PART-A (1 Mark)
PART-A (1 Mark)
MATHEMATICS
MATHEMATICS
1. 1. xx22 + bx + a = 0 + bx + a = 0 x x22 + ax + b = 0 + ax + b = 0 x = x = – 1 – 1 x + 1 =x + 1 = (x + 1) (x + 1)22 + a(x + 1) + b = 0 + a(x + 1) + b = 0 x x22 + (a + 2)x + 1 + a + b = 0 + (a + 2)x + 1 + a + b = 0 x x22 + + bbx x + + a a = = 00 aanndd 1 1 + + a a + + b b = = aa b = – 1b = – 1 a = – 3 a = – 3 ___ ____________ ___ a + b = – 4 a + b = – 4 2. 2. 33x/yx/y = = tt 33t t –– 3 3 tt = 24 = 24 8t = 3 × 248t = 3 × 24 t = t = 99 SSoo,, 33x/yx/y = t = t 33x/yx/y = 9= 9
33x/yx/y = 3= 322 x = 2y.x = 2y.
y y – – x x y y x x y y y y 3 3 = 3. = 3. 3. 3. 22 )) 1 1 n n (( n n 6 6 )) 1 1 n n 2 2 )( )( 1 1 n n (( n n 3 3 1 1 n n 2 2 = = kk n =n = 2 2 1 1 – – k k 3 3 1 1 2 2 1 1 – – k k 3 3 100 100 33 3k3k 201201 11 kk 3 3 1 1 20 20 1 1 k k 67.67.
Number of odd integers = 34. Number of odd integers = 34.
ANSWER KEY
ANSWER KEY
HINTS & SOLUTIONS (YEAR-2010)
HINTS & SOLUTIONS (YEAR-2010)
Q Quueess. 1 . 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 110 0 111 1 112 2 113 3 114 4 1155 Ans. Ans. CC DD BB AA BB BB AA DD DD DD BB AA CC DD CC Q Quueess. . 116 6 117 7 118 8 119 9 220 0 221 1 222 2 223 3 224 4 225 5 226 6 227 7 228 8 229 9 3300 Ans. Ans. DD AA DD BB AA DD AA AA BB BB AA AA DD CC CC Q Quueess. . 331 1 332 2 333 3 334 4 35 35 336 6 337 7 338 8 339 9 4400 Ans. Ans. DD CC AA DD DD AA CC BB CC BB
4. 4. 22+ b+ b22= 39= 3922, b, b22 + h + h22 = 40 = 4022, h, h22 + + 22 = 41= 4122 Add Addinging 2( 2(22+ b+ b22+ h+ h22) = 39) = 3922 + 40 + 4022 + 41 + 4122 2 2 2 2 2 2 c c b b == 2 2 41 41 40 40 39 3922 22 22 = = 2 2 )) 1 1 40 40 (( 40 40 )) 1 1 – – 40 40 (( 22 22 22 = = 2 2 2 2 )) 40 40 (( 3 3 22 = = 2 2 4802 4802 = = 24012401 = 49. = 49. 5.
5. It has to be an isosceles triangle.It has to be an isosceles triangle.
= = 2 2 1 1 × 1 × 1 4 4 1 1 – – x x22 = = 44xx – –11 4 4 1 1 22 Perimeter = 1 + 2x
Perimeter = 1 + 2x odd which is always irrational. odd which is always irrational.
6. 6. == 2 2 1 1 × 12 × 6 × 12 × 6 = 36 = 36 7. 7. A A11 = = AA55== 2 2 2 2 a a 360 360 60 60 = = 24 24 a a22 ,, AA22 = = AA44 = = 2 2 2 2 a a 3 3 360 360 60 60 == 8 8 a a 3 3 22 A A33 = = 2 2 2 2 a a 5 5 360 360 240 240 == 6 6 a a 25 25 22
Area that can be grazed
Area that can be grazed == 24 24 a a 2 2 2 2 + + 8 8 a a 3 3 2 2 2 2 + + 6 6 a a 25 25 22 = 5 = 5aa22..
8. 8. dt dt = = c.c. v = v = r r hh 3 3 1 1 22 tan tan = = h h r r = = 3 3 h h33 tan tan dt dt dv dv = = hh22 tan tan dt dt dh dh ct + k =ct + k = 3 3 tan tan h h33 w whheenn t t = = 00, , h h = = HH k =k = 3 3 tan tan H H33 ct =ct = 3 3 (h (h33 – H – H33) tan) tan w whheenn t t = = 221 1 , , h h == 2 2 H H c = c = 33 6 6 H H33 8 8 7 7 – – h = 0 h = 0 tan tan )) H H (– (– 3 3 3 3 = = HH – –8877 tt 6 6 3 3 3 3 24 24 7 7 = = 72 72 8 8 7 7 t t t = 24t = 24
More time in minutes does it empty the vessel is 3 More time in minutes does it empty the vessel is 3 9.
9. Water + solid = 1000Water + solid = 1000 Water is Water is 10001000 100 100 99 99 = 990 = 990 water evaporated is x. water evaporated is x. so so 100100 x x – – 1000 1000 x x – – 990 990 = 98 = 98 99000 – 100 x = 96000 – 98x 99000 – 100 x = 96000 – 98x 1000 = 2x 1000 = 2x x = 500x = 500 10. 10. xy = 10 and yz = 4 xy = 10 and yz = 4 z =z = yy 2255xx also also x x 5 5 2 2 a = 12 a = 12 a =a = x x 30 30 an andd x x 30 30 (b) = 15 (b) = 15 b =b = 2 2 x x an andd 2 2 x x (c) = 25 (c) = 25 c =c = x x 50 50 Area = x Area = x x x 50 50 = 50 = 50
PHYSICS
PHYSICS
1 11.1. T =T = 22 gg First distance of com f
First distance of com f rom suspension point will increase then decrease.rom suspension point will increase then decrease.
TT..
12.
12. when sliding has startedwhen sliding has started (KVPY / (KVPY / 2010 2010 / / SA)SA) ttiilll l aacccceelleerraattiioon n oof f bblloocck k iis s zzeerroo F F – – f f kk = ma = ma
F – f F – f ss = 0 = 0 f f ss = F = F 13. 13. Mg = 40 ×Mg = 40 × 1000 1000 49 49 × × 1 1 500 500 M = M = 98 98 10 10 10 10 5 5 49 49 40 40 = 100 kg. = 100 kg. 14. 14. 2 2 r r GMm GMm = = r r mv mv22
No change because distance between them will be from
S
Siinnccee,, iigg = = 00 PR = QS PR = QS
Still it will be a balanced
Still it will be a balanced W.S.B.W.S.B. So, again i So, again igg = 0. = 0. 16. 16. 22 R R KQq KQq = = 0 0 4 4 1 1 .. 22 R R Qq Qq
Since initially net force on Q was zero by symmetry Since initially net force on Q was zero by symmetry So,
So, FF11FFReRemainingmaining111100
1 1 11 11 maining maining Re Re FF F F So, So, 22 0 0 r r Qq Qq .. 4 4 1 1
towards the position of the removed charge. towards the position of the removed charge.
17. 17. PPii = = ii 2 2 ii R R V V S
Siinnccee,, RRf f < R < Rii ke keepepiningg V = coV = consnstantantt V = V = VVf f P Pf f = = f f 2 2 R R V V S Siinnccee,, RR P P.. 18. 18.
the combination will behave as parallel slab so light get
the combination will behave as parallel slab so light get laterally displaced without any spectrum.laterally displaced without any spectrum. 19. 19. 2200ººCC 8800ººCC S S as T as T d d = m.s.d = m.s.d dd = ms d = ms d
Since, average S of body which is initially at 80ºC is higher then body initially at temperature 20ºC so 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
temperature decreases of earlier will be less then temperatur e increases of letter.e increases of letter. S
20. 20. ttxx = = 3 3 T T + + 330000 Q Q = = mmss tt x x = 3 = 3TT S = S = .. m m Q Q S = S = .. m m Q Q Since, unit of
Since, unit of is joule in both system is joule in both system X X TT m = m m = m00kgkg mm00kgkg Q = Q Q = Q00JJ QQ00.J.J tt x x TT S Sxx = = x x 0 0 0 0 tt m m Q Q = = 11440000 SSTT = = mm TT Q Q 0 0 0 0 = = mm00 ttxx Q Q 3 3 S STT = 3 × 1400 = 4200 J-kg = 3 × 1400 = 4200 J-kg –1 –1KK –1 –1
CHEMISTRY
CHEMISTRY
21.21. Aqueous Aqueous solution solution containing containing more number more number of paof particles have rticles have more elevatmore elevation in ion in boiling boiling point.point. 22.
22. 1414Si : 1sSi : 1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p22
23.
23. CaCOCaCO33 (s) (s) CaO (s) + CO CaO (s) + CO22 (g) (g)
Number of mole Number of mole 100 100 25 25 100 100 25 25 Amount of CO Amount of CO22 = = 100 100 25 25 × 44 = 11 gram. × 44 = 11 gram. 24.
24. As As we we move ‘lmove ‘left to eft to right’ right’ in in 22ndnd period, atomic radii decreases due to increase in ef period, atomic radii decreases due to increase in ef fective nuclear charge.fective nuclear charge.
25.
25. In BClIn BCl33 octet rule octet rule is not satisfy.is not satisfy.
T
Total number of 6 elecotal number of 6 electrons in outermost trons in outermost shell of B after bonding.shell of B after bonding. 26.
26. MnOMnO22 + 4HCl + 4HCl MnCl MnCl22 + 2H + 2H22O + ClO + Cl22 Cl
Cl22 gas produces. gas produces. 27. 27. CC44HH77BBr r CH CH22 = CH – CH = CH – CH22 – C – CHH22 – – Br Br H H H H|| || Br Br – – C C – – C C – – C C C C – – H H || || || || H H H H H H H H
Number of covalent bond = 12. Number of covalent bond = 12.
28. 28. pH = 2pH = 2 [H[H ]]ii = 10 = 10 M M pH = 5 pH = 5 [H[H++]] f f = 10 = 10 –5 –5 M M ii f f ]] H H [[ ]] H H [[ = = 22 5 5 10 10 10 10 = = 1000 1000 1 1 So, H
So, H++ concentration decreases thousand fold. concentration decreases thousand fold.
2 299. . FFoor r 11stst jar : jar : Number of moles of H Number of moles of H22 (g) = (g) = 2 2 2 2 =1 mole. =1 mole. Number of molecules of H Number of molecules of H22 (g) = 6.02 × 10 (g) = 6.02 × 102323.. For 2
For 2ndnd jar : jar :
Number of moles of N Number of moles of N22 (g) = (g) = 28 28 28 28 =1 mole. =1 mole. Number of molecules of N Number of molecules of N22 (g) = 6.02 × 10 (g) = 6.02 × 102323..
So, both jar have same number of molecules. So, both jar have same number of molecules.
30.
30. and and
(
(cciiss)) ((ttrraannss)) cis and trans are ster
PART-B (5 Mark)
PART-B (5 Mark)
MATHEMATICS
MATHEMATICS
1.
1. Let total amount is nLet total amount is n22
T
Total borrowed amount = otal borrowed amount = (2u + 1) (2u + 1) 1010 n
n22 – (2u + 1) 10 < 10 – (2u + 1) 10 < 10
T
Trurue e foforr. . n n = = 66 u = u = 11 So, the left amount = 6. So, the left amount = 6.
2. 2.
Let
Let AD ADE is equilateral E is equilateral and D is mand D is mid point of AB and E is id point of AB and E is mid point mid point of ACof AC
[given condition is true for above assumption] [given condition is true for above assumption]
Area of Area of quad. quad. ADPE ADPE = = Area of Area of quad. quad. DPFBDPFB =
= Area Area of of quad. quad. EPFCEPFC
Area of Area of ABC = 12 sq. units ABC = 12 sq. units
3.
3. From the question if m = 1111 or From the question if m = 1111 or
((ii)) mm==111111..1111
is always divisible by n = 11 which is coprime
is always divisible by n = 11 which is coprime with 10with 10 (ii) by choosing a = k 10
(ii) by choosing a = k 10bb (10 (10cc – 1) when k is any natural number we can option any natural number k. – 1) when k is any natural number we can option any natural number k.
The problem seen to have an err
The problem seen to have an err or which may be due to memoror which may be due to memor y retersion constraints.y retersion constraints.
DESCRIPTIVE TYPE QUESTIONS
PHYSICS
PHYSICS
4. 4.
Assume to
Assume to be spherical be spherical concave.concave.
cos cos00 = = R R )) H H R R (( (i) P
(i) P.E. = mg(R – .E. = mg(R – R cosR cos) = mgR(1 – cos) = mgR(1 – cos)) (ii) mgH – mgR (1 – cos
(ii) mgH – mgR (1 – cos) = kinetic energy) = kinetic energy
(iii) m(g sin (iii) m(g sin)R = (mR)R = (mR22)) 2 2 2 2 dt dt d d ttPQPQ = = 4 4 16 16 1 1 g g 2 2 4 4 T T 2 2 0 0 = = 16 16 1 1 g g 2 2 2 2 0 0 (iv) (iv) N – mg = N – mg = R R mV mV22 N = mg + N = mg + R R mV mV22 where V = where V = 22ghgh..
5. 5. (i) For (i) For RRPP A A V V V V estimated estimated RR R R R R RR RR R R V V V V A A A A V V R R R R R R R R R R RR RR RR RR R R P P R RPP = = V V = = V V V V R R R R RR RR + R+ R A A (ii)
(ii) RRPP = = A A V V V V RR R R R R R R .. R R R R S
Siinnccee,, RR A A < < R < < R < < R < < RVV R
Restest = = RRRRRRRR RR A A – –~~ RR V V V V P P R R = 0 = 0 (iii) (iii) RQRQ V V == V V A A V V A A R R R R R R R R )) R R R R (( R Restimatedestimated = = V V A A V V A A R R R R R R R R )) R R R R (( RQ RQ = = R –R – V V A A V V A A R R R R R R R R )) R R R R (( = = V V A A V V A A V V V V A A 2 2 R R R R R R R R R R RR RR RR RR RR RR R R = = V V A A V V A A A A 2 2 R R R R R R R R R R RR RR R R = = A A V V A A V V 2 2 R R R R RR RR R R R R RQ RQ – –~~ RR A A . . (iii) (iii) V V V V V V A A A A 2 2 R R R R R R R R R R RR RR R R = 0 = 0 After
(i) (i) 10 10 1 1 20 20 1 1 V V 1 1 20 20 1 1 10 10 1 1 V V 1 1 = = 20 20 1 1 2 2 V = V = 3 3 20 20 (ii) f (ii) f LMLM = –5 cm = –5 cm 3 3 20 20 + d = 10 + d = 10 d = 10 – d = 10 – 3 3 20 20 = = 3 3 10 10 cm cm (iii) (iii) (A) I
(A) Istst reference with lens reference with lens
V = V = 3 3 0 0 2 2 (B) Then mirror, (B) Then mirror, X Ximim = = –X–X0 M0 M (C) Again by lens, (C) Again by lens, 40 40 3 3 V V 1 1 = = 10 10 1 1 40 40 3 3 10 10 1 1 V V 1 1 = = 40 40 3 3 4 4 V = V = 7 7 40 40 cm cm
It means right of lens at a distance It means right of lens at a distance
7 7 40 40 cm. cm.
CHEMISTRY
CHEMISTRY
7.7. ((II)) ((AA) ) BBoottttllee--3 3 ddooees s nnoot t rreeaacct t wwiitth h HHCCl l oor r NNaaOOHH.. (B) Bottle-2 reacts only with NaOH.
(B) Bottle-2 reacts only with NaOH.
(C) Bottle-4 reacts with both NaOH or HCl. (C) Bottle-4 reacts with both NaOH or HCl. (D) Bottle-1 r
(D) Bottle-1 r eacts with HCl eacts with HCl onlyonly..
((IIII)) BoBottttlele-4 is h-4 is higighlhly soy solulublble in de in disistitilllled wed watater der due to zue to zwiwittetter ion fr ion forormatmatioion.n. 8.
8. ((ii)) BBaallaanncceed d eeqquuaattiioon n aarre e ::
((aa)) 3 3 CCu u + + 8 8 HHNNOO33 3 Cu(NO 3 Cu(NO33))22 + 2NO + 4H + 2NO + 4H22OO ((bb)) 2 2 CCuu22 Cu Cu2222 + + 22
((iiii)) MMoolle e oof f 22 = = 254 254 54 54 .. 2 2 = 0.01 = 0.01 Mole of Cu
Mole of Cu22 = 2 × mole of = 2 × mole of 22 = 0.02 = 0.02 Mole of Cu
Mole of Cu22 = mole of = mole of Cu = 0.02Cu = 0.02 wt. = 0.02 × 63.5 = 1.27 g wt. = 0.02 × 63.5 = 1.27 g % purity = % purity = 2 2 27 27 .. 1 1 × 100 = 63.5% × 100 = 63.5% 9. 9. ((ii)) 2255000 0 × × 44..118 8 = = 110044550 0 kkJJ
((iiii)) MMoolle e oof f ssuuccrroosse e rreeqquuiirreed d == 66 3 3 10 10 6 6 .. 5 5 10 10 10450 10450 = 1.866 = 1.866 wt. of sucrose required wt. of sucrose required 1.866 × 342 = 638.172 g1.866 × 342 = 638.172 g
1 mole of C1 mole of C1212HH2222OO1111 12 mole of CO 12 mole of CO22
1.866 moles of C1.866 moles of C1212HH2222OO1111 1.866 × 12 moles of 1.866 × 12 moles of COCO22
22.392 moles of CO 22.392 moles of CO22 1 mole of CO1 mole of CO22 22.4 22.4 22.392 moles of CO 22.392 moles of CO22 22.4 × 22.392 22.4 × 22.392.. 501.58 501.58 1
100. . ((aa)) Difference Difference in flower colour is in flower colour is most likely due most likely due to environmental factorsto environmental factors (b)
(b) Perform cross breeding between the plants from Perform cross breeding between the plants from Chandigarh and those from Shimla to find outChandigarh and those from Shimla to find out whether we get any pink flower or flowers with any shade of color
whether we get any pink flower or flowers with any shade of color between pink and white in the F1between pink and white in the F1 generation
generation (c)
(c) Grow the plants from Grow the plants from Chandigarh in Shimla and check whether they still produce white fChandigarh in Shimla and check whether they still produce white f lowers of lowers of bear pink flowers.
bear pink flowers. 1
111. . ((aa)) In experiment A, ethanol fermentation occ In experiment A, ethanol fermentation occ urs producing COurs producing CO22, turning lime water , turning lime water milky. Since acid ismilky. Since acid is not produced the dye colour does not change.
not produced the dye colour does not change. In experiment B, lactic acid ferm
In experiment B, lactic acid ferm entation takes place, which produces acid but does not produce COentation takes place, which produces acid but does not produce CO22.. Hence dye colour changes to yellow but the lime water does not tur
Hence dye colour changes to yellow but the lime water does not tur n milky.n milky. In experiment C, since the lime water tur
In experiment C, since the lime water tur ns milky, ethanol fermentation is occurring. In addition, sincens milky, ethanol fermentation is occurring. In addition, since removal of air did not affec
removal of air did not affec t the reaction, the fermentation is anaerobic and yeast must be the organismt the reaction, the fermentation is anaerobic and yeast must be the organism in the flask.
in the flask. (b)
(b) In RBC’s lactic acid fermentation occurs.In RBC’s lactic acid fermentation occurs. 1
122. . ((aa)) The result of the radio-carbon dating was correc The result of the radio-carbon dating was correc t.t. Reason :
Reason : V Vehicles running on the highway beside the house emittehicles running on the highway beside the house emitt ed carbon dioxide from ed carbon dioxide from the combustthe combustionion of petrol or diesel, which are fossil fuels. The carbon in this carbon dioxide, coming from living material 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
that has been converted into petroleum millions of years ago, would get illions of years ago, would get assimilatassimilated into the tissues of theed into the tissues of the plant as it uses carbon dioxide from the surrounding atmospher
plant as it uses carbon dioxide from the surrounding atmospher e for photosynthesis. Therefore tisse for photosynthesis. Therefore tissues of ues of the plant, when used for r
the plant, when used for r adio-carbon dating, would show the age of adio-carbon dating, would show the age of the plant to be mthe plant to be m any thousands of any thousands of years old.
years old. (b)
(b) A A simple simple experiment experiment to to test test the the validity validity of of this this explanatexplanation ion would would be be to to collect collect seeds seeds from the from the plant plant andand grow them in a plot of land away from the highway or other sources of
grow them in a plot of land away from the highway or other sources of carbon dioxide coming from carbon dioxide coming from thethe burning or fossil fuels. Radio-carbon dating of plants growing from these seeds show them as young burning or fossil fuels. Radio-carbon dating of plants growing from these seeds show them as young plants.
