Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI

IIT – JEE - 2013
(ADVANCED PATTERN)

FULL TEST – 2

MARKS: 240
PAPER - II

SECTION - I PHYSICS

PART I: Single Correct Answer Type
This section contains 8 multiple choice questions
...

(+3, –1)

1
...
Find the magnitude of the electric force on any sphere due to the other two
...


In the shown arrangement of the meter bridge if AC corresponding to null deflection of
galvanometer is x, what would be its value if the radius of
the wire AB
is doubled?
R2

R1

G
A

(a) x

C

x

(b)

x
4

B

(c) 4x

(d) 2x

3
...
Its longest edge is thrice its shortest edge
...


The ratio of the magnetic field at the centre of a current carrying coil of radius a to the field at a
distance 3a on its axis is
(a) 20 10
(b) 2 10
(c) 10 10
(d) 10

5
...
It is placed in a uniform magnetic


field B0 such that B0 is perpendicular to the plane of the loop
...


A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b
respectively
...


Two thin long parallel wires separated by a distance b are carrying a current I ampere each
...


A circular loop of radius 2m is kept in a magnetic field of strength 2T x
(plane of loop is perpendicular to the direction of magnetic field
...
A conductor of length 4m is
x

1
sliding with a speed 2 ms as shown in the figure
...
Each question has four choices (A),
(B), (C) or (D) out of which ONE or MORE are correct
...


A milliammeter of range 10 mA and resistance 9  is joined in a circuit as shown
...
It gives
a full scale deflection again for a current I ' when A and C are used as terminals
...
9 

0
...
1 A

10
...
The
magnitude of the magnetic field is
(a) maximum at the axis of the wire
(b) minimum at the axis of the wire
(c) maximum at the surface of the wire
(d) minimum at the surface of the wire

11
...
The current is distributed
uniformly over its cross-section
...

R
L
C
Equal currents will flow through L and C at time t 0
...


(a) t 0  RC
(c) t 0 

L In  2 
R

(b) t 0  RCIn  2 
(d) t 0  LR

V

S

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

PART III: PARAGRAPH TYPE
This section contains 3 multiple choice questions relating to ONE paragraph
...

(+4, –1)

PARAGRAPH – I
A person wants to roll a solid non-conducting spherical ball of mass m and radius r on a surface
whose coefficient of static friction is 
...
By some arrangement he makes a
current I to pass through the coils either in the clockwise direction or in the anti-clockwise direction
...

Assume  is sufficient enough to ensure pure rolling motion
...

y

B

I

x

13
...






j
(c)  nIr 2 B ˆ





ˆ
(d) nIr 2 B k









Angular acceleration of the ball after it has rotated through an angle    180o , is
(a)

15
...
The answer to each question is a single digit
integer, ranging from 0 to 9 (both inclusive)
(+4,0)
16
...
What will be the current, in ampere, through the 2  resistance for this position of
the battery
...


A uniform disc of radius r and mass m is charged uniformly with the charge
q
...
A uniform magnetic field is present in a circular region  a  r 
but varying as kt 3 as shown in figure
...
(Given r = 1 m, m = 18 kg, q = 1C,   0
...


A small conducting loop of radius a and resistance per unit length  , is
pulled with velocity  perpendicular to a long straight conductor
carrying a current I0
...
{Given that x >> a}
...


A long solenoid of radius 2 R contains another coaxial solenoid of
half the radius
...
At a same instant, the
currents in both solenoids starts increasing linearly with time
...
Due to
increasing currents, a charged are the same
...
Then find k
...
Each question contains statements given in two columns, which have
to be matched
...
Any given statement in Column-I can have correct matching
with ONE OR MORE statement(s) in Column- II
...

(+8, 0)
20
...

(c) A wire carrying constant current 
(d) I = Io cos t 

Column 2
(P) Electric field is present
(Q) Magnetic field is present

(R) Induced electric field is present
(S) Magnetic moment is present
...


(d)
2

1  2q 
q2
F

40  2R 2 40R 2
 Fnet  3F  3

Fnet  F2  2F2 cos 60

q2
4R 2

R
R

E

R

F

 Fnet 
2
...


3q 2

AC
will remain uncharged
...


  0 

R min 



3 2
0

  3 0 
2
0

 min
A max

R max
9
R min

(c)

 I
Bcentre  0 , Baxis 
2a

 0Ia 2



2 a 2  9a 2
0 I
2a
Ia 2

So, the desired ratio is



2 10a 2

32



32

 10 

 10 10

32



5
...


6
...
So number of turns per unit length is n 

Ndx
ba
If dB is the field due to this element at the centre, then
 NI dx
dB  0
2b  a x
dN 

b

 B   dB 
a

7
...


0 NI
b
log e  
2b  a 
a

F 0 I 2

 2b

(d)

Bv
2
R Half    2     4

where
R eq  R Half 

 2 


 2  4  2  8A
 I
2
So, if Fm is the magnetic force, then
I

Fm  BI

9
...
01A

VA  VB  I  Ig



  0
...
9 

0
...
1
Similarly I ' I g  0
...
1
1
 I '  0
...
1  Ig 10



10
...




(b,c)

11
...




(a)  



    B , where  is the dipolement

   nI r 2 ˆ  Bi
j ˆ

ˆ
    nIr 2 B k

 



14
...
(Do not confuse Icm with
the current I) and
f  ma cm  mr
…
...


(a)
d 5  nIB 

 
 cos 
d 7  m 


2 5  nIB 
 
 sin 
2 7 m 

17
...


(1)

18
...


a   P  ; b   P, Q,S ;c   Q  ; d   Q, R 

(8)

19
...
Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct
...


During electrophilic substitution reaction of the compound

A

B

C

, the ring that

is most likely to undergo electrophilic substitution is
(a) Ring A
(b) Ring B
(c) Ring A and C
(d) All the three rings
2
...

(2) CCl3 is meta directing group due to reverse hyperconjugation
...

(4) CH = CH2 group is o,pdirecting group
of these statements
(a) (1) and (2) are correct
(b) (1) and (3) are correct
(c) (1), (2) and (4) are correct
(d) (3) and (4) are correct

3
...
68 hours
(b) 2
...
4 hours
(d) 10
...


At pH = pKin + 1 (where Kin= dissociation constant of the indicator), the ratio [In]/ [HIn] in the
 10 
solution is  
...


Whichof the following species is iso-structural with XeF4?

(a) TeF5
(b) I 3
(c) BrF4

6
...


Addition polymers include
1
...
polyethylene
3
...


In how many elements does the last electron have the quantum numbers of n = 4 and
l = 1?
(a) 4
(b) 6
(c) 8
(d) 10

PART II: Multiple Correct Answer(s) Type
This section contains 4 multiple choice questions
...

(5, –2)
9
...
H 2SO4 at 200 0 C

10
...
An electrochemical cell stops working only when
(a) electrode potential of the two half cells becomes equal
...

(c) whole of the metal used as anode is consumed
...


11
...


Which of the following statement(s) is/are correct :
(a) Hydrolysis of ethyl acetate in acid medium is a first order reaction
...

(c) The rate of a chemical change is always directly proportional to concentration
...


PART III: PARAGRAPH TYPE
This section contains 3 multiple choice questions relating to three paragraphs with two
questions on each paragraph
...

(+4, –1)


...
NaOH [R] 1
...
NaIO 4
2
...
H 
C11H12O2
C11H12O2
C13H19BrO

O

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

13
...


CHO
CHO

O

(d)

The structural formula of intermediate product [R] is :
O

(a)

O
CHO

OH

(b)

O
O

O

(c)

(d)
O

15
...
The answer to each question is a single digit
integer, ranging from 0 to 9 (both inclusive)
(+4, 0)

16
...
0
...
3010
...
5 atmospheres?

17
...
The overall order of the reaction

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

18
...


19
...
The molecular weight of the gas Y is

PART V: Matrix Match Type
This section contains 1 question
...
The statements in Column-I are labelled A, B, C and D, while the statements in
Column - II are labelled p, q, r, s and t
...
The appropriate bubbles corresponding to the
answers to these questions have to be darkened as illustrated in the following example: If the correct
matches are A - p , s and t ; B - q and r ; C - p and q ; and D - s and t
...


Column – I (Compounds/class
of compounds)
(A) ortho silicates
(B) Aluminium chloride (anhydrous)
(C) CuSO45H2O
(D) KMnO4

Column – II (Structural features)
(p)
(q)
(r)
(s)

d-d transition
Intramolecular H-bonding
sp 3 – p overlap
p  – p or p  – d overlap

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

PAPER – II (SOLUTION)
1
...


(b)

2
...


The electroplating reaction would be
Cr3+ + 3e 
 Cr
Let the current is passed for t hours
...
68 hours

(b)

4
...


For XeF4,
N 84

6
2
2

Structure is octahedral with 2 lone pairs, i
...
square planar
...


For I 3 ,
N 7  2 1

5
2
2
For BrF4 ,
N 7  4 1

6
2
2
It is iso-structural with XeF4 as it also has two lone pairs
...


b

11
...


12
...


7
...


b

9
...


(b), (c), (d)

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

OH
KMnO 4 


CHO
CHO



OH NaIO 4 

[Q]
14
...


a

O

OH


...
H 3O 
O

OH

HBr


OH

Br
[S]

OH

NaH 



Br

O

 Br 


Br  

O

[S]
16
...
0 log K  0
...
5
n
x
x
 p1/n
...
5)1/1  1
m
m

17
...


1
1s2*1s22s2 *2s2 2px2 2py2 =  2pz2*2py2
= * 2pz2
10  8
B order = 
1
2

19
...


3RT
2RT
U mp 
M
M
3R  400
2R  60
From questions

 4 g mol1 M = 4
...
Each question has four choices (A),
(B), (C) and (D) out of which ONLY ONE is correct
...
P
...
P
...
P
...
P
...
P
...
Then a, b, c are in
(A) A
...

(B) G
...

(C) H
...


(D) none of these

4

Four dice are rolled
...
Each question has four choices (A),
(B), (C) and (D) out of which ONE or MORE are correct
...
Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct
...
P
...
P
...
5
...
P
...
The answer to each question is a single digit integer,
ranging from 0 to 9 (both inclusive)
...



Find the value of kl
...
Find the value of  b  a 
...

If the domain and range of f  g  x   are  a, b  and  c, d  respectively  a  b, c  d  , then find the
g  x    x 1

value of

;

b d

a c

CENTERS : MUMBAI / AKOLA / NASHIK / PUNE / DELHI / KOLKATA / LUCKNOW / GOA

PART V: Matrix Match Type
This section contains 1 question
...
The statements in Column-I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t
...
The appropriate bubbles corresponding to the answers to these
questions have to be darkened as illustrated in the following example: If the correct matches are A - p ,
s and t ; B - q and r ; C - p and q ; and D - s and t
...
P
...

a b c



a b c
, , are in H
...


...
P
...
P
...

2
1
1
a c



c  a b  c a  b b  c  a  b 
2



2  b  c  a  b     a  c 



2 ab  ac  b 2  bc  



2 ab  2b 2  bc  



2b



2b   a  c  2 ac    a  c  2b 



a   c  3b  3 ac











a c





2







2

a c



a c

2

a c



a c

2



a c



a c



2





2

2

 a  c  0 





which is not independent of a , b and c
...

a b

since x1  x2  x3  c we get
a b

 a  b



2

 a 2  b2

ab

c



2ab
c
a b


...
P
...
The nfumber of possible outcomes in which
does not appear on any die is 54 , so that tens digit of 10! 11!
...
Therefore,
the number of possible outcomes in which at least one die shows a 2 is
64  54  1296  625  671
...
When two flags
1
are used, the number of signals that can be generated is 6 P2
...
Hence, the number of different
signals that can be generated is
6

6
Sol

P  6 P2  6 P3  6 P4  6 P5  6 P6  6  30  120  360  720  720  1956
...

x 





Putting 2 n  3r  m,i
...
, r 
2n

C1
3

 2n  m 



1
 2n  m  , the coefficient of x m is
3

 2n !
1

 1

 2n  3  2n  m   !  3  2n  m   !

 




 2n !
1
 1

 3  4 n  m   !  3  2n  m   !

 


7
[Hint:

dy 


= k2  tan  = k2  cot     = k2

2

dx  x  0


  = cot1 k2 = sin1
2


 

1
1  k4

8
[ Hint : x2y = c3
x2

dy
dy
2y
+ 2xy = 0 
= 
dx
dx
x

equation of tangent at (x,y)

 B

]



2y
(X  x)
x

Y–y= 

Y = 0, gives , X =

3x
=a
2

and
Now

9
Sol

X = 0 , gives , Y = 3y = b
a2b =

27 2
27 3
9x 2
x y
c  (C) ]

...


...


Let x0 be the number of objects to the left of the first object chosen, x1 the number of
objects between the first and the second, x2 the number of objects between the second
and the third and x3 the number of objects to the right of the third object
...
(1)

The number of solution of (1)
= coefficient of y n3 in

1  y  y

2










...
y  y 2  y 3 
...


= coefficient of y n3 in y 2 1  y  y 2  y 3 
...

 n 53Cn 5  n 2C3 
n3

Also
12
Sol

n3

C3 

 n  2  n  3 n  4 
6

C2 

n2

C3

The third term in the expansion of the given expression is
T3 



5

C2

 x   xlog x 
5 2

10

2

   xlog x   10x32log

 10 x3

10

10

x

Since T3  106 , we therefore get
10 x3 2log10 x  106



x3 2log10 x  105

Taking logarithms and putting y  log10 x, we get

 3  2 log10 x  log10 x  log10 105  5log10 10  5


2 y2  3y  5  0



2y2  2y  5y  5  0



2 y  y  1  5  y  1  0



 2 y  5  y  1  0
...

2

13 to 15
Let t 2n 1  x
then, t n 1  t 3n 1
x  2n  x
...
1210]
[Sol
...
]




17
Sol
...
1
Roots of equation ax3 + x – 1 – a = 0 are 1, , 
 = – 1 and  =


lim
x

1


=

x



(1  a)x 3  x 2  a
(e1x  1)( x  1)

lim

1


=

lim
x

a(1   x  x  x 2
1  x

1


=

1 a
a

( x  1)[(1  a)x 2  ax  a]
(e1 x  1)( x  1)

lim
x

k = 1 and  = 1

1


a(1  x )(1   x )
1  x

=
=

lim
x

lim
x

1


1


ax 2  ax  a
( e1 x  1)

a(1 – x) =

a(    )


and so k = 1

18

[Ans
...


[Ans
...


2

6
6
The period of sin x  cos x is , period of cot 2 x is  and the period of sin 3x is

3
2
Hence 2, LCM of all the three is the period of the gieven function
...

2
4

(D)

sin   x   0





the required is

period of 23{x}  sin   x  is period of x  1
...


A

2
...


8
...


A, B, C, D

13
...


A

20
...


C OR D

4
...


C

6
...


A

10
...


A, B, D

12
...


(9)

(16)

18
...


(24)

17
...


(A)

2
...


(A)

4
...


(B)

6
...


(D)

8
...


(A,B,D)

10
...


(A,C,D)

12
...


(D)

14
...


(B)

16
...


(4)

18
...


(3)

20
...


D

2
...


A

4
...


C

6
...


A,C

8
...


A,B,C,D

10
...


A

12
...


A

14
...


14

16
...


3

18
...


0

20
...


D

2
...


C

4
...


8
...


A, C

10
...


A, B, C, D

14
...


A

16
...


(2)

20
...


D

6
...


B

12
...


A

(8)

19
...


(B)

2
...


(B)

4
...


(C)

6
...


(B)

8
...


(A,B,C,D)

10
...


(A, C, D)

12
...


(C)

14
...


(A)

16
...


(1)

18
...


(4)

20
...


C

2
...


C

4
...


A

6
...


B

8
...


A,B,C

10
...


A,B,C

12
...


C

14
...


D

16
...


1

18
...


4

20

---