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CODE-B

JEE(MAIN) – 2017 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 02nd APRIL, 2017)
PART A – PHYSICS
1
...
AT time t = 0, it is at its position
of equilibrium
...
E
...
E
...
The atmospheric pressure in the room
remains 1 × 105 Pa
...
5 × 1025
(2) –2
...
61 × 1023
(4) 1
...
(2)

T/4 T/2

T

t

KE

(3)

0

T/2

T

T t

KE

Sol
...
(2)

T

t

Vmax

V=0



Sol
...


KE

(2)

1
mA 2 2 cos2 t
2

Equilibrium
position

V=0

(N is number of moles)
P0V0 = NiR × 290


...
(2)
[Tf = 273 + 27 = 300 K]

from equation (1) and (2)
Extreme
position

Nf – Ni =

P0 V0
PV
 0 0
R  300 R  290

Time taken to reach the extreme position from
equilibrium position is

T

...

V = A cos t
K
...
=

1 2
mv
2

(m is the mass of particle and v is the velocity of
particle

difference in number of moles 

P0 V0  10

 290  300 
R 


Hence nf – ni is
= 

P0 V0  10


 6
...
5 × 1025

1

JEE(MAIN)-2017
Which of the following statements is false ?
(1) A rheostat can be used as a potential divider
(2) Kirchhoff's second law represents energy
conservation
(3) Wheatstone bridge is the most sensitive when
all the four resistances are of the same order
of magnitude
...

Ans
...


4
...
25 × 10–2 m
rise of water, h = 1
...
80 m/s2 and the simplified relation
rhg
× 103 N/m, the possible error in surface
T=
2
tension is closest to :
(1) 2
...
15% (4) 1
...
(4)

Sol
...


R2

R1
G

T r h


0
T
r
h





R1 R 2

R3 R 4



T  102 
...
01 


100
T  1
...
45  102 

= (0
...
689)

On balancing condition

5
...
(1)

= (1
...
489 %
T

 1
...
The bandwidth (m)
of the signal is such that m << c
...


(2)

(2) 10%

modulated wave ?

G

(1) m + c
(3) m

R4

R3

Sol
...
526

...
So
4th statement is false
...
(3)

On balancing condition

R1 R 3

R2 R

(2) c – m

Three frequencies are contained
m + c, c – m & c

CODE-B
A diverging lens with magnitude of focal length
25 cm is placed at a distance of 15 cm from a
converging lens of magnitude of focal length
20 cm
...
The final image formed is :
(1) real and at a distance of 40 cm from the
divergent lens
(2) real and at a distance of 6 cm from the
convergent lens
(3) real and at a distance of 40 cm from convergent
lens
(4) virtual and at a distance of 40 cm from
convergent lens
...
(3)
Sol
...


I

m  2
m 
 

4  3  

For maxima & minima
dI m  2
m 
   2 0
d 4  3   

2 R 2 
2
m
 2 

3  
3
 2

or

2 R 2

3




2 3

R2 2

or

2
3

R
2







f = –25 cm

8
...

The moment of inertia of a uniform cylinder of
length  and radius R about its perpendicular
bisector is I
...


(1) 1

An electron beam is accelerated by a potential
difference V to hit a metallic target to produce
X-rays
...
If  min is the smallest
possible wavelength of X-ray in the spectrum, the
variation of log  min with log V is correctly
represented in :

(2)

3
2

(3)

3
2

(4)

log min
(1)
log V
log min

(2)
log V

log min
(3)
log V

3
2

log min

Ans
...


I

(4)

m 2 mR 2

12
4

or

I

log V

m  2
2
 R 
4 3


Also

m = R2



R2 


...
(3)
Sol
...

A radioactive nucleus A with a half life T, decays
into a nucleus B
...

At sometime t, the ratio of the number of B to that
of A is 0
...
Then, t is given by :
(1) t = T log (1
...
3

(2) t =



NB

...
3 NA
NA



 
ˆ
ˆ
 p  3E k  p E  k 
ˆ
0  Ek  3p  p 
x

py
px





t

10
...
3 n 1
...
When

ˆ
subjected to an electric field E1  Ei , it

ˆ
experiences a torque T1  k
...
The angle  is :
(1) 60°

4

(2) 90°

(3) 30°

 3

tan  =

(4) 45°

3

Sol
...


n 1
...
3) = t or t 
1
...
In a common emitter amplifier circuit using an
n-p-n transistor, the phase difference between the
input and the output voltages will be :
(1) 135°
(2) 180°
(3) 45°
(4) 90°
Ans
...
3



y

x

T
log(1
...
3
(4) t = T log 2

Ans
...
At time t

N
NA  0
1
...
So from



9
...
(1)


e
  nV  n
hc


put voltage are out of phase
...
However
increase in voltage across RC is in opposite sense
...
It is observed
that
Cp – Cv = a for hydrogen gas
Cp – Cv = b for nitrogen gas
The correct relation between a and b is :

(1) a = 14 b
1
(3) a =
b
14
Ans
...
CP – CV = R
where CP and CV are molar specific heat capacities
As per the question
a=
a = 14 b

R
2

b=

R
28

CODE-B
13
...
It is dropped in a copper calorimeter of mass

Ans
...
10 = (5 × 10–3) (15 + R)

100 gm, filled with 170 gm of water at room
temperature
...


system is found to be 75°C
...
1 cal/gm°C
(1) 1250°C

(2) 825°C

(3) 800°C
Ans
...
There is negligible friction at
the pivot
...
The angular acceleration
of the rod when it makes an angle  with the
vertical is :
z

Sol
...
1)(T – 75) = (100)(0
...


A body of mass m = 10–2 kg is moving in a medium
and experiences a frictional force F = –kv2
...
(1)
1 2 1 2
mvf  mv0
Sol
...




speed is v0 = 10 ms–1
...
(3)
(1)

10

dv
 v2 = – 100 k  dt
10
0

17
...
The ratio of the wavelengths r = 1/2,
is given by :

1 1
–
= 100 k (10)
5 10
k = 10–4
15
...
The value of the

–3E

resistance to be put in series with the galvanometer
to convert it into to voltmeter of range 0 – 10 V is:
(1) 2
...
005 × 103 

(3) 1
...
045 × 103 

3
4
4
(3) r 
3
(1) r 

1
3
2
(4) r 
3
(2) r 

5

JEE(MAIN)-2017
Current

Ans
...
using E =

1 =

2 =

hC
E
q=

3hC
E

q =  I dt

(3) 9

1
(4)
9

=

1
 10  0
...
5 coloumb
2

q=


R

= 2
...




Force mg volume  density  g
Stress 


area
A
Area
3

Stress =

L g
L2



Stress  L
In a coil of resistance 100 , a current is induced

by changing the magnetic flux through it as shown
in the figure
...
)

In a Young's double slit experiment, slits are
cm away
...
The least
distance from the common central maximum to the
point where the bright fringes due to both the
wavelengths coincide is :
(1) 9
...
6 mm

(3) 1
...
8 mm

Ans
...
For common maxima
n11 = n22
n1 × 650 = n2 × 520

n1 4

n2 5
Time 0
...
5 mm, and the screen is placed 150

Ans
...
(1)


R



1
(2)
81

time

 = change in flux

1
1
2 = r = 3
A man grows into a giant such that his linear
dimensions increase by a factor of 9
...


0
...


Sol
...


10A

hC


yd
 n
D
4  650  109  1
...
5  103
y = 7
...


A magnetic needle of magnetic moment

23
...
7 × 10 –2 Am 2 and moment of inertia

In the above circuit the current in each resistance
is :

2V

7
...
01 T
...
98 s

(2) 8
...
65 s

(4) 8
...
5 A
Ans
...
(3)

I
MB
I = 7
...
7 × 10–2 Am2
By substituting value in the formula
T =
...
65 sec
Answer option 3
The variation of acceleration due to gravity g with
distance d from centre of the earth is best
represented by (R = Earth's radius):

Sol
...
(2)

Sol
...


6V

Taking voltage of point A as = 0
Then voltage at other points can be written as
shown in figure
Hence voltage across all resistance is zero
...
The
2
collision is head on, and elastic
...


A 2
(1)   3
B
(3)

A 1
(2)   2
B

A 1

B 3

(4)

After collision
M
M
m V1
Sol
...
(4)

mv = mv1 +

d

(4) 0
...


2V

by law of collision
v  v1
e 2
u1  u 2
u = v2 – v1


...
(2)

7

JEE(MAIN)-2017
By equation (1) and (2)
v
4v
v1 
;
v2 
3
3

1 

h
;
p1

2 

From work energy theorem
1
WF  K
...

K is the bulk modulus of the material of the cube
and  is its coefficient of linear expansion
...
The temperature should be raised by :
3
(2) 3PK
(1)
PK

P
(3)
3K

P
(4)
K

Ans
...
Due to thermal expansion

v
 3T
v

P
3K
26
...
If the particle starts from rest, the
work done by the force during the first 1 sec
...
5 J
(4) 22 J
Ans
...
F = 6t = ma
 a = 6t



T 



dv
 6t
dt
v

1

 dv   6t dt
0

0

v = (3t2)1 = 3 m/s
0
8

28
...


1
1 9  0  = 4
...
An observer is moving with half the speed of light
towards a stationary microwave source emitting
waves at frequency 10 GHz
...
3 GHz
(2) 15
...
1 GHz
(4) 12
...
(1)
Sol
...
3 GHz
In the given circuit diagram when the current
reaches steady state in the circuit, the charge on
the capacitor of capacitance C will be :
r
E
r1

C
r2
(1) CE

r2
 r  r2 

(2) CE

(3) CE

(4) CE

r1
 r1  r 

r1
 r2  r 

Ans
...
It steady state, current through AB = 0
r
E
r1

B

A

C
D

r2

 VAB = VCD



 VAB     r2   VCD
 r  r2

 QC = CVAB
 r 
 CE  2 
 r  r2 

C

CODE-B
29
...


circuit across a potential difference of 1
...


A body is thrown vertically upwards
...

The minimum number of capacitors required

v

to achieve this is :
(2)
(1) 24
(3) 2

t

(2) 32
(4) 16

v

Ans
...
To hold 1 KV potential difference



minimum four capacitors are required in series

v



C1 

1
for one series
...
(1)

So for Ceq to be 2F, 8 parallel combinations
are required
...
Velocity at any time t is



given by



8 Parallel

V0
g
v = u + at
v = v0 + (–g)t

1 KV

 Minimum no
...


Let k be an integer such that triangle with
vertices (k, –3k), (5, k) and (–k, 2) has area 28
sq
...
Then the orthocentre of this triangle

Sol
...
(1)
Sol
...
(2)
3

n2 = 121 (using (1) in (2))
or
n = 11
 1 1
33
...

(2) invertible
...

(4) surjective but not injective
Ans
...
f : R   – ,  ,
 2 2

E

H



()

B
(5, 2)

D

f(x) =

x
x  R
1 x2

C(–2, 2)

 1
 orthocentre is  2, 
 2

32
...
(1)

(  1) 

 5k2 + 13k – 46 = 0
or
5k2 + 13k + 66 = 0 (no real solution exist)
 k=

 (x 2  (2r 1)x  (r 2  r)) 10n

 f (x) 

(1  x 2 ) 1  x  2x –(x  1)(x 1)

(1  x 2 ) 2
(1  x 2 )2

–
x = –1

+

–
x=1

If, for a positive integer n, the quadratic
equation,

y

sign of f(x)
1, 1
2

x(x + 1) + (x + 1) (x + 2) +
...
(1)
10

(2) 12
(4) 10

(0,0) x = 1

x

–1, – 1
2

 From above diagram of f(x), f(x) is
surjective but not injective
...
(2)
Sol
...


Sol
...


(1) a singleton
(2) an empty set



(4) a finite set containing two or more elements

1 1 1



D  1 a 1  0  a = 1
a b 1

and at a = 1

2

1

0

x2
dx
4

5
2

For any three positive real numbers a, b and c,
9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c)
...
(1)

2

 (1  x) dx   (3  x) dx  
0

x + ay + z = 1

Sol
...


Then :
(1) a, b and c are in G
...

(2) b, c and a are in G
...

(3) b, c and a are in A
...

(4) a, b and c are in A
...


Ans
...
(15a)2 + (3b)2 + (5c)2 – (15a) (5c) – (15a) (3b)
– (3b) (5c) = 0

but at a = 1 and b = 1

First two equationsare x + y + z =1 
 There is nosolution
...


1
[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2]=0
2

The area (in sq
...
(1)

5
2

(2)

59
12

(3)

3
2

(4)

7
3

x}

 b=

5c
c
, a=
3
3

a + b = 2c
 b, c, a

in A
...

11

JEE(MAIN)-2017
A man X has 7 friends, 4 of them are ladies
and 3 are men
...
Assume
X and Y have no common friends
...
(2)

38
...


x

Point of intersection with y–axis (0, 1)



(x 2  5x  6)(1)  (x  6)(2x  5)
(x 2  5x  6)2

y= 1 at point (0, 1)
 Slope of normal is –1
Hence equation of normal is x + y = 1
1 1
  ,  satisfy it
...


P  2, 3  and has foci at (± 2, 0)
...


2x
3y

1
1
3

Hence  2 2,3 3  satisfy it
...
If f(x) = ax2 + bx + c is such
that a + b + c = 3 and
f(x + y) = f(x) + f(y) + xy,  x, y  R,
10

then

 f(n)

is equal to :

n 1

(1) 255
(2) 330
(3) 165
Ans
...
f(x) = ax2 + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3
Now
f(2) = 7
f(3) = 12
Now
Sn = 3 + 7 + 12 +
...
(1)

3 + 7 +
...
(2)

On subtracting (2) from (1)
tn = 3 + 4 + 5 +
...
(3)

x2 y2
 1

1 3



x6
Sol
...
(2)
a 2 b2
On solving (1) and (2)
a2 = 8 (is rejected) and a2 = 1 and b2 = 3

4 Men

1 1
(4)  ,  
 2 3

2, 3 



1 1
(3)  , 
2 2





3 Ladies

Total number of ways
4C · 3C · 3C · 4C
4
3
3
4
0
3
3
0 + C1 · C2 · C2 · C1
4C · 3C · 3C · 4C + 4C · 3C · 3C · 4C
+ 2
1
1
2
3
0
0
3
= 485
The normal to the curve y(x – 2)(x – 3) = x + 6
at the point where the curve intersects the
y-axis passes through the point :
1 1
 1 1
(2)   ,  
(1)  , 
 2 3
 2 2

Ans
...


x2 y2
– 1
a 2 b2
foci is (±2, 0) hence ae = 2,  a2e2 = 4
b2 = a2(e2 – 1)

...
Equation of hyperbola is

(n 2  5n)
2

Sn = tn =

(n 2  5n)
 2

1  n(n  1)(2n  1) 5n(n  1) 
Sn  


2
6
2


S10 = 330

CODE-B
42
...
Let c be a
i j

Sol
...
Then a·c

tan  =

is equal to :

AC 1 AB 1
=
=
AP 2 AP 4

1


 AC  AB 

2


B

1
(1)
8

25
(2)
8

(3) 2

(4) 5
C



Ans
...


1
2



 
Now : | c  a | = 3

c2

+

a2

tan   tan 
1– tan  tan 






ˆ j ˆ
a  2i  ˆ  2k , b  ˆ  ˆ and | a |  3
i j
 
ˆ
 a  b  2i  2ˆ  k
j ˆ

Sol
...

Then the maximum area (in sq
...
5

(3) 10

Ans
...
Total length = r + r + r = 20

 
4 + 9 – 2a·c  9



 
a·c  2
43
...
Let C be the mid-point of AB and
P be a point on the ground such that AP = 2AB
...
(4)

6
(2)
7

1
(3)
4

2
(4)
9

dA
0
dr

10 – 2r = 0, r = 5

A = 50 – 25 = 25
13

JEE(MAIN)-2017
3
4

45
...


 dx

The integral  1  cos x is equal to :-

+ C, where C is a constant of integration, then
the ordered pair (a, b) is equal to :-


4

(1) –1
Ans
...
I4 + I6 =

 dx
I
 1  cos x


...


 dx
I
 1– cos x




 (tan

4

x  tan 6 x) dx =

1
tan5 x + c 
5

 tan

4

x sec 2 x dx

1
,b=0
5

a=

Let  be a complex number such that
2 + 1 = z where z = –3
...
(2)
Sol
...
(2)

Sol
...
+
(21C10 – 10C10) is :(1) 220 – 210
(2) 221 – 211
(3) 221 – 210
(4) 220 – 29
Ans
...
(21C1 + 21C2
...
10C10)

49
...
+
2

21 C

10 )

+ (21C11 +
...
(2)


4

(1)

1
5

Ans
...


 1 
 5 




(1)  – ,0  (2)  – ,1  (3)  ,0  (4)  , – 1 

(4) 4

3
4

Sol
...
I4 + I6 = a tan5x + bx5

=

1 21
[2 – 2] – (210 – 1)
2

= (220 – 1) – (210 – 1) = 220 – 210

CODE-B
50
...
Line PQ ;

x 1
y2
z 3
=
=
1
4
5

Let F( + 1, 4 – 2, 5 + 3)

1
(1)
4

1
(2)
24

1
(3)
16

1
(4)
8

P(1,–2,3)

Ans
...


Q
F lies on the plane
2( + 1) + 3(4 – 2) – 4(5 + 3) + 22 = 0

1
1
1
·1·
=
8
2
16

 –6 + 6 = 0



Sol
...
(1)

{Let cos2 x = t}

F (2, 2, 8)

PQ = 2 PF = 2 42
...




1  t 
 t  = 2(2t – 1) + 9
Sol
...

3
3

Ans
...
Normal vector

2

7
 1
cos 4x = 2    – 1 = –
9
 3
52
...
(3)
15

JEE(MAIN)-2017
54
...
(2)
 6x x 
 1
0, 
Sol
...
(3x 3 / 2 ) 
–1 (3x3/2)

3 / 2  = 2 tan
 1  (3x ) 

As 3x3/2

 4  y   y   
 y 2  y  2  1 

 22  4  15  0
4  16  120 4  2 34

4
4


2  34
34  2

2
2



9
x
1  9x 3

34  2
2 2
which is not in options therefore it must be
bonus
...

Given curves are y = 4 – x2 and y = |x|
r

9
1  9x3
The radius of a circle, having minimum area,
which touches the curve y = 4 – x2 and the lines,
y = |x| is : g(x) =



55
...
(Bonus or 4)

(0, 4)

(2) 2  2  1
(4) 4  2 –1

(0,4–r)



(1) 4  2  1

4

2

Sol
...
The circle shown is of least area
...




 r  r 2

 r–4=±r 2

r=

CODE-B
56
...

If 10 balls are randomly drawn, one–by–one,
with replacement, then the variance of the
number of green balls drawn is :-

58
...
, 10), then the probability that
their sum as well as absolute difference are both
multiple of 4, is :-

6
12
(2)
(3) 6
(4) 4
(1)
25
5
Ans
...
We can apply binomial probability distribution
Variance = npq
3
2
12
×
=
5
5
5
The eccentricity of an ellipse whose centre is

If two different numbers are taken from the set

(1)

7
55

(2)

6
55

(3)

12
55

(4)

14
45

= 10 ×

57
...
If one of its directices is
2
x= –4, then the equation of the normal to it at

Sol
...
, 10 }
n(s) = 11C2 (where 'S' denotes sample space)
Let E be the given event
 E  {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
n(E) = 6



 3
 1,  is : 2
(1) x + 2y = 4
(3) 4x – 2y = 1
Ans
...
(2)

(2) 2y – x = 2
(4) 4x + 2y = 7

1
2

59
...
Eccentricity of ellipse =

 P(E) =

6
55

For three events A, B and C,
P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs) =

(0, 0)

x = –4

a
=–4
e

 a=4×



Now, –

x=4

1
=2
2

 1
 b2 = a2(1 – e2) = a2  1   = 3
 4
 Equation of ellipse

x2
y2
1
+
4
3
3x
x
2y

+
× y' = 0  y' = –
4y
2
3
3 2
1
y ' |(1,3 / 2)     
4 3
2
 3
 Equation of normal at  1, 
 2
3
y–
= 2(x – 1)  2y – 3 = 4x – 4
2
 4x – 2y = 1

1
and
4

P(All the three events occur simultaneously) =

1

...
(3)
P(exactly one of A or B occurs)
= P(A) + P(B) – 2P(A  B) =

1
4

P(Exactly one of B or C occurs)
= P(B) + P(C) – 2P(B  C) =

1
4

P(Exactly one of C or A occurs)

17

JEE(MAIN)-2017
Ans
...
Given A = 

 4 1 

Adding all, we get

3
4

2P(A) – 2P(A  B) =

 P(A) – P(A  B) =

Now, P(A  B  C) =

 16 9
3A2 = 

 12 13 

3
8

1
16

 24 36 
12A = 

 48 12 

(Given)
 72 63
 3A2 + 12A = 

 84 51 

 P(A  B  C)
= P(A) –  P(A  B) + P(A  B  C)

60
...


Which of the following compounds will
significant amont of meta product during
mono-nitration reaction ?

OH

63
...
(3)
Sol
...


(1) (III) < (II) < (I)
(2) (II) < (I) < (III)
(3) (I) < (III) < (II)
(4) (II) < (III) < (I)
Ans
...
For any S N1 reaction reactivity is decided by
ease of dissociation of alkyl halide

R – X  R   X 

+



(ii) Aniline acts as base
...

Since stability of cation follows order
...
H2SO4
+ conc
...
(3)
Sol
...




 p – H 3CO – C 6 H 4 – CH 2

(iii) Anilinium ion is strongly deactivating
group and meta directing in nature so it
give meta nitration product in significant
amount
...


The increasing order of the reactivity of the
following halides for the S N1 reaction is

64
...
h = 6
...
1091 × 10–31 kg ; charge of electron
e = 1
...
854185 × 10–12 kg–1 m–3 A2)
(1) 1
...
76Å

(3) 0
...
(4)
Sol
...
53 n2Å
Radius of II Bohr orbit = 0
...
12 Å

19

JEE(MAIN)-2017
pKa of a weak acid (HA) and pKb of a weak
base (BOH) are 3
...
4, respectively
...
2
(2) 6
...
0
(4) 1
...
(2)
Sol
...


Mass due to 1H is = 75 
1H

68
...
4
As given salt is of weak acid and weak bas

1
1
 pH = 7 + pK a – pK b
2
2

Which of the following , upon treatment with
tert-BuONa followed by addition of bromine
water, fails to decolourize the colour of
bromine ?
O
C6H5
(1)

(2)

Br

Br
O

O

1
1
(3
...
4)
2
2

(3)

(4)



Br

= 6
...
(1)
Sol
...


Br

Ans
...
Ans
...


O-tBu

Br



66
...


So mass gain by person =7
...
2

=7+

10
 7
...
4%) ; Carbon (22
...
0%) ; and Nitrogen (2
...
The weight
which a 75 kg person would gain if all 1H atoms
are replaced by 2H atoms is
(1) 15 kg
(2) 37
...
5 kg
(4) 10 kg
Ans
...
Mass in the body of a healthy human adult
has :-

67
...
4%, Carbon = 22
...
0% and Nitrogen = 2
...


CODE-B
In the following reactions, ZnO is respectively
acting as a/an :
(a) ZnO + Na2O  Na2ZnO2
(b) ZnO + CO2  ZnCO3
(1) base and acid
(2) base and base
(3) acid and acid
(4) acid and base
Ans
...
Although ZnO is an amphoteric oxide but in
given reaction
...


H
H

base

Br
CH3

Et
(II)
CH3
H
CH3

Br
H

salt

acid

Et
(IV)

A metal crystallises in a face centred cubic
structure
...


(1) 2a

+ 7H2O

 4MgCO3
...
5H2O + 2HCO3–



3-Methyl-pent-2-ene on reaction with HBr in
presence of peroxide forms an addition product
...
(4)

Sol
...

71
...
(1)
Sol
...


+2

H
H

CH3

(B) ZnO + CO2  ZnCO3
base

Br
H 3C

Et
(I)

(A) ZnO + Na2O  Na2ZnO2
acid

CH3

CH3

73
...
Activation energy of R1
exceeds that of R2 by 10 kJ mol–1
...
314 J mol–1K–1)

Sol
...
(4)

(4) 4

Ans
...
From arrhenius equation,
 Ea

5

1

K  A
...
e  Ea 1 / RT


...
(2)

K 2  A
...


( Ea 1  Ea 2 )
RT

(As pre-exponential factors of both reactions
is same)
 K  E  E a2
10,000
ln  2   a1

4
RT
8
...
(1)
Sol
...

(b) The diameter of the dispersed particle is not
much smaller than the wavelength of the
light used
...

(d) The refractive indices of the dispersed phase
and dispersion medium differ greatly in
magnitude
...
(2)
As per NCERT book (fact)
76
...


[Ag(NH3)2]OH
Tollens reagent

CHO

CO2H
+



H /CH 3OH
(esterification)

(2)

O

O

C

(3)
OCH3
CH3MgBr

CH3
(4)
H 3C–C–CH 3
OH

22

Ans
...
(1) Ester in presence of Aqueous KOH solution
give SNAE reaction so following reaction takes
place
HOCH2

CH2OH
HOCH2
O
Aq
...
(1)
Sol
...

79
...


80
...
8 kJ mol–1
(2) +144
...
8 kJ mol–1
(4) –144
...
(3)
CO2(g)+2H2O()  CH4(g)+2O2(g);rH°= 890
...
5 –285
...
(3)

rH° = –285
...
3 kJ mol–1
Based on the above thermochemical equations,

Sol
...

KOH (SN AE ) reaction takes place &
-Hydroxy carbonyl compound is formed
which give ve Tollen's test So this compound
behave as reducing sugar in an aqueous KOH
solution
...
5 kJ mol–1

1

0
X eF4  O2 F2  X eF6  O2

O



77
...


0

 ( f H)Re actan ts

890
...
5)  2(–285
...
3  965
...
8 kJ / mol

Br

H
C6H 5

C6H 5

BuOK


(1)    C6 H5 CH  O t Bu  CH2 CH 6 H5
(2) C6H5CH=CHC6H5
(3) (+)C6H5CH(OtBu)CH2H5
(4) (–)C6H5CH(OtBu)CH2C6H5
Ans
...
Elimination reaction is highly favoured if
(a) Bulkier base is used

NO  One unpaired electron is present in *
molecular orbital
...


O2  Two unpaired electrons are present in *
molecular orbitals
...
H2SO4
...
'X' is :4

(1) C6H5COONa
(2) HCOONa

1
2

i = 1    1



(3) CH3COONa
(4) Na2C2O4



Sol
...
(4)

The freezing point of benzene decreases by
0
...
2 g of acetic acid is added to
20 g of benzene
...
12 K kg mol–1)
(1) 64
...
4%
(3) 74
...
6%
Ans
...
In benzene
2CH3COOH  (CH3COOH)2

83
...


B2  Two unpaired electrons are present in 
bonding molecular orbitals
...


82
...
(2)

 0
...
12 
0
...
527
2

 = 0
...
5%
Which of the following molecules is least
resonance stabilized ?
(1)

(2)

O

(2) CO
(4) B2
(3)
Ans
...




N

(4)
O

CODE-B
–

Number of Cl present in ionization sphere =
O
Sol
...
02

2
mole of complex
0
...
H2O
86
...
1 M solution of
CoCl 3
...
2 × 1022
ions are precipitated
...


CHO

(1)

pyridine is aromatic

(1) [Co(H2O)4 Cl2]Cl
...
3H2O
(3) [Co(H2O)6]Cl 3



(4) [Co(H2O)5Cl]Cl 2
...
(4)

Sol
...
1
= 0
...
2  10 22
6
...
02 moles

COOH

Molarity  volume(ml)
1000

Moles of ions precipitated with excess of
AgNO3 =

CHO

(3)

CHO

(4)
CHO

Ans
...
DIBAL – H is electrophilic reducing agent
reduces cynide, esters, lactone, amide,
carboxylic acid into corresponding Aldehyde
(partial reduction)

25

JEE(MAIN)-2017
87
...


Given
Eo
Cl


F   10; SO2   100; NO3  50
4

(1) only

(2) both

SO2
4

and

2

7

4

(1) Cr

Ans
...
NO3–: The maximum limit of nitrate in drinking
water is 50 ppm
...
Such as
methemoglobinemia
...
36 V, E Cr 3 / Cr  0
...
33 V, E o  / Mn2  1
...

Cr
MnO

the anion/anions that make / makes the water
sample unsuitable for drinking is / are :

NO3

2

(4) Cl–

Ans
...
(i)

...
36 V



SO4 : above 500 ppm of SO4 ion in drinking
water causes laxative effect otherwise at
moderate levels it is harmless

E o  / Mn 2  1
...


–

E o O 2 / Cr 3  1
...
(iii)

E o 3 / Cr  0
...
(iv)

2

7

–

F : Above 2ppm concentration of F in
drinking water cause brown mottling of teeth
...


2–

90
...
e 10 ppm) make water
unsuitable for drinking purpose :

The group having isoelectronic species is :–

2–
+
2+
(1) O , F , Na , Mg
–

–

+

1 gram of a carbonate (M2CO3) on treatment
with excess HCl produces 0
...
the molar mass of M2CO3 in g mol–1 is :-

(2) O , F , Na , Mg

(1) 1186

88
...
3



(3) 118
...
86
Ans
...
Given chemical eqn
M2CO3 + 2HCl 2MCl + H2O + CO2
1gm
0
...

1
 0
...
3 gm/mol

26

–

(3) O2– , F , Na , Mg2+
–

–

+

2+

Ans
...

–2

O

F

Atomic number = 8

–

9

ions

–

No

---