CONCEPT RECAPITULATION TEST - II Paper 2

Question Paper 18 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

Contributed by

Swati Kant

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CONCEPT RECAPITULATION TEST - II

Paper 2

Time Allotted: 3 Hours Maximum Marks: 198

Please read the instructions carefully. You are allotted 5 minutes
specifically for this purpose.
 You are not allowed to leave the Examination Hall before the end of
the test.

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 08) contains 8 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Section-A (09 to 14) contains 3 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.

Section-A (15 to 20) contains 6 multiple choice questions which have more than one correct

answer. Each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

ALL INDIA TEST SERIES

FIITJEE

JEE (Advanced), 2014

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s
2

Planck constant h = 6.6 10

34
J-s
Charge of electron e = 1.6  10

19
C
Mass of electron m
e
= 9.1  10

31
kg
Permittivity of free space 
0
= 8.85  10

12
C
2
/N-m
2
Density of water 
water
= 10
3
kg/m
3

Atmospheric pressure P
a
= 10
5
N/m
2

Gas constant R = 8.314 J K

1
mol

1
CHEMISTRY

Gas Constant R = 8.314 J K

1
mol

1

= 0.0821 Lit atm K

1
mol

1

= 1.987  2 Cal K

1
mol

1

Avogadro's Number N
a
= 6.023  10
23

Planck’s constant h = 6.625  10

34
Js
= 6.625  10
–27
ergs
1 Faraday = 96500 coulomb
1 calorie = 4.2 joule
1 amu = 1.66  10
–27
kg
1 eV = 1.6  10
–19
J

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,
N=9, Na=11, Mg=12, Si=14, Al=13, P=15, S=16,
Cl=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,
Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33,
Br=35, Ag=47, Sn=50, I=53, Xe=54, Ba=56,
Pb=82, U=92.
Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16,
F=19, Na=23, Mg=24, Al = 27, Si=28, P=31, S=32,
Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,
Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108,
Sn=118.7, I=127, Xe=131, Ba=137, Pb=207, U=238.

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I
SECTION – A
Single Correct Choice Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. A ball is dropped from a height of 80 m on a floor. At each collision, the ball losses half of its
speed. Then the graph of speed versus time is
(A)

20

4

t (sec)

Speed (m/s)

40

6

8

(B)

20

4

t (sec)

Speed (m/s)

40

6

8


20

(C)

20

4

t (sec)

Speed (m/s)

40

6

8

(D)

20

4

t (sec)

Speed (m/s)

40

6

8

2. A projectile is required to hit a target whose co-ordinates relative to horizontal and vertical axes
through the point of projection are (, ). If the gun velocity is
2g

. Then the condition when it
is impossible to hit the target.
(A) 3 > 4 (B) 4 > 3
(C) 3 > 4 (D) 4 > 3
Space for Rough work
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3. A particle moves from rest at A on the surface of a smooth circular
cylinder of radius r as shown. At B its leaves the cylinder. Then the
equation relating  and  is
(A) 3sin  = 2cos  (B) 2sin  = 3cos 
(C) 3sin  = 2cos  (D) 2sin  = 3cos 

B

A





Smooth

4. The gravitational field in a region is given by E =


ˆ ˆ
2i 3j
 N/kg. The equation of the line on which
no work is done by the gravitational when the particle is moved
(A) 2y + 3x = 5 (B) 5y + 2x = 3
(C) 3y + 2x = 5 (D) 3y + 5x = 2

5. The distance of centre of mass from O of a uniform plate having quarter
circular inner and outer boundaries of radii a and b from centre O.
(A)
3 3
2 2
4 b a
3 b a
 

 
 
 
(B)
3 3
2 2
4 2 b a
3 b a
 

 
 
 

(C)
3 3
2 2
4 b a
b a
 

 
 
 
(D)
3 3
2 2
4 2 b a
3
b a
 

 
 
 

a

b

O

6. Two concentric conducting spheres of radii R and 3R carrying charges Q and 2Q respectively. If
the charge on inner sphere is doubled. The potential difference between inner and outer spheres
will
(A) becomes two times (B) becomes four times
(C) be halved (D) remains same

7. In an L-C circuit shown in the figure, C = 1F, L = 4H. At time t = 0, charge in
the capacitor is 4C and it is decreasing at a rate of
5
C/s. Choose the
correct statements.
(A) maximum charge in the capacitor can be 6C
(B) maximum charge in the capacitor can be 8C

C

+ 

q

(C) charge in the capacitor will be maximum after time 2 sin

1
(2/3) sec
(D) None of these
Space for Rough work
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8. A massive planet of radius R has a diametrical hole as shown in
figure, such that hole does not effect the spherecity of the
planet. The planet has uniform density. Two equal mass A and
B [m << M] are simultaneously released, from positions shown.
If the collisions are elastic then the total distance traveled by
mass A after being released at the time of 6
th
collision is : [
Mmass of planet, m mass of A and B neglect the
gravitational interaction between mass A and mass B.]
(A) 8.5 R (B) 5R
(C) 6.5 R (D) 9.5 R

Comprehension Type

This section contains 3 groups of questions. Each group has 2 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct.

Paragraph for Questions 9 and 10

In the circuit shown AB is a 10  uniform slide wire 50 cm long. E
1
is 2
V accumulator of negligible internal resistance R
1
and R
2
are 15  and
5  respectively. When k
1
and k
2
are both open, the galvanometer
shows no deflection when AJ = 31.25. When k
1
and k
2
are both closed
the balance length AJ = 5 cm

9. The emf of the cell E
2

(A) 0.5 V (B) 1 V
(C) 1.5 V (D) 2 V

A

G

E
2
K
2
R
2
= 5


K
1
R
1
= 15


E
1
= 2 V
B

J

10. The internal resistance of the cell E
2

(A) 7.5  (B) 8 
(C) 10  (D) 2 
Space for Rough work

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Paragraph for Question Nos. 11 to 12

Two uniform circular disc of masses M and 4M and radii R and 2R
respectively are connected by a massless spring of force constant k and
placed in a vertical plane on consecutive steps of a stair as shown.
Initially spring is compressed and system is released from rest. Assume
that friction is enough to prevent slipping.

R

11. When spring is compressed by x, what is the friction force from stair on the mass 4M
(A) Kx/2 left (B) kx/2 right,
(C) kx/3 left, (D) kx/3 right

12. What is the angular frequency of oscillation of the system
(A)
5k
6M
(B)
5k
4M

(C)
k
5M
(D) none of these

Paragraph for Question Nos. 13 to 14

A ring of radius R is made of a thin wire of material of density , having cross–section area a and Young’s
Modulus y. The ring rotates about an axis perpendicular to its plane and through its centre. Angular
frequency of rotation is .

13. The tension in the ring will be
(A)
2 2
a R
2
 
(B) a

R
2

2

(C) 2a

R
2

2
(D)
2 2
a R
4
 
.

14. The ratio of kinetic energy to potential energy is
(A)
2
y
R

 
(B)
2
2y
R

 

(C)
2
y
2 R

 
(D)
2
y
4 R

 
.
Space for Rough work

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Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

15. A reflecting surface is represented by the equation
y =
2L x
sin
L

 
 

 
, 0  x  L
A ray travelling horizontally becomes vertical after
reflection. The co-ordinate of the points where this
ray is incident is

y

x

(A)
L 2L
,
4
 
 
 

 
(B)
L 3L
,
3
 
 
 

 

(C)
3L 2L
,
4
 
 
 

 
(D)
2L 3L
,
3
 
 
 

 

16. Two balls A and B of masses m and 2m respectively
are lying on a smooth surface as shown in the figure.
Ball A hits the ball B (which is at rest) with a velocity
v = 16 m/s. Such that B just reaches the highest point
of inclined plane. (g = 10 m/s
2
)

A

B

v

5m

(A) e = 3/4 (B) e = 7/8
(C) v
A
= 4 m/s (D) v
A
= 8 m/s
where e is the coefficient of restitution and v
A
is velocity of the ball A after collision.

17. A particle of mass m is executing SHM as shown in
diagram (1) and (2). In diagram (2) particle contains
charge q such that Eq = mg. If their velocities are
same at mean position and let A
1
and A
2
be their
amplitudes and T
1
and T
2
be their time periods
respectively then

k

m

(i)

(ii)

m

q

E

(A) A
1
= A
2
(B) A
1
< A
2

(C) T
1
= T
2
(D) T
1
> T
2

Space for Rough work

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18. Two objects of mass m and 4 m are at rest at an infinite separation. They move towards each
other under mutual attraction. If G is the universal gravitational constant. Then at separation r.
(A) The sum of energy of the two objects is negative.
(B) Their relative velocity of approach is
1/ 2
10Gm
r
 
 
 
in magnitude
(C) The sum of kinetic energy of the objects is
2
4Gm
r
.
(D) The sum of angular momenta of both the objects is zero about any point.

19. Figure shows roughly how the force F between two adjacent
atoms in a solid varies with separation ‘r’
(A) OQ is equilibrium separation
(B) Hooke’s law is obeyed around P
(C) The potential energy of the atom is the gradient of the graph
at all points
(D) The energy to separate the atoms completely is obtained
from the area enclosed below the axis or r.
F
O
P
Q
r

20. Figure (Five straight lines numbered 1, 2, 3, 4 and 5) shows
graph of change in temperature T versus heat supplied Q for
different processes performed on a gas of one mole.
(A) line 4 and 5 (coinciding with Q and T axis respectively)
represent isothermal and adiabatic processes respectively
(B) line 2 represents isochoric process for diatomic gas
(C) line 1 and 3 represent isobaric process for a diatomic gas
and isochoric process for a monoatomic gas respectively
(D) line 2 represents isobaric process for a monoatomic gas.

T
1
2
3
4
5
Q
10T
0
2T
0
5RT
0
3
7
10T
0

Space for Rough work

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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
Straight Objective Type
This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A), (B),
(C) and (D), out of which ONLY ONE is correct.

1. Which among the following is the strongest oxidizing agent?
(A)
4
IO

(B)
4
6
XeO


(C)
3
4
PO

(D)
2
4
SO


2. The compound which undergoes S
N
1 reaction most rapidly is:
(A)
Br

(B)
Br

(C)

Br

(D)

Br

3. At 20
o
C, the osmotic pressure of a solution of 100 ml of which contains 6.33 g of the colouring
substance of the blood haematin is 243.4 kP
a
. What is the molecular formula of the haematin; if
the elementary composition of haematin is C = 64.4%, H = 5.2%, N = 3.8%, O = 12.6% and Fe
8.8%?
(A) C
34
H
33
N
4
O
5
Fe (B) C
33
H
32
N
4
O
5
Fe
(C) C
34
H
30
N
3
O
5
Fe (D) C
32
H
28
N
4
O
5
Fe
4. Calculate E
o
for the reactions:



2 4
ZnY 2e Zn s Y
  
 


, where
4
Y

is the completely deprotonated anion of EDTA, the
formation constant for
2
ZnY

is 3.2 × 10
16
and E
o
for
2
Zn Zn 2e
 
  is 0.76V.
(A) - 1.25 V (B) 0.48 V
(C) + 0. 68 V (D) - 0.27 V

5. The freezing point of 0.20 M solution of weak acid HA is 272.5 K. The molality of the solution is
0.263 mol kg
-1
. Find the pH of the solution on adding 0.25 M sodium acetate solution:
(K
f
of water = 1.86 K kgmol
-1
).
(A) 5.4 (B) 3.2
(C) 6.1 (D) 4.11
6. Aqueous solution of a (Y) salt is alkaline to litmus. On strong heating, it smells up to give a glassy
material. When conc. H
2
SO
4
is added to hot concentrated solution of (Y), white crystal of a weak
acid separates out. Hence, the compound (Y) is:
(A) Na
2
SO
4
.10H
2
O (B) Ca
2
P
6
O
11
.10H
2
O
(C) Na
2
B
6
O
11
(D) Na
2
B
4
O
7
.10H
2
O
Rough Work

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7. The major product of the following reaction is:

CH
3
OH
CH CH
2
H



(A)
CH
3
O
CH
3

(B)
OH
CH
3
CH
3

(C)

CH
2
CH
3
O

(D)

OH
CH
3
CH
3

8.
O
O
N
K
+
     
2
H /H O
X Y Z Glycine


   

Compound X, Y, Z are:
(A)
Chloroacetic acid,
N CH
2
C
O
OH
O
O
,
OH
O
O
OH

(B)
Bromo acetic acid,
N CH
2
C
O
OH
O
O
,
O
O
O

(C)

Ethyl chloro acetic acid,
N CH
2
C
O
OC
2
H
5
O
O
,
O
O
OH
OH

(D)

None of the above
Rough Work

Page 10

CONCEPT RECAPITULATION TEST - II Paper 2

CONCEPT RECAPITULATION TEST - II Paper 2

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