KVPY SA Stream Solution 2011
Short Description
KVPY SA Stream Solution 2011...
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HINTS & SOLUTIONS (YEAR-2011) ANSWER KEY Ques.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans.
A
C
A
B
B
D
A
A
A
C
B
C
B
B
C
A
A
C
D
A
Ques. 21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Ans.
B
C
C
D
C
B
B
D
A
C
D
C
D
D
A
D
A
A
B
D
Ques. 41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans.
D
D
B
B
C
B
D
C
B
C
D
A
C
C
A
B
A
B
B
D
Ques. 61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
Ans.
C
B
B
A
B
B
B
C
A
A
C
B
A
A
B
D
C
A
C
C
PART-A (1 Mark)
MATHEMATICS
1.
( x b)( x c ) ( x c )( x a) ( x a)( x b) + + (a b)(a c ) (c a)(c b) (b c )(b a)
P(x) = Let
f(x) = P(x) – 1 f(a) = 1 + 0 + 0 – 1 = 0 f(b) = 0 + 1 + 0 – 1 = 0 f(c) = 0 + 0 + 1 – 1 = 0
f(x) is a polynomial of degree atmost 2, and also attains same value (i.e., 0) for 3 distinct values of x (i.e. a,b,c).
2.
f(x) is an identity with only value equal to zero. f(x) = 0
x R P(x) = 1, x R
Using cauchy schwartz’s inequality (a2 + b2) (x2 + y2) (ax + by) 2 equality holds at
a x
=
b y
ay – bx = 0
Aliter : a2 + b2 = 81
......(i)
x2 + y2 = 121
......(ii)
ax + by = 99 2
2
2
......(iii) 2
(a + b ) (x + y ) = 81 × 121
......(iv)
[(i) × (ii)]
(ax + by)2 = 992
......(v)
[squaring (iii)]
(iv) – (v) (ay – bx)2 = 0 ay – bx = 0
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3.
x
Then,
x x2 +
x x3 +
1
1 x
a and x 2 1 b 3 x
2
a2 x 1
x
+2 = a2
2
1
.........(i)
3
= a3
x
1 3 x = a 3 x x 1
.........(ii)
3
add equation (1) and (2)
2 1 3 1 x 3 x 2 2 3 x x x b +
x3
4.
1
= a2 + a3
x
3 1 x 2 + 2 + 3a = a 2 + a3 x 1 x2
= a3 + a2 – 3a – b – 2
|a – b| = 2, |b – c| = 3, |c – d| = 4 a–b=±2 b–c=±3 c–d=±4 possible values of (a – d) are ± 9, ± 5, ± 3 |a – d| = 9, 5, 3, 1 Sum of all possible values are 18
5.
Given 0 < r < 4 in all the obtain (Base) x =
9 5
the option having least base will give the largest x.
So, in option B base 1
r
is minimum for 0 < r < 4.
17
Aliter : (1 + x =
4 5
x log (1 + ) = log (1.8)
x=
log (1.8) log (1 )
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For x to be maximum log (1 + ) should be minimum r
= r 17 6.
or
2r
or
r 17
1
or
in 0 < r < 4
r
is minimum.
Angle bisector theorem x 6
=
y z
20 = 6y
y=
xz = 6y
10 3
6
x
2
10cm y
cm
z
ALITER : Let
ACB = BC = 6 cos , AB = 6 sin BD : CD = AB : AC = 6 sin : 6 = sin : 1 BD =
CD =
6 cos
1 sin
sin
6 cos 1 1 sin
.........(i)
.........(ii)
BD = CD sin
ADC = 10 cm 2 Now, area of 1 × 6(CD) sin = 10 2 CD sin = 10/3 cm
7.
Slant height = 13
by = r 2.5 = 13
10 13
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8. 20 x 12 = 8 x
x2 – 20x + 96 = 0 x = 8, 12
ALITER :
A(0,12)
C(20,8) B(0,0)
= 4 52
42
1
AC =
20 2
(122 + x2) + (82 + (20 – x) 2)
E(x,0)
D(20,0)
= 4 26 = 202 + 42
2x2 – 40x + 400 + (12 2 – 202) + (82 – 42) = 0 2x2 – 40x + 144 + 12.4 = 0 x2 – 20x + 72 + 24 = 0 x2 – 20x + 96 = 0 x = 12, 8
9.
AP = 2 sin 60° =
10.
3
d = 2 3
5123 – (2533 + 2593) = 5123 – [(512) (253 2 + 2592 – 253.259)] = 512 (5122 – ((512)2 – 3(253)(259)) = 512 (3.253 – 259) = 29. 3. 253 . 7.37 = 29. 3. (11) . (23). 7.37 So, number of distinct prime divisors are 6.
11.
Top layer has (13 × 13) balls Simillary one layer below top layer will have (14 × 14) balls and we have 18 lesens to t otal number of ball N = (13)2 + (14)2 + .............+ (30)2 N=
30 31 61 6
–
12 13 25 6
N = 8805
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12.
Let distance is 6d Mud
:
Tar
:
Stream
Distance
d
:
3d
:
2d
Speed
3V
:
5V
:
4V
time
d 3V
:
3d 5V
:
10
:
18
:
2d 4V 15
(Note : order is changed in questions)
13.
3 step
14.
The clock well show 1 in an hour for 19 tim e for 11 hours it will show the incorrect time for (19 × 11) time. The last 12 th hour will always show the incorrect time so total incorrect tim e (19 × 11 + 60) min = 269 min there are 24 hours in a day to = 269 × 2 = 538 min 538 min =
269 30
= 9 hou hours rs
the fraction day when the clock shows correct tim e is = 1 = 1
3 8
9 24 5 8
15. Righ Rightt 15 14 13 12 11 10
Wrong ong 15
Una Unattemp tempte ted d 0 4 8 12 16 20
6 cases only
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PHYSICS 16.
By mechanical energy conservation KEi + Ui = KEf + Uf O + U i = 0 + Uf Ui = Uf hi = h f So D will lie on line AB
17.
Since toy is not accelerating so net external for ce on toy is zero. So (A)
18. |S1| + |S2| = H 1 2
gt 2 + ut –
ut = H .
2
gt 2 = H
....(I)
u2
H=
1
....(II)
2g
u
t = 2g
S2 = ut – =
3 4
1 2
gt 2 = 4
u 1 u2 g 2 2g 2 4g
3u2 8g
H
19.
By concept of centre of mass 36x = 9(20–x) 36x = 180 – 9x 45x = 180 x = 4m 20.
All three will will be in thermal equilibrium with air of room. so temperature of the three will be same
21.
Pressure of gas is same everywhere in the vessel.
22.
To travel from P to Q in minimum time, she should travel on path PCQ.
23.
i = 45° C C = 45° for minimum sin 45 = 1
=
2 = 1.42
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24. In this case only half part of lens is used to form the image so intensity will reduce
25.
R= R=
R=
R =
A
A
2 V
(2)2 V
= 4R
26. E1 = Electric field due to +Q E2 = Electric field due to –2Q There resultant is 0 at this point
27.
mgh / t
=
750
100
300 10 6 60 = 750
=
300 750
100
100
= 40%
28.
6A from Q to P
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29.
pb 314 82
2e –1 +
4 + X210 He 2 82
So 82 proton and 128 Neutron 30.
PV = NKT 105 × 100 = N × 1.38 × 10 –23 × 273 N 3 × 1027
CHEMISTRY 31.
Since pressures of the gases are same in both the containers. So, the final pressure will not change
32.
Reativity towards Friedel-Crafts alkylation is pr oportional to electron density in the benzene ring.
33.
n=2 l = 0, 1
0 m = 0, 1 1
34.
Average Kinetic Energy depends only on temperature temperature (KE)avg =
35.
Ideal gas
3kT per molecule 2
H = 0 S > 0
(randomness increases)
36.
Cr 2O3 + N2 + 4H2O (NH4)2 Cr 2O7
37.
Accord Acc ording ing to the graph gra ph solu bility bili ty 40° is appr ox. 200 g per 100 ml. ml . For 50 ml, amount amo unt is 100 g approx.
38.
Aldehyde, ketones with acetyl acetyl group CH3 –
39.
40.
1 n
2
1 16
– show Iodoform test.
n, number of half-lives = 4 = 2 hrs. half-life half-life = 30 min.
ZnS ZnSO 4 during roasting, sulphide ore is converted into sulphate.
41.
42.
Cl 2 + 2KBr → 2KCl + Br 2 reddish brown
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43.
(i) and (iv) are hetro aromatic and the resonance form of azulene (iii) is arom atic.
(ii) Is nonaromatic.
44.
Simple nomenclature of alkane.
45.
The major product formed in the reaction :
Product is :
Sol.
( A) i (B ) ii ( C ) ii i ( D ) iv (C) The given reaction is S N2 which occurs at sp 3 carbon with good leaving group.
PART-II (2 Marks)
MATHEMATICS 61.
f(x) = ax 2 + bx + c, given,
f ( 1) = 0
f(1 f(1) = 0 a+b+c=0
and
4 0 < f (6 ) < 5 0 40 < 36a + 6b + c < 50 40 < 35a + 5b < 50 8 < 7a + b < 10
7a + b = integer = 9 and
.......(1)
6 0 < f (7 ) < 7 0 60 < 49a + 7b + c < 70
60 < 48a + 6b < 70 10 < 8a + b < 11.6 8a + b = 11
.......(2)
solving equation (1) and (2) a = 2, b = –5, so c = 3 f(x) = 2x2 – 5x + 3 f(50) = 4753 t= 4
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62.
We can write the expression as
2 1 1 = r 1r 1 r 2 1
2011 r 2
r 2
1 1 1 r 1 r 1
=
Putting r = 2, 3, ............, 2011 1
= 2010 + 1 +
= 2011 +
2
1 2012
1 2011
1 1 2 2011 2012 1
1 ) 2
this lies between (2011, 2011
63.
Let initially 2 bases have radii 5 & r. and finally bases have radii (1.21 × 5) & r.
Ratios of volumes =
h
V2 =
3
h
V1 =
V2 V1
3
V2 V1
1.21
((6.05)2 + 6.05 r + r 2)
(52 + 5r + r 2)
1.21
( 6.05 )2
6.05 r r 2 1.21 5 2 5r r 2
36.602 36.6025 5 + 6.0 6.05 5 r + r 2 = 30.25 + 6.05r + 1.21 r 2 .21r 2 = 6.3525 r 2 =
r=
64.
6.3525 .21 11 2
cm. = 55 mm
Let speed of B = V km/hr. Let speed of A = 3V km/hr.
V = 30 km/hr Given 4r = 2 × 60 km/hr Distance covered by then after 10 min. = 2 × 10 = 20 k m So, remaining distance = (30 – 20) km = 10 km. Time table by by B to cover cover 10 km = 65.
10 30
= 20 min
A
B
C
10 hr
20 hr
30 hr
Exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. So, So,
first first we we say say, A and and B are are open open for for 1 hou hourr, then then B & C and and then then C & A.
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1 1 1 1 1 1 + + = 22 10 20 20 30 30 10 60 First then
second then
Third then th
22 part In three hours the tank will be f illed 60 Now, for minimum tim e, the rest tank m ust be filled with A and B taps.
9 1 1 10 20 60
th
38 Part of tank will take 5 hour more So, the rest 60 So, the tank will be filled in 8 th hour.
PHYSICS
66.
kx1 + 1vg = vg kx1
v = g g 1 kx2 + 2vg = vg kx 2
v = g g 2 kx1 ( 1)g
kx 2 ( 2 )g
x1 – x1 = x2 – x2 x1 – x2) = x1 – x2 =
2x1 1x 2 x1 x2
8
67.
4
1 2
F.dx
1
1
2
2
mv 2f mv i2
× 3 × 8 –
1 2
× 1.5 × 4 =
1 2
× 0.5 (vf 2 – 3.16)2
24 – 6 = 0.5 (v f 2 – 3.162) 36 = vf 2 – 3.16 2 vf 6.8 m/s
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68.
vi = vf 5 2
×1 × V +
5
× 0.5 × 4v =
2
5 2
P 5V
3V = 5PV P=
3 5
= 0.6
69.
sin(90 – ) = sinr 4
cos = 3
cos =
2 3
x/2 h2
1 2
h x
2
2
x 4
2
4
= 3 53
2 1 4
=
= 3
1 49 16
4 16
8 3 53
70.
Pi = i 2R 2
100 10 P = 1 16 4
Pf = i2R 2
100 100 10 9 = 9 = P 12 12 16 12 CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892
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CHEMISTRY 71.
100 mL 0.1 M CH 3COOH 50 mL 0.4 M CH3COONa
CH3COO
Ch3COOH = 1 pH = pKa + log 1 = pKa = 4.76 72.
9-structural isomers are possible. CH3CH2CCH, CH2 =CH–CH=CH2 ,
CH3 –CC–CH3 CH2 =C=CH–CH3
73.
74.
75.
Co+3 Ni+2 Cr +3 Fe+2 Co+3
: 3d6 45° 45° stro strong ng fiel field d lig ligan and d (NH (NH3) 8 : 3d 45° SFL (NH3) : 3d3 45° WFL (H2O) : 3d6 45° 45° WFL (H2O) will be be diamagnetic (i)
Reac Reacti tion on quot quotri rien entt Q =
[H] 2 [H2 ] [ 2 ]
=
(0.4 )2 0.1 0.2
=8
Q< K reaction proceeds in the forward direction.
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