CONCEPT RECAPITULATION TEST - IV [ANSWERS, HINTS & SOLUTIONS CRT–IV]

other 9 Pages

Joint Entrance Examination (Main)

Concept Recapitulation Test

Contributed by

Baldev Mehan

AITS-CRT-IV-PCM (Sol)-JEE (Main)/14
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1

ANSWERS, HINTS & SOLUTIONS
CRT–IV
(Main)

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

2. 2.4 km downstream from the point directly opposite to 0

3. rate of change of speed
= tangential component of acceleration
(component of acceleration in the direction of velocity)

2
ˆ ˆ
3i 3j 6 4
ˆ ˆ
a.v (2i j) . 2m / s
5 5
 
 
 
    
 
 
 

4. Force is always perpendicular to velocity

5. elementary mass = x dy
x = R cos, y = R sin
dy = R cos d
elementary mass =  (Rcos) (Rcos d)
= R
2
cos
2
 d

/2
2 2
0
an
/2
2 2
0
Rcos R cos d
ydm
y
.dm
R cos d


  
 
  





4
R
3



6. Let the block is in translational equilibrium and ready to
topple.
 = F
N = mg
Taking torque about C.O.M. = 0

a a a
F N
2 2 2
     
 
     
     
f
2F = N

F
N

mg


= F


R

x
y

y

x


n
a


a


t
a


v

For maximum



1N

3N

9N

10N

23N

A triangle is possible with sides 4 unit, 9 unit and 10 unit. Hence minimum
resultant is zero.
3N

1N

9N

10N

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N
F
2


N
2

f
but
N
 
f

N
N
2
 

1
2
 

7.
o
o
s
V V
V V
 


 

 
f f
.
Here, V
o
= u + at
V
s
= 0
  varies linearly with ‘t’.

8. As long as cylinder is completely inside, B is constant in magnitude, then its value decreases
linearly with displacement and finally becomes zero.

9.

 
2
2
n
5
V
R 5 5m
1
a
5
  

10. Applying work-energy theorem w.r.t wedge

19.
 
0 0 0 0
P
kq kq kq 3kq
V
2a a a / 2 2a
    .
20. Pitch =
1.5
0.3
5
 mm
Least count =
0.3
0.006
50
 mn

21.
1 1 2 2
m v m v


2 1 1 2
v m v / m
  (1)

2 2
2 1 1 2
2
2
a 1 1
m g m v m v
2 2
kx
2 2
1
   (2) From work energy theorem
Substituting (1) in (2) and solving

2
2 2
1
1 1 2
m (ka 4 m ga)
v
4m (m m )
 


= 10 cm/s

22. Mechanical energy loss happens due to viscous/frictional forces without which the liquid will
oscillate. This is released as heat
1/ms
at t = 0
at t = 1sec

5

1/ms
2/ms
1m/s
2

1m/s
2

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23.
2
3
GMm
mr
r
 

3 2
GM r
 
(GM = gr
2
)

3 2
2
r .
g
R



3 2
2 2
(4nR) .
g
n .R



2
g 64 nR
 
.

24. Frequencies are in odd ratio hence it is a closed organ pipe.

v
freq (2n 1)
4L
 

2v
(frequency) 100Hz
4L
  


L = 1.7 m.

25. For maximum elongation charges on the blocks must be equal to
Q
2
on each block.

2
2
0 0
Q
1
2
kx
4 ( x)
 
 
 

 


0 0
Q 2( x) 4 kx
   .

26. Number of emitted electrons
12 4
5
10 2 10 25
10

  
 = 5 × 10
4
electrons
q = + ne = 8 × 10
–15
C.

27.
3 6
R 2
3 6

   


1 2
1 1 1
R R R
  
1 2
2 2 2
1 2
dR dR
dR
R R R
 
% error =
R
100%
R


=
1 2
1 1 2 2
R R
1 1
R 100 100 %
R R R R
 
 
    
 
 

=
1 1
2 1 2 %
3 6
 
  
 
 

=
4
%
3

28. As V increases, E increases, 
min
decreases, 
K
remains constant so (
K
– 
min
) increases

29. Increasing the filament current increases the intensity because more electron
now strike the target. Decreasing the potential difference increases the minimum wavelength.

30.


5
9
1.6 1 1.8 10
900 10


  

= 12
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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.

CH
3
HC
OEt
OEt
(p) (Acetal)
3
H O
3
CH CHO



: differentiated by Na, Fehling and Tollen’s test

 
3
H O
3 2 2 3
CH CH O CH CH EtOH; differentiated by Na,Fehling
andTollen's test
q

    

2.

Ag/AgX/X
E

=

Ag/Ag
E

+ Kspln
F
RT

3. Compounds with more covalent nature, was unable to respond to chromyl chloride test ‘SnCl
4
’,
due to mole polarizing power of Sn
+4
the compound is more covalent.

5.
 
   
2
3
c
2
2 2
SO
K
SO O


 
c
2
1
K 100
O
 
[O
2
] = 10
– 2

2
n
10
10




n = 0.1

6.
2
2
1
2
1
1
21
T
VP
T
VP
T
V2P
nnn 

8. x
A
= 0.3 , x
B
= 0.7

T
B
B
T
A
A
P
P
y,
P
P
y 

2
7
xP
xP
2
3
4.0
6.0
y
y
BB
AA
B
A




9. RbCl and KCl have rock salt structure
5.0aa
KClRbCl







5.0rr2rr2
ClKClRb






Hence

A33.1r
K



11. Intramolecular aldol condensation.

12. Perkins’s reaction
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13. Bromination of benzylic carbon followed by hydrolysis

14. Hoffmann elimination, least stable alkene will be the major product.

16. -Hydroxy compound is easily dehydrated.

17.
2 2
2AB (g) 2AB(g)+B (g)
1 0 0
1-x x x/
2



Total moles = 1 – x + x + x/2 = 1 + x/2.

  
 
2
p
2
x 2 x
p
k =
1+x/2
1-x
 
 
 
[

x is small, 1-x = 1 and 1 + x/2 = 1].

 
1
3
3
p p p
x p
k = Þ x = 2k P = 2k P
2

19. At constant pressure, the addition of inert gas shifts the equilibrium towards larger no. of moles.

20. The hybridisation of B in
3
BBr
is
2
sp
that C in
2
CS
is sp, that of S in
2
SO
is
2
sp
while that of
S in
4
SF
is
3
sp d
. Hence
2
CS
has the maximum bond angle of 180°.

27.
3 2 3 2
NaHCO NaOH Na CO H O
  

28. For a bcc lattice

o
X 0.4
0.732 Y 0.546A
Y 0.732



   

Now, for fcc lattice,

Z
0.414 Z Y 0.414
Y

 

   

= 0.546 × 0.414
= 0.226
o
A

29.
C
O
H
C
O
Ph
O
H
C
O
Ph
(I)
O
C
H
Cl
C HCl
Cl HO
(II)
C
C
O
O
C
H
O
O
H
(III)
H C
O
O
H
H
H
(V)

30. Hint: It is Acyl Cleavage

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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

SECTION – A

1. f(x) = f(–x)
 x
n
= –(–x)
n

2.
a
a
x
log x
lim
x

 0 and
x
x
a
a
lim
log x

 

3. g(x) = cos (x) – ||x| – 1|, x  R

4.
dv rdh 2hdr
r
dt dt dt
 
  
 
 

5. f(x) =
4 3 2
ax bx cx
dx
4 3 2
  
f(0) = 0, f(–1) =
a b c
d 0
4 3 2
   

6. h(x) = f
2
(x) + (f(x))
2

h(x) = –2x(f(x))
2
g(x)

7. Let F be the fuel charges. Then,
2
3v
F
16

Let train covers  kms. Then, total cost for running the train,
3v
C 300
16 v
 
  

For
dc
0
dv

and
2
2
d c
0
dv


 v = 40 km/hr

8. For maximum or minimum, f(x) = 0
 –x
2
+ 8x – (12 + t) = 0
 t < 4

9. AD divides BC in ratio AB : AC
So, P.V. of D =




ˆ ˆ ˆ ˆ ˆ ˆ
AB 2i 5j 7k AC 2i 3j 4k
AB AC
    

 
 

10. Required vector
1 2
r x b x c
 

 
and
 
2
r a a
3
 
  






1 1
ˆ ˆ ˆ
r 2i j x 2 k 4 x
      

, where x
1
 R

12.


2
r a x a b c
 

   
,


3
r b x b c a
 
 
  
,


1
r c x c a b
 

   

13. Equation of line x + 2y + z –1 + (–x + y – 2z – 2) = 0
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x + y – 2 + (x + z – 2) = 0
(0, 0, 1) lies on it
  = 0,  = –2
For point of intersection, z = 0

14. Equation of lines, (y – mx) + , (z – c) = 0
(y + mx) + 
2
(z + c) = 0
It meets at x–axis, hence 
1
= 
2

 cy = mzx

15. Slope of reflected ray =
3
4

 Equation of incident ray is (y + 4) =
 
3
x 2
4
 
i.e., 4y + 3x + 22 = 0

16. Required area = 4(Area of S
2
OS
4
) =
1
1
4 ae 8be
2
 

17.
   
1 1 1 1
r r 1 r r 1
sin tan tan r tan r 1
r r 1 1 r r 1
   
   
   
   
   
   
  
   

18.
2 2 2
1 1
AG 2b 2c a a
3 3
    ;
2 2
1
BG b 4c
3
 

  
 
1
GAB
AG BG AB 5 13
R
4 12
 


19. z
n
– 1 = (z – 1)(z – z
1
)(z – z
2
) ….. (z – z
n – 1
)
Differentiating w.r.t. x

n 1
n
n 1
nz 1 1 1
.....
z 1 z 2 z z
z 1


   
  


20.
   
r
1 1 1
T
2 1 3 5 ..... 2r 1 1 3 5 ..... 2r 1
 
 
 
         
 

22.
 
1 cosx cosx
sinx 2cosx 0 sinx cosx 0 0
0 0 sinx cosx
    


tan x = –2 or tan x = 1

23. Required area =
 
 
2
x 2
0
2 2x x dx
 


(0, 1)

y

(2, 4)

x


x

y


O

(2, 0)

26. Combined mean = 16; d
1
= 9; d
2
= –6
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



2 2 2 2
1 1 1 2 2 2
2
1 2
n d n d
67.2
n n
    
  


27. 1  1 + sin
3
x  2
I
2
2 2
 
 

28.
dy
xy
dx
x e


lnx
ydy dx
x

 


 
2
y lnx c
  

29. Let A(, 0) and B(, 0) be the two points
OT
2
= OA·OB =  =
c
a

30.
 
18
1 x
f x
1 x





 
 
18
1 x 1
f x 1
x
 
 
Page 9