13-05-15 sr.iit-iz-co-spark(outgoing) ph-iii jee adv gta-10 (2014 p2) key & solutions free

7/25/2019 13-05-15 Sr.iit-iz-co-spark(Outgoing) Ph-III Jee Adv Gta-10 (2014 p2) Key & Solutions free

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NARAYANA IIT ACADEMY

INDIASEC:SR.IIT-IZ-CO-SPARK(OUTGOING) JEE ADVANCED DATE : 13-05-15TIME : 3:00 2014_P2 MODEL MAX MARKS : 180

KEY&SOLUTIONS

PHYSICS KEY CHEMISTRY KEY MATHEMATICS KEY

1 C 21 C 41 B

2 B 22 D 42 D

3 D 23 A 43 A

4 C 24 B 44 B

5 A 25 D 45 B

6 A 26 A 46 C

7 B 27 D 47 C

8 C 28 C 48 A

9 C 29 C 49 C

10 A 30 A 50 B

11 B 31 B 51 C

12 A 32 D 52 A

13 C 33 C 53 B

14 D 34 A 54 A

15 A 35 A 55 A

16 B 36 A 56 D

17 B 37 C 57 D

18 C 38 D 58 D

19 A 39 D 59 A

20 D 40 C 60 C


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Narayana IIT Academy 13-05-15_SR.IIT-IZ-CO-SPARK(OUTGOING)_GTA-10_2014-P2_Key &Sols

Paper-2 Page 2

Solutions

PHYSICSSingle correct choice Type

1. (C)Sol: 5 0.1l cm

2 0.01r cm 2V r l

% error in volume=2 2 0.01 0.1

100 100 3%2 5

r l

r l

2. (B)

Sol: 0 00

,appV V

f fV V

assuming car moving towards us .

345 1268 60 30 /

345 C

c

V m sV

3. (D)

Sol: 2 10,kq kqE Va a

10V

aE

4. (C)

Sol:0.5

. 0.01. . 50

PitchL C mm

C S D

Zero error is positiveReading . . . . . .X M S R C S R Zero Error L C

1 0.5 46 4 0.01 0.92X mm

2 0.5 48 4 0.01 0.94X mm 3 0.5 44 4 0.01 0.90X mm

Best reading= 0.92mean

X mm

5. (A)Sol: 10 100total i f w vQ mS mL mS mL

Putting 11 , 0.5, 1, 540, 80w V fm kg S S L L

6. (A)

Sol: 21

. . 04 4 2

i i f f

kQq kQqU K E U K E mv

a a

Given 64

kQ qV V v

a m


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Paper-2 Page 3

7. (B)

Sol: 0 .... 12

iB

r

Where2

e ei

T

.

Also 2L I mr

2

L

mr

Putting values in (1) , 034

eB L

mr

8. (C)Sol: Let rate of heat supply = P

Heat to boil 1Q P (9 min) = 100 10 90wmS m

Heat to evaporate 2 540vQ Pt mL m

54054min

9 90

tt

9. (C)Sol: At point of leaving contact

2 2 2 1 cos4

Rv gh g R

And 2

cos 0mv

Mg NR

2 5cos cos

6

v

Rg

10. (A)

Sol:

/v r

V

Rest

hl

2 21 1

2 2

MV I Mgh

2 22

2

1sin

2 2 2

MR vMV Mgl

R

Putting 030 and 10 /v m s 15l m


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Paper-2 Page 4

Section-II

Comprehension Type

Paragraph for Questions No.s 11 to 12

11. (B)12. (A)

Sol:3

;2 4 4

new New b

V T TV T F plVg

0T Vg

0

0

3 4

4 3

l

l


13. (C)14. (D )

CM

2V

V1V

CM

Sol: i fL L about CM of rod

2

1 ...... 12 2 12

L L MLmV mV

1 2 ...... 2i f mV mV mV

2

2 2 2 21 2

1 1 1 1...... 3

2 2 2 2 12i f

mLE E mV mV mV

Solving the equations

1

4

4V V

and

12

4

V

L


15. (A)16. (B)

Sol: 2 2 2 2

0 0 0

; 0 ;2 2 2

c hV R X X X V R V R h h

2 20 0 0 0

0; 4 22 2 2

q q q qmgh R h h R mg mg

2 2 2 22 2 ; 2 2mgh mg R h h mgR h R h h R

2 2 2 2 2 2 2 2 2 42 2 ;4 4 4 ;4 4 ;3 43

RR h R h R h R h Rh h h Rh h Rh h


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Paper-2 Page 5

For equilibrium 0Net

F 2 2 2 2

0

1 ; 2 12

q x xmg mg mg

R X R X

2 2

11

2

x

R X

2 2 2 2 2

2 2

14 ;3

2 3

x RX R X X R X

R X

Section-III

17. (B)

1m

2m

Sol :

1maxsf

1kf

2maxsf

2kf

1 6m kg 18 12 16 12 2 4m

1 2m kg 6 4 24 18 2 6m

1 7m kg 21 14 12 9 2 3m

1 1m kg 3 2

If all three stationary T=15N

2T

30

For Case I :

In 1 2max max& 15s sP F f Both static : 1,2P In 1R f static 2f kinetic : 1,4R

In 2Q f static , 1f kinetic : 2,3Q

(Check option)18. (C)19. A

(P) speed constant & moving in circle2

1 2 3 3 2& 0cmv

F F F f F F R

(for equilibrium in vertical) & 1F

must be static friction.

(Q) 1 2&F F will cancel each other.1F friction (static) & 1 2 3 cF F F F

(as circular motion)

(R) 1 2 3 cF F F F

1F static friction

(S) Car at rest 1 2 3 0F F F


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Paper-2 Page 6

20. D

1

8p

K

2P

2 2 2

22 2

1 1 1 1 1; ;

8 4 8 822

2

p p p p

p

V V V PmP

Pm V V V m

m

1/221 1 1 1

; ; ; 4 28 4 8 4 2

p p p p p

p

P m P P

P m P P

CHEMISTRY21. C

3 2 2 36 3 2Ca P HCl CaCl PH

4 3 2PH I KOH PH KI H O

4 2 3 2 23 3 3P NaOH H O PH NaH PO

Red phosphorus + NaOH soln. no reaction

22. D

1 2 1D

d

461

d

23. A

Potential difference of electrical double layer formed in a colloidal solution is called

Zeta potential.

24. B

25. D


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Paper-2 Page 7

Complexes Hyd. Of central

Metal cation

No. of

stereoisomers

(A) 2 3Cd gly H O NH

3

sp 2

(B) 2 2 2Pt Br H O 2

dsp 2

(C) 3

3Cr en

2 3d sp 2

(D) 2 2Co Br NO en

2 3

d sp 3

26. A

mRT (m = molarity, R = 0.082 L atm/mol. K, 25 298T C K )

3 45 10 2 100.082 298

M mRT

wM

m V

wm

M V

4

4

105 10

2 10

m

Each glucose unit (mw = 180)

Bonded to the starch chain with bss of 21 18H O mm giving a effective molecular

mass 180 18 162

So approximate average number of glucose45 10

309 300162

27. D

t

AReaction is zero order

28. C

5 3PCl Ag AgCl PCl


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Paper-2 Page 8

3PCl is water soluble while AgCl is insoluble in dil 3HNO .

29. C

A) Ionization isomerism B) Co-ordination isomerism

C) No isomerism D) Linkage isomerism

30. A2

2

13.613.6sep

zE

n

2 2n z

2

2

13.6 213.6

2nKE E eV

8 622.18 10 2.18 10 /

2

V m s

2

2 2

h h hJ n

31. B

2Cu give ppt with excess of KI

2 3excess brownCu KI CuI KI

32. D

2Cu gives deep blue colour solution in excess of 3NH .

33. C

7 60.1 10 0.08 10OH

7 70.1 10 0.08 10

89 10

4

3 10

4 log3pOH

34. A

7 3

4

10 0.1 10

0.1 3 10 3XOH

35. A


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Paper-2 Page 9

+

mol.massNE=

H

Assume B is monocarboxylic aid and NE = MW the COOH accounts for 45 g/ml of

molecular wt, leav 60 45 15 /g mol for the remainder 3due to CH so b is

3CH COOH A has a second COOH replacing an H of 3CH COOH, it is malonic acid

2CHCOOH

COOH

10452 /

2g mol

36. A

Due to H-bonding it is weak acid & have higher pKa.

37. C

P: 0.0591

1cell anode cathodeE pH pH

0.059cellE x y

;cellE ve x y

Q: 1

2

0.0591log

2cell

PE

P

cellE ve 1 2P P

R:0.0591

log1cell

yE

x

,cell

E ve y x

S: 2

1

0.0591log

2cell

PE

P

2 1cellE ve P P


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Paper-2 Page 10

38. D

(P) D-xylose

OH

OH

HO H

H

H

CHO

2CH OH

Oxidation

OH

OH

HO H

H

H

COOH

P.O.S

COOH

MESO

(Q) D-Ribose

OH

OHH

H

CHO

2CH OH

OHH 2 epimer of D-Arabinose, give same osazone as D-arabinoseC

(R) D-threose

OH

OHH

H

CHO

2CH OH

OH

OHH

H

H C OH

2CH

: :O

Smallest aldose (4C) able to form cyclic hemiacetal.

(S) D-galactose = it is C-4 epimer of D-glucose

OHH

CHO

2CH OH

OHH

OHH

OHH

5

6

1

2

3

44

Epimer

C

OH H

CHO

2CH OH

OH H

OHH

OHH

5

1

2

3

4

(D-glucose) (D-galactose)

39. (D)

3 2 3 2 22Co NO Al O CoO NO g O g


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Paper-2 Page 11

Cool2 2 3NO g NO g N O s

4 2 7 2 2 3 22NH Cr O s N g Cr O s H O g

4 4 4 3 3 2Na NH HPO CuSO NaPO NH g H O g

40. C

(P) Pyrcuric acid||

3

O

CH C COOH

have carbonyl group i.e. show 2, 4 DNP

test.

(Q) Aspirine is

COOH

3OCOCH

& give 2CO gas with 3NaHCO

(R) Acrylic acid 3CH CH CH COOH

Due to bond , it declourise 2 4/Br CCl (Unsaturation test).

(S) Benzyl alcohol is

OH

& react with Na metal evolve 2H gas.


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Paper-2 Page 12

Part-III: Mathematics

Single correct choice Type

41. (B)Sol: Let 1 2 ,1 3 , 3 2p p p be a point on the first line and , 1 5 ,k q q q be a point on

the line. If they are co-ordinates of point of intersection then

3 2p q and 1 3 1 5p q 1 2p k q

1p q and 2k

Point of intersection will be 3, 4, 1

42. (D)Sol: From given condition odd elements of the domain can associate with unequal odd

elements of the co-domain and even elements of the domain can associate withunequal even number of the co-domain

Number of ways if X is odd =1 1 1 1

4! 1 91! 2! 3! 4!

Number of ways if X is even =1 1 1 1

4! 1 91! 2! 3! 4!

43. (A)

Sol: 21

2 nM A A I ;

21

2 nN A A I

4 31 1

.4 4

n nM N A I A A I

nI 3 0A 1M N

44. (B)

Sol: 1 1

33 2 3 3

1 1 0

3 3 3 sin 1 1 4 sin 1 4 sin 2 sinx x x

x x x

t t t t dt t t dt t tdt t tdt

1

3 2

130

3 3 3 sin 1

limsin sin 2

x

x

x

t t t t dt

x x x

3

30

40 0

2 sinsin 2 1

lim . . lim . .sin sin 2 5 sin sin 2 5

x

x x

t tdt x x x x x x

x x x x x x


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Paper-2 Page 13

45. (B)Sol: 2 3 2z i is a circle and 5 3z i is another circle

The distance between their centres is equal to sum of their radii thats why Z will be theirpoint of contact .

16 7 61

5 5 5Z i Z

46. (C)

Sol:

4 2 2 4

3 3 3 31 1 . 1 1k

kT

k k k k

2 2

3 32 2

3 33 32 2

3 3

1 11

1 14

1 1

k k k

k k

k k

2 22 23 33 3

1 11 1 0 1 1

4 4nS n n n n

2

3

22/3 2/3

3

1 1 1 11 1

4 2n n

x

S SLim

n n nn

47. (C)

Sol:2

1

1I

x

ln2

1

xdx

x

Put1

xt

2

dtdx

t

0 1 1 1 1

1 0 0 0 0

1 1 1 2 21I ln dt lnt t dt lntdt ln t dt lntdt

t t 48. (A)

Sol: Let 1 1,P x y & 2 2,Q x y then the roots of the equation 2

1 4 0mx ax

2

2 2

2 4 10

m ax x

m m

are 1x and 2x

And the roots of the equation 2 21 4 4

4 0y a a

y a y ym m m

and 1y and 2y

Required circle is

2 2

2 2

2 4 4 1 4

0

m a a a

x y X ym m m m

(1)2 2 18 4 5 0x y x y ..(2)

From (1) and (2)

2

4 1 44, 5

a a

m m m

and

2

2 410

m a

m

1

3m and

1

3a

49. (C)


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Paper-2 Page 14

Sol: 4 4 4 1 4 0x y z xyz 4 4 2 2 4 2 2 2 22 1 2 2 2 4 0x y x y z z x y z xyz

2 2 22 2 2 1 2 0x y z xy z

2 1z xy z and 2 2x y 1z 1x 1y

1x 1y 1 1 1z x y

1 1x y

1,1,1 1, 1,1 1,1, 1 1, 1, 1A B C and D

2 2AB i j

2 2AC i k

2 2AD j k

2 2 01 1 8

2 0 2 8 86 6 3

0 2 2

Volume

50. (B)Sol:

0,0S

1,1P

Q

R

From properties of parabolaRS PS and RQ PR

: 2 3PR x y

0 3, 3RS x y R

Now:

: 2 9

9, 9

RQ x y

PQ x y Q

1 2PS l

2 9 2QS l

Length of latus-rectum= 1 21 2

4 4 9 2 2 36 2 18 2

10 59 2 2

l l

l l


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Paper-2 Page 15

Section-II

Comprehension Type

Paragraph for question Nos : 51-52

51. (C)52. (A)Sol: Consider the function 3 2f x x x x

f x is an increasing function such that 0 0,1f x x x

And 0 1f x x x

Now , if , , 0,1a b c

,b a c b and a c (not possible)

If , , 1,a b c

,b a c b and a c (not possible)

1a b c ir 0a b c

Paragraph for question Nos : 53-5453. (B)54. (A)

A

B D C E Sol: EAD and CAD

tan , tan and tan are in GP ,

2 22

2 2

sin sin sin sintan

cos cos cos sin

2 2 2 2 2 2 2 2sin cos sin sin cos sin cos sin 2sin 0 or 2tan 1

0

4

cos 5 24

AD AE

Paragraph for question Nos : 55-56

55. (A)56. (D)


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Paper-2 Page 16

Sol: 5

2

x

x

f x dx ax b x R

5 2f x f x a x R

' '5 2f x f x a x R '

f x is a periodic function but f x is polynomial 'f x is a polynomial

'f x k (a constant)

'f x kx c

Now 0 3f and 1 5f

2 3f x x

Section-III

Matching List Type

57. (D)58. (D)

Sol: The co-normal points will be 0,0 5 ,52

and 5 , 52

centroid 5 ,03

Circumcenter25

,0 ;4

Radius =25

4; Area of

25

2ABC

59. (A)

Sol:3 3 32 2 2

3a b c

a b c

Let32

3x

x

, ,x a b c

3 3 2 0x x 0, 3a b c ab bc ca and 2abc

3 3 3 3 3 33 0 6a b c abc a b c 2 2 2 2 6a b c ab bc ca

60. (C)

Sol: (A) 0 1

2 1

1 1 2 3 2 1 1 2 0x x x x x dx x x x x x dx

(B) If 0f x has two pairs of equal roots than y f x has 1 point where it is not

differentiable .

(C) 8 41

8 4lim ....... 1

1 1xL

x x

8 4

8 41

8 4lim ....... 2

1 1xx x

Lx x

1 2

2 8 4 4 2L L (D) 4 3 260 2.2.3.5 60 2 2 3 5x ax bx cx x x x x

31 60 8 2a b c

13-05-15 sr.iit-iz-co-spark(outgoing) ph-iii jee adv gta-10 (2014 p2) key & solutions free

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