IIT JEE-Advanced 2014 Solved papers by Triumph Academy

Transcription

TRIUMPH ACADEMY
SOLUTIONS TO JEE(ADVANCED)-2014
CODE
5
PAPER-1
Note: Front and back cover pages are reproduced from actual paper back to back here.
Time: 3 Hours
Maximum Marks: 180
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
A.
1.
2.
3.
4.
5.
6.
7.
8.
9.
B.
10.
11.
12.
13.
C.
14.
15.
INSTRUCTIONS
General
This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by the
invigilators.
The question paper CODE is printed on the left hand top corner of this sheet and on the back cover page of this booklet.
Blank space and blank pages are provided in the question paper for your rough work. No additional sheets will be provided
for rough work.
Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadget of any
kind are NOT allowed inside the examination hall.
Write your name and roll number in the space provided on the back cover of this booklet.
Answers to the questions and personal details are to be filled on an Optical Response Sheet, which is provided separately.
The ORS is a doublet of two sheets – upper and lower, having identical layout. The upper sheet is a machine-gradable
Objective Response Sheet (ORS) which will be collected by the invigilator at the end of the examination. The upper sheet
is designed in such a way that darkening the bubble with a ball point pen will leave an identical impression at the
corresponding place on the lower sheet. You will be allowed to take away the lower sheet at the end of the examination
(see Figure-1 on the back cover page for the correct way of darkening the bubbles for valid answers).
Use a black ball point pen only to darken the bubbles on the upper original sheet. Apply sufficient pressure so that the
impression is created on the lower sheet. See Figure -1 on the back cover page for appropriate way of darkening the
bubbles for valid answers.
DO NOT TAMPER WITH / MUTILATE THE ORS SO THIS BOOKLET.
On breaking the seal of the booklet check that it contains 28 pages and all the 60 questions and corresponding answer
choices are legible. Read carefully the instruction printed at the beginning of each section.
Filling the right part of the ORS
The ORS also has a CODE printed on its left and right parts.
Verify that the CODE printed on the ORS (on both the left and right parts) is the same as that on the this booklet and put
your signature in the Box designated as R4.
IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET / ORS AS APPLICABLE.
Write your Name, Roll No. and the name of centre and sign with pen in the boxes provided on the upper sheet of ORS. Do
not write any of this anywhere else. Darken the appropriate bubble UNDER each digit of your Roll No. in such way that
the impression is created on the bottom sheet. (see example in Figure 2 on the back cover)
Question Paper Format
The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections.
Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE
OR MORE THAN ONE are correct.
Section 2 contains 10 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9 (both
inclusive)
Please read the last page of this booklet for rest of the instructions.
TA Ltd., TA House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.TA.com.
JEE(ADVANCED)2014-Paper 1-PHYSICS-2
PART I : PHYSICS
SECTION – 1 (One or More Than One Options Correct Type)
This section contains 10 multiple choice type questions. Each question has four choices (A), (B), (C) and (D) out
of which ONE or MORE THAN ONE are correct.
*1.
A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s-1.
He is told that the air in the tube has been replaced by another gas (assume that the column remains filled
with the gas). If the minimum height at which resonance occurs is (0.350  0.005)m , the gas in the tube is
(Useful information: 167RT  640 J1/ 2 mole 1/ 2 ; 140RT  590 J1/2 mole 1/ 2 . The molar masses M in
grams are given in the options. Take the values of

10
7
 
(A) Neon  M  20,
20 10 


10 9 
 
(C) Oxygen  M  32,
32 16 

2.
10
for each gas as given there.)
M

10 3 
 
(B) Nitrogen  M  28,
28 5 


10 17 
 
(D) Argon  M  36,
36 32 

At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating
current I(t) = I0 cos (t), with I0 = 1A and  = 500 rad/s starts flowing in it with the initial direction shown
7
in the figure. At t =
, the key is switched from B to D. Now onwards only A and D are connected. A
6
total charge Q flows from the battery to charge the capacitor fully. If C = 20 F, R = 10  and the battery
is ideal with emf of 50 V, identify the correct statement(s).
B
D
A

50 V
C= 20F
R= 10 
(A) Magnitude of the maximum charge on the capacitor before t =
7
is 1  10-3C.
6
7
is clockwise.
6
(C) Immediately after A is connected to D, the current in R is 10 A.
(D) Q = 2  103C.
(B) The current in the left part of the circuit just before t =
3.
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that
covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the
capacitor is C while that of the portion with dielectric in between is C1. When the capacitor is
charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets
charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2. Choose
the correct option/options, ignoring edge effects.
E
E
1
(A) 1  1
(B) 1 
E2
E2 K
(C)
Q1 3

Q2 K
(D)
C 2 K

C1
K
Q1
E1
Q2
E2
JEE(ADVANCED)2014-Paper 1-CHEMISTRY-3
*4.
One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the waves in the string
is 100 ms1. The other end of the string is vibrating in the y direction so that stationary waves are set up in
the string. The possible waveform(s) of these stationary waves is (are)
x
50t
x
100t
(A) y(t) = A sin
cos
(B) y(t) = A sin
cos
6
3
3
3
5x
250t
5x
(C) y(t) = A sin
cos
(D) y(t) = A sin
cos 250 t
6
3
2
5.
A transparent thin film of uniform thickness and refractive index n1 = 1.4 is
coated on the convex spherical surface of radius R at one end of a long solid
glass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays of
light parallel to the axis of the cylinder traversing through the film from air
to glass get focused at distance f1 from the film, while rays of light
traversing from glass to air get focused at distance f2 from the film. Then
(A) |f1| = 3R
(B) |f1| = 2.8 R
(C) |f2| = 2R
(D) |f2| = 1.4 R
n1
n2
Air
6.
Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the
temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same
material, each of length L and diameter 2d. The way these wires are connected is given in the options.
How much time in minutes will it take to raise the temperature of the same amount of water by 40K?
(A) 4 if wires are in parallel
(B) 2 if wires are in series
(C) 1 if wires are in series
(D) 0.5 if wires are in parallel.
7.
Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3
are connected as shown in the figure. The current in resistance R2 would
be zero if
(A) V1  V2 and R1  R 2  R 3
(B) V1  V2 and R1  2R 2  R 3
(C) V1  2V2 and 2R1  2R 2  R 3
(D) 2V1  V2 and 2R1  R 2  R 3
V1
R1
R2
V2
R3
8.
Let E1 (r), E 2 (r) and E 3 (r) be the respective electric fields at a distance r from a point charge Q, an
infinitely long wire with constant linear charge density , and an infinite plane with uniform surface charge
density . If E1 (r0 )  E 2 (r0 )  E 3 (r0 ) at a given distance r0, then
(A) Q  4r02
(C) E1 (r0 / 2)  2E 2 (r0 / 2)
9.

2
(D) E 2 (r0 / 2)  4E 3 (r0 / 2)
(B) r0 
A light source, which emits two wavelengths 1 = 400 nm and 2 = 600 nm, is used in a Young’s double
slit experiment. If recorded fringe widths for 1 and 2 are 1 and 2 and the number of fringes for them
within a distance y on one side of the central maximum are m1 and m2, respectively, then
(A) 2 > 1
(B) m1 > m2
(C) From the central maximum, 3rd maximum of 2 overlaps with 5th minimum of 1
(D) The angular separation of fringes for 1 is greater than 2
*10.
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle
 with the horizontal floor. The coefficient of friction between the wall and the ladder is 1 and that
between the floor and the ladder is 2. The normal reaction of the wall on the ladder is N1 and that of the
floor is N2. If the ladder is about to slip, then
1
mg
(A) 1 = 0 2  0 and N2 tan  =
2
mg
(B) 1  0 2 = 0 and N1 tan  =

2
2
mg
(C) 1  0 2  0 and N2 =
1  1 2
(D) 1 = 0 2  0 and N1 tan  =
mg
2
SECTION – 2: (Only Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).
11.
During Searle’s experiment, zero of the Vernier scale lies between 3.20  10-2 m and 3.25  102 m of the
main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.
When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 
10-2 m and 3.25 102 m of the main scale but now the 45th division of Vernier scale coincides with one of
the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8  107
m2. The least count of the Vernier scale is 1.0  105 m. The maximum percentage error in the Young’s
modulus of the wire is
*12.
Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with
respect to the horizontal respectively as shown in the figure. The speed of A is 100 3 ms1. At time t = 0
s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity
perpendicular to the line of motion of A. If at t = t0, A just escapes being hit by B, t0 in seconds is
A
30
*13.
B
60
A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to the final state f
along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system
along the paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to
the system along the path iaf, ib and bf are Qiaf , Qib and Qbf respectively. If the internal energy of the
system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio Qbf / Qib is
a
f
P
i
b
V
14.
Two parallel wires in the plane of the paper are distance X0 apart. A point charge is moving with speed u
between the wires in the same plane at a distance X1 from one of the wires. When the wires carry current of
magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast,
if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is
X
R
R2. If 0  3, the value of 1 is
X1
R2
15.
To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses
dimensional analysis and assumes that the distance depends on the mass density  of the fog, intensity
(power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to
S1/n. The value of n is
*16.
A rocket is moving in a gravity free space with a constant
acceleration of 2 ms–2 along + x direction (see figure). The
length of a chamber inside the rocket is 4 m. A ball is
thrown from the left end of the chamber in + x direction
with a speed of 0.3 ms–1 relative to the rocket. At the same
time, another ball is thrown in  x direction with a speed of
0.2 ms–1 from its right end relative to the rocket. The time in
seconds when the two balls hit each other is
a = 2 ms–2
0.3 ms
–1
0.2 ms
–1
x
4m
17.
A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990  resistance, it
2n
can be converted into a voltmeter of range 0 – 30 V. If connected to a
 resistance, it becomes an
249
ammeter of range 0 – 1.5 A. The value of n is
*18.
A uniform circular disc of mass 1.5 kg and radius 0.5 m is
initially at rest on a horizontal frictionless surface. Three
forces of equal magnitude F = 0.5 N are applied
simultaneously along the three sides of an equilateral
triangle XYZ with its vertices on the perimeter of the disc
(see figure). One second after applying the forces, the
angular speed of the disc in rad s–1 is
F
X
O
F
Y
Z
F
*19.
A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about
its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are
attached to the platform at a distance 0.25 m from the centre on its either sides along
its diameter (see figure). Each gun simultaneously fires the balls horizontally and
perpendicular to the diameter in opposite directions. After leaving the platform, the
balls have horizontal speed of 9 ms–1 with respect to the ground. The rotational speed
of the platform in rad s–1 after the balls leave the platform is
*20.
Consider an elliptically shaped rail PQ in the vertical plane with
OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulled along the
rail from P to Q with a force of 18 N, which is always parallel to
line PQ (see the figure given). Assuming no frictional losses, the
kinetic energy of the block when it reaches Q is (n×10) Joules. The
value of n is (take acceleration due to gravity = 10 ms–2)
Q
4m
90°
O
3m
P
PART - II: CHEMISTRY
SECTION – 1 (One or More Than One Options Correct Type)
This section contains 10 multiple choice type questions. Each question has four choices (A), (B), (C) and (D) out
of which ONE or MORE THAN ONE are correct.
*21.
The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are
(A) tert-butanol and 2-methylpropan-2-ol
(B) tert-butanol and 1, 1-dimethylethan-1-ol
(C) n-butanol and butan-1-ol
(D) isobutyl alcohol and 2-methylpropan-1-ol
*22.
An ideal gas in a thermally insulated vessel at internal pressure = P1, volume = V1 and absolute
temperature = T1 expands irreversibly against zero external pressure, as shown in the diagram. The final
internal pressure, volume and absolute temperature of the gas are P2, V2 and T2, respectively. For this
expansion,
Pext = 0
Pext = 0
Irreversible
P2, V2, T2
P1, V1, T1
Thermal insulation
(A) q = 0
(C) P2V2 = P1V1
(B) T2 = T1
(D) P2V2 = P1V1
*23.
Hydrogen bonding plays a central role in the following phenomena:
(A) Ice floats in water.
(B) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions.
(C) Formic acid is more acidic than acetic acid.
(D) Dimerisation of acetic acid in benzene.
24.
In a galvanic cell, the salt bridge
(A) does not participate chemically in the cell reaction.
(B) stops the diffusion of ions from one electrode to another.
(C) is necessary for the occurrence of the cell reaction.
(D) ensures mixing of the two electrolytic solutions.
*25.
For the reaction:
I  ClO3  H 2SO 4 
 Cl   HSO4  I 2
The correct statement(s) in the balanced equation is/are:
(A) Stoichiometric coefficient of HSO4 is 6.
(B) Iodide is oxidized.
(C) Sulphur is reduced.
(D) H2O is one of the products.
*26.
The reactivity of compound Z with different halogens under appropriate conditions is given below:
mono halo substituted derivative when X 2  I 2
OH
X2
di halo substituted derivative when X 2  Br2
C(CH3 )3
Z
tri halo substituted derivative when X 2  Cl 2
The observed pattern of electrophilic substitution can be explained by
(A) the steric effect of the halogen
(B) the steric effect of the tert-butyl group
(C) the electronic effect of the phenolic group
(D) the electronic effect of the tert-butyl group
*27.
The correct statement(s) for orthoboric acid is/are
(A) It behaves as a weak acid in water due to self ionization.
(B) Acidity of its aqueous solution increases upon addition of ethylene glycol.
(C) It has a three dimensional structure due to hydrogen bonding.
(D) It is a weak electrolyte in water.
28.
Upon heating with Cu2S, the reagent(s) that give copper metal is/are
(A) CuFeS2
(B) CuO
(C) Cu2O
(D) CuSO4
*29.
The pair(s) of reagents that yield paramagnetic species is/are
(A) Na and excess of NH3
(B) K and excess of O2
(C) Cu and dilute HNO3
(D) O2 and 2- ethylanthraquinol
30.
In the reaction shown below, the major product(s) formed is/are
NH2
acetic anhydride

 product  s 
CH 2 Cl2
NH2
O
(A)
H
(B)
N
NH2
CH3
H
O
CH3 COOH
N
NH2
CH3
CH3 COOH
O
O
O
(C)
(D)
H
N
CH3
NH3 CH3 COO
H
O
H
N
CH3
H2 O
N
O
O
CH3
O
O
SECTION – 2: (Only Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).
31.
Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number of BLACK coloured
sulphides is
*32.
The total number(s) of stable conformers with non-zero dipole moment for the following compound is(are)
Cl
Br
CH3
Br
Cl
CH3
33.
Consider the following list of reagents:
Acidified K 2Cr2 O7 , alkaline KMnO4 ,CuSO 4 , H 2 O2 ,Cl2 ,O3 , FeCl3 , HNO3 and Na 2S2 O3 .
The total number of reagents that can oxidise aqueous iodide to iodine is
34.
A list of species having the formula XZ4 is given below.
XeF4 ,SF4 ,SiF4 , BF4 , BrF4 , Cu  NH3  4 
2
2
2
,  FeCl4  , CoCl4 
2
and  PtCl 4  .
Defining shape on the basis of the location of X and Z atoms, the total number of species having a square
planar shape is
35.
Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are
independently reacted with NaBH4 (NOTE: stereoisomers are also reacted separately). The total number of
ketones that give a racemic product(s) is/are
*36.
In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1 and ms = –1/2 is
*37.
If the value of Avogadro number is 6.023  1023 mol–1 and the value of Boltzmann constant is
1.380 1023 JK 1 , then the number of significant digits in the calculated value of the universal gas constant
is
38.
MX2 dissociates in M2+ and X ions in an aqueous solution, with a degree of dissociation () of 0.5. The
ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of
freezing point in the absence of ionic dissociation is
39.
The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of
the peptide shown below is
O
H
O
O
H
O
N
N
O
N
N
O
H
N
N
N
CH2
H
H
N
O
H
CH2
O
O
*40.
A compound H2X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml –1. Assuming
no change in volume upon dissolution, the molality of a 3.2 molar solution is
JEE(ADVANCED)2014-Paper 1-MATHEMATICS--9
PART - III: MATHEMATICS
SECTION – 1 : (One or More than One Options Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE or MORE THAN ONE are correct.
x
41.
Let f : (0, )  R be given by f (x) =

e
 1
 t  
 t
1/ x
dt
, then
t
(A) f (x) is monotonically increasing on [1, )
1
(C) f (x) + f   = 0, for all x  (0, )
 x
(B) f (x) is monotonically decreasing on (0, 1)
(D) f (2x) is an odd function of x on R
42.
Let a  R and let f : R  R be given by f (x) = x5  5x + a, then
(A) f (x) has three real roots if a > 4
(B) f (x) has only one real roots if a > 4
(C) f (x) has three real roots if a <  4
(D) f (x) has three real roots if  4 < a < 4
43.
For every pair of continuous functions f, g : [0, 1]  R such that max{ f (x) : x  [0, 1]} = max{g(x) : x 
[0, 1]}, the correct statement(s) is(are)
(A) (f (c))2 + 3f (c) = (g(c))2 + 3g(c) for some c  [0, 1]
(B) (f (c))2 + f (c) = (g(c))2 + 3g(c) for some c  [0, 1]
(C) (f (c))2 + 3f (c) = (g(c))2 + g(c) for some c  [0, 1]
(D) (f (c))2 = (g(c))2 for some c  [0, 1]
*44.
A circle S passes through the point (0, 1) and is orthogonal to the circles (x  1)2 + y2 = 16 and x2 + y2 = 1.
Then
(A) radius of S is 8
(B) radius of S is 7
(C) centre of S is (7, 1)
(D) centre of S is (8, 1)
45.
 

Let x , y and z be three vectors each of magnitude

2 and the angle between each pair of them is . If
3

 



a is a non-zero vector perpendicular to x and y  z and b is a non-zero vector perpendicular to y and
 
z  x , then

   

   
(A) b  b  z  z  x 
(B) a   a  y  y  z 
 
   

   
(C) a  b    a  y  b  z
(D) a   a  y  z  y 




46.
From a point P(, , ), perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = 
x, z = 1. If P is such that QPR is a right angle, then the possible value(s) of  is(are)
(A) 2
(B) 1
(C) 1
(D)  2
47.
Let M be a 2  2 symmetric matrix with integer entries. Then M is invertible if
(A) the first column of M is the transpose of the second row of M
(B) the second row of M is the transpose of the first column of M
(C) M is a diagonal matrix with non-zero entries in the main diagonal
(D) the product of entries in the main diagonal of M is not the square of an integer
JEE(ADVANCED)2014-Paper - MATHEMATICS--10
48.
Let M and N be two 3  3 matrices such that MN = NM. Further, if M  N2 and M2 = N4,
then
(A) determinant of (M2 + MN2) is 0
(B) there is a 3  3 non-zero matrix U such that (M2 + MN2)U is the zero matrix
(C) determinant of (M2 + MN2)  1
(D) for a 3  3 matrix U, if (M2 + MN2)U equals the zero matrix then U is the zero matrix
49.
Let f : [a, b]  [1, ) be a continuous function and let g : R  R be defined as
0
if
x  a,

 x

g  x    f  t  dt if a  x  b,
 a
 b
xb
 a f  t  dt if
Then
(A) g(x) is continuous but not differentiable at a
(B) g(x) is differentiable on R
(C) g(x) is continuous but not differentiable at b
(D) g(x) is continuous and differentiable at either a or b but not both


50.
  
Let f :   ,   R be given by f(x) = (log(sec x + tan x))3. Then
 2 2
(A) f (x) is an odd function
(B) f (x) is a one-one function
(C) f (x) is an onto function
(D) f (x) is an even function
SECTION – 2 : (One Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both
inclusive).
*51.
Let n1 < n2 < n3 < n4 < n5 be positive integers such that n1 + n2 + n3 + n4 + n5 = 20. Then the number of
such distinct arrangements (n1, n2, n3, n4, n5) is _________
*52.
Let n  2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment.
Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of
red and blue line segments are equal, then the value of n is _________
53.
Let f : R  R and g : R  R be respectively given by f (x) = |x| + 1 and g(x) = x2 + 1. Define h : R  R by
 max  f  x  , g  x  if x  0
.
h x  
 min  f  x  , g  x  if x  0
Then number of points at which h(x) is not differentiable is __________
*54.
Let a, b, c be positive integers such that
b
is an integer. If a, b, c are in geometric progression and the
a
arithmetic mean of a, b, c is b + 2, then the value of
55.
a 2  a  14
is ________
a 1
 


Let a , b , and c be three non-coplanar unit vectors such that the angle between every pair of them is .
3
 
   

p 2  2q 2  r 2
If a  b  b  c  pa  qb  rc , where p, q and r are scalars, then the value of
is ______
q2
JEE(ADVANCED)2014-Paper 1-MATHEMATICS-11
56.
The slope of the tangent to the curve (y  x5)2 = x(1 + x2)2 at the point (1, 3) is ______
1
57.
58.
The value of
2
 d
4 x3  2 1  x 2
 dx
0


5
  dx is ____________

1 x
x
 ax  sin  x  1  a 1
The largest value of the non-negative integer a for which lim 

x 1  x  sin  x  1  1 



1
is ______
4
*59.
Let f : [0, 4]  [0, ] be defined by f (x) = cos1(cos x). The number of points x  [0, 4] satisfying the
10  x
equation f (x) =
is ___________
10
*60.
For a point P in the plane, let d1(P) and d2(P) be the distances of the point P from the lines x  y = 0 and x +
y = 0 respectively. The area of the region R consisting of all points P lying in the first quadrant of the plane
and satisfying 2  d1(P) + d2(P)  4, is ________
*****
JEE(ADVANCED)2014-Paper 1
JEE(ADVANCED)-2014
PAPER 1 CODE 5
ANSWERS
PART - I: PHYSICS
1.
D
2.
C, D
3.
A, D
4.
A, C, D
5.
A, C
6.
B, D
7.
A, B, D
8.
C
9.
A, B, C
10.
C, D
11.
4
12.
5
13.
2
14.
3
15.
3
16.
2
17.
5
18.
2
19.
4
20.
5
21.
A, C, D
22.
A, B, C
23.
A, B, D
24.
A
25.
A, B, D
26.
A, B, C
27.
B, D
28.
B, C, D
29.
A, B, C
30.
A
31.
6
32.
3
33.
7
34.
4
35.
5
36.
6
37.
4
38.
2
39.
1
40.
8
41.
A, C, D
42.
B, D
43.
A, D
44.
B, C
45.
A, B, C
46.
C
47.
C, D
48.
A, B
49.
A, C
50.
A, B, C
51.
7
52.
5
53.
3
54.
4
55.
4
56.
8
57.
2
58.
2
59.
3
60.
6
JEE(ADVANCED)2014-Paper 1
HINTS AND SOLUTIONS
PART - I: PHYSICS
1.

1 RT
4 M
1 RT
for gases mentioned in options A, B, C and D, work out to be 0.459 m, 0.363 m
4 M
0.340 m & 0.348 m respectively. As  = (0.350  0.005)m ; Hence correct option is D.
Calculations for
2.
As current leads voltage by /2 in the given circuit initially, then ac voltage can be represent as
V = V0 sin t
 q = CV0 sin t = Q sin t
where, Q = 2  10-3 C
3
I0 and hence current is anticlockwise.
2
7
 Current ‘i’ immediately after t =
is
6
V  50
i= c
= 10 A
R
 Charge flow = Qfinal – Q(7/6) = 2  106C
Hence C & D are correct options.

3.
At t = 7/6 ; I = 
As E = V/d
E1/E2 = 1 (both parts have common potential difference)
Assume C0 be the capacitance without dielectric for whole capacitor.
C
2C
k 0  0 C
3
3
C 2 k

C1
k
Q1 k
 .
Q2 2
4.
Taking y(t) = A f(x) g(t) & Applying the conditions:
1; here x = 3m is antinode & x = 0 is node
25 multiple of fundamental frequen2; possible frequencies are
 odd
Hz
4 v 3
cy.
fundamental
Thewhere,
correct options
are= A, C, D.
5.
For air to glass
1.5 1.4  1 1.5  1.4


f1
R
R
 f1 = 3R
For glass to air.
JEE(ADVANCED)2014-Paper 1-14
1 1.4  1.5 1  1.4


f2
R
R
 f2 = 2R
6.
7.
8.
V2
V2
V2
4
t1 
t2
R
R/2
R /8
t1 = 2 min.
t2 = 0.5 min.
H=
V1 
R 1 (V1  V2 )
 V1R 3  V2 R1
R1  R 3
V2 
R 3 (V1  V2 )
 V2 R1  V2 R 3
R1  R 3
Q




2
40 r0 2 0 r0 20
Q


r 
r 
r 
E1  0  
, E2  0  
, E3  0  
2
 2   0 r0
 2  0 r0
 2  2 0
r 
r 
E1  0   2E 2  0 
2
 
2
D

d
 2 > 1  2 > 1

9.
10.
Also m11 = m22  m1 > m2
D
D
Also 3   (600 nm)  (2  5  1)   400 nm
d
 2d 

Angular width  
d
Condition of translational equilibrium
N1 = 2N2
N2 + 1N1 = Mg
mg
Solving N 2 
1  1 2
 2 mg
1  1 2
Applying torque equation about corner (left) point on the floor

mg cos   N1 sin   1 N1 cos 
2
1  1 2
Solving tan  
2 2
N1 
11.
FL
since the experiment measures only change in the length of wire
A
Y


100 
100
Y

From the observation 1 = MSR + 20 (LC)
Y
2 = MSR + 45 (LC)
 change in lengths = 25(LC)
and the maximum permissible error in elongation is one LC
Y
(LC)

100 
 100  4
Y
25(LC)
12.
13.
14.
The relative velocity of B with respect to A is perpendicular to line
of motion of A.
 VB cos30  VA
 VB  200 m/s
And time t0 = (Relative distance) / (Relative velocity)
500
=
 5 sec
VB sin 30
Ub = 200 J, U i  100 J
Process iaf
Process
W(in Joule)
ia
af
Net
300
 Uf = 400 Joule
Process ibf
Process
W(in Joule)
ib
100
bf
200
Net
300
Q bf 300


2
Qib 150
Case – I
X0
30
U(in Joule)
0
200
200
Q(in Joule)
U(in Joule)
50
100
150
Q(in Joule)
150
300
450
500
Case – II
I
I

VA
X0
I
I
P
X0/3
P
X0/3
15.
B1 
1   0   3I 
 
2  2   x 0 
R2 
R1 
mv
qB1

mv
qB2
R 1 B 2 1/3


3
R 2 B1 1/9
d  x Sy Fz
 [L]  [ML3 ]x [MT 3 ]y [T 1 ]z
 x + y = 0, –3x = 1, –3y –z = 0

VB
60
1
1
, y  , z  1
3
3
1
 y
n
 n=3
 x
16.
Maximum displacement of the left ball from the left wall of the chamber is 2.25 cm, so the right ball has to
travel almost the whole length of the chamber (4m) to hit the left ball. So the time taken by the right ball is
1.9 sec (approximately 2 sec)
17.
1.5
RG
RG
RS
RS
i
V
R
0.006 
30
4990  R
R = 10
18.
i RG  0.006
i RS  1.494
Since RG and RS are in parallel, iGRG = iSRS
 2n 
0.006 R  1.494 

 249 
 n=5
  I
3 FRsin30° = I
MR 2
I
2
=2
 = 0 + t
 = 2 rad/s
F
R
30°
Rsin30°
F
F
19.
Since net torque about centre of rotation is zero, so we can apply conservation of angular momentum of the
system about center of disc
Li = Lf
0 = I + 2mv (r/2); comparing magnitude
0.5
 0.45  0.5  0.5 

2
   0.05  9 
2
2


=4
20.
Using work energy theorem
Wmg + WF = KE
– mgh + Fd = KE
–1×10×4 + 18(5) = KE
KE = 50
n=5
SECTION – 1:
21.
Isomeric alcohols of C4H10O are
CH3
C CH3
CH CH3 , H3 C
CH CH2 OH , H3C
2-methylpropan-1-ol
OH
OH
or
tert-butanol
sec-butyl alcohol
isobutyl alcohol
or
or
2-methylpropan-2-ol
butan-2-ol
OH , H3C
H3C CH2 CH2 CH2
n-butanol
or
butan-1-ol
CH3
CH2
22.
Since container is thermally insulated. So, q = 0, and it is a case of free expansion therefore W = 0 and
E  0
So, T1 = T2
Also, P1V1 = P2V2
23.
(A) Ice has cage-like structure in which each water molecule is surrounded by four other water molecules
tetrahedrally through hydrogen boding, due to this density of ice is less than water and it floats in
water.

(B) R  NH 2  H  OH  R  N H 3  OH 
 I

 R 3 N  H  OH   R 3  N H  OH 
 II
The cation (I) more stabilized through hydrogen boding than cation  II  . So, R–NH2 is better base
than (R)3N in aqueous solution.
(C) HCOOH is stronger acid than CH3COOH due to inductive effect and not due to hydrogen bonding.
(D) Acetic acid dimerizes in benzene through intermolecular hydrogen bonding.
O
H3C
O
C
C
O
25.
H
H
CH3
O
The balanced equation is,
ClO3  6I   6H 2SO4 
 3I 2  Cl  6HSO4  3H 2 O
OH
26.
I
I2
Rxn i 

CMe3
OH
OH
Br
Br2
Rxn  ii 

CMe3
CMe3
Br
OH
Cl
Cl
Cl2

Rxn  iii 
CMe3
Cl
27.
(A) H3BO3 is a weak monobasic Lewis acid.
 B  OH    H 
H 3 BO 3  H  OH 
4
… (i)
(B) Equilibrium (i) is shifted in forward direction by the addition of syn-diols like ethylene glycol which

forms a stable complex with B  OH  4 .
H
O
H
H
O
O
O
H
B
B
H
O
O

O
O
O
O
H
H
H
O
 4H 2 O
O
 stable complex 
(C) It has a planar sheet like structure due to hydrogen bonding.
(D) H3BO3 is a weak electrolyte in water.
28.

(A) 2CuFeS2  O 2 
 Cu 2S  2FeS  SO 2
o
1100 C
(B) 4CuO 
 2Cu 2 O  O 2

2Cu 2 O  Cu 2S 
 6Cu  SO2

(C) Cu 2S  2Cu 2 O 
 6Cu  SO2
1
720o C
(D) CuSO 4 
 CuO  SO 2  O 2
2
1100o C
4CuO 
 2Cu 2 O  O 2
29.

2Cu 2 O  Cu 2S 
 6Cu  SO2
(A) sodium (Na) when dissolved in excess liquid ammonia, forms a blue coloured paramagnetic solution.
(B) K  O 2 

KO 2
and KO2 is paramagnetic.
 potassium superoxide 
(C) 3Cu  8HNO3 
 3Cu  NO3  2  2NO  4H 2 O
 dilute 
Where “NO” is paramagnetic.
OH
(D)
O
CH2 CH3
CH2 CH3
O 2  air  

OH
H 2 O2
O
 2  ethyl anthraquinol 
Where “H2O2” is diamagnetic.
30.
Only amines undergo acetylation and not acid amides.
O
NH2
NH C
O
O
C
H3 C
CH3
C
O
OH
 CH 3COOH


C
NH2
C
O
NH2
O
SECTION – 2:
31.
Black coloured sulphides PbS, CuS, HgS, Ag 2 S, NiS, CoS
* Bi2S3 in its crystalline form is dark brown but Bi2S3 precipitate obtained is black in colour.
32.
Cl
Cl
Me
BrBr
Cl
CH3
Br
Br
Br
Me

Cl
Br
Cl
33.
Cl Br
Cl
(unstable)
(stable   0)
Br
Me
Br
Me
Cl Me
(unstable)
60
600
600
CH3
0
Me
Me
ClBr
Me
Br
Me
MeBr
600
0
60
Br
Cl Cl
Cl
(stable  0)
(unstable)
Br
Me
Me
Cl
Cl
(stable   0)
K 2 Cr2 O7 , CuSO4 , H 2 O2 , Cl 2 , O3 , FeCl3 , HNO3
K 2 Cr2 O7  7H 2SO 4  6KI 
 4K 2SO4  Cr2 SO 4 3  3I2  7H 2 O
2CuSO 4  4KI 
 Cu 2 I 2  I2  2K 2SO 4
H 2 O 2  2KI 
 I 2  2KOH
Cl2  2KI 
 2KCl  I 2
O3  2KI  H 2 O 
 2KOH  I2  O 2
2FeCl3  2KI 
 2FeCl2  I2  2KCl
8HNO3  6KI 
 6KNO3  2NO  4H 2 O  3I2
2KMnO 4  KI  H 2 O 
 KIO3  2MnO 2  2KOH
34.
XeF4  Square planar
BrF4  Square planar
 Cu  NH3  4 
2
 Square planar
 Pt Cl4 2  Square planar
SF4  See  saw
SiF4  Tetrahedral
BF4  Tetrahedral
 FeCl4 2  Tetrahedral
CoCl 4 2  Tetrahedral
35.
(1)
O
||
*
Will not give a racemic mixture on reduction with NaBH4
CH3  CH 2  C H  C  CH 3
|
CH3
(2)
(3)
O
||
CH 3  CH 2  CH 2  CH 2  C  CH 3
O
||
Will give a racemic mixture on reduction with NaBH4
CH3  CH  CH 2  C  CH3
|
CH3
(4)
CH3 O
H3C
C
C
CH3
CH3
(5)
O
||
CH 3  CH 2  C  CH 2  CH 2  CH 3
(6)
O
||
CH3  CH 2  C  CH  CH3
|
CH3
36.
n4
  0,1, 2,3
| m  | 1  1
1
2
For   0, m   0
  1, m  1, 0,  1
ms  
  2, m  2,  1,0,  1, 2
  3, m  3,  2, 1,0,  1, 2, 3
1
So, six electrons can have | m  | 1 & ms   37.
2
R  kN A
k
R
NA
 1.380 1023  6.023 10 23
 8.31174
 8.312
38.
 M 2   2X 
MX 2 
1 

39.
2
  0.5
i  1  2
i=2
This peptide on complete hydrolysis produced 4 distinct amino acids which are given below:
O
(1) H2N
CH2
C
O
OH
(2) HO
C
NH2
Glycine
 natural 
CH2
O
O
(3) HO
O
(4) HO
C
C
C
OH
NH
NH2
CH2
Only glycine is naturally occurring amino acid.
40.
Here, Vsolution  Vsolvent
Since, in 1 solution , 3.2 moles of solute are present,
So, 1 solution  1 solvent (d = 0.4g/ml)  0.4 kg
moles of solute
3.2
So, molality  m  

8
mass of solvent  kg  0.4
41.
f ' x  
2e
 1
 x  
 x
x
Which is increasing in [1, )
1
Also, f  x   f    0
x
2x
  
g  x   f 2x 
2 x
2 x
g  x  

2
 1
 t  
e  t
x
t
e
 1
 t  
 t
t
dt
dt   g  x 
Hence, an odd function
42.
Let y = x5 – 5x
(–1, 4)
–1
1
(1, 4)
43.
Let f (x) and g (x) achieve their maximum value at x1 and x2 respectively
h (x) = f (x)  g (x)
h (x1) = f (x1)  g (x1)  0
h (x2) = f (x2)  g (x2)  0
 h (c) = 0 where c  [0, 1]  f (c) = g (c).
44.
Given circles
x2 + y2  2x  15 = 0
x2 + y2  1 = 0
Radical axis x + 7 = 0
… (1)
Centre of circle lies on (1)
Let the centre be ( 7, k)
Let equation be x2 + y2 + 14x  2ky + c = 0
Orthogonallity gives
 14 = c  15  c = 1
… (2)
(0, 1)  1  2k + 1 = 0  k = 1
Hence radius = 7 2  k 2  c  49  1  1 = 7
Alternate solution
Given circles x2 + y2 – 2x – 15 = 0
x2 + y2 – 1 = 0
Let equation of circle x2 + y2 + 2gx + 2fy + c = 0
Circle passes through (0, 1)
 1 + 2f + c = 0
Applying condition of orthogonality
–2g = c – 15, 0 = c – 1
 c = 1, g = 7, f = –1
r  49  1  1  7 ; centre (–7, 1)
45.
  

a is in direction of x   y  z 
     
i.e.  x  z  y   x  y  z

 1   
 a  1  2   y  z  
 2


 
a  1  y  z 
… (1)
 
   
Now a  y  1  y  y  y  z 
 
= 1 (2  1)  1 = a  y
… (2)
    
From (1) and (2), a  a  y  y  z 

   
Similarly, b  b  z  z  x 
 
       
Now, a  b   a  y  b  z  y  z    z  x  
   
=  a  y  b  z 1  1  2  1
   
=  a  y b  z .
 
 
 
 
46.
47.
48.
x y z 1
 
 r , Q (r, r, 1)
1 1
0
y z 1
x
Line 2: 

 k , R (k,  k,  1)
1 1
0

PQ = (  r) î + (  r) ĵ + (  1) k̂

and   r +   r = 0 as PQ is  to L1
 2 = 2r   = r

PR = (  k) î + ( + k) ĵ + ( + 1) k̂

and   k    k = 0 as PR is  to L2
k=0
so PQ  PR
(  r) (  k) + (  r) ( + k) + (  1) ( + 1) = 0
  = 1,  1
For  = 1 as points P and Q coincide
  =  1.
Line 1:
a c 
Let M  
 (where a, b, c  I)
c b
then Det M = ab – c2
if a = b = c, Det(M) = 0
if c = 0, a, b  0, Det(M)  0
if ab  square of integer, Det(M)  0
M2 = N4  M2  N4 = O  (M  N2) (M + N2) = O (As M, N commute.)
Also, M  N2, Det ((M  N2) (M + N2)) = 0
As M  N2 is not null  Det (M + N2) = 0
Also Det (M2 + MN2) = (Det M) (Det (M + N2)) = 0
 There exist non-null U such that (M2 + MN2) U = O
49.
Since f(x)  1  x  [a, b]
for g(x)
LHD at x = a is zero
x
 f  t  dt  0
and RHD at (x = a) = lim
a
x a
Hence g(x) is not differentiable at x = a
Similarly LHD at x = b is greater than 1
g(x) is not differentiable at x = b
x a
50.
= lim f  x   1
x a
  
f (x) = (log (sec x + tan x))3  x    , 
 2 2
f ( x) =  f (x), hence f (x) is odd function
  
Let g (x) = sec x + tan x  x    , 
 2 2
  
 g (x) = sec x (sec x + tan x) > 0  x    , 
 2 2
 g (x) is one-one function
Hence (loge(g(x)))3 is one-one function.
  
and g(x)  (0, )  x    , 
 2 2
 log(g(x))  R. Hence f (x) is an onto function.
51.
When n5 takes value from 10 to 6 the carry forward moves from 0 to 4 which can be arranged in
4
4
4
4
C3
C1
C2
C4
4
C0 



=7
4
3
2
1
Alternate solution
Possible solutions are
1, 2, 3, 4, 10
1, 2, 3, 5, 9
1, 2, 3, 6, 8
1, 2, 4, 5, 8
1, 2, 4, 6, 7
1, 3, 4, 5, 7
2, 3, 4, 5, 6
Hence 7 solutions are there.
52.
Number of red lines = nC2  n
Number of blue lines = n
Hence, nC2  n = n
n
C2 = 2n
n  n  1
 2n
2
n  1 = 4  n = 5.
53.
x2  1

 x  1
hx   2
x 1
x 1

, x   ,  1
, x   1, 0
, x   0, 1
, x  1,  
Hence, not differentiable at x = –1, 0, 1
–1
54.
b c
  (integer)
a b
b2 = ac  c 
b2
a
a bc
 b2
3
a + b + c = 3b + 6  a – 2b + c = 6
b2
2b b 2 6
 6  1


a
a a2 a
a  2b 
2
6
b 
  1   a = 6 only
a
a 
55.
  
a  b  c 1
   
  
a  b  b  c  pa  qb  rc
  
 
 
a  b  c  p  q a  b  r a  c


 
  
1
And a b c  =
2

q r
 
p    a b c 
2 2
p
r
q  0
2
2
p q

  r  a b c 
2 2
 p = r = –q
p2  2q 2  r 2
4
q2
56.
….. (1)
….. (2)
….. (3)
 dy

2 (y  x5)   5x 4 
 dx

= 1 (1 + x2)2 + (x) (2 (1 + x2) (2x))
dy
Now put x = 1, y = 3 and
 m.
dx
2 (3  1) (m  5) = 1 (4) + (1) (4) (2)
12
m5=
4
m=5+3=8
dy
m8.
dx
1
1
57.

0
4x 3
I
d2
1 x2
2
dx
II

d

=  4x 3
1 x2
dx


=  4x 3  5 1  x 2


5

dx
1
5
1
0
0
   12x

4
2
d
1 x2
dx

5

dx
1

5 1
 12   x 2 1  x 2   2x 1  x 2
0
 
0
0
1
 2x 

1
 
= 0 – 0 – 12[0 – 0] + 12 2x 1  x 2
5


 

5

dx 

dx
0
1
6

2
1

x

= 12   

6


0
 1
= 12 0   = 2
 6

58.

1 x
x
 ax  sin  x  1  a 1
lim 

x 1  x  sin  x  1  1 


 sin  x  1

a 

 x  1

lim 

x 1  sin  x  1
 1 

  x  1

 a = 0, a = 2
a=2
59.

1
4
1 x 
2

1
1
 1 a 

 
4
4
 2 
f : [0, 4]  [0, ], f(x) = cos–1(cos x)
10  x
 point A, B, C satisfy f  x  
10
Hence, 3 points

1
0
60.
y = cos–1(cosx)
A
B

2
C
4
3 10
x
y  1
10
2  d1(p) + d2(p)  4
For P(, ),  > 
 2 2  2  4 2
y=x
22 2
P(, )
2
2

 Area of region =  2 2  2 


= 8 – 2 = 6 sq. units

  
x= 2
x=2 2
D.
16.
17.
Marking Scheme
For each questions in Section1, you will be awarded 3 marks if you darken all the bubbles(s)
corresponding to the correct answer(s) and zero mark if no bubbles are darkened. No negative marks
will be awarded for incorrect answers in this section.
For each question in Section 2, you will be awarded 3 marks if you darken only the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will
be awarded from incorrect answer in this section.
Appropriate way of darkening the bubble for your answer to be evaluated:
a
The one and the only one acceptable
a
Part darkening
a
a
Answer will not be
evaluatedno marks, no
negative marks
Darkening the rim
a
Cancelling after darkening and
darkening another bubble
Attempt to Erase after darkening
a
Figure-1 : Correct way of bubbling for a valid answer and a few examples of invalid answers.
Any other form of partial marking such as ticking or crossing the bubble will be considered invalid.
5
0
0
4
5
2
3
1
0
0
0
0
0
1
1
1
1
1
1
2
2
2
3
3
4
2
4
2
3
3
3
4
4
4
4
5
5
5
5
5
6
6
6
6
6
6
7
7
7
7
7
7
7
9
9
9
9
8
9
8
9
9
Figure-2 : Correct Way of Bubbling your Roll Number on the ORS. (Example Roll Number : 5045231)
Roll Number
Name of the Candidate
I have read all instructions and shall abide
by them.
……………………………………………
Signature of the Candidate
I have verified all the information filled by
the candidate.
……………………………………………
Signature of theinvigilator
Note:
For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 2014 are also
given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics
from class XI have been marked with ‘*’, which can be attempted as a test. For this test the time allocated in
Physics, Chemistry & Mathematics are 22 minutes, 24 minutes and 25 minutes respectively.
TRIUMPH ACADEMY’S SOLUTIONS TO
JEE(ADVANCED)-2014
CODE
8
PAPER-2
P2-14-8
2207548
Note: Front and back cover pages are reproduced from actual paper back to back here.
Time: 3 Hours
Maximum Marks: 180
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
INSTRUCTIONS
A.
1.
2.
3.
4.
5.
6.
7.
8.
9.
B.
10.
11.
12.
13.
C.
14.
15.
16.
General
This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by the
invigilators.
The question paper CODE is printed on the left hand top corner of this sheet and on the back cover page of this booklet.
Blank space and blank pages are provided in the question paper for your rough work. No additional sheets will be provided
for rough work.
Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadget of any
kind are NOT allowed inside the examination hall.
Write your name and roll number in the space provided on the back cover of this booklet.
Answers to the questions and personal details are to be filled on an Optical Response Sheet, which is provided separately.
The ORS is a doublet of two sheets – upper and lower, having identical layout. The upper sheet is a machine-gradable
Objective Response Sheet (ORS) which will be collected by the invigilator at the end of the examination. The upper sheet
is designed in such a way that darkening the bubble with a ball point pen will leave an identical impression at the
corresponding place on the lower sheet. You will be allowed to take away the lower sheet at the end of the examination
(see Figure-1 on the back cover page for the correct way of darkening the bubbles for valid answers).
Use a black ball point pen only to darken the bubbles on the upper original sheet. Apply sufficient pressure so that the
impression is created on the lower sheet. See Figure -1 on the back cover page for appropriate way of darkening the
bubbles for valid answers.
DO NOT TAMPER WITH / MUTILATE THE ORS SO THIS BOOKLET.
On breaking the seal of the booklet check that it contains 28 pages and all the 60 questions and corresponding answer
choices are legible. Read carefully the instruction printed at the beginning of each section.
Filling the right part of the ORS
The ORS also has a CODE printed on its left and right parts.
Verify that the CODE printed on the ORS (on both the left and right parts) is the same as that on the this booklet and put
your signature in the Box designated as R4.
IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET / ORS AS APPLICABLE.
Write your Name, Roll No. and the name of centre and sign with pen in the boxes provided on the upper sheet of ORS. Do
not write any of this anywhere else. Darken the appropriate bubble UNDER each digit of your Roll No. in such way that
the impression is created on the bottom sheet. (see example in Figure 2 on the back cover)
Question Paper Format
The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections.
Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE
is correct.
Section 2 contains 3 paragraphs each describing theory, experiment and data etc. Six questions relate to three paragraphs
with two questions on each paragraph. Each question pertaining to a particular passage should have only one correct
answer among the four given choices (A), (B), (C) and (D).
Section 3 contains 4 multiple choice questions. Each questions has two lists (Lits-1: P, Q, R and S; List-2, : 1, 2, 3,
and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY one is correct.
Please read the last page of this booklet for rest of the instructions.
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com.
(2)
PART I : PHYSICS
SECTION – 1 : (Only One Option Correct Type)
which ONLY ONE option is correct.
1.
Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R
and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from
the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then
P
P
R
P
R
R
2Q
Q
4Q
R/2
2R
Sphere 1
(A) E1 > E2 > E3
*2.
Sphere 2
Sphere 3
(A) E3 > E1 > E2
(C) E2 > E1 > E3
(D) E3 > E2 > E1
A glass capillary tube is of the shape of truncated cone with an apex angle  so that its two ends have cross
sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius
of its cross section is b. If the surface tension of water is S, its density is , and its contact angle with glass
is , the value of h will be (g is the acceleration due to gravity)
2S
(A)
cos(   )
bg
(B)
2S
cos(  )
bg
(C)
2S
cos(   / 2)
bg
(D)
2S
cos(   / 2)
bg
h
3.
If cu is the wavelength of K X-ray line of copper (atomic number 29) and Mo is the wavelength of the K
X-ray line of molybdenum (atomic number 42), then the ratio cu/Mo is close to
(A) 1.99
(B) 2.14
(C) 0.50
(D) 0.48
*4.
A planet of radius R =
1
 (radius of Earth) has the same mass density as Earth. Scientists dig a well of
10
R
on it and lower a wire of the same length and of linear mass density 103 kgm1 into it. If the wire
5
is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the
radius of Earth = 6  106 m and the acceleration due to gravity of Earth is 10 ms2)
(A) 96 N
(B) 108 N
(C) 120 N
(D) 150 N
depth
(3)
*5.
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting
the surface. The force on the ball during the collision is proportional to the length of compression of the
ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most
appropriately? The figures are only illustrative and not to the scale.
(A)
(B)
K
K
t
t
(C)
(D)
K
K
t
t
6.
A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum
speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio
u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly
(A) 3.7 eV
(B) 3.2 eV
(C) 2.8 eV
(D) 2.5 eV
*7.
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is
fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it
slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is
A
(A) always radially outwards.
(B) always radially inwards.
(C) radially outwards initially and radially inwards later.
(D) radially inwards initially and radially outwards later.
90
8.
B
During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed
at 40.0 cm using a standard resistance of 90 , as shown in the figure. The least count of the scale used in
the metre bridge is 1 mm. The unknown resistance is
R
90 
40.0cm
(A) 60  0.15
(B) 135  0.56
(C) 60  0.25
(D) 135  0.23
9.
Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings
of temperature 300 K. Take Stefan-Boltzmann constant  = 5.7×10–8 Wm–2 K–4 and assume that the energy
exchange with the surroundings is only through radiation. The final steady state temperature of the black
body is close to
(A) 330 K
(B) 660 K
(C) 990 K
(D) 1550 K
10.
A point source S is placed at the bottom of a transparent block of height
10 mm and refractive index 2.72. It is immersed in a lower refractive
index liquid as shown in the figure. It is found that the light emerging
from the block to the liquid forms a circular bright spot of diameter
11.54 mm on the top of the block. The refractive index of the liquid is
(A) 1.21
(B) 1.30
(C) 1.36
Liquid
Block
S
(D) 1.42
(4)
SECTION – 2 : Comprehension type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the
three paragraphs with two questions on each paragraph. Each question has only one correct answer among
the four given options (A), (B), (C) and (D).
Paragraph for Questions 11 & 12
The figure shows a circular loop of radius a with two long parallel wires
(numbered 1 and 2) all in the plane of the paper. The distance of each wire
from the centre of the loop is d. The loop and the wires are carrying the
same current I. The current in the loop is in the counterclockwise direction
if seen from above.
Q
S
d
d
Wire I
Wire 2
a
P
R
11.
When d  a but wires are not touching the loop, it is found that the net magnetic field on the axis of the
loop is zero at a height h above the loop. In that case
(A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h  a
(B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h  a
(C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h  1.2a
(D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h  1.2a
12.
Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position
shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its
new position will be (assume that the net field due to the wires is constant over the loop)
 I2 a 2
 I2 a 2
30 I2 a 2
30 I2 a 2
(A) 0
(B) 0
(C)
(D)
d
2d
d
2d
In the figure a container is shown to have a movable (without friction) piston on top. The container and
the piston are all made of perfectly insulating material allowing no heat transfer between outside and
inside the container. The container is divided into two compartments by a rigid partition made of a
thermally conducting material that allows slow transfer of heat. The lower compartment of the container
is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2
moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are
3
5
5
7
CV  R, CP  R, and those for an ideal diatomic gas are CV  R, CP  R .
2
2
2
2
*13.
Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final
temperature of the gases will be
(A) 550 K
(B) 525 K
(C) 513K
(D) 490 K
*14.
Now consider the partition to be free to move without friction so that the pressure of gases in both
compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
(A) 250 R
(B) 200 R
(C) 100 R
(D) –100 R
(5)
A spray gun is shown in the figure where a piston pushes air out of a
nozzle. A thin tube of uniform cross section is connected to the nozzle.
The other end of the tube is in a small liquid container. As the piston
pushes air through the nozzle, the liquid from the container rises into the
nozzle and is sprayed out. For the spray gun shown, the radii of the piston
and the nozzle are 20 mm and 1 mm respectively. The upper end of the
container is open to the atmosphere.
*15.
If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of
(A) 0.1 ms–1
(B) 1 ms–1
(C) 2 ms–1
(D) 8 ms–1
*16.
If the density of air is a and that of the liquid  , then for a given piston speed the rate (volume per unit
time) at which the liquid is sprayed will be proportional to
a

(A)
(B) a 
(C)
(D) 

a
SECTION – 3: Match List Type (Only One Option Correct)
This section contains four questions, each having two matching lists. Choices for the correct combination of
elements from List-I and List-II are given as option (A), (B), (C) and (D) out of which one is correct.
*17.
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is
at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person.
In the following, state of the lift’s motion is given in List I and the distance where the water jet hits the
floor of the lift is given in List II. Match the statements from List I with those in List II and select the
correct answer using the code given below the lists.
List I
List II
Lift is accelerating vertically up.
P.
1. d = 1.2 m
Lift is accelerating vertically down 2. d > 1.2 m
Q.
with an acceleration less than the
gravitational acceleration.
Lift is moving vertically up with
R.
3. d < 1.2 m
constant speed.
Lift is falling freely.
S.
4. No water leaks out of the jar
Code:
(A) P-2, Q-3, R-2, S-4
(B) P-2, Q-3, R-1, S-4
(C) P-1, Q-1, R-1, S-4
(D) P-2, Q-3, R-1, S-1
18.
Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = -2a, a, +a and +2a,
respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the
signs of these charges are given in List I. The direction of the forces on the charge q is given in List II.
Match List I with List II and select the correct answer using the code given below the lists.
(+0, b)
List I
List II
q
Q1, Q2, Q3 Q4 all positive
P.
1. +x
Q1, Q2 positive; Q3, Q4 negative
Q.
2. x
Q1, Q4 positive ; Q2, Q3 negative
R.
3. +y
Q1, Q3 positive; Q2, Q4 negative
S.
4. y
Q1
(2a, 0)
Code:
(A) P-3, Q-1, R-4, S-2
(C) P-3, Q-1, R-2, S-4
Q2
(a, 0)
Q3
(+a, 0)
(B) P-4, Q-2, R-3, S-1
(D) P-4, Q-2, R-1, S-3
Q4
(+2a, 0)
(6)
19.
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r
and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in
List II and select the correct answer using the code given below the lists.
List I
List II
P.
1. 2r
Q.
2.
r/2
R.
3.
r
S.
4.
r
Code:
(A) P-1, Q-2, R-3, S-4
(C) P-4, Q-1, R-2, S-3
*20.
(B) P-2, Q-4, R-3, S-1
(D) P-2, Q-1, R-3, S-4
A block of mass m1 = 1 kg another mass m2 = 2kg, are placed together (see figure) on an inclined plane
with angle of inclination . Various values of  are given in List I. The coefficient of friction between the
block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m 2
and the plane are equal to  = 0.3. In List II expressions for the friction on the block m2 are given. Match
the correct expression of the friction in List II with the angles given in List I, and choose the correct option.
The acceleration due to gravity is denoted by g.
[Useful information: tan (5.5)  0.1; tan (11.5)  0.2; tan (16.5)  0.3]
m1
m2

List I
P.
 = 5
Q.
 = 10
R.
 = 15
S.
 = 20
Code:
(A) P-1, Q-1, R-1, S-3
(B) P-2, Q-2, R-2, S-3
(C) P-2, Q-2, R-2, S-4
(D) P-2, Q-2, R-3, S-3
1.
2.
3.
4.
List II
m2g sin 
(m1+m2) g sin 
m2g cos
(m1 + m2)g cos
(7)
PART II : CHEMISTRY
*21.
Assuming 2s – 2p mixing is NOT operative, the paramagnetic species among the following is
(A) Be2
(B) B2
(C) C2
(D) N2
*22.
For the process
H 2 O    
 H 2O  g 
at T = 100C and 1 atmosphere pressure, the correct choice is
(A) Ssystem  0 and Ssurrounding  0
(B) Ssystem  0 and Ssurrounding  0
(C) Ssystem  0 and Ssurrounding  0
(D) Ssystem  0 and Ssurrounding  0
*23.
For the elementary reaction M  N, the rate of disappearance of M increases by a factor of 8 upon
doubling the concentration of M. The order of the reaction with respect to M is
(A) 4
(B) 3
(C) 2
(D) 1
24.
For the identification of -naphthol using dye test, it is necessary to use
(A) dichloromethane solution of -naphthol.
(B) acidic solution of -naphthol.
(C) neutral solution of -naphthol.
(D) alkaline solution of -naphthol.
*25.
Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the
figure.
[Figure]
I
II
and
The correct order of their boiling point is
(A) I > II > III
(C) II > III > I
26.
III
and
(B) III > II > I
(D) III > I > II
The major product in the following reaction is
[Figure]
O
Cl

CH3
(A)
1. CH 3 MgBr, dry ether, 0 C


2. aqueous acid
(B)
O
H3C
H2C
CH3
(C)
OH
CH3
CH3
(D)
O
CH3
CH2
O
CH3
(8)
27.
Under ambient conditions, the total number of gases released as products in the final step of the reaction
scheme shown below is
Complete
Hydrolysis
XeF6 
 P  other product
OH  / H 2 O
Q
slow disproportionation in OH  / H 2 O
products
(A) 0
(C) 2
(B) 1
(D) 3
28.
The product formed in the reaction of SOCl2 with white phosphorous is
(A) PCl3
(B) SO2Cl2
(C) SCl2
(D) POCl3
*29.
Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a
(A) reducing agent, oxidising agent
(B) reducing agent, reducing agent
(C) oxidising agent, oxidising agent
(D) oxidising agent, reducing agent
30.
The acidic hydrolysis of ether (X) shown below is fastest when
[Figure]
acid
OR 


(A)
(B)
(C)
(D)
OH  ROH
one phenyl group is replaced by a methyl group.
one phenyl group is replaced by a para-methoxyphenyl group.
two phenyl groups are replaced by two para-methoxyphenyl groups.
no structural change is made to X.
SECTION – 2 : Comprehension type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the
X and Y are two volatile liquids with molar weights of 10 g mol1 and 40 g mol1 respectively. Two cotton plugs,
one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as
shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K.
Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X.
Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
(9)
L = 24 cm
Cotton wool
soaked in Y
Initial formation of
the product
d
Cotton wool
soaked in X
*31.
The value of d in cm (shown in the figure), as estimated from Graham’s law, is
(A) 8
(B) 12
(C) 16
(D) 20
*32.
The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is
due to
(A) larger mean free path for X as compared to that of Y.
(B) larger mean free path for Y as compared to that of X.
(C) increased collision frequency of Y with the inert gas as compared to that of X with the inert gas.
(D) increased collision frequency of X with the inert gas as compared to that of Y with the inert gas.
Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed
in each step for both schemes.
1. NaNH2  excess 
2. CH CH I 1 equivalent

3
2 
H 
X
3. CH3 I 1 equivalent 
Scheme  1
4. H 2 , Lindlar 's catalyst
HO
M
H
1. NaNH2  2 equivalent 
2.
OH
Br

Y
3.H O ,  mild 
N
Scheme  2
3
4. H2 , Pd /C
5. CrO3
33.
The product X is
(A) H3 CO
(B)
H
H
H
(C)
H
H3 CO
(D)
CH3 CH2 O
H
H
H
34.
H
CH3 CH2 O
The correct statement with respect to product Y is
(A) It gives a positive Tollens test and is a functional isomer of X.
(B) It gives a positive Tollens test and is a geometrical isomer of X.
(C) It gives a positive iodoform test and is a functional isomer of X.
(D) It gives a positive iodoform test and is a geometrical isomer of X.
(10)
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and
square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral
complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which
dissolves in excess of S. The reactions are summarized in the scheme given below.
SCHEME:
Q
R
Tetrahedral 
M1 
 Square planar
excess
excess
Q
R
Tetrahedral 
excess M 2 
excess  Tetrahedral
S,stoichiometricamount
S
White precipitate 
excess  precipitate dissolves
35.
36.
M1, Q and R, respectively are
(A) Zn 2 ,KCN and HCl
(B) Ni 2 , HCl and KCN
(C) Cd 2 , KCN and HCl
(D) Co2  , HCl and KCN
Reagent S is
(A) K 4  Fe  CN  6 
(B) Na 2 HPO4
(C) K 2 CrO4
(D) KOH
SECTION – 3: Match List Type (Only One Option Correct)
This section contains four questions, each having two matching lists. Choices for the correct combination of
elements from List-I and List-II are given as option (A), (B), (C) and (D) out of which one is correct.
37.
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and
select the correct answer using the code given below the lists
{en = H2NCH2CH2NH2; atomic numbers: Ti = 22, Cr = 24; Co= 27; Pt = 78)
List – I
List - II
P.
[Cr(NH3)4Cl2]Cl
1.
Paramagnetic and exhibits ionization isomerism
Q.
[Ti(H2O)5Cl](NO3)2
2.
Diamagnetic and exhibits cis-trans isomerism
R.
[Pt(en)(NH3)Cl]NO3
3.
Paramagnetic and exhibits cis-trans isomerism
S.
[Co(NH3)4(NO3)2]NO3
4.
Diamagnetic and exhibits ionization isomerism
Code:
(A)
(B)
(C)
(D)
P
4
3
2
1
Q
2
1
1
3
R
3
4
3
4
S
1
2
4
2
(11)
*38.
Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct
answer using the code given below the lists.
List – I
List - II
p – d  antibonding
P.
1.
Q.
2.
d – d  bonding
R.
3.
p – d  bonding
S.
4.
d – d  antibonding
Code:
(A)
(B)
(C)
(D)
P
2
4
2
4
Q
1
3
3
1
R
3
1
1
3
S
4
2
4
2
(12)
39.
Different possible thermal decomposition pathways for peroxyesters are shown below. Match each
pathway from List I with an appropriate structure from List II and select the correct answer using the code
given below the lists.
P
CO 2 
Q
O
CO 2 
O
R
O
R'
R
(Peroxyester)
R   R 'O 
R   R 'O  
 R   X   carbonyl compound 
RCO 2  R 'O 
R   X '  carbonyl compound 
 CO 
2
S
RCO 2  R 'O  
R   R 'O 
 CO 2 
List – I
P.
List – II
Pathway P
O
1.
O
Q.
Pathway Q
2.
O
C6 H5 CH2
O
CH3
O
C6 H5
R.
Pathway R
O
O
3.
CH3
CH3
O
O
C6 H5 CH2
S.
Pathway S
CH3
CH2 C6 H5
O
4.
O
C6 H5
O
C6 H5
CH3
CH3
Code:
(A)
(B)
(C)
(D)
P
1
2
4
3
Q
3
4
1
2
R
4
3
2
1
S
2
1
3
4
(13)
40.
Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes
(I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.
List – I
List - II
1. Scheme I
P.
H
H
 i  KMnO 4 , HO , heat  ii  H , H2 O
 iii  SOCl2  iv  NH3
? 
 C 7 H 6 N 2 O3
OH
2. Scheme II
 i  Sn / HCl  ii  CH3 COCl  iii  conc. H 2 SO 4
iv HNO v dil. H SO , heat vi HO 
 
 
3 
2
4
? 
 C6 H 6 N 2 O 2
Q.
OH
NO 2
3. Scheme III
 i  red hot iron, 873 K  ii  fu min g HNO 3 , H 2SO 4 , heat
  2
3 
2
2
4 
? 
 C6 H 5 NO3
iii H S.NH iv NaNO , H SO v hydrolysis
R.
NO 2
4. Scheme IV
 i  conc. H 2SO4 , 60 C
 ii  conc. HNO3 , conc. H 2SO4 iii  dil. H 2SO 4 , heat
? 
 C 6 H5 NO 4
S.
CH3
Code:
(A)
(B)
(C)
(D)
P
1
3
3
4
Q
4
1
4
1
R
2
4
2
3
S
3
2
1
2
PART III : MATHEMATICS
41.
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at
least one more than the number of girls ahead of her, is
1
1
(A)
(B)
2
3
2
3
(C)
(D)
3
4
*42.
In a triangle the sum of two sides is x and the product of the same two sides is y. If x2  c2 = y, where c is
the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
3y
3y
(A)
(B)
2x  x  c 
2c  x  c 
(C)
3y
4x  x  c 
(D)
3y
4c  x  c 
(14)
*43.
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that
each envelope contains exactly one card and no card is placed in the envelope bearing the same number and
moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can
be done is
(A) 264
(B) 265
(C) 53
(D) 67
*44.
The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P, Q
and the parabola at the points R, S. Then the area of the quadrilateral PQRS is
(A) 3
(B) 6
(C) 9
(D) 15
*45.
The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation
p(p(x)) = 0 has
(A) only purely imaginary roots
(B) all real roots
(C) two real and two purely imaginary roots
(D) neither real nor purely imaginary roots
46.
The following integral
/ 2
17
  2cosec x 
dx is equal to
/ 4


log 1 2


(A)

log 1 2
u
2 e e
u
16

du
 e
(B)
0

log 1 2
(C)
u
 e u
17

du
0


log 1 2
 e
u
e
u
17

du


2 eu  e  u

(D)
0
47.

16

du
0
The function y = f (x) is the solution of the differential equation
dy
xy
x4  2 x
 2

in (1, 1) satisfying
dx x  1
1 x2
3
2

f (0) = 0. Then

48.
f  x  dx is
3
2
(A)

3

3 2
(B)

3

3 4
(C)

3

6 4
(D)

3

6 2
Let f : [0, 2]  R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0) = 1.
x2
Let F  x  
 f  t  dt
for x  [0, 2]. If F(x) = f (x) for all x  (0, 2), then F(2) equals
0
(A) e2  1
(C) e  1
*49.
(B) e4  1
(D) e4
Coefficient of x11 in the expansion of (1 + x2)4 (1 + x3)7 (1 + x4)12 is
(A) 1051
(B) 1106
(C) 1113
(D) 1120
(15)
*50.
For x  (0, ), the equation sinx + 2sin2x  sin3x = 3 has
(A) infinitely many solutions
(B) three solutions
(C) one solution
(D) no solution
SECTION – 2 : Comprehension Type (Only One Option Correct)
This section contains 3 paragraph, each describing theory, experiments, data etc. Six questions relate to the
Paragraph For Questions 51 and 52
Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box
3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the
number on the card drawn from the ith box, i = 1, 2, 3.
51.
52.
The probability that x1 + x2 + x3 is odd, is
29
(A)
105
57
(C)
105
53
105
1
(D)
2
(B)
The probability that x1 , x2 , x3 are in an arithmetic progression, is
9
10
(A)
(B)
105
105
11
7
(C)
(D)
105
105
Let a, r, s, t be non-zero real numbers. Let P(at2, 2at), Q, R(ar2, 2ar) and S(as2, 2as) be distinct points on the
parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0).
*53.
The value of r is
1
(A) 
t
(C)
*54.
1
t
(B)
t2 1
t
(D)
t2 1
t
If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is
t
(A)
2

1
2
(B)
2t 3


a t2 1
(C)
t3


a t2 1
2
2
2t 3

a t2  2
(D)

2
t3
(16)
1 h
Given that for each a  (0, 1), lim
h 0
t
a
1  t a 1 dt
exists. Let this limit be g(a). In addition, it is given that the
h
function g(a) is differentiable on (0, 1).
1
The value of g   is
2
(A) 

(C)
2
55.
(B) 2

(D)
4
1
The value of g    is
2

(A)
2

(C) 
2
56.
(B) 
(D) 0
SECTION – 3 : Matching List Type (Only One Option Correct)
This section contains four questions, each having two matching list. Choices for the correct combination of
elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which ONE is correct.
57.
Match the following:
List  I
The number of polynomials f (x) with non-negative integer
coefficients of degree  2, satisfying f(0) = 0 and
(P)
List  II
(1)
8
(2)
2
(3)
4
(4)
0
1
 f  x  dx  1 , is
0
The number of points in the interval   13 , 13  at which
f(x) = sin(x2) + cos(x2) attains its maximum value, is
(Q)
2
3 x2
 1  e 
(R)
x
dx equals
2
 1/2
 1  x  

cos 2 x  log 
 dx

 1  x  
 1/2
equals
 1/2

1 x 
 cos 2 x  log 
 dx 

1 x  

 0


(S)

Codes:
(A)
(B)
(C)
(D)
P
3
2
3
2
Q
2
3
2
3
R
4
4
1
1
S
1
1
4
4
(17)
58.
Match the following:
List  I
List  II
3
. Then
2
(1)
1
Let A1, A2, ….. , An(n > 2) be the vertices of a regular

polygon of n sides with its centre at the origin. Let ak be the
position vector of the point Ak, k = 1, 2, ….. n. If
n 1  
n 1  
ak  ak 1 =
a  ak 1 , then the minimum
k 1
k 1 k
(2)
2
(3)
8
(4)
9
Let y(x) = cos(3cos1x), x [1, 1], x  
(P)
d 2 y  x
dy  x  
1  2
x
 x 1
 equals
2
y  x  
dx 
dx

(Q)

 

 

value of n is
If the normal from the point P(h, 1) on the ellipse
*(R)
2
2
x
y

 1 is perpendicular to the line x + y = 8, then the
6
3
value of h is
Number of positive solutions satisfying the equation
1 
 1 
1 
1  2 
tan 1 
  tan 
  tan  2  is
 2x 1 
 4x 1 
x 
(S)
Codes:
P
4
2
4
2
(A)
(B)
(C)
(D)
Q
3
4
3
4
R
2
3
1
1
S
1
1
2
3
Let f1 : R  R, f2 : [0, )  R, f3 : R  R and f4 : R  [0, ) be defined by
59.
 x
f1  x    x
e
if
if
x0
sin x if
; f 2  x   x 2 ; f3  x   
if
x0
x
if
x0
 f 2  f1  x  
and f 4  x   
x0
 f 2  f1  x    1 if
List  I
(P)
(Q)
(R)
(S)
f4 is
f3 is
f2 o f1 is
f2 is
(1)
(2)
(3)
(4)
P
3
1
3
1
Q
1
3
1
3
R
4
4
2
2
S
2
2
4
4
x0
List  II
onto but not one-one
neither continuous nor one-one
differentiable but not one-one
continuous and one-one
Codes:
(A)
(B)
(C)
(D)
x0
(18)
 2k  
 2k  
Let zk = cos 
  i sin 
 ; k = 1, 2, …. , 9.
 10 
 10 
*60.
List  I
For each zk there exists a zj such zk . zj = 1
(P)
There exists a k  {1, 2, ….. , 9} such that z1 . z = zk
has no solution z in the set of complex numbers
(Q)
1  z1 1  z2 .... 1  z9
(R)
10
1
(S)

equals
 2k  
cos 
 equals
k 1
 10 
9
List  II
(1)
True
(2)
False
(3)
1
(4)
2
Codes:
(A)
(B)
(C)
(D)
P
1
2
1
2
Q
2
1
2
1
R
4
3
3
4
S
3
4
4
3
(19)
ANSWERS
PAPER-2 [Code – 8]
JEE(ADVANCED) 2014
ANSWERS
PART I : PHYSICS
1.
C
2.
D
3.
B
4.
B
5.
B
6.
A
7.
D
8.
C
9.
A
10.
C
11.
C
12.
B
13.
D
14.
D
15.
C
16.
A
17.
C
18.
A
19.
B
20.
D
PART II : CHEMISTRY
21.
C
22.
B
23.
B
24.
D
25.
B
26.
D
27.
C
28.
A
29.
A
30.
C
31.
C
32.
D
33.
A
34.
C
35.
B
36.
D
37.
B
38.
C
39.
A
40.
C
41.
A
42.
B
43.
C
44.
D
45.
D
46.
A
47.
B
48.
B
49.
C
50.
D
51.
B
52.
C
53.
D
54.
B
55.
A
56.
D
57.
D
58.
A
59.
D
60.
C
(20)
PART I : PHYSICS
1.
For point outside dielectric sphere E 
Q
4 0 r 2
For point inside dielectric sphere E = E s
r
R
Exact Ratio E1 : E2 : E3 = 2 : 4 : 1
2.
If R be the meniscus radius
R cos ( + /2) = b
R
Excess pressure on concave side of meniscus =
2S
R
2S 2S


hg 
 cos    
R
b
2

2S


 h
cos    
bg
2

3.
4.
5.
6.
Cu  ZMo  1 


 Mo  ZCu  1 
 + (/2)
b

/2
2
Inside planet
r 4
g i  g s  Gr
R 3
Force to keep the wire at rest (F)
= weight of wire
R
4
 4
  9 
    dr   Gr    G    R 2
3
 3
  50 
4R /5
Me
Here,  = density of earth =
4
R 2e
3
R
Also, R  e ; putting all values, F = 108 N
10
d  KE 
dt
 mv
dv
dt
hc

1
u2
 12
hc
u2

2
 = 3.7 eV
7.
Initially bead is applying radially inward normal force.
During motion at an instant, N = 0, after that N will act radially outward.
(21)
8.
Alternatively
x
90
100  x
 R  60 
R
dR 0.1 0.1


R
40 60
 dR = 0.25
dR
100

dx
R (x) (100 – x)
 dR 
100
0.1 60
(40) (60)
= 0.25
9.
Rate of radiation energy lost by the sphere
= Rate of radiation energy incident on it
   4r 2  T 4  (300) 4   912  r 2
 T  11 102  330 K
10.
 sin C 
C
r
C
h

b
    2.72 
11.
Liquid
r 5.77

 3
h
10
 C  30
tan  C 
1
 1.36
2
The net magnetic field at the given point will be zero if.


Bwires  Bloop
 2
0 I

a

0 Ia 2
2 (a 2  h 2 )3/ 2
2 a 2  h 2
a2  h2
 h  1.2 a
The direction of magnetic field at the given point due to the loop is normally out of the plane. Therefore,
the net magnetic field due the both wires should be into the plane. For this current in wire I should be along
PQ and that in wire RS should be along SR.
12.
13.
14.
  MBsin   Ia 2  2 
 I2a 2
0 I
sin 30  0
2 d
2d
3
Heat given by lower compartment = 2  R  (700  T) ...(i)
2
7
Heat obtained by upper compartment = 2  R  (T  400) ...(ii)
2
equating (i) and (ii)
3 (700 – T) = 7 (T – 400)
2100 – 3T = 7 T – 2800
4900 = 10 T  T = 490 K
5
Heat given by lower compartment = 2  R  (700  T)
2
at eq. TK
400 K
700 K
...(i)
(22)
7
Heat obtained by upper compartment = 2  R  (T  400)
2
By equating (i) and (ii)
5(700  T)  7(T  400)
3000 – 5T = 7T– 2800
6300 = 12 T
T = 525K
 Work done by lower gas = nRT = – 350 R
Work done by upper gas = nRT = +250 R
Net work done  100 R
15.
16.
By A1V1 = A2V2
 (20) 2  5  (1)2 V2
...(ii)
 V2  2 m/s 2
1
1
a Va2   V2
2
2
For given Va
a
V 

17.
In P, Q, R no horizontal velocity is imparted to falling water, so d remains same.
In S, since its free fall, aeff = 0
 Liquid won’t fall with respect to lift.
18.
P: By Q1 and Q4, Q3 and Q2 F is in +y
Q: By Q1 and Q4, Q2 and Q3 F is in +ve x.
R: By Q1 and Q4, F is in +ve y
By Q2 and Q3, F is in –ve y
But later has more magnitude, since its closer to
(0, b). Therefore net force is in – y
S: By Q1 and Q4, F is in +ve x and by Q2 and Q3, F is in
–x, but later is more in magnitude, since its closer to
(0, b). Therefore net force is in –ve x.
19.
(+0, b)
Q1
(2a, 0)
Q2
(a, 0)
Q3
(+a, 0)
Q4
(+2a, 0)
 1
1
1 
    1 


f
 R1 R 2 
f=R
f = 2R
f = 2R
1
1 1
Use
 
f eq f1 f 2
(P)
1
1 1 2
R
   ; f eq 
f eq R R R
2
(Q)
1
1
1
1


 ; f eq  R
f eq 2R 2R R
(R)
1
1
1
1


  ; feq = - R
f eq
2R 2R
R
(S)
1
1
1
1
 

; feq = 2R
f eq R 2R 2R
(23)
20.
Condition for not sliding,
fmax > (m1 + m2) g sin 
N > (m1 + m2) g sin 
0.3 m2 g cos   30 sin 
6  30 tan 
1/5  tan 
0.2  tan 
 for P, Q
f = (m1 + m2) g sin 
For R and S
F = fmax = m2g sin 
m1g cos 
m2g cos 
m1
m2
(m1+ m2) g sin 
(m1+ m2) g cos 

PART II : CHEMISTRY
21.
Assuming that no 2s-2p mixing takes place
 A  Be2  1s 2 ,  *1s 2 , 2s 2 ,  *2s 2  diamagnetic 
 B  B2 
22.
0
1s 2 ,  *1s2 , 2s 2 ,  *2s 2 , 2p 2z , 2p0x
(diamagnetic )
 2p y
1
 C  C2 
1s2 ,  *1s 2 , 2s 2 ,  * 2s2 , 2p 2z , 2p1x ,
 D N2 
1s 2 ,  *1s2 ,  2s2 ,  *2s 2 , 2p 2z ,  2p2x ,
 2p y
2
 2p y
* 2p0x
, * 2p 0z
* 2p0y
(paramagnetic)
* 2p0x
, * 2p 0z
* 2p0y
(diamagnetic)
 H 2 O  g  is at equilibrium. For equilibrium
At 1000C and 1 atmosphere pressure H 2 O    
Stotal  0 and

Ssystem  Ssurrounding  0
Ssystem  0and Ssurrounding  0
n
23.
r1 1  M 
 
n  3
r2 8  2M n
24.
N=N-Ph
OH
OH
Ph N N Cl
alkaline solution
25.
III > II > I
More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point.
26.
OH
Cl
OMgX
CH 3MgBr, dry ether, 0 C

CH3 
Cl
aqueous acid
CH3 
CH3
CH3
O
CH3
(24)
27.
XeF6  3H 2 O 
 XeO3  3H 2 F2
OH 
HXeO 4
OH  / H 2 O (disproportionation)
XeO64 s   Xe g   H 2 O   O2 g 
28.
P4s   8SOCl 2  
 4PCl3   4SO2 g   2S2 Cl 2g 
29.
KIO4  H 2 O 2 
 KIO3  H 2O  O 2
NH 2 OH  3H 2 O2 
 HNO3  4H 2 O
30.
31.
32.
When two phenyl groups are replaced by two para methoxy group, carbocation formed will be
more stable
rX
d
40


2
rY 24  d
10
d = 48 – 2d
3d = 48
d  16cm
As the collision frequency increases then molecular speed decreases than the expected.
Solution for the Q. No. 33 to 34.
HO
C
NaNH
H
2

CH CH I
3
2


O
M
H3 CO
H , Lindlar 's catalyst
CH I
2


3
O

O
H
H
X
OH
Br
NaNH 2
H 
1 equivalent 
C


NaNH 2 1eq 
O
N
O
OH
H2 , Pd /C
H O


3

mild 
CrO
3

Y
OH
X and Y are functional isomers of each other and Y gives iodoform test.
(25)
Solution for the Q. No. 35 to 36.
HCl  Q 
KCN  R 
Ni 2 
  Ni  CN 4 
 NiCl4 2  
excess
excess
M 
1
Tetrahedral
 ZnCl4 
2
Tetrahedral
2
square planar
HCl Q 
KCN  R 

Zn 2  
  Zn  CN  4 
excess
excess
 M2 
2
Tetrahedral
KOH
 S
KOH
Zn  OH  2 
  Zn  OH  4 
excess
White ppt.
37.
(P)
(Q)
(R)
(S)
2
Soluble
[Cr(NH3)4Cl2]Cl  Paramagnetic and exhibits cis-trans isomerism
[Ti(H2O)5Cl](NO3)2  Paramagnetic and exhibits ionization isomerism
[Pt(en)(NH3)Cl]NO3  Diamagnetic and exhibits ionization isomerism
[Co(NH3)4(NO3)2]NO3  Diamagnetic and exhibits cis-trans isomerism
38.
P.

d – d  bonding
Q.

p – d  bonding

p – d  antibonding

d – d  antibonding
R.
S.
39.
(P) – 1; (Q) – 3; (R) – 4; (S) – 2
O
(P)

C6 H 5  C H 2  CO 2  CH 3O 
O
H 5 C6 H2 C
O
CH3
(Q)
O

O
H5 C6 H 2 C
O
CH3
O
CH3
|
C6 H5  C H 2  CO 2  Ph  CH 2  C  CH 3
|
CH3
CH2C6H5

Ph  C H 2  CH 3  CO  CH 3
(26)
O
(R)
C6 H5
O
CH3
O
CH3
O
|


C6 H5  C O 2  CH3  C  CH 3
|
CH3
Ph  CH 3  CO  Ph
-CO2
C6 H5
C6 H5
O
(S)

C6 H 5  C O2  CH 3O
O
C6H5
40.
C6 H5  CO2
CH3
O
(P)
NH2
NO2
red hot
3CH  CH 

iron873K
H S.NH
NaNO2


H SO
2
3


selective reduction 
fuming HNO /H SO
3
2
4



2
4
NO2
NO2
+
N2 HSO4-.
OH
Hydrolysis


NO2
(Q)
NO2
OH
OH
O2 N
O2 N

N O2
Conc.H 2SO 4


600 C
dil.H 2SO 4




OH
OH
SO3 H
NHCOCH3
NHCOCH3
NH2
NO2
OH
OH
SO3H
(R)
OH
OH
NHCOCH3
NO2
conc.H SO



HNO3

2
4


CH3COCl
Sn/ HCl
SO 3H
SO3 H
dil. H 2SO 4 
NHCOCH3
NH2
NO2
NO2
OH 


(S)
NO2
NO2
H O
KMnO 4


OH/ 
CH3
NO2
3


COO
NO2
NO2
SOCl2

COOH
NH
3


COCl
CONH2
(27)
41.
Either a girl will start the sequence or will be at second position and will not acquire the last position as
well.
 3 C1  2 C1  = 1 .
Required probability =
5
2
C2
42.
x=a+b
y = ab
x2  c2 = y
a 2  b2  c 2
1

   cos 120 
2ab
2
2
 C =
3
abc

R=
,r=
4
s
1
 2  
4  ab sin   
2
r
4
 3 
2



xc
R s  abc 
 yc
2
3y
r

.
R 2c  x  c 
2


43.
Number of required ways = 5! 4  4!  4 C2  3!  4 C3  2!  1 = 53.
44.
1

Area =  1  4  3   2  15
2


(2, 4)
(–1, 1)
(–1, –1)
(2, –4)
45.
P (x) = ax2 + b with a, b of same sign.
P(P(x)) = a(ax2 + b)2 + b
If x  R or ix  R
 x2  R
 P (x)  R
 P(P(x))  0
Hence real or purely imaginary number can not satisfy P(P(x)) = 0.

2
46.
17
  2 cosec x 
dx

4
(28)


 u = ln 1  2 , x =
u=0
4
2
eu  e  u
 cosec x + cot x = eu and cosec x – cot x = e–u  cot x 
2
(eu – e–u) dx = –2 cosec x cot x dx

Let eu + e–u = 2 cosec x, x =
 

eu  e  u
e
17


 e
ln 1 2

ln 1 2
=
u
16

 e u

du
du



2 eu  e u

 e u
2 cosec x cot x
0
= 2
u

16

du
0
47.
dy
x
x 4  2x
 2
y
dx x  1
1 x2
This is a linear differential equation
x
1
ln x 2 1
 2 dx
I.F. = e x 1  e 2
 1 x2
 solution is
x  x3  2
y 1 x2  
 1  x 2 dx
2
1 x
or y 1  x 2    x 4  2x  dx =
x5
 x2  c
5
f (0) = 0  c = 0
 f (x) 1  x 2 
x5
 x2
5
3/2
Now,
3/2
f  x  dx 

 3/2
3/2
= 2
x2

1 x2
0

 3 /2
 /3
dx = 2 
0
x2
1 x2
dx (Using property)
sin 2 
cos d (Taking x = sin )
cos 
 /3
 3 
3
  sin 2 

= 2  sin 2 d  2  
= 2    2 
.
  

4 0
2
6
0
 8  3 4
 /3
48.
F(0) = 0
F(x) = 2x f(x) = f(x)
f(x) = e x
2
c
2
f(x) = e x ( f(0) = 1)
x2
F(x) =
x
 e dx
0
2
F(x) = e x  1 ( F(0) = 0)
 F(2) = e4 – 1
49.
2x1 + 3x2 + 4x3 = 11
(29)
Possibilities are (0, 1, 2); (1, 3, 0); (2, 1, 1); (4, 1, 0).
 Required coefficients
= (4C0  7C1  12C2) + (4C1  7C3  12C0) + (4C2  7C1  12C1) + (4C4  7C1  1)
= (1  7  66) + (4  35  1) + (6  7  12) + (1  7)
= 462 + 140 + 504 + 7 = 1113.
50.
sin x + 2 sin 2x – sin 3x = 3
sin x + 4 sin x cos x – 3 sin x + 4 sin3 x = 3
sin x [–2 + 4 cos x + 4(1 – cos2 x)] = 3
sin x [2 – (4 cos2 x – 4 cos x + 1) + 1] = 3
sin x [3 – (2 cos x – 1)2] = 3
 sin x = 1 and 2 cos x – 1 = 0


 x  and x 
2
3
which is not possible at same time
Hence, no solution
51.
Case I : One odd, 2 even
Total number of ways = 2  2  3 + 1  3  3 + 1  2  4 = 29.
Case II: All 3 odd
Number of ways = 2  3  4 = 24
Favourable ways = 53
53
53
Required probability =

.
3  5  7 105
52.
Here 2x2 = x1 + x3
 x1 + x3 = even
Hence number of favourable ways = 2C1  4C2 + 1C1  3C1 = 11.
53.
Slope (QR) = Slope (PK)
2a
  2ar
2at  0
 t
a
at 2  2a
 ar 2
t2
 1

 r 
t
t2 1
t
 2
 
  r
t
t 2
 1  r2 
 2

t

54.
Tangent at P: ty = x + at2 or y =
x
 at
t
x 2a a


t
t t3
2a a
Solving, 2y = at +

t t3
Normal at S: y 


a t2 1
y=
55.
1
g    lim
 2  h0
2
2t 3
1 h

t 1/ 2 1  t 
1/ 2
dt
h
(30)
1
1
=
1
dt

tt
2
dt

1  1
t  
4  2
= sin1 1  sin1( 1) = .
0
0
2
1

t 2
1
= sin 

 1 
 2 0
56.
We have g (a) = g (1  a) and g is differentiable
1
Hence g    = 0.
2
57.
(P) f (x) = ax2 + bx,
1
 f  x  dx  1
0
 2a + 3b = 6
 (a, b)  (0, 2) and (3, 0).


(Q) f (x) = 2 cos  x 2  
4

For maximum value, x 2 
 x2 = 2n +

4

9
, 
as x    3,
4
4
x= 
2
(R)

 2n
4
 3x 2
3x 2
0  1  ex  1  e x
13  .
2

dx

3x 2 dx  8 .



0
1/ 2
(S)
58.
 1 x 
cos 2x ln 
 dx  0 as it is an odd function.
 1 x 
1/ 2

(P) y = cos (3 cos1 x)
3sin  3 cos 1 x 
y =
1 x2
1  x 2 y   3sin  3 cos 1 x 

x
2
y   1  x 2 y   3 cos  3 cos 1 x  .
1 x
  xy + (1  x2) y =  9y
1
  x 2  1 y   xy   9 .
y
3
1  x2
2
n
2

ak  ak+1 = r 2 cos
n
n 1
n 1
 
  a k  a k 1   a k  a k 1
(Q) (ak  ak+1) = r 2 sin
k 1
k 1
2
2
 r2 (n  1) sin
 r 2  n  1 cos
n
n
(31)
2
8
1  n =
n
4k  1
 n = 8.
h 2 12
(R)
 1, h =  2
6 3
2x y
Tangent at (2, 1) is
  1 x + y = 3.
6 3
1
2
 1 
1
(S) tan 1 
 tan 1 2
  tan
4x  1
x
 2x  1 
 3x  1 
1 2
tan 1  2
  tan
x2
 4x  3x 
 3x2  7x  6 = 0
2
x =  , 3.
3
tan
59.
 x 2 , x  0
f2 (f1) = 
2x
, x 0
e
f4: R  [0, )
f 2  f1  x  
, x0
f4(x) = 
f 2  f1  x    1 , x  0
 x 2
, x0
= 
2x
e  1 , x  0
y =f1(x)
y = f3(x)
y = f2(x)
0
y =f2(f1(x))
y = f4(x)
60.
(P) zk is 10th root of unity  zk will also be 10th root of unity. Take zj as zk .
z
(Q) z1  0 take z = k , we can always find z.
z1
(R) z10  1 = (z  1) (z  z1) … (z  z9)
 (z  z1) (z  z2) … (z  z9) = 1 + z + z2 + … + z9  z  complex number.
(32)
Put z = 1
(1  z1) (1  z2) … (1  z9) = 10.
(S) 1 + z1 + z2 + … + z9 = 0
 Re (1) + Re (z1) + … + Re (z9) = 0
 Re (z1) + Re (z2) + … + Re (z9) =  1.
9
2k
 1   cos
2.
10
k 1
(33)
D.
17.
Marking Scheme
For each question in Section 1, 2 and 3 you will be awarded 3 marks if you darken only the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus
one (1) mark will be awarded.
Appropriate way of darkening the bubble for your answer to be evaluated:
a
The one and the only one acceptable
a
Part darkening
a
a
Answer will not be
evaluatedno marks, no
negative marks
Darkening the rim
a
Cancelling after darkening
a
Erasing after darkening
Figure-1 : Correct way of bubbling for a valid answer and a few examples of invalid answers.
Any other form of partial marking such as ticking or crossing the bubble will be invalid.
5
0
0
4
5
2
3
1
0
0
0
0
0
1
1
1
1
1
1
2
2
2
3
3
4
4
2
2
3
3
3
4
4
4
4
5
5
5
5
5
6
6
6
6
6
6
7
7
7
7
7
7
7
9
9
9
9
8
9
8
9
9
Figure-2 : Correct Way of Bubbling your Roll Number on the ORS. (Example Roll Number : 5045231)
Roll Number
Name of the Candidate
I have read all instructions and shall abide
by them.
……………………………………………
Signature of the Candidate
I have verified all the information filled by
the candidate.
……………………………………………
Signature of the invigilator

IIT JEE-Advanced 2014 Solved papers by Triumph Academy

Transcription

Similar documents

Respect - Status - Stability

Solid State Chemistry and Allied Areas

Instructions for Admission to Preschool (Session

Announcement of opening of branch office in Delhi,India

FAX COVER SHEET No. of Pages Date To

UPAI study

REPORT ON DELHI BOOK FAIR

Verification of current employment Fax Cover Sheet Metropolitan

Sharam Aur Besharam

CUSTOMER/CREDIT APPLICATION CONTACT INFORMATION