key & solutions physics chemistry maths...nov 02, 2019 · narayana iit/neet academy 25.11.2019...

Narayana IIT/NEET Academy 25.11.2019 XI_REG_ADV_Jee-Adv_2013-P1key & hints

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XI -REG_Adv. Date: 25.11.2019Time: 3 hrs Max.Marks: 180

KEY & SOLUTIONS

PHYSICS CHEMISTRY MATHS

1 A 21 C 41 B

2 A 22 C 42 A

3 B 23 A 43 A

4 D 24 C 44 C

5 D 25 B 45 A

6 C 26 B 46 A

7 B 27 A 47 D

8 D 28 C 48 B

9 A 29 B 49 C

10 C 30 A 50 B

11 ABC 31 ABCD 51 ABC

12 AD 32 ABC 52 AC

13 CD 33 CD 53 BD

14 ABD 34 AC 54 AD

15 ABC 35 BCD 55 ABCD

16 5 36 4 56 2

17 7 37 4 57 5

18 6 38 2 58 8

19 2 39 1 59 4

20 5 40 1 60 5


p

PHYSICS

1 A

2 A

3 B4 D

5 D

6 C

7 B

8 D


p

9 A1 1 2 2 1 1 2 2m u m u m v m v

10 C

Ball 2 with wall – 1st (Ball 2 rebounds)Ball 2 with ball 1 – 2nd (Both balls rebounds)Ball 2 with wall – 3rd (ball 2 rebounds)

11 ABC12 AD13 The friction force will aid rotation but opposes translation. The friction force is reduced if

reduced.14 ABD15 A L

when A is constant vector. According to given condition, it is cross product so is

perpendiculat to L

and also to A

Forther more sindl

Aldt

the component L

in the direction of A

(i.e., cos )l will not change

which time otherwise sinAL will not satisfied. The magnitude of L

does not change because A L

isperpendicular to

16 5

17 718 6

19 2

20 5


p

CHEMISTRY21. (C)

Z > 1 real gas shows positive deviation from ideal behavior when repulsive forces dominate

22. (C)

2

83

27 27C C C

a aT P V b

Rb b

23. (A)

pn V

RT

Slope of line tanP

RT

16.42

tan 0.0821tan

PT

R

For B, 200 3T K

For A,200

3T K

24. (C)

For cyclic process : dU = 0

w q

25. B

,gH U n RT 2,gn H U

, 0,gw n RT w q ve

26. B

(i) (c) w = –P. 2 8.314 600V nR T = – 9.97 kJ

(ii) (b) . 0total AB BC p mH H H nC T

72 800 200

24200

R

R

27. A

28. (C)

3totalP P

0.3180.106

3P

3 34 4.76 10pK P

29. (B)


p

3

2

2 4c

NOK

N O

If 2NO at y – axis

2 4N O at x-axis

.y Kc x

Parabolic curve

30. (A)

Q = 1 atm > Kp

31. (ABCD)

For Boyle’s law

T = constant

PV = K = constant

32. (ABC)

0U q w w w

vmw n C dT

2 1

2 1

2 1

1

ext

nRw T T

T Tw nR P

P P

33. (CD)

At1 100

1001A

RV R

1 600200

3B

RV R

VB > VA so expansion of gas takes place

VB = 200 × 0.0821 – 16.42 L

34. (AC)

gH U n RT

(a) 2gn U H

(b)1

2gn U H

(c)1

2gn U H

(d) 0gn U H

35. (BCD)


p

36. 4

37. 4

38. 2

U q w

300 400 0.5 100

2

V

V L

39. 1

40. 1

MATHS41. (B)

For required circle, P(1, 8), and O(3, 2) will be the end point of its diameter

2 2

1 3 8 2 0

4 10 19 0

x x y y

x y x y

42. (A)Here, equation of chord of contact w.r.t P is

4 20. 9

5x y

5 4 20 45 ....x y i

and equation of chord bisected at the point Q(h,k) is2 29 9xh yk h k

2 2xh ky h k ii From Equa (i) and (ii), we get

2 2

5 4 20 45

h k h k

2 2

20

4 5

g hand

h k h k

2 2

20 9

4 5

h h

h k h k

Or 2 220 0 4 5h k h k

Or 2 220 36 45x y x y

43. (A)As, the two circles intersect in two distinct points


p

Distance between centres lies between 1 2| |r r and 1 2| |r r

i.e., 2 2| 3 | 4 1 1 3 | 3 |r r

| 3 | 5 | 3 | 8 2r r r or r 2 8r

44. (C)Given, |z| = 1, arg iz z e

But1

zz

1arg arg

11

zz

z

45. (A)46. (A)47. (D)

Given that , cos sin iz i e

15 15 2 12 1

1 1

152 1

1

Im Im

Im

mn i

m m

i m

m

z e

e

sin sin 3 sin 5 ... sin 29

29 15 2sin sin

2 22

sin2

sin 15 sin 15 1

sin 4sin 2

48. (B)We have,

2 2 2225 9 25 75 45 15 0a b c ac ab bc

2 2 215 3 5 15 5 15 3 3 5 0a b c c c a b b c

2 2 2115 3 3 5 5 15 0

215 3 ,3 5

a b b c c a

a b b c

and 5c = 5c

15 3 5

1 5 3, 5 , 3 .

a b c

a b csay

a b c

49. (C)

Let Sn = cn2

2 21 1 2nS c n cn c cm

12n n n nT cn c T S S

22 2 2 22 4 4nT cn c c n c n


p

Sum

22 2 24 . 1 2 1

2 16n

c n n nT nc c n n

2 2 22 1 2 1 3 6 1

3

c n n n nc c n n

2 2 2 24 6 2 3 6 6 4 1

3 3

nc n n n nc n

50. (B)

Here, 2 2 2 2 2 2 22 0a b c p ab bc cd p b c d

2 2 2 2 2 2 2 2 22 2 2 0a p abp b b p bcp c c p cdp d

2 2 20ap b bp c cp d [since, sum of squares in never less than zero]

2 2 20ap b bp c cp d

b c dp

c b c

a,b,c,d are in GP

51. (ABC)52. (AC)53. (BD)

Put z = x + iy and simplify.54. (AD)55. (ABCD )

56. (2) The circle and coordinate axes can have 3 common points, if it passes through origin. [p = 0]

If circle is cutting one axis and touching other axis. Only possibility is of touching other axis.

57. (5) The line 5x – 2y + 6=0 meets the Y-axis at the point (0,3) and therefore the tangents has to passthrough the point (0, 3) and required length

2 21 1 1 1

2 2

6 6 2

0 3 6 0 6 3 2

25 5

x y x y

58. 8

Use AM GM

59. 4

60. 5

i.e., 1 2 1 2| | | | |z z z z

5 53 3 2 2

2 2

93 2

2

9 9 5| 3 2 | 2

2 2 2

z i z i i i

z i i

z i

5 53 |2 6 5 | 5

2 2z i or z i

key & solutions physics chemistry maths...nov 02, 2019 · narayana iit/neet academy 25.11.2019...

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