CONCEPT RECAPITULATION TEST - IV Paper 1 [ANSWERS, HINTS & SOLUTIONS]

other 9 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

HHP

Contributed by

Heena Hari Pandey

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

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

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

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

1. There will be no change.
 h = h

h
A

2.  =
0



v
A
– v
B
=
0 0
(b) (a)
 

 

 W = Q(v
A
– v
B
) =
0
Q
(b a)




3.  = 0 + 1  10 = 10 rad/sec
2

 v = r = 1  10 = 10 m/s

0
3
q(v r )
B
r





|B |
=
0
2
qv
4 r



B =
7
2
10 0.1 10
(1)

 
= 10
-7
T

4. conservation of momentum

M 2gh m 2gh
 =MV
1
+ mV
2
(1)

1 2
V V
1
2 2gh

  (2)

2
V 3 2gh


2
2
V
h' 9h
2g
 

5.
1 2
1 1 1
C C C
 
=
0 0
x a b x
A A
 

 

 C =
0
A
a b



6. FBD of ‘B’ and ‘C’

a B

2g

T
a

C

3g

T

 T – 2g = 2a . . . (i)
and 3g – T = 3a . . . (ii)
T =
12g
5

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for A

a

A

mg

T=2T

 T = mg
2 
12g
5
= mg  m = 4.8 kg.

7. W =
/ 2
0
mgR
f.Rd
2

 
 

= 1 joule.
9. a =




v g a m g a
m
   
= 20 m/s
2

t =
2h
a
= 1 sec
11. Angular momentum =
nh
2


kinetic energy 
2
1
n

De-broglie wavelength  =
h
mv
so as v increases  decreases.

14. Relative velocity of B w.r.t. A along vertical direction should be zero
20 sin  = 10
 sin  =1/2
x = 20 cos 30 
1
2
=
5 3
m
15. Change in angular momentum = angular impulse
L = 5
I =
2
ma
3

K =
2 2
2
L 75
2I 2ma



Section -C
1. Conservation of energy
2
1 3
mv mg mg m 3 2
2 2 2 2
 
  
 
 
 
 
 g

v 8g



60

2. Potential across capacitor is zero, hence energy stored is zero.

3. v
A
– v
B
=  E. dx = 10 (1)
= 10 volt.

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4. Z =
2 2
R X

=
2
9 X


but cos  =
R 3
Z 5


X = 4 .

5.
E
2
= E – ir
 i =
E
2r
. . . (i)
2E = i(3 + r) . . . (ii)
 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. K
c
= 4 × 10
-19

 
1
2
Ag 2CN Ag CN
0.03 0.1 0

 
 

 



 
1
2
Ag CN Ag 2CN
t 0 0.3 0 0.04
0.03 x x 0.04 2x

 
 

 

 



 
2
19
c
x 0.04 2x
K 4 10
0.03 x


  

 
 

0.04 2x 0.04
 

0.03 x 0.03
 

 
2
19
c
x. 0.04
K 4 10
0.03

  

x = 7.5 × 10
-18

2. PV = nRT
As PV constant
n
1
T
1
= n
2
T
2

n
1
. 300 = n
2
. 400
n
2
=
1
300
n
400

1
3
n
4

4. Energy of
e
in H-atom = - E
Energy of
e
in Li
+2
= - 9E
Energy supplied by photon =
2
1
IE mv
2


IE |E |


2
p
1
E 9E mv
2
  

 
p
2 E 9E
v
m


9.
4 2 3 3 2
FeSO Fe O SO SO

  
12.
o
2
1
H e H E 0.0 v
2
 
  

2
H /H
0.059 1
E E 0 log
1
H


  
 
 

0.059 7 0.413
    

So,
2
2
H /H Ni /Ni
E E
 

Ni will deposit at cathode

2 2
2H O 2e H 2OH
 
  
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2
0.059
E 0.83 log OH
2

 
  
 

0.83 0.059 7
   

= - 0.417
13. Both have Cr
+3
(d
3
), and C.N. is 6. Hence d
2
sp
3
.
Pd
+2
(d
8
) complexes with C.N. is 4 are square plannar and magnetic moment is zero.
14.
3 2 2
FeCl FeCl Cl

 
Dil. CuCl
2
is blue due to [Cu(H
2
O)
4
]
2+
PH
3
is obtained by heating white P with NaOH
All are reducing because of weak P – H bond.
SECTION – C

1.
A =
H
2
C
CH
3
Br
C =
CH
3
Br
Br
CH
3

B =
Br
CH
3
Br
CH
3
E =
CH
3
Br
CH
2
Br
H
D =
inactive

2.
3
KNO

22
2
1
OKNO 
223
2
1
ONaNONaNO 

23
)(2 NOPb

2
22
4 ONOPbO 




2
3
2
3
, NOSrNOCa ,

34
NONH

OHON
22
2
23
)(NOPb do not respond to the test.

23
)(2 NOCa

22
42 ONOSrO 



2
3
2 NOMg

22
42 ONOMgO 

3.
O S
O
O
H S OH
O

4.
b b
T K m i
   

2.08 = 0.52 × 1 × i
i = 4



3
6
K Fe CN
 

 

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5.
Cl Cl
Cl Cl



OH OH
OH OH
O
O
 
X
 
Y
 
Z

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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.
 
 
 
 
1 1
2 1
I x 1 1 1 dx x 1 0 1 dx
 
 
      
 
 

2. Here,
 
 
1
f x , if 0 x
2
1 1
g x f , if x 1
2 2
3 x , if 1 x 2

 


 

  
 

 

  



4. Let
A
CAD
3
   

By m – n theorem,
(1 + 1) cot  = 1 cot 2 – 1cot 

A

B

C

D





2


6. Here, z
2
– z
3
= i(z
1
– z
3
)
 ACB
2

 


13
AC
2
 ; AC = BC

9. (0, 0) lies on the director circle

10.
1/ 3 1/ 2
2/ 3 1/ 3 1/ 2
x 1 x 1
x x
x x 1 x x

 
  
  

12.
1 1
3
a b 1 1 1
a b
x y z 3 ;
2 x y z 2
 

 

 
 
     
 
 

14. Here, f
–1
(x) =
a x a
 

SECTION – C

1. Equation of tangent is




y f' x
     

Also,


 
2
f '
f ' 1
   
 
 

Thus,
2 2
dy 1 y x
dx 2 xy
 


 
 

Thus, equation of curve is x
2
+ y
2
– 2x = 0

2. Equation of normal is
 
n
n 1
1
y a x a
na

   

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4. Here, 
1
= d(a
1
– a
2
)(a
2
– a
3
)(a
3
– a
1
) = –2d
4

Also, 
2
= –2d
4

5. sin
2
x[sin
2
x + sin x + 2] = 0
 sin x = 0, where x = 0, , 2, 3
Page 9