phase test answers

8/11/2019 Phase Test Answers

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IITJ INTERNAL TESTPHASE IBATCHES:

ANSWERS IIT MAINS

QP CODE: 120762.1

SR.NO

PHYSICS C.CODE CHEMISTRY C. CODE MATHS C.CODE

1. A P110201 C C110202 C M1101092. C P110403 B C110108 B M1205093. A P110507 C C110706 B M1102034. A P110320 D C110306 C M1205085. B P110508 C C111202 D M1102026. B P110409 D C110803 A M1207017. A P110314 A C111003 A M1208038. D P110405 C C111102 B M1106069. D P110403 B C110103 C M111402

10. C P110310 C C110704 A M11130811. C P110403 B C110305 D M11130312. C P110401 C C111207 B M11140213. B P110504 B C110903 B M11141114. C P110403 C C111006 B M11140315. D P110403 C C111101 C M11140716. B P110320 B C110103 A M11070117. C P110310 C C110702 A M11071418. A P110405 A C110306 B M11071819. C P110326 D C111206 C M11072820 B P110502 B C110902 D M11071221. A P110404 B C111006 D M110822

22. C P110415 B C110202 B M11080223. C P110326 D C110107 A M11070424. A P110320 B C110112 D M11082025. C P110416 A C110703 A M11080926. B P110404 C C111204 A M11082127. B P110404 B C110808 B M11081528. C P110325 C C111006 A M11070229. C P110404 C C110304 B M11082330. C P110318 B C110107 C

Answers & Solutions

Physics1. A (Concept code: P110201)

1. 1 2

1 2

R R

R R

+

+

=A

A

2. C (Concept code: P110403)

2. 1

2

T m(g a)

T m(g a)

=

+


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Two Year CRP(1416)-Phase Test-1_PCM-IITJEE/16


2

3. A (Concept code: P110507)

3. W =0.2

3 12

0.1

F dx (5x 16x )dx 8.1 10 = + =


4.

r

3

45

r

6

ris same in both cases.

5. B (Concept code: P110508)

5. 1

2

5glv

v 4gl=


6. 2 2 2resa a a 2a cos(180 )= + +

= 2a sin2

(180 )

a

a


7. AB =ucos60t

cos30

8. D (Concept code: P110405)8. NQ= (60 + 15) 10 = 750 N.

NR= 150 tan 53 = 200 N

9. D (Concept code: P110403)

9.o o

T w

sin 120 sin 120=

10. C (Concept code: P110310)10. ucos usin gt =


11.2

tan3

= g

O

3g

2

12. C (Concept code: P110401)12. a g=

2 26 0 2 gS= +


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3


13. Ek=4

2

Pdt = 46J

14. C (Concept code: P110403)14. 2Fsin mg =

T = F cos

F

F

15. D (Concept code: P110403)15. > 1


16. T a

= 0 2 1m m

a 2a=


17. tan=

2 2

2

V sin

tan2g

2V sin cos

g

=

18. A (Concept code: P110405)18. If A is pushed with acceleration = g tan, the block B will be at rest wrt A and it has no

slipping tendency.


19. s = 21

gt2

= 21

10 0.2 0.2m.2

= Block haves contact will the floor.


20. Energy cost due to air friction = 21

1 (20) 1 10 182

= 20J.


21. 4g T = oa T

4 & 2gsin30 2a2 2

=

6a = 2g


22. At y = a, x = 2a, dydx

= 1 = 45 af= gcos 45 = g2

.


23. Dist. =

2

2 21(10 2) g (2)2

+



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4

24. tan =V

1V

=

25. C (Concept code: P110416)25. V1cos1= V2cos2


26. k 2 k3 =

k =3

k2


27. Fnet=2 2M a a 2 ma+ =


28. a =Vdv

dx

dv a

dx v= = constant.

29. C (Concept code: P110404)29. W = m geff


30.2v

ar

=

Chemistry

1. C (Concept Code: C110202)

1. ( )4 10 3 4 2P O 6MgO 2Mg PO+

2. B (Concept Code: C110108)

2. ( ) ( )2 2 2 3C 2s 2p e C 2s 2p + 2p3is a half filled electronic configuration.

3. C (Concept Code: C110706)3. All are isoelectronic anions. The ion with least number of protons, has the largest size.

4. D (Concept Code: C110306)

4. ( ) ( )2 5 2 5 2F 2s 2p F 2s 2p F+

z2pz2p z2p z2p

5. C (Concept Code: C111202)5. Let the pressure of the gas in the 9 L vessel be x atm.

P1V1= P2V2x 9 = (x + 1) 6x = 2, The pressure in smaller vessel = 2 + 1 = 3 atm.



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5

7. A (Concept Code: C111003)7. Atomic hydrogen is formed in the hydrogen torch by applying very high voltaget on

hydrogen molecules. The combination of these hydrogen atoms releases a large amountof energy in forms of heat which is sufficient for welding purposes.


8. Moles of HCl =200 0.4

1000

= 0.08

Moles of NaOH = 0.08Mass of NaOH = 0.08 40 = 3.2 g

9. B (Concept Code: C110103)9. 2r = n

or,2 r

= n

10. C (Concept Code: C110704)10. The amount of energy required to remove one electron from a neutral gaseous atom is

called ionization energy.

11. B (Concept Code: C110305)11. Species Bond angle

4NH+ 109o28

2NO+ 180o

3CH+ 120o

4PH+ Less than 109o28

12. C (Concept Code: C111207)12. Let mass of CH4= mass of He = x g

Moles of CH4= x16

and moles of He = x4

Mole fraction of CH4=

x

16x x

16 4

1

5+=

Partial pressure of CH4=1

5800 = 160 mm Hg


13. ( )3 2 2 32Mg N 6H O 3Mg OH 2NH+ +

14. C (Concept Code: C111006)14. The octet of one oxygen atom is not satisfied in (C). Hence, it is the most unstable

structure.



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6

15.

( )4 2 7 2 3 2 22NH Cr O Cr O 4H O N + +

12+ gain of 6e ' s 6+

-6 0loss of 6 electrons

n-factor = 6 E = M6

16. B (Concept Code: C110103)16. The ionization energy of H or H-like species is given as:

2

2

ZI.E 13.6 eV

n

=

17. C (Concept Code: C110702)17. Be and Al are diagonally related elements.

18. A (Concept Code: C110306)


20. B (Concept Code: C110902)20. Be2+(2s02p0) makes four vacant orbitals available for coordination with H2O.


21. 3 2I I I +

2 2 2 2I H O H I H O ++ + +

22. B (Concept Code: C110202)22. Mol. mass of Na2CO3.xH2O = (106 + 18x)

Meq. of HCl in 20 mL solution = 19.8 0.1 = 1.98Meq. of HCl in 100 mL solution = 1.98 5 = 9.9Meq of Na2CO3.xH2O = Meq of HCl = 9.9

( )

W1000 9.9

E

0.71000 9.9

106 18x

2

=

=+

On solving, x = 2


23. The orbitals with n + = 5 are 5s, 4p and 3d which can contain maximum of 2, 6 and 10

electrons respectively.Total no. of electrons = 18

24. B (Concept Code: C110112)24. Spin value = n

n = no. of unpaired electrons.

The number of unpaired electrons present in 2O+ , O2, 2O

and 22O are respectively 1, 2, 1

and zero.


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7

25. A (Concept Code: C110703)

25. ( ) ( ) ( )1 2I.E I.E1 6 o 2 5Na 3s Na 2p 3s Na 2p+ + For I.E2the electron is removed from an ion which has inert gas configuration.

26. C (Concept Code: C111204)26. Excluded volume per molecule of a real gas

= 4 molecular volume = 4 34 r3

= 316 r3


27. NaOH CO HCOONa+


28. 2 2 4 2PbS 4H O PbSO 4H O+ +

29. C (Concept Code: C110304)29. XeF4has square planar(symmetrical) structure.

30. B (Concept Code: C110107)30. Zn(Z = 30) = 1s22s22p63s23p64s23d10

No. of electrons with = 1 is 12.

Mathematics

1. The solution of( ) ( ) ( )

3x 1 x 2 x 4

0x 2

is [ ] { }x 1,4 2 . Hence, the integral solutions

are { }x 1,3,4 .

2.( )

dydy asin t sin t 1 cos t 1 cos tdt

dxdx a 1 cos t 1 cos t 1 cos t sin t

dt

+ += = = = +

3.( )

( ) ( )22log 2 2log x 1

2 x 5 x 1 x 5 x 1 x 5

> + > + > + 2x 3x 4 0 x 1 > < or x 4>

Also, x 1 0 > for log to be defined. Hence, x 4> .

4.

( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( )

2

22

2

2 2

22

1 t 2 2t 2tdy

1 tdy 1 t xdtdxdx 2t y1 t 2t 1 t 2tdt

1 t

+

+ = = = = +

+

5. 4 2 2 31

ab log 6 2ab log 6 log 3 1 log 22ab 1

= = = + =


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8

6.( )7 8

7

dx dx

1x x 1 x 1x

= + +

Put7

11 t

x+ = .

7. ( )2 2

/3 /3 /3 2 2

2 2 2 2/4 /4 /41 sin x cos xdx dx sec x cosec x dx

sin x cos x sin x cos x

+= = +

( ) ( )/3

/4

1 2tan x cot x 3 1 1

3 3

= = =

8.

4 2

3x x x

3 3

1sin

1x1 1 1 1 xx sin x x sin

1 0x x x xlim lim lim 11 1 0 11 x 1 1

x x

++ + +

= = = =++ + +

9.A

tan6 tan42 tan66 tan78B

=

( ) ( ) ( ) ( )tan6 tan 60 6 tan 60 6 tan18 tan 60 18 tan 60 18tan54 tan18

+ + =

( ) ( )tan 3 6 tan 3 181

tan54 tan18

= =

10. ( )1 cos x 1 sin1 sin cos x sin1

11. ( ) ( )22 2 2cos 4 2 2 2cos 2 2 2 cos2 2 2cos2+ + = + = + =

( ) ( )22 2sin 2 sin 2sin= = =

12.3 cos20 sin20 2sin60 cos20 sin20

3 cos ec20 sec 201sin20 cos20

sin402

= =

( ) ( ) ( )2 sin80 sin40 sin20 2 2sin30 cos50 sin 40 2 sin40 sin 404

sin 40 sin40 sin40

+ + + = = = =

13. ( )2212sin 9sin 4 3sin 2 =

14. ( )3 1 1 2 2

cos 0.22 3 2 3

+ =

15. ( ) ( )3sin x.sin3x3sin x sin3x 3 1

.sin3x cos2x cos 4x 1 cos6x4 8 8

=

=


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9

16.5

x4m 3

=+

There is no positive integral value of m for which the value of x is an integer.

17. The smallest value of 2 2 + is the square of the perpendicular distance of the origin fromthe line.

18. Equation of PQ is ( ) ( )y 1 1 x 4 = orx y 5+ = .

The point Q is5 15

,4 4

.

2 25 15 11 2

PQ 4 14 4 4

= + =

Q y = 3x

P (4, 1)

19. After reflection about the line y x= , point

P becomes Q ( )1,4 .After translation through a distance of 2

units, point Q becomes R ( )3, 4 .

OR OR' 5= = 4

tan3

=

The point R is given by

( ) ( )( )5cos 45 ,5sin 45 + +

( ) ( )5 5

cos sin , cos sin2 2

= +

,

where 4sin5

= and 3cos5

= .

R '

y

R (3, 4)

Ox

45o

20. Let P ( ),2 be any point on the curve. Equation of normal at P is given by( )y 2 x = . It passes through the point ( )21,30 . Hence,

( )3

19 30 0 = . Solving, we get 5 = .

21. ( ) ( )1 2 1 2C 3,7 , C 3,0 , r 10, r 3 = =

1 2 1 2C C 7 r r = > + Hence, 4 common tangents.

22. The required circle is the one having (2, 0) and (0, 4) as the end-points of a diameter.


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10

23. Use the formula

1 2 3 1 2 3ax bx cx ay by cy,a b c a b c

+ + + + + + + +

for the

incentre.

C(1, 1)

y

B (1, 1)

A(0, 0)

y = 1

y= x y= x

x

24. ( )2C 0,0 . Equation of AB is given by

1 2S S 0 = , i.e. x y 4+ = .

2 2 2

0 0 4C P 2 2

1 1

+ = =

+

2 2 1 cos 2

4 4 22

= = = =

C1

A

B

C2P

25. Let P ( )h,4 2h be any point on the line 2x y 4+ = . The chord of contact is given by

( )hx 4 2h y 1+ = or ( ) ( )h x 2y 4y 1 0 + = . This represents a family of lines passing

through the point of intersection of the lines x 2y 0 = and 4y 1 0 = , which is1 1

,2 4

.

26. Let the equation of the required circle be 2 2x y 2gx 2fy c 0+ + + + = . According to the givenconditions, we have

2 2a b 2ga 2fb c 0+ + + + = and ( ) ( ) 2 22g 0 2f 0 c p c p+ = =

Hence,2 2 2

a b 2ga 2fb p 0+ + + + = . Replace g by x and f by y to get the equation ofthe locus of the center.

27. The center of the second circle lies on the common chord of the two circles.

28. Solve graphically.

29. ( )2

2 2 2nAB 4AM 4 4 2 8 n2

= = =

Hence, required sum

( ) ( )2 22 8 1 2 8 2 14 8 22= + = + = .

B

M

A

C

30. ( )p q

p q ~q p ~ q ~ q p p ~ q ( )p q T T F F T F TT F T T T T FF T F T T T F


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11

F F T T F F T

p Q p q q p p qT T T T T

T F F T F

F T T F F

F F T T T

QP CODE: 120762.2

SR.NO


1. C P110403 B C110305 D M1113032. C P110401 C C111207 B M1114023. B P110504 B C110903 B M1114114. C P110403 C C111006 B M1114035. D P110403 C C111101 C M1114076. B P110320 B C110103 A M1107017. C P110310 C C110702 A M1107148. A P110405 A C110306 B M1107189. C P110326 D C111206 C M11072810. B P110502 B C110902 D M11071211. A P110404 B C111006 D M11082212. C P110415 B C110202 B M11080213. C P110326 D C110107 A M11070414. A P110320 B C110112 D M11082015. C P110416 A C110703 A M11080916. B P110404 C C111204 A M11082117. B P110404 B C110808 B M110815

18. C P110325 C C111006 A M11070219. C P110404 C C110304 B M11082320 C P110318 B C110107 C -21. A P110201 C C110202 C M11010922. C P110403 B C110108 B M12050923. A P110507 C C110706 B M11020324. A P110320 D C110306 C M12050825. B P110508 C C111202 D M11020226. B P110409 D C110803 A M12070127. A P110314 A C111003 A M12080328. D P110405 C C111102 B M11060629. D P110403 B C110103 C M111402

30. C P110310 C C110704 A M111308


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12

QP CODE: 120762.3

SR.NO


1. A P110404 B C111006 D M1108222. C P110415 B C110202 B M1108023. C P110326 D C110107 A M110704

4. A P110320 B C110112 D M1108205. C P110416 A C110703 A M1108096. B P110404 C C111204 A M1108217. B P110404 B C110808 B M1108158. C P110325 C C111006 A M1107029. C P110404 C C110304 B M11082310. C P110318 B C110107 C -11. A P110201 C C110202 C M11010912. C P110403 B C110108 B M12050913. A P110507 C C110706 B M11020314. A P110320 D C110306 C M12050815. B P110508 C C111202 D M110202

16. B P110409 D C110803 A M12070117. A P110314 A C111003 A M12080318. D P110405 C C111102 B M11060619. D P110403 B C110103 C M11140220 C P110310 C C110704 A M11130821. C P110403 B C110305 D M11130322. C P110401 C C111207 B M11140223. B P110504 B C110903 B M11141124. C P110403 C C111006 B M11140325. D P110403 C C111101 C M11140726. B P110320 B C110103 A M11070127. C

P110310 C

C110702 A

M110714

28. A P110405 A C110306 B M11071829. C P110326 D C111206 C M11072830. B P110502 B C110902 D M110712


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13

QP CODE: 120762.4

SR.NO


1. B P110404 C C111204 A M1108212. B P110404 B C110808 B M1108153. C P110325 C C111006 A M110702

4. C P110404 C C110304 B M1108235. C P110318 B C110107 C -6. A P110201 C C110202 C M1101097. C P110403 B C110108 B M1205098. A P110507 C C110706 B M1102039. A P110320 D C110306 C M12050810. B P110508 C C111202 D M11020211. B P110320 D C110803 A M12070112. C P110310 A C111003 A M12080313. A P110405 C C111102 B M11060614. C P110326 B C110103 C M11140215. B P110502 C C110704 A M111308

16. C P110403 B C110305 D M11130317. C P110401 C C111207 B M11140218. B P110504 B C110903 B M11141119. C P110403 C C111006 B M11140320 D P110403 C C111101 C M11140721. A P110404 B C110103 A M11070122. C P110415 C C110702 A M11071423. C P110326 A C110306 B M11071824. A P110320 D C111206 C M11072825. C P110416 B C110902 D M11071226. B P110409 B C111006 D M11082227. A

P110314 B

C110202 B

M110802

28. D P110405 D C110107 A M11070429. D P110403 B C110112 D M11082030. C P110310 A C110703 A M110809


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14

FIITJEEINTERNAL TESTBatches:Two Year CRP(1416)

IIT JEE 2016

PHASE IPAPER I

ANSWERS

QP CODE: 120763.1

SECTION I (PHYSICS) SECTION II (CHEMISTRY) SECTION III (MATHS)

PART A PART A PART A

Q.N. Ans. Con.Code Q.N. Ans. Con.Code Q.N. Ans. Con.Code

1. C P110320 1. C C110111 1. B M111305

2. D P110320 2. C C110207 2. D M111304

3. B P110404 3. A C110304 3. B M110722

4. B P110414 4. B C110704,

C110703

4. A M110730

5. D P110414 5. C C111102 5. D M110731

6. B P110507 6. C C111203 6. C M110710

7. B P110507 7. C C110809 7. C M110712

8. C P110507 8. B C111003 8. B M110708

9. C P110513 9. C C110103 9. A M110723

10. D P110320 10. B C110305 10. B M110202

11. BC P110404 11. AC C110808,

C110908

11. AB M110801

12. AB P110413 12. BD C110103 12. A BCD M110802

13. CD P110502 13. AB C110307 13. ABCD M110715

14. BC P110404 14. ACD C111101 14. BCD M111402

15. ABCD P110404 15. ABCD C110703 15. ACD M110723PART C PART C PART C

1. 4 P110316 1. 8 C110112 1. 1 M111411

2. 8 P110507 2. 8 C110305 2. 5 M110811

3. 4 P110204 3. 4 C111208 3. 9 M110813

4. 2 P110502 4. 5 C110802,

C110902

4. 8 M110710

5. 5 P110513 5. 5 C111102 5. 1 M110707


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15

Answers & Solutions

Physics

Part A


1.pk

V

=1 1

V sin i V cos j

pk, p 2V V j=

p pk pk,p 1 2 1 V V V V sin i (V V cos )j= = +

But 2 1V V cos 0 = cos =2

1

V

V

p 1V V sin i,=

2 2p 1 2V V V=

V2

V1


2. y = uyt +2

y

1a t

2 2

1y ut gt

2= (1)

dy(u gt)

dt= (2)

From eq. (2), slope dy/dt first decrease and then it will increase in (ve) direction. So, (D)option is correct.

3. B (Concept code: P110404)3. F = N1+ f1= N1+ m1g (1)

N1= N2+ f2= N2+ m2g (2)N2= f3= m3g (3) N1= (m2+ m3)g

m3Fm2

m1N2

f3

N1

f2f1

N1

4. B (Concept code: P110414)4. T1= T2+ m1R1w

2 (1)T2= m2R2w

2 (2)2

1 1 1 2 2

22 2 2

T (m R m R )w 9

T 8m R w

+= =

m1T1

m1R1w2

T2 M2T2 m2R2w2

5. D (Concept code: P110414)5. Mv2/l= T mgcos

T = mv2/l+ mgcosAt lowest pointcos= 1 = 0

v = uTL= mu2/l+ mg

l

l

mgsin

T

V

mgsinmgu

6. B (Concept code: P110507)6. Force is conservative and displacement zero so work done is zero.


7. For a varying force, we must integrate: W F dr=

=3m

y

0

F dy = (2N/m3)


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16

3m3m3 3 4

0 0

1 81y dy (2N / m ) y N m 40.5J

4 2

= = =

8. C (Concept code: P110507)8. At the highest elevation vyvanishes, since the cannon ball has been going upward and has

stopped its vertical motion as it is about to start downward. Vxis constant during the motion,so

At highest point Ek=2 2

x

1 1m | v | mu

2 2=

So work done by gravity = 2y1

mu2


9. Elongation in the spring =Mg

k

And mgx = 21

kx2


10.2 1 2 2

V t V t1

V V t V V t

+ = 1 2

1 2

2v vv

v v=

+

11. BC (Concept code: P110404)11. F T = 10 a (1)

T = 10 a (2)F = 20a a = 1T = 10 N, aa= ab= 1, ac= 0

b T

c

T a FT

TTT

12. AB (Concept code: P110413)

12. a v a v

13. CD (Concept code: P110502)13. Work energy theorem

14. BC (Concept code: P110404)14. N = W F cos N < W

F sin= fk < 90, sin< 1F > fk

N

W

fk

N

W

Fsin

fk

Fcos

F

15. ABCD (Concept code: P110404)15. Incline may accelerate in any direction.

Part C

1. 4 (Concept code: P110316)

1. At maximum height R =2 2

0u cos

g

Maximum height of projectile H =2 2

0u sin

2g


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17

Given: R = 2H2 2 2 2u cos u sin

2g 2g

=

tan= 1 = 45.

2. 8 (Concept code: P110507)2. aBR= 1/2 = 0.5 m/s

2VBR= 0 + 0.5 2 = 1 m/s

VRG= 2m/sVBG= vBR+ vRG= 2 + 1 = 3m/s

KQ=21 2 (3)

2 = 9 J

Kp=21 2 (1)

2 = 1 J

KQ KP= 9 1 = 8 J


3. ( ) ( )1 Area 3i j k i j k2

= + +

4. 2 (Concept code: P110502)4. Work energy theorem

WG+ WF= k

mgd + 21

F d mv2

=

5 10 0.2 + 100 0.2 = 21

5 v2

On solving, we get v = 2 m/s


5. Conservation of mechanical energymgH = mgh + 2

1mv

2

0.45 mg = 3mg 2R 1

mv2 2

+

9 30R = v2 (i)

AC =R 3

2

For reaching at B let particle takes t sec.

v R 3t

2 2 = (ii)

2v 3 1 Rt g t2 2 2

= (iii)

60

A

B

R/2

C60

V

On solving, v2= 15R by eq. (i)

9 30R = 15R R =9

10045

= 20 cm

= 1/5 meter.


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18

Chemistry

Part A

1. C (Concept Code: C110111)1. (i) The energy difference between 4s and 3d subshells is the lowest in comparison to

other given subshell.(ii) Due to maximum exchange energy associated with 3d5in comparison to np3.

2. C (Concept Code: C110207)2. H3BO3is a monobasic acid

( )3 3 2 4H BO H O B OH H + + +

3. A (Concept Code: C110304)3. In OF2, the bond pair moments are cancelled by the lone pair moments. In other case

both type of moments are reinforced with each other.

O

F F

4. B (Concept Code: C110704, C110703)4. The first electron gain enthalpy of N is positive and that of oxygen is negative. The

second electron affinity of sulphur is endothermic and the I.E1of Na is also endothermic.

5. C (Concept Code: C111102)5. The balanced equation is:

2 24 2 4 2 22MnO 5C O 16H 2Mn 10CO 8H O + ++ + + +

6. C (Concept Code: C111203)6. The variation of Z with P at Boyle temperature is given as:

Z

P

1


7. 2 3 2 22NaOH 2NO NaNO NaNO H O+ + +


8. 2 2MO H M H O+ + Ca is more electropositive than hydrogen. Hence it is not reduced by it.


9. Orbital angular momentum = ( )h

12

+

= Azimuthal quantum number


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19

The values of for the electrons given in (A), (B), (C) and (D) are respectively 0, 1, 2 and

0.

10. B (Concept Code: C110305)10. Species Hybridization

2ICl+ sp3

2

ICl sp3d23SO sp3

2NH sp3

XeF2 sp3d

BeF2 sp23CO sp2

11. AC (Concept Code: C110808, C110908)

11. ( )3 3 2 22Ca HCO CaCO CO H O + +

2 3

3 2 2

Na CO No decomposition product

2NaNO 2NaNO O

+

2KNO No decomposition product

12. BD (Concept Code: C110103)

12.34 8

9 19

hc 6.6 10 3 10E 4.16eV

300 10 1.6 10

= = =

Metals having work function less than 4.16 eV will show photoelectric effect.

13. AB (Concept Code: C110307)13. BF3Due to p- pback bonding

23CO Due to -bond formation

14. ACD (Concept Code: C111101)15. ABCD (Concept Code: C110703)

Part C

1. 8 (Concept Code: C110112)1. The atoms are Li, Fe, Al, Cl, Ni, Ti, S and P.

2. 8 (Concept Code: C110305)2. The compounds are SO2, SO3. SF4, PF5, XeF2, POCl3, ClF3and IF7

3. 4 (Concept Code: C111208)

3. 2 2 4

4 4 2

SO SO CH

CH CH SO

r n M 2 16 2 1 2 1 1

r n M 3 64 3 4 3 2 3= = = = =

x 1

y 3 = and x + y = 3 + 1 = 4

4. 5 (Concept Code: C110802, C110902)4. The required substances are KO2, Ca3N2, Mg2C3, CaH2, Ca3P2.

5. 5 (Concept Code: C111102)

5. Meq. of 4MnO = Meq of metal


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20

300 0.4 =W W W

1000 1000 n 1000E M / n M

= =

or, 300 0.4 = mole n 1000 = 0.04 n 1000n = 3Let the oxidation number of the metal ion after oxidation is +xx 2 = 3or x = 5

Mathematics

Part A

1. B M1113051. cos cos sin sin + = +

On squaring, 2 2cos cos 2cos cos + + 2 2sin sin 2sin sin= + +

( )cos2 cos2 2cos + = +

2. D M111304

2. 2cos sin 1 + =

( )

( )( )

( )

2

2 2

2 1 tan / 2 2 tan / 21

1 tan / 2 1 tan / 2

+ =

+ +

1

tan / 2 1, tan / 23

= =

7cos 6sin + = 6 ,2

3. B M110722

3. ( ) ( )2 23x 8xy 3y 3x y x 3y+ = + and2 23x 8xy 3y 2x 4y 1+ + ( ) ( )3x y 1 x 3y 1= + +

The first pair intersects at (0, 0) and the second pair intersects as1 2

,5 5

Equation of one diagonal is y = 2x

Slope of second diagonal is1

2and it will pass through

1 1,

10 5

So equation of second diagonal is 2x 4y 1 =

4. A M110730

4. Let m and m

2

be the slopes of the lines. Then

2 32h a

m m , mb b

+ = = We have

( ) ( )32 3 6 3 2m m m m 3m m m+ = + + +

3 2

3 2 2

8h a a 6ha

bb b b

= +

( ) 2a b 8h6

h ab

+ + =

5. D M1107315. Let lines make and with x axis

Then 1 2tan m , tan m = =


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21

1 2 22h 2 tan 2

m mb sin cossin

+ = = =

2 2

1 2 2

a tan cosm m

b sin

+ = =

( ) ( ) ( )2 2 2

1 2 1 2 1 2tan tan m m m m 4m m = = +

2 4

2 2

4

1 sin cossin cos

( )

( )

2 2

2 2

4 sin cos

4sin cos

= =

tan tan 2 =

6. C M110710

6. Angle made by the lines with the x axis in the positive directions are ( )o 190 + and

( )o 290 + respectively.

Angle between the lines ( ) ( )o o1 290 90= + + or ( ) ( )o o2 190 90+ +

1 2= or 2 1

7. C M1107127. Let 1BC L 11x 6y 14 0 = + + =

2CA L 9x y 12 0 = + =

3AB L 2x 5y 17 0 = + =

Solving the equation, the vertices are obtained as

( ) ( )A 1, 3 ,B 4,5 and ( )C 2, 6

( )1L A 11 1 6 3 14 0= + + >

( )2L C 2 2 5 6 17 0= + <

A

P (k, k )

B C

Since P and A are to lie on the same side of BC, 211k 6k 14 0,+ + > .(I)

Since P and B are to lie on the same of CA,

2

9k k 12 0+ < ------(II)Since P and C are to lie on the same side of AB, 2k + 5k 2 17 < 0 (III)From (I), (II), (III) we see that k should satisfy the inequality

1 86 9 129k

5 2

+< <

Number of integer values of k = 3

8. B M1107088.

B

C

A (1, 0)

x + 2y = 1x = 0

0, 1/2

x 2 = 1

0, 1/2


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22

Required area ( ) ( )1

BC OA2

=

( ) ( )

1 11 1

2 2= =

9. A M110723

9. ( ) ( ) ( )sin sin sin 2sin sin + = + 2 2 2sin sin 2sin sin sin = +

( )22sin sin sin 0 = ( ) ( )sin sin sin sin sin sin 0 + + = Since 0 , ,< <

Since, 0 < , sin , sin > 0sin sin sin 0 + +

so, sin sin sin 0 =

or, ( )1 sin 1. sin sin 0 + + =

Hence ( )x sin y sin sin 0 passes through 1, 1 + + =

10. B M110202

11. AB M11080111. Let P (x, y) be any point of the circle on the chord joining A (1, 2) and B (2, 1)

The slope ofy 2

APx 1

=

and slope of

y 1BP

x 2

+=

Since they include an angle of4

so,

y 1 y 2

x 2 x 1 tan 1y 2 y 1 41x 1 x 2

+ = =

++

( ) ( ) ( ) ( )y 1 x 1 x 2 y 2 + ( ) ( ) ( ) ( )x 1 x 2 y 2 y 1 = + +

2 2x y 5 0+ = or 2 2x y 6x 2y 5 0+ + =

12. A BCD M11080212. Let slope of OA is m,

Then om 1

tan451 m

=

+

m 1

11 m

=

+

m=0 or undefinedSolving y = 3 and 2 2x y 10x 6y 30 0+ + =

C

B(5, 3)

D

A

2

2

y = x +

m

O

45o

Two vertices are (3, 3) and (7, 3)other diagonal is by solving x = 5 and 2 2x y 10x 6y 30 0+ + = Other two vertices are (5, 1) and (5, 5)

13. ABCD M110715


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23

13. Equation of any line through P (3, 4) making an angle with the positive direction of x

axis isx 3 y 4

rcos sin

= =

(i)

where r is the distance of any point on the line from P.

Therefore, coordinates of any point on the line (i) are ( )3 r cos , 4 r sin+ + ..(ii)If Eq. (ii) represents R, then

3

3 r cos 6 r PRcos+ = = = If Eq. (ii) represents S, the 4 r sin 8+ =

r 4 / sin PS = = Hence, PR 3sec , PS 4cosec= =

( )2 3sin 4cos3sin 4cosPR PS

sin cos sin2

+ + + = =

and ( ) ( )2 2 2 23 / PR 4 / PS cos sin 1+ = + =

( ) ( )

2

9 161

PSPR+ =

14. BCD M111402

14. We have,1 sin 1 cos

x , ycos sin

+ = =

Multiplying, we get( ) ( )1 sin 1 cos

xycos sin

+ =

1 sin cos sin cos sin cos

xy 1cos sin

+ + + =

1 sin cos

cos sin

+ =

and ( ) ( )1 sin sin cos 1 cosx y cos sin + =

2 2sin sin cos cos

cos sin

=

( )sin cos 1

xy 1cos sin

= = +

Thus, xy x y 1 0+ + =

y 1

xy 1

=

+and

1 xy

1 x

+=

15. ACD M11072315. Reflected ray is a line which passes through the inter section of the lines 3x y 8 0+ =

and 2x 7y 8 0+ = is ( )3x y 7 2x 7y 8 0+ + + =

( ) ( ) ( ) ( )2 3 x 7 1 y 8 9 0 1 + + + + =

Slope of the line (1) is( )

( )

2 3

7 1

+

+

Slope of 3x y 9 0+ = is 3

Slope of 2x 7y 8 0+ = is2

07


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24

If is the angle between the above lines ,

23

197tan6 13

17

+ = =

+

19

13=

For the reflected ray,

( )( )

2 3 2

197 1 7

2 2 3 131 7 7 1

+ + + = +

+ +

260,

53

=

Hence the reflected ray corresponds to26

53

Equation of the reflected ray is 107x 129y 269 0 =

Solving 3x y 9 0+ = and 2x 7y 8 0+ = ,We get the point P as55 6

,19 19

Part C

1. 1 M111411

1. On adding ( )23 cos 4 4sin2

a 1 sin22

+ = = +

On subtracting ( )2 4b 1 sin2 ab cos 2 1= =

2. 5 M110811

2. Equation of chord of contact (QR) is 26x 8y r 0+ =

( )

2 2

2 2

6.6 8.8 r 100 r PM

106 8

+ = =

+

and( )

2 2

2 2

0 0 rrOM106 8

+ = =

+

Then ( ) ( ){ }2 2QR 2.QM 2 OQ OM= = 4

2 r2 r100

=

Area of QPR =1

2. QR. PM

( )24

2100 r 1 r

say .2 r .2 100 10

=

( )

( )

32 2

2r 100 r

z say1000

= =

( ) ( ) ( ){ }2 32 2 2dz 1 r .3 100 r . 2r 100 r .2r

dr 1000= +

( ){ }

22

2 22r 100 r

100 r 3r 1000

=

For maximum or minimumdz

0,dr

= then we get

P(6, 8)

(0, 0)

Q

RO

r

r

M


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25

( )r 5 r 10 as P is outside the circle=

and2

2

r 5

d zve

dr=

=

Area of triangle is also maximum at r = 5

3. 9 M110813

3. The given circle2 2

1S x y 14x 4y 28 0= + + + = and 2 22S x y 14x 4y 28 0= + + = Centres and radii of circles 1S and 2S are

( )1 1C 7, 2 , r 49 4 28 5 = + =

and ( )2 2C 7, 2 , r 49 4 28 = + + = 9 respectively

Here ( ) ( )2 2

1 2d C C 7 7 2 2= = + +

1 2212 r r 5 9 14= > + = + =

1 2d r r > +

Hence circles dont touch or cut.Length of external common tangent

( )22

ex 2 1L d r r =

( )2

212 5 9 212 196= + =

16 4= = =

18 + =

4. 8 M110710

4. Let be the acute angle between AD and BE.

Then 2 1 1tan1 2 3 = =+

Now o o o180 90 360 + + + = o90 = +

( )tan tan = +

2 22b 2a

tan tan 2b 2aa b32b 2a1 tan tan 3ab

1 .a b

+ + + = = =

29ab 2c =

E

AA

CB

c

Da/2 a/2

5. 1 M110707

5. ( )1

ax bz cy area of ABC2

+ + =

( ) 2a 3

2014 a2 4

=

3

2a=length of altitude=2014


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26

QP CODE: 120763.2SECTION I (PHYSICS) SECTION II (CHEMISTRY) SECTION III (MATHS)



1. B P110404 1. A C110304 1. B M110722

2. B P110414 2. BC110704,

C1107032. A M110730

3. D P110414 3. C C111102 3. D M110731

4. B P110507 4. C C111203 4. C M110710

5. B P110507 5. C C110809 5. C M110712

6. C P110507 6. B C111003 6. B M110708

7. C P110513 7. C C110103 7. A M110723

8. D P110320 8. B C110305 8. B M110202

9. C P110320 9. C C110111 9. B M111305

10. D P110320 10. C C110207 10. D M111304

11. CD P110502 11. AB C110307 11. ABCD M110715

12. BC P110404 12. ACD C111101 12. BCD M111402

13. ABCD P110404 13. ABCD C110703 13. ACD M110723

14. BC P110404 14. ACC110808,

C11090814. AB M110801

15. AB P110413 15. BD C110103 15. ABCD M110802

PART C PART C PART C

1. 4 P110204 1. 4 C111208 1. 9 M110813

2. 2 P110502 2. 5C110802,

C1109022. 8 M110710

3. 5 P110513 3. 5 C111102 3. 1 M110707

4. 4 P110316 4. 8 C110112 4. 1 M111411

5. 8 P110507 5. 8 C110305 5. 5 M110811


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27



Q.N. Ans. Con.CodeQ.

N.Ans. Con.Code Q.N. Ans. Con.Code

1. B P110507 1. C C111203 1. C M110710

2. B P110507 2. C C110809 2. C M110712

3. C P110507 3. B C111003 3. B M110708

4. C P110513 4. C C110103 4. A M110723

5. D P110320 5. B C110305 5. B M110202

6. C P110320 6. C C110111 6. B M111305

7. D P110320 7. C C110207 7. D M111304

8. B P110404 8. A C110304 8. B M110722

9. B P110414 9. BC110704,

C110703 9. A M110730

10. D P110414 10. C C111102 10. D M110731

11. ABCD P110404 11. ABCD C110703 11. ACD M110723

12. BC P110404 12. ACC110808,

C11090812. AB M110801

13. AB P110413 13. BD C110103 13. ABCD M110802

14. CD P110502 14. AB C110307 14. ABCD M110715

15. BC P110404 15. ACD C111101 15. BCD M111402

PART C PART C PART C1. 5 P110513 1. 5 C111102 1. 1 M110707

2. 4 P110316 2. 8 C110112 2. 1 M111411

3. 8 P110507 3. 8 C110305 3. 5 M110811

4. 4 P110204 4. 4 C111208 4. 9 M110813

5. 2 P110502 5. 5C110802,

C1109025. 8 M110710


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28




1. C P110513 1. C C110103 1. A M110723

2. D P110320 2. B C110305 2. B M110202

3. C P110320 3. C C110111 3. B M111305

4. D P110320 4. C C110207 4. D M111304

5. B P110404 5. A C110304 5. B M110722

6. B P110414 6. BC110704,

C1107036. A M110730

7. D P110414 7. C C111102 7. D M110731

8. B P110507 8. C C111203 8. C M110710

9. B P110507 9. C C110809 9. C M110712

10. C P110507 10. B C111003 10. B M110708

11. AB P110413 11. BD C110103 11. ABCD M110802

12. CD P110502 12. AB C110307 12. ABCD M110715

13. BC P110404 13. ACD C111101 13. BCD M111402

14. ABCD P110404 14. ABCD C110703 14. ACD M110723

15. BC P110404 15. ACC110808,

C11090815. AB M110801

PART C PART C PART C

1. 8 P110507 1. 8 C110305 1. 5 M1108112. 4 P110204 2. 4 C111208 2. 9 M110813

3. 2 P110502 3. 5C110802,

C1109023. 8 M110710

4. 5 P110513 4. 5 C111102 4. 1 M110707

5. 4 P110316 5. 8 C110112 5. 1 M111411


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29

FIITJEEINTERNAL TESTBatches:Two Year CRP(1416)

IIT JEE 2016

PHASE IPAPER II

ANSWERS

QP CODE: CODE: 120764.1SECTION I (PHYSICS) SECTION II (CHEMISTRY) SECTION III (MATHS)



1. D P110420 1. B C110113 1. C M110202

2. B P110308 2. C C111101 2. CM110705,

M110706

3. A P110409 3. D C110702 3. A M110731

4. D P110316 4. C C110306 4. B M110814

5. D P110301 5. C C111204 5. D M110801

6. B P110504 6. C C110908 6. C M111301

7. A P110325 7. DC111004,

C1110067. C M111403

8. A P110327 8. B C110113 8. D M110602

9. A P110310 9. C C110104 9. B M110724

10. C P110310 10. B C110109 10. A M110724

11. C P110402 11. B C110305 11. C M110802

12. B P110403 12. A C110204 12. C M110802

13. A P110420 13. C C110207 13. D M111402

14. C P110420 14. D C110207 14. B M111402

15. ABCD P110403 15. BCD C111208 15. AD M110725

16. CD P110306 16. ABC

C110107,

C110108 16. ACD M110819

17. AD P110404 17. ACD C110306 17. BCD M111304

18. AB P110201 18. AC C110205 18. ACD M110707

19. ACD P110325 19. AB C110904 19. BD M110202

20. ABD P110416 20. B C110006 20. AD M110808


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30

Answers & Solutions

Physics1. D (Concept code: P110420)

1. Kx =Mg

3 ; x =

Mg

3k


2. 2.45 = 21

gt2

t =1

2

3. A (Concept code: P110409)3. 2 a + 1 2 + 1 4 = 0.


4. 2R = V 60 =175

606

1750 = 2R

R =1750

2

T = 60 sec and T should be same for both.

jeep jeep

cheetah cheetah

v r

v r=

cheetah jeepR 60 60(2 )

v v 1 105R 1750

= =

= 82.37 km/hr.


5. Average velocity =

2

2R

H2net displacement

total time T / 2

+ = .


6. Avg. power =

21 10(2)2 1W

20

= .


7.dv

a vdx

=


8. avs

vt

=


9. v u a t= + = o oa v

0g

+



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31

10. x =

2

2 0o

v1 1ut at 0 a

2 2 g

+ = +

11 C (Concept code: P110402)11. F.B.D.


12. T cos = mg


13. For stable equilibrium,2

2

d U0

dr> .


14. Wext= 2A B

02A2A

BB

=2 2 2B B B

04A 2A 4A

=

.

15. ABCD (Concept code: P110403)15. T2sin 2= mg and T2cos 2=mg

T1sin 1= T2sin 2and T1cos 1= mg + T2cos 2

16. CD (Concept code: P110306)16. Sign of velocity of acceleration indicate their direction in the co-ordinate axis. If they are

of same sign, speed increases.

17. AD (Concept code: P110404)17. Newtons third law state that force applied by one object on other is exactly of same

magnitude that applied by other on one.

18. AB (Concept code: P110201)18. C2= A2+ B2+ 2AB cos

19. ACD (Concept code: P110325)

19. Concept of slopeds

vdt

= and2

2

d sa

dt=

20. ABD (Concept code: P110416)20. Fnet= ma

Chemistry

1. B (Concept Code: C110113)1. The orbitals in (A), (B), (C) and (D) are respectively 4d, 3s, 3p and 4f,

No. of radial nodes = (n - - 1)

3s has maximum number(2) radial nodes.



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32

2.

22 2Cu S Cu SO

+ +

6e

2e


3. Except the valence electrons, the remaining electrons are called core electrons.Hydration of the ions decreases on moving down the group. Hence, the radii of hydratedions decrease on moving down the group.

4. C (Concept Code: C110306)4. Orbitals having maximum amount of directional character(p-orbital) undergo overlapping

by maximum extent. So, the order should bep > sp3> sp > s

Since the hybrid orbitals are formed by such a manner that electron density remainsmaximum along one direction and minimum along other direction.

Atomic orbital

Hybrid orbital

Due to high electron density the hybrid orbitals undergo better overlapping along onedirection. The atomic orbitals undergo equal overlapping along both directions.

5. C (Concept Code: C111204)5. The van der Waals equation for one mole of real gas is given as

( )2a

P V b RTV

+ =

At high P,2

a

Vis neglected, making P(V - b) = RT

6. C (Concept Code: C110908)6. MgCl2.6H2O does not form anhydrous salt on heating. It undergoes partial hydrolysis

during heating.

2 2 2 2 2MgCl .6H O MgCl .H O 5H O

+

HCl+Mg

Cl

OH7. D (Concept Code: C111004, C111006)

8. B (Concept Code: C110113)8. The probability of finding electrons at a certain region is represented by 2. 2is zero at

nucleus as there is no electron present at the nucleus.

9. C (Concept Code: C110104)9. By supplying 12.75 eV energy, the maximum excitation of electrons takes place from the

ground state to fourth orbit(n = 4)

No. of photons =( ) ( )n n 1 4 4 1 4 3

62 2 2

= = =


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33

10. B (Concept Code: C110109)10. E4 1= E4 3+ E3 1

4 1 4 3 3 1

4 1 4 3 3 1

o4 1

hc hc hcor,

1 1 1 1 1 5or,

8000 2000 8000

8000 1600A5

= +

= + = + =

= =

11. B (Concept Code: C110305)11. Hybridization is sp3as Cl forms three sigma bonds and contains a lone pair.

The structure of 3ClO is Cl

O OO

12. A (Concept Code: C110204)12. in +4 oxidation state chlorine contains an unpaired electron and a lone pair. E.g. ClO2.


13. ( )3 4 2 3 4 22Na AsO 3CaCl Ca AsO 6NaCl+ +

Initial moles800 0.4 1000 0.5

0 01000 1000

= 0.32 = 0.5Moles after 0 0.02 0.16 0.96reactionTotal moles of Cl= (0.02 2) + 0.96 = 1

Molarity of Cl=1 1

1800 1.8

1000

= = 0.55 M


14. All the 34AsO ions are precipitated as calcium arsenate. Hence its molarity is zero.

15. BCD (Concept Code: C111208)15. The gases have same molecular mass.

16. ABC (Concept Code: C110107, C110108)16. It contains four unpaired electrons. Hence magnetic moment =

( ) ( )n n 2 4 4 2 2 6+ = + = B.M.

17. ACD (Concept Code: C110306)17. p-dback bond formation takes place in (Y).

18. AC (Concept Code: C110205)

19. AB (Concept Code: C110904)


20. 2 2 2 2H O I H I H O ++ + +


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34

Mathematics

1. C M110202

1. ( ) ( ) ( )1/ 2 2

3 2 xlog 3 log 2 log 834 9 10+ =

( ) ( ) x1/2 log 8324 9 10+ =

( ) ( ) x1 log 10

83 83=

x1 log 10 x 10= =

2. C M110705 & M110706

2. Slope of8

OB6

=

Slope of3

OC4

=

BOC2

=

OBC is right angled at O

Circumcentre = mid point of hypotenuse =

11

1, 2

Orthocentre = vertex O (0, 0)

Distance121 5 5

14 2

= + =

unit

B(6, 8)

(4, 3)

O

C

3. A M110731

3. Here,( )2

1 12 h ab

tan m tana b

= +

1252 64

tan2 3

= +

1 1tan5

=

Here,1

m5

=

4. B M1108144. The common chord must be a diameter of the second circle. The equation to the

common chord is 1 2S S 0 = i.e. ( )6x 14y c d 0+ + + = . The centre of the second circle

(1, 4) lies on it. ( ) ( )6 1 14 4 c d 0+ + + = c d 50 + =

5. D M110801

5. Let the given lines be 1 1 1 1L a x b y c 0= + + = and 2 2 2 2L a x b y c 0= + + = .Suppose L1 meets the coordinates axes at A and B and L2 meets at C and D. Thencoordinates of A, B, C, D are


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1 1

1 1

c cA , 0 , B 0,

a b

,

2

2

cC ,0 ,

a

and 2

2

cD 0,

b

Since A, B, C, D are concyclic, therefore OA . OC = OD . OB

y

X XC O A

D

y'

B

1 2 2 1

1 2 2 1

c c c c

a a b b

=

1 2 1 2a a b b=

6. C M1113016. 4n =

( ) ( )tan 2n 1 tan 2n tan cot2

= = =

( )tan tan2 tan3 .... tan 2n 1

( )( ) ( )( )tan tan 2n 1 tan2 tan 2n 2 ..... tan n

( )1.1.1....1 tan n=

tan n tan 14

= = =

7. C M111403

7.3

sin cos , sin cos2 2

= + =

( )sin 3 sin , + =

( )sin 5 sin =

( )4 43

3 sin sin 32

+ +

( )6 62 sin sin 52

+ +

{ } { }4 4 6 63 cos sin 2 cos sin= + +

{ } { }2 2 2 23 1 2sin cos 2 1 3sin cos= = 1

8. D M110602


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36

8.x

x

1 1lim 0

7 7

= =

(not an indenterminant form)x

xlim1 1 1

= = (not an indeterminate form)

x

xlim 5 5

= = (not an indeterminant form)

( )

x

x1lim 1 1 0 1x

+ = + =

(indeterminant form)

9. B M1107249. Centroid is

acos t bsin t 1x

3

+ +=

3x 1 acos t b sin t = + (i)

andasin t bcos t 0

y3

+=

3y asin t bcos t= (ii)

Squaring and adding Eqs. (i) and (ii), we get( ) ( ) ( ) ( )

2 2 2 23x 1 3y acost bsin t asin t bcos + = + +

or ( ) ( )2 2 2 23x 1 3y a b + = +

10. A M110724

10. Let (x, y) be coordinate of vertex C and ( )1 1x , y be coordinates of centroid of the triangle

1x 2 2

x3

+ = and 1

y 3 1y

3

+=

1x

x3

= and 1y 2

y3

=

The centroid ( )1 1x , y lies on the line 2x 3y 1+ = 1 12x 3y 1+ =

x y 22 3 1

3 3

+ =

2x 3y 9, + = which is the required locus of the vertex C.

11. C M110802

11. Particle covers an arc of k unit on a circle of radius k unit and the moves on the next

circle.

2

2

1

(1, 0)x

O


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The particle starts at (1, 0) (i.e. it covers a distance of 1 radian) and to cross the positivedirection of x axis for the first times particle has to cover a distance2 r 2 1 2 = = radian

2 rad

2 2 3.141rad

= =

6.28= i.e. on the 7thcircle

12. C M11080212. Particle moves 1 radian on each circle starts from (1, 0)

To cross x axis again should cover rad

rad

3.141rad

= =

i.e. will be on 4thcirclepoint (4, 0)

13. D M111402

13. 2 4 6cos cos cos7 7 7

2 4 8cos cos cos 2

7 7 7

=

2 4 8cos cos cos

7 7 7

=

3

3

2 2 2sin 2 . sin 2 sin

17 7 7

2 2 2 82 sin 8 sin 8sin

7 7 7

+ = = = =

14. B M111402

14.5 7

sin sin sin18 18 18

5 7cos cos cos

2 18 2 18 2 18

=

8 4 2cos cos cos

18 18 18

=

2 4cos cos cos

9 9 9

=

3here

2 1

= +

3

12

=

1

8=

15. AD M110725

15.2

1 2 6

3 1 4 0

4

=


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2 2 8 0 + = ( ) ( )4 2 0 + =

4, 2 =

16. ACD M110819

16. 2 21c x y 2x 4y 4 0= + = ..(i)

and2 2

2c x y 2x 4y 4 0= + + + + = ..(ii)Radical axis is 1 2c c 0 =

x 2y 2 0+ + = which is L= 0Centre and radius of 1C 0= are (1, 2) and 3.

Length of from (1, 2) on L = 0

is1 4 2 7

51 4

+ +=

+radius

L is also the common chord of C1and C2

Centres of 1C 0= and 2C 0= are (1, 2) and ( )1, 2

Slope of Line joining centres of circles 1C 0= & 2C 0= is 12 2 2 m1 1 = = (say)

and slope of L = 0 is 21

m2

= (say)

1 2m m 1= Hence L is perpendicular to the line joining centres of C1and C2.

17. BCD M111304

17.3

cos5

=

4sin

5 =

and5

cos13

=

12sin

13 =

( )cos cos cos sin sin + =

4 5 4 12. .

5 13 5 13=

33

65=

( )sin sin cos cos sin + = +

4 5 3 12. .

5 13 5 13= +

56

65=

( )2 1 cossin2 2

=

{ }1 cos cos sin sin2

+ =

3 5 4 121 . .

5 13 5 13

2

+ =

631

652

=

1

65=


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39

( )cos cos .cos sin .sin = +

3 5 4 12. .

5 13 5 13= +

63

65=

18. ACD M110707

19. BD M110202

20. AD M11080820. Circle possible in IV quadrant. Equation of circle is

( ) ( )2 2 2x c y c c + + =

Its passes through (3, 6)

( ) ( )2 2 23 c c 6 c + =

2c 18c 45 0 + =

( ) ( )c 15 c 3 0 =

x' x

y'

y

c

(c, c)c

(3, 6)

c = 3, 15From Eq. (i) equation of circles are

2 2x y 6x 6y 9 0+ + + =

and 2 2x y 30x 30y 225 0+ + + =


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1. A P110409 1. D C110702 1. A M110731

2. D P110316 2. C C110306 2. B M110814

3. D P110301 3. C C111204 3. D M110801

4. B P110504 4. C C110908 4. C M111301

5. A P110325 5. DC111004,

C1110065. C M111403

6. A P110327 6. B C110113 6. D M110602

7. D P110420 7. B C110113 7. C M110202

8. B P110308 8. C C111101 8. CM110705,

M110706

9. C P110310 9. B C110109 9. A M110724

10. A P110310 10. C C110104 10. B M110724

11. B P110403 11. A C110204 11. C M110802

12. C P110402 12. B C110305 12. C M110802

13. C P110420 13. D C110207 13. B M111402

14. A P110420 14. C C110207 14. D M111402

15. AD P110404 15. ACD C110306 15. BCD M111304

16. AB P110201 16. AC C110205 16. ACD M110707

17. ACD P110325 17. AB C110904 17. BD M11020218. ABD P110416 18. B C110006 18. AD M110808

19. ABCD P110403 19. BCD C111208 19. AD M110725

20. CD P110306 20. ABCC110107,

C11010820. ACD M110819


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41




1. D P110301 1. C C111204 1. D M110801

2. B P110504 2. C C110908 2. C M111301

3. A P110325 3. DC111004,

C1110063. C M111403

4. A P110327 4. B C110113 4. D M110602

5. D P110420 5. B C110113 5. C M110202

6. B P110308 6. C C111101 6. CM110705,

M110706

7. A P110409 7. D C110702 7. A M110731

8. D P110316 8. C C110306 8. B M110814

9. A P110310 9. C C110104 9. B M110724

10. C P110310 10. B C110109 10. A M110724

11. C P110402 11. B C110305 11. C M110802

12. B P110403 12. A C110204 12. C M110802

13. A P110420 13. C C110207 13. D M111402

14. C P110420 14. D C110207 14. B M111402

15. ACD P110325 15. AB C110904 15. BD M110202

16. ABD P110416 16. B C110006 16. AD M110808

17. ABCD P110403 17. BCD C111208 17. AD M110725

18. CD P110306 18. ABCC110107,

C11010818. ACD M110819

19. AD P110404 19. ACD C110306 19. BCD M111304

20. AB P110201 20. AC C110205 20. ACD M110707


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42




1. A P110325 1. DC111004,

C1110061. C M111403

2. A P110327 2. B C110113 2. D M110602

3. D P110420 3. B C110113 3. C M110202

4. B P110308 4. C C111101 4. CM110705,

M110706

5. A P110409 5. D C110702 5. A M110731

6. D P110316 6. C C110306 6. B M110814

7. D P110301 7. C C111204 7. D M110801

8. B P110504 8. C C110908 8. C M111301

9. C P110310 9. B C110109 9. A M110724

10. A P110310 10. C C110104 10. B M110724

11. B P110403 11. A C110204 11. C M110802

12. C P110402 12. B C110305 12. C M110802

13. C P110420 13. D C110207 13. B M111402

14. A P110420 14. C C110207 14. D M111402

15. ABD P110416 15. B C110006 15. AD M110808

16. ABCD P110403 16. BCD C111208 16. AD M110725

17. CD P110306 17. ABC C110107,C110108

17. ACD M110819

18. AD P110404 18. ACD C110306 18. BCD M111304

19. AB P110201 19. AC C110205 19. ACD M110707

20. ACD P110325 20. AB C110904 20. BD M110202

phase test answers

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