CONCEPT RECAPITULATION TEST - I Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT–I] 2014

other 11 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

SAS

Contributed by

Srinivasan Aayushman Sood

AITS-CRT-I(Paper-1)-PCM (Sol)-JEE (Adv)/14
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1

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

Q. No. PHYSICS CHEMISTRY MATHEMATICS
1.
C C D
2.
A D A
3.
B D B
4.
B C A
5.
A B B
6.
D D A
7.
B C

C
8.
A. C A
9.
B B C
10.
C B A
11.
A, B, C A, C A, B, C
12.
A, B, C, D B, C A, B, C, D
13.
B, C, D A, C A, D
14.
A, B, C A, B, C B, C
15.
A, B, C A, C B, C
1.
2 2 4
2.
1 1 2
3.
5 4 6
4.
1 3 2
5.
2 6 3

ALL INDIA

TEST SERIES

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

SECTION – A

1.
rel
U g t 10 1
   

= 10 m/s

1
1
S g(1) 5m
2
 

2
rel 1 re re
1
S S u a t
2
  
 
= 5 + 10  3 + 0 = 35 m

2. sin  =
h


mgh =  mg cos (/2)

h cos
2
 



tan  =
2



/2


/2



3. Centre of mass does not move in the absence of external force so
m
1
x
1
= m
2
x
2

1 2
2 1
x m
x m


4.
2 2
1
dA
r d r
A
dt 2 dt 2

   

L =
2
Mr


L = 2MA

5. F
2
Gm(M m)
0
r

 

2
dF G
(M 2m) 0
dm r
  

M = 2m
m = (M/2)

6.
A g
s
A




s
g




7.
h
2
0
gh r (2 rdh) gh
    


2
2
h
gh r 2 r g
2
    

h
r 2
2

r = h

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8. Avg. speed
2 2
r mr
(v t) (v t)
3
t



2
r
v 5 9
 


r
v 2m/ s


9.
avg
1 t 1
v v
3v
2 2 2
v
t 4
   
 

10.
m m
g T a
2 2
  …(i)
T cos 60 =
ma
cos 60

…(ii)
Solving (i) and (ii) acceleration of ring =
2g
9

11.
2 2
2
2kq mv
mg T
R R
  
…(i)
mgR +
2 2
1 1
mu mgR mv
2 2
  …(ii)
if T = 0 
2
min
2Kq
u 5gR
mR
 

12.
dU
F 5(2x 4)
dx
    

At mean position F = 0  x = 2m

min
U 20 J
 
a = 50  2(x  2)
 = 10 rad/sec
T = /5 sec

13. If the volume immersed initially is (V/3). Then
V
g mg
3
  …(i)
If the volume immersed when the system accelerates is V then

g mg
V' g mg
2 2
 
   
 
 

V
V'
3


14. The potential of the two surface will be equal when the whole charge Q flows from inner to other
shell.

SECTION –C

1. Consideration refraction at glass-water interface

Ruv
122










r
)2/3()3/4(
r2
3
v3
4





r
(4/5)r

(27r/20)

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
r6
1
r2
3
v3
4

 v = 
r
5

Now refraction at water air surface
u = -
5
r9
r
5
4
r 
















1212
uv








)3/4(1
r93
54
v
1

r27
20
v
1

v = -
20
r27

so height above centre = 2r -
20
r27

=
20
r27r40

=
20
13
r = 13 cm.

2. Mg  T = Ma …(i)
T  mg = ma …(ii)
T =
2Mmg
(M m)



9
T 2Mmg 2mg
2 10
m
A A(M m)
A 1
M
   

 

 
 

 M = 1.86 Kg

m

M

3. Relative motion between block and table
will start when

2 2
m r sin (mg m r cos )
      
…(i)

t
  
…(ii)
solving (i) and (ii)

t 500 22.4 sec
  .

4 cm

3 cm





m

2
r

4.  =
g
R

= 1 rad/sec

5. N
1
+ N
2
= mg . . . (1)
N
1
= N
2
. . . (2)
N
2
=
 
2
mg
1
 

N
1
=
 
2
mg
1

 

N
1
N
2
N
1
N
2
mg

Torque about centre of mass
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(N
1
+ N
2
)R =
2
1
mR
2


 =


1 2
2 N N
mR
 
=


 
2
2 1 g
1 R
  
 


2
=
2
0
2
  

 =
2
0
2



 =


 
2 2
0
1 R
2 2 1 g
  
   

Hence number of turns
N =
2


=


 
2 2
0
R 1
4 1 g.2
  
   

N =


 
2 2
0
R 1
8 g 1
  
  

C
C
h
h
e
e
m
m
i
i
s
s
t
t
r
r
y
y PART – II

SECTION – A

1. In ‘A’  bond is at bridge head which is not stabilized in the given structure;
In ‘B’ the product is
CN
ACOH

;
In ‘D’ the product is

2.
O


H
2
N OH
O
NH
2
OH
IMPT


OH
NH
OH
2
H O


N
OH
N
OH
2
H



2
H O


N
2
H O
H



N
OH

5. CH
3
Cl is formed by S
N
2 while
CH
3
CHClC
2
H
5
is formed by S
N
i.

6.






   
5 3 2
PCl g PCl g Cl g
Let initially no. of moles x
at equilibrium volume V x 1 x x

 
   



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 
2 2
c
x
K
x 1 V


 
, also K
p
= K
c
(RT)
g
n


on reducing volume to
V
2
, initial concentration of PCl
5
doubles and its degree of dissociation is
increased but K
p
does not change, as it is characteristic constant for a reaction at constant
temperature.

7.


2 6 2 3
6
Al Cl 12H O 2H Al OH 6HCl
 
  
 
;

10. In blast furnace,
In combustion zone, CO is produced ultimately which reduce Fe
2
O
3
to Fe
3
O
4
in reduction zone,
which is uppermost zone.

11. O
O
3 2 5
CH COOH C H OH
O
O
P =
Q =
R =
S =
Oxidation by XO
2 5 3
C H OH CH CHO


;

12. Due to hydrogen bonding Gauche conformation of
H
2
C CH
2
NH
2
NH
2
and

H
2
C CH
2
F
OH

are more stable than their anti-conformations.
In case of ClCH
2
CH
2
Cl, on increasing temperature, % of Gauche conformation increases. Hence
dipole moment increases. In case of option (D) boat conformation is most stable.

13. Mn appears colourless in reducing flame in Borax Bead Test.





3
4 4
6 6
3
Prussianblue
Fe K Fe CN Fe Fe CN K
 
   
   
   

2 2 3 2 2 5 2 4 2
Na S O .5H O Na S Na SO H O

  

14. At pH = pI, amino acid exists as Zwitter Ion, they are highly soluble in polar solvents.

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15. pH of 10
–8
M HCl (aq) is 6.96 and pH of 10
–8
M NaOH (aq) is 7.04

SECTION – C
1.  = CRT × i
hence i = 1.1; so  = 0.1
hence [H
+
] = c = 0.1  0.1 = 10
–2
M
pH = 2

2. It has only one CH
2
OH group.

3. Three equivalents to remove three hydrogens from CH
3
group and one for attack on C of CHO
group.

4. BaCrO
4
– Yellow, soluble in dil HNO
3

Hg
2
CrO
4
– Red, soluble in conc. HNO
3

ZnS – White, soluble in Conc. HNO
3

BaSO
4
– White, insoluble in all mineral acids
BaS
2
O
3
, CH
3
COOAg and AgNO
2
all are white solid and are soluble in dilute HNO
3
solution.

5. Except
Cl
and
Cl

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

SECTION – A

1. [y + [y]] = 2 cosx  [y] = cosx
y =
1
3
[sinx + [sinx + [sinx]]] = [sinx]
 [sinx] = cosx
Number of solution in [0, 2] is 0
Hence total solution is 0.
 both are periodic with period 2

2. (x  0)
2
+ (y  k)
2
= k
2

 x
2
+ (y  k)
2
= k
2

2x + 2 (y  k)
dy
0
dx



dy x
dx y k
 


y  k = 
xdx
dy

k = y +
xdx
dy

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 x
2
+
2
2
xdx xdx
y y y
dy dy
 
   
   
 
   
   
 

 x
2
+
2 2
2 2 2
dx dx 2xydx
x y x
dy dy dy
   
  
   
   

x
2
= y
2
+
2xydx
dy

(x
2
+ y
2
)
dy
2xy 0
dx
 

3. Given  <  <  <  also sin  = sin  = sin  = sin  = k
and , , ,  are smallest positive angles
  =   ,  = 2 + ,  = 3  
as sin  = sin  and  > 
sin  = sin  and  > 
sin  = sin  and  > 
Putting these values in the given expansion, we have given expression
=
2 sin cos 2 1 sin 2 1 k
2 2
 
 
     
 
 

4. x
3
– x < – a
2
+ a –
2
3 3

f(x) = x
3
– x
f(x) = 3x
2
– 1 = 0
x = 
1
3

In positive region minimum value of f(x) =
1 1 2
3 3 3 3 3
  
So, – a
2
+ a –
2
3 3
> –
2
3 3

a
2
– a < 0
a  (0, 1)

5.
3x 2 4x 4 5x 1 4x 5
      
…(1)
v u p q
u
2
– v
2
= x + 6 = p
2
– q
2

u – v = p – q …(2)
solving (1) and (2)
we get,
2 4x 4 2 5x 1
  

 x = 3

6. Limit be equal to y
logy =
n
1
n 1 n 2
lim
log log ...
n
n n
 

 
 
   
 
 
   
   
 

=
n
n
r 1
1
n r
lim
log
n
n



 

 
 
 
 
 

=
 
1
n
n
r 1
0
1
r
lim log log dx
1
1 x
n
n


 
  


 
 



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=  . 2
1
log2
2
 

 
 
= [log4 – loge] =  log
4
e
 
 
 

logy = log
4
e

 
 
 

7. 0  argz 
4

, represent the region of complex plane lying in the first quadrant and bounded by x-
axis and the line y = x
|2z – 4i| = 2|z – 2i|
least value of |z – 2i| is length of perpendicular from (0, 2) to y = x, which is
2

So the least value of
2
|2z – 4i| is 4

8. Let f(x) = 2x
3
+ 3x
2
– 12x + 3
 f(x) = 0 has three real roots (, , )
 +  +  = –
3
2

Centroid = , ,
3 3 3
 
  
 
 
 
  
=
1 1 1
, ,
2 2 2
 
  
 
 
which lies on x = y = z.

9. If a > b > c and a
2
= b =
c
then b < 1  cot
–1
x < 1  x > cot1

10. a  4sinA 
a
4
sinA

 R  2
so for any point (x, y) inside the circumcircle, x
2
+y
2
< 4
 |xy| < 2

11. a, b and A are given in ABC,
 cos A =
2 2 2
b c a
2bc
 

 c
2
 2bc cos A + (b
2
 a
2
) = 0
 c
2
 (2b cos A)c + (b
2
 a
2
) = 0
which is quadratic in c and gives two values of c and let these are c
1
and c
2

 c
1
+ c
2
= 2b cos A
and c
1
 c
2
= (b
2
 a
2
)

2 2
1 2 1 2
c c 2c c cos2A
 
= (c
1
+ c
2
)
2
 2c
1
c
2
(1 + cos 2A)
= 4b
2
cos
2
A  2 (b
2
 a
2
) (2 cos
2
A)
= 4b
2
cos
2
A  4b
2
cos
2
A + 4a
2
cos
2
A
= 4a
2
cos
2
A

12.
1 1 1
2 5 5 8 8 11
    
  
n terms
=
5 (n 1)3 2 (n 1)3
5 2 8 5 11 8
3 3 3 3
    
  
     
=
 
3n 2 2 3n 2 2 n
3
3n 2 2
3 3n 2 2
   
 
 
 

Page 9
AITS-CRT-I(Paper-1)-PCM (Sol)-JEE (Adv)/14
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10

=
n n
n
2 3n 2 3n
 
 

13.
2
sin
24 R 2




 R 2 1 cosec
24

 
 
 
 

(A) is true


2
sin cos
2 2
1 cosec
sinA
 
 

 
 
  

(D) is true

14. x
2
+ x + 1 = |[x]|
Hence only 2 solutions exist
i.e. x = – 1 and another lie – 2 < x < – 1

15. sin x + cos x = 1 – a sin x cos x
 a
2
sin
2
x cos
2
x – 2(a + 1) sin x cos x = 0
 sin2x
 
2
a
sin2x 2
a 1
2
 

 

 
= 0
Hence sin2x = 0  a  R
 x =
n
2

, n  I
And sin2x =
 
2
4
a 1
a

 a  (–, 2 – 2
2
]  [2 + 2
2
, )

SECTION – C

1. a =
2 2
r 1 r 1
1 1
, b
r (2r 1)
 
 


 

a =
2 2 2 2
1 1 1 1
...
1 2 3 4
    
and b =
2 2 2 2
1 1 1 1
...
1 3 5 7
    

b =
2 2 2 2 2 2 2
1 1 1 1 1 1 1
... ...
1 2 3 4 2 4 6
   
         
   
   

b =
2 2 2 2 2 2 2
1 1 1 1 1 1 1
... 1 ...
1 2 3 4 2 2 3
   
         
   
   

b = a 
1
4
a
b =
3
a
4


a 4
b 3


2.
1
0 0
f (x)dx tf (t)dt
 



 

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CONCEPT RECAPITULATION TEST - I Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT–I] 2014

CONCEPT RECAPITULATION TEST - I Paper 1 [ANSWERS, HINTS & SOLUTIONS CRT–I] 2014

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