CONCEPT RECAPITULATION TEST - II Paper 2 [ANSWERS, HINTS & SOLUTIONS CRT –II]

other 13 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

Contributed by

Swati Kant

AITS-CRT-II-(Paper-1)-PCM(S)-JEE(Advanced)/14

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ANSWERS, HINTS & SOLUTIONS
CRT –II
(Paper-1)

Q.
No.
PHYSICS CHEMISTRY MATHEMATICS
1.
C C B
2.
A D A
3.
C D B
4.
D A C
5.
B D A
6.
D C D
7.
C B D
8.
B C C
9.
C C A
10.
A C A
11.
B, D D A, B, D
12.
A, B, C A, B, C, D A, B, C, D
13.
A, B, C B A, B
14.
B, D A, B, C, D A, B, C
15.
A, B, D B, C, D A, D
1.
1 2 2
2.
7 6 3
3.
4 1 4
4.
3 2 0
5.
1 4 6

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

3.
2
e
dm
F v Av
dt
  

2
( Av )H
2

 

7.
2
H
mv
N qEcos
R
  
…(i)

H
N 0


(
H
N

horizontal component of normal force)
qER(1cos ) =
2
1
mv
2

qE

N



8.
3 2
F
0.02
A
10
n 10 kg / ms 10 poise
2
dv
1
dx
 
 
 
 
   
 
 
 

9. Velocity of bike V
b
= 30 m/ s
Let velocity of car be V
c

f
obs
= f
actual

cso
bso
VV
VV



120 Hz = 100 Hz 
c
V

330
30330

 V
C
= 30 m/s
Distance between initial and final pulse of horn
= V
so
 t - V
C
t {t is duration of emission of sound}
= (330 – 30)  6 = 1800 m
This wave train travels with 330 m/s and cross bike rider moving with velocity of 30 m/s towards
it.
time taken =
bso
VV
trainwaveoflength

=
1800 180
5
330 30 36
 

sec
11. Kinetic friction acts opposite to relative velocity w.r.t. contact surface.

12. X
L
= 2X
C

C
X

= 4X
C

L
X

= 2X
L


L
X


C
X

= 2X
L
 4X
C
= 2[X
L
 2X
C
] = 0

14. Since vertical Impulse is acting on A
Hence momentum of A & B will constant in horizontal direction only.
and M.E. of A & B will remain constant

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SECTION –C
1. Angular momentum Conservation

2
2 2
0
1 MR
MR mR '
2 2
 
   
 
 

0
M
'
M 2m
 
  
 

 

U =
0
M
'R R
M 2m
 
  
 

 

2. M.E. Conservation
mg(4) 
2 2
k 1
x mv
2 2

kx = 55
mg + kx = ma
5520 = 2a
a = 17.5 m/sec
2

3.  =
qBR
t
mg
 
 
 
= 0.4

5. t =


2 2
dE B C M
1000sec
B mg




= 1
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C
C
h
h
e
e
m
m
i
i
s
s
t
t
r
r
y
y PART – II

SECTION – A

1. In (C), the tautomeric form will be

O
C
6
H
5
C
6
H
5
H
5
C
6
H
5
C
6
OH
C
6
H
5
C
6
H
5
H
5
C
6
H
5
C
6

The tautomeric form, contains sp-hybridised carbon in ring, making it unstable.
 tautomerism is not shown by (C).
2.
2AgBr 2OH


 
OHOH
Hydroquinol
   
2
2Ag s 2H O  
OO
 
2Br aq


Hydroquinone

3. PV

= constant

1
105V
P Constant
100

 
 
 
 

1
P
105
1
P 100

 
  
 
 

7
5
1
P
100
or
P 105
 

 
 

( for N
2
or O
2
= 7/5)
P
1
= 0.93P,
 7% decrease in pressure.
4. O
2–
is replaced by X
-8/3
, so formula of spinel is MgAl
2
X
3
and deficiency by one anion.
5.
W RT
M V
  
RT
Slope
M

3 3
0.082 293
M
4.65 10 10
 

 
 

 5.16  10
6
g





C


6.
 
 
5 4
PBr s PBr Br


 
 


2
2 5 2 3
N O s NO NO sp
 
 
 


3 2
5 4 6
PCl s PCl PCl sp d
 
 
 


3
2 6 2 4
Cl O s ClO ClO sp
 
 
P
Cl
Cl
Cl
Cl
Cl
Cl

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7.
H
(I)
H
(II)
H
(III) (IV)
(Conjugate base)
(Aromatic)
(Non-aromatic)
(anti-aromatic) (anti-aromatic)
H

(III) and (IV) both are anti-aromatic, but (IV) has more resonating structures than (III).
So, stability of conjugate bases: (I) > (II) > (IV) > (III)
And pk
a

order is: (I) < (II) < (IV) < (III)

8. pH = 9.71, 
10
H 1.95 10
 
 
 
 

2 3
PrNH HCl PrNH Cl
 
 



 
2
a
3
Pr NH H
K
PrNH


 
 

 
 



2
a
3
Pr NH
K
H PrNH
 

   
   



2
3
PrNH
0.1
PrNH

 
 
 

9.
CH
2
COOH
Cl
dil
aq. NaOH

 
A
CH
2
OH
COOH
H


O
O
CH
2
2 2 7
acidic
Na Cr O

COOH
COOH
8 4 3
Pr oduct(C H O )
O
O
O



10.
C
Me
CH
 
2 2
H /Ni B
p 2 catalyst

CH
Me
CH
2
H


CH
Me
CH
3
Ring expansion

Me
CH
3
H



Me
Me

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11. CH
3
CH
3
CH
3
CH
3
Trans - 1, 2 - dimethyl cyclohexane
Trans - 1, 3 - dimethyl cyclohexane

No plane of symmetry in both the above isomers, hence optically active.
12.
 
4
NO BF



B FF
F
F
No. of  bonds in
 
4
BF 4



= B. O. of
NO 3.0

 , i.e. one sigma bond and two  bonds.
 It has 5 sigma and two , bonds.
 NO
+
is diamagnetic and
4
BF

is also diamagnetic
 B—F bond energy is lower in
4
BF

than in BF
3
, due to presence of back bonding in BF
3
.
13.


2 2
2
o
10
2H /H 2H /H
2
p H
0.059
E E log
2
H
 

 
 
 

14.




o o o
f f
H H products H reac tants
    
= –166 – (–51) = – 115 KJ mol
–1

S
o
= 266 – 243 = 23J = 0.023 KJmol
-1
K
-1

H
o
= –ve
S
o
= +ve
 G
o
= H
o
- TS
o

G
o
will be –ve at all temperature and  the reaction is favour at temperature.
15. Formula of complex is : [Co(SCN)
2
(NH
3
)
4
]
3
[Co(ox)
3
]
 Linkage isomerism is due to presence of
SCN


 Optical isomerism is exhibited due to presence of [Co(ox)
3
]
3–
which has asymmetric structure.
 Geometrical isomerism is exhibited by [Co(SCN)
2
(NH
3
)
4
]
+
part
SECTION – C
1.
 
3 2 2 3 2
Black ppt
AgNO H S O Ag S
 

2.
Sb
F
F
F
F
F
F
Sb
F
F
F
F
F

Sb-hybridisation is sp
3
d
2
= total 6 hybrid orbitals per sb-atom.
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3.
O
O
3
RCO H

O
O
O
 
Pr oduct X

The number of sp
2
– carbon is X are 10, the sum 1 + 0 = 1
4.
 
 
n n
s
nM nX M X ne L.H.S
 
  
For,
 
n n
M X ne nM nX R.H.S
 
  

10
0.059 0.01
E log
n 0.1

 
 n = 2
5.
2 2
1 1
A B AB
2 2
 

A A B B A B
1 1
H e e e
2 2
  
   
– 100 =
x 0.5x
x
2 2
 

 x = 400 KJmol
–1

x = 4  10
2
KJmol
–1

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M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s PART – III

1.
ˆ ˆ
u xi 3xj
 

, |u| = 2x, x > 0
Now
2
2
ˆ
u u 2i u i
  

   


2 2
2 2
2 x x 2 3x x 1 3x
    

2 2
4 x x x 1 4x 2x 1
    

Squaring, 16x
2
(x
2
– x + 1) = 16x
4
+ 4x
2
+ 1 – 16x
3
– 4x + 8x
2

 16x
2
= 12x
2
– 4x + 1

4 32 1 2
x
8 2
   
 
 x =
2 1



u 2 1 2 1
   
 a = 2, b = 1

2. 2000x
6
+ 100x
5
+ 10x
3
+ x – 2 = 0

 


3
2
6
2
x 10x 1
2000x 2 0
10x 1

  





 
6
6
2
x 1000x 1
2 1000x 1
10x 1

  


 1000x
6
– 1= 0 or
2
x
2
10x 1
 


 x = –(10x
2
– 1)

2
1
x
10
 which is not possible
 20x
2
+ x – 2 = 0

1 161
x
40
 
 , m = –1, n = 161, r = 40
 (m + n + r) = 200

3. Let P(a cos , b sin )
Slope of
b
cp tan
a
 

2 2
a b
tan sin2
2ab

  



P()

y

C

(0, 0)



x


2 2
2ab
sin2 tan
a b
  


Using sin 2  1 and
2
2
2
b
e 1
a
 

2
2
b
2
a
tan 1
b
1
a
 


2
2
2 1 e
tan 1
e

 

 e
4
+ 4e
2
tan
2
 – 4 tan
3
  0
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 
2
2sin
e 2tan sec tan
1 sin

     
 

4.
 
    
 
1
k
1 k 1 k k 1 k 2
T cos
k k 1 k k 1

 
  
 
 
 
 

Let
1
x
k

,
1
y
k 1



 
   
2
2
2
1 k 1 1 k k 2
1 y 1
k 1 k 1
k 1
  
    
 


1 1
k
1 1
T cos cos
k 1 k
 
 
 
 
  
, substituting n = 2, 3, 4

1 1
n
1 1
S lim cos cos
n 1 2 2 3 6
 

  
 
    
 
  

n
120
S
6 k
 
 
 k = 720

5. EAC =  – 
EAB =  + 
It is given that tan( – ), tan  and tan( + ) form a G.P.
Thus
   
2 2
2
2 2
tan tan
tan tan tan
1 tan tan
  
        
  

tan  = 1,  = 45º, thus
A

B

D

C

E







AD = DE =
5 2
, so that area (ABC) =
1
BC AD CD AD 50tan
2
    

Now, cot , cot( – ), cot  form an A.P.
2 cot(45º – ) = 1 + cot   cot  = 3

 
50
ABC
3
 

6. ABCD is a trapezium and its area =
 
1
a b h
2

Where
a b
EF
2

 (mid parallel)
Area = 2r(EF)
Now equation of EF is y = –x + c ….. (1)
From equation (1) passes through (r, r)
 c = 2r
x + y = 2r, hence E = (2r, 0) and F(0, 2r)

2 2
EF 4r 4r 2 2r
  

 Area ABCD = 2r ·
2
2 2r 4 2r


2
4 2 r 900 2
 , r
2
= 225  r = 15

7. A
2
– 2A + 2I = 0
Divide B by A
2
– 2A + 21
 B = (A
2
– 2A + 21)(f(A)) + A – I
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
0 1
B A I
1 0
 
  
 

 

8. FG = p
DE = q
HI = r
In AFG
P AG AF
sinA sinB sinC
 

bp
AG
a
 ,
cp
AF
a

A

B

I

C

D

F

G

H

E

Now, incircle of ABC is the ex–circle opposite to A for AFG the semi perimeter of AFG is

AFG
1 bp cp Sp
S p
2 a a a

 
   
 
 


AFG
A Sp A
r S tan tan
2 a 2

 
Since r = (s – a)
A
tan
2


sp
s a
a
  
p a
1
a s
 

Similarly
q b
1
b s
 

r c
1
c s
 

9. Since the coefficient of
ˆ
k
in both
A

and
B

are the same, the only way that
A

and
B

can be
parallel is that f(t) = –f(t), f(t) = g(t)
The first differential equation, f(t) = –f(t)
 f(t) = p cos t + q sin t. (p, q  R) =
 
2 2
p q sin t
  
,
p
tan
q
 

 
2 2
max
f t M p q
  

   
2 2 2 2
A f t g t 1 p q 1
     


A

is constant

10. A upon simplification (a + b + c) (a – b) (b – c) (c – a)
The only way this can be 0 is (a + b + c) = 0
 fixed point is (1, 1)
11.
   
x x
0 0
t f x t dt f t dt sinx cosx x 1
     
 


   
x
0
x t f t dt


=
 
x
0
f t dt sinx cosx x 1
   



     
x x x
0 0 0
f t dt t f t dt f t dt sinx cosx x 1
     
  

Again differentiating
x f(x) + f(x) + f(x) – xf(x) – f(x) > f(x) – sin x – cos x
f(x) = f(x) – sin x – cos x
Solving this linear differential equation
We get, f(x) e
–x
= e
–x
cos x + c
If x = 0, f(0) = 0  f(x) = e
x
– cos x
Page 10

CONCEPT RECAPITULATION TEST - II Paper 2 [ANSWERS, HINTS & SOLUTIONS CRT –II]

CONCEPT RECAPITULATION TEST - II Paper 2 [ANSWERS, HINTS & SOLUTIONS CRT –II]

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