1. Let total amount is n2
Total borrowed amount = (2u + 1) 10 n2– (2u + 1) 10 < 10
True for. n = 6 u = 1 So, the left amount = 6.
2.
Let ADE is equilateral and D is mid 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
Area of ABC = 12 sq. units 3. 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 error which may be due to memory retersion constraints.
PHYSICS
4.
Assume to be spherical concave.
cos0=
5.
estimated V R
R
Restimated=
V
(A) Istreference with lens
V = 3
0
2 (B) Then mirror,
Xim= –X0 M (C) Again by lens,
40
It means right of lens at a distance 7 40cm.
CHEMISTRY
7. (I) (A) Bottle-3 does not react with HCl or NaOH.
(B) Bottle-2 reacts only with NaOH.
(C) Bottle-4 reacts with both NaOH or HCl.
(D) Bottle-1 reacts with HCl only.
8.(i) Balanced equation are :
(a) 3 Cu + 8 HNO3 3 Cu(NO3)2+ 2NO + 4H2O (b) 2 Cu2 Cu22+2
(c) 2 Na2S2O3+2 Na2S4O6+ 2 Na
(ii) Mole of2= 254 54 .
2 = 0.01
Mole of Cu2= 2 × mole of2= 0.02 Mole of Cu2= mole of Cu = 0.02 wt. = 0.02 × 63.5 = 1.27 g
% purity = 2 27 .
1 × 100 = 63.5%
9. (i) 2500 × 4.18 = 10450 kJ
(ii) Mole of sucrose required = 6
3
10 6 . 5
10 10450
= 1.866
wt. of sucrose required 1.866 × 342
= 638.172 g
1 mole of C12H22O11 12 mole of CO2
1.866 moles of C12H22O11
1.866 × 12 moles of CO2
22.392 moles of CO2
1 mole of CO2 22.4
22.392 moles of CO2 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 flowers of bear pink flowers.
11. (a) In experiment A, ethanol fermentation occurs producing CO2, turning lime water milky. Since acid is not produced the dye colour does not change.
In experiment B, lactic acid fermentation takes place, which produces acid but does not produce CO2. Hence dye colour changes to yellow but the lime water does not turn milky.
In experiment C, since the lime water turns milky, ethanol fermentation is occurring. In addition, since removal of air did not affect 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 correct.
Reason : Vehicles running on the highway beside the house emitted carbon dioxide from the combustion of petrol or diesel, which are fossil fuels. The carbon in this carbon dioxide, coming from living material that has been converted into petroleum millions of years ago, would get assimilated into the tissues of the plant as it uses carbon dioxide from the surrounding atmosphere for photosynthesis. Therefore tissues of the plant, when used for radio-carbon dating, would show the age of the plant to be many thousands of years old.
(b) A simple experiment to test the validity of this 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.
KVPY [HINTS & SOLUTION] - 2011
One Mark Questions MATHEMATICS
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).
f(x) is an identity with only value equal to zero.
f(x) = 0 x R P(x) = 1, x R 2. Using cauchy schwartz's inequality
(a2+ b2) (x2+ y2) (ax + by)2 equality holds at
x
add equation (1) and (2)
Sum of all possible values are 18 5. Given 0 < r < 4 in all the obtain
(Base)x= 5 9
the option having least base will give the largest x.
So, in option B base
x = log(1 ) For x to be maximum
log (1 +) should be minimum
r is minimum.
6. Angle bisector theorem
6
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 total number of ball
N = (13)2+ (14)2+ ...+ (30)2
12. Let distance is 6d
Mud : Tar : Stream
(Note : order is changed in questions)
13. 3 step
14. The clock well show 1 in an hour for 19 time for 11 hours it will show the incorrect time for (19 × 11) time.
The last 12thhour will always show the incorrect time so total incorrect time
(19 × 11 + 60) min = 269 min
there are 24 hours in a day to = 269 × 2 = 538 min 538 min = 30
269 = 9 hours
the fraction day when the clock shows correct time
is = 24
1 9
= 8
5 8 13
15.
Right Wrong Unattempted
15 15 0
14 4
13 8
12 12
11 16
10 20
6 cases only
PHYSICS
16. By mechanical energy conservation KEi+ Ui= KEf+ Uf
O + Ui= 0 + Uf Ui= Uf
hi= hf
So D will lie on line AB
17. Since toy is not accelerating so net external force on toy is zero. So (A)
18.
|S1| + |S2| = H gt2
2
1 + ut – 21gt2 = H ut = H . ....(I) H = u2g
2
....(II)
t = 2ug
S2= ut – 21gt2
= 42ug 21g4ug 38ug 2 2 2
= 43H
19.
By concept of centre of mass 36x = 9(20–x)
36x = 180 – 9x 45x = 180 x = 4m
20. All three 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
24.
25. R = A There resultant is 0 at this point
27. = mgh750/t100 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
proportional to electron density in the benzene ring.
33. n = 2
34. Average Kinetic Energy depends only on temperature (KE)avg=
2 kT
3 per molecule
35. Ideal gas H = 0
S > 0
(randomness increases)
36. (NH4)2Cr2O7 Cr2O3+ N2+ 4H2O
37. According to the graph solubility 40° is approx.
200 g per 100 ml. For 50 ml, amount is 100 g approx.
38. Aldehyde, ketones with acetyl group CH3– – show Iodoform test.
40. ZnS ZnSO4 during roasting, sulphide ore is converted into sulphate.
41.
42. Cl2+ 2KBr 2KCl + Br2 reddish brown
43. (i) and (iv) are hetro aromatic and the resonance form of azulene (iii) is aromatic.
(ii) Is nonaromatic.
44. Simple nomenclature of alkane.
45. The given reaction is SN2 which occurs at sp3carbon with good leaving group.
PART-II
Two Marks Questions MATHEMATICS
solving equation (1) and (2)a = 2, b = –5, so c = 3 f(x) = 2x2– 5x + 3
f(50) = 4753 t = 4
62. We can write the expression as
this lies between (2011, 2011 2 1)
63. Let initially 2 bases have radii 5 & r. and finally bases have radii (1.21 × 5) & r.
Exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour.
So, first we say, A and B are open for 1 hour, then B
First then second then Third then
In three hours the tank will be filled
th
Now, for minimum time, the rest tank must be filled
with A and B taps.
So, the rest
th
So, the tank will be filled in 8thhour.
PHYSICS
66.
kx1+1vg =vg v = gkx1g
g
72. 9-structural isomers are possible.
CH3CH2CCH, CH3–CC–CH3 Co+3 will be diamagnetic (i)
75. Reaction quotrient Q = [H ][ ]
reaction proceeds in the forward direction.