Kvpy Sa Stream 2007
Short Description
Kvpy Sa Stream solution 2007...
Description
HINTS & SOLUTIONS (YEAR-2007) ANSWER KEY Ques.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
B
C
D
C
B
D
D
B
A
B
D
C
D
D
B
Ques.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
B
C
B
C
C
B
C
C
A
C
A
C
B
A
C
Ques.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans.
B
A
C
C
D
B
A
B
B
D
A
C
C
A
D
Ques.
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans.
A
A
A
D
B
A
D
A
B
D
D
D
D
B
B
Ques.
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
Ans.
C
A
B
A
C
A
A
D
A
B
D
D
A
C
C
PART-I (1 Mark) MATHEMATICS
1.
n [2.8 + (n – 1) 4] 2 s1(n) = s2(n) s1(n) =
s2(n) =
n [2.17 + (n – 1)2] 2
n n [2.8 + (n – 1) 4] = [2.17 + (n – 1)2] 2 2 2(n – 1) = 18 n–1=9 n = 10 s1(10) = 5[16 + 36] = 2 60 = s 2(10)
2.
(sin(2x)) 4 = 1/8 Range x [0, 2] Let 2x = y given range 0 x 2 0 2x 4 0 y 4 4 (sin(2x)) = 1/8 From above graph we can say, say, Total Eight solution 3.
3.r 2 – 8.r 5
0 4.r 2 – 3.r 7 3r 2 – 3r – 5r + 5 > 0 3r(r – 1) –5(r – 1) > 0 (3r – 5)(r – 1) > 0 (3r – 5)(r – 1) > 0 (– , 1) (5/3, ) r
and
4r2 – 3r + 7 > 0
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
as
D 0 can be
O
So option (D) is right 8.
Case - I Here a = 1
Case – II
3
Altitute
a=
9.
2 3
2
a=1
.
Let AB = a then BE = a tan CE
= tan tan CF CF = a cot – a Now, In GHF tan =
HF GH
=
1 tan 2 cot
Solving we get tan =
10.
2 3
sin =
2 13
R = 5, c = 6, b = 6 S=
= R=
12 x , = 2
s(s – a )(s – b)( s – c )
12 x x x 12 – x . . . 2 2 2 2 abc 4
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 85
5
⇒
36 x 4 (12 x )x 2 (12 – x ) 4
(12)2 – x 2 =
⇒
36 5
2
11.
⇒
36 = x2 (12) – 5
⇒
x = 48/5
2
Let AP = x 1, PQ = x 2 Now,
ar ( ADE) ar ( ABC)
=
1 2
2
=
2
12.
x1
( x1 x 2 )
1 2
2x12 = x 12 + x 22 + 2x 1x2 x12 – x22 – 2x1x2 = 0
x1 x2
=
2 44 2
x1 = x2
2 1 1 1 4
cos2 + cos2 = 3/2
and
sin sin =
1 cos 2 1 cos 2 + = 3/2 2 2
and
2sin sin =
cos 2 + cos 2 = 1
and
cos ( – ) – cos ( + ) =
1 2 1 2
...(ii)
2 cos ( + ) cos ( – ) = 1 ...(i) From (i) and (ii) cos ( + ) cos ( – ) = cos ( – ) – cos ( + ) ⇒
+ = 13.
cos ( + ) =
1 2
cos ( – ) = 1
3
1 Area = [x + 2x + 3x] = 2
3x2 4
3x2
⇒
3x =
⇒
x= 4 3
⇒
and
4
Area = 12 3
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 86
14.
Let wealth A, B, C are x, y, z Given : A = B + C x=y+z ....(i) A distribute half of his wealth wealth to B & C
x 2 1 2 : 1 i.e. & 2 3 3 Now, A =
x 2
B=y+
x 3 2
C=z+
x 3 2
2
1
Now, B = A + C y+
x 3
x x + z + 6 2
= x
y–z=
3
3y – 3z = x
. . . ( i i)
Let t be fraction that A should distribute and the ratio of distribution is 1 : 2 i.e.
1 3
<
2 3
.
Now, A = (1 – t)x tx
B=y+
3 2tx
C=z+ z+ z+
2tx 3 tx 3
3
= y +
tx 3
+ (1 – t) x
= y + x – tx
3z + tx = 3y + 3x – 3tx 4tx = 3 (y – z) + 3x 4tx = 3x + x 4tx = 4x t = 1. So, A would distribute his whole wealth t o B and C. Fraction is 1. 15.
Put 3 times water of 110 ml to container and and take 13 times 25 ml water from container then container has 5 ml water.
16.
Distance = speed × time Distance covered in car = 50 × 4 = 200 km He move for 1 hr = 20 km Time take while retraining from town to village time =
Dis tan ce 200 = = 5 hr Speed 40
Total journey = 200 + 200 + 20 Total time taken = 4 + 1 + 5 = 10 average speed =
420 10
= 42 km/hr
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 87
17.
Given 4 × 4 × 4 cubes is mode faces 64 1 × 1 × 1 cubes total cubes = 64, white = 20, Red = 44 To find minimum number of visible white box Counting total visible faces of unit cube Total number of f aces of small cube on bigger cube except boundry cubes = 4 × 6 = 24 Counting boundry cube = 16 + 8 + 8 = 32 Total visible faces = 56 But we have 44 Red cube minimum of number of white faces cubes which are visible = 56 – 44 = 12
18.
Multiple of 3 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 36 then n = 5, 8, 11, 14, 17, 20, 23...... Multiple of 5 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 then n = 8, 13, 18, 23, 28 ...... Now, common two digit numbers are 23, 38, 52, 67, 82 , 97.
19.
if start with H1
2 ways
if start with H 2 2 ways if start with W 1
2 ways
if start with W 2
2 ways
m=8 in case of circle H1
W2
H1
H2
H2
W2
W1 20.
W1
Total students = 300 One student read = 5 newspapers Number of newspapers read by 300 students = 5 × 300 = 1500 newspaper Number of different newspapers =
1500 60
= 25.
PHYSICS
T
21.
Hence n = 2
a
mg–T = ma T = m(g–a)
mg
= 6000 (10–2) = 4.8 × 104 N
22.
A
u
C v
5 5 Between A and B v2 = u2 = +2a(2s) Between A and C v2 = u2 + 2as
from (i) and (ii) v = 24.
v
B .....(i) .....(ii)
u2
v2 2
238 92 U
206 82 Pb 824 He 601e
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 88
CHEMISTRY
31.
Using the formula, N = N 0
1 2
n
where, No = initial amount of radioactive substance N = Amount of substance left after ‘n half lives
Total time (t) No. of half lives (n) =
n=
Half life period
21.2 years 5.3 years
n=4
1 so, N = 20 2 32.
4
=
20 = 1.25g 16
Zn(s) + H2SO4(aq) Zinc
ZnSO4(aq) + H2(g)
Sulphuric acid
Zinc sulphate
Hydrogen gas
(salt)
Zn(s) + 2NaOH (aq) Na2ZnO2 (aq) + H2(g) Zinc
Sodium hydroxide
33.
Sodium
Hydrogen gas
zincate
Acetylene (CH CH) has a triple bond.
34.
Ne has 10 electrons and oxygen has 8 electrons.
35.
KMnO4 being strong oxidising agent will oxidise Cl – ion present in HCl to form chlorine gas. 2KMnO4 + 16 HCl 2KCl + 2MnCl2 + 8H2O + 5Cl2
36.
Order of strength for halogen acids is H I > HBr > HCl > HF
38.
Concentrated sulphuric acid is a dehydrating agent and it burns the organic compounds like sucrose so the colour of the solution turns black. Conc . H2SO 4 C12H22O11(s) 11H2O(g) 12C + 11H
Sucrose
Carbon
W ater vapour
(Black)
Oxidation 39.
ZnO+C Zn + CO Reduction
40.
Monomer of Teflon is Tetrafluoroethene(CF 2=CF2)
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 89
PART-II (2 Mark) MATHEMATICS 51.
x = x3 + y4 y = 2xy Two cases
;
y = 2x y
1.
case -1
y = 0 & x R
2.
case - 2
x=
1 & y R 2
Now, equation x = x 3 + y4 Taking case - I x = x3 Solution x = 0, 1, –1 ∴ pairs = (0, 0), (1, 0), (– 1, 0) Taking case ca se 2
1 = 2 3 8
3 y=± 8 52.
3
1 + y4 2
= y4
1/ 4
1 3 1/ 4 Solution set = 2 , 8 ,
f(x) degree (n)
1 3 1/ 4 , 2 8
f(1) =
2 f(3) = f(x) = axn + bxn – 1 + cx n – 2 ....... If degree 0 f(x) = a = constant f(1) = f(3) but it is not t rue If degree 1 f(x) = ax + b f(1) = a + b =
2
f(3) = 3a + b = Two variables & two unknown a & b can be found uniquly one polynomial used only For n>1 Let n = 2 ax 2 + bx + c We have 3 variable & only 2 equations can be formed from given condition Hence infinite such polynomial can be formed 53.
p(x) = Q(x) . (x2 – 3x + 2) + (2x – 3) p(1) = – 1 p(2) = 1 at least one root between 1 & 2
1 –1
1
2
–1
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 90
N
x
D
x
C y
2y
M
54.
y B
2x
A
In MCN, tan =
y 2y and In ADN, tan (90 – ) = x x
cot =
2y x
tan =
x 2y
2y2 = x 2
x = y
2 1
B ( 2 , 3) S
R
x
x
x P
O(0, 0)
55.
x
(6, 0) A
Q
a 2
tan =
x=
3 2
=
3a
x a
=
3 – x
....(i)
2 – a
and tan =
x 6 – a – x
=
3 6 – 2
....(ii)
[from (i)]
2
2x 3
a=
and
6x –
2 x = 18 – 3a – 3x
9x –
2 x = 18 – 3 .
x= 2
Q’
[from (ii)]
2x 3
P
102º
56.
R’
r A
r
O
B
Since R can be any point between A & R & hence its corresponding point Q will lie on the arc AQ. Hence PRA can not be uniquely determined.
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 91
y
3x 57.
A
2x
B
C
D
5y
Given
AB CD
=
3 2
1 BC = 5 AD Let AB = 3x CD = 2x , BC = y, y, AD = 5y 3x + y + 2x = 5y 5x = 4y To find AC : BD = (3x + y) : (y + 2x)
3x y 3 x 5x / 4 = 2x y 2 x 5x / 4 3 5/4
58.
12 5
17
2 5/4
=
Let and and
r = radius of sphere R = radius of cone H = height of cone
4 3
1
r 3 =
3
85
=
13
R2H
4r 3 = R2H Now,
....(i)
k4r 2 = R2 + R R 2 k(4r 2) = R2 + R R 2
H2
H2
16r 6 2 4kr 2 = R R R R4 6 1 R 2 16r r 2 Let 2 = x R6 R
r 2 4k 2 = 1 R
4kx = 1 +
16k 2x2 – 8kx + 1 = 1 + 16x 3 x=0 16k2x – 8k = 16x 2
k = 0,
1 16 x 3
2x2 – 2k 2x + k = 0 4k4 – 8k = 0 (2)1/3
sinc since e cone cone is uniq unique ue ther theref efo ore it has equa equall root roots s
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 92
59.
Let original four digit number is = m2 and new 4-digit number is = n 2 then m 2 – n2 = 1111 (m + n)(m – n) = 1 × 1111 or 11 × 101
(
w
h
e
r
e
,
m
,
n
,
)
Case - I
m + n = 1111 m–n=1 m = 556 n = 555 But m 2 = is a 4-digit number Case - II m + n = 101 m – n = 11 m = 56 and n = 44 So, there is only one such 4-digit number. 60.
m , n are integer 1/3 < m/n < 1 ...(i) m n
m n
=
n 0
m = +ve integer m 1 m=2 Using equation (i) n = 3, 4, 5
=
m
n
2
=
3
,
2 4
,
2 5
PHYSICS 61.
= 0t –
1 2 t 2
= 60 × 5 –
1 × 8 × 25 2
= 300 – 100 = 200 radian or n =
100 2
~ 32 rev./s 62.
Energy released = mL V = volume of water × density ×L V = 100 × (10 3)2 ×1×103 × 22.5 × 10 5 J = 22.5 × 1016 J Number of bombs =
63.
Req =
22.5 1016 1014
2250 ~ 2000
5r 5 1 5 6 6 6
V 10 volt
So, I =
V R
10 6 5
12 A
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 93
64.
time taken in rotataing angle is t, then
45 103
t=
8
15 10 5 s
3 10 t = 15 ×10 –5 second 360
T=
1 T
N= 65.
N
27º
15 10 5 1 4 10
3
1000 = 250 rev/s 4
Focal length, f = 1m, size of object, h 1 = 1000m, size of image, h 2 = 0.2 m image is real, so –
h2 h1
v u
v 2 10 4 u
1 1 1 f v u
1=–
1 2 10
4
u
1 u
u = – 5000 m or u = –5 km
CHEMISTRY 66.
N1V1 = N2V2 36 × V1 = 3 × V 2 V2 V1
= 12
67.
4Fe + 3O2 + nH2O 2Fe2O3.nH2O Iron Iron Oxy Oxygen gen Water Water Hydr Hydrat ated ed Ferr Ferric ic oxid oxide e (Rust)
68.
Solubility of KNO3 at 90º C = 200 g/100 m l Solubility of KNO3 at 30º C = 40 g/100 ml Decrease in solubility when KNO 3 solution is cooled from 90ºC to 30ºC = 200 – 40 = 160 g/100ml
69.
Yeast
4 2C2H5OH CH3COOH C6H12O6 – 2CO
KMnO
2
70.
Zn + 2H 2SO4(conc.) ZnSO4 + SO2(g) + 2H2O (CH3COO) 2 Pb + SO 2
PbS (Black Precipitate)
Cu + SO 2
CuS
+
O2
(Black mirror)
CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
Page # 94
View more...
Comments
