1

CODE Phase Test – I

NARAYANA I I T / N E E T A C A D E M Y

PHASE TEST - 2 XII-NEW-REG BATCHES :: PAPER – I & II :: Date: 02.07.17

PAPER – I KEY PHYSICS CHEMISTRY MATHEMATICS

1. (1) 2. (3) 3. (2) 4. (0) 5. (2) 6. (2) 7. (6) 8. (8) 9. (A), (B), (C) 10. (A), (C) 11. (A), (B), (C) 12. (B), (C) 13. (A), (D) 14. (BC) 15. (A), (D) 16. (C), (D) 17. (AD) 18. (A), (B) 19. A-S, B-S, C-P, D-S 20. A – QR, B – PS, C – QR,

21. (3) 22. (4) 23. (5) 24. (1) 25. (0) 26. (8) 27. (0) 28. (3) 29. (C), (D) 30. (A), (B) 31. (B), (C), (D) 32. (C), (D) 33. (A), (B), (D) 34. (D) 35. (BCD) 36. (ABD) 37. (B)

38. (A), (B), (C) 39. (A – Q,R), (B – RS), (C – R,T), (D – PT) 40. (A – PQS), (B – T), (C – R,Q), (D – R)

41. (5) 42. (8) 43. (4) 44. (1) 45. (0) 46. (6) 47. (2) 48. (4) 49. (B), (C) 50. (B,C) 51. (A), (B) 52. (A), (B), (C), (D) 53. (AC) 54. (BC) 55. (A), (D) 56. (A), (B) 57. (B), (C), (D) 58. (A), (B) 59. A-Q, B-P, C-S, D-T 60. A – Q, B – P, C-R, D-S

PAPER – II KEY PHYSICS CHEMISTRY MATHEMATICS

1. (7) 2. (2) 3. (9) 4. (2) 5. (4) 6. (2) 7. (3) 8. (6) 9. (A), (B), (D) 10. (B), (C) 11. (A) (B), (D) 12. (B), (C) 13. (A), (B) 14. (BD) 15. (A), (C) 16. (A), (D) 17. (A) 18. (C) 19. (C) 20. (B)

21. (3) 22. (0) 23. (3) 24. (3) 25. (4) 26. (8) 27. (3) 28. (4) 29. (B), (C), (D) 30. (A) (B), (D) 31. (A), (B), (D) 32. (ACD) 33. (BD) 34. (A), (D) 35. (A), (C), (D) 36. (C) 37. (D) 38. (A) 39. (D) 40. (B)

41. (7) 42. (0) 43. (0) 44. (4) 45. (3) 46. (3) 47. (5) 48. (6) 49. (A) (B) (C), (D) 50. (ABC) 51. (A), (C) 52. (A), (B), (D) 53. (A), (B) 54. (A), (C) 55. (A), (C) (D) 56. (A), (D) 57. (C) 58. (C) 59. (A) 60. (D)

2

HINTS & SOLUTION PAPER - I

PHYSICS (PAPER I)

1. (1) Slope of line PQQR tanQR

vtan

1

Slope of line AB at P

dv 1at Pds tan

dv 1ds v

2p

vdv v 1a 1m / sds v

2. (3) Finally all the blocks will move with common velocity say v. From conservation of linear momentum: 3×2 = (3+2+1)v v = 1 m/s

Work done by friction: f f iW KE KE

221 13 2 1 t 3 2 3J2 2

3. (2) When force on the byllet is zero, then 5 30 600 2 10 t t 3 10 s

Impulse t

0F.dt

= 33 10 5

0600 2 10 t dt

33 105 2

0

2 10 t600t2

= 0.9 = 0.42 (2) N = 2

4. (0) F f r I

21 aF f r mr2 r

Solving the above equations : f = 0

5. (2)

21 1

22 2

m 2r / 3L I 2L I 2m r / 3

6. (2) Spring constant for combination 3 9K 3K K2 2

Now 0 09 9mg Kx or 9K Kx2 2

0x 2cm

3

7. (6) A B t ln8 3 ln 2

ln 2 ln 2 t 3ln 2 t 6 min1 2

8. (8)

9. (ABC) Both the blocks would feel same pseudo force = ma in downward direction. Since

masses are same, there is no acceleration relative to pulley. So only acceleration of blocks is acceleration of pulley. There could be some initial velocities of two blocks. In that case speeds of blocks relative to pulley will be same. But if there is no initial velocities of blocks, then both blocks will move with same velocity along with pulley.

Tension in the string connecting blocks : T m g a mg 10. (AC)

4

11. (ABC)

12. (B,C) Work done by the spring on block = loss in spring potential energy 2 2 2 21/ 2 ka 1/ 2 kb 1/ 2 k a b This is also work done against friction mg a b

So,

2 21 k a b k2 a bmg a b 2mg

13. (AD) PE = 2KE

And 2

nnr2

14. (BC) Two or more higher nuclei combine to form a heavier nucleus in fusion reaction. 15. (AD) A BF mg F T B AF F mg T Buoyancy mg Buoyancy does not depend upon atmospheric pressure.

String is under tension, mg > B material liq

5

16. (CD) The angle of contact at the free liquid surface inside the capillary tube will change such that the vertical component of the surface tension forces just balance the weight of the liquid column.

17. (AD)

18. (AB)

19. (A-S); (B-S); (C-P); (D-S)

20. (A-QR); (B – P,S); (C-Q,R); For (P) and (r) : in this case final surface area is greater than

initial surface area, and hence surface energy increase the expansion of internal energy. Hence, it causes cooling. For (Q); Reverse of the above reason.

CHEMISTRY (PAPER I) 21. (3) Ca, Mg, Al

22. (4) weightgm atom = atomic weight

23. (5) No. of waves = No. of orbit = 5 25. (0) Hybridization of each carbon atom is sp2 = 3 hybrid orbitals per carbon Total hybrid orbitals = 3 x 2 = 6

6

Each carbon and hydrogen has one unhybrid orbital, so No.of unhybrid orbitals = 2 + 4 = 6 X – Y = 6 – 6 = 0 26. (8)

110

49 1.6 10 898

M percentageby mass of solute densitymolecular mass

M

27. (0) In Cr(CO)6, there is no unpaired electron. 28. (3)

29. (C), (D) N = n-factor x M = 2 X 2 = 4, Moles = milli moles/1000 30. (A), (B) Bohr’s model can be applied on one electron one nucleus atom or ions.

31. (B), (C), (D) 32. (CD) Apply Le-Chatelier Principle 33. (A,B,D) 34. (D) Due to d-d-transition 35. (B, C, D) Anodes are of impure Cu and Cu-strip is cathode. The electrolyte is acidified

CuSO4 and the net process is, Anode: 2Cu Cu 2e Cathode: 2Cu 2e Cu s Impurities from the blister Cu deposited as anode mud. 36. (A, B, D)

CH2

N CH2

CH2

CH2 N

CH2

CH2

COO-

COO-

OOC

OOC

.. ..

37. (B) 38. (ABC) 39. (A - Q,R); (B-R,S); (C-R,T); (D – P,T) 40. AP,Q,S; BT; CR,Q, DR

(P) Siderite - FeCO3

(Q) Malachite - CuCO3.Cu(OH)2

(R) Bauxite - AlOx(OH)3-2x

(S) Calamine - ZnCO3 (T) Argentite - Ag2S

MATHEMATICS (PAPER I)

41. (5) 2 2 2

21 a 1 3a a 8 POMN a2 4 2 4 OMN 3 Q

7

42. (8)

43. (4) 2 21 2

1 21 2 1 2

at a at a 1l a t t 1 lt t t t

2

2

4 aa 1 la a

Hence the result.

44. (1) 2 2x 4t 1

9 4

2 2 2x 9t 9 x 9 9t Above statement satisfy when 29 9t 0 t 1 k 1 45. (0)

46. (6) By total prob theorem the req probability

3 3 2 3 125 5 5 10 25

47. (2) 3 0 1 01 ; 20 3 0 13

48. (4) 2n 1 9A A

8

49. (BC) 1 1 1 14 2 2 4 69 3 4 9

3

1 1 1 1 1 12 4 2 1 1 2 3

9 3 1 1 4 9

32 2 2 648 , 2 81; 9

50. (BC) 0 0 1

A 0 1 01 0 0

, A 1 , 0 0 1

Adj.A 0 1 01 0 0

Hence 1A A

51. (AB) 1 TA B A B A B 1TA B

1 1A B A B A B A B

52. (ABCD)

4n E 9 9 3P En S 5 5 5 5 5

53. (AC) 2a 3b c 3a 4b 2c 0 54. (B,C) By hit and trial, (1, 0) and (-1 , 0) are two rational points on the circle. If there is a third

point, then the co-ordinates of centre (or circumcentre of the triangle passing through three rational points) will be rational. But the centre of the given circle is not a rational point. (B) and (C) are correct.

55. (AD) Mid point of AE must be point of intersection of diagonals of parallelogram Let E = (h, k)

h 1 k 2,2 2

must lie on 4x 8y 7 0

(h, k) is also lie on AB whose equation is x + y = 3; 15 9h ,k2 2

56. (AB) 2 22

2564 256 256 64t 40t

57. (BCD)

58. (AB) 2

2

256 2g16y 2fy c 0y y

4 3 2y 2fy cy 32gy 256 0

9

i

4 2ft

59. (A-Q); (B-P); (C-S); (D-T)

60. (A-Q) (B – P)(C-R) (D-S)

(A) 15

716

6

C 1Req probC 2

(B) Req prob 1 14 1 1 1 1 8115 15 2 2 2 2 15

(C) 14

616

6

C 7Req probC 30

(D) 8 7 7Req prob 115 30 30

PAPER - II

PHYSICS (PAPER II)

1. (7) 2

CMRot

2 2TotCM CM

1 IK 2f 1 1K MV I2 2

For a solid sphere, 22I MR5

2 2 2 2CM CM

1 1 2 1I MR MV2 2 5 5

2CM

2 2CM CM

1 1MV 25 5f 1 1 1 1 7MV MV2 5 2 5

2. (2) For translational motion, N = Mg F = Ff For cube to topple, resultant torque about O. Fa Mg(a/2) or F Mg/2 n = 2

a

Ff

F a

N

Mg O 3. (9) From conservation of angular momentum,

(L/2)mv + mv(L/2) = 2 22L L L2m m m

12 2 2

10

or mvL = 2 2 2 2mL mL mL 2mL

6 4 4 3

= (3v/2L) = (3 6)/(2 1) = 9 rad/s

4. (2) 22 2

Gm M m GF mM mr r

for F to be maximum dF/dm = 0 (as M and r are constants)

2

2Gd mM mr 0

dm

i.e. M – 2m = 0 M/m = 2

5. (4) Initial angular momentum of boy about the axis of rotation of merry-go-round is = mvr = 30 3 2 = 180 Js Initial angular momentum of merry-go-round = 0 As external torque is zero, angular momentum of (merry-go-round + boy) = constant

180 = (120 + mr2) = 2

180 180 3120 30 2 120 120 4

3 3x 4 x = 4

6. (2) o o

o

FL FLF / AY YA/ L A

l l l

= constant Y(R2) = constant YR2 = constant

2 2B B s SY R Y R

2 11sB

10s B

YR 2 10 2R Y 10 10

7. (3) 2

2PEm

2 2

2 2

1 1

2 4E P PE P P

2 14E E = E1 + 3E1 E/E1 = 3

2 1E E E , 1 1E E , 13E 8. (6) Let v be the velocity of ball after

collision, since collision is elastic, Or relative velocity of separation = relative velocity of approach. v – 1 = 4 + 1 v = 6 m/s (away from the wall)

4 m/s 1 m/s 1 m/s

v

9. (A), (B), (D) CONCEPTUAL 10. (B), (C) Height of liquid in vessel C is maximum. Therefore, force on the base of vessel C is

maximum [F = (Po + hg)A]. Net force on all the three vessels = weight of liquid, which is same for all the three vessels. 11. (A), (B), (D) Even if no external force were acting on the system, torque can act on the

system and hence angular momentum does not remain conserved. The change in KE and PE can be caused by internal forces. Hence, answers (A), (B) and (D)

are correct. 12. (B), (C) Av 0 , B

ˆv vi , C

ˆv 2vi

C Aˆv v 2vi

B C

ˆv v vi

C Bˆv v vi

A

2v

B

C

v

v v

v

v = R

y

x v

11

13. (A), (B) For r > R, the gravitational field is F = GM/r2 F1 = 21

GMr

and F2 = 22

GMr

21 2

22 1

F rF r

For r < R, the gravitational field is, F = (GM/R3)r

F1 = (GM/R3) r1 and F2 = (GM/R3) r2 1 1

2 2

F rF r

14. (B), (D) We know that transfer of momentum will be minimum when target is massive and transfer of kinetic energy will be maximum when target and projectile are having same mass. It means statement (A) and (C) are correct, but statement (B) and (D) are incorrect because when target is very light, then after collision it will move with double the velocity of projectile and when collision is oblique and m2 at rest with 1 2 ,m m after collision the ball move perpendicular to each other.

15. (A), (C) If point mass m is pulled through a height h then work done

W = mgh Similarly for a chain we can consider its centre of mass at the

middle point of the hanging part i.e. at a height of L/(2n) from

the lower end and mass of the hanging part of chain Mn

L/2n

Centre of mass

So work done to raise the centre of mass of the chain on the table is given by

2

M LW gn n

[As W = mgh] or 22MgLW

n

(L/n)

L

Taking surface of table as a reference level (zero potential energy)

Potential energy of chain when 1/nth length hanging from the edge 22MgLn

Potential energy of chain when it leaves the table 2

MgL

Kinetic energy of chain = loss in potential energy

22

12 2 2

MgL MgLMvn

22

1 112 2

MgLMvn

Velocity of chain 2

11v gLn

16. (A), (D) When equilibrium is reached, the two spheres along with the

spring will be arranged as shown. From equilibrium of lower sphere, FB + FS = mg 2Vg + FS = 3Vg FS = Vg (1)

kx = 3

34 4 R gR g x3 3k

From equilibrium of upper sphere B sF' F m 'g = FS + Vg (2) Putting FS from equation (1) in equation (2), we get BF' 2 Vg

k

3 R

R

12

which is the maximum buoyant force that can act on upper sphere.

17. (A) Excess pressure is always on the concave side of the surface. For angle of contact of 2 / 3, the liquid should have a convex surface. So the excess pressure should be in the upward direction.

18. (C) Difference in pressure between the inner and outer surface is i 0p p

Force = 2i 0p p r

Since excess pressure is given, i 0p p p

2F p r So, choice (c) is correct and the rest are incorrect. 19. (C) From the graph and the fact that nlp (=number of neutrons/number of protons) ratio for

manesioum is 27/12, which is greater than 1 (= unit slope). 20. (B) 230 214 0 e 4 2

90 83 2TH Bi x 1 y He z (gamma ray)

CHEMISTRY (PAPER II) 21. (3) 22. (0) In Cr(CO)6, there is no unpaired electron. 23. (3)

29. (B, C, D) Anodes are of impure Cu and Cu-strip is cathode. The electrolyte is acidified

CuSO4 and the net process is, Anode: 2Cu Cu 2e Cathode: 2Cu 2e Cu s Impurities from the blister Cu deposited as anode mud. 30. (A, B, D)

CH2

N CH2

CH2

CH2 N

CH2

CH2

COO-

COO-

OOC

OOC

.. ..

36. (C) Kp value depends on temperature and nature of reactants only. 39. (D) [Cu(NH3)4]+2 is a square planar complex.

40. (B) One unpaired e– in [Cu(NH3)4]+2 but no unpair e– in [Fe(CN)4]–2

. MATHEMATICS (PAPER II)

41. (7) It is the altitude to the least side. 42. (0)

13

43. (0) 1 2x x 0 y mx a is a horizontal line m = 0. 44. (4)

45. (3)

sin A sin A2

sin A sin A

sin A3

sin A

46. (3) 1001 1P 100 2

3 2

47. (5) 2 2

2 2

3 2 1 5 53 3 1

48. (6) 1 1 1

D 10!11!12! 11 12 1312 11 13 12 14 13

10!11!12! 2

2 23

D 4 2900 29 2 510!

K 3 32 18 K / 3 6

14

49. (ABCD) Now applying 3 3 1 2C C g C f C

f g 0f g 0f g 0

f x y f x g y g x f y

50. (ABC) 2 1 0

A 1 2 10 1 2

51. (AC)

m m

1 4k 1 k 1

1x P 2k 12k

1 21 1x x 72 2

52. (ABD)

2

n3

2 n 1C .n

22 C

53. (AB) use parametric form

54. (AC) C2 is the director circle of C1. So the equation of C2 is 22 2x y 2.2 . Again C3 is

the director circle of C2. Hence the equation of C3 is 22 2x y 2. 2.2 .

55. (ACD) The standard form is 2

x y 1 12 22

x y 1 02

Here 11S 0 for the point P(1, 2) 56. (AD) conceptual 57. (C) 2 2

0 0 0B I and A A

20 0det A A .....10terms

0det 10A 1000; X0 0 1 2 49 0B B .......B B 0 1A X C has no solution. 58. (C)

15

59. (A) 60. (D)

**********