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".
Time : 2 Hr. Date : 12-10-2014 Max. Marks : 240
ALL INDIA IJSO(STAGE-I) TEST SERIES
1. If
a2b2
is a prime number then
a2b2
=...(A)
ab
(B)
ab
(C) 1 (D) 02.
(
8
n
3
n)
is always divisible by(A) 8 (B) 3 (C) 5 (D) 2
3. If
4xy is a multiple of 3, then
(
4
x
2
7
xy
2
y
2)
is divisible by(A) 3 (B) 7 (C) 6 (D) 9
4. The highest power of 3 contained in (1000 factorial) is
(A) 490 (B) 492 (C) 498 (D) 500
5. The values of n n
Z for which
n219n92
is a square are(A)
8,11
(B)
8,11
(C)
8, 11
(D)
8, 11
6. The pairs of positive integers (m, n) for which 2m3n is a perfect square is
(A)
1,2
(B)
4,2
(C)
3,1
(D)
6,3
7. The remainder when
2222
5555 is divided by 7 is(A) 4 (B) 5 (C) 6 (D) 0
8. The rational roots of 3 2
2x 3x 11x 6 0 are (A) 1 ,1,2 2 (B) 1 , 2,3 2 (C) 1 , 1,2 2 (D) 1 , 2, 3 2
9. If 4p525qp3125rp2625sp3125t0then the roots of the equations
t
sx
rx
qx
px
x
5
4
3
2
are(A) A.P. (B) G.P. (C) H.P. (D) A.G.P.
10. The condition that the roots of x3px2qxr 0 be in G.P. is
(A) p r2 q2 0 (B) p r3 q3 0 (C) p r4 2q3 0 (D) p r3 2q 0
11. Roots of the equation 4 3 2
10 26 10 1 0
x x x x are
(A) (+3, +2) (B)
32 2,2 3
(C)
3, 2
(D)
1 2,2 3
12. Solution of the system of equations
2 4 4 3 9 9 4 16 16
log log log 2
log log log 2
log log log 2
x y z y z x z x y (A) 1; 34; 26 3 3 7 x y z (B) 3 32 z ; 8 27 y ; 3 2 x (C) 2; 27; 32 3 8 3 x y z (D) 2; 34; 32 3 3 3 x y z
13. Real (x, y) that satisfy x3y37 &x2y2xyxy 4 are
(A)
2, 1 ,
1,2
(B)
2, 1 ,
1, 2
(C)
2,1 , 1,2
(D)
2,1 , 1, 2
14. If a, b, c are +ve real numbers representing the sides of a triangle then which of the following is correct. (A)
2
(
)
(
)
23
(
)
ca
bc
ab
c
b
a
ca
bc
ab
(B) 3
abbcca
abc
2 4
abbcca
(C)
2 2 ab bc ca a b c ab bc ca (D) none of these15. In aABC , the incircle touches the sides BC, CA and AB respectively at D, E and F. If the radius of the incircle is 4 units and if BD, CE and AF are consecutive integers, the sides of ABC
16. A circle passes through the vertex C of a rectangle ABCD and touches its sides AB and AD at M and N. If the distance from C to the line segment MN is equal to 5 units, find the area of the rectangle ABCD O A N D M B C (A) 20 (B) 36 (C) 25 (D) 15
17. In an acute angled triangle there are points D, E, F are points on BC, CA, AB such that ADBC ; AE = EC and CF bisects C internally. Suppose CF meets AD and DE in M and N respectively. If FM = 2; MN = 1; NC = 3, the perimeter of ABC is
A F B D C E M N (A) 6 3 (B) 18 3 (C) 12 3 (D) 24 3
18. If a transversal cuts the sides BC, CA, AB of a ABC at X, Y, Z then BX CY AZXC YA ZB. . equals
19. In a ABC , D is the mid-point of BC. If ADB45º and ACD30º ; the value of BAD is
(A) 60º (B) 90º (C) 30º (D) 45º
20. ABCD is a cyclic quadrilateral with AC DB and AC meets BD at E. Then 2 2 2 2
EA EB EC ED is C B A D O E (A) R2 (B) 3R2 (C) 4R2 (D) 2R2
Where R is the radius of circumscribed circle]
21. A boy leaves his house at 9:30 a.m. for his school. The school is 2 km away and classes start at 10:00 a.m. If he walks at a speed of 3 km/h for the first kilometre, at what speed should he walk the second kilometre to reach just in time ?
(A) 6 km/h (B) 8 km/h (C) 9 km/h (D) 12 km/h
22. A cylinder of mass 10 g weighs 7 g in water. If its area of cross-section is 0.75 cm2, its length will be
(A) 4 cm (B) 3 40 cm (C) 3 28
(D) known only in terms of its density
23. A bullet in motion hits and gets embedded in a solid resting on a frictionless table. What is conserved :
(A) Momentum and K.E. (B) Momentum only
(C) K.E. alone (D) None of these
24. The statement for a given pair of forces (in figure) can be written as :
W= mg Rn
(A) It is not an action-reaction pair (B) It is an action-reaction pair
25. An object of mass m is dropped from a height h, it penetrates in sand upto a depth x. Find the average resistance (force) exerted by the sand :
(A) mg x h (B) mg (1 + x h ) (C) mgh (D) none of these
26. What is the relation between FA and FB. FA and FB are the magnitudes of magnetic field at A and B?
(A) FA = FB (B) FA > FB (C) FA < FB (D) Nothing can be said 27. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive
index of water is 4/3 and the fish is 12 cm below the surface, the radius of the circle is :
(A) 12 × 3 × 5 cm (B) 12 × 3 × 7 cm (C) 12 × 5/2 cm (D) 12 × 7 3
cm
28. A step down transformer reduces 220V to 11V. The primary coil draws 5 A current and secondary coil supplies 90A. Efficiency of the transformer will be :
(A) 4.4% (B) 20% (C) 33% (D) 90%
29. Which of the following correctly depicts the graphical variation in case of a spherical mirror ?
(A) v-1 u-1 y x O (B) v-1 u-1 y x O (C) u-1 v-1 y x O (D) u-1 y x O v-1
30. A body is moving on a circular path if its velocity (linear velocity) is incresed by 10% and angular velocity is increased by 20%, then what will be the percent change in centripetal accleration ?
(A) 10% (B) 20% (C) 30% (D) 32%
31. The result of mixing equal masses of ice at – 10ºC and water at 60ºC (the specific heat of ice = 0.5 cal g–1 ºc–1) :
(A) temperature 0°C, 16 11
of total mass of ice melts
(B) temperature 0°C, 11 16
of total mass of Ice melts
(C) temperature 10°C, 16 11
of total mass of ice melts (D) data given are not sufficient
32. Water falls from a height of 100 m. Find the change in temperature of water at the bottom :
(A) 2.4ºC (B) 0.24ºC (C) 24ºC (D) 0ºC
33. A wire 88 cm long bent into a circular loop is placed perpendicular to the magnetic field of flux density 2.5 Wb m–2. Within 0.5 s, the loop is changed into a square and flux density is increased to 3.0 Wb m–2.
The value of e.m.f. induced is :
(A) 0.018V (B) 0.016V (C) 0.020V (D) 0.012V
34. A pebble is dropped into a well of depth h. The splash is heard after time t. If c be the velocity of sound, then : (A) t = h 2 gc (B) t = c + gh (C) t = c – v (D) t = g h 2 + c h
35. An infinite wire bent in the form of L carries current . What is the magnetic field at the point O ?
(A) zero (B) d 4 0 (C) d 2 4 0 (D) d 2 4 0
36. The focal length of a concave mirror is 20 cm. Determine where an object must be placed to form an image magnified two times when the image is
real-(A) 30cm from the mirror (B) 10cm from the mirror (C) 20cm from the mirror (D) 15cm from the mirror
37. An oscillator crosses the mean possition at t = 0. Its displacement at the end of 3s is twice that after 1s. The period of the oscillation is :
(A) second (B) 12 second (C)
12 second (D) 6 second
38. Light incident on a rotating mirror M is reflected to a fixed mirror N placed 22.5 km away from M. The fixed mirror reflects it back to M (along the same path) which in turn reflects the light again along a direction that makes an angle of 27° with the incident direction. The speed of rotation of the mirror is:
(A) 250 revolutions s–1 (B) 500 revolutions s–1
(C) 1000 revolutions s–1 (D) 125 revolutions s–1
39 . A real inverted image in a concave mirror is represented by (u, v, f are coordinates)
(A) (B) (C) (D)
40. According to the quantum theory, a photon of electromagnetic radiation of frequency v has energy E = hv where h is known as planck’s constant. According to the theory of relativity, a particle of mass m has equivalent energy E = mc2, where c is speed of light. Thus a photon can be treated as a particle
having effective mass m = 2 C hv
. If a flash of light is sent horizonatally in earth’s gravitational field, then
photons while traveling a horizontal distance d would fall through a distance given by
-(A) 2 2 c 2 gd (B) mc h (C) h mcd2 (D) zero
41. Which set of quantum number is not consistent with the theory? :
(A) n = 2, = 1, m = 0, s = – 1/2 (B) n = 4, = 3, m = 2 s = – 1/2 (C) n = 3, = 2, m = 3, s = + 1/2 (D) n = 4, = 3, m = 3, s= + 1/2 42. Match column (A) with column (B) and select the correct option.
Column (A) Column (B)
(a) Resistant to corrosion (i) Galena
(b) Pig iron (ii) Galvanised iron (c) Fe2O3 (iii) Impure iron
(d) PbS (iv) Haematite
(A) a(iii), b(ii),c(iv), d(i) (B) a(ii), b(iii), c(iv), d(i) (C) a(iv), b(i), c(iii), d(ii) (D) a(i), b(iv), c(ii), d(iii)
43. Elements upto atomic number 103 have been synthesized and studied. if a newly discovered element is found to have an atomic number 106. Its electronc configuration will be :
(A) [Rn] 5 f 14, 6d4, 7s2 (B) [Rn] 5 f 14, 6d1, 7s2 , ,7p3
(C) [Rn] 5 f 14, 6d6, 7s0 (D) [Rn] 5 f 14, 6d5, 7s1
44. The quantum levels upto n = 3 has :
(A) s- and p-levels (B) s, p, d, f -levels
(C) s, p, d–levels (D) s–levels
45. Aluminium is not extracted by carbon reduction process. This is due to the following reason : (A) At the temperature of the furnace, Al is oxidised by CO2
(B) Melting point at Al is very high (C) Melting point at Al is very low (D) Al reacts with carbon
46. Cu2S + 2Cu2O 6Cu + SO2
In which process of metallurgy of copper, above equation is involved ?
(A) Roasting (B) Reduction (C) Bessemerisation (D) Purification
47. The electron identified by quantum numbers n and , (i) n = 4, =1 (ii) n = 4, = 0 (iii) n = 3, = 2 (iv) n = 3, = 1 can be placed in order of increasing energy from the lowest to highest :
(A) (iv) < (ii) < (iii) < (i) (B) (ii) < (iv) < (i) < (iii) (C) (i) < (iii) < (ii) < (iv) (D) (iii) < (i) < (iv) < (ii)
48. In the reaction, H2O2 + NaCO3 Na2O2 + O2 + H2O, the substance underrgoing oxidation is (A) H2O2 (B) Na2CO3 (C) Na2O2 (D) None of these
49. In the above question, the velocity acquired by the electron will be :
(A)
V/m
(B)
eV/m
(C)
2eV/m
(D) None of these.50. Iodine has highest oxidation number in the compound :
(A) KIO4 (B) IF5 (C) KI2 (D) KI
51. Which is a redox reaction?
(A) 2CuI2 2CuI + I2 (B) NaCI + AgNO3 AgCI + NaNO3
52. 171 g of can sugar (mol.wt. = 342) are dissovled in 1000 g of water at 30ºC. If the density of solution is 1.1 g/mL, then :
(A) Molarity < Molality (B) Molarity = Molality (C) Molality < Molarity (D) None of these 53. The empirical formula of a compound is CH2O. If its VD is 30, its molecular formula is :
(A) CH2O (B) C2H4O2 (C) C3H6O3 (D) CH3OH
54. An aqueous solution of urea containing 18 g urea in 1500 cm3 of solution has a density of 1.052 g/cm3. If
the molecular weight of urea is 60, then the molality of solution is :
(A) 0.2 (B) 0.192 (C) 0.064 (D) 1.2
55. Equal volumes of 0.1 M AgNO3 and 0.2 M NaCl are mixed. The concentration of NO3– ions in the mixture
will be :
(A) 0.1 M (B) 0.05 M (C) 0.2 M (D) 0.15 M
56. The species having octahedral shape is :
(A) SF6 (B)
3
BF (C) PCI5 (D) 3
3 BO
57. How many - bond are there in the nitrogen molecule :
(A) One (B) Three (C) Two (D) None of these
58. The type of bonds present in CuSO4.5H2O are...only . :
(A) Electrovalent and covalent (B) Electrovalent and co-ordinate (C) Electrovalent , covlent and co-ordinate (D) Covalent and co-ordinate
59. In nuclear fission and fusion:
(A) Fission refers to the absorption of a neutron by the nucleus and the emission of an alpha particle
(B) Fusion reactions be illustrated by the reaction, H Li 242He
7 3 1
1 + energy
(C) The critical mass for nuclear fission depends on the temperature and pressure of the system (D) Fusion occurs at lower temperatures than fission
60. The temperature of an ideal gas is increased from 140 K to 650 K. If at 140 K the root mean square velocity of the gas molecules is V, at 560 K it becomes :
(A) 5V (B) 2V (C) V/2 (D) V/4
61. The organelles that contain their own genetic material are:
(A) Mitochondria, Vacuoles (B) Plastids, Golgi complex (C) Mitochondria, Plastids (D) Ribosomes, Nucleolus
62. The enclosures in which heat is trapped to maintain higher temperature, are called
(A) white houses. (B) warm houses. (C) blue houses. (D) green houses
63. The stage in which separation of sister chromatids occurs is
(A) anaphase. (B) telophase. (C) metaphase. (D) prophase. 64. Haversian canals occur in
(A) humerus. (B) pubis (C) scapula. (D) clavicle.
65. Bone marrow is absent in the bones of
(A) fish. (B) bird. (C) reptile (D) frog.
66. A fern commonly used as biofertiliser is
(A) Lycopodium. (B) Marsilea. (C) Azolla. (D) Adiantum.
67. Camel is best adapted to desert habitat as
(A) it can drink 50 litres of water at a time which is evenly distributed in all its tissues. (B) it excretes very small amount of water during urination
(C) it can regulate its body temperature at a wider range (D) all are correct
68. Aerenchyma is found in which plants ?
(A) sciophytes (B) hydrophytes (C) mesophytes (D) epiphytes
69. In the lunch, you ate boiled green vegetables, a piece of cooked meat, one boiled egg and a sugar candy. Which one of these foods may have been digested first ?
(A) Boiled green vegetables (B) The piece of cooked meat
(C) Boiled egg (D) Sugar candy
70. In the given food chain 'Plants Sheep Man', 5 J of energy is available to man. The energy that was available at producer level is
(A) 50 J. (B) 500 J. (C) 5 J. (D) 0.5 J.
71. Producers prepare their own food like
(A) blue green algae. (B) amoeba. (C) rhizopus. (D) yeast.
72. Which one does not produce any digestive enzyme ?
(A) Pancreas (B) Liver (C) Stomach (D) Duodenum
73. Tunica albuginea is the covering of
(A) lungs (B) testis (C) kidneys (D) heart
74. An organism with two identical alleles of a gene in a cell is called
75. Mendel’s law do not explain principle which is :
(A) Segregation of genes (B) Dominance
(C) Linkage (D) Independent assortment
76. Role of isolation in evolution is
(A) magnification (B) maintenance of species
(C) evolutionary divergence (D) extermination of species
77. Soil is composed of (A) mineral + water + air
(B) mineral + organic matter + air (C) mineral + organic matter + air + water (D) organic matter + water
78. The link between kreb’s cycle and glycolysis is
(A) citric acid (B) acetyl - co - A (C) Succinic acid (D) Fumaric acid
79. Compensation point refers to the intensity of light at which (A) Rate of respiration = Rate of photosynthesis
(B) Rate of respiration > Rate of photosynthesis
(C) Rate of respiration < Rate of photosynthesis (D) None of the above.
80. Prevention of a disease is more desirable than its cure because
(A) some of the body functions may be damaged during the effect of the disease. (B) the person suffering from the disease will not be bedridden.
(C) the disease can not be communicated to others during the course of treatment. (D) body does not look good during this condition.
ANSW ER KEY
HINTS & SOLUTIONS
1. a2 – b2 = (a – b) (a + b) ..(i)
a2 – b2 a prime equation
= – 1
(a – b) = 1 [ (a – b) < (a + b)]
The only adivisor of a prime equation are 1 and itself Equation (i) becomes (a2 – b2) = 1(a + b)
or a2 – b2 = (a + b)
eg. 32 – 22 = 5 (which coprime)
32 – 22 = 3 + 2, 3, 2
N
2. 8n – 3n = (8 – 3) (8n–1 + 8n–2 3 + 3n–1)
(89 – 3–n) is divisible by 5
If n is either even or odd
xn – yn = (x – y) (xn–1 + xn–2 + ... + yn–1) If n is odd xn+ yn = (x + y) (xn–1 – xn–2 + xn–3y2 –....+ yn–1) 3. (4x – y) is a multiple of 3 (4x – y) = 3m y = (4x – 3m)
on putting value of y in (4x2 + 7xy – 2y2)
4x2 + 7xy – 2y2 = 4x2 + 7x(4x – 3m) – 2(4x – 3m)2
= 4x2 +28x2 2 + 48mx
ALL INDIA IJSO(STAGE-I) TEST SERIES
MOCK TEST PAPER # 4
DATE : 28-09-2014
Que s. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. B C D C D B B B A B B B A B B C C D C C Que s. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans. A A B A B B D D C D A B A D A A B A A A Que s. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. C B D C D C A D C A A A B B B A C C B B Que s. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans. C D A A B C D B D B A B B A C C C B A A4. if p is a prim number and the longest exponent of p such that pe divides n! then
e =
10 1 i pi n[x] is the largest integer x In this case p = 3, n = 1000 p n = 3 1000 = 333 2 p n = 2 3 1000 = 11111 3 p n = 3 3 1000 = 37 4 p n = 4 3 1000 = 12 5 p n = 5 3 1000 = 4 6 p n = 6 3 1000 = 1 7 p n = 7 3 1000 = 0
Highest power of 3 contained in 1000! = 333 + 111 + 37 + 12 + 4 + 1 + 0 = 498
5. n2 + 19n + 92 = m2 : n is a non negative integer. Then
n2 + 19n + 92 – m2 = 0 solving for n = 2 ) 7 m 4 19 ( 2 (4m2 – 7) is a square ie 4m2– 7 = p2 when p N (2m – p) (2m + p) = 7
(2m + p) being positive is 7 and (2m – p) = 1 ; 4m = 8 m = 2 Thus we have n2 + 19n + 92 = 4 n2 + 19n + 88 = 0 (n + 8) (n + 11) = 0 n = – 8 or n = 11 6. Let 2m + 3n = k2
7. 2222 3 mod7 (2222)3 27 mod 7 and 27 – 1 (2222)3 = – 1 mod 7 (2222)5553 = –1 mod 7 (2222)2 = 9 mod 7 (2222)5555 = –9 mod 7
But –9 5 mod 7 ; (2222)5555 5 mod 7 8. Let the runs be if the form
2 p (where (p – q) = 1 and q > 0) since q/2, q must be 2 or 1 and p/6 *1, 2, 3, 6 f 2 1 = f(–2) = f(3) = 0
Hence root are 2 1
, –2 and 3 9. Let the root be
a 2d, a – d, a, a + d, a + 2d sum of roots = 5a = – p a = 5 p
But a is a root of the given equation hence 5 5 p + p 4 5 p + q 3 3 p + r 2 5 p + t = 0 or 4p5 – 25ap2 + 125 rp2– 625p + 3125 t = 0
10. Let the root of the given equation be ap, a, p a
= = ap a p a = r a3= r ; a = r1/3But a is root of the given equation (r1/3) p(2/3) + qr1/3 – r = 0 pr2/3 = qr1/3 p2r2 = q2r p3r – q3 = 0 is required condition 11. Dividing throughtout by x2 2 1 x – 10 x 1 + 26 = 0
x = 2 32 6 = 3 2 2 and x + x 1 = 4 ; x2 4x + 1 = 0 x = 2 12 4 = 2 3
12. Log2x + log4y + log4z = 2 log3y + log9z + log9x=2 log4z + log16y + log16z = 2
(log4x) log24 + log6y + log4z = 2
x2yz = 16 (i)
similarly the other equation riduce to (ii) y2zx = 81 and
z2xy = 256 ...(iii)
solving
divide (i) by (iii)
y x = 81 16 ; x = 81 y 16
Dividing (ii) by (iii)
z y = 256 81 ; z = 81 256 y 2 2 ) 81 ( ) 16 ( y2 . y. 81 256 y = 16 y4 = 3 3 ) 16 ( ) 81 ( , y = 8 27 ; x = 81 8 27 16 = 3 2 ; z = 3 32 13. Let x + y = , xy = and hence x2 + y2 = 2 – 2 x2 + y3 = (x + y) (x2 – xy + y2) (2 – 3) = 7, 3 – 3a = 7 and x2 + y2 + x + y + xy = 4 2 – 2 + + = 4 = 2 – + = 4 = 2 + – 4 2– 3– 3 + 12 = 7 – 22 – 32– 12 + 7 = 0 f(1) = 22 + 32– 12 + 7 = 0 hence (– 1) (factor of f()) f() = ( – 1)(2a2 + 5a – 7)
x,y are root of 4t + 14t + 19 = 0
whose discriminent < 0. Hence no real roots. hence solution are (2, –1) and (–1, 2)
14. 2 b a2 2 > ab , 2 c b2 2 > bc and 2 a c2 2 > ca [A.M. > G.M.] a2 + b2 > 2ab ; b2 + c2 > 2bc ; c2 + a2 > 2ca ; a2 + b2 + b2 + c2 + c2 + a2 > 2 (ab + bc + ca) a2 + b2 + c2 > ab + bc + ca ab + bc + ca < a2 + b2 + c2 ..(i)
In ABC, with sides BC = a, CA = b AB = c, we have b2 + c2 – a2 = 2bc cos A b2 + c2 – a2 < 2bc [ cos A < 1] similarly c2 + a2 – b2 < 2ca and a2 + b2 – c2 < 2ab adding these three
a2 + b2 + c2 < 2(ab + bc + ca) ..(ii)
from (i) and (ii) we get
ab + bc + ca < a2 + b2 + c2 < 2(ab + bc + ca) ...(iii)
OR 1 < ca bc ab c b a2 2 2 < 2
Adding 2(ab + bc + ca) throughout in (iii) 3(ab + bc + ca) < (a + b + c)2 < 4(ab + bc + ca)
15. A C B x+2 x x x+1 x+2 x+1 F D E Suppose that BD = x CE = x + 1 and AF = x + 2 Then CD = CE = x + 1 ; AE = AF = x + 2 and BF = BD = x Hence a = BC = x + x + 1 = 2x +1 b = CA = x + 1 + x + 2 = 2x + 3 c = AB = x + x + 2 = 2x + 2
x2 + 8x – 6x – 48 = 0
x(x + 8) – 6(x + 8) = 0 (x + 8) (x – 6) = 0
x = 6; hence a = 13, b = 15, c = 14
16. Let O be the centre of the circle and P be the root of the perpendicular from C to MN.
Then OM is perpendicular to AB , ON is perpendicular to AD. Hence OM = ON = radius of the circle. so AMON is a square MNC = 2 1 MON = 45° CMP + CNP = 135° = CMP + CMB = CNP + CND Hence CNP = CMB and CMP = CND Hence in CMP and CND CMP = CND CPM = CDN = 90° CMP ~ CND ; A to CNP ~ CNB CM CN = CB CP ; CN CM = CD CP
multipling these equations
1 = CD CB CP2 CB CD = CP 2 = 52 – 25
17. FN = 3 = NC. So N is the midpoint of FC and E is the midpoint of AC.
Hence ND || AB and so D is the mid point of BC. Thus AD is attitude as well as median. Hence AB = AC. AD is also angle
bisector of A AFM ~ DNM
since AMF = DMN and FAM = NDM
so MD AM = MN FM = 1 2
Thus M is the centroid of DABC CF passes through M. CF is also a median i.e. angle bisector CF is also a median. Hence CA = CB ;
consequently ABC is equilat eral CF = 6 = altitude of an equilateral .
side of the equilateral = 3 12
18. A Qz B C x Y R P h2 h1 h3
Let h1, h2, h3 be the length of 1s AP, BQ, CR respectively from A, B, C on the transersal
BQX ~ CRX XC BX = 3 2 h h .. (i) Similarly YA CY = 1 3 h h ..(ii) and ZB AZ = 2 1 h h ...(iii) Hence XC BX . YA CY . ZB AZ = 1
but are obs OAC that XC BX is negative Hence XC BX . YA CY . ZB AZ = – 1 19. 5° 60° 15° 120° 45° 30° L C D B A
Draw BLAC and join L to D BCL = 30°, we get CBL = 60°
20. Let O be the centre of the circle AOB + COD = 2 (ACD + CBD) = 2(ACD + 90 – ACB)
= 180°
Let AOB = Q then
21. v1 = 3 km/h, s = 2 km
t = 9:30am to 10am = 30 min = 0.5 h = 2 1
h Time taken to complete the first half distance
t1 = 3 1 v 2 / s h
Time remained to move the second half distance, t2 = t – t1 = 2 1 – 3 1 h = 6 1 h speed = time ce tan dis = 6 / 1 1 = 6 km/h
He should walk with a speed of 6 km/h to reach time.
36. v = 2u u 1 v 1 = f 1 u = – 30 cm
39. For real inverted image by concave mirror. V = – ve , u = – ve f = – ve f u & f V are positive 1 is right answer
41. = 2 then m can be –2, –1, 0, +1, +2 only. 44. For n = 3, may have values 0(s), 1(p) and 2(d).
45. Aluminium cannot be reduced easily with carbon at moderate. It requires high temperature at which al may combine with carbon to form carbide.
46. During the process of bessemerization SO2 gas is evolved. 2Cu2S + 3O2 2Cu2O + 2SO2
Cu2S + 2Cu2S 6Cu + SO2
52. d 1000 ) 171 1000 ( 342 171 m = 0.429 × 1.4 = 0.47 m = 1 342 171 = 0.5 M = 0.18 53. Mol. wt. of (CH2O)n = 30 × 2 = 60 n = 30 2 60 Empirical formula wt. (CH2O) = 30 54. m = ) 18 052 . 1 1500 ( 60 1000 18 = 0.19 55. mM of AgNO3 = 0.1 × V mM of NaCl = 0.2 × V mM of NO3 – = 0.1 × V and total V = 2V [NO3 –] = V 2 V 1 . 0 = 0.05
56. S atom in SF6 in sp3d2 -hybridized state and shows octahedral shape
57. N N or N2 has 1 and 2-bonds.
59. Fusion involves joining of two lighter nuclei.
60. urms = M RT 3 = M 140 R 3 at 140 K urms = M 560 R 3 at 560 K urms = 2 × urms .
