CONCEPT RECAPITULATION TEST - II [CONCEPT RECAPITULATION TEST - II]

other 13 Pages

Joint Entrance Examination (Main)

Concept Recapitulation Test

ARK

Contributed by

Alex Ratan Khare

AITS-CRT-II-PCM (Sol)-JEE (Main)/14
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ANSWERS, HINTS & SOLUTIONS
CRT–II
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS
1.
B C C
2.
A B A
3.
A C C
4.
A C B
5.
B B B
6.
C B C
7.
C D A
8.
C B D
9.
A B D
10.
C A A
11.
D B B
12.
A C A
13.
B D A
14.
C A B
15.
A C C
16.
D C D
17.
D B D
18.
A C C
19.
B C B
20.
A B C
21.
B D C
22.
B C D
23.
D A A
24.
B C D
25.
A B B
26.
A D C
27.
D A D
28.
B A D
29.
D C A
30.
C B A
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.  Net force on the system is
F = 3 mg sin   
k
3 mg cos 
 a
cm
=
k
3mgsin 3 mgcos
3m
   
= 4 m/s
2

2. Consider an arbitrary moment when the wedge has travelled a distance x
into region II.
The area of the top surface inside the region II = ax sec 
Force on it = ax sec  P = ax sec  [P =1]
Component of the force opposite velocity = ax sec . sin  = ax tan .
If it further moves by dx then the work done = ax tan dx

b
2
0
0
1
mv atan xdx
2
 


 v
0
=
abh
M


h

x

3. S
A
= v
0
t 
2
A
1
gt
2

S
B
= v
0
t 
2
B
1
gt
2

S
A
 S
B
= L    t =
 
 
B A
2 L
g

  


4. N/N
0
= e

t

5. First law of motion.

6. For vertical oscillation time period
T
1
= 2
m
k
. . . (1)
For the transverse oscillation period of simple pendulum T
2
= 2
g


But 2T
1
= T
2
 k =
4mg


8. 
2
3GMm 1
mv 0
2R 2
 

 v =
3gR

9. Tension in the string T = mg  F
B
= 3vg  vg = 2vg
Balancing torque
kxR/2 = TR
 x = 2T/k = 4vg/k
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10. II and IV quadrant field will be added.

11.
6
P
4
1600 x 10
v
4 x 10


 = 4 m/s

6
Q
4
1600 x 10
v
2 x 10


 = 8 m/s

 
2 2
A B A Q P
1
v v gh
2
       

h = 3.6 mm

12.
mg T ma
  ----- (1)

2
r
mr a
T x
2 r

----- (2)
From (1) and (2),
2g mg
a & T mg
3 3
   

1
3
 

13. J
1
x 0.1 = 5 x I
J
1
= 50 mA
I = 50 + 5 = 55 mA

S
1

0.5
B

C

D

0.1
0.2 0.3
5 mA
I
I
J
1

5 mA

16. Let V
x
and V
y
be components of velocity. At contact point A
and B,
the velocity along normal should be same.

o o o
y x
ucos60 V cos30 V cos60
  …. (1)

x
2 2
x y
v 2u
v V V

 

2u
u
60
o

60
o

30
o

V
y

V
x

19.
2
1
qEy mv
2

Magnetic force will not do any work.

20. Conceptual.

21. Now downward force on the right
block is more.

mg

T

T

mg

22. I
C
= I
0
+ M(OC)
2
= I
0
+ M(OB
2
+ BC
2
) = I
B
+ M(BC)
2

23. Coulombic force between them remains same.
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24. Circuit is forming a wheatstone bridge R
eq
= 2R
For maximum power transfer 2R = r.



R

R

R

R

R

4
R

r

25.  =
3
4

2
B
 =
2
d 3 dB
dt 4 dt

 
i =
2
3
R 4R




26. C =
0 0
RT 5RT
M 3M



dx = C. dt =
L 0
0
0
T T
5R
T x dt
3M L
 

 

 
 
 
 

L 0
2L 3M
t
5R
( T T )



A B
T
L
T
0
L
x
dx

27. Equivalent circuit is

T T
90 V 20
2R R

  V = 50
0
C

P
R
R
T
V

2R
T
R
T
(90
0
C) (30
0
C)

28. Leftward Force =
2I P
C t




29. Conceptual, torque of pseudo force.

30. Effective length = (r
2
+ r
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.
1 1 2 2
P V P V
 


 
7
1 2 1
5
2 1 2
P V
32
P V
 
   

   
   

   

 
7
5 7
5
2 2 128
   .

2.
 
 
4 3
NH Cl NH g HCl
t 0 0.20 0 0
At eqm. 0.2 x x x






3
2
P NH HCl P
K P P P P K
    
=
0.36 0.6 atm
 .
NH
3
formed
PV 0.6 10
RT 0.0821 600

 


= 0.122 mole = moles of NH
4
Cl decomposed.


Moles of NH
4
Cl left = 0.2 – 0.122 = 0.078.

7.
 
2 4 7 2 4 2 4 2 4 7
X
Na B O H SO Na SO H B O
  

2 4 7 2 3 3
H B O 5H O 4H BO
 

3 3 2 3 2
2H BO B O 3H O

 

 
2 3 2
X
B O 6Na 2B 3Na O
  

8.
o o o
G H T S
    

o
2
H O H OH H 13.7 kcal
 
  



o o
o
H G 13.7 19.14
S
T 298
   
  
= 0.1102 kcal mol
-1
= 110.2 cal k
-1
mol
-1

9. In bcc,
3 a
r
4

Edge length = a


Edge length not covered by atom = a – 2r

3 2 3
a a a
2 3
 

   
 
 
 


percentage fraction not covered =
2 3
a
2
100 0.134 100
a
 

 
 
   = 13.4%

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10.
3
3
1
Al 3e Al K
Al


  
 
 



3
3
Al OH Al 3OH
 




pH = 14, [H
+
] = 1 × 10
-14

(OH

) = 1 M
K
sp
= [Al
3+
][1]
3



3 33
Al 1 10 M
 
  

o
3
0.0591 1
E E log
3
Al

 
 
 

33
0.0591 0.0591
1.66 log10 1.66 33
3 3
      

= - 2.31 V

11.
2
Cl
4
h
CH



MgClH
3
C
2
Br /h

Br
KCN

C
N
3
H O


COOH
CH
3
Cl
CH
3
CH
2
CH
3

13.
   
2
1 1 2 2
3
1 2
M V M V
CO H
n n
 


1
1
M 50
4 49.35
M 1.974
1 2


   

B
x d 10 x 1.25 10
M 1.974
m 106
   
  

x 16.7
 

15. H
2
O is weak field ligand while CN

is a strong field ligand. In complex (i), the distribution is
3 2
2g g
t e
,
i.e. all electrons are unpaired while in (ii) complex, the distribution is
5 0
2g g
t e
, i.e. two t
2g
orbitals are
paired while one is unpaired.

16.
C
O
Cl
C
C
O
Cl
O
Cl
4
LiAlH

OH
OH
OH
2 4
Conc. H SO



17.
3
H O


O
OH
 OH
O
H
(A)

O
(B)

Aldehyde and ketone can be differentiated by Fehling’s solution.

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18.
H


OH
O
OH
2
+
O

O

O
OH

19. The rate can be compared by stability of carbocation.

C
C
OCH
3
(I)
(II)
C
NH
(III)

N has more tendency to donate electron than O.

21. H = E + PV  H = E + nRT
H
2
= E
2
+ nRT
2

H
1
= E
1
+ nRT
1

(H
2
– H
1
) = (E
2
– E
1
) + nR(T
2
– T
1
)

H E nR T 4 8.314 50
        

= 1660 J

22.
2
2
H S 2H S
 




2 2
CdS Cd S
 




CdS
2 2
sp
K Cd S
 
   
   



27 14 2
8 10 1 10 S
  
 
   
 

2 13
S 8 10 M
 
 
  
 

2
2
2
H S
K H S
 
   

   

2
13 22
H 8 10 1 10
  
 
    
 

2
10
H 1.25 10
 
 
  
 

[H
+
] = 1.11 × 10
-5

pH 5
 
.

23.
   
 
2 4 2
N O g 2NO g
1 0
1 x 2x



Total moles = 1 – x + 2x = 1 + x
Since PV = nRT, V  n
% of
 
2
2x
NO 100 50
1 x
  


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1
x
3
 

 
D Theoretical VD
1 x
d Experimental VD
  

4 46 46 3
d 34.5
3 d 4

    

28. Species
NO


NO


CN


CN


Bond order 2 3 3 2
Bond order
1
Bondlength

Bondlength of
NO

> CN
+
due to greater number of antibonding electrons.

29.
f f
T i K m 86 0.001
     

0.0054 i 1.86 0.001
   

0.0054
i 2.9 3
1.86 0.001
   


i.e. there are 3 ions in the solution of complex.
So, [Pt(NH
3
)
4
Cl
2
]Cl
2
is the correct choice.

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.




2 2
2 3a 4b c 3 4a 3b c
10
a b
    




2 2
18a 17b 5c
10
a b
 




2 2
18 17
a b c
5 5
2
a b
 



2. b
2
= 4  2 = 8,
2ae 40

a
2
e
2
= 10 = b
2
+ a
2

 a
2
= 2
Centre = (1, 5)
Equation of transverse axis  x – 3y + 14 = 0
Equation of conjugate axis  3x + y – 8 = 0
Equation of hyperbola =
   
2 2
3x y 8 x 3y 14
20
1 4
   
 

3.
 
4 2
x 14x 25
f x
5
x 2
x
 

 
=
2 2
2
25
x x 14
x
5
x 2
x
 
 
 
 
 

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=
2
2
5
x x 4
x
5
x 2
x
 
 
 
 
 
 
 
 
=
2
5
x x 2
x
 
 
 
 

 x
2
– 2x – 5 = 0

2 4 20
x
2
 
 =
2 2 6
2

=
1 6


4. 6 places can be selected in
9
C
6
ways and 6 can be placed only at 5 places, except the right most
of other 6 selected. Remaining numbers i.e. 7, 8, 9 in 3! ways
Hence number of ways =
9
6
C 5 3!
 

5. z = 18 + 26i,
r 10 10
 ,
13
tan
9
  ,
0,
2

 
 


 

0,
3 6
 
 



 


 
 
1/ 3
1/ 3
z 10 10 cos isin 10 cos isin
3 3
 
 
     
 
 

3
2
3t t
tan
1 3t

 

, t tan
3

 

 
 


1
tan
3 3



1/ 3
z
= 3 + i, a = 3, b = 1

6. Let parabola is y
2
= 4ax
 = at
1
t
2
,  = a(t
1
+ t
2
), Q = (at
1
t
3
, a(t
1
+ t
3
)), R  (at
2
t
3
, a(t
2
+ t
3
))
Let T = (h, k) then h =
 
3 3
1 2
at t
t t
2 2

 
2k = a(t
1
+ t
2
) + 2at
3
=
4ah
 


The locus id 4ax – 2y + 
2
= 0

7. We observe that  = 
because of perpendicularity of two secant
Now SOR + SOQ – SOP =  ( = )
 (arg(z
3
) – arg(z) + (arg(z
2
) – arg(z)) – (arg(z
1
) – arg(z)) =

 arg(z
3
) + arg(z
2
) =  + arg(z
1
) + arg(z)
= arg(z
2
z
3
) =  + arg(zz
1
)
 z
2
z
3
= –zz
1

 zz
1
+ z
2
z
3
= 0

R(z
3
)

Q(z
2
)

P(z
1
)

S(z)

O(0)





8.
 
 
1
1
n
1
n
n 1
1 tan t
C dt
n
sin t






L =
1
1
2
n
1
n n
n
n 1
tan t
lim n C lim n dt
sin t


 

 

(  0)
Page 9
AITS-CRT-II-PCM (Sol)-JEE (Main)/14
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10

L =
 
1
1
1
n
n 1
n
tan t
dt
sin t
lim
1
n





(Applying Leibniz rule)

9. We have to find non negative integral solution of equation 2x + y + z = 21
Note that x can take a maximum value of 10 and a minimum value of 0.
We rearrange equation so that we get an integer equation with y and z as variables, x as a
constant
y + z = 21 – 2x
The number of non negative integer solution is
21 – 2x + 2 – 1
C
1
=
22 – 2x
C
1
= 22 – 2x
We now add the number of solutions so obtained for all the possible values of x.
The total number of solution is therefore
 
10
x 0
22 2x 132

 


10. As a > 0  R.H.S. > 0 hence x > 0
Applying AM  GM we get,

2
1
x
1
x
x
x
1
a
2
a
x
1
ax












L.H.S.  2









x
1
x
2
1
a
But
x
1
x   2  x > 0
 2









x
1
x
2
1
a  2 a  a  1  L.H.S.  2 a and equality holds if x = 1.

11. The required probability is
12
2
51
2
12 13
2 2
51 51
2 2
C
1
4
C
11
p
50
C C
1
4
C C

 
 

12.
 
2
13
x 2 3 y
3
 
  
 
 
= latus rectum = 3
Other conic is
 
 
2
2
2 2
y 2
x 3
1
7
7
2


 
 
 
 
which an ellipse is
Latus rectum =
2
2b 2 49 7
a 4 7 2

 


Positive difference
7 1
3
2 2
 

13. B = adj (2A)  |B| = |2A|
n – 1

1024 = 4
n – 1
2
n(n – 1)

2
10 n n 2
2 2
 


n = 3

Page 10

CONCEPT RECAPITULATION TEST - II [CONCEPT RECAPITULATION TEST - II]

CONCEPT RECAPITULATION TEST - II [CONCEPT RECAPITULATION TEST - II]

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