2013 PAPER-I€¦ · 2013_PAPER-I Master JEE CLASSES JEE-ADVANCE-2013-P1-Model Kukatpally,...

2013_PAPER-I

Master JEE CLASSES Kukatpally, Hyderabad.

JEE-ADVANCE-2013-P1-Model Max. Marks: 180

IMPORTANT INSTRUCTIONS: 1) This booklet is your Question Paper. 2) Use the Optical Response Sheet (ORS) provided separately for answering the questions 3) Blank spaces are provided within this booklet for rough work. 4) Write your name, roll number and sign in the space provided on the back cover of this booklet. 5) You are allowed to take away the Question Paper at the end of the examination. OPTICAL RESPONSE SHEET: 6) Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the corresponding place on the Candidate’s sheet. 7) The ORS will be collected by the invigilator at the end of the examination. 8) Do not tamper with or mutilate the ORS. Do not use the ORS for rough work. 9) Write your name, roll number and code of the examination center, and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your roll number. DARKENING THE BUBBLES ON THE ORS 10) Use a BLACK BALL POINT PEN to darken the bubbles on the ORS. 11) Darken the bubble COMPLETELY. 12) The correct way of darkening a bubble is as: 13) The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way. 14) Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or “un-darken” a darkened bubble.

Time: 03:00hour’s IMPORTANT INSTRUCTIONS Max Marks: 180 PHYSICS:

Section Question Type +Ve Marks

- Ve Marks

No.of Qs

Total marks

Sec – I(Q.N : 1 – 10) Questions with Single Correct Choice 2 0 10 20

Sec – II(Q.N : 11 – 15) Questions with Multiple Correct Choice 4 -1 5 20

Sec – III(Q.N : 16 – 20) Questions with Integer Answer Type 4 -1 5 20

Total 20 60 CHEMISTRY:


- Ve Marks

No.of Qs

Total marks



Sec – III(Q.N : 36 – 40) Questions with Integer Answer Type 4 -1 5 20

Total 20 60 MATHEMATICS:


- Ve Marks

No.of Qs

Total marks



Sec – III(Q.N : 56 – 60) Questions with Integer Answer Type 4 -1 5 20 Total 20 60

JEE-ADVANCE-2013-P1-Model

space for rough work

PHYSICS: Max. Marks: 60 SECTION - I

SINGLE CORRECT ANSWER TYPE This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. A hemisphere is lying on a surface as shown. For small oscillations of this without

slip, the time period is (centre of mass lies at a distance of 38R from centre of curvature

where R is radius)

A) 26215

Rg

B) 1328

Rg

C) 26225

Rg

D) 2

of curvature is at origin as shown. The acceleration due to gravity is given by

2g x j

units. For small oscillations of the wire its time period is given by


2. A semicircular uniform wire of radius R= can rotate freely about x-axis. The centre

y

x

A) 3 B) 2 C) 6 D)

3. In the figure are shown four tunnels ACB, ADB, AEB and AFB, which are supposed

to be bored through the earth between two places A and B. The tunnel AFB runs

straight from A to B and the point C is the center of the earth.

A particle released from place A reaches the place B in time intervals TC, TD,TE and TF

respectively. Assume that the particle does not collide with the walls of the tunnel

anywhere and passes the corner smoothly. Which of the following relations do these

time intervals bear?


C

A B BA

CD

BA

C

E BA

C

F

A) TC = TD = TE = TF B) TC = TF> TD = TE

C) TC = TF< TD = TE D) TE> TC = TF> TD

4. If a smooth bead of mass m oscillates on the parabolic surface 2Y kx , which is in a

vertical plane, its time period of small oscillations is :

m

O x

y

A) 2kg

B) 2 kg

C) 22 km

D) 2 km


Rm

a

A) 2 21

2g a

R B) 1

2gR

C) 12

g aR D) 1

2g a

R

6. In the given figure, the block is attached with a system of three ideal springs A,B,C.

The block is displaced by a small distance x from its equilibrium position vertically

downwards and released. T represents the time period of small vertical oscillations of

the block. Then (Pulleys are ideal)

5. A smooth wedge is having a cylindrical cut of radius R. A bead of mass mis placed

inside the wedge. If the wedge moves with an acceleration ‘a’. Find the frequency of

small oscillation of the bead about the position of stable equilibrium of the bead

relative to the wedge


m2K

K

B

K

A

C

A) 1122

mTK

B) The deformation of the spring A is (2/7) times the displacement of the block

C) The deformation of the spring C is (1/7) times the displacement of the block

D) The deformation of the spring B is (4/7) times the displacement of the block

7. The potential energy of a particle of mass 1 kg in motion along the x-axis is given by

U = (10 – 10 cos 6x) J. The period of small oscillations is (2 10)

A) 31 s B)

23 s C)

32 s D)

41 s


8. a solid uniform cylinder of mass M performs small oscillations in horizontal plane if

slightly displaced from its mean position as shown in figure. Initially springs have its

natural lengths and cylinder does not slip on ground during oscillations. If force

constant of each spring be K, find time-period of oscillations

A) 32

mK

B) 3mK

C) 223

mK

D) 32 mK

9. In a uni-dimensional potential field the potential energy of a particle depends on the

co-ordinate x as 0 2 cosx U axU , 0U and a are constants. Find the time period of

small oscillations that the particle performs about the equilibrium position.

A) 20

2 mU a

B) 02

ma U C) 2

0

22 mU a

D) 20

mU a


displaced by an angle 6

shown in fig.) and released from rest. Then the frequency of oscillation of ball B is

A) 2

B) 1

C) 3

D) 4

10. Two identical balls A and B, each of mass 0.1kg, are attached to two identical

massless springs. The spring mass system is constrained to move inside a rigid smooth

pipe bent in the form of circle as shown in figure. The pipe is fixed in a horizontal

plane. The centres of the balls can move in a circle of radius 0.06m. Each spring has

a natural length of 0.06 m and spring constant 0.1N/m. Initially, both the balls are

rad with respect to the diameter PQ of the circle (as


SECTION - II MULTIPLE CORRECT ANSWER(S) TYPE

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 11. A simple pendulum has a bob of mass m and swings with an angular amplitude .

The tension in the thread is T. At a certain time, the string makes an angle with the

vertical

A) cos ,T mg for all values of

B) cos ,T mg only for

C) T mg , for 1 1cos 2cos 13

D)T will be larger for smaller values of

surface as shown in the figure. Just as the oscillating object reaches its extreme

position, another object of mass 2m is dropped on to oscillating object, which sticks to

it. For this situation mark out the correct statement (s)

An object of mass m is performing simple harmonic motion on a smooth horizontal .


user

Typewritten text

12.

A) Amplitude of oscillation remains unchanged

B) Time period of oscillations changed

C) The total mechanical energy of the system does not change

D) The maximum speed of the oscillating object changes

13. A simple pendulum consists of a bob of mass m and a light string of length l as shown

in the given figure. Another identical ball moving with the small velocity 0v collides

with the pendulum bob and sticks to it at t = 0. For this new pendulum of mass 2m,

mark out the correct statement (s) ( is angle made with vertical)


A) Time period of the pendulum is 2 lg

.

B) The equation of motion for this pendulum is 0 sin2

v g tlgl

.

C) The equation of motion for this pendulum is 0 cos2

v g tlgl

.

D) Time period of the pendulum is 22 lg

.


14. Two blocks connected by a spring rest on a smooth horizontal plane as shown in the

given figure. A constant force F starts acting on block 2m as shown in the figure.

Which of the following statements are not correct ?

A) Length of the spring increases continuously if 1 2m m .

B) Blocks start performing SHM about centre of mass of the system, which moves

rectilinearly with constant acceleration.

C) Blocks start performing oscillations about centre of mass of the system with

increasing amplitude

D) Acceleration of ‘m2’ is maximum at initial moment of time only

15. A horizontal platform with a mass ‘m’ placed on it is executing SHM along y-axis. If

the amplitude of oscillation is 2.5 cm, the minimum period of the motion for the mass

not to be detached from the platform is 210 /g m s


A) 10 s

B)

10s C)

10s D) 2

10s

SECTION - III INTEGER ANSWER TYPE

This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive). 16. A uniform rod of mass m and length L is hinged at one of its end with the ceiling and

other end of the rod is attached with a thread which is attached with horizontal ceiling

at point P. If one end of the rod is slightly displaced horizontally and perpendicular to

the rod and released. If the time period of small oscillation is 2 sin θ2πx g . Find x.


17. One end of an ideal spring is fixed to a wall at origin O and axis of spring is parallel to

the x axis. A block of mass m =1 kg is attached to the free end of the spring and it is

performing SHM. Equation of position of the block in coordinate system shown in

figure is 10 3sin 10x t , where ‘t’ is in seconds and x in centimetres. The force

constant of spring is 4n2. Where n =....

. 18. A uniform plank of mass m, free to move in the horizontal direction only, is placed at

the top of a solid cylinder of mass 2 m and radius R. The plank is attached to a fixed

wall by means of a light spring of spring constant k. There is no slipping between the


cylinder and the plank, and between the cylinder and the ground. Find the time period

of small oscillations of the system is xmk

where x =....

19. In the given diagram the threads, spring and pulleys are mass less, no friction

anywhere

m

m

K

Find the angular frequency of the system (In rad/sec)

[given K = 20 Nm–1 and m = 4 kg] space for rough work

20. Consider a hypothetical field in which force on any particle at position ‘x’ varies

according to the relation 0x ˆF F 1 i4

where F is in newton, x is in meter and F0 is a

positive constant. A particle of mass m is released from rest from x = 12m at t = 0 in

this field. If 0

m2F

= 6 second and then find value of ‘t’ in second when the particle

reaches the origin first time.


CHEMISTRY: Max. Marks: 60

SECTION - I SINGLE CORRECT ANSWER TYPE

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

21. + H2 Ni,high temp.highpressure

(A). Which of the following can be isolated as the product

of this reaction.

2 3CH C CH CH ?

A) 2 2 4Cu Cl NH OH B) dil alkaline 4KMnO solution C)

A) Toluene with 2Cl in the presence of 3FeCl B) Benzene with 3CH Cl in the presence of 3AlCl C) Toluene with 2Cl in the presence of sunlight D) Benzene with 2Cl in the presence of 3FeCl

22.

A) B) C) D)

Which of the following reaction can be used to differentiate CH2 CH CH CH2

&

Br2 water D) HBr Benzyl chloride can be obtained as one of the products prepared by reacting


user

Typewritten text

23.

A) B)

C) D) All have equal Resonance energy per ring Cl

O2N NO

2 2Br H O

2eqPh C CH oduct is

A) Br

CH

B)

CH

Br

24. Which of the following will have maximum resonance energy per ring?

25. 2

Which IUPAC name is correct for given compound?

A) 1-chloro-2,4-dinitrobenzene

B) 4-chloro-1,3-dinitrobenzene

C) 2,4-dinitrochlorobenzene

D) 2-chloro-1,5-dinitrobenzene

26. Pr ?


C)

CHBr2

O

D)

H

O

H

27. 3 ;HCl HIleq major major

CH C CH A B then find A and B in the reaction sequence?

A) A is

3 2CH C CH

Cl , B is

3 2CH CH CH I

Cl

B) A is

3CH CH CH

Cl , B is

3 3CH C CH

Cl

I

C) A is

3 2CH C CH

Cl , B is

3 3CH C CH

Cl

I

D) CH3 CH=CHClA= , B= CH3 CH2 CH

I

Cl


28.

CH =CH2

KMnO4Br2A B Compound A and B respectively are :4CCl

A) o bromo styrene, benzoic acid

B) p bromo styrene, benzaldehyde

C) m bromo styrene, benzaldehyde

D) styrene dibromide, benzoic acid

3AlCl forms

CH3

CH3O

which on reaction with 'B' forms

CH3

CH3

'A' and 'B' are:

29. Benzene on reaction with 'A' and Anhydrous


A) 3 2CH CH CH C N

3CH

& Zn HgHCl

B) 3 22

Zn HgCH CH CH COCl &

HCl

C) 3 3 2

Zn HgCH C CH COCl &

HCl

O

O

CH3CH3

& Zn Hg

HCl(D)

30. Reaction-I : NO2r1C6H6

Reaction-II : NO2r2C6D6

A) rate 2 1r r

B) rate 1 2r r

C) rate 1 2r r

D) 1 2&r r are not comparable


SECTION - II MULTIPLE CORRECT ANSWER (S) TYPE

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 31. Hydrogenation of the following compound in the presence of poisoned palladium

catalyst does NOT give:

H

HCH3

H3CH C C

A) an optically active compound

B) an optically inactive compound

C) a racemic mixture

D) a diastereomeric mixture


A)

2N .Cl

3 2H PO

COONa

+ NaOH CaOB)

OH

+ Zn dust C)

SO3H

D)dil.HCl,

33. Which of the following reaction(s) give alkylation product:

A) +

OH

H

B)

+ H

C)

NO2

+ 3 2CH CH Cl 3AlCl D)

+ 3AlClMe3C Cl

O

32. Which of the following will form Benzene as final product?


(Hint: Suppose in one structure = 1,2-dibromo, then in other structure also it should be 1,2-dibromo)

A) K and L B) M and N C) L and M D) L and N

A) O O (Rotational energy barrier)

B) (Dipole moment)

C) Friedel crafts reaction fails for nitrobenzene

D) 6 6 6 6 6 6C T C D C H (rate towards iodination)

34. Out of the following four structures given, which structures when given IUPAC name will have bromine atoms located at same IUPAC numbering in there respective IUPAC names.

35. Identify the true statements from following?



This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive). 36. How many out of the following reactions, can give aromatic product?

1) S2CH CH M

2) 2 2CH NCH CH N

3) RedHOtFe(or)Cu tube3CH CH O

4) m CPBACH CH P

37. CaC2

H2OA

red hot iron873 K

B

O

O

OAlCl3

C(i) Zn-Hg/HCl

(ii) H3PO4W

Find the number of oxygen atoms in W.


39. Examine the structural formulas shown below and find out how many compounds can

not give Friedel Crafts reaction.

CH3 NH2HO F

OHO3NH C N

, , , , , , ,

38. Among the following, the number of aromatic compound(s) is


4KMnO

CH3 CH3CH3

CH3

CH3CH3

CH3CH3

CH=CH 2PhPh

Ph

C CH

, , ,

, , ,

40. Examine the structural formulas shown below and find out how many compounds will

show side chain oxidation reaction with acidic


MATHS: Max. Marks: 60 SECTION - I

SINGLE CORRECT ANSWER TYPE This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

41. In a triangle if 1 2 3r > r > r , then

A) a>b>c B) a<b<c C) a>b and b<c D) a<b and b>c

42. If in a triangle 2

1 3

rr = ,r r

where r,r1,r2,r3 have their usual meanings, then

A) 090A B) 090B C) 090C D) None of these

43. If in aABC 2 1 3 1 2 3r - r r - r = 2r r , then the ABC is

A) Acute angled B) obtuse angled C) equilateral D) right angled

44. The product of the distances of the incentre from the angular points of a ABC is:

A) 4 R2 r B) 4 Rr2 C) a b c Rs

D) a b c sR

45. A circle of radius 4 unit is inscribed in an equilateral triangle ABC. Sides AB and AC

of triangle are produced and a circle is drawn touching the sides BC and produced part

of sides AB and AC. The area of the circle drawn is

A) 16π sq. unit B) 64π sq. unit C) 144π sq. unit D) 256π sq. unit


46. The sides of a triangle ABC satisfy the relations a +b-c =2 and 2ab-c2 = 4 Then the

radius of the circle escribed opposite to the angle A is

A) 1 B) 3 C) 32

D) 13

47. In triangle ABC, if 1 2 3r = 3,r = 10,r = 15 , then c=

A) 5 B) 12 C) 13 D) 13/2

48. In a le1 2 3Δ ABC,r = 2r = 3r .D is mid point of BC and ADC = θ then cosθ =

A) -7/25 B) 3/25 C) 125

D) 1225

49. Let H be the ortho centre of leΔ ABC, then angle subtended by the side BC at the

centre of in circle of leΔ BHC is

A) 2 2A B)

2 2B C

C) 2 2

B C D)

2 4B C

50. Three Circles with radii a,b,c touch one another externally tangents at contact points

meet at P then the distance of P to the contact point of any two circles is

A) abca b c

B) abca b c

C) a b cabc D) a b c

abc


SECTION - II MULTIPLE CORRECT ANSWER(S) TYPE

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

51. Let

1 2 3 1 2 3A A A r r r B) 1 2 3 4A A A R r

C) 1 2 3

1 1 1 1A A A A

D) 2

1 2 3 1 2 3

1 1 1. .s

A A A r r r

52. Three equal circle each of radius r touch one another. The radius of the circle touching

all the three given circles DEF is

A) 2 3 r B) 2 3

3r

C) 2 3

3r

D) 2 3 r

53. If a right angled ABC of maximum area is inscribed within a circle of radius R,

represents area of triangle ABC and 1 2 3r,r ,r ,r represent inradius and exradii, and s is

the semi perimeter then

A) 2R B) 1 2 3

1 1 1 2 1r r r R

C) 2 1r R D) 1 2s R

A, A1 , A2 , A3 be the areas of the inscribed and escribed circles of ABC, then:

A)


54. In a triangle ABC, C2

, then

A) r3 = r2 + r1 + r B) 2 2 2 2a b sin A B a b sin A B

C) 2cosC cos B 2cos A c bc b a ab ac

D) r + 2R = s

55. In a leΔ ABCwhich of the following results are true.

A) 1 2

3

ab rr rr

B) 2 3

1

bc r r rr

C) 1 3

2

ac r r rr

D) cos cos cos 1 rA B CR


This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive).

56. AD, BE and CF are the perpendiculars from the angular points of a ABC upon the

opposite sides. The perimeters of the DEF and ABC are in the ratio krR

then find

the value of k.


57. With usual notation in a ABC, if R = 1k

r r r r r rr r r r r r

1 2 2 3 3 1

1 2 2 3 3 1

where k has the value

equal to ?

58. In a triangle ABC, a : b : c = 4 : 5 : 6 and the ratio of the radius of the circumcircle to

that of the incircle is K, then the value of 16/K is

59. The value of 1 2 3

b c c a a br r r

is

60. Consider a triangle ABC and let a,b and c denote the lengths of the sides opposite to

vertices A,B and C, respectively. Suppose a=6, b=10 and the area of triangle is 15 3. if

ACB is obtuse and if r denotes the radius of the incircle of the triangle, then find the

value of r2.


KEY SHEET

PHYSICS 1 C 2 C 3 D 4 A 5 A

6 A 7 A 8 A 9 A 10 B

11 BCD 12 ABCD 13 AB 14 ACD 15 B

16 3 17 5 18 7 19 2 20 4

CHEMISTRY 21 C 22 A 23 C 24 C 25 A

26 C 27 C 28 D 29 B 30 C

31 ACD 32 ABCD 33 ABD 34 D 35 ABCD

36 3 37 1 38 5 39 5 40 5

MATHS 41 A 42 C 43 D 44 B 45 C

46 B 47 C 48 A 49 B 50 B

51 ABCD 52 BC 53 ABCD 54 ABCD 55 ABCD

56 1 57 4 58 7 59 0 50 3

Master JEE CLASSES Kukatpally, Hyderabad.

JEE-ADVANCE-2013-P1-Model Max.Marks:180

Page 1

SOLUTIONS

PHYSICS

1. In the position shown torque, sinmgAC mg AC Where C is centre of mass and is angle between AC and AO. For small oscillations AC R OC

3 58 8R RR

2 2 2

25 2 3 58 5 8 8A CR R RI I m MR M M

58 ARI mg I

58 A

mg RI

2 5

8 A

mg RI

2. Let R dθ be an element of wire located at an angular position of θ wrt – ve y axis. When slightly rotated by angle torque due to gravity is d d F r

2 cos .sinMdmg r Rd x RR

2 cosMdR x R

, for small

as sinx R 2

2 2

2

cos sinMR R d

------------------- (1)

But 2I MR ------------------- (2)

Where 2

2

ddt

2 2T

3. Conceptual 4. Conceptual 5.

y

mg

ma

2 2 2 ym y m g aR

Page 2

6. Let the deformation in each spring be 1 2 3,x x and x respectively as shown Let the block be displaced by x. Then 1 2 32 .....x x x x i 3 2 12 ;2 ;2 2 ....F Kx F Kx F Kx ii Also ...eqF K x iii

From equation (i), (ii) and (iii)

42eq

F F F RK K K K

211eq

KK

1122

mTK

1eqF K x Kx

1

211K x Kx

1

211

x x

2

2eq

KxF K x

2211 2

KxK x

2

411

x x

32eqF K x Kx

3

2 211K x Kx

1

111

x x

7. U = 10 – 10 cos 6 x

dUF 60sin 6 360xdx

(for small x)

2360a x 3601

2 1T S3

(A) 8. Net torque, 2 2F R F R or, 22 2 82 R R KRK

Page 3

2

2 22

3 82

dMR KRdt

I

16 3 3, 23 16 2

K m mso Tm K K

9. 2

20, , U cdU so a x xdx m

F

2

2 02

0

, 2U c mTm U c

10. Conceptual 11.

A B

C

l

T

2

cos mvT mgl

2

2 (cos cos )mv mgl

2

cos 2 (cos cos )mvT mg mgl

(3cos 2cos )T mg 3cos 2cos 1

1 1cos (2cos 1)3

12. Period of oscillation changes as it depends on mass and becomes three times. The amplitude of oscillation does not change, because the new object is attached when the original object is at rest. Total energy does not change as at extreme position the energy is in the form of potential energy stored in spring which is independent of mass, and hence maximum; KE also does not change but as mass changes the maximum speed changes.

Page 4

13. The time period of simple harmonic pendulum is independent of mass, so it would be

same as that 2 1/T g . After collision, the combined mass acquires a velocity of 02v ,

as a result of this velocity, the mass (2m) moves up and at an angel 0 (say) with vertical, it stops, this is the extreme position of bob.

2

00

20 2 1 cos2 2

vm mgl

2

20 01 cos 2sin8 2vgl

0 0sin2 4

vgl

If 0 is small, 0 0 00sin

2 2 2v

gl

14. Conceptual 15. Acceleration 2a x

Maximum acceleration 2

2 2 25100

a AT

When block and platform are separated, 10a g

2 2

22

4 1. sec40 100 10

T TT

16. Consider a small angular displacement Torque about the rotational axis

17. Conceptual

18. 2 2 21 1 12 2 2 PE mv kx I

2 2 21 1 32 2 8

mv kx mv

2 247

kE v xm

constant

Page 5

Which gives 2 47

km

or 7mTk

19. If mass m on inclined plane moves by a distance x, the hanging mass moves by distance x/2

Using Energy Equation

2

2 20 0

1 1 1 v xE k (x x ) mv m mg xsin mg U2 2 2 2 2

At mean position 0mgKx mgsin2

Differentiating with respect to time we get

4ka x 05m

4k 4 20 2 rad / sec5m 5 4

20. Mean position (F = 0) is at x = 4 m. At a position l with respect to mean position x = (l + 4)

Acceleration, 0

0

( 4)F [1 ] F4am 4m

ll

Therefore motion is SHM and 2 0F4m

or, time period, 0

4mT 2F

= 2 × 6 = 12s.

x = +12 m is extreme position & x = + 4m is mean position. Therefore amplitude of motion A = (+12) – (+4) = 8m.

Origin is at a distance 4m = A2

from mean position.

Minimum time taken to reach from l = + A ( 1 = 90°) to l = – A2

( 2 = 210°) is

(210) (90)t 12360

= 4sec.

CHEMISTRY

21. Ortho position is less crowded

22. 1 – ynes gives red precipitate with 2 2 4Cu Cl NH OH

23. In the presence of light free radical reaction taken place.

24. Maximum number of electrons participating in resonance per ring.

25. Electron density more on Anisole

26. H2O attack at more substituted carbon.

27. Electrophilic addition taken place following markownikoff rule.

Page 6

37.

W=

O

38. Conceptual

39. Friedel craft’s reaction fails for strongly deactivating groups, aniline.

40. Conceptual

MATHS

41. We have 1 2 3r r r s a s b s c

Dividing each by , then 1 1 1s a s b s c

s-a<s-b<s-c (subtracting s from each) -a<-b<-ca>b>c

42. Given that 2

1 3

rrr r

s a s cs s b

s a s b s s c ab s a b c 2 2 2a b c the triangle is right angled at C.

43.

2 .s b s a s c s a s b s c

2s a s b s a s cs a s b s c s a s b s c

2 2

b as a

28. Conceptual

29. Directive influence of more activating group is deciding the product.

30. Isotopic effect is absent in nitration

31. syn addition takes place converting alkyne to cis-alkene

32. Conceptual

33. Friedel craft’s alkylation

34. Conceptual

35. First ring is aromatic

36. x=2 , y=3 so answer is 5

Page 7

222 2 2 2b a c a s a b a c a s a

2b a c a b c a

2 2 2 22 2 2 2bc ab ac a b c a bc ca ab

2 2 2a b c ABC is a rt. d

44. 1. IA = r3 cosec A2

= 3

A2

4r4 sin

= 34 r . Rr

= 4 r2 R B

45. Here r=4 R=8

Trigonometry

A=2sinA=2(8)sin600= 316 8 32

Now, 11tan 12 3 12

2 3Ar s

Area of drawn circle = 144 square units 46. 2a b c and 22 4ab c

a b c and

2

1

3 24

33 2

rs a

47. 1 30 2.1 1 1 10 3 23 10 15

r

15 2 3 1 13c

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48.

49. H-Orthocentre

4 4 180;

2 2 2 2C B B CS S

50.

Required distance is distance between Incenter and side 51. (a,b,c,d) 2; / ;A r r s Page 9

2 21 1 1 2 2 2; ; ; ;A r r A r r

s a s b

23 3 3;A r r

s a

1 2 3 1 2 3A A A r r r a is true Also 1 2 3 4r r r r R is true 1 2 3 4A A A r r (b) is also true

Now, 1 2 3

1 1 1A A A

21 2 3

1 1 1 1 1 1 1 1. .r r r r r A

(c) is also true

Also 1 2 31 2 3

1 1 1 1 1 1 1 1 1.r r r rA A A

2 3 1 3 1 2

2 3

11.

r r r r r rr r r

2 2 2

1 2 3

1 1 1 1.s b s c s a s c s a s b r r r

1 2 3

1.. .

s a s b s cs a s b s c r r r

2

1 2 3 1 2 2

1 1.. .

s ss a s b s c r r r r r r

hypotenuse is 2R. Then area is maximum when its is isosceles triangle With each side = 2R

1 2 2 2 2 12

S R R

21 2 . 22

R R R

2

2 12 1Rr r R

S R

1 2 3

1 1 1 1 1 2 12 1r r r r RR

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(d) is correct 52. Conceptual 53. For a right angled triangle inscribed in a circle of radius R the length of the

54. r3 = r1 + r2 + r

s c s a s b s

2s a b

2s a b cs s c s a s b

s(s – c) = (s – a)(s – b) s2 – (a + b)s + ab = s2 – cs s(a + b – c) = ab (a + b + c)(a + b – c) – 2ab = 0 (a + b)2 – c2 – 2ab = 0 a2 + b2 = c2 C = /2 C = /2 R = c/2 c = 2R r = (s – c) tan(C/2) = s – c r = s – 2R r + 2R = s Multiplying both sides by abc in option C 2ab cos C + ac cos B + 2bc cos A = c2 + b2

b2 + a2 – c2 + 2 2 2

2 2 2 2 2a c b b c a c b2

2 2a b c C = /2 a2 + b2 = c2 2 2 2sin A sin B sin C 0 (a = 2R sin A) sin(A – B)(sin2A + sin2B – sin2C) = 0 (sin2A + sin2B)sin(A – B) = sin2(A + B)sin(A – B) (A + B = - C) = sin(A + B)(sin2A – sin2B) (a2 + b2)sin(A – B) = (a2 – b2)sin(A + B) (sin A = a/2R

55.

56. Note that DEF is a pedal triangle whose sides are R sin 2A, R sin 2B and R sin 2C. Page 11

ratio =R Aa b c sin2 = 4

2R AR A

sinsin =

2 84

2 2

2

. sin coscos

A A

A = 4 sin A2

= rR

57. Numerator of RHS = 2s .4Rk

and denominator of RHS = s2

(using r1 = s tan 2A etc.)

numeratordeno atormin

of RHS = 4Rk

= LHS = R k = 4

58. We have abcR4

and rs

R abc s.r 4

= 2

abc.s4

=

abc.s4s s a s b s c

But a : b : c = 4 : 5 : 6 (given)

a b c k4 5 6 (let)

a = 4k, b = 5k, c = 6k

1 1 15ks a b c 4k 5k 6x2 2 2

4k 5k 6kR15k 15k 15kr 4 4k 5k 6k

2 2 2

= 3

3

30k 30.8 1615 8 15 10 15 12 7.5.3 7k

2 2 2

= K

16 16 7K 16 / 7

59. 1 2 3

b c c a a br r r

s a s b s cb c c a a b

s a b c s b c a s c a b

= s b c c a a b ab ac bc ba ac bc

= 0 0

60. 1 sin2

ab c

02 2 15 3 3sin 1206 10 2

C Cab

2 2 2 cosa b abc C

Now, rs

22

225 3 36 10 14

2

r

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