2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
MATHEMATICS
1. For 1x , if 2 2 22 4y x y
x e , then 2
1 log 2e
dyx
dx is equal to
1) log 2e x
2) log 2 log 2
e ex x
x
3) log 2ex x
4) log 2 log 2
e ex x
x
Key: 4
Sol: 2 2 22 4y x y
x e 2 2 4 2 2y n x n x y
2
1 2
x ny
n x
1
2
11 2 2
1 2
n x x nxy
n x
21 2 2
1 2x n x n
y n xx
2. The sum of the distinct real values of , of rwhich the vectors,
ˆ, ,i j k i j k i j k are co-planer, is:
1) 2
2) 0
3) -1
4) 1
Key: 3
Sol:
1 1
1 1 0
1 1
2 1 1 1 1 1 0
3 1 1 0
3 3 2 0
3 1 3 1 0
21, 2 0
1 , 2 Sum of distinct solutions =-1
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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3. Let S be the set of all points in ( , at which the function, f ( x ) = m i n
sin ,cosf x x x is not differentiable. Then S is a subsest of which of the following?
1) 3 3
, , ,4 4 4 4
2) 3 3
, , ,4 2 2 4
3) , , ,2 4 4 4
4) ,0,4 4
Key: 1
Sol:
4. The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first
and the second of these terms, the three terms now from an A.P. Then the sum of the
original three terms of the given G.P. is
1) 36
2) 24
3) 32
4) 28
Key: 4
Sol: Let terms are a
r, a, .ar G P
3 512 8a a
8
4.12,8 . .r A Pr
824 4 8r
r
1
2,2
r r
r=2 (4, 8, 16)
116,8,4
2r
Sum=28
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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5. The integral cos loge x dx is equal to :
(where C is a constant of integration)
1) sin log cos log2 e e
xx x C
2) cos log sin log2 e e
xx x C
3) cos log sin loge ex x x C
4) cos log sin loge ex x x C
Key:2
Sol: cosI nx dx
cos sinI nx x nx dx
cos sin cosnx x nx x nx dx
sin cos2
xI nx nx C
6. Let 1 2 3 ...
k
kS
k
. If 2 2 2
1 2 10
5...
12S S S A , then A is equal to
1) 303
2) 283
3) 156
4) 301
Key:1
Sol: 1
2K
KS
2 1
2k
KS
2 5
12kS A
10 5
12Ki
A
2 2 2 210 1 2 3 .......... 11 5
2 4 12Ki
KA
11 12 23 5
16 3
A
5
5053
A , A=303
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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7. Let S={1,2,3,….100}. The number of non-empty subsets A of S such that the product of
elements in A is even is
1) 50 5020 2 1
2) 1002 1
3) 502 1
4) 502 1
Key: 1
Sol: 1,2,3......100S
=Total non empty subjects-subsets with product of element is odd
100 502 1 2 1
100 502 2 50 502 2 1
8. If the sum of the deviations of 50 observations from 30 is 50, then the mean of
these observation is :
1) 50
2) 51
3) 30
4) 31
Key:4
Sol: 50
1
30 50i
i
x
50 30 50ix 50 50 30ix
Mean 1 50 30 5030 1 31
50
xx
n
9. If a variable line, 3x+4y–=0 is such that the two circles x2+y2–2x–2y+1= 0 and
x2+y2–18x–2y+78 = 0 are on its opposite sides, then the set of all values of is the
interval :-
1) [12, 21]
2) (2, 17)
3) (23, 31)
4) [13, 23]
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Key:1
Sol: Centre of circles are opposite side of line 3 4 27 4 0
7 31 0
7,31
Distance from 1S
3 41 ,2 12,
5
Distance from 2S
27 4,21 41,
5
So 12,21
10. A ratio of the 5th term from the beginning to the 5th term from the end in the
binomial expansion of
1/31/3
12
2 3
is:
1) 1
31:4 16
2) 1
31: 2 6
3) 1
32 36 :1
4) 1
34 36 :1
Key: 4
Sol:
410 4110 3
4 13
1/3510 41
5
410 1/34 1
3
12
2 34. 36
12
2 3
C
T
T
C
11. let C1 and C2 be the centres of the circles x2+y2–2x–2y–2 = 0 and x2+y2–6x–
6y+4= 0 respectively. If P and Q are the points of in tersection of these circles,
then the area (in sq. units) of the quadrilateral PC1 C2 is :
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1) 8
2) 6
3) 9
4) 4
Key:4
Sol:
Area = 1
2 .4 22
12. In a random experiment, a fair die is rolled until two fours are obtained in succession.
The probability that the experiment will end in the fifth throw of the die is equal to :
1) 5
150
6
2) 5
175
6
3) 5
200
6
4) 5
225
6
Key: 2
Sol:
3 2
12 3 3 5
1 5 2 5 175
6 6 6 6
C
13. If the straight line, 2x–3y+17 = 0 is perpendicular to the line passing
through the points (7, 17) and (15, b), then b equals
1) -5
2) 35
3
3) 35
3
4) 5
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Key: 4
Sol: 17 2
18 3
5
14. Let f and g be continuous functions on [0, a] such that f(x) = f(a– x) and
g(x)+g(a–x)=4,then 0
a
f x g x dx is equal to
1) 0
4a
f x dx
2) 0
2a
f x dx
3) 0
3a
f x dx
4) 0
a
f x dx
Key:2
Sol: 0
a
I f x g x dx
0
a
I f a x g a x dx
0
4a
I f x g g x dx
0
4 1a
I f x dx
0
2a
I f x dx
15. The maximum area (in sq. units) of a rectangle having its base on the x-axis and its
other two vertices on the parabola, y = 12–x2 such that the rectangle lies inside the
parabola, is :
1) 20 2
2) 18 3
3) 32
4) 36
Key:3
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: 22 12f a a a
2' 2 12 3f a a
Maximum at a=2 Maximum area a=f(2)=32
16. The Booleean expression p q p q p q is equivalent to :
1) p q
2) p q
3) p q
4) p q
Key:3
Sol:
17.
3
4
cot tanlim
cos4
x
x x
x
is
1) 4
2) 8 2
3) 8
4) 4 2
Key: 3
Sol: 3
4
cot tanlim
cos4
x
x x
x
2
4
1 tanlim
cos 4x
x
x
2
4
1 tan2 lim
cos 4x
x
x
2 2
24
cos sin 12 lim
cos sin cos2
x
x x
x x x
4
4 2 lim cos sin 8x
x x
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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18. Considering only the principal values of inverse functions, the set
1 10 : tan 2 tan 34
A x x x
1) is an empty set
2) contains more than two elements
3) contains two elements
4) is a singleton
Key:4
Sol: 1 1tan 2 tan 3 4x x
2
51
1 6
x
x
26 5 1 0x x
1x or 1
6x
1
6x 0x
19. An ordered pair(a,b) for which the system of linear equations 1 2x x z
1 3x y z 2 2x y z has a unique solution is
1) (1,-3)
2) (-3,1)
3) (2,4)
4) (-4,2)
Key:3
Sol: For unique solution
1 1
0 1 1 0
2
1 1 0
0 1 1 0 2
2
20. The area (in sq. units) of the region bounded by the parabola, 2 2y x and the lines,
y=x+1, x=0 and x=3, is
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1) 15
4
2) 15
2
3) 21
2
4) 17
4
Key:2
Sol:
Req. area = 3
2
0
1 15 152 .5.3 9 6
2 2 2x dx
21. If be the ratio of the roots of the quadratic equation in x, 2 23 4 2 0m x m m x ,
then the least value of m for which 1
1
is
1) 2 3
2) 4 3 2
3) 2 2
4) 4 2 3
Key:2
Sol: 2 23 4 2 0m x m m x
1 1
, 1
, 2 2
23
2 2
2 2 2
4 3 2 4 6,
3 3 9 3
m m m
m m m m
24 18m , 4 18m , 4 3 2
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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22. If the vertices of a hyperbola be at (-2,0) and (2,0)and one of its foci be at (-3,0), then
which one of the following points does not lie on this hyperbola?
1) 4, 15
2) 6,2 10
3) 6,5 2
4) 2 6,5
Key:3
Sol:
Ae=3, 3
2e , 2 9
4 14
b
, 2 5b
2 2
14 5
x y
23. If zR
z
is a purely imaginary number and 2z , then a value of is:
1) 1
2) 2
3) 2
4) 1
2
Key:2
Sol: 0
z z
z z
2 2 0zz z z zz z z a
2 2
z , 2a
24. Let 4, 4P and 9,6Q be tow points on the parabola, 2 4y x and let X be any point on
the are POQ of this parabola, where O is the vertex of this parabola, such that the area of
PXQ is maximum. Then this maximum area (in sq. units) is :
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1) 125
4
2) 125
2
3) 625
4
4) 75
2
Key:1
Sol: 2 4y x 2 ' 4yy
1
' 2yt
, 1
2t
Area=
11 1
41 125
9 6 12 4
4 4 1
25. The perpendicular distance from the origin to the plane containing the two lines
2 2 5
3 5 7
x y z and
1 4 4
1 4 7
x y z , is
1) 11
6
2) 6 11
3)11
4) 11 6
Key:1
Sol:
3 5 7
1 4 7
i j k
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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ˆˆ ˆ35 28 21.7 12 5i j k
ˆˆ ˆ7 14i j k
ˆˆ ˆ2i j k
1 2 2 2 1 15 0x y z
25 11 0x z
11 11
4 1 1 6
26. The maximum value of 3cos 5sin6
for any real value of is
1) 19
2) 79
2
3) 31
4) 34
Key:1
Sol: 3 1
3cos 5 sin cos2 2
y
5 3 1sin cos
2 2
max
75 119
4 4y
27. A tetrahedron has vertices 1,2,1 , 2,1,3 , 1,1,2 0,0,0P Q R and O . The angle between
the faces OPQ and PQR is
1) 1 9cos
35
2) 1 19cos
35
3) 1 17cos
31
4) 1 7cos
31
Key:1
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: ˆ ˆˆ ˆ ˆ ˆ2 2 3OP OQ i j k i j k
ˆˆ ˆ5 3i j k
ˆ ˆˆ ˆ ˆ ˆ2 2PQ PR i j k i j k
ˆˆ ˆ5 3i j k
28. Let y y x bet he solution of the differential equation, log , 1e
dyx y x x x
dx . If
2 2 log 4 1ey , then y(e) is equal to
1) 2
4
e
2) 4
e
3) 2
e
4) 2
2
e
Key:2
Sol: dy y
nxdx x
1
dxxe x
xy x nx C 2 21
.2 2 2
x xnx
2
2 4
x xxy nx C , for 2 2 2 2 1y n
0C 2 4
x xy nx
4
ey e
29. Let
1 0 0
3 1 0
9 3 1
P
and ijQ q be two 3 3 matrices such that 5
3Q P I . Then 21 31
32
q q
q
is
equal to
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1) 15
2) 9
3) 135
4) 10
Key:4
Sol:
1 0 0
3 1 0
9 3 1
P
2
1 0 0
3 3 1 0
9 9 9 3 3 1
P
3
1 0 0
3 3 3 1 0
6.9 3 3 3 1
P
2
1 0 0
3 1 0
13 3 1
2
nP n
n nn
5
1 0 0
5.3 1 0
15.9 5.3 1
P
53Q P I
2 0 0
15 2 0
135 15 2
Q
21 31
32
15 13510
15
q q
q
30. Consider three boxes, each containing 10 balls labelled 1,2, .. .. ,10. Suppose one
ball is randomly drawn from each of the boxes. Denote by ni, the label of the
ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls
can be chosen such that n1 < n2 < n3 is
1) 82
2) 240
3) 164
4) 120
Key:4
Sol: Number of ways = 310 120C
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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PHYSICS
31. Two light identical springs of spring constant k are attached horizontally at the two ends
of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its centre
'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed
to rigid supports as shown in figure. The rod is gently pushed through a small angle and
released. The frequency of resulting oscillation is:
1) 2 6
2
k
m
2) 1 2
2
k
m
3) 1
2
k
m
4) 1 3
2
k
m
Key:1
Sol:
2 cos2
lK
2
2
KlC
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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2
2
1 1 22 2
12
KlC
fMlI
1 6
2
Kf
M
32. A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer
radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of
the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of
the system for heat flowing along the length of the cylinder is:
1) 1 2K K
2) 1 2
2
K K
3) 1 22 3
5
K K
4) 1 23
4
K K
Key:4
Sol:
1 1 2 2
1 2
eq
K A K AK
A A
2 2
1 2
2
3
4
K R K R
R
1 23
4
K K
33. A travelling harmonic wave is represented by the equation y (x, t) = 10–3 sin (50 t + 2x),
where x and y are in meter and t is in seconds. Which of the following is a correct
statement about the wave?
The wave is propagating along the
1) negative x-axis with speed 25ms–1
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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2) The wave is propagating along the positive x-axis with speed 25 ms–1
3) The wave is propagating along the positive x-axis with speed 100 ms–1
4) The wave is propagating along the negative x-axis with speed 100 ms–1
Key:1
Sol: siny a t kx
wave is moving along –ve x-axis with speed
50
25 / sec2
v v mK
34. A straight rod of length L extends from x = a to x=L + a. The gravitational force is exerts
on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2, is given by:
1) 1 1
Gm A BLa L a
2) 1 1
Gm A BLa a L
3) 1 1
Gm A BLa L a
4) 1 1
Gm A BLa a L
Key:2
Sol: dm= (A + Bx2)dx
2
GMdmdF
x
2
2
a L
a
GMF A Bx dx
x
a L
a
AGM Bx
x
1 1
GM A BLa a L
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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35. A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light
gets reflected and the amplitude of the electric field of the incident light is 30V/m, then
the amplitude of the electric field for the wave propogating in the glass medium will be:
1) 10 V/m
2) 24 V/m
3) 30 V/m
4) 6 V/m
Key:2
Sol: 1
96
100refractedP P 2 22 1
96
100t iK A K A 2 22 1
96
100t ir A r A
22 96 130
100 3tA 26430 24
100tA
36. The output of the given logic circuit is :
1) AB
2) AB
3) AB AB
4) AB AB
Key:2
Sol:
Y A B A
A AB A AB
A A B
A AB AB
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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37. In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the
energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is:
1) 23
8
Q
C
2) 23
4
Q
C
3) 21
8
Q
C
4) 25
8
Q
C
Key:1
Sol: 21
2iV CE
2 21
2 4 2 4f
CE CEV
c
2 21 3 3
2 4 8E CE CE
38. A particle of mass m moves in a circular orbit in a central potential field 21
2U r kr .
If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels
vary with quantum number n as:
1) 2
2
1,n nr n E
n
2) 1
,n nr n En
3) ,n nr n E n
4) ,n nr n E n
Key:4
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: 2
dV mvF kr
dr r
2
nhmvr
2r n 2r n
2 2 21 1
2 2E kr mv r
n
39. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series
across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2
respectively, then:
1) P1 = 9 W, P2 = 16 W
2) P1 = 4 W, P2 = 16W
3) P1 = 16 W, P2 = 4W
4) P1 16 W, P2 = 9W
Key:3
Sol: 2
1
220
25R
2
2
220
100R
1 2
220L
R R
21 1P i R
22 2 4P i R W
2 2
2 2
220 220
25220 220
25 100
400
1625
W
40. A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A
meteorite of the same mass, falling towards the earth, collides with the satellite
completely inelastically. The speeds of the satellite and the meteorite are the same, just
before the collision. The subsequent motion of the combined body will be :
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1) in a circular orbit of a different radius
2) in the same circular orbit of radius R
3) in an elliptical orbit
4) such that it escapes to infinity
Key:3
Sol: ˆ ˆmvi mvj 12mv 1
2
GMy
R
41. Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and
outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass
such that its moment of inertia about its axis is also I, is:
1) 12 cm
2) 18 cm
3) 16 cm
4) 14 cm
Key:3
Sol: Conceptual
42. A passenger train of length 60m travels at a speed of 80 km/hr. Another freight train of
length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger
train to completely cross the freight train when : (i) they are moving in the same
direction, and (ii) in the opposite directions is
1) 5
2
2) 25
11
3) 3
2
4) 11
5
Key:4
Sol: Conceptual
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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43. An ideal gas occupies a volume of 2m3 at a pressure of 3×106 Pa. The energy of the gas
is:
1) 3× 102
2) 108 J
3) 6×104 J
4) 9×106 J
Key:4
Sol: 1
2 2
fEnergy nRT PV
63 10 22
f
Considering gas is monoatomic i.e. f = 3 E. = 9 × 106 J
Option-(4)
44. A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal.
What is the modulation index?
1) 0.6
2) 0.5
3) 0.3
4) 0.4
Key:1
Sol: 160m cE E
100 160mE
60mE
60
100m
C
E
E
0.6
45. The galvanometer deflection, when key K1 is closed but K2 is open, equals q0 (see
figure).On closing K2 also and adjusting R2 to 5 , the deflection in galvanometer
becomes 0
5
. The resistance of the galvanometer is, then, given by [Neglect the internal
resistance of battery]:
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1) 12
2) 25
3) 5
4) 22
Key:4
Sol: Case – I
0220g
g
Ei C
R
_________ (i)
Case – II
055 55
2205
gg g
g
CEi
R R
R
_________ (ii)
05
225 1100 5g
CE
R
_____________ (ii)
0220 g
EC
R
______________ (i)
225 1100
51100 5
g
g
R
R
5500 25 225 1100g gR R
200 4400gR 22gR Ans : - 4
46. A person standing on an open ground hears the sound of a jet aeroplane, coming from
north at an angle 60° with ground level. But he finds the aeroplane right vertically above
his position. If is the speed of sound, speed of the plane is :
1) 2
3
2)
3) 2
4) 3
2
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Key:3
Sol: AB = Vp × t
BC = Vt
cos60AB
BC
1
2p
V t
Vt
2p
VV
47. A proton and an -particle (with their masses in the ratio of 1:4 and charges in the ratio
of (1:2) are accelerated from rest through a potential difference V. If a uniform magnetic
field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r of the
circular paths described by them will be :
1) 1: 2
2) 1: 2
3) 1:3
4) 1: 3
Key:1
Sol: KE q V
2mq V
rqB
m
rq
1
2
pr
r
48. A point source of light, S is placed at a distance L in front of the centre of plane mirror
of width d which is hanging vertically on a wall. A man walks in front of the mirror along
a line parallel to the mirror, at a distance 2L as shown below. The distance over which the
man can see the image of the light source in the mirror is :
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1) 3d
2) 2
d
3) d
4) 2d
Key:1
Sol:
3d
49. The least count of the main scale of a screw gauge is 1 mm. The minimum number of
divisions on its circular scale required to measure 5 m diameter of wire is :
1) 50
2) 100
3) 200
4) 500
Key:3
Sol: Pitch
Least count=Number of division on circular scale
3
6 105 10
N
N = 200
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50. A simple pendulum, made of a string of length l and a bob of mass m, is released from a
small angle 0 . It strikes a block of mass M, kept on a horizontal surface at its lowest
point of oscillations, elastically. It bounces back and goes up to an angle 1 . Then M is
given by :
1) 0 1
0 12
m
2) 0 1
0 12
m
3) 0 1
0 1
m
4) 0 1
0 1
m
Key:3
Sol: Before colision After collision
0 1 12 1 2 1v gl cos v gl cos
By momentum conservation
0 12 1 2 1mm gl cos MV m gl cos
0 12 1 1 mm gl cos cos MV
And
1
0
2 11
2 1
mV gl cose
gl cos
0 12 1 1 mgl cos cos V ____________ (i)
0 12 1 1 mm gl cos cos MV _________ (ii)
Dividing
0 1
0 1
1 1
1 1
cos cos M
mcos cos
By componendo divided
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1
1
00
sin1 2
1 sin2
cosm M
m M cos
0 1 0 1
0 1 0 1
MM
m
51. What is the position and nature of image formed by lens combination shown in figure?
(f1, f2 are focal lengths)
1) 70 cm from point B at left; virtual
2) 40 cm from point B at right; real
3) 20
3cm from point B at right , real
4) 70 cm from point B at right, real
Key:4
Sol: For first lens
1 1 1
20 5V
20
3V
For second lens
20 14
23 3
V
1 1 1
14 53
V
V = 70 cm 52. In the figure shown, a circuit contains two identical resistors with resistance R = 5 and
an inductance with L = 2mH. An ideal battery of 15 V is connected in the circuit. What
will be the current through the battery long after the switch is closed?
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1) 6A
2) 7.5A
3) 5.5A
4) 3A
Key:1
Sol: Ideal inductor will behave like zero resistance long time after switch is closed
2 2 15
65
I AR
53. Determine the electric dipole moment of the system of three charges, placed on the
vertices of an equilateral triangle, as shown in the figure:
1) 2
i jql
2) 32
j iql
3) 3ql j
4) 2ql j
Key:3
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Sol:
1P q d 2P qd Resultnat 2 cos30P 3
2 32
pd pd
54. The position vector of the centre of mass r
cm of an symmetric uniform bar of negligible
area of cross-section as shown in figure is :
1) 13 5
8 8rcm Lx Ly
2) 11 3
8 8rcm Lx L y
3) 3 11
8 8rcm Lx L y
4) 5 13
8 8rcm Lx Ly
Key:1
Sol:
52 2 132
4 8cm
mLmL mL
X Lm
2 052
4 8cm
Lm L m m
LY
m
55. As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in
such a way that the segments LP and QM are along the x-axis, while segments PS and
QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magnetic field
at O is 10–4 T, and the two wires carry equal currents (see figure), the magnitude of the
current in each wire and the direction of the magnetic field at O will be ( 0 = 4×10–7
NA– 2) :
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1) 40 A, perpendicular into the page
2) 40 A, perpendicular out of the page
3) 20 A, perpendicular out of the page
4) 20 A, perpendicular into the page
Key:4
Sol: Magnetic field at ‘O’ will be done to ‘PS’ and ‘QN’ only i.e. B0 = BPS + BQN Both inwards Let current in each wire = i
0 00 4 4
i iB
d d
or 7
4 02
2 1010
2 4 10
i i
d
i = 20A
56. In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the
variation dR
dl of its resistance R with length l is
1dR
dl l . Two equal resistances are
connected as shown in the figure. The galvanometer has zero deflection when the jockey
is at point P. What is the length AP?
1) 0.25 m
2) 0.3m
3) 0.35 m
4) 0.2 m
Key:1
Sol: For the given wire : ,dl
dR Cl
where C = constant.
Let resistance of part AP is R1 and PB is R2
1
2
'
'
RR
R R or R1 = R2 By balanced WSB concept.
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Now dl
dR cl
1
1/21
0
.2.R C l dl C l
1
1/22
0
2 2R C l dl C l Putting R1 = R2
2 2 2C l C l 2 1l 1
2l
i.e., 1
0.254
l m m
57. For the given cyclic process CAB as shown for a gas, the work done is :
1) 1 J
2) 5 J
3) 10 J
4) 30 J
Key: 3
Sol: Since P–V indicator diagram is given, so work done by gas is area under the cyclic diagram.
W Work done by gas = 1
2 × 2 × 4 × 5 J
= 10 J
58. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a
potentiometer of length 1 m and resistance 5 . The value of R, to give a potential
difference of 5 mV across 10 cm of potentiometer wire, is :
1) 490
2) 480
3) 395
4) 495
Key:3
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Sol: Let current flowing in the wire is i.
4
5i A
R
If resistance of 10 m length of wire is x
then 0.1
0.5 51
x
Dv = P. d. on wire = i. x
3 45 10 0.5
5R
2410 or 5 400
5R
R
395R
59. A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V.
Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of
2500 V. The ratio of de-Broglie wavelengths A
B
is close to :
1) 10.00
2) 14.14
3) 4.47
4) 0.07
Key:2
Sol: K.E. acquired by charge = K = qV
2 2
h h h
p mK mqV
2 4 . .2500
2 50. .502
B B BA
B A A A
m q V m q
m qm q V
= 2 × 7.07 = 14.14
60. There is a uniform spherically symmetric surface charge density at a distance R0 from the
origin. The charge distribution is initially at rest and starts expanding because of mutual
repulsion. The figure that represents best the speed V(R(t)) of the distribution as a
function of its instantaneous radius R (t) is :
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1)
2)
3)
4)
Key:1
Sol: At any instant 't' Total energy of charge distribution is constant
i.e. 2 2
2
0
10
2 2 2
KQ KQmV
R R
2 22
0
1
2 2 2
KQ KQmV
R R
2
0 0
2 1 1.
2
KQV
m R R R
2
0 0
1 1 1 1KQV C
m R R R R
Also the slope of v-s curve will go on decreasing Graph is correctly shown by option(1)
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CHEMISTRY
61. Iodine reacts with concentrated HNO3 to yield Y along with other products. The
oxidation state of iodine in Y, is :-
1) 5
2) 3
3) 1
4) 7
Key: 1
Sol: 2 3 3 2 210 2 10 4I HNO HIO NO H O
In 3HIO oxidation state of iodine is +5
62. The major product of the following reaction is:
1)
2)
3)
4)
Key: 3
Sol:
DIBAL-H will reduce cyanides & esters to aldehydes.
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63. In a chemical reaction, 2 2K
A B C D , the initial concentration of B was 1.5 times
of the concentration of A, but the equilibrium concentrations of A and B were found to be
equal. The equilibrium constant(K) for the aforesaid chemical reaction is :
1) 16
2) 4
3) 1
4) 1
4
Key: 2
Sol: 2 2A B C D 0 00 1.5 0 0t a a 0 01.5 2 2eqt t a x a x x x
At equilibrium [A] = [B]
0 0 0
0 0 0 0
1.5 2 0.5
0.5 0.5 0.5eq
a x a x x a
t t a a a a
22
0 02 2
0 0
0.54
0.5 0.5C
a aC DK
A B a a
64. Two solids dissociate as follows
1
2; PA s B g C g K x atm
2
2; PD s C g E g K y atm
The total pressure when both the solids dissociate simultaneously is :-
1) 2 2x y atm
2) x y atm
3) 2 x y atm
4) x yatm
Key: 3
Sol:
1
1 2 1 1 2
.
..... 1
P B CA s B g C g K x P P
P P x P P P
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2
2 2 1 2 2
.
..... 2
P C ED s C g E g K y P P
P P y P P P
Adding 1 and 2 X+y = (P1 + P2)
2 Now total pressure T C B EP P P P
1 2 1 2 1 22P P P P P P
2TP x y
65. Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous
solution of Y. If molecular weight of X is A, then molecular weight of Y is :-
1) A
2) 3A
3) 4A
4) 2A
Key: 2
Sol: For same freezing point, molality of both solution should be same. x ym m
4 1000 12 1000
96 88x yM M
96 12
, 3.274 88yM M A
Closest option is 3A
66. Poly-b-hydroxybutyrate-co-bhydroxyvalerate( PHBV) is a copolymer of___.
1) 3-hydroxybutanoic acid and 4-hydroxypentanoic acid
2) 2-hydroxybutanoic acid and 3-hydroxypentanoic acid
3) 3-hydroxybutanoic acid and 2-hydroxypentanoic acid
4) 3-hydroxybutanoic acid and 3-hydroxypentanoic acid
Key: 4
Sol: PHBV is a polymer of 3-hydroxybutanoic acid and 3-Hydroxy pentanoic acid.
67. Among the following four aromatic compounds, which one will have the lowest melting
point ?
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1)
2)
3)
4)
Key: 1
Sol: M.P. of Napthalene 80°C
68. cannot be prepared by
1) 3 2HCHO PhCH CH CH MgX
2) 2 3 3PhCOCH CH CH MgX
3) 3 3 2PhCOCH CH CH MgX
4) 3 2 3CH CH COCH PhMgX
Key: 1
Sol:
69. The volume of gas A is twice than that of gas B. The compressibility factor of gas A is
thrice than that of gas B at same temperature. The pressures of the gases for equal
number of moles are :
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1) 2 3A BP P
2) 3A BP P
3) 2A BP P
4) 3 2A BP P
Key: 1
Sol: 2A BV V
3A BZ Z
3. .
.A A B B
A A B B
P V P V
n RT n RT
2 3A BP P
70. The element with Z = 120 (not yet discovered) will be an/a :
1) transition metal
2) inner-transition metal
3) alkaline earth metal
4) alkali metal
Key: 3
Sol: Z = 120 Its general electronic configuration may be represented as [Nobal gas] ns2 , like other
alkaline earth metals.
71. Decomposition of X exhibits a rate constant of 0.05 mg/year. How many years are
required for the decomposition of 5 mg of X into 2.5 mg ?
1) 50
2) 25
3) 20
4) 40
Key: 1
Sol: Rate constant (K) = 0.05 g/year means zero order reaction
21/2
550
2 2 0.05 /
a gt year
K g year
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72. The major product of the following reaction is :
1)
2)
3)
4)
Key: 4
Sol:
73. Given
Gas H2 CH3 CO2 SO2
Critical 33 190 304 630
Temperature/K On the basis of data given above, predict which of the following gases
shows least adsorption on a definite amount of charcoal ?
1) H2
2) CH4
3) SO2
4) CO2
Key: 1
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Sol: Smaller the value of critical temperature of gas, lesser is the extent of adsorption. so least adsorbed gas is H2
74. For diatomic ideal gas in a closed system, which of the following plots does not correctly
describe the relation between various thermodynamic quantities ?
1)
2)
3)
4)
Key: 2
Sol: At higher temperature, rotational degree of freedom becomes active.
7
2PC R (Independent of P)
5
2VC R (Independent of V)
Variation of U vs T is similar as CV vs T.
75. The standard electrode potential E and its temperature coefficient dE
dT
for a cell are
2V and –5×10–4 VK–1 at 300 K respectively. The cell reaction is
2 2Zn s Cu aq Zn aq Cu s
The standard reaction enthalpy , H at 300 K in kJ mol–1 is, [Use R = 8jK–1 mol–1 and
F = 96,000 Cmol–1]
1) –412.8
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2) –384.0
3) 206.4
4) 192.0
Key: 1
Sol: Chiefly NO2 , O3 and hydrocarbon are responsible for build up smog.
76. The molecule that has minimum / no role in the formation of photochemical smog, is:
1) CH2=O
2) N2
3) O3
4) NO
Key: 2
Sol: Conceptual
77. In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out
of :
1) Platinum
2) Carbon
3) Pure aluminium
4) Copper
Key: 2
Sol: In the Hall-Heroult process the cathode is made of carbon.
78. Water samples with BOD values of 4 ppm and 18 ppm, respectively, are :
1) Highly polluted and Clean
2) Highly polluted and Highly polluted
3) Clean and Highly polluted
4) Clean and Clean
Key: 3
Sol: Clean water would have BOD value of less than 5 ppm whereas highly polluted water
could have a BOD value of 17 ppm or more.
79. In the following reactions, products A and B are :
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1)
2)
3)
4)
Key: 4
Sol:
80. What is the work function of the metal if the light of wavelength 4000 Å generates
photoelectrons of velocity 6×105 ms–1 form it ?
(Mass of electron = 9×10–31 kg Velocity of light = 3×108 ms–1 Planck's constant = 6.626
× 10–34 Js Charge of electron = 1.6 × 10–19 JeV–1)
1) 0.9 eV
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2) 4.0 eV
3) 2.1 eV
4) 3.1 Ev
Key: 3
Sol: ºhv hv 2
0
1 1 1
2mv he
21
2hv mv
34 8
231 5
10
6.626 10 3 10 19 10 6 10
4000 10 2
193.35 10 2.1J eV
81. Among the following compounds most basic amino acid is :
1) Lysine
2) Asparagine
3) Serine
4) Histidine
Key: 4
Sol: Histidine
82. The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are :
1) 2,xz yz zd d and d
2) ,xy xz yzd d and d
3) 2 2xy x yd and d
4) 2 2 2x y zd and d
Key: 4
Sol: 3 6K Co CN 3 6
183Co Ar d
83. The hardness of a water sample (in terms of equivalents of CaCO3) containing
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10–3MCaSO4 is:
(molar mass of CaSO4 = 136 g mol–1)
1) 100 ppm
2) 50 ppm
3) 10 ppm
4) 90 ppm
Key: 1
Sol: 3ppmof CaCO
3 310 10 100 100 ppm
84. The correct order for acid strength of compounds
3 2 2,CH CH CH C CH and CH CH
1) 2 2 3CH CH CH CH CH C CH
2) 3 2 2HC CH CH C CH CH CH
3) 3 2 2CH C CH CH CH HC CH
4) 2 22CH C CH CH CH CH CH
Key: 2
Sol: 3 2 2CH CH CH C CH CH CH (Acidic strength order)
85. Mn2(CO)10 is an organometallic compound due to the presence of :
1) Mn – Mn bond
2) Mn – C bond
3) Mn – O bond
4) C – O bond
Key:2
Sol: Compounds having at least one bond between carbon and metal are known as organometallic compounds.
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86. The increasing order of reactivity of the following compounds towards reaction with
alkyl halides directly is :
1) (B) < (A) < (D) < (C)
2) (B) < (A) < (C) < (D)
3) (A) < (C) < (D) < (B)
4) (A) < (B) < (C) < (D)
Key: 2
Sol: Nucleophilicity order
87. The pair of metal ions that can give a spinonly magnetic moment of 3.9 BM for the
complex [M(H2O)6]Cl2, is :
1) Cr2+ and Mn2+
2) V2+ and Co2+
3) V2+ and Fe2+
4) Co2+ and Fe2+
Key:2
Sol:
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88. In the following reaction Aldehyde + Alcohol HCl Acetal
Aldehyde Alcohol
HCHO tBuOH
CH3CHO MeOH
The best combinations is:
1) HCHO and MeOH
2) HCHO and tBuOH
3) CH3CHO and MeOH
4) CH3CHO and tBuOH
Key: 1
Sol:
1
rate steric crowding of aldehyde
t-butanol can show formation of carbocation in acidic medium
89. 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution.
The amount of NaOH in 50 mL of the given sodium hydroxide solution is :
1) 40 g
2) 20 g
3) 80 g
4) 10 g
Key: 0 (add)
Sol: 2 2 4 2 2 4 22 2H C O NaOH Na C O H O
eqm of 2 2 4 eqH C O m NaOH
50 0.5 2 25 1NaOHM
2NaOHM M
Now 1000ml solution 2 10 gram NaOH
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50 ml solution = 4 gram NaOH
90. A metal on combustion in excess air forms X, X upon hydrolysis with water yields H2O2
and O2 along with another product. The metal is :
1) Rb
2) Na
3) Mg
4) Li
Key:1
Sol: 22 excessRb O RbO
2 2 2 2 22 2 2RbO H O RbOH H O O