KVPY SA Stream Solution 2010
Short Description
KVPY SA Stream Solution 2010...
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HINTS & SOLUTIONS (YEAR-2010) ANSWER KEY Ques.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
C
D
B
A
B
B
A
D
D
D
B
A
C
D
C
Ques.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
D
A
D
B
A
D
A
A
B
B
A
A
D
C
C
Ques.
31
32
33
34
35
36
37
38
39
40
Ans.
D
C
A
D
D
A
C
B
C
B
PART-A (1 Mark)
MATHEMATICS
1.
x2 + bx + a = 0
x2 + ax + b = 0 x = – 1
x + 1 =
2
(x + 1) + a(x + 1) + b = 0 x2 + (a + 2)x + 1 + a + b = 0 x2 + bx + a = 0
and
1+a+b=a
b=–1 a=–3 ______ _________ ___ a+b=–4
2.
3x/y = t
3.
3t –
t 3
= 24
8t = 3 × 24
t= 9
So,
3x/y = t
3x/y = 9
3x/y = 32
x = 2y.
2n 1 = k 3
n=
3 3k 201
1 k
x y 3y = 3. x – y y
n(n 1)(2n 1) 2 6n(n 1) 1
3k – 1 100 2
3k – 1 2 20 1 3
1 k 67.
Number of odd integers = 34.
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4.
2 +
b2 = 392, b2 + h2 = 402, h2 + 2 = 412 Adding Adding 2(2 + b2 + h2) = 392 + 402 + 412
2
2
b c
2
=
=
39 2 40 2 412 = 2 3( 40)2 2 2
=
( 40 – 1)2 40 2 ( 40 1)2 2
4802 = 2
2401
= 49. 5.
It has to be an isosceles triangle.
=
1 1 1 2 4 x 2 – 1 × 1 x – = 2 4 4
Perimeter = 1 + 2x odd which is always irrational.
6.
=
1 × 12 × 6 2
= 36
7.
2
2
2 2 60 a 60 3a = a , A = A = = 3a A1 = A5 = 2 4 360 2 360 2 24 8
2
2 240 5a = 25a A3 = 360 2 6
a2 3a2 25a 2 2 2 Area that can be grazed = 24 + 8 + = 5a2. 6
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8.
dv = c. dt v=
=
3
1 2 r h 3
tan =
r h
h3 tan
dv dh = h2 tan dt dt when
t = 0, h = H
when
t = 21 , h =
H 2
ct + k = k=
c=
h3 tan 3
H3 tan
3
ct =
3
(h3 – H3) tan
– 7 H3 8 6
3
h=0
3
3 – 7 7 H t 7 = t 3 8 24 8 72 6
(–H3 ) tan =
t = 24
More time in minutes does it empty the vessel is 3 9.
Water + solid = 1000 99
1000 = 990 100 water evaporated is x.
Water is
so
990 – x 100 = 98 1000 – x
99000 – 100 x = 96000 – 98x 1000 = 2x
x = 500
10.
and and
z=
30 (b) = 15 x
b=
xy = 10 and yz = 4
y
2x 5
x 2
2 x a = 12 5
also
and and
x (c) = 25 2
a=
c=
30 x 50 x
50 = 50 x
Area = x
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PHYSICS
11.
T = 2
g
First distance of com f rom suspension point will increase then decrease.
T.
12.
when sliding has started (KVPY / 2010 / SA) till acceleration of block is zero F – f s = 0 f s = F
13.
F – fk = ma
49 × 500 1000 1
Mg = 40 ×
M=
40 49 5 10 = 100 kg. 10 98
14.
GMm r 2
=
mv 2 r
No change because distance between them will be from centre to centre distance which is unchanged.
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15.
Since, ig = 0 PR = QS Still it will be a balanced W.S.B. So, again i g = 0.
16.
KQq R
2
=
1
.
40
Qq R2
Since initially net force on Q was zero by symmetry
F1 FRe maining 11 0
So,
FRe maining11 F1
So,
17.
1
.
Qq
40 r 2
Pi =
towards the position of the removed charge.
Vi2 Ri
Since, Rf < Ri keep keepin ing g V = cons constan tantt V = Vf Pf =
V2 R f
Since, R P . 18.
the combination will behave as parallel slab so light get laterally displaced without any spectrum. 19.
20ºC
80ºC
S as T d = m.s.d d = ms d Since, average S of body which is initially at 80ºC is higher then body initially at temperature 20ºC so temperature decreases of earlier will be less then temperatur e increases of letter. So,
Tf > 50ºC.
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20.
Q = ms
tx = 3 T + 300 tx = 3 T S=
Q m.
Q m.
S=
is joule in both system Since, unit of X m = m 0kg Q = Q0J t x Sx =
T m0kg Q0.J T
Q0 m0 t x = 1400
ST =
Q0 3Q = m0 T m0 t x
ST = 3 × 1400 = 4200 J-kg –1K –1
CHEMISTRY 21. 22.
Aqueous solution containing more number of particles particles have more elevation elevation in boiling point.
14
Si : 1s2 2s2 2p6 3s2 3p2
CaCO3 (s) CaO (s) + CO 2 (g)
23.
Number of mole
25 100
25 100
25 × 44 = 11 gram. 100
Amount of CO2 =
24.
As we move ‘left ‘left to right’ in 2nd period, atomic radii decreases due to increase in ef fective nuclear charge.
25.
In BCl3 octet rule is not satisfy.
Total number of 6 elec electrons trons in outermost shell of B after bonding. 26.
MnCl2 + 2H2O + Cl2 MnO2 + 4HCl Cl2 gas produces.
27.
C4H7Br CH2 = CH – CH 2 – CH CH2 – Br H H H H | | | | H – C C – C – C – Br | | H H Number of covalent bond = 12.
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28.
pH = 2 pH = 5
[H ] f
[H ]i
=
[H+]i = 10 –2 M [H+]f = 10 –5 M
10 5
1 = 1000 10 2
So, H+ concentration decreases thousand fold. 29.
For 1st jar : Number of moles of H 2 (g) =
2 =1 mole. 2
Number of molecules of H2 (g) = 6.02 × 10 23. For 2nd jar : Number of moles of N 2 (g) =
28 =1 mole. 28
Number of molecules of N2 (g) = 6.02 × 10 23. So, both jar have same number of molecules.
30.
and
(c (cis)
(trans)
cis and trans are ster stereoisomeric eoisomeric pair.
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DESCRIPTIVE TYPE QUESTIONS PART-B (5 Mark)
MATHEMATICS 1.
Let total amount is n 2 Total borrowed amount = (2u + 1) 10 n2 – (2u + 1) 10 < 10 True rue for for. n = 6
u =1
So, the left amount = 6.
2.
Let ADE ADE is equilateral and D is m id point of AB and E is mid point of AC [given condition is true for above assumption]
Area of quad. ADPE = Area of quad. DPFB = Area of quad. EPFC
3.
ABC = 12 sq. units Area of
From the question if m = 1111 or (i)
m = 111.11
is always divisible by n = 11 which is coprime with 10 (ii) by choosing a = k 10b (10c – 1) when k is any natural number we can option any natural number k. The problem seen to have an err or which may be due to memor y retersion constraints.
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PHYSICS
4.
Assume to be spherical concave.
cos0 =
(R H) R
(i) P.E. P.E. = mg(R – R cos ) = mgR(1 – cos ) (ii) mgH – mgR (1 – cos ) = kinetic energy d2
(iii) m(g sin)R = (mR ) 2
dt 2
0 2 1 g 16
T 2 tPQ = 4
4
0 2 = 2 g 1 16
(iv)
N – mg =
mV 2 R
mV 2 N = mg + R
where V =
2gh .
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5.
(i) For RP
RR V R A Restimated R R V
RP =
(ii)
V
=
RR V
RP R
RR V RR A R AR V R RV
+ R A
R RV
R.R V R A RP = R R R V
Since,
R A < < R < < R V Rest =
RR V R RV
~ R R A –
RP = 0 RQ V =
(iii)
(R R A )R V R A R R V
Restimated =
RQ = R –
=
(R R A )R V R A R R V
(R R A )R V R R A R V
R 2 RR A RR V RR V R AR V R R A R V
=
R 2 RR A R AR V R R A R V
R 2 RR A R A = RV RV RQ – ~ R A . (iii)
R 2 RR A R AR V R RV RV After solving,
= 0
R=
R AR V .
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6.
(i)
1 V
V=
1
20
1
1
10
V
1 10
1 20
=
2 1 20
20 3
(ii) f LM = –5 cm 20 3
+ d = 10
d = 10 –
20 3
10
=
3
cm
(iii)
(A) Ist reference with lens 20
V=
3
(B) Then mirror, Xim = –X0 M (C) Again by lens, 1 V 1 V
3
40
V=
1 10
40 7
1
=
10 3
40
=
4 3 40
cm
It means right of lens at a distance
40 cm. 7
CHEMISTRY 7.
8.
(I)
(A) Bottle-3
(II)
(B) Bottle-2 reacts only with NaOH. (C) Bottle-4 reacts with both NaOH or HCl. (D) Bottle-1 r eacts with HCl only. only. Bott Bottle le-4 -4 is hig highl hly y solu solubl ble e in dis disti till lled ed wat water er due due to zwi zwitte tterr ion for format matio ion. n.
(i)
does not react with HCl or NaOH.
Balanced equation are : (a) 3 Cu + 8 HNO3 3 Cu(NO3)2 + 2NO + 4H 2O (b) 2 Cu2 Cu22 + 2 (c) 2 Na2S2O3 + 2 Na2S4O6 + 2 Na
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(ii)
Mole of 2 =
2.54 254
= 0.01
2 = 0.02 Mole of Cu2 = 2 × mole of Mole of Cu 2 = mole of Cu = 0.02 wt. = 0.02 × 63.5 = 1.27 g % purity =
9.
(i) (ii)
1.27 2
× 100 = 63.5%
2500 × 4.18 = 10450 kJ Mole of sucrose required =
10450 103 5.6 10 6
= 1.866
wt. of sucrose required 1.866 × 342 = 638.172 g 1 mole of C12H22O11 12 mole of CO 2 1.866 moles of C 12H22O11 1.866 × 12 moles of CO2 22.392 moles of CO 2 1 mole of CO 2 22.4 22.392 moles of CO 2 22.4 × 22.392 . 501.58
10.
(a) Difference in flower colour is most likely due to environmental factors (b) Perform cross breeding between the plants from Chandigarh and those from Shimla to find out whether we get any pink flower or flowers with any shade of color between pink and white in the F1 generation (c) Grow the plants from Chandigarh in Shimla and check whether they still produce white f lowers of bear pink flowers.
11.
(a) In experiment A, ethanol fermentation occ urs producing CO 2, turning lime water milky. Since acid is not produced the dye colour does not change. In experiment B, lactic acid ferm entation takes place, which produces acid but does not produce CO 2. Hence dye colour changes to yellow but the lime water does not tur n milky. In experiment C, since the lime water tur ns milky, ethanol fermentation is occurring. In addition, since removal of air did not affec t the reaction, the fermentation is anaerobic and yeast must be the organism in the flask. (b) In RBC’s lactic acid fermentation occurs.
12.
(a) The result of the radio-carbon dating was correc t. Reason : Vehicles Vehicles running on the highway beside the house emitt ed carbon dioxide from the combustion combust ion of petrol or diesel, which are fossil fuels. The carbon in this carbon dioxide, coming from living material that has been converted into petroleum m illions of years ago, would get assimilated assimilat ed into the tissues of the plant as it uses carbon dioxide from the surrounding atmospher e for photosynthesis. Therefore tissues tiss ues of the plant, when used for r adio-carbon dating, would show the age of the plant to be m any thousands of years old. (b) A (b) A simple experiment to test the validity of this explanation explanation would be to collect seeds from the plant and grow them in a plot of land away from the highway or other sources of carbon dioxide coming from the burning or fossil fuels. Radio-carbon dating of plants growing from these seeds show them as young plants.
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