CONCEPT RECAPITULATION TEST - IV Paper 2 [ANSWERS, HINTS & SOLUTIONS]

Question Paper 6 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

HHP

Contributed by

Heena Hari Pandey

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

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

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

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

2. T sin  = mg/2
T = T cos 
=
mg mg
cot
2 2tan
 


T sin


T cos


T

T

3.
2
1
mv pt
2

(P = const)
 v =
2Pt
m

a =
dv 2P 1
dt m
2 t

F = ma =
mP 1
2t v


5.
1 2 3 4 5
F F F F F F
    
     

2 5 2 4
F F and F F
 
   

1 3 2 1
F F 2F cos30 2F cos60
    
 

F
3
=
2
2
Gm
4a
; F
2
=
2
2
Gm
3a
; F
1
=
2
2
Gm
a

m

m

m

m

m

m

F
1

F
2

F
3

F
4

F
5

F =
2
2
Gm 5 1
4
a
3
 

 
 
= m
2
a
 =
3
Gm 5 1
a 4
3
 

 
 

T = 2
 
3
4 3a
Gm 5 3 4


6.
   
d
g L x gL xu
dt
       

7.
x
p
v 5

,
y
p
v 5


5 2
= 7.1 m/s approx.

8.
6
P
4
1600 10
v
4 10





= 4 m/s

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

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

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11-12. Let the point of suspension is origin(O), then C.M. lies on
vertical line when it is in equilibrium position.
x
cm
=
m(0) m(L/ 2) m(L)
3m
 
= L/2 and
y
cm
=
m( L/ 2) m( L) m( L/ 2) 2L
3m 3
     

 tan  =
L/ 2
2L/ 3
= ¾   = tan
-1
(3/4) or 37
0

d
2
=
2 2
2L L
3 2
   

   
   

d =
5L
6




/2

m

5

/6

2

/3


/2

y
x

O

Moment of inertia I = Mk
2
=
2 2 2 2
m
4m
12 4 2 3
 
  
 
 
   
= 3m
2

(Assume full square and calculate moment of inertia about origin and subtract the moment of
inertia of fourth rod.)
Time period of a physical pendulum = 2
I
mgd
= 2
6
5g

sec

13-14. F = qvB sin  = qvB ( = 90)
So, B =
F
qv
=
20
19 3
3.2 10
1.6 10 4 10



  
= 5  10
-7
T
Now,
p q
7
0
I I
5 10
2 5 2

 

  
 

 

 I = 4 amp.
If the distance of point R from third current carrying current is X, then
B
R
= 0

7
0
2 2.5
5 10 0
4



  
 

so x =  1 m

16. P(2r)dr = 2r
dv
dr
 2(r+dr) 
dv
dr

P(2r)dr =  2 
dv
dr
 
 
 
dr
P(2r) =  2 
0
2
2v r
R

 
 
 

P =
0
2
2 v
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
8. (D) is incorrect, since Gabriel method is used to prepare 1
o
aliphatic amines.
10. Bottles 1, 2, 3, 4 have Pb(NO
3
)
2
, HCl, Na
2
CO
3
and CuSO
4
.
When mixing:
Bottle 1 + Bottle 2  PbCl
2

Bottle 1 + Bottle 3  PbCO
3

Bottle 1 + Bottle 4  PbSO
4

Bottle 2 + Bottle 3  CO
2

Bottle 2 + Bottle 4  No visible reaction
Bottle 3 + Bottle 4  CuCO
3

12.
2 2
N O
2 3
Li Li O Li N


 
2
H O
LiOH
2
Li N

  
2
H O
3
LiOH NH
  

Solution for the Q. No. 15 & 16
OH
NO
2
 
A
2 5
C H I

OC
2
H
5
NO
2
 
B
Sn HCl

OC
2
H
5
NH
2
 
C
2
Ac OH AcOH

OC
2
H
5
NH C
O
Me
 
D
 
A 
OSO
2
Ph
NO
2
NaF/DMSO

F
NO
2
 
E
 
F

20. (P – 2) oxalic acid dehydrate, loses H
2
O when treated at 105
o
C.
(Q – 3) when heated at 200
o
C or when refluxed with H
2
SO
4
at 90
o
C, it decomposes into CO
2
, CO,
HCOOH and H
2
O.
(R – 4) malonic acid on heating with P
2
O
5
loses H
2
O and forms carbon suboxide.
(S – 1) oxalic acid is oxidised by KMnO
4
to CO
2
.

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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

HINTS AND SOLUTION

1. Put x
2
= t

 
2 2 2
t t t 2 t
1
I e te e 2t e dt
2
 
  
 


2. We have
a b
k
2 3
 
and
2
8
sec A
5



2
5
cos A
8



2 2 2
5
c 10k , k
2


4.
 
 
1 1
1 2 3
1 2
tan tan
a 1 b
tan 1
1 b a
1 tan tan
  
 
       
 
  




7.
 
x if x [1, 2)
f x
x if x (2, )
 



 


8. Clearly, A = (–2 cos 60º, 2 sin 60º)
B = (2 cos 60º, –2 sin 60º)
Tangent at A is x(–2 cos 60º) + y(2 sin 60º) = 4
and tangent at B is x(2 cos 60º) + y(–2 sin 60º) = 4

30º

2

B

60º

2

A



P 3, 1

2

x

x


9. Since, parabola is above x–axis
D = a
2
– 4  0
Thus, a
max
= 2

10. Equation of tangent at (0, 1) to the parabola is y – 1 = a(x – 0)

11.-12.
az bz c 0
  
..... (1)

az bz c 0
  
..... (2)
Eliminating
z
from (1) and (2)



2 2
b a z ca bc
  
Adding (1) and (2)
 


 
a b z a b z c c 0
     

This is of form
Az Az B 0
  

13.-14 Here,
2x 2x
A
2x 2x
e 1 e 1
1
e
2
e 1 e 1
 
 

 
 
 
 

So, f(x) = e
2x
+ 1; g(x) = e
2x
– 1

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15.-16.
 
 
/ 2
0
f x sinx cost f t dt sinx

  


 f(x) – P sin x = sin x
or f(x) = (P + 1) sin x
Where
 
/ 2
0
P cost f t dt

 


=
 
/ 2
0
cost P 1 sint dt

 


=
 
/ 2
0
P 1
sin2t dt
2

 




P 1
P
2
 
  P
2


 

 
f x 1 sinx 2
2

 
  
 
 
 

 sin x = 2 – 
   [1, 3]

/ 2
0
2
sinx dx 3
2


 



4
3
 

20. (P) 2(F(x) – f(x)) = f
2
(x)
 f(x) – f(x) = f(x) f(x)

 


   
f x 1
f ' x 1
1 f x 1 f x
  
 

 f(x) is an increasing function
(Q) 21  f(5) – f(–2)  28
(R)
2
x 2
y
2x 3x 6


 


1 1
y ,
13 3
 
 
 
 

(S) y = e
–x
cos x
 y
1
= –e
–x
sin x – e
–x
cos x = –e
–x
sin x – y
y
2
= –e
–x
cos x + e
–x
sin x – y
1

 y
4
+ 4y = 0
 k = 2
Page 6