Sec: Sr. IIT_IZ Jee-Main Date: 24-12-18 Time: 09:00 AM –12:00 Noon GTM-1 Max.Marks:360
KEY SHEET
PHYSICS
1 2 2 3 3 2 4 3 5 3
6 3 7 2 8 2 9 4 10 4
11 2 12 2 13 2 14 4 15 2
16 1 17 2 18 1 19 1 20 4
21 3 22 2 23 4 24 4 25 2
26 4 27 2 28 4 29 4 30 2
CHEMISTRY
31 3 32 2 33 3 34 2 35 1
36 1 37 2 38 4 39 2 40 2
41 2 42 4 43 3 44 4 45 2
46 3 47 1 48 1 49 1 50 3
51 2 52 3 53 3 54 3 55 4
56 4 57 3 58 2 59 4 60 3
MATHS
61 2 62 3 63 1 64 3 65 2
66 2 67 2 68 3 69 4 70 3
71 2 72 3 73 2 74 1 75 2
76 3 77 2 78 1 79 3 80 3
81 1 82 3 83 3 84 3 85 2
86 1 87 4 88 1 89 3 90 3
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 2
SOLUTIONS PHYSICS
1. Work function
hC hC hC
2 2= − =
2. As seen by B: F = mg + T
As seen by A F = mg + T - ma
3. fv 0.97v 0.97 440 427 Hz= = =
4.
The shown figure represents(C) By snell’s law:
2
1
sin i
sin r
=
For 1 4 1i ; r and 1= = =
12
4
sin
sin
=
5.
4Adh gh r
dt 8
=
6. 360 200 div→
0 18
15 200 div360
→
1nm
LC200
=
1
x 10 mm200
=
31
F 100 10 5mN20
−= =
7. B,gg cosˆ ˆa g sin i j
= − −
2
P,gˆ ˆa g sin i g cos j= − −
P,gg cos ˆa j
= −
2
P,gˆ ˆu cos i usin j= + 4 45 45
( )usin
Tg cos /
=
2 45
2
8.
2
0
4T
r 2
=
( )
2
22
0
4T 1 Q
r 2 4 r=
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 3
( )208TQ 4 r
r
=
( )2 02TQ 8 r
r
=
0Q 8 r 2T r=
9. f Fsin=
mg cos
F
= + =2 02
ll
mgcos
F
=2
mg Fcos= −
For maximum
du
d=
0
10. Conceptual
11. The direction of velocity is along the tangent
Hence
dy
2xdx
=
0 4 2
tan53 2x 2x x m3 3
= = =
2
2 2 4y x m
3 9
= = =
Hence ( )2 4ˆ ˆ ˆ ˆL r P M i j 15cos53i 15sin53i3 9
= = + +
( )2 4ˆ ˆ ˆ ˆ2 i j 9i 12j 83 9
= + + =
12. For perfectly reflecting mirror, the force exerted by the light of power P is
( )2 power
Fc
=
For equilibrium, ( )2 power
F mgc
= =
6mg c
Power 30 10 W2
= =
As only 30% of the power is given to the mirror
So, 100
P ' 30 100MW30
= =
13. Conceptual
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 4
14.
mag 2F N mg+ =
2 magN mg F= −
1 magN mg F= +
( )2 2 2 magF N F mg F= = −
( )1 1 1 magF N F mg F= = +
2 1F F
2Bi
−=
15.
2
2 2
1tan g
2 u cos
= −
( )
221
tan g 1 tan2 2g
= − +
2 4tan 4 tan 1 0
− + + =
If particle will not hit the target
( )2b 4ac 0−
416 4 1 0 4 3
− +
16. L mvr r=
( )A B Cmeanr r r
A B CL L L
17. Index error in u = +1cm
U=8cm
Index error in v= 1cm−
v= 17+1=18 cm
1 1 1
f 5.53cmf 18 18= + =
18. Conceptual
19.
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 5
mg sin f kx cos 0− − =
N kx sin mg sin 0− − =
f N=
20. Pressure at point 0 B 0
2T 2TB P P P
3R 3R= + − =
Pressure point 0 C 0
2T 2T 4T 32TC P P P
R3R R 3R
2
= + + + − =
Ratio of point C 0 B 0P P &P P 16− − =
21. Conceptual
22.
When suddenly current injected into the left loop flux through the IInd loop increase with respect to time. So,
according to Lenz law a induced current develop in the IInd loop.
M B clockwise =
23. Consider system of two disks to be short dipole.
3
2kdpdE
r=
dp dq dA= =
3 3
2k 2kE dp A
r r= =
22
3 3
0 0
2 RE R
4 r 2 r
= =
24. conceptual
25. C C S Sg 1m V m V− … (i)
( )C C C S Sg 2m V m m V+= …………(ii)
( )C S S C C S Sg2 2m m V m V m V'+ = − + …………(iii)
C Cg
S2S
2m VV '
m=
26. Internal loss 2i r=
ir R
=
+
Battery life time=battery cap
current
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 6
27. conceptual
28.
R
1
LC =
2 R from diagram
Circuit will behave as capacitance circuit
current leads voltage
29. E S E 1
GS G S R G 4R
G S
= + +
++
3RS
HR 3S
=−
R S
G 3S=
30. 2 20I
I cos 30 cos 302
=
09I
32=
CHEMISTRY
31. 3
O
CH3
OH
O
CH3
OH2
OCH3
O O OH
H +
⎯⎯→
H −tauto
32. 2
33. 3
34. 2
Higher the value of + on carbonyl carbon more prone the molecule will be for reactivity
35. 5.3 g and 4.2 g
Using PhH mmoles of Na2CO3 = 2.5 x 0.2 = 0.5
Using MeOH, mmoles of Na2CO3+mmoles of NaHCO3 = 2.5 x 0.4 = 1.0
Mmoles of NaHCO3= 0.5 and mmoles of Na2CO3= 0.5
For 3
WNaHCO 1000 0.5
84 =
W = 0.042g in 10 mL
W = 4.2 g in 1 litre
For 2 3
WNa CO 1000 0.5
106 =
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 7
W = 0.053 g in 10 mL
W = 5.3 g in 1 litre
36. 2NH−
Pyramidal
37. 2
+ 2 2CH Cl + 3AlCl
CH2 Cl + 3AlCl
2 2Ph CH
38. i) n 3 l 2 m 0 n 1/ 2= = = = + Allowed
ii) n 2 l 2 m 1 n 1/ 2= = = = + not allowed
(l as to be 0 .... n - 1)
iii) n 4 l 3m 1 n 1/ 2= = = − = −
iv) n 1 l 0 m 2 n 1/ 2= = = − = − not allowed
l(m l)=
v) n 3 l 2 m 3 n 1/ 2= = = = + not allowed
l(m l)=
39. 2CO
2 2NO,O ,O−
are paramagnetic according to
MOT, unparired e−
are in * *
px py,
40. 2
As carboxylic acid is preffered functional group in the given molecule so it is given lower locant
41. pH = 9.35 , pOH = 14-9.35=4.65
x
pOH 4.74 log 4.65100
= + =
x
log 0.09100
= −
x
0.813100
=
x 81.3mmoles=
millimoles of ( )4 42NH SO 40.65 132 5.365g= =
42. 4
43. It is the defination fo Nucleoside
44. 100
2(g) (g) (g)XY XY Y+
time t = 0600 0 0(Pressure in mmHg)
time t = teq(600-x) x x (Pressure in mmHg)
given: 2(g)XY XY(g) Y(g)P P P 800+ + =
( )600 x x 800 x 200 − + + = =
2
XY(g) Y(g)
P
XY (g)
P PK 100
P
= =
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 8
45. 120.6kcal mol−
|
(s) 2 2 2
CH3
5C 4H CH C CH CH+ → = − =
H ? =
( ) ( )R p
H Bondenergy Bondenergy = −
(s) (g) H H C C C H C C5 C 4 H 2 H 8H 2 H− − = − −= + − − −
5 171 4 104 2 147 8 98.8 2 83 + − − −
120.6Kcalmol−=
46. 3
Terylene and Dacron are same name of polymer obtained by polymerisation of ethylene glycol and
Terepthalic acid. It is an ester polymer
47. Sol: 0.12264nm
o150 150
1.5 1.22A 0.122nmV 100
= = = = =
48. 26 min
0
C2.303log kt
C= −
40% to completion means 60 % unreacted
( )42.303log0.6 3.3 10 t−= −
t 26min =
49. 1
NH
O
O
O
O
N
O
O
CH2 Br⎯⎯⎯→KOH − N K
Br CH2 Cl
50. 99.1%
9.8
9.8g Cu 2 0.308 mole63.5
−= =
1
1000 3 0.311mole96500
− =
If 0.311 mol electrons provided by the current 0.308 mol was used to deposit copper. The current
efficiency is 0.308
100 99.1%0.311
=
51. The aE of forward reaction is more then that of backward reaction
9K decreases with temperature. Hence H must be –ve as evident from the Eq.
9
2
d ln K H
dT RT
=
( ) ( )a aH E FR E BR = −
( )a aE FR E (BR)
52. 3
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 9
HPh
CH3
Ph Br
CH3
is
PhPh
3CH
3H C
H
Br Br
Ph
Ph
H
3CH
3CH
⎯⎯⎯→rotate ⎯⎯⎯→2
alc
KOHE Ph
CH3Ph
CH3
53. 40, 20
Let x and y be the molar mass of A and B respectively then
( )
f
f
T 8 1000 80m 1 .....(1)
K x 2y 100 x 2y
= = − =
+ +
And
( )
f
f
T 10 1000 100m 1 .....(2)
K 2x y 100 2x y
= = − =
+ +
Solving (1) and (2) x = 40 ; y = 20
54. 4 6Na XeO
In the compound 4NaHXeO , Xenon is in +6 oxidation state. If it disproportionate to give ‘Xe’ (zero)
then Xe (VIII) must be formed. In this reaction 4 6Na XeO . 28H O get precipitated.
55. 4
Ranitidine is antihistamine so. Ans is D. Rest all are artificial sweetners.
56. behaves as strong electrolyte Below CMC no micellization takes place.
Sodium oleate ionizes almost completely, aqueous solution
57. 2
58. 2
2 3Ph - CH - CH ⎯⎯⎯→2Br /hν
3Ph - CH - CH
*
Br , so Ans is B. C & D forms diastereomers
59. 2O −
ions are present at octahedral voids 2Na O has antiflourite structure in which 2O − ions are
present at ccp and Na+ ions are occupy all the tetrahedral voids which are located at body diagonals.
60. Extraction is done by electrolysis
MATHS
61. 2 3 4 2 2 3 2 2 21 + + = + + + = + + + +
2 2 2 2 3 & y 3 x= + + + + = + = +
( ) ( )3 2
y 3 y 3 1 − = − +
3 2y 10y 33y 37 0 − + − =
62. R1 , R2, R3
contain 1 person each
1 1 1C C C2 .2 .2 3! P=
8.6 48
316 16
= =
63. x y 1 0 + − = & ( ) ( )6 x 2 8 y 2 0+ + − + = represent some time
2 6− =+
2 2 8 2 10 0− = − −+ =
64. ( )2013 22 2 2k
1 2 1 2
k 0
2Z Z 2014 Z Z 2014 2
2014=
+ = + = =
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 10
67. 1D 0 tan 3
3
−
1b
tan 0 02a
−−
( ) ( )3
1af 0 0 2 co t 0 R− =
Minimum integral value is 2
68. ( )( )2 2 2
0 1 21 x x 1 x x ........ a a x a x ......+ + + + + = + + +
1 2a 2 1 1 a 1 1 1 3= = + = + + =
3 4a 1 1 1 3 a 1 1 1 3......= + + = = + + =
( )2010a 1 1 1 3.....2 2009 3 6029= + + = + =
69. k
1 1 12cos k.sin sin k sin k
2 2 2t
1 12sin 2sin
2 2
+ − −
= =
n
1 1sin n sin n
2 2S
12sin
2
+ − −
=
n
n
Slt 0
n→= as nS is a finite quantity for every n
70. ( ) ( )22 22g x x g 3 3= =
( )( )22 1f g 3 f (3)
3= =
2 1 1f (1) f (2) f (3) f (2)
3 3 3= + = − =
2 1f (0) f (1) f (2) f (0) 1
3 3= + = + =
71. ( ) ( ) ( ) ( ) ( )( )2g ' 0 a 5, g '' 0 a 3, g ' 0 g '' 0 a 2 a 1 0 a 1, 2+ − = + + = + − = = −
72. AC a c,BC b c,OA b,OB a,OC a b c= + = + = = = + − Use cosine rule and ABC & OAC of
ABC & OAC
73. conceptual
74. conceptual
75. ( )2 2cot cot 1 1 cot + = −
2 22
2
sin coscot cosec
sin
− =
cot cos2 = −
Now cos2 tan2−
2
2 tancot
1 tan
− −
−
2
1 2 tan
tan 1 tan
− −
−
( )
2 2
2
1 tan 2 tan
tan 1 tan
− + −
−
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 11
( )
( )22
22
1 tan1 tancot
1 tantan 1 tan
+ − = =
− −
sec2 cot= −
1
.cotcos 2
= −
1
.cot 1cot
= =−
79. f(g(x) = x
( )( ) ( )' ' 1f g x g x =
( )( ) ( )' 1 ' 1 1f g g =
( ) ( )' 0 ' 1 1f g =
( )1
' 13
g =
G.E = 1 + 3 + 5 = 9= ( )( )
2
1
' 1g
80. 2ac
0 ba c
= =+
Now , ( )
( )( ) ( )
43 3 2 23 33
3 3 3 3 3
a c a c a ac ca ct a c ......(I)
b 8a c 8a c
+ + − ++= = + =
Now, ( )2 2 2a c a 2ac c 0− = − +
( )2 2a ac c ac........ II − +
Also, ( ) ( )4 2 2a c 2 ac a c 16a c ........ III+ +
(II) (III)..
( ) ( )4 2 2 3 3a c a ac c 16a c + − +
( ) ( )
4 2 2
3 3
a c a ac c2
8a c
+ − +
t 2
81.
( ) 1
n
1 1 1 11 1 1 1 1 ....... 1
3 4 5 n 12 3 4 n2 .........
n 2 3 4 n
+ + + + + + + + + + +
( ) ( )1
nn nn
n Sn 1 S n n 1 n
n
+ + + −
82. ( ) ( )r x a b y 2b c= + + −
r.a 1 x 1= =
( )r xa b x 2y yc= + + −
2 2r 13y 14y 5= + +
min
4r
13=
87. x y2 2cos sina 2 b 2
+ + + +
+
Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s
Sec: Sr. IIT_IZ Page 12
= 2cos 1eq2
−−
→
x cos ysin p + = → 2eq
(1) = (2)
88. Centre of circle ( )a,0 and radius 2a
Equation of circle ( )2 2 2x a y 4a− + =
( )2 2 2 2x a x y 2ax 3a 0− + − − = and
2y 4ax= solving 2 2x 4ax 2ax 3a 0+ − − =
2 2x 2ax 3a 0+ − =
x 3a,a= − and y 2a=
Length of AB 4a=
89. Equation of tangent to the hyperbola : 2 2 2y mx m a a= −
Let ( )1 1,P x y be locus
2 2 2y mx m a a− = −
S.B.S
( )2 2 2 2 2
1 1 1 12 0m x a y x m y a− − + + =
2 2
1 1 11 2 1 22 2 2 2
1 1
2;
x y y am m m m
x a x a
++ = =
− −
0 1 2
1 2
tan 451
m m
m m
−=
+
( ) ( ) ( )2 2 2
1 2 1 2 1 2 1 21 4m m m m m m m m + = − = + −
2 2 2
1 1 1 1
2 2 2 2 2 2
1 1 1
21 4
y a x y y a
x a x a x a
+ + + = −
− − − 90. Let the two unknown items be x and y, then
Mean 1 2 6 x y
4 45
+ + + += =
x y 11+ = ……. (i)
and variance = 5.2
( )2 2 2 2 2
21 2 6 x ymean 5.2
5
+ + + +=
( )22 241 x y 5 5.2 4 + + = +
2 241 x y 106+ + =
2 2x y 65+ = ……. (ii)
Solving Equations (i) and (ii) for x and y, we get
x 4, y 7 or x 7, y 4= = = =