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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3. A (Concept code: P110507)
3. W =0.2
3 12
0.1
F dx (5x 16x )dx 8.1 10 = + =
4. A (Concept code: P110320)
4.
r
3
45
r
6
ris same in both cases.
5. B (Concept code: P110508)
5. 1
2
5glv
v 4gl=
6. B (Concept code: P110409)
6. 2 2 2resa a a 2a cos(180 )= + +
= 2a sin2
(180 )
a
a
7. A (Concept code: P110314)
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. C (Concept code: P110403)
11.2
tan3
= g
O
3g
2
12. C (Concept code: P110401)12. a g=
2 26 0 2 gS= +
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13. B (Concept code: P110504)
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. B (Concept code: P110320)
16. T a
= 0 2 1m m
a 2a=
17. C (Concept code: P110310)
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. C (Concept code: P110326)
19. s = 21
gt2
= 21
10 0.2 0.2m.2
= Block haves contact will the floor.
20. B (Concept code: P110502)
20. Energy cost due to air friction = 21
1 (20) 1 10 182
= 20J.
21. A (Concept code: P110404)
21. 4g T = oa T
4 & 2gsin30 2a2 2
=
6a = 2g
22. C (Concept code: P110415)
22. At y = a, x = 2a, dydx
= 1 = 45 af= gcos 45 = g2
.
23. C (Concept code: P110326)
23. Dist. =
2
2 21(10 2) g (2)2
+
24. A (Concept code: P110320)
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24. tan =V
1V
=
25. C (Concept code: P110416)25. V1cos1= V2cos2
26. B (Concept code: P110404)
26. k 2 k3 =
k =3
k2
27. B (Concept code: P110404)
27. Fnet=2 2M a a 2 ma+ =
28. C (Concept code: P110325)
28. a =Vdv
dx
dv a
dx v= = constant.
29. C (Concept code: P110404)29. W = m geff
30. C (Concept code: P110318)
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.
6. D (Concept Code: C110803)
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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. C (Concept Code: C111102)
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. B (Concept Code: C110903)
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.
15. C (Concept Code: C111101)
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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)
19. D (Concept Code: C111206)
20. B (Concept Code: C110902)20. Be2+(2s02p0) makes four vacant orbitals available for coordination with H2O.
21. B (Concept Code: C111006)
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. D (Concept Code: C110107)
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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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. B (Concept Code: C110808)
27. NaOH CO HCOONa+
28. C (Concept Code: C111006)
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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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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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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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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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
PHYSICS C.CODE CHEMISTRY C. CODE MATHS C.CODE
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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QP CODE: 120762.3
SR.NO
PHYSICS C.CODE CHEMISTRY C. CODE MATHS C.CODE
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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QP CODE: 120762.4
SR.NO
PHYSICS C.CODE CHEMISTRY C. CODE MATHS C.CODE
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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FIITJEEINTERNAL TESTBatches:Two Year CRP(1416)
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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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Answers & Solutions
Physics
Part A
1. C (Concept code: P110320)
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. D (Concept code: P110320)
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. B (Concept code: P110507)
7. For a varying force, we must integrate: W F dr=
=3m
y
0
F dy = (2N/m3)
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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. C (Concept code: P110513)
9. Elongation in the spring =Mg
k
And mgx = 21
kx2
10. D (Concept code: P110320)
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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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. 4 (Concept code: P110204)
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. 5 (Concept code: P110513)
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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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. C (Concept Code: C110809)
7. 2 3 2 22NaOH 2NO NaNO NaNO H O+ + +
8. B (Concept Code: C111003)
8. 2 2MO H M H O+ + Ca is more electropositive than hydrogen. Hence it is not reduced by it.
9. C (Concept Code: C110103)
9. Orbital angular momentum = ( )h
12
+
= Azimuthal quantum number
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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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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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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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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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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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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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( )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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QP CODE: 120763.2SECTION 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. 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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QP CODE: 120763.3SECTION I (PHYSICS) SECTION II (CHEMISTRY) SECTION III (MATHS)
PART A PART A PART A
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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QP CODE: 120763.4SECTION 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 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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PHASE IPAPER II
ANSWERS
QP CODE: CODE: 120764.1SECTION 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. 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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Answers & Solutions
Physics1. D (Concept code: P110420)
1. Kx =Mg
3 ; x =
Mg
3k
2. B (Concept code: P110308)
2. 2.45 = 21
gt2
t =1
2
3. A (Concept code: P110409)3. 2 a + 1 2 + 1 4 = 0.
4. D (Concept code: P110316)
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. D (Concept code: P110301)
5. Average velocity =
2
2R
H2net displacement
total time T / 2
+ = .
6. B (Concept code: P110504)
6. Avg. power =
21 10(2)2 1W
20
= .
7. A (Concept code: P110325)
7.dv
a vdx
=
8. A (Concept code: P110327)
8. avs
vt
=
9. A (Concept code: P110310)
9. v u a t= + = o oa v
0g
+
10. C (Concept code: P110310)
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10. x =
2
2 0o
v1 1ut at 0 a
2 2 g
+ = +
11 C (Concept code: P110402)11. F.B.D.
12. B (Concept code: P110403)
12. T cos = mg
13. A (Concept code: P110420)
13. For stable equilibrium,2
2
d U0
dr> .
14. C (Concept code: P110420)
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.
2. C (Concept Code: C111101)
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2.
22 2Cu S Cu SO
+ +
6e
2e
3. D (Concept Code: C110702)
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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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. C (Concept Code: C110207)
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. D (Concept Code: C110207)
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. B (Concept Code: C110006)
20. 2 2 2 2H O I H I H O ++ + +
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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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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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( )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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QP CODE: 120764.2SECTION 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. 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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QP CODE: 120764.3SECTION 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. 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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QP CODE: 120764.4SECTION 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. 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