CONCEPT RECAPITULATION TEST - I [ANSWERS, HINTS & SOLUTIONS]

other 13 Pages

Joint Entrance Examination (Main)

Concept Recapitulation Test

Contributed by

Anees Batta

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ANSWERS, HINTS & SOLUTIONS
CRT – I
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. C D A
2. B B C
3. C C C
4. D C B
5. D C A
6. A A B
7. C B C
8. D D D
9. B B A
10. C A B
11. C C C
12. B B B
13. D B B
14. D A C
15. A B B
16. C B C
17. B B A
18. B D B
19. B B C
20. B B B
21. C C C
22. A C B
23. D D A
24. B B A
25. B C B
26. C C C
27. D D C
28. B B C
29. C C B
30. B A C

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

1.
6
3 14 10 5
1

  
 

15.7
 
ohm metre.

2 r L R
r L R
  
 
    
 
 
ohm metre.

01 01
15 7 2
1 1 5
01
 
 
     
 
 

ohm metre



15.7 02 01 002
       ohm – metre

15.7 032
   
 m

5024
  
 m

2.
P
100 3 0 1 2 0 2 3 0 3 4 0 1
P

            

3 4 9 4 2%
        

3.
0
t
0
A N e

 

t
0
3Ae 3 N e

 

 
0
t t
1
1
e e
e
 

  

0
1
t t
 

.

4. At
o
1 0
v sin
ˆ
t ,v v cos i
g

  


2 0
ˆ ˆ
v v k gtj
 


2 1 0 0
ˆ ˆ ˆ
v v v k gt j v cos i
     
 

5.
 
2
2
f a Z b .
 

(D)


6.


k mg x
dv
v
dx m
 
 

0
0
x
0
v o
k mg
vdv xdx
m
 
 
  
 
 
 

 
0 0
m
x v
k mg
 
 

7.


 
2
1 2
max
2
min
1 2
I I
I
I
I I

  


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2
1
2
I
1
I
1
 
 

 
 
 
 
if I
1
> I
2

8.
c
100
I sin 50t
4
5 2

 
 
 
 

L
100
I sin 50t
4
5 2

 
 
 
 

I = 20sin(50t)

5 2


/4

5 2


/4

/4

10 2

/4
I=Ic+I
L

9. 2mg sin30 – T = 2m(a)
T = ma

mg
a g / 3 & T
3
  


Net force =
Result of these forces






0
ˆ ˆ ˆ
T j Tcos60 j Tsin60 i
     

3 3
ˆ ˆ
Ti Tj
2 2
  

 
mg
ˆ ˆ
3i 3j
6
  
As net force in pulley = 0,
force by clamp
 
mg
ˆ ˆ
3i 3j
6
 

T

0
30

T

60
0

10. The circuit can be simplified as:

0
0
 ,peak current =20 Ampere.

5

.1H
.004F

100 sin [50t] volts.

.004F

5


1
5
c
 


L 5  

12. When u decreases v increases. For plane mirror u decreases.


the image in plane mirror moves towards the left.

13.
0 0
0
I
1 1
4 . sin45 sin45
4 R 2R

 
 
  
 
 

 

0 0
I I
1
2
2R
2 R
 
 
 
 
 
 


 


14.
1
10
V 4
4 4
 

volts = 5 volt

2
100 4
10
100 4
V
100 4
4
100 4








= 4.9 volts
1 2
V V 0.1
   volt

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15. In both cases
applying com,

0
mv Mv


0
mv
V
M
 

1
2
V
1:1
V
 

16. In steady state :-

P Q
E E
V V R
2R 2
   

Q S
E E
V V R
2R 2
   

S R
V V


P R
V V E
 

A
Q CE
 

B
CE
Q
2


17.


P 0
V V 10 10 10 10
     
= - 200 volts.

18. For the two blocks to move together

 
min
F F
M m g 2Mg
  


As
min
,
  
the two blocks move together.


Velocity of the system (M+m) is

F
v at t
2M
 
 
 
 

2
F F
P Fv F t t.
2m 2M
   

19.
2
P
Axt mcT
4 r
 


2
4 r mcT
t
PA



20.


A b 2 T
   

.

21. After collision,
the rod starts rotating about its centre of mass.
Now the torque of mg about the centre of mass = 0, hence  is constant.

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22.
0
x

elongation in equilibrium

0
macos mgsin kx
     (1)
For x = elongation from equilibrium
net force


0
mgsin macos k x x
     

kx
 
(i)

kx
acc
m
  

m
T 2
k
  

N

θ

K(x
0
+ x)
ma

Mg sin


Mg cos

23.
 
3
0 T
4
r n n g 6 rV .
3
   



2
0
T
2r n n
V g
9






2
0
3
g
2r n n g
4
P r .n.g.
3 9

 


 
5
2
0
8 r n
n n g
27

 

55.
x 2cm 0.02m
  

24. 
k x 100 .02
   

= 2rad

25.
0 0
2T 2T
P gh P
r 2r
    

T ghr.
 

26.
M
G
GM 2GM
4
5R
2R 5R
2

  

27. The apparent frequency initially decreases & then increases in one time period.

28.
e e
GM m GM m
E
12R 4R
   
e
GM m
6R


29.


  
2 2 2 2
v u 2gh 0 40 2 10 h
    

h 80m
 

where h is maximum height reached by ball A
v = u – gt  t = 4 sec,

A B
80
t' 2sec
u u
 


A
u' gt' 20m / s
 

B
u' 40 2 10 20m / s
   

U
A
=0

40m/sec
= u
B

h =80 m

A(t=4sec)

B
u
A

u
B

(T=t+t)

Thus the speed of each ball gets interchanged (for equal mass & elastic collision)

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30.
2
1 2 1
T T m r
  



2
2 1 2
T m r r
  



2
1 1 2
T m 2r r
   
Thus
1 2
T T


m

T
1
T
2

T
2

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

1. M
wt
C
6
H
5
Cl = 112.5
M
wt
C
6
H
5
CH
3
= 92
Given:

6 5
C H Cl
x 0.5


So mass% composition reaction of C
6
H
5
Cl is
% C
6
H
5
Cl =
112.5
100
204.5
 = 55%

2. As density increases, volume of gas decreases so intermolecular forces increase and diffisubility
decreases.

3. For 
432

n = 4, l = 3, m = 2
Angular momentum =
 
h 12h 3h
1
2 2
  
  
 
Angular node =

= 3
So total node =
3h
3


=
3h 3
 


4.
o
250 C
3 4 4 2 7 2
Pyrophosphoric acid
2H PO H P O H O 
OH P O P OH
O
OH OH
O

5.
P
B
B
P
P
B
O
+
O
+
O
+
O
+
O
+
O
+
H
H
H
H
H
H







3 p—p back bonding between B and O and
3 p—d back bonding between P and O is possible.

7.
 
2 2 7 2 4 3 4 2
Chromic anhydride
Red Crystal
K Cr O 2H SO 2CrO 2KHSO H O
   

9.
2
NH

behaves as a base and E
2
elimination takes place.
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10. OH
3
NaOH
CCl

OH
CCl
3
H

OH
CHO

11.
 
 
2 2 3 3 2 2 3 2
black ppt
white
Na S O AgNO Ag S O Ag S
  

12.
c
8a
T
27Rb
 so
c
1
T
b

; and b = 4  N
A
 V  b  r

13. AB type structure in which Zn
+2
is tetrahedrally surrounded by S
–
ion, S
–2
is present in half of
tetradedral acid.

14. It is an extensive property.

15.


 
 
1 2
K t K t
1
1 2
0
2 1
d B
K
A K e K e
dt K K
 
 
  
 

 

16.
 
 
2
2 3
3
2
6
Fe H O OH H O
ksp
Fe H O



 
 
 
 

 
 
, Ksp should be high.

17.
1 2 2
a 1 a C
H K C K 2.5

 
  
 

18. G
0
(net) =
0 0
1 2
G G
  

Total e
–
exchanged = (a + b)

0 0
G nFE
  



0 0 0
3 1 2
a b FE aFE bFE
    

 
0 0
0
1 2
3
aE bE
E
a b




19.
2 2 7 2 4 2 3 2
K Cr O K CrO Cr O O

  

20.
CH
3
C
6
H
5
O
M

PPh
3

CH
3
C
6
H
5
3
POPh (Wittig reaction)


21. Nitrogen is less electronegative than oxygen and on 3
rd
reactive site it is sp
3
hybridised.

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22. O
C
O
CH
3
(1) NaOI

O
COOH


O
2
CO


23. As metallic character increases down the group so acidic nature of oxide decreases.

26.
Ni
4-
Cl
Cl
Cl NH
3
Cl
NH
3
-2
Ni
4-
Cl
Cl
Cl Cl
NH
3
NH
3
-2
and
Cis
Trans

27.
3
2 3 2 4
As O 5H O 2AsO 10H 4e
  
   

29.
2
2BeO 2C Be C 2CO

  

30. Only lithium nitride with formula LiN
3
in first group and Mg with Mg
3
N.

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

2. Since the equations are consistent
 D = 0

3 3 3
(a 1) (a 2) (a 3)
(a 1) (a 2) (a 3) 0
1 1 1
   
    


Put u = (a + 1), v = a + 2, w = a + 3
u  v = 1, v  wz 1, w  u = 2
u + v + w = 3a + 6

3 3 3
u v w
u v w 0
1 1 1


(u  v) (v  w) (w  u) (u + v + w) = 0
(1) (1) (2) (3a + 6) = 0
i.e. a = 2.

3. 1234 = 3n + 1, so 1234
567
= 3m + 1,89 = 3k – 1, so 89
1011
= 3p – 1.
Hence 1234
567
+ 89
1011
= 3n 1234 is even,
So 1234
567
= 4q. 89 = 4r + 1, so 89
1011
= 4t + 1.
Hence 1234
567
+ 89
1011
= 4l + 1. So it is 9x + 1.

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4. Here, tan A + tan B =
2
c
ab
and tan A . tan B = 1
tan A . tan B = 1  tan A = cot B  A = 90
0
– B  A + B = 90
0
 C = 90
0

 sin A =
a
c
, sin B =
b
c

 sin
2
A + sin
2
B + sin
2
C =
2 2
2 2
a b
1 2
c c
  

5. I
1
=
  
101
2 2 4x
100
dx
5 2x 2x 1 e


  

=
   
 
 
 
101
2
2 4 1 x
100
dx
5 2 1 x 2 1 x 1 e
 

    


 2I
1
=
101
2
100
dx
5 2x 2x

 

= I
2


1
2
I
I
=
1
2
.

6. Since f(x) and g(x) are one-one and onto and are also the mirror images of each other with
respect to the line y = 2. It clearly indicates that h(x) = f(x) + g(x) will be a constant function and
will always be equal to 4.

7. Equation of the curve passing through the points A, B, C, D is
(3x + 4y – 24)(4x + 3y – 24) – xy = 0
If it represents a parabola then the quadratic terms must form a perfect square.
  = 1 or 49.

8. Locus is a hyperbola with eccentricity =
1 2
1 2
r r
2
r r




9. In an equilateral triangle r =
2
R

Also exradius r
1
= 4R sin
2
A
cos
2
B
cos
2
C
= 4R R
2
3
2
3
.
2
3
.
2
1

 r, R, r
1
are in A.P.

10.
 
2
x / 2
2
2 2
x
0
t
lim dt
x 1 t



=


form

 
2
x /2
2
2
0
2
x
t
dt
1 t
lim
x



=
4
4
x
x
x
4 x
lim
2x


=
5
5
x
x
lim
2x 8


=
1
2
.

11. (3m + 1)(3n – 1) = 3n
2
– 1, so if p divides 3n – 1, then it also divides 3n
2
– 1
and hence also (3n
2
– 1) – n(3n – 1) = n – 1, and hence also (3n – 1) – 3(n – 1) = 2.
So 3n – 1 = 0, ± 1 or ± 2.
But n must be an integer, so 0, 1, – 2 do not work. – 1 gives n = 0 and hence m = 0, which is a
solution. 2 gives n = 1 and hence m = 0, which is a solution.

Page 9
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10

12. Two lines can meet in a point

6
C
2
= 15
line and circle meet in two points = (
6
C
1

4
C
1
)  2 = 48.
Two circles meet in 2 points =
4
C
2
 2 = 12
total number of points = 48 + 15 + 12 = 75.

13. 1 + cos3x+ 1 –
7
cos 2x
2 6
 
 
 
 
 
 
 
 
 
 
 
= 0

2
3x 2
2cos 1 cos 2x 0
2 3

 
   
 
 

2 2
3x
2cos 2sin x 0
2 3

 
  
 
 

3x
cos 0
2

,
sin x 0
3

 
 
 
 

x =
3

,  and x =
3

,
2 7
,
3 3
 
, …..
 x =
3

is the common value which satisfies both
 x = 2n +
3

= (6n + 1)
3


15.
2
4 2 1
(x 1)
dx
1
(x x 1)cot x
x


 
  
 
 

=
2
2 1
2
1
1
x
dx
1 1
x 2 1 cot x
x
x

 

 
 
   
   
   
   


=
2 1
dt
(t 1)cot t




Put x 
1
t
x


2
1
1 dx dt
x
 
 
 
 

=
du
u



Let cot

1
t = u

2
1
dt du
(1 t )
 

=  ln|u| + c
=  ln
1
1
cot x c
x

 
 
 
 

16.
70 70 70
23 7 2 32
3 9 27
     
     
     
     

Page 10

CONCEPT RECAPITULATION TEST - I [ANSWERS, HINTS & SOLUTIONS]

CONCEPT RECAPITULATION TEST - I [ANSWERS, HINTS & SOLUTIONS]

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