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NARAYANA IIT ACADEMY INDIA

Sec: Sr. IIT-IZ-CO SPARK Jee-Advanced Date: 16-05-18 Time: 09:00 AM to 12:00 Noon 2015_P1 Max.Marks:264

KEY SHEET

PHYSICS 1 1 2 2 3 3 4 5 5 9

6 7 7 5 8 4 9 ABD 10 ABCD

11 ABD 12 BC 13 B 14 BCD 15 ABCD

16 ABD 17 ABCD 18 AB 19

A-RT; B-S; C-P; D-Q

20

A-PQRT; B-QS; C-PQRS; D-PRT

CHEMISTRY

21 2 22 2 23 8 24 4 25 3 26 0 27 2 28 1 29 AB 30 AB 31 B 32 AB 33 ABC 34 AC 35 ABD

36 ABCD 37 BCD 38 BD 39

A-PR; B-R; C-PT; D-QS

40

A-S; B-P; C-Q; D-R

MATHS 41 3 42 3 43 2 44 4 45 1

46 3 47 7 48 1 49 ABCD 50 BD

51 ABCD 52 D 53 ABCD 54 AC 55 AC

56 AD 57 BD 58 ABC 59

A-PQ B-PQ C-PR D-PQS

60

A-S B-R C-Q D-R

Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s

Sec: Sr. IIT-IZ

SOLUTIONS PHYSICS

1. Complete the pyramids OBAD and 1OB AD for this ‘C’ is the geometric centre . The closed Figure enclose ‘6q’ charge electric flux through all 6 faces is

0

6q

. Through each

face 0

q

So n=1

2.

04mK22mK

12mK

2 12 sin 2 .............(1)mk mk

2 02 cos 4 ...........(2)mk mk 2 1 02 2 4mk mk mk ( squaring and add 1 and 2) 2 1 02 0.8 2 2.8k k k eV 2 1 02 0.8 2 2.8k k k eV Energy released = 1 2 0 3.6 1 2.6k k k eV

3 R2 2

0 00E 4 R 4 G 4 r p r dr

4

00 2

G p pEr

r2 2 2

1 0E 4 r 4 G 4 r dr

21 0E G S r

Potential inside the earth

i i S SV V V V V V

r R

i i 0RV E dr E dr

4R /2 R

0 0 2R

RG dr G drr


Sec: Sr. IIT-IZ

3

031G R12

To escape from field of earth

3

2031G R 1m mv24 2

3031G RV

12

4. Ans : 3

Sol : sin 0 &x cm cmce F a V remains vertical

Constraint equation 22g

r lV

From L C E

rA

O

0 AV r

gV

2r

U K

2 21 12 2 22 g

l lmg MV I

2 2

22 2 1 12 2 2 62 2

l Mlmgl M

2 2 2 2 2 22 28 12 482 2

Ml ml Mlmgl

6 2 2 2

5

g

l

5. Conceptual

6. 1 71 3 8.023829H Li u

4 42 2 8.005206He He u

0.018623m u

0.018623 931.5 mevEu


Sec: Sr. IIT-IZ

=+17.35 Mev

7. Conceptual

8. Image formation due to convex lens

1 1 1

36 30v

30 36 1806

cm

This image will act like a vertical object for mirror and after reflection from mirror its

image (shown by 2I ) will be formed at 80cm below optical axis of convex lens.

For concave lens, this image will be object at a position of 15cm below the lens for

final image formed by concave lens.

1 1 120 15 f

1 5

300f

Also

1 1 11f R R

Or 5 3 21

300 2 R

60R

9. 1 0.1258

Smallest divisionon main scale cm cm

5 divisions of the vernier scale = 4 divisions of the main scale =4x0.125 = 0.5cm

I division of vernier scale 0.5 0.15

cm

Least count of vernrer = 1 main scale divisions -1 vernier scale divisions = 0.125 –

0.1=0.025cm

Least count of screw gauge = pitch/100

If pitch of the screw gauge is twice the least count of vernier calipers then the least

count of


Sec: Sr. IIT-IZ

Screw gauge 0.05 0.005100

mm

Also, least count of linear scale of screw gauge =2x0.025cm=0.05m 2 0.05 0.1 1pitch cm cm mm

1 0.01100mmLeast Count mm

Hence (C ) is correct .

11. Conceptual

12. from LCAM

20 0 0I I Mr 0 0

20

II mr

2

0 02

0

rr r

I dVa r VI mr dr

On integrating 2rV

And t fV R from LCAM 1f and 1tV

2 2 3g r tV V V

13. Electric field is discontinuous at the locations of charges Hence (B)

14.

1 1 1 1 1 125 20u f

1 1 1 1 10020 25 100

V cm

From O to I intensity increases and then decreases at

x = 90 cm and 110cm intensity is same (d) Radius at

x = 200cm is equal to radius of lens.

15. The no. of turns in the spiral coil. Per u nit radial width ' ' Nnb a

Consider a concentric

ring of radius ‘r’ and radial thickness dr. No. of turns in that thickness N drb a

.


Sec: Sr. IIT-IZ

The flux linked with that ring 2r B

Emf induced 20

2.sinN d tdr r Bb a dt T

2

02 2. cosr N te dr B

b a T T

Total emf = b

a

e

2

20 2 2cosb

a

B N t r drb a T T

3 320 2 2cos

3

b aB N tb a T T

2 22

0 02 2 2cos cos

3

b ab a tE B N E tT T T

Current through the coil EiR

0 2cosE

i tR T

Current will be max when 30, , ,2 2T Tt T ……

16. At a given temperature the average Kinetic energy per molecule is given (f/2) kT

which is same for all Diatomic gases hence option (A) is correct. RMS velocity of gas

molecules at same temperature is inversely Proportional to the square root of molar

mass of gas hence option (B) is correct.

. From gas law pv=NKT we

Can see that option (D) is also correct.

17. Following from the graph in the question O<i<1 resistance of diode 5DR for 1i

there after

0. 0DV RI

(dynamic). Consider option (a).

When I = 1 A; 5DR ; and 5 10BV i = 15 volt

There after slope of 101

BV

( since 0DR )

30 15 15 102.5 1 1.5

So option (a) correct.

Option (b) slope 2 10Vi

from graph so it is correct


Sec: Sr. IIT-IZ

Option c) when 215 , 5 , 10B DV V V V V V till this value of ,B DV V is linear there after 5DV V (

remains constant) option ( c) is correct

Option (D) when 215 ; 5 ; 10 .B DV V v V V V

There after for every 5V rise of ,BV 5 v rise in 2V appears so finally

when 230 , 25BV V V V

Option (d ) is correct.

18. Kmax = E – W

A AT 4.25 W

B A BT (T 1.50) 4.70 W

WB–WA = 1.95 eV

de–Broglic wavelength

h

2km

1k

B A

A B

KK

TA = 2 eV

WA = 2.25 eV WB = 4.2 eV

19. Conceptual

20.


Sec: Sr. IIT-IZ

CHEMISTRY

24. Key: 4

Solution : Form A: 3 4Cr NH Cl Br cl

Form B : 3 24Cr NH Cl Br

X= 1

Y = 0

Z = 3 Total 4

25. Key: 3

Solution : 4NO HSO

Isoelectronic with 2N B.O =3

26. ANS: 0

Sol: 'r

R is Identical to , , 12 2 2 3x y zp p p n

,Y 21 ,2 ,3 3S S S n

1 2 0n n .

27. ANS; 5 0 28.5 28500G kj j

28500 0.2996500cellE

0.059 0.29HcellE Xp 5Hp

28. ANS: 1

Sol: 2m , mole =0.2g Gmw=100

0.279=1.5x1.86x100

x x10 1.5i

00.5 c

.

30. Key: A,B,

34. a) Covalent chlorides do not give test b) Due to more solubility product some PbCl2 will remain in solution and black ppt

will be formed in second group also during qualitative analysis c) 3

4Cu CN

d) rosy red ppt

36. key: abcd


Sec: Sr. IIT-IZ

Solution : NCERT

37. Ans : BCD

SOL : CONCEPTUAL

39. Key: A-pr; b-r; c-pt; d-qs

Solution : Given processes are involved in their extraction processes.

40. Conceptual

MATHS

41. 1( ) ( )f x f x x

42. (i) 111 2 ,

11x y y k

2x y is divisible by 11

(ii) 213 2

13x y y k

x – 2y is divisible by 13

6 12 111

x y 6 1

11x y

6 12 1

13x y

6 13 1

13x y y

6 113x y

.

6 143x y k

6 6 143 5 138 5 5x y k y k y k

k y is divisible by 6

6 6 138 5 6x y k 28x y

43. 2 2211 10 11r r

42 100

5021

r

r

44.

2017

1 40350 1

xI dxx

=

1 2017 2017

4035 40350 11 1

x xdx dxx x

Put 1xt


Sec: Sr. IIT-IZ

2017 2017

1 4035 40350 1 1

x tI dx dxx t

1 2017 2016

1 240350

( )1

x xI dx Ix

1

2

1II

45 Here for x = 1, f(x) is zero which lies in [-1, 1) so if f(x) does not contain any value in

[-1, 1), (x-1) should be cancelled 1x k has a root 1 1 1 0k .

k = 0

For k = 0, 1 1 [1, )1

xf x xx

46 1 2 21 2 2 1 0( ) ....n n n

n n nlet f x a x a x a x a x a x a

11 1'( ) 1 ....n n

n nxf x na x n a x a x

22 1 21 1'( ) '' 1 ...n n

n nxf x f x n a x n a x a

'' ' 0af f

0, ' 0, ''( ) 0f f f

min 3value of k is 47 Let P denotes the chances of single bacteria to die, then

1 1 1. . . .4 2 4

P P P P P P

3 2 22 4 1 0 1 3 1 0P P P P P P

3 132

13 3 5 131 12 2

P

P

48. MULTI CORRECT

49. conceptual

50. 2 3 101 ..... 0a

10 2 9 3 3 4 7 5 6, , , ,

2 3 4 5Reso 6 7 2 8Re


Sec: Sr. IIT-IZ

6 5 10 7 4 2 9 8 31 1 1Re 12 2 2

(b) 11

2 101 ......1

x x x xx

2 4 6 8 10 12 14 15 18 20x x x x x x x x x x

= 2 4 6 8 10 3 5 7 9x x x x x x x x x x

10

1

k

k

x

Put x = 7

77 10

71

1 01

k

k

(c) put x=1 11 310

1

1 1 11 1

k

k

i iii i

(d) put x= 1110

1

1 01

k

k

51. conceptual

52. : 2 3 2 2 32x ydy x x y dx xy dx y dx

2 2 3 2 2 3x ydy y xdx x dx x ydx xy dx y dx

xy 2 2xdy ydx x x y dx y x y dx

2

2 1 1

y yddxx xxy y

x x

Let 1yx 2 1 1

tdt dxxt t

2

1 1 1 log4 1 4 1 2 1

dt dt dt x ct t t

log 41 2 log

1 1t x kt t

53.

12

2sin1

xx

1

1

1

( 2 tan ), 12 tan , 1 1

2 tan , 1

x xx x

x x

1 12 tan 5 tan ,x f x g x x x


Sec: Sr. IIT-IZ

1for x

1. . 3tan 1i e x x for x

54. Let A a bc d

, then a bc d

a cb d

= 1 00 1

2 2 1a b

ac+bd=0 2 2 1c d

det 1

11a b

A Ic d

ad a d bc

55. Conceptual

56.

consider the function f(x)= 3 21 3 2 5xe x x xx

57. The parabolas are 2 2 2 2 2 24sin ( sin ) 4cos ( cos )y x and y x , hence the locus is

2 2cos sin 0 1 0x x .

58.

59. Key: (A)-p,q (B)-p,q (C)-p,r (D)-p,q,s

60.

Key: (A)-s (B)-r (C)-q (D)- r

(A) tan tan 3A C sin sin 3cos cosA C A C

(B)

Similarly,


Sec: Sr. IIT-IZ

So,

But (using )

Also, in an acute angled

(Using )

But

So,

Therefore,

So,

The minimum value of is 8.

narayana iit academy - narayanadwarka.files.wordpress.com · 06.05.2018 · narayana iit academy...

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