CONCEPT RECAPITULATION TEST - II Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT –II]

other 12 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

Contributed by

Swati Kant

AITS-CRT-II-(Paper-2)-PCM(S)-JEE(Advanced)/14

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ANSWERS, HINTS & SOLUTIONS
CRT –II
(Paper-2)

Q.
No.
PHYSICS CHEMISTRY MATHEMATICS
1.
A B D
2.
B D B
3.
C A B
4.
C A B
5.
D D B
6.
A D B
7.
A A B
8.
A C C
9.
A D B
10.
A D A
11.
D A B
12.
A B A
13.
B D A
14.
A B C
15.
B, D B, C, D A, C
16.
B, C A, B A, D
17.
A, C A, B A, B, C
18.
B, C, D A, B, D B, D
19.
B, D A, D A, D
20.
A, B, C, D A, C, D A, C, D

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

2.  =  tan  
2
2 2
1
g
2 u cos



 =  tan  
 
2
2
1
g 1 tan
2 2g

 


tan
2
  4 tan  +
4
1

 

 

 
= 0





(, )



u

2g

u

If particle will not hit the target.
(b
2
 4ac) < 0
16  4
4
1

 

 

 
< 0
4 > 3

3. At B, mg sin  =
2
B
mv
r
. . . (1)
Using energy considerations

2
B
1
mv
2
= mgr(cos   sin ) . . . (2)

B

A





Smooth

N

v
B

From (1) and (2)
mg sin  = 2mg(cos  sin )
2cos = 3 sin 

7. i =
5A

2
2
2
m
q
q 1
Li
2C 2C 2
   q
max
= 6 C

8. Time period does not depend on amplitude SHM and both particle will exchange velocity at every
collision.

15. The slope of curve at such a point will be  1

dy x
2cos 1
dx L

 
  
 
 

 x =
L
3
or
2L
3

y

x

135


45


16.
1
P m 16 2m 0 16m
    

' '
f A B
P mv 2mv
 

2
B
1
2mv 2mgh
2
 

B
v 10
 m/s
16 m =
'
A
mv 2m 10
 


'
A
v 4m/s
 

' '
B A
A B
v v
e
v v



=
10 ( 4) 14 7
16 0 16 8
 
 


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17. If v be the velocity at mean position in two cars then
2 2 2
1 2
1 1 1
mv KA KA
2 2 2
 
and
1 2
m
T T 2
K
  
A
1
= A
2

18. Initially potential and kinetic both energies zero and from conservation of mechanical energy total
energy of the two object zero
Further, decrease in P.E. = increase in K.E.

2
r
G(m)(4m) 1
v
r 2
 

r
10Gm
v
r

Total K.E.
2
G(m)(4m) 4Gm
r r
 

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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.
4
6
XeO

(perxenate ion) because oxidation state of Xe is +8 and it is highly unstable compound.
2.
Br

Due to formation of resonance stabilized carbocation.
3.
293.4
CRT,So,C 0.1 mol / L
RT 8.31 293

    


If 100 ml contains 6.33 g of haematin.


1 L will contain 63.3 g of haematin.
Molecular mass of haematin = 633 u
Estimating the empirical formula of compound
C : H : N : O : F =
64.6 5.2 8.8 12.6 8.8
: : : :
12 1 14 16 56

34 : 33 : 4 : 5 :1


So, empirical formula is C
34
H
33
N
4
O
5
Fe, whose mass corresponds to the molecular mass.
4.
2 2 4 2
o o
Zn /ZnY / Y Zn /Zn
0.059
E E logK
2
    
 

2
f
4
f
ZnY
1
K K
K
Y


 
 
  
 
 

2 2 4
o
10
16
Zn / ZnY / Y
0.059 1
E 0.76 log
2
3.2 10
  
  


= - 1.25 V
5.
f f
T 0.5K iK m
  

f
f
T
0.5
i 1.022
K m 1.86 0.263

   


HA H A
1 0 0
1
 
 
   

i 1 0.022
     

K
a
= C
2
= 0.2 × (0.022)
2
= 9.6 × 10
-5

 pK
a
= 4.0177
On adding 0.25 M NaA, buffer is formed.
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

 
a 10
Salt
pH pK log 4.0177 0.0969
Acid
    

= 4.1146
6.
2 4 7 2 3 3
Strongbase
Weak acid
Alkaline
Na B O 7H O 2NaOH 4H BO
  


2 4 7 2 2 3
Glassy bead
Na B O 2NaBO B O

 


 
2 4 7 2 4 2 2 4 3 3
White crystals
Na B O H SO 5H O Na SO 4H BO
   

7.
CH
3
OH
CH CH
2
H


CH
3
OH
CH CH
3
Ring
expansion

O
CH
3
H
H
3
C
H


O
CH
3
H
3
C

8.
O
O
N
K
+
||
2 2 5
O
KCl
Cl CH C OC H

    
N
O
O
CH
2
C
O
OC
2
H
5
 
X
 
Y
2
2 5
H /H O
2 2
C H OH
Glycine
H N CH COOH


   
O
O
OH
OH
 
Z

9.
 
 
Burning
2 3 2
A
B
3M N M N
 

 


 
 
3 2 2 3
2
B D
C
M N 6H O 3M OH 2NH
  
All statement with respect to NH
3
are correct.
10. Ba and Ca quite readily liberate hydrogen.
In cold water Mg decomposes water only on heating. So, element (A) may be Ca and Ba not Mg.



 


 
2 2
Lime water Baryta water
Ca OH & Ba OH
, both givess milkiness with CO
2
.
11.
2
1 1 2
P
H 1 1
2.303log
P R T T
 

 
 
 

1
H 11.99 kJmol

 
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12
 
2
2
1
b
3
8.314 353 100
RT M
K 8.64 Kkgmol
1000 H 1000 11.99 10

 
  
  



b b
T K 1 m
   

 
1 1000
1 8.64 1
100 100
    
 = 0.158
13. Two structural isomers of (A) are:

Me
Me
Me
OH
&
Me
Me
OH
Me

There are four geometrical isomers for each strucutre, i.e. total number of stereoiosmers = 8.
14. Convert aldehyde and ketone to alcohol

(1)
OH
(5C )

(2)
OH
OH
(6C )

(3)
(2C )
OH
OH

15.
OOH
H
H
CH
2
OH
H
OH
OH
H
H
O
O
H
CH
2
OH
H
OH
OH
H
H
OH
H

16. (A)
N
H
H
H
N
F
F
F

3 3
NH NF
  

(B) (I) is 3
o
allylic with extended conjugation.
(II) is 3
o
allylic, (III) is 3
o
free radical.
(C) Correct order is:


3
RS CH O OH Basic strength
  
 
(D) Correct order is:


I Br Cl F Basic strength
   
  
17. (A)
4 3 3 3 2
NH NO NaOH NH NaNO H O
   

3 2 2 3 2
7NaOH NaNO 4Zn 4Na ZnO NH 2H O
    
(B)
4 2 2 3 2
NH NO NaOH NaNO NH H O
   

3 2 2 3 2
3Zn 5NaOH NaNO 3Na ZnO NH H O
    
18.
rms
3RT 3PV 3P
v
M M d
  
19. In metal carbonyls,
M CO




, the C – O bond length increases compared to that in CO
molecules due to synergic bonding between metal and carbonyl ligand.
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Facial and meridenal isomers
 
n
3 3
Ma b

contains plane of symmetry.
20.
 
3 2 3
CH COO H O CH COOH OH
0.1 1 h 0.1h 0.1h
 
 








 
2 5
h
0.1h 0.1h
K 0.1h h 7.48 10
0.1 1 h

    


5 1 6
OH Ch 7.48 10 10 7.48 10
   
 
      
 

14
9
W
5
K
10
H 1.33 10
7.48 10
OH

 


 
    
 

 
 

And pH = 8.8 approx.

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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. We express x in terms of a new variable y

1
x 1
y
 

y 1
x
y



       
5 3 2
5 3 2
6 y 1 5 y 1 y 1 4 y 1
1
y
y y y
   
   
=
5 4 3 2
a b c d e
f
y
y y y y
    

Multiplying both side by y
6
we get 6y(y + 1)
5
+ 5y
3
(y + 1)
3
+ y
4
(y + 1)
2
+ 4y
5
(y + 1) – y
6

= ay + by
2
+ cy
3
+ dy
4
+ ey
5
+ fy
6

Differentiating both side and then satisfying y = 1
Gives the value a + 2b + 3c + 4d + 5e + 6f = 910

2.
   
 
50
n 51 n
n 1
f x x n


 


 
 
 
 
50
n 1
ln f x n 51 n ln x n

  


Differentiating both side w.r.t. x

 
 
 
50
n 1
f ' x n 51 n
f x x n









 
f ' 51
1275
f 51


3. a
2
= b
2
+ c
2
– 2bc cos A = (b – c)
2
+ 2bc(1 – cos A)
 =
1
bcsinA
2

4
2bc
sinA



   
2
2
4
a b c 1 cosA
sinA

    = (b – c)
2
+
A
4 tan
2

Since A and  are fixed hence a is minimum when b = c
Hence 2bc = 2b
2
=
4
sinA


4.


 


 
f 3x 3y f 3x 3y
sin 2x 3y sin 2x 2y
 

 
 x, y  R

 
2x
f x ksin
3
 

 
 


 
2
f ' 0 k 1
3
 

3
k
2



 
3 2x
f x sin
2 3
 

 
 

3 3
f '
2 4

 

 
 

Equation of tangent is
3 3 1
y x
4 2 2

 
  
 
 

5.
i j
1 i j m
aa
  

=
2
m m
2
i i
i 1 i 1
1
a a
2
 
 
 
 

 
 
 
 
 
 
 

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=
   
2
m 2 2m
1 1
2
a 1 r a 1 r
1
2 1 r
1 r
 
 
 
 
 
  
 
  
  

=
 
 
2
2
m 2m
1
a
1 r 1 r
2 1 r 1 r 1 r
 
 
 
 

 
    

=




m 1 m
1 1
a 1 r a 1 r
r
1 r 1 r 1 r

   
 
   
     

=
m 1 m
r
S S
1 r



6. Use L’Hospital’s Rule twice

7. g(x) =


 
3
2
f x dx
x x 1


=
   
2 2 3
A B C D E
dx
x x 1
x
x 1 x 1
 
   
 

 
 


=
 
 
2
B D E
Alnx cln 1 x
x 1 x
2 x 1
    



Since g(x) is a rational function, hence logarithmic function must be absent
 A = C = 0

 
   
2 2 3
B D E
g x dx
x
x 1 x 1
  
 


 f(x) = (B + D)x
3
+ (3B + D + E)x
2
+ 3Bx + B
 B + D = 0, f(0) = 1 gives B = 1  D = –1
f(x) = (2 + E)x
2
+ 3x + 1
f(x) = 2(2 + E)x + 3
f(0) = 3

8. Denoting the two curve by S
1
and S
2
the equation of curve S passing through intersection of S
1

and S
2
can be S = S
1
+ S
2
= 0. Since S is circle
 Coefficient of x
2
= Coefficient of y
2
, Coefficient of xy = 0 we get  = 1
S = (a + a)(x
2
+ y
2
) – 2(g + g)x – 2(f + f)y + 2c = 0
The centre is
g' g f ' f
,
a' a a' a
 
 
 
 
 
which is P
PA
2
+ PB
2
+ PC
2
= 3(Radius)
2
= 3(PD)
2

9.


x
sinx sin
180

 
x
sinx sin
180

 
  
 
 
or
x
sin 2
180

 
 
 
 

x
x
180

   ,
x
x 2
180

  

180
x
180


 
,
360
x
180


 

On comparing, m = 360, n = 180, p = 180, q = 180
 (m + n + p + q) = 900

10.


2 2
x
A C x Bx
lim 2
B
x A C
x

 

 
 
 
 

For existence of limit A = C
2

Page 9
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10

Hence,
C A
 ,
B
2
A C




B
4
C


For
C A
  limit does not exist

11. The visible are of P(2, –6) is the area of ABP ABP has base AB of length 4 and its height is the
distance from AB to point P. Which is 10, since AB is parallel to x–axis.
Thus area =
1
b h 20
2
  

12. Similarly area = QBC =
1 1
b h 4 4 8
2 2
     

13. LHD =
2
h 0
sinh tanh cosh 1
2h ln(2 h) tanh
lim
h

   
 
 
  
 
 
  

=
 
2
2
h 0
sinh tanh 1 cosh
h
h h
h
lim 0
2h ln 2 h tanh


  

  

RHD =
2 2
h h
h 0
e 1 0 e 1
lim h 0
h h

  
  

 L
1
 y = 0, L
2
 x = 0

14.
1
2 cot cot 1
2 4 2 2
 
  
 
     
 
 
 
 

=
2sin
4
1
2 sin sin
4 2 2


 
  
 

 
 
 
 

=
2
1
cos cos
4 4

 
 
  
 
 

 
 
min
2
1 1 2 2 1 3 2 2
1
1
2
       


15. Four point A, B, C, D are coplanar if three vector
AB

,
AC

,
AD

are coplanar or STP of there
vectors is 0



AB AC AD 0
  
  


 
 
1 0 0
3 f t 1 0
2 f ' t 2

 t  R
 f(t) = 2f(t)
 f(t) = ce
2t

Similarly for


ˆ ˆ
B i j


differential equation will be f(t) –2 f(t) = –4
On solving f(x) = 2 + ce
2x
, c  R
Page 10

CONCEPT RECAPITULATION TEST - II Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT –II]

CONCEPT RECAPITULATION TEST - II Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT –II]

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