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Narayana IIT Academy 02-11-14_Sr.IIT-(IZ3) SPARK_JEE-ADVANCED_(2013_P1) PTA-3

Sr.IIT-(IZ3)PAPER - I_Solutions

NARAYANA IIT ACADEMY INDIA

Sec: Sr.IIT-IZ3-CO-SPARK PTM-3 Dt: 02-11-14 PAPER - I

KEY SHEET PHYSICS

1 A 2 B 3 B 4 A 5 C 6 C 7 D 8 B 9 C 10 C 11 ABCD 12 BD 13 CD 14 AB 15 BC 16 2 17 2 18 1 19 9 20 2

CHEMISTRY 21 D 22 C 23 D 24 B 25 C 26 C 27 D 28 B 29 B 30 B 31 A 32 ABD 33 ABC 34 ABD 35 AD 36 5 37 1 38 4 39 8 40 3

MATHS 41 C 42 A 43 B 44 D 45 A 46 D 47 B 48 C 49 D 50 A 51 AB 52 BC 53 AB 54 ACD 55 AB 56 2 57 9 58 1 59 3 60 6

SOLUTIONS

PHYSICS Section-1

1 Conceptual

2 Conserving the momentum of two balls before and after collision in horizontal direction

1 6 (1 1)

3 /v

v m s



Now at the instant of maximum deflection ball will be moving horizontal with same speed as of trolley

(i.e, velocity of balls w.r.t. trolley will be 0)

Let it be 0v

02 3 (4 2) v

Now conserving energy

2

2 2

1 2 3 2 10 1.5(1 cos )2

1 12 42 2

v v

0cos 0.8 37

3 conceptual

4 T cos =mg

T’-mg 2

cos 0mvr

(since v=0)

2

2

' cos

' 3cos2

T mg

TT

' 34

TT

5 For the decreasing speed of the disk, there are two accelerations n ta and a as shown



6 2 21 1

2 2mv I mgh

22

21 1 22 2 5

mR vmv mghR

or

Clearly h becomes free from mass of the sphere 7 The angular impulse of the system about O=0 0 8 For rotational equilibrium ,

0 0

Or 2 1sin cos 03 3 2

lmg l mg

Or 1tan4

9 'mvR I mv R

Or 2 ' '

2mR vmVR mv R

R

' 2 / 3v v

10 22 / 2d l l

2 2

2

3 12ml mlI md

253

ml

SECTION-2 11 Graph (a) indicates two displacements at a given time,which is impossible

Graph (b) indicates two velocities at given time, which is impossible

Graph (c) indicates speed can be negative , which is impossible



Graph (d) indicates distance-travelled increases then decreses, which is impossible 12 For collision position of A = position of B

R distance travelled by a distance travelled by B

212

2

R vt vt at

Rta

they collide after time 2 Rta

2B

B

v v at v aRv v

13 2

2,/ 2

f fR mgR ga g andm I I mR R

00v at and t From above equations, we get

0av

Here a

becomes free from

2 2 20

1 1 1 tan2 2 2

W K mv I I cons t

14 sin sinta r k l kl

Also sindv kldt

Or sindvv klds

Or sindvv klRd

Or 0 0

sinv vdv kl d

R

Or 2

2 1 cosv klR

15

(b,c) if cmv is the velocity of C.M , then cmv v R For no slipping , , 0cmv R v



SECTION-3

16 From CWF theorem, W E

Work done by gravity singW mgx

cosf kW mg x

Work done by spring force

2

2

2

12

1sin cos2

1 02

s

k

W kx

W mgx mg x kx

E mv

Or 2 21 1sin cos .....(1)2 2kmv mgx mg x kx

17 We have

2 2

,x y yv v vR T

g g

After first collision

''x x

y y

v evv v

Distance covered before 2nd collision 1 .'2

x yx

ev vTd vg



After second collision

'' ''' '

x x

y y

v vv ev

2

1 2

1 2

2

2 '' '' 2

(1 2 )

22 2

(1 2 ) 1;2 1 0

x y x y

x y

x y

v v ev evd

g gev v

d d eg

v vRbut d dg

e e e e

1 1 8 1 3

4 4e

Rejecting –ve value e =1/2

18 By newton’s second law 2 1F F ma 1 2F F ma 5 1 2 3N For rotational equilibrium , taking moment of forced about centre of mass, we get

1 02 2l lF Fl y

3 5 0.2 02 2l l

19 2 6

2 12L mL pp

mL

Rod will rotate about its c.m., one half exerts centrifugal force on the other , there fore 2

2 4m LF



20 Given , diameter= 6cm

20 106 y y

Or 20 10 6y y Or 20 10 60y y Or 30 60y Or 2y

CHEMISTRY SECTION-1

23. Enolate ion formation is an RDS. 24. Strong electron releasing group on benzene ring do not undergo Cannizaro reaction.



25.

26. Bridged head carbon (2, 2, 1 system) cannot form the double bond. 28. Intramolecular H bond keeps –NH2 group in the plane in the second compstructure due to this it has double bond character.

N+

N

N+

N+

O

O-O

-

O

HH

O O-

SECTION-2

32. Ph is formed as a major product

33.

SECTION-3

36. are possible 37. Chiral centre of alcohol is not disturbed.



38.

ClCl

Cl

CH3Cl

are gives white precipitate with Aqueous AgNO3

MATHEMATICS SECTION-1

41 Equation of AB is 0 112 0

y x

2 2 0x y PA PB AB .

Thus PA PB is maximum if points A,B and P are collinear. Hence, solving 2 2 0 4 3 9 0x y and x y



We get point 6 17,5 5

P

42 Let d be common difference of A.P, then 2b a d and 6 .c a d Clearly ( ) 3

2 3 0.b a c a

a b c

So, the straight line 0,ax by c passes through (2,-3), which also satisfies 2 2 13x y

43 since 3 .1 4 1 0, 3 sin cos 1 0so

3 1 1sin cos2 2 2

1sin6 2

7 11 4 26 6 3

maximum value of sin is 0.

44 d) Since point , 2 lies inside the circle

22 2 4 0

2 0 ....(1)

And also point , 2 lies in the smaller segment made by the line so that

3 4 2 8 0

16 ......(2)7

(Since centre of circle (0 ,0) and point , 2 lies in the opposite sides of the given line )

From (1) and (2),

45 Let P be 1 2 cos , 2 sin and C is (1,0)

Circum centre of triangle ABC lies on midpoint of PC

2 1 2 cos 1 2 2 sinh and k



2 22 1 2 2h k

2 22 1 1 0h k

2 22 2 4 1 0x y x

46 d)

Let 2 2, 2 2f x y x y gx fy c

2 20, 2 , ,0 2f y y fy c f x x gx c

Since 0, 0f y has y=3 as it’s repeated root

2 6, 9 3 9f c f and c

Also roots of 1 3 1,0 0 1, ,2 4 2

f x are x g c

Twovalues of c are different

47 Given 1 1 1, ,a b c are in A.P

, ,a b c are in H.P

101 101 101 101

2a c ac b

. . .A M G M H M

101 101 1012 0b a c

product of the roots of given equation

101 101 1012 0

1b a c

48 2 31 4

1 4 2 3 4 1 3 2

1 1 1 1;a aa a so ora a a a a a a a

4 3 2 1

1 1 1 1 ....(1)a a a a



2 3 1 4

2 3 1 4

3( ) ;a a a aAlso soa a a a

3 2 4 1

1 1 1 13 ...(2)a a a a

From 1 and 2

2 1 3 2 4 3

1 1 1 1 1 1a a a a a a

1 2 3 4

1 1 1 1, , . .so and arein H Pa a a a

49 Real part of z = constant (say k)

2 02

z z k z z k

Writing in standard form we get ( ) 0z z ,where and are arbitrary real numbers. 50 0 0 0 0 0sin18 cos18 2 sin (45 18 ) 2 sin 63 0 0,sin 63 sin 45Now 0 02 sin 63 1. , 2 sin 63 2Also

02 sin 63 1.So For point of intersection

sin 1 (4 1)2 2 2x x n

4 1,x n where n I SECTION-2

51 We have

11 21

1 2

11 21

1 2

22 1 122

zz z iz i z zzz z iz i

z z

1

1 2

2zz z

is purely real

1 2

1

z zz

is purely real 2

1

zz

is purely real.

52 The given points are collinear if

2 2 1 2 2 1

1 2 1 0 2 1 4 2 0 04 6 2 1 4 4 0

k k k kk k k k

k

2 2 1 3 3 1,R R R R R R

4 2 1 4 2 4 2 0k k k

1 2 4 8 4 0k k

1 2 1 0 1 1/ 2k k k or k



53 Solving all the three equations we get

2

2

2

1 (2 ) 22 1 0, 0

(2 ) 4.1 01

1 1

m a mm amm R D

aaa or a

54 The given equation is 2 2sec ( 2) 1 0a b a 2 2tan ( 2) 0,a b a which holds if and only if 20, tan ( 2) 0a a b 2tan 2 0b

0, ,2 2

b

( , ) (0,0), 0, , 0,2 2

a b

1 10 ( 0), ( 0)2 2 2

y x y x

and 1 ( 0)2 2

y x

2 , 2 ,2y x y x y x 55 Here circle equation is 2 2 2 sin (cos 1) 0x y x So sin will be defined for sin 0 [0, ] …(1) Also ,length of intercept on x-axis 22 2 sin cos 1 2g c sin cos 0

54 4 ……(2)

From (1) and(2)

,4

Section-3 56 Let circle 2 2 2 2 0.....( )x y gx fy c A It is passing through (1,t),(t,1) and (t,t) Then 21 2 2 0 .....t g ft c i



2 1 2 2 0 ......t gt f c ii

22 2 2 0 .......t gt ft c iii (ii)-(i)and(iii)-(ii) Then 22 ( 1) 2 (1 ) 0 0 1 2g t f t or g f and t f

( 1)2

tf g

From (iii), 22 ( 1) ( 1) 0 2t t t t t c c t From (A), 2 2 2 2( 1) ( 1) 2 0 ( ) 2 0x y t x t y t x y x y t x y , Which is of the form 0.S L Hence always pass through points of intersection of

2 2 0 2 0.x y x y and x y On solving we get 1 1. , 1, 1x and y So a b 57

Common chord of both the circle is x=1 Now ,we have to find the ratio of areas of equilateral triangles ANB and CQD. Now in triangle OPN, ON = OP cosec 030 2 2 4. Area of triangle NAB

12

MN.AB =MN. AM=MN. MN tan 030

= 2 0 2 1 9( ) tan 30 (4 1)3 3

ON OM sq. units.

Now in triangle NLQ 0cos 30 4 3.NQ NL ec Since area of triangle

0

1 . .2

. tan 30

CQD QM CD QM CM

QM QM

2 0 2 1 57 24 3( ) tan 30 (3 4 3)3 3

MN NQ sq units.

So, ratio of area of triangles 57 24 39

58 The given circle 2 2, 6 0........( )S x y x y x y i

Has centre at 1 1,2 2

C



According to the required conditions, the given point ( 1, 1)P must lie inside the given circle

i.e 22( 1, 1) 0 . .( 1) 1 1 1 6 0S i e

i.e 2 2 0 . 2 2 0i e i.e 1 2 ......( )iii And also P and C must lie on the same side of the line , 2 0 ......L x y x y iii

i.e 1 1, 1, 12 2

L and L

must have the same sign

Now since 1 1 1 1, 2 02 2 2 2

L

we have 1, 1 1 1 2 0 . 1L i e Inequalities (ii) and (iv) together the permissible value of 1 1as

59 We have 0

(1 )n

n n rr

rx C x

Integrate under the limit 0 to x

Then 0 0 0

(1 )nx x

n n rr

rf x dx f C x dx

1 1

0

(1 ) 11 1

n n rnr

r

x C xn r

Integrating again between the limits 0 and x

1 1

0 0 0

(1 ) 11 1

n n rnx xr

r

x C xf dx f dxn r

2 21 (1 ) 11 2 1 2

n n rrx C xx

n n r r

Integrating again between the limits 0 and x

3 21 1 (1 ) 1

1 2 3 2

nx xxn n n

3

0 ( 1)( 2)( 3)

n rnr

r

C xr r r

Put 1x to both the sides, we get



1 1 1 111 2 3 2n n n

3

0

( 1)( 1)( 2)( 3)

n rnr

r

Cr r r

3

0

1 ( 1)2( 3) ( 1)( 2)( 3)

n rnr

r

Cn r r r

0

3! ( 1) .( 1)( 2)( 3)2( 3)

3!

r nnr

r

Cr r rn

30

3! ( 1) .2( 3)

r nnr

rr r

Cn C

30

3!( 3) ( 1) 32

nnr r

rr r

CnC

60 From the given equation we have,

2sin cos 2sin cos2 2 2 2

x y x y x y x y

sin cos cos 02 2 2

x y x y x y

sin 2sin sin2 2 2

x y x y

Which holds if either sin 0 sin 0 sin 02 2 2

x y x yor or

Also since 1 1, 1.x y x y So, the r equired solution is Either 0 0 0x y or x or y If 0, 1 0, 1x y and if y x

And if 102

x y y x y x x

So that the required pairs ,x y are

1 1 1 10, 1 , 1,0 , , , ,2 2 2 2

which are 6 number

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