (when the two hemispheres are pulled apart by the same force) to separate the hemispheres is
27. A ball is dropped vertically from height h and is bouncing elastically on the floor (see figure). Which of
the following plots best depicts the acceleration of the ball as a function of time.
t
h
(A)
t
acce
lera
tion
(B)
t
acce
lera
tion
(C)
t
acce
lera
tion
(D)
t
acce
lera
tion
28. A particle starts moving along a line from zero initial velocity and comes to rest after moving distance d. During its motion it had a constant acceleration f over 2/3 of the distance, and covered the rest of the distance with constant retardation. The time taken to cover the distance is (A) 2d/3f (B) 2 d/3f (C) 3d/ f (D) 3d/2f
29. A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible
mass as shown. When a 50 Kg mass is attached to the other end of the piston, it moves down. If the air in the enclosure is cooled from temperature T to T – T, the piston moves back to its original position. Then T/T is close to (Assuming air to be an ideal gas, g = 10 m/s2, atmospheric pressure is 105 Pascal),
L
(A) 0.01 (B) 0.02 (C) 0.04 (D) 0.09 30. Force F applied on a body is written as ˆˆ ˆF (n.F)n G , where n̂ is a unit vector. The vector G is equal
to
(A) n̂ F (B) ˆ ˆn (n F) (C) ˆ(n F) F/ | F | (D) ˆ ˆ(n F) n
31. Fluoxymesterone, C20H29FO3, is an anabolic steroid. A solution is prepared by dissolving 10.0 mg of the steroid in 500 mL of water. How many moles of Fluoxymesterone are present in 1 mL of solutions.
32. What volume of 2 10–4 M Ba(OH)2 must be added to 300 mL of a 1 10–4 M HCl solution to get a solution in which the molarity of hydronium (H3O
+) ions is 5 10–11 M ? (A) 375 mL (B) 300 mL (C) 225 mL (D) 450 mL 33. An electron is accelerated from rest and it has wavelength of 1.41 Å by how much amount potential
should be dropped so that wavelength associated with electron becomes 1.73 Å. (A) 25 V (B) 50 V (C) 75 V (D) 12.5 V
34. In which of the followng arrangements, the sequence is not strictly according to the property written against it ?
(D) MgF2 < CaF2 < SrF2 < BaF2 – Lattice energy. 35. The hybridisation and shape of BrF3 molecule are: (A) sp3d and T shaped (B) sp2d2 and tetragonal (C) sp3d and bent (D) none of these 36. Rate of a chemical reaction does not depend upon : (A) temperature of reaction (B) concentration of reactants (C) moles of reactants (D) catalyst
37. Standard enthalpy of vapourisation vap Hº for water at 100ºC is 40.66 kJ mol–1. The internal energy of vaporisation of water at 100ºC (in kJmol–1) is :
(Assume water vapour to behave like an ideal gas). (A) + 37.56 (B) – 43.76 (C) + 43.76 (D) + 40.66
38. Which species among the following have smallest ionic radius :Al3+, K+ , Mg2+ , Rb+ (A) Al3+ (B) K+ (C) Mg2+ (D) Rb+ 39. A gas cylinder contains 1 mole oxygen gas at 2.46 atm pressure and 27ºC. The mass of oxygen that
would escape, if first the cylinder was heated to 127ºC and then the valve was held open until the gas pressure was reduce to 1 atm, the temperature being maintained at 127ºC, is :
(A) 22.4 g (B) 11.2 g (C) 20.8 g (D) 9.6 g 40. Which one of the following alkali metals gives hydrated salts? (A) Li (B) Na (C) K (D) Cs 41. Which of the following statement is correct ? (A) The complete transfer of electron takes place in the inductive effect. (B) Inductive effect increases with increase in distance. (C) The resonance structures are hypothetical structure and they do not represent any real molecule. (D) The energy of resonance hybrid is always more than that of any resonating structure.
Order of rate of electrophilic addition reaction with HBr will be : (A) IV> I > III > II (B) I > II > III > IV (C) I > III > II > IV (D) IV > I > II > III 45. Write the IUPAC name of following compound
58. Blood carries the CO2 in 3 forms the correct percentages of CO2 in these forms are
As Carbamino-haemoglobin in RBC As bicarbonates Dissolved form in plasma
(A) 20-25% 70% 7%
(B) 70% 20-25% 7%
(C) 20-25% 7% 70%
(D) 7% 20-25% 70%
59. Hypothalamus is the important part of the brain concerned with homeostasis of the body. The functions
of hypothalamus are
(A) Secretion of posterior pituitary hormones (B) Regulation of body temperature
(C) Regulation of water balance (D) All of these
60. Benedict's solution is not reduced by
(A) Glucose (B) Fructose (C) Maltose (D) Sucrose
PART- B
Two Mark Questions
MATHEMATICS 61. The number of solid cones with integer radius and integer height each having its volume numerically
equal to its total surface area is (A) 0 (B) 1 (C) 2 (D) infinite
62. Consider a semicircle of radius 1 unit constructed on the diameter AB, and let O be its centre. Let C be a point on AO such that AC : CO = 2 : 1. Draw CD perpendicular to AO with D on the semicircle. Draw OE perpendicular to AD with E on AD. Let OE and CD intersect at H. Then DH equals
(A) 1
5 (B)
1
3 (C)
1
2 (D)
5 – 12
63. What is the sum of all natural numbers n such that the product of the digits of n (in base 10) is equal to
36n10n2 ? (A) 12 (B) 13 (C) 124 (D) 2612
64. Let ABC be a triangle in which AB = BC. Let X be a point on AB such that AX : XB = AB : AX. If AC =
65. If a 3-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number (which is a 3-digit number) is divisible by 4 and 5 ?
(A) 1
45 (B)
29180
(C) 1160
(D) 14
PHYSICS
66. Persons A and B are standing on the opposite sides of a 3.5 m wide water stream which they wish to cross. Each one of them has a rigid wooden plank whose mass can be neglected. However, each plank is only slightly longer than 3m. So they decide to arrange them together as shown in the figure schematically. With B (mass 17 kg) standing, the maximum mass of A, who can walk over the plank is close to,
3.5 m
3 m A B
(A) 17 kg. (B) 65 kg (C) 80 kg (D) 105 kg 67. A uniform metal plate shaped like a triangle ABC has a mass of 540 gm. The length of the sides AB,
BC, and CA are 3 cm, 5 cm and 4 cm, respectively. The plate is pivoted freely about the point A. What mass must be added to a vertex, so that the plate can hang with the long edge horizontal ?
(A) 140 gm at C (B) 540 gm at C (C) 140 gm at B (D) 540 gm at B 68. A 20 gm bullet whose specific heat is 5000 J(kg–°C) and moving at 2000 m/s plunges into a 1.0 kg
block of wax whose specific heat is 3000 J (kg–°C) . Both bullet and wax are at 25°C and assume that (i) the bullet comes to rest in the wax and (ii) all its kinetic energy goes into heating the wax. Thermal temperature of the wax in °C is close to.
69. On a pulley of mass M hangs a rope with two masses m1 and m2 (m1 > m2) tied at the ends as shown in
the figure. The pulley rotates without any friction, whereas the friction between the rope and the pulley is large enough to prevent any slipping. Which of the following plots best represents the difference between the tensions in the rope on the two sides of the pulley as a function of the mass of the pulley ?
m1 m2
M
T1 T2
(A)
M 0
T1
– T
2
(B)
M 0
T1
– T
2 (C)
M 0
T1
– T
2
(D)
M 0
T1
– T
2
70. A smaller with side b (depicted by dashed lines) is excised from a bigger uniform cube with side as shown below such that both cubes have a common vertex P. Let X = a/b. If the centre of mass of the remaining solid is at the vertex O of smaller cube then X satisfies.
b
P
O
(A) x3 – x2 – X – 1 = 0 (B) x2 – x –1 = 0 (C) x3 – X2 – X – 1 = 0 (D) X3 – X2 – X + 1 = 0
CHEMSITRY
71. 10 g of a hydrocarbon (not necessarily alkane) on analysis was found to contain 1 g hydrogen. If all H-atoms from 1 mole of hydrocarbon are removed and converted into H2 gas, then the gas produced has mass of 8 g. Then, molecular formula of hydrocarbon is :
(A) C2H4 (B) C3H4 (C) C4H8 (D) C6H8 72. Given below are a set of half-cell reactions (acidic medium) along with their E° (V with respect to normal hydrogen electrode) values.
2 + 2e– 2¯ E° = 0.54 Cl2 + 2e– 2Cl¯ E° = 1.36 Mn3+ + e– Mn2+ E° = 1.50 Fe3+ + e– Fe2+ E° = 0.77 O2 + 4H+ + 4e– 2H2O E° = 1.23 Among the following, identify the correct statement : (A) Chloride ion is oxidised by O2 (B) Fe2+ is oxidised by iodine (C) Iodide ion is oxidised by chlorine (D) Mn2+ is oxidised by chlorine
73. The bond dissociation energy of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal mol–1 respectively. The enthalpy of formation for HCl gas will be
(A) – 44.0 kcal (B) – 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal 74. The correct stability order of following is :
(I) (II) (III) (IV)
(A) I > II > III > IV (B) III > IV > II > I (C) II > IV > III > I (D) IV > III > II > I 75. Member of which of the following pair of isomers are not position isomers ?
(A) & (B) &
(C) & (D) &
BIOLOGY
76. Select the mismatched pair (A) Crossing over - Recombinase (B) Alignment of chromosomes at the equatorial plate - Metaphase (C) Separation of sister chromatids - Anaphase-I (D) G2 - Period of cytoplasmic growth and
DNA content is 4C
77. Apical dominance in higher plants is due to (A) Presence of Auxin (B) Enzyme activity and metabolism (C) Carbohydrates (D) Photoperiodism
78. Which of the following statement is absolutely correct? (A) Human kidneys can produce urine nearly fourteen times concentrated than the initial filtrate formed. (B) ADH can affect the kidney function by its dilator effects on blood vessels (C) ANF can cause vasodilation (dilation of blood vessels) and thereby decrease the blood pressure (D) NaCl is transported by the descending limb of Henle’s loop which is exchanged with the ascending
limb of vasa recta
79. Fight or flight reaction causes activation of the (A) Parathyroid glands, leading to increased metabolic rate (B) Kidney, leading to suppression of renin angiotensin-aldosterone pathway (C) Adrenal medulla, leading to increased secretion of epinephrine and norepinephrine (D) Pancreas leading to a reduction in the blood sugar levels 80. ATP is (A) Adenosine D-ribose -3- phosphate (B) Adenosine L-ribose -3- phosphate (C) Adenine D-ribose -3- phosphate (D) Adenine L-ribose -3- phosphate
1. x4 – x2 + 2x – 1 = 0 X2 – (x –1)2 = 0 (x2 + x – 1) (x2 – x + 1) = 0 x2 + x – 1 = 0 has two real roots. 2.
M
N
B
D
C
L
K
A
P
Note : Area of APN = Area of PDN Area of APK = Area of PBK Area PCL = Area of PBL Area of PCM = Area of PDM Hence . Area (PKAN) + Area (PLCM) = Area (PMDN) + Area (PLBK) Hence Area (PLCM) = 36 + 41 – 25 = 52 3. y = P(x) = a0 + a1 a2 x2 + ……+anxn
a0,a1,a2,a3,……….an I 2 = P(2) ….(1) 5 = P(4) ….(2) By (1) & (2)
3 = a1 (4 – 2) + a2 (42 – 22) + a3(43 –23) + …….. + an(4n – 2n) Clearly RHS is even and LHS is odd no polynomial exists.
4. 165 = 3 × 5 × 11 x + y divides xn + yn
32 + 1 divides 3211 + 111
Hence N1 is multiple of 33, simultaneously unit digit in N1 is 9 so it not the multiple of 5
Hence HCF of N1 & N2 is 33 5. 152 × 518
= 9 × 520
2010log 9 5 = 3
10 102log 20log 5
= 2 4771 20 1– 0.3010
= 14 {characteristic} Hence the number have 15 digits At worst all digit can have value 9 Hence sum should less then 135 And the last digit should be 5 Hence Sum should greater than Or equal to 6
6. t2 = a + b
t3 = t2 . t = at2 + bt
= a(at + b) + bt t3 = (a2 + b) .t + ab
t3 = 4t + 3 when a = 1 and b = 3 t3 = 10t + 3 when a = 3 and b = 1 t3 = 6t + 5 when a = 1 and b = 5
but t3 can never be equal to 8t + 5 for any point of a and b.
7. If , are roots of p(x) = 0 Roots of g(x) = 0 are x3 = , Hence only 2 real roots
further g(x) b– 4
a2
xR.
Hence I & III are correct
8. 1yx 20
20 x0 – y0 fixed
x, y arbitrary
1yx 22 , Let x = con
y = sin for men z = (x – x0)2 + (y – y0)2
z = x2 + x02 + y2 + y02 – 2 (x x0 + y y0) put x - cos , y = sin z = 2
9. Subtracting the given equations we get 5x + 3y = 100
x = 20 – 5y3
y is multiple of 5, let y = 5k x = 20 – 3k k = 0, 1, 2, ……6 Hence numbers of solutions are 7. 10. Let a2 = m & b2 = N then m > 0 and N > 0
Now given condition is M + N > 1 and M2 + N2 < 1
(M,N) lies inside circle x2 + y2 < 1 and above line x + y > 1 (M, N) lies in shaded region and number of points in shaded region are infinite, so number of pair (a,b) are also infinite.
11.
G
M R Q
P
N
L
Let QR= p, PR = q, PQ =r Given p2 +q2 = 5r2
Now 22
22
2PM
3QN
GMQG
= 9
PM9
QN4
22
=
222222 pq2r241
qp2r241
.491
= 22
QM4
p
Hence Angle QGM is 90º. 12. Let x1 < x2 < x3 ……x11
median of x1, x2 …..x10 is 2
xx 65
Now the new set of number are x1, x2,
….x5, 2
xx 65 , x6, ….x10
Hence median is 2
xx 65 < x6
median decreases 13. (a – 8)2 – (b – 7)2 = 5 (a – b –1) (a + b – 15) = 5 I1 I2 Four cases I1 I2 5 1 1 5 –5 –1 –1 –5 Case-1 a – b – 1 = 5 & a + b – 15 = 1 a = 11, b = 5 Case-2
a – b – 1 = –5 & a + b – 15 = –1 a = 5, b = 9 Case-3
a – b – 1 = 1 & a + b – 15 = 5 a = 11, b = 9 Case-4
a – b – 1 = –1 & a + b – 15 = –5 a = 5, b = 5
Perimeter = 4 + 4 + 6 + 6 = 20 14. Let the number be N If we write c after the last digit now number
is 10N + c Now c|10N + c c = 1, 2, ….., 9 c|10N c = 1, 2, ….., 9 c|N for c = 4, 7, 9 Hence N is LCM of (4, 7, 9) = 252
16. r = 1.128cm leff = l + r = 63.2 + 1.128 = 64.3
17.
H
M N
h
2r
2r
A
Pressure at Point A PA = 0hg Pressure at M PM = PA + W Hg 0h + W H)g
18. (i) 2
c
Va
R and
2 RT
V
(ii) ac' = 8ac
2
1V2R
= 2V
8R
V1 = 4V
T' = 2 (2R)
4V
= RV
= T2
19. During slope speed will be increase, so
distance covered will increase parabolically
with the time. Then speed will remain constant so further graph will be a straight line with greater slope than the first straight line portion of graph. So if change in speed at the sharp corners is not considered, the best answer should be (B) rather than C.
V
t
time
distance
However if kink is considered and Collision is inelastic then graph will be (C). But in either case answer cannot be D
20. Since bucket and water both are in state of free fall so water will not come out of the hole. (D)
34. Lattice energy decreases as the size of the metal increases.
MgF2 = – 2906 kJ mol– ;
CaF2 = –2610 kJ mol–1
SrF2 = – 2459 kJ mol– ;
BaF2 = – 2367 kJ mol–1
37. H = E + n(g) RT 40.66 × 1000 = E + (1) × 8.314 × 373. E = 37.56 kJ mol–1
38. Rb+ and K+ has more number of shells than Mg2+ and AI3+ . AI3+ and Mg2+ are isoelectronic but AI3+ has higher nuclear charge so AI3+ < Mg2+. Hence Al3+ is the smallest one.
K+ = 1.38 Å, Mg2+ = 0.72 Å, Al3+ = 0.535 Å and Rb+ = 1.64 Å
39. 1 1
1
n TP =
2 2
2
n TP
1 3002.46
=
2n 4001
n2 = 0.3
Mass of oxygen left = 0.3 × 32 = 9.6 g
Mass of oxygen escaped = 1 × 32 – 9.6 = 22.4 g.
42. Position of nuclei does not changes in case of resonating structures. 43. R is CH3–CH2–CH2–CH2– or CH3 – –
CH2 – CH3
44. Rate of electrophilic addition reaction stability of carbocation.
45.
3-Bromo-2-methyl-4-sulphohexanoic acid
PART- B MATHEMATICS
61. Leh Height of cone = h
Radius of base = r
And slant height = ; = = 22 hr Given volume = surface area.
22 rrhr31
r33rh r3–rh31
3–h3r
hr 22
9h6–h9r
hr 22
22
3hr2–
9hr
h222
2
9–r
546
9–r
r6h
22
2
Since h and r must be integers, and r2 – g must be a factor of 54
Since O < k2 – 1 < 6 k = 2 is only such value; for which h is integer. So, r = 2 × 3 = 6i
h = 6 + 36 = 8
10 is the only possible values for r and h.
62.
H E
AB
D r = 1
O C
ACCO
= 21
OC = 13
AC = 23
OD = 1
OCD : OD2 = OC2 + CD2
1 = 19
+ CD2
CD = 89
Let ADC =
tan = ACCD
= 2
8 =
1
2
cos = DEDH
= DA2DH
DH =DA
2cos { cos =
CDAD
}
= 2 2
3 2 1 1
12
= 1
2
DH = 1
2
63. Product of digits of natural number will be a non negative integer so, n2 – 10n – 36 0 n (–, 5 – 61 ] (5 + 61 , ) but n IN so n 13; where n N case-1 for all 2 digit natural numbers max value of product of digits = 9 × 9 = 81
so n2 – 10n – 36 81 n [5 – 142 , 5 + 142 ] but n is taken as a 2 digit natural no.; so 13 n < 17; product of digits = 3, 4, 5 or 6 for 13, 14, 15 and 16 respectively checking n = 12 product of digits = 1 × 3 = 3 and 132 – 10 × 13 – 36 = 3 so 13 satisfies the given condition Hence it is a solution chcking for n = 14 product = 1 × 4 = 4 142 – 10 × 14 – 36 = 196 – 140 – 36 = 20 > 6 and n2 – 10n – 36 is increasing function for n > 5; rest of the 2 digit integers won't satisfy the given condition case-2 for all 3-digit integers max product = 9 × 9 × 9 = 729 The smallest 3 digit no. is 100 f(n) = n2 – 10n – 36; f(100) = 1002 – 10 × 100 – 36 = 8964 > 729 and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy n = 13 is the only answer.
64.
B
x
D
C b A
a–b
b
Let AC = b & AB = BC = a
Given ba
bab
b2 = a2 – ab
2
51ab
Let D be foot of perpendicular from B
sin (ABD) = a2
b=
415
ABD = 18º ABC = 36°
65. Let abc be required numbers which is
divisible by 20 Clearly one digit must be 0 and one digit must be even, other can be any
23. ,d fo|kFkhZ xq:Roh;Roj.k g dks Kkr djus ds fy, ,d iz;ksx djrk gSA fo|kFkhZ ,d LVhy dh xsan dks izkjfEHkd xfr u ls Åij Qsadrk gS vkSj vyx&vyx le;ksa t ij xsan }kjk r; dh x;h Å¡pk;h h dks ekirk gSA g ds eku dks 'kh?kzrk ls Kkr djus ds fy, fo|kFkhZ dks ,d xzkQ isij ij dkSulk xzkQ vkjsf[kr djuk pkfg, \
(A) h ds lkis{k t (B) h ds lkis{k t2 (C) h ds lkis{k t (D) h/t ds lkis{k t
24. ikuh ls Hkjh ,d dVksjh esa dkyh ehpZ ds pw.kZ dks ,d leku :i ls fNM+dk x;k gSA vc dVksjh esa fLFkr ikuh ds lrg ds chpksa chp lkcqu ?kqys nzo dh ,d cw¡n fxjkbZ tkrh gSA blds rqjUr ckn nzo dh lrg dSlh fn[kkbZ nsxh\
(A)
(B)
(C)
(D)
25. fp=k esa 2a Hkqrk okyk ,oa M nzO;eku ds ,d ycM+h dk Bksl ?ku ,d {kSfrt ry ij foJkekoLFkk esa gSA ;g ?ku vius
AB ds lkis{k eqDr :i ls ?kw.kZu dj ldrk gSA m (<< M) nzO;keku dh ,d xksyh {kSfrt pky v ls ABCD ds Bhd foijhr okyh lrg ls 4a/3 ÅWpkbZ ij Vdjkdj ?ku dks dks.kh; osx iznku dj nsrk gSA Vdjkus ds ckn xksyh ds
HBr ds lkFk bysDVªkWuLusgh ;kSxkRed vfHkfØ;k dh nj dk Øe gksxk % (A) IV> I > III > II (B) I > II > III > IV (C) I > III > II > IV (D) IV > I > II > III
(A) 2-esfFky-3-czkseks-4-lYQksgsDlsuksbd vEy (B) 3-czkseks-2-esfFky-4-lYQksgsDlsuksbd vEy (C) 4-czkseks-5-dkcksZDlhgsDlsu-3-lYQksfud vEy (D) 5-dkckZsDlh-4-czkseks-3-lYQksfud vEy
Hence N1 is multiple of 33, simultaneously unit digit in N1 is 9 so it not the multiple of 5
Hence HCF of N1 & N2 is 33 5. 152 × 518
= 9 × 520
2010log 9 5 = 3
10 102log 20log 5
= 2 4771 20 1– 0.3010
= 14 {characteristic} Hence the number have 15 digits At worst all digit can have value 9 Hence sum should less then 135 And the last digit should be 5 Hence Sum should greater than Or equal to 6
6. t2 = a + b t3 = t2 . t = at2 + bt = a(at + b) + bt t3 = (a2 + b) .t + ab t3 = 4t + 3 when a = 1 and b = 3 t3 = 10t + 3 when a = 3 and b = 1 t3 = 6t + 5 when a = 1 and b = 5 but t3 can never be equal to 8t + 5 for any point of a and b.
7. ;fn , ds ewy p(x) = 0 g(x) = 0 ds ewy x3 = ,
vr% dsoy 2 okLrfod ewy gSA
iqu% g(x) b– 4
a2
xR.
8. 1yx 20
20 x0 – y0 fixed
x, y vpj
1yx 22 , ekuk x = con
y = sin z = (x – x0)2 + (y – y0)2
z = x2 + x02 + y2 + y02 – 2 (x x0 + y y0) x - cos , y = sin z = 2
y is multiple of 5, let y = 5k x = 20 – 3k k = 0, 1, 2, ……6 vr% gyksa dh la[;k 7 gSA
10. ekuk a2 = m vkSj b2 = N rc m > 0 rFkk N > 0 vc fn;k x;k izfrcU/k M + N > 1 vkSj M2 + N2 < 1
(M,N) lies inside circle x2 + y2 < 1 and above line x + y > 1 (M, N) lies in shaded region and number of points in shaded region are infinite, so number of pair (a,b) are also infinite.
11.
G
M R Q
P
N
L
ekuk QR= p, PR = q, PQ =r fn;k x;k p2 +q2 = 5r2
vc 22
22
2PM
3QN
GMQG
= 9
PM9
QN4
22
=
222222 pq2r241
qp2r241
.491
= 22
QM4
p
vr% QGM dk dks.k 90º. 12. ekuk x1 < x2 < x3 ……x11
ekf/;dk x1, x2 …..x10 = 2
xx 65
Now the new set of number are x1, x2,
….x5, 2
xx 65 , x6, ….x10
vr% ekf/;dk 2
xx 65 < x6
ekf/;dk gkkleku 13. (a – 8)2 – (b – 7)2 = 5 (a – b –1) (a + b – 15) = 5 I1 I2 pkj fLFkfr I1 I2 5 1 1 5 –5 –1 –1 –5 fLFkfr-1 a – b – 1 = 5 & a + b – 15 = 1 a = 11, b = 5 fLFkfr-2
a – b – 1 = –5 & a + b – 15 = –1 a = 5, b = 9 fLFkfr-3
a – b – 1 = 1 & a + b – 15 = 5 a = 11, b = 9 fLFkfr-4
a – b – 1 = –1 & a + b – 15 = –5 a = 5, b = 5
ifjf/k = 4 + 4 + 6 + 6 = 20 14. ekuk la[;k N gSA If we write c after the last digit now number
is 10N + c vc c|10N + c c = 1, 2, ….., 9 c|10N c = 1, 2, ….., 9 c|N for c = 4, 7, 9 Hence N is LCM of (4, 7, 9) = 252 blfy, vadksa dk ;ksx = 9 15. f(xy) = f(x) + f(y) f(x) = knx vc, 24 = k(n12)
16. r = 1.128cm leff = l + r = 63.2 + 1.128 = 64.3
17.
H
M N
h
2r
2r
A
Pressure at Point A PA = 0hg Pressure at M PM = PA + W Hg 0h + W H)g
18. (i) 2
c
Va
R and
2 RT
V
(ii) ac' = 8ac
2
1V2R
= 2V
8R
V1 = 4V
T' = 2 (2R)
4V
= RV
= T2
19. During slope speed will be increase, so
distance covered will increase parabolically with the time. Then speed will remain constant so further graph will be a straight line with greater slope than the first straight line portion of graph. So if change in speed at the sharp corners is not considered, the best answer should be (B) rather than C.
V
t
time
distance
However if kink is considered and Collision is inelastic then graph will be (C). But in either case answer cannot be D
20. Since bucket and water both are in state of free fall so water will not come out of the hole. (D)
n (–, 5 – 61 ] (5 + 61 , ) ijUrq n IN blfy, n 13; tgk¡ n N case-1 for all 2 digit natural numbers max value of product of digits = 9 × 9 = 81 so n2 – 10n – 36 81 n [5 – 142 , 5 + 142 ] but n is taken as a 2 digit natural no.; so 13 n < 17; product of digits = 3, 4, 5 or 6 for 13, 14, 15 and 16 respectively checking n = 12
vadksa dk xq.ku = 1 × 3 = 3 vkSj 132 – 10 × 13 – 36 = 3 so 13 satisfies the given condition vr% gy gSA chcking for n = 14 xq.kuQy = 1 × 4 = 4 142 – 10 × 14 – 36 = 196 – 140 – 36 = 20 > 6 and n2 – 10n – 36 is increasing function for n > 5; rest of the 2 digit integers won't satisfy the given condition case-2 for all 3-digit integers max product = 9 × 9 × 9 = 729 The smallest 3 digit no. is 100 f(n) = n2 – 10n – 36; f(100) = 1002 – 10 × 100 – 36 = 8964 > 729 and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy n = 13 dsoy mÛÙkj gSA
64.
B
x
D
C b A
a–b
b
ekuk AC = b rFkk AB = BC = a
fn;k gSba
bab
b2 = a2 – ab
2
51ab
Let D be foot of perpendicular from B
sin (ABD) = a2b
=4
15
ABD = 18º ABC = 36° 65. Let abc be required numbers which is
divisible by 20 Clearly one digit must be 0 and one digit must be even, other can be any
Case-I when two digits are 0,0
x 0 0 x = {1,2,.....,9}
9 rjhds Case-II when one O and remaining two are distinct x y
Hence number of ways = 4 × 8 × 2 × 2 = 128 ways
Case-III when two are even 0 and same and other 0.