iit jee - mains model test paper -1 (physics, chemistry, maths) - solutions
TRANSCRIPT
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7/24/2019 IIT JEE - Mains Model Test Paper -1 (Physics, Chemistry, Maths) - Solutions
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Max Marks:36
KEY & HINTS
PHYSICS
1 2 3 4 5 6 7 8 9 10
C C B B B A A D B B
11 12 13 14 15 16 17 18 19 20
A B A D B B A A B C
21 22 23 24 25 26 27 28 29 30
C A A A C B D B C A
MATHS
31 32 33 34 35 36 37 38 39 40
A D C B D B D B A C
41 42 43 44 45 46 47 48 49 50
B C D C B C C D B C
51 52 53 54 55 56 57 58 59 60
B B B B B A C B D D
CHEMISTRY
61 62 63 64 65 66 67 68 69 70
C D A D B B A A C B
71 72 73 74 75 76 77 78 79 80
A B D B D B C C B B
81 82 83 84 85 86 87 88 89 90
D D A A A B C D B D
PHYSICS SOLUTIONS
1. Conceputal
2. Conceputal
3. Conceputal
4.
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Page 2
5. L-conservation 2
2 12 2 6cm cm
L ML MvL LMv v v
P-conservation 22 2 6
cm
L L Lv v v
3v
L
6. v vnRTPdV nC dT dV nC dTV
32
dV dT V T
2 3V CT , where C is a constant.
7. Induced electric field will have lines of force in the form of concentric circles and every segment has
their own enduced emf by Faraday law of electromagnetic induction.
8. Conceptual
9. Conserving moment of system along rod.
By work energy theorem :
10. Let v be the velocity of the engine and V be the velocity of sound. The frequency of the sound heard
as the engine approaches the observer is given by
The frequency of the sound heard as the engine recedes from the observer is given by
It is given that
The difference in the frequencies of the sound heard is
11. Conceptual
12. Conceptual
13. Conceptual
14. for wire of radius b is
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15. 2r
r ( )2
ll
2r l
2r
l
16.
17. If there will be no current in G. As the potential difference across resistance equal to then
18. FBD of cylinder and block are as shown by Newtons laws
402Tfs= 8a (1)
Tfs= 4a/2 (2)
Subtracting equation (2) from equation (1)
403T = 6a
T = (406a)/3Also, by = I, we get
T R + fsR =3a
I2R
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T + fs= 3Ia/2R2
(406a)/3 + (4012a)/3 = 3Ia/2R2 a = 80/(18 + 9I/2R
2)
As = I= I (3a/2R)
= I= I (3a/2R)
2
3I 80
9I2R18
2R
3 80 80
N m18 9 9
19. Conceptual
20. mv2/r = GMm/r2
mvr = nh/2
= h/mv
min= h2/2GMm
2
21. For V = 0, x = amplitude
AmplitudeE
A
For x = 0, V = Vmax
maxEVB
Vmax= A
E
AB
BE
A
BT 2
A
Dimensions of Ax2
= dimensions of BV2
Dimensional formula for1 2
2
2
A [LT ][T ]
B [L ]
.
22. Current 0 0 0V V sin t V cos t
iR
0 0
V V 2 sin( t 45 )
R
T2
0 0rms T
0
i dt V 2i
Rdt
.
23. i + eA = 60
i + e = 120
and i 30 = ei = 75, e = 45
i = 45, e = 75.
24. When the oil is poured and fraction of ice in the water decreases. Volume of ice melted into water >
volume of water displaced by ice So water level rises Overall volume of ice will decrease as it melts.
So oil's upper level falls.
25. Velocity of efflux is 2g
2
y y
1s u t gt
2 ; x xs u t
26. ForF
a =M + m
no slipping between m and M
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For no toppling of m
27. Conceptual
28. To get for curved surface
Use west d=3cm
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29. Energy stored in the inductor = heat produced
=
=
Hence (C) is the correct answer.
30.
Magnitude of induced emf
= constant
So, it a case of charging of a capacitor in C R circuit. Hence of
Where
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MATHS SOLUTIONS31) Two persons between A and B be filled in 6.5 ways. A,B be rearranged in 2 ways. Now taking these
4 members as a group the 5 persons can be rearranged in 120 ways
32) Let the ratio of roots be :1, then
2 2 21 p l
q m
33)
Cos= -4/5, then sin= 3/5
34) Use l-hospital rule
35)
0
2
lim
,2
y
f x y f xf x kx
y
f x kx
xf x k Find k and
36)
2 22 2
2 2 1 1 5 5a a a
37) Put tan2
xt
38) 2 2cos cos sin
axax ee bx dx a bx b bx
a b
39) Given integral =1 2
2
0 1
x dx xdx
40) 0.08dy f x x
41) B is the image of (5, 7) in the line x+3y=16
42)
Clearly 1 2 1 2 1 20, 0c c a a b b , hence obtuse angle bisector is2 7 1
2 5 2
x y x y
43) Req area = 11r S for the point C
44) Tangent with slope 1 is 2x2y +1 =0, hence the point of contact is (1/2, 1)
45)21 4
2c m
m
46) External angle bisector is the tangent at /4
47) Conceptual
48) Use the formula of cos(A+B+C)
49) Eliminate r from the given equations
50)
1 11 1cos cos2 2
x x
51)
b=5 and 2 2 2b a c and c-a = 1
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52) length of the altitude = 1 2
1 2
2r r
r r
53) 21 2, 2 , 2x w w
54)
Use1
1
1z
z
2
2
25z
z
55) The given lines are concurrent implies k = 1
56)
P = 2 1 2
3,1, 1 15 , ,3 3 3
57) Conceptual
58) 1 6 111
35 25
gof g
59)
x x is satisfied by all integers and negative real numbers
60)
Replace x by x+99
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CHEMISTRY SOLUTIONS
61. Conceptual
62. Conceptual
63. 4 45HIO 5HIO 2Glu cos e 5HCOOH HCHO; Fructose 3HCOOH 2HCHO CO
64. Conceptual
65.
OHCONaClHClCONagmgm
225.36210 6
32 22
Eq.wt of Na2CO3= 106/2 = 53 gm
66. For adiabatic changes dq = 0 hance 0Tdq
S
67. Conceptual
68.
HInI
PP nKinH
log
During colour change HInIn . So PH= 5
69. Conceptual
70. As C2has max. bond order = 2, hence max. B.E & min. bond length.
71. Lets total no. of moles of the gas be n, of which n1are in the larges sphere & n2in the smaller sphere
after the stop cock is opened
N = n1+ n2& pV = nRT
RT
Vp
RT
Vp
RT
pV
2
|
1
|
1 2
12
2|
2
2
TT
pTp
72. Lattice energy of hydrides of alkalimeter decrease down the gp, hence stability of hydrides &reactivity of GPIA elements with H2decreases.
73. RTEiiieAK
where i = 1,2,3..
(1)RTEaAeK
(2)
Putting 1 toRTE
RTERTE
eA
eAeA
K
KKK
2
31
22
2
3
33
1
2
2
3
3
1
.
....3.3
321 231
2
2
3
3
1 ...3 EEE
RTeA
AA
(3)
Now comparison 2 with 3 we get the ans.
74.
OHBaSOClCrOSOKSOHOCrKBaClAd
24Re
2242427222 32232
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OHNaClCrONaNaOHClCrO
BYellowC24222 224
COONaCHPbCrOCrONaPbCOOCHCYellow
344223 2
75.M
E C tw
316
500 2 160 10 1600.1 J KJ
(minus sign is used as heat is evolved)
76. Due to the absence of d-orbitals, F2does not combine with F-to from
3F ion.
77. Conceptual
78. Conceptual
79. Conceptual
80. Conceptual
81. Conceptual
82. Conceptual
83. Conceptual
84. Conceptual
85. Conceptual
86. Conceptual
87. Conceptual
88. Conceptual
89. Conceptual
90. Conceptual