Answer Key of Nwtra35g01,Nwtra35a01-A04, Nwtra35c01-c02, Camp35, Nwtw35a01-A02, Nwtwa35a01-A07 & Nwtwa35c01-c02_reshuffling Test-1_paper-1 & 2- Test Date 23-03-14 (1)

FIITJEE Ltd., North West Delhi Centre, 31-32-33, Central Market, West Avenue Road, Punjabi Bagh (West), New Delhi - 110026, Ph: 011-45634000

BATCHES – NWTRA35G01 ,NWTRA35A01-A04 , NWTRA35C01-C02 , CAMP35, NWTW35A01-A02 , NWTWA35A01-A07 & NWTWA35C01-C02

Reshuffling Test – 1 Paper – 1

PHYSICS, CHEMISTRY & MATHEMATICS QP CODE: 120627.1

ANSWERS Reshuffling Test – 1

ANSWER KEY Physics Chemistry Mathematics

1 A P110305 1 C C110108 1 D M111301 2 B P110605 2 C C110705 2 D M110909 3 D P110308 3 C C110306 3 D M110204 4 D P110415 4 B C111102 4 A M110302 5 C P110404 5 C C111206 5 B M111004 6 C P110507 6 C C110908 6 C M111205 7 D P110604 7 D C111006 7 A M110903 8 B P110605 8 C C110606 8 D M110410 9 B P110413 9 D C110401 9 B M111203

10 C P110708 10 B C110502 10 D M110304 11 D P110718 11 C C111906 11 C M110410 12 C P110804 12 C C111505 12 C M110101 13 A P110905 13 D C111607 13 D M110505 14 C P110905 14 B C111303 14 C M111301 15 A P110905 15 C C111404 15 B M111403 16 D P111101 16 B C111804 16 B M110305 17 A P111101 17 C C111705 17 B M110413 18 C P111026 18 D C110501 18 B M111301 19 A P111213 19 C C110601 19 A M110733 20 B P111207 20 C C110606 20 B Not Available 21 D P111213 21 C C111706 21 D M111301 22 C P111005 22 D C111301 22 A M111502 23 A P110613 23 B C110804 23 D M110808 24 B P111005 24 A C110306 24 C M110501 25 A P111303 25 D C111606 25 B M110701 26 B P111306 26 C C111906 26 A M110504 27 C P111307 27 C C111301 27 A M110826 28 B P111103 28 D C111713 28 B M110503 29 A P111103 29 C C111801 29 B M110109 30 A P110723 30 D C111403 30 D M110414

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2

Hints & Solutions SECTION – 1 (Physics)

1. A (Concept Code: P110305)

Sol. Maximum acceleration = 60 2040 30

−−

= 4.

2. B (Concept Code: P110605)

Sol. o1tan 45e

θ = ⇒ θ ≥

3. D (Concept Code: P110308)

Sol. 21(4 2)t (1 2)t 02

− + − =

T = 4, = 4 × 4 + 12

× (4)2

= 24 m 4. D (Concept Code: P110415)

Sol. dV 3 12tdt 2

−⎛ ⎞= ⎜ ⎟⎝ ⎠

⇒ 23t 6tV

2⎛ ⎞−

Δ = ⎜ ⎟⎝ ⎠

= 0

5. C (Concept Code: P110404) Sol. T = 8 – μ2 × 3 × 10 = 2 f1 = T – 1 = 1 ; f2 = μ2 × 3 × 10 = 6 6. C (Concept Code: P110507) Sol. W = 40 × 10 × 0.4 × 20 7. D (Concept Code: P110604)

Sol. K.E. = 2(6 8)

2 4××

= 288


Sol. { }2 21 IuW (2mu) (mu)2m 2

= − = , I = 3mu

9. B (Concept Code: P110413) Sol. Force = Slope of curve 10. C (Concept Code: P110708) Sol. VA = VC + VC – VP = 2VC – VP 11. D (Concept Code: P110718) Sol. W × 0.3 + 500 × 0.5 = T1 × 0.7

and T1 = T2 = W 5002+

12. C (Concept Code: P110804)


3

Sol. (0.2 + xo) × 10 = 4 ⇒ xo = 0.2 m below 0. 13. A (Concept Code: P110905) Sol. g ∝ ρr 14. C (Concept Code: P110905)

Sol. 2r gg

r h 2⎛ ⎞ =⎜ ⎟+⎝ ⎠


Sol. F = ρghA – mg = 20N↓ 16. D (Concept Code: P111101) Sol. F ∝ r2 17. A (Concept Code: P111101) Sol. ΔV = (100 cm3) × (4 × 10–5) × (100) 18. C (Concept Code: P111026) Sol. Surface energy ∝ Surface area. 19. A (Concept Code: P111213)

Sol. W = 2 1 3 2

2

(T T )(T T )nR

T− −

− 10 kJ−


Sol. 5 × 20 × 1 = 2 × 20 × 1 m 802+ ×

m = 1 kg ice melts So, mass of water = 5 + 1 = 6 kg. 21. D (Concept Code: P111213)

Sol. Use gas equation PV = nRT 22. C (Concept Code: P111005) Sol. N – mg = ma ⇒ N = mg + ma ⇒ N = mg + M3g = 4W 23. A (Concept Code: P110613)

Sol. 3dm 3 10 50kg / sec

dt 60×

= = , ( )d mgh dmP ghdt dt

= =

= 50 x 10 x 25 = 12500 W 24. B (Concept Code: P111005) Sol. When stones are there inside the boat. the volume of water displaced is equal to the

weight of the boat + stones. When stones are unloaded into the water the volume of water displaced is equal to the volume of stones unloaded plus (weight of boat) / (density of water). This time the displaced volume of water is smaller than in the first case. Therefore the water level falls.



4

Sol. 34 2 2 2V

27(2 2 2)− × × −

= =+ ×

m/s


Sol. 2 1M2π

λ = =π

; 2λ = 50 cm

27. C (Concept Code: P111307)

Sol. P

Q

SinV 14V 2Sin

2

π⎛ ⎞×⎜ ⎟⎝ ⎠= =π⎛ ⎞×⎜ ⎟

⎝ ⎠


Sol. (g +a) b

1 g⎛ ⎞ρ− =⎜ ⎟ρ⎝ ⎠

b

ag a

ρ=

ρ +

29. A (Concept Code: P111103) Sol. Phase diff. = (π – 2α)

Cos α =

a13

a 3=

30. A (Concept Code: P110723) Sol. Force produces zero torque about the centre which is the point of rotation because

r Fτ = × , It τ = 0 the L is constant.

SECTION – 2 (Chemistry) 1. C Sol. Fe = [Ar]184s23d6 Fe+ = [Ar]184s13d6 (Contains 5 unpaired electrons) Fe3+ = [Ar]184s03d5 (Contains 5 unpaired electrons) 1. C110108 2. C Sol. Li and Mg display diagonal relationship. 2. C110705 3. C Sol. The central atoms of SF4, CF4, XeF4 and SeF4 undergo sp3d, sp3, sp3d2 and sp3d

hybridization respectively. 3. C110306 4. B Sol. Meq of metal ion = Meq of 4MnO− solution


5

W 1000 N VE

26or, 1000 400 1 5 200052 / n

∴ × = ×

× = × × =

or, n = 4 ∴ Change in oxidation number of 4 units. 4. C111102 5. C

Sol. R.M.S speed of CH4 = 3RT 3R 800 150RM 16

×= =

Most probable speed of He = 2RT 2 R 300 150RM 4

× ×= =

5. C111206 6. C Sol. Plaster of paris(CaSO4).H2O sets on standing to a hardmass. 6. C110908 7. D Sol. H2O2 easily decomposes to H2O and O2 even in presence of dust particles. 7. C111006 8. C Sol. ( ) ( )A g 2B gΔ⎯⎯→ a 0 a – x 2x For 1:1 molar ratio, a – x = 2x ⇒ x = a/3

2.303 a 2.303 3t log logk a a / 3 k 2

= =−

= 1635.3 sec = 27.25 min

8. C110606 9. D Sol. ( ) ( ) ( )X s 2Y g 3Z g+ KC = [Y]2[Z]3 [2Z]3 [Y’]2 = [Y]2[Z]3

[ ] [ ]22 YY '

8∴ =

or, [Y’] = [ ]Y2 2

9. C110401 10. B Sol. At second equivalence point the only species present at appreciable concentration is

24HPO − .

aa 32 KKp p 12 8pH 10

2 2+ +

= = =

10. C110502 11. C


6

Sol. It is a neutralization reaction. 11. C111906 12. C Sol. X is HBO2 which is obtained by removing one water molecule from H3BO3. 12. C111505 13. D Sol. The compound in (D) after hydrolysis does not polymerise beyond two molecules. 13. C111607 14. B Sol. II and III are stabilized by resonance and II is aromatic. IV (1o) is more stable than 1(2o). 14. C111303 15. C

Sol. 23 2 3 3 2 2 3 3

ClCH CH CHCH CH CH CHCH Cl CH CHCHCH⎯⎯⎯→Δ

+

CH3 CH3 CH3

Cl

15. C111404 16. B Sol. Orientation of product takes place according to the activating group(OH). 16. C111804 17. C Sol. The initially formed 3o carbocation undergoes rearrangement to another 3o carbocation

for more no. of hyperconjugative structures. Then ring expansion followed by methyl shift takes place.

17. C111705 18. D Sol. It is most soluble in NH4OH due to complex formation, followed by KNO3. AgNO3 and

NaCl exert common ion effect. Therefore, it is more soluble in low conc. NaCl than that in higher conc. AgNO3.

18. C110501 19. C

Sol. Rate = [ ]2d I13 dt

[ ]2d Idt

∴ = 3 × Rate = 3 × 1.2 × 10–3 = 3.6 × 10–3 mol L–1s–1.

19. C110601 20. C

Sol. 1/2

0.693kt

=

1C

2

k 0.693 / 10 1K 0.1k 0.693 / 1 10

= = = =

20. C110606 21. C


7

Sol. It has maximum number of hyperconjugative structure. 21. C111706 22. D Sol. In(III), negative charge is present on oxygen and in (I) it is present on carbon. So (III) is

more stable than (I). (II) is most unstable due to more number charged species. 22. C111301 23. B Sol. 2 3Na CO No gas is producedΔ⎯⎯→ 2 3 2Li CO CO isevolvedΔ⎯⎯→ ↑ 23. C110804 24. A Sol. Due to pπ - pπ back bond formation in BF3. 24. C110306 25. D Sol. Due to weak C – I bond in comparison to other C – X bonds. 25. C111606 26. C Sol. At equilibrium, ΔSsys. + ΔSsurr. = 0 26. C111906 27. C Sol. Due to presence of bulky methyl group, 2NO− does not exhibit –R effect in (C). 27. C111301 28. D Sol. X = CH3CH2CH2CH3, Y = Trans-2-butene 28. C111713 29. C

Sol. It gains aromaticity by forming the cation , , and Cl–.

29. C111801 30. D Sol. (D) gives four geometrical isomers as it is not symmetrical like the other compounds. 30. C111403

SECTION – 3 (Mathematics) 1. D M111301 Sol. The given series is geometric with an initial term of 1 and a common ratio of 2cos ,θ so

its sum is

2n2 2

n 0

1 15 cos1 cos sin

∞

=

= θ = =− θ θ∑

Therefore 2 1sin ,5

θ = and


8

2 2 3cos2 1 2sin 15 5

θ = − θ = − =

2. D M110909 Sol. Let ( )B a, b= and ( )A a, b= − − . Then 24a 7a 1 b+ − = and 24a 7a 1 b− − = −

So 2 2 50AB 2 a b 2 5 24

= + = =

3. D M110204

Sol. We have 10sin1 0 log 1> = and 103 3sin 1 0log2 2π π= − < . Since 54, 10

2π

π < < and

105 5log 1 sin2 2π π< = Also, 10

7 7log 0 1 sin2 2π π> > − =

Since both functions are continuous we have at least one real roots in each of the

intervals 3 3 51, , ,2 2 2π π π⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and 5 7,2 2π π⎛ ⎞

⎜ ⎟⎝ ⎠

4. A M110302 Sol. ( ) ( )1 2 1 2 2 3 3 3 1 2 1 1 2 39z z 4z z z z z z z z z z z z 12+ + = + =

Or, ( )1 2 3 1 2 3 1 2 3 1 2 3z z z z z z 12 or, z z z z z z 12+ + = + + =

Or, 1 2 3 1 2 312z z z z z z 2

3 2+ + = + + = =

×

5. B M111004

Sol. Here x4

= cos t + sin t … (1)

and y7

= cos t − sin t … (2)

Squaring and adding equation (1) and (2)

2 2x y

16 7+ = 2 (sin2 t + cos2 t)

⇒ 2 2x y 1

32 14+ = , which is an ellipse

eccentricity e = 2

2

b 14 31 132 4a

− = − = .

6. C M111205 Sol. Total number of ways 6 6 .......= × × to n times n6= . Total number of ways to show only even numbers 3 3 .........n= × × times n3 ∴ The required number of ways n n6 3= − . 7. A M110903 Sol. Any point on the parabola is (x , x2+ 7x + 2) Its distance from the line y = 3x –3 is given by

P = ( )2 23x x 7x 2 3 x 4x 5

9 1 10

− + + − + +=

+=

2x 4x 510

+ +


9

(as x2 +4x + 5 > 0 for all x ∈ R)

dPdx

= 0 ⇒ x = -2

The required point ≡ (-2, -8). 8. D M110410 Sol. ( 1 – x4) (1 + 9C1 x+ 9C2 x2 + 9C3 x3 …………9C9x9)

Coefficient of x7 = 9C7 – 9C3 = – 48 9. B M111203 Sol. Total number of ways of posting the letters = 36 − 3C1 26 + 3C2 = 540. 10. D M110304 Sol. | iz + ( 3 – 4i)| ≤ | iz| + |3 – 4i| = |z| +5 < 4 + 5 = 9. 11. C M110410

Sol. In the expression 10

2/3 1/3 1/2

x 1 x 1x x 1 x x

+ −⎛ ⎞−⎜ ⎟− + −⎝ ⎠

To simplify, for first term put x = p3 and for second term put x = q2, then it will become (x1/3 – x-1/2)10 Tr+1 = 10Cr(x1/3)10-r ( x –1/2)r For term independent of x;

⇒ 10 r r

3 2x x− −⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= x0 ⇒ 10 r r 03 2−

− =

⇒ 5r = 20 ⇒ r = 4 ⇒ T5 = 10C4 12. C M110101 Sol. Given ,α β are roots of equation 2x 2x 3 0− + = 2 2 3 0⇒α − α + = ………….(1) and 2 2 3 0β − β + = ………….(2) 2 3 22 3, 2 3⇒α = α − ⇒ α = α − α ( )2 2P 2 3 3 5 2= α − α − α + α − = 2 2 2 3 2 1,−α + α − = − = by Eq. (1)

Similarly, we can have Q 2= Sum of roots 3= and product of roots = 2

So, the required equation is 2x 3x 2 0− + = 13. D M110505 Sol. Let d be the common difference of the A.P., then 3 = a10 = 2 + 9d ⇒ d = 1/9 ∴ a4 = 2+3d = 7/3

Next, let D be the common difference of the A.P. 1 2 10

1 1 1, ,...h h h

Then

10

1 1 1 19D D3 h 2 54= = + ⇒ = −

77 1

1 1 7 186D hh h 18 7

= + = ⇒ =

Hence, a4 . h7 = (7/3) (18/7) = 6.


10

14. C M111301

Sol. 2 2 2

2 21 2 2 2 2 2

4a a cos 2p psec cosec cos sin

α+ = +

α + α α + α

= a2 (sin2 2α + cos2 2α) = a2 And ( )2 2 4 2 2 4 2

1 2p p a sin 2 cos 2 1/ 4 a sin 4= α α = α

( )2 2 2

1 2 21 22 2 2

2 1 1 2

p pp p 4 4cosec 4p p p p sin 4

+⎛ ⎞∴ + = = = α⎜ ⎟ α⎝ ⎠

15. B M111403 Sol. The given expression is equal to 2 o 2 o o ocos 10 cos 50 cos50 cos10+ −

( )2o o o ocos10 cos50 cos10 cos50= − +

( ) ( )2o o o o12 sin20 sin30 cos60 cos402

= + +

( )o o1 1 11 cos40 cos402 2 2

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

34

=

16. B M110305 Sol. z1 + z2 = − ( )1 2z z+

z1 + z2 = 1 2z z− − 1 1 2 2z z z z⇒ + = − −

i1 1

2 2

z z 1 ez z

π+= − =

+

17. B M110413

Sol. n

r rn

r 1 r 1

a C n r 11 1 1a rC− −

− ++ = + = +

n 1r+

=

18. B M111301 Sol. The expression is equal to ( ) ( )4 4 6 63 cos sin 2 cos sinα + α − α + α

= ( ){ } ( ) ( )2 32 2 2 2 2 2 2 2 2 23 cos sin 2cos sin 2 cos sin 3sin cos sin cos⎡ ⎤α + α − α α − α + α − α α α + α⎢ ⎥⎣ ⎦

= ( ) ( )2 2 2 23 1 2sin cos 2 1 3sin cos 1− α α − − α α = 19. A M110733 Sol. x2 − y2 + 2y − 1 = 0 ⇒ Equations of lines are y = x + 1, y = − x + 1 ⇒ angle bisectors are y = 1 and x = 0

⇒ area of triangle = 1 2 22× × = 2 sq. units


11

20. B Not Available

Sol: σ2 = 2 2n

2

r 1

1 n 1 1 n(n 1)(2n 1) (n 1)rn 2 n 6 4=

+ + + +⎛ ⎞− = ⋅ −⎜ ⎟⎝ ⎠

∑

= (n + 1) 22n 1 n 1 (n 1)(4n 2 3n 3) n 1

6 4 12 12+ + + + − − −⎛ ⎞− = =⎜ ⎟

⎝ ⎠

Hence, B is correct answer 21. D M111301 Sol. According to given condition

Sinα + sinβ = – a and cosα + cosβ = –c

Or , 2 sin cos a and 2 cos cos c2 2 2 2

α + β α −β α + β α −β= − = −

Or , tan a2 c

α + β⎛ ⎞ =⎜ ⎟⎝ ⎠

Now

sin( α+ β) = 2 22

2 tan2ac2

a c1 tan2

α + β⎛ ⎞⎜ ⎟⎝ ⎠ =α + β +⎛ ⎞+ ⎜ ⎟

⎝ ⎠

22. A M111502

Sol. Note that cos A sinB cosBsinA sinCcot A cotBsinA sinB sinA sinB

++ = =

Hence cot C sin A sinB cosCcot A cotB sinC sinC

⎛ ⎞= ⎜ ⎟+ ⎝ ⎠

By the Laws of the Sines and Cosines,

( )2 2 22 2 2

2

9 a b 9ccot C a b a b c 5cot A cotB c c 2ab 918c

+ −+ −⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

23. D M110808 Sol. Let the line y x= − intersect the

circle and the line y 6= at M and K, respectively, and let the line y x= intersect the circle and the line y 6= at N and L, respectively. Quadrilateral PMON has four right angles and MP = PN, so PMON is a square. In addition, MK = KJ=6 and KO = 6 2 . Hence

r MO MK KO 6 6 2= = + = +

Py

N

L K

O

M

J

24. C M110501 Sol. We have b a r, c a 2r,= + = + and d a 3r,= + where r is positive real number. Also

2b ad= yields ( ) ( )2a r a a 3r+ = + , or 2r ar= . If follows that r a= and d a 3a 4a= + = .

Hence a 1d 4=


12

25. B M110701

Sol. Line AC has slope 12

− and y – intercept

(0, 9), so its equation is 1y x 92

= − +

Since the coordinates of A’ satisfy both this equation and y x,= if follows that

( )A ' 6,6= . Similarly, line BC has equation

y 2x 12,= − + and ( )B' 4, 4= . Thus

( ) ( )2 2A 'B ' 6 4 6 4 2 2= − + − =

y

y=x

C

B'x

B (0,12)

A (0,9) A'

(2 , 8)

26. A M110504 Sol. 18logx k, 21logy k, 28logz k= = = 18 21 28x y z= = y z x3,3 log x, 3log y, 7log z are in 18 log x = k 21 log y = k 28 log z = k

k kk18 28213, 3 , 3 , 7k k k21 28 18

21 28 183, , ,6 7 4

7 13, , 4, 42 2

are in A.P

27. A M110826 Sol. The given circle is (x − 1)2 + (y − 1)2 = 1.

In the figure MC = 1 × 32

⇒ locus of the point M is

(x − 1)2 + (y − 1)2 = 34

.

B

M1

1

A

C(1,1)

28. B M110503

Sol. 1 2 3aa , a a, a arr

= = =

If ( )1 2 3 1a , a , a a 0> are in G.P. ⇒ 1 3 29a 5a 14a+ ≤

⇒ a9 5ar 14ar+ ≤


13

⇒ 9 5r 14r+ ≤

⇒ 25r 14r 9 0− + ≤ ⇒ 25r 5r 9r 9 0− − + ≤

∴ The internal is 91,5

⎡ ⎤⎢ ⎥⎣ ⎦

29. B M110109 Sol. 2 2x 18x 30 2 x 18x 30 15+ + = + + + Let 2x 18x 30 y+ + = y 2 y 15⇒ = + y 6⇒ = − (rejected),y=10 Or 2x 18x 20 0+ + = ⇒ product of real roots=20 30. D M110414 Sol. 4n2 15n 1− − ( )n1 15 15n 1+ − − n 2 n 3 n n

2 3 nC 15 C 15 ............... C 15= + + + = divisible by 225.


14

QP CODE: 120627.2 ANSWERS

Reshuffling Test – 1


1 C P110507 1 C C110908 1 C M111205 2 D P110604 2 D C111006 2 A M110903 3 B P110605 3 C C110606 3 D M110410 4 B P110413 4 D C110401 4 B M111203 5 C P110708 5 B C110502 5 D M110304 6 D P110718 6 C C111906 6 C M110410 7 C P110804 7 C C111505 7 C M110101 8 A P110905 8 D C111607 8 D M110505 9 C P110905 9 B C111303 9 C M111301 10 A P110905 10 C C111404 10 B M111403 11 D P111101 11 B C111804 11 B M110305 12 A P111101 12 C C111705 12 B M110413 13 C P111026 13 D C110501 13 B M111301 14 A P111213 14 C C110601 14 A M110733 15 B P111207 15 C C110606 15 B Not Available 16 D P111213 16 C C111706 16 D M111301 17 C P111005 17 D C111301 17 A M111502 18 A P110613 18 B C110804 18 D M110808 19 B P111005 19 A C110306 19 C M110501 20 A P111303 20 D C111606 20 B M110701 21 B P111306 21 C C111906 21 A M110504 22 C P111307 22 C C111301 22 A M110826 23 B P111103 23 D C111713 23 B M110503 24 A P111103 24 C C111801 24 B M110109 25 A P110723 25 D C111403 25 D M110414 26 A P110305 26 C C110108 26 D M111301 27 B P110605 27 C C110705 27 D M110909 28 D P110308 28 C C110306 28 D M110204 29 D P110415 29 B C111102 29 A M110302 30 C P110404 30 C C111206 30 B M111004


15




1 D P110718 1 C C111906 1 C M110410 2 C P110804 2 C C111505 2 C M110101 3 A P110905 3 D C111607 3 D M110505 4 C P110905 4 B C111303 4 C M111301 5 A P110905 5 C C111404 5 B M111403 6 D P111101 6 B C111804 6 B M110305 7 A P111101 7 C C111705 7 B M110413 8 C P111026 8 D C110501 8 B M111301 9 A P111213 9 C C110601 9 A M110733 10 B P111207 10 C C110606 10 B Not Available 11 D P111213 11 C C111706 11 D M111301 12 C P111005 12 D C111301 12 A M111502 13 A P110613 13 B C110804 13 D M110808 14 B P111005 14 A C110306 14 C M110501 15 A P111303 15 D C111606 15 B M110701 16 B P111306 16 C C111906 16 A M110504 17 C P111307 17 C C111301 17 A M110826 18 B P111103 18 D C111713 18 B M110503 19 A P111103 19 C C111801 19 B M110109 20 A P110723 20 D C111403 20 D M110414 21 A P110305 21 C C110108 21 D M111301 22 B P110605 22 C C110705 22 D M110909 23 D P110308 23 C C110306 23 D M110204 24 D P110415 24 B C111102 24 A M110302 25 C P110404 25 C C111206 25 B M111004 26 C P110507 26 C C110908 26 C M111205 27 D P110604 27 D C111006 27 A M110903 28 B P110605 28 C C110606 28 D M110410 29 B P110413 29 D C110401 29 B M111203 30 C P110708 30 B C110502 30 D M110304


16




1 D P111101 1 B C111804 1 B M110305 2 A P111101 2 C C111705 2 B M110413 3 C P111026 3 D C110501 3 B M111301 4 A P111213 4 C C110601 4 A M110733 5 B P111207 5 C C110606 5 B Not Available 6 D P111213 6 C C111706 6 D M111301 7 C P111005 7 D C111301 7 A M111502 8 A P110613 8 B C110804 8 D M110808 9 B P111005 9 A C110306 9 C M110501 10 A P111303 10 D C111606 10 B M110701 11 B P111306 11 C C111906 11 A M110504 12 C P111307 12 C C111301 12 A M110826 13 B P111103 13 D C111713 13 B M110503 14 A P111103 14 C C111801 14 B M110109 15 A P110723 15 D C111403 15 D M110414 16 A P110305 16 C C110108 16 D M111301 17 B P110605 17 C C110705 17 D M110909 18 D P110308 18 C C110306 18 D M110204 19 D P110415 19 B C111102 19 A M110302 20 C P110404 20 C C111206 20 B M111004 21 C P110507 21 C C110908 21 C M111205 22 D P110604 22 D C111006 22 A M110903 23 B P110605 23 C C110606 23 D M110410 24 B P110413 24 D C110401 24 B M111203 25 C P110708 25 B C110502 25 D M110304 26 D P110718 26 C C111906 26 C M110410 27 C P110804 27 C C111505 27 C M110101 28 A P110905 28 D C111607 28 D M110505 29 C P110905 29 B C111303 29 C M111301 30 A P110905 30 C C111404 30 B M111403


17

Paper-2 PHYSICS, CHEMISTRY & MATHEMATICS

CODE: 120628.1 ANSWERS


ANSWER KEY Physics Chemistry Mathematics Part – A Part – A Part – A

1 C P110309 1 B C110101 1 C M111502 2 D P110310 2 D C110307 2 B M110406 3 C P110806 3 C C110504 3 B M111210 4 D P111012 4 D C111804 4 A M111201 5 D P110610 5 C C111507 5 A M110101

6 B P111302, P111305 6 D C111401 6 A M110504

7 A P111026, P111209 7 B C110402 7 B M110902

8 B P110815 8 C C111302 8 B M111214 9 B P110815 9 D C110603 9 D M110101 10 B P110815 10 C C110603 10 A M10101 11 C P110815 11 B C110603 11 B M110107 12 A P111105 12 B C111902 12 C M110902 13 A P111105 13 A C111903 13 D M110910

14 B P111105, P111222 14 B C111905 14 B M110902

15 ABC P111207 15 ABCD C110908 15 ABCD M1105010 16 AC P111105 16 ABCD C111804 16 ABC M110415 17 AB P111010 17 AB C111509 17 AC M110406 18 CD P111002 18 ABCD C111204 18 BD M111103 19 ACD P110912 19 BC C111102 19 AD M110719 20 AD P110917 20 BCD C110503 20 BC M110810

Part – C Part – C Part – C 1 3 P110606 1 8 C110402 1 2 M111422 2 1 P110705 2 9 C111301 2 2 M110304 3 4 P110807 3 5 C110809 3 1 M111426 4 8 P111309 4 6 C110107 4 3 M110510


18

Hints & Solutions Physics PART – A 1. C (Concept code: P110309)

Sol. Vavg = DisplacementTime taken

2. D (Concept code: P110310)

Sol. 2

2 21 xy x tan g2 u cos

= θ −θ

3. C (Concept code: P110806) Sol. Net force is equal to mass multiplied by acceleration and net torque about cm should be

zero. 4. D (Concept code: P111012) Sol. F = (mg + kv) v decreases as ball goes up Constant downward force + decreasing downward force. 5 D (Concept code: P110610) Sol. VB = VBW + VW w.r.t wedge

4

4

4 2 = approach velocity = – rebound velocity

6. B (Concept code: P111302, P111305) Sol. y0 2πf = 4fλ 7. A (Concept code: P111026, P111209) Sol. Wgas + Watm + WS = 0 8. B (Concept code: P110815) Sol. V dT (γHg – γglass) = πr2 dh 9. B (Concept code: P110815)

Sol. COTME ⇒ Kf = mg2

and 2

21 m mg2 3 2

ω =

⇒ 3gw =

(i)

(f)IAR

10. B (Concept code: P110815)

Sol. COTME ⇒ Kf = mg2

11. C (Concept code: P110815)

Sol. Vcm = w2


19

12. A (Concept code: P111105)

Sol. Time taken to move from mean to extreme (t) = 2A 2Am(F / m) F

=

∴ Time period = 2Am4t 4F

=

13. A (Concept code: P111105) Sol. Acceleration = constant ⇒ velocity time curve should be straight line. 14. B (Concept code: P111105, P111222)

Sol. T = Time taken to reach from (–A) to (0) – Time taken to reach from A( A) to2−⎛ ⎞− ⎜ ⎟

⎝ ⎠

= 2A 2(A / 2)F / m F / m

−

15. ABC (Concept code: P111207) Sol. e = a

And ρ = mv

16. AC (Concept code: P111105) Sol. The force experienced by pendulum will be same, while they will move in opposite

direction. (phase diff. = π rad) 17. AB (Concept code: P111010) Sol. The velocities at upper end and lower end of the lower pipe are not equal, so it will not

be fully filled. 18. CD (Concept code: P111002) Sol. Free body diagram.

Po πr2

4Po πr2

N

19. ACD (Concept code: P110912) Sol. For the angular momentum to remain conserved, the final velocity of the particle must

not be zero. 20. AD (Concept code: P110917)

Sol. 2 23

GMV [3R r ],2R−

= − inside the earth.

PART – C 1. 3 (Concept code: P110606) Sol. 1

2

1

2

2 V2

2v 2I2 2−⎛ ⎞= −⎜ ⎟− −⎝ ⎠

⇒ v2 = 6


20

2. 1 (Concept code: P110705)

Sol. I1 = Icm + 2Rm

2⎛ ⎞⎜ ⎟⎝ ⎠

and I2 = 2

cmRI m2

⎛ ⎞+ ⎜ ⎟⎝ ⎠

cm R/2

R/2


Sol. cm2

F FRa and2m mR3

= α =


Sol. 1 24 23 4

=

Chemistry

PART – A 1. B Sol. Kinetic energy = e × stopping potential

∴ Stopping potential = 19

19

K.E 6.4 10e 1.6 10

−

−

×=

×= 4 V

1. C110101 2. D Sol. The bond order of 2

2O + is highest among the given species. ∴ it contains the strongest covalent bond 2. C110307 3. C Sol. aKp = - log(2 × 10–4) = 3.7

[ ][ ]

aK Salt xpH p log , or 3.4 3.7 logAcid 0.1

= + = + , on solving x = 0.05.

3. C110504 4. D Sol. The activating order of the group attached to benzene ring is:

( ) ( )3 2 2 3 2 2

HyperconjugationReffectHigh charge densityLow charge densitycarbonyl carboncarbonyl carbon

NHCOCH CH CONH CONHCH COCH NH↓+ ↓

> > >

4. C111804 5. C Sol. X = B3N3H3Cl3 5. C111507 6. D Sol. In both compounds the CH3 groups are at different positions. 6. C111401 7. B


21

Sol. ΔG0 = - RT nK

or, ΔH0 – TΔS0 = - RT nK

or, 0 0H 1 SnK

R T R−Δ Δ

= × +

y = mx + c

nK

1/T

A

0

Intercept = 0S

RΔ = 4 ⇒ ΔS0 = 4 × 2 = 8 Cal K–1 mol–1.

7. C110402 8. C Sol. Acidic strength: Penolic OH > 1o – OH > 2o – OH >3o – OH 8. C111302 9. D Sol. ΔH = HP – HR = 20 – 40 = - 20 kJ 9. C110603 10. C Sol. Energy of product + Ea(b) of uncatalysed reaction = 140 ∴ Ea(b) = 140 - 20 = 120 kJ 10. C110603 11. B Sol. Let activation energy of the forward reaction for the catalysed reaction be '

aE . ∴ '

aE = Ea – x Where x = amount of energy lowered down by the catalysed reaction. k(catalysed reaction) = k(uncatalysed reaction)

'a aE /R 500 E /R 300

'a a a

Ae AeE E E xor,500 300 300

− × − ×∴ =

−= =

∴ Since Ea = 100 kJ, x will be 40 kJ 11. C110603 12. B Sol. A B B C C AW W W W→ → →= + +

= 0 + (-2.303 nRT C

B

Vlog

V+ [-P(VA - VC)]

= 2.303 nRT B

C

VlogV

+ P(VC – VA)

= 2.303 PBVBB

C

VlogV

+ P(2V – V)


22

= 2.303 × 3PV Vlog2 V

+ PV

= 2.303 × 3 PV 1log2

+ PV

= 2.303 × 3 PV × -0.3010 + PV = - 2PV + PV = -PV 12. C111902 13. A Sol. C → A is isobaric path

∴ q = nCPΔT = [ ] C CA AA C

P VP V5 51 R T T R2 2 R R

⎡ ⎤× − = −⎢ ⎥⎣ ⎦

5 PV 2PV 5R PV2 R 2

−⎡ ⎤= = −⎢ ⎥⎣ ⎦

A → B is isochoric path

q = nCVΔT = [ ] B B A AB A

P V P V3 31 R T T R2 2 R R

⎡ ⎤× − = −⎢ ⎥⎣ ⎦

3 3PV PV 3R 2PV 3PV2 R 2

−⎡ ⎤= = × =⎢ ⎥⎣ ⎦

13. C111903 14. B Sol. Moles of gas at A = Moles of gas at B

A A B B

1 2

2 11 2

P V P VRT RT

PV 3PVor, T 3TT T

=

= ⇒ =

When moles at A and C are compared, T2 will be equal to 2 T1. 14. C111905 15. ABCD Sol. ( ) ( )Cold

2 2 22 22Ca OH 2Cl CaCl Ca OCl 2H O+ ⎯⎯⎯→ + +

( ) ( )

( ) o

Hot2 2 3 22 2

30 C2 2 22

6Ca OH 6Cl 5CaCl Ca ClO 6H O

Ca OH Cl CaOCl H O

+ ⎯⎯⎯→ + +

+ ⎯⎯⎯→ +

15. C110908 16. ABCD

Sol. 33 3 3

AlClCH C C Cl CH C C CH C CO⎯⎯⎯⎯→− − − − − ⊕ − ⊕ +

CH3 CH3 CH3

CH3 CH3 CH3O O


23

Both the above electrophiles form products and polysubsitution of

C(CH3)3

takes place

at the

para position. As (CH3)3C – CO+ is a deactivating group, polysubstitution of

O = C - C(CH3)3

does not take place.

16. C111804 17. AB Sol. Graphite is black whereas BN is white. Graphite is a poor conductor but BN does not

conduct at all. 17. C111509 18. ABCD

Sol. For isochoric curve, RP T y mxV b

= × ≡ =−

For isobaric curve, PV – Pb = RT or, PV = RT + Pb

Ror, V T b y mx CP

= + ≡ = +

Gas shows positive deviation. So attractive forces overcome by repulsive forces. 18. C111204 19. BC Sol. H3PO3 is a dibasic acid and H3PO2 is a monobasic acid. 19. C111102 20. BCD Sol. These ions undergo hydrolysis in water

4 2 4

2

2

NH H O NH OH H

CN H O HCN OH

Fe H O FeOH H

+ +

− −

+ + +

+ +

+ +

+ +

20. C110503 Part – C 1. 8 Sol. ( ) ( ) ( )2C s CO g 2CO g+ P 0 P – x 2x P – x + 2x = P + x = 6

PP 6 P 4and x 22

⇒ + = ⇒ = =


24

( ) ( )

2

2 2CO

PCO

p 4 16K 8atmp 2 2

∴ = = = =

1. C110402 2. 9 Sol. The carbocation has 6 and the double bond has 3 hyperconjugation structures. 2. C111301 3. 5 Sol. X = NaHCO3, Y = HCOONa 3. C110809 4. 6 Sol. n = 4, = 0, 1 → 4s and 4p subshells m = 0, +1 → 4s, 4px and 4py s = ± ½ → 2 2 2

x y4s , 4p and 4p 6 electrons= 4. C110107 Mathematics PART – A 1. C M111502

Sol. Expression=2 2 2 2 2 2 2 2 2 2 2 2b c a c a b a b c a b c

2abc 2abc 2abc 2abc+ − + − + − + +

+ + =

( ) ( )2 2a b c 2 ab bc ca 11 2.38 92abc 2.40 16

+ + − + + −= = =

2. B M110406

Sol. Center is 0 0z z i2+

and Diameter is 0z

3. B M111210 Sol. The first and last digits must be both odd or both even for their average to be an integer.

There are 5 . 5 = 25 odd – odd combinations for the first and last digits. There are 4 .5 =20 even – even combinations that do not use zero as the first digit. Hence the total is 45.

4. A M111201

Sol. The slope of the line is ab

− , We may assume that a 0> and b 0< so that a b≠ . We

must also have a c b≠ ≠ . If c 0= , there are 3 ways to choose a and 3 ways to choose b.

However, the lines x y 0, 2x 2y 0− = − = and 3x 3y 0− = are the same. Thus the number of lines in this case is 23 2 7− = . If c 0,≠ there are 3 ways to choose a, 3 ways to choose b and 4 ways to choose c. Thus the number of lines in this case is 23 4 36= . The total is 7 36 43+ = .

5. A M110101


25

Sol. Multiplying the given equation by 3

ca

, we get

2 2 2

23 2

b c b cx x c 0a a

− + =

⇒ 2

2 2

bc bca x b x c 0a a

⎛ ⎞ ⎛ ⎞− − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ – 2

bc x ,a

= α β

⇒ (α + β)αβ x = α, β

⇒ x = 1 1,( ) ( )α + β α α + β β

.

6. A M110504 Sol. Let the common ratio be r. Then

4

2

a log 3ra log 3+

=+

8

4

a log 3a log 3+

=+

4 8

2 4

log 3 log 3log 3 log 3

−=

−

2 2

2 2

1 1log 3 log 32 3

1log 3 log 32

−=

−

13

=

7. B M110902

Sol. The roots of ( ) ( )2 2n n x 2n 1 x 1 0+ − + + = are 1n

and 1

n 1+. Hence

1992 1992

n nn 1 n 1

1 1 1 1992A B 1n n 1 1993 1993= =

⎛ ⎞= − = − =⎜ ⎟+⎝ ⎠∑ ∑

8. B M111214 Sol. We have ( )( )( ) ( )1!.2!.3!......9! 1 1.2 1.2.3 1.2...9= =

9 8 7 6 5 4 3 2 1 30 13 5 31 2 3 4 5 6 7 8 9 2 3 5 7= = The perfect square divisors of that product are the numbers of the form 2a 2b 2c 2d2 3 5 7

with 0 a 15, 0 b 6, 0 c 2≤ ≤ ≤ ≤ ≤ ≤ , and 0 d 1≤ ≤ . Thus there are

( )( )( )( )16 7 3 2 672= such number 9. D M110101 Sol. 0α + β + γ = q rαβ + βγ + γδ = αβγ = − 3 q r 0α + α + = 3 q rα = −α −


26

5 3 2q rα = −α − α ( )5 2q q r rα = − −α − − α

2 2q q r= + α + γ − γ

( ) ( )5 2 2 2 2q 3qr rα = α + β + γ + − α + β + γ∑

5qr= 10. A M10101

Sol. 31

q r r qα α

= = −− α − +α − αβγ

11. B M110107 Sol. 3 2 33 0, 3 c, dα + β = α + α β = − α β =

1 3 c 8dd 3c

+ = − ⇒ α = −β α

12. C M110902 13. D M110910 14. B M110902 SOLUTION_12 – 14

1 2m 1 2mh ,k2l 2m− +

= =

( )

1 k 2m l2k 2 2h k 1

−= =

− −

(l,m) lies on 2y 4ax= 2m 4al=

( )

21 k 24a2k 2 2h k 1

⎛ ⎞−⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

( ) ( )2

2 3 1x 8a y 3y 2 y x 2a2 8a

⎛ ⎞= − + ⇒ − = +⎜ ⎟⎝ ⎠

vertex is 32a,2

⎛ ⎞−⎜ ⎟⎝ ⎠

length of latus rectum = 18a

curve in symmetric about line 3y2

=

15. ABCD M1105010 Sol. b is the H.M. of a and c and their G.M. = ac G.M > H.M ⇒ ac > b And A.M. of an and cn > G.M. of an and cn

nnnn

ca2

ca>

+

( ) nnnn b2ac2ca >>+ Put n = 100, 3, 5, 2


27

16. ABC M110415 Sol. Putting x = ω in the equation, 0 = a0 + a1 ω +a2ω2 +a3 + . . . .... (1) Putting x = ω2 in the equation, 0 = a0 + a1ω2 +a2ω + a3 + . . . .... (2) Putting x = 1 in the equation, 3n = a0 + a1 +a2 + a3 + . . . .... (3) adding (1), (2) and (3), 3n = 3(a0 +a3 +a6 + . . . .) .... (a) ⇒ a0 + a3 +a6 + . . . . = 3n-1 (option C) subtracting (2) from (1), 0 = (ω - ω2) (a1 –a2 + a4 –a5 + . . . ) Since ω - ω2 ≠ 0 , a1 +a4 + a7 +. . . = a2 + a5 +a8 + . . . . . . (4) Also from (3) – (a), a1 + a2 +a4 +a5 + . . . . = 3n – 3n-1 = 2.3n-1 . . . .(5) From (4) and (5) , a1 +a4 + a7 +. . . = a2 + a5 + a8 + . . . = 3n-1 = a0 + a3 + a6 + . .

17. AC M110406 Sol. Clearly the inscried triangle is equilateral.

⇒ 2i2 0 3

1 0

z z e ,z z

π−=

−

2i3 0 3

1 0

z z ez z

π−−

=−

⇒ z2 = -1 +i(2 + 3 ) and z3 = -1 + i(2 - 3 ) z1(2 +2i)

z2

z3

z0(2i)

18. BD M111103

Sol. Equation of the ellipse is 2 2

2 2

x y 1a b

+ =

which is of the from 2 2

2 2

x y 1a b

+ = with 2 2 2 2a 2 , b 1= =

Here ( )2 2 21 2 1 e= −

⇒ 3e2

= ( )( )2 2 2b a 1 e= −∵

Foci of the ellipse are ( )3, 0 and ( )3, 0−

∴ Eccentricity of the hyperbola 23

=

2 2 2 2 2 24 1b a 1 3b a b a3 3

⎛ ⎞⇒ = − ⇒ = ⇒ =⎜ ⎟⎝ ⎠

∴ Equation of hyperbola is 2 2

2 2

x 3y 1a a

− =

Also the hyperbola passes through a focus of the ellipse, i.e. ( )3, 0

∴ 22

3 1 a 3a

= ⇒ = .

Hence, the equation of hyperbola is 2 2x 3y 3− =

A focus of hyperbola is ( ) ( )2ae, 0 3 , 0 2, 03

⎛ ⎞= × =⎜ ⎟⎝ ⎠

19. AD M110719


28

Sol. P lies on the line 2x + 3y + 1 = 0 ........(1) |PA – PB| is maximum if P, A, B are collinear. Equation of the line AB is x + y = 2 ........(2) P is the point of intersection of (1) and (2) which is (7, – 5) Equation of perpendicular bisector of AB is y = x .......(3) For minimum value of |PA – PB| ,

P is the point of intersection of (1) and (3) which is 1 1,5 5

⎛ ⎞− −⎜ ⎟⎝ ⎠

20. BC M110810 Sol. Let L1 : y = mx Putting y = mx in the equation of circle , we get x2 + m2 x2 – x + 3mx = 0 ⇒ x[ (1+ m2)x –1 + 3m] = 0

⇒ x = 0, 21 3m1 m−+

So points of intersection of L1 and the circle are

P1(0, 0), P2( )

2 2

m 1 3m1 3m ,1 m 1 m⎡ ⎤−−⎢ ⎥+ +⎣ ⎦

Also points of intersection of L2 and the circle are Q1(1, 0) and Q2(2, –1). By the given condition; P1P2 = Q1Q2

2 ⇒ (1 + m2) (3m –1)2 = 2(1 + m2)2 ⇒ (3m –1)2 = 2(1+ m2) ⇒ 7m2 – 6m – 1 = 0

⇒ (7m + 1) (m –1) = 0 ⇒ m = 1, 17

−

So L1 : y = x, y = 1 x7

−

i.e. x – y = 0, 7y + x = 0. Part – C 1. 2 M111422 Sol. 4 [1 + cot2 π(a + x)] + a2 – 4a = 0 4cot2π (a + x) + (a – 2)2 = 0 ⇒ a – 2 = 0 and cot2π (a + x) = 0

⇒ a = 2. 2. 2 M110304 Sol. 2 2

1 2 1 2z 2z 2 z z− = −

( ) ( )2 21 2z 4 z 1 0− − =

1z 2= 3. 1 M111426 Sol. 10x6xcos3 21 +−− is defined if x2 –6x + 10 ≤ 1 if x2 –6x + 9 ≤ 0 if |x –3|2 ≤ 0


29

if x –3 = 0 if x = 3 4. 3 M110510

Sol. Given expression is 1 2a 1 2a 312 b c a 2 b c a 2

⎛ ⎞= + −⎜ ⎟+ − + −⎝ ⎠∑ ∑

= 1 1 3(a b c)2 b c a 2

⎛ ⎞+ + −⎜ ⎟+ −⎝ ⎠∑

Now, as (a + b + c) = (b c a)+ −∑ Applying A.M. ≥ H.M.

Minimum value of the expression = 1 39 32 2× − =


30




1 C P110806 1 C C110504 1 B M111210 2 D P111012 2 D C111804 2 A M111201 3 D P110610 3 C C111507 3 A M110101

4 B P111302, P111305 4 D C111401 4 A M110504

5 A P111026, P111209 5 B C110402 5 B M110902

6 B P110815 6 C C111302 6 B M111214 7 C P110309 7 B C110101 7 C M111502 8 D P110310 8 D C110307 8 B M110406 9 B P110815 9 C C110603 9 A M10101

10 C P110815 10 B C110603 10 B M110107 11 B P110815 11 D C110603 11 D M110101 12 A P111105 12 A C111903 12 D M110910

13 B P111105, P111222 13 B C111905 13 B M110902

14 A P111105 14 B C111902 14 C M110902 15 AB P111010 15 AB C111509 15 AC M110406 16 CD P111002 16 ABCD C111204 16 BD M111103 17 ACD P110912 17 BC C111102 17 AD M110719 18 AD P110917 18 BCD C110503 18 BC M110810 19 ABC P111207 19 ABCD C110908 19 ABCD M1105010 20 AC P111105 20 ABCD C111804 20 ABC M110415



31




1 D P110610 1 C C111507 1 A M110101

2 B P111302, P111305 2 D C111401 2 A M110504

3 A P111026, P111209 3 B C110402 3 B M110902

4 B P110815 4 C C111302 4 B M111214 5 C P110309 5 B C110101 5 C M111502 6 D P110310 6 D C110307 6 B M110406 7 C P110806 7 C C110504 7 B M111210 8 D P111012 8 D C111804 8 A M111201 9 C P110815 9 B C110603 9 B M110107

10 B P110815 10 D C110603 10 D M110101 11 B P110815 11 C C110603 11 A M10101

12 B P111105, P111222 12 B C111905 12 B M110902

13 A P111105 13 B C111902 13 C M110902 14 A P111105 14 A C111903 14 D M110910 15 ACD P110912 15 BC C111102 15 AD M110719 16 AD P110917 16 BCD C110503 16 BC M110810 17 ABC P111207 17 ABCD C110908 17 ABCD M1105010 18 AC P111105 18 ABCD C111804 18 ABC M110415 19 AB P111010 19 AB C111509 19 AC M110406 20 CD P111002 20 ABCD C111204 20 BD M111103



32




1 A P111026, P111209 1 B C110402 1 B M110902

2 B P110815 2 C C111302 2 B M111214 3 C P110309 3 B C110101 3 C M111502 4 D P110310 4 D C110307 4 B M110406 5 C P110806 5 C C110504 5 B M111210 6 D P111012 6 D C111804 6 A M111201 7 D P110610 7 C C111507 7 A M110101

8 B P111302, P111305 8 D C111401 8 A M110504

9 B P111105, P111222 9 B C111905 9 B M110902

10 A P111105 10 B C111902 10 C M110902 11 A P111105 11 A C111903 11 D M110910 12 C P110815 12 B C110603 12 B M110107 13 B P110815 13 D C110603 13 D M110101 14 B P110815 14 C C110603 14 A M10101 15 AD P110917 15 BCD C110503 15 BC M110810 16 ABC P111207 16 ABCD C110908 16 ABCD M1105010 17 AC P111105 17 ABCD C111804 17 ABC M110415 18 AB P111010 18 AB C111509 18 AC M110406 19 CD P111002 19 ABCD C111204 19 BD M111103 20 ACD P110912 20 BC C111102 20 AD M110719


Documents

Answer Key of Nwtra35g01,Nwtra35a01-A04, Nwtra35c01-c02, Camp35, Nwtw35a01-A02, Nwtwa35a01-A07 & Nwtwa35c01-c02_reshuffling Test-1_paper-1 & 2- Test Date 23-03-14 (1)