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CODE : 8
IIT - JEE 2015 (Advanced)
(2) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
PART I : PHYSICS
Section 1 (Maximum Marks : 32)
This section contains EIGHT questions
...
For each question, darken the bubble corresponding to the correct integer in the ORS
...
The energy of a system as a function of time t is given as E(t) = A 2 exp(t), where
= 0
...
The measurement of A has an error of 1
...
If the error in the measurement
of time is 1
...
[4]
E(t) = A2 exp (t)
(E) = 2A (A) exp (t) + A2 exp ( t)
(E) = 2A exp ( t) A + A2 exp (t) (t)
(E) = 2A exp (t) A + A2 exp (t) tt
(E)
A
2
(t)20
...
20
...
0250 + 0
...
040
% error = 4%
(E)
2
...
The moments
R
R
of inertia of the individual spheres about axes passing through their centres are IA and IB,
I
n
respectively
...
[6]
x
x
R
V = 4 x2 x
(m) = 4(x) x2
...
Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,
/3, 2/3 and
...
The
value of n is
3
...
For a radioactive material, its activity A and rate of change of its activity R are defined as
dN
dA
A=
and R =
, where N(t) is the number of nuclei at time t
...
Their rates of
R
n
change of activities at t = 2 are RP and RQ, respectively
...
[2]
P and Q have same activity at t = 0
...
A monochromatic beam of light is incident at 60° on one face of an equilateral prism of
refractive index n and emerges from the opposite face making an angle (n) with the
d
normal (see the figure)
...
The value of m is
dn
5
...
In the following circuit, the current through the resistor R (=2) is I Amperes
...
[1]
The given Ckt reduces to simple Ckt like this
2
2
1
2
6
2
2
6
4
6
...
5
4
12
4
4
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (5)
I
2
2
2
6
4
...
5
12
6
...
An electron in an excited state of Li2+ ion has angular momentum 3h/2
...
The value of
p is
7
...
A large spherical mass M is fixed at one position and two identical point masses m are
kept on a line passing through the centre of M (see figure)
...
All three masses interact only through their mutual gravitational
interaction
...
The value of k is
288
8
...
Each questions has FOUR options (A), (B), (C) and (D)
...
A parallel plate capacitor having plates of area S and plate separation d, has capacitance
C1 in air
...
the
C
ratio 2 is
C1
(A) 6/5
9
...
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as
shown in the figure)
...
The gas is then heated very slowly to temperature T2,
pressure P2 and volume V2
...
Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)
1
P1V1
4
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
7
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is P1V1
3
17
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
P1V1
6
10
...
2V1P1V1 = 3P1V1
[B is correct]
2 2
PV
3V
4
(C) 1 1 = P2 1
P2 = P1
T1
4T1
3
Sx
1
P2S = Kx
P2
= kx 2 = w
2
2
1 4
w = P1 (V2 V1 )
2 3
2
4
[C is incorrect]
P1 (3V1 V1 ) = P1V1
3
3
4
3 4
4
3
4
9
(D) P1V1 (P2 V2 P1V1 ) = P1V1 P13V1P1V1 = P1V1 P1V1
3
2
3
2
3
2 3
8P V 27P1V1
35
= 1 1
= P1V1
[D is incorrect]
6
6
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is
7
(8) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
11
...
54
236
Considering 92 U to be at rest, the kinetic energies of the products are denoted by KXe,
KSr, Kx (2 MeV) and Ky (2 MeV), respectively
...
5 MeV, 8
...
5 MeV, respectively
...
(A)
Q Value of the reaction
Q = 94 8
...
5 236 7
...
Heavier particle will
have less kinetic energy
...
In plotting stress versus strain curves for two materials P and Q, a student by mistake puts
strain on the y-axis and stress on the x-axis as shown in the figure
...
(A), (B)
P
Strain
Strain
Q
P
Q
Stress
Stress
8
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (9)
For same strain stress in Q is more than P
...
(D) is not correct
...
P is more ductile
(A), (B)
13
...
If P(r) is the pressure at r(r < R), then the correct option(s) is(are)
P(r3R / 4) 63
(A) P(r = 0) = 0
(B)
P(r2R / 3) 80
P(r3R / 5) 16
P(rR / 2) 20
(C)
(D)
P(r2R / 5) 21
P(rR / 3) 27
13
...
Two spheres P and Q of equal radii have densities 1 and 2, respectively
...
They float in equilibrium with the sphere P in L1 and
sphere Q in L2 and the string being taut (see figure)
...
(A), (D)
(UP WP)
( 1 )
( 2 )
P 1
Q a
2
1
p
2
1 1
a 2 2 1
…(i)
(D) VP VQ 0
T
P
For Eq
...
(ii)
=
From (i) and (ii) option (A) follows
Also it is obvious from (i) & (ii) p &1 have opposite signs
(WQ UQ)
D is true
...
In terms of potential difference V, electric current I, permittivity 0, permeability 0 and
speed of light c, the dimensionally correct equation(s) is(are)
(A) 0I2 = 0V2
(B) 0I = 0V
(C) I = 0cV
(D) 0cI = 0V
15
...
Consider a uniform spherical charge distribution of
radius R1 centred at the origin O
...
If the electric
field inside the cavity at position r is E(r) , then the
correct statement(s) is (are)
(A) E is uniform, its magnitude is independent of R2 but its direction depends on r
(B) E is uniform, its magnitude depends on R2 and its direction depends on r
(C) E is uniform, its magnitude is independent of but its direction depends on
(D) E is uniform and both its magnitude and direction depend on
10
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (11)
16
...
Option (D) only is correct
...
The light guidance in the structure takes place due to successive total
internal reflections at the interface of the media n1 and n2 as shown in the figure
...
The numerical aperture (NA) of the structure is defined as sin im
...
For two structures namely S1 with n1 = 45 / 4 and n2 =3/2, and S2 with n1 = 8/5 and
n2 = 7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the
correct option(s) is (are)
(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of
16
refractive index
...
(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index
4
...
17
...
9
16 = 9
4
16
3
15
5 = 153 = 9
16
165 16
3 15
Option (B) :
NA1
NA2
=
3
4 = 3 15 =
6
4 6
15
=
15
5 = 3 15
4
20
3
Option (B) is incorrect
...
Option (D) : is incorrect at (C) is correct
...
If two structures of same crosssectional area, but different numerical apertures NA1 and
NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined
structure is
NA1NA 2
(A)
(B) NA1 NA 2
(C) NA1
(D) NA2
NA1 NA 2
18
...
[smaller NA]
sin i2 < NA2
sin i1 > sin i2
NA of combination = smaller NA
Section 3 (Maximum Marks : 16)
This section contains TWO paragraph
...
ONE OR MORE THAN ONE
of these four option(s) is(are) correct
For each question, darken the bubble(s) corresponding to all the correct option(s) in the
ORS
Marking scheme :
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened
0 If none of the bubbles is darkened
2 In all other cases
PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive xdirection,
as shown in the figure
...
A uniform magnetic field B is applied on the strip along the positive ydirection
...
This results in
accumulation of charge carriers on the surface PQRS and appearance of equal and opposite
charges on the face opposite to PQRS
...
Charge accumulation continues until the magnetic force is balanced by the
electric force
...
13
(14) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
19
...
Their lengths are the
same, widths are w1 and w2 and thicknesses are d1 and d2, respectively
...
V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively
...
(A), (D)
I
I
J=
= nevj
evd =
mwd
wd
IB
FB = evdB =
mwd
V
eV
E=
FG =
w
w
Under equation condition
IB
eV
IB
=
V=
nwd
w
nde
Vd is same in two strips
...
Consider two different metallic strips (1 and 2) of same dimensions (length , width w
and thickness d) with carrier densities n1 and n2 respectively
...
Then
V1 and V2 are the potential differences developed between K and M in strips 1 and 2,
respectively
...
5V1
(D) If B1 = 2B2 and n1 = n2, then V2 = V1
20
...
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both
inclusive
...
Marking scheme :
+4 If the bubble corresponding to the answer is darkened
0 In all other cases
14
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (15)
21
...
II
...
IV
...
[4]
CHO
(I)
CO, HCl
Anhydrous AlCl /CuCl
3
CHO
Cl
(II)
OH
H2O
CH Cl
100C
CH OH
(III)
H2
Cl
Pd BaSO 4
C H
O
O
(IV)
C
2
O
O
C
H O
O
DIBAL H
Me
Toluene, 78C H2O
C
H
+ MeOH
22
...
[3]
23
...
[6]
NH2
Cl
NH2
NH2
NH2
NH2
NH2
+3
Co
NH2
Cl
trans
+3
Co
NH2
Cl
Cl
cis
24
...
The number of moles of
boron containing product formed is
24
...
25
...
01 M) is 10 times smaller than
the molar conductivity of a solution of a weak acid HY (0
...
If 0 0 , the
X
Y
difference in their pKa values, pKa(HX) pKa(HY), is (consider degree of ionization of
both acids to be << 1)
25
...
A closed vessel with rigid walls contains 1 mol of
238
Considering complete decay of 92 U to
pressure of the system at 298 K is
206
82 Pb,
238
92 U and
1 mol of air at 298 K
...
[9]
238
4
0
206
92 U 82 Pb8 2He6 1
Initial no
...
Final no
...
In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by MnO
...
[8]
Rate of change of [H+] is 8 times the rate of change of MnO
4
16
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (17)
28
...
[4]
H
H
H
HO
CH3
CH3
H 2O
H2O
CH3
CH3
CH3
CH3
1,2 methyl shift
OH
CH3
HO
Aq
...
Each questions has FOUR options (A), (B), (C) and (D)
...
Under hydrolytic conditions, the compounds used for preparation of linear polymer and
for chain termination, respectively, are
(A) CH3SiCl3 and Si(CH3)4
(B) (CH3)2SiCl2 and (CH3)3SiCl
(C) (CH3)2SiCl2 and CH3SiCl3
(D) SiCl4 and (CH3)3SiCl
29
...
When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2
...
(A), (B), (C), (D)
17
(18) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
31
...
The relationship of interatomic potential V(r) and interatomic distance r for the
gas is given by
(A)
(B)
(C)
(D)
31
...
In the following reactions, the product S is
(A)
(B)
(C)
(D)
32
...
The major product U in the following reactions is
(A)
(B)
(C)
(D)
33
...
In the following reactions, the major product W is
(A)
(B)
(C)
(D)
19
(20) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
34
...
The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is
(are)
(A) The number of Cl = O bonds in (ii) and (iii) together is two
(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
(C) The hybridization of Cl in (iv) is sp3
(D) Amongst (i) to (iv), the strongest acid is (i)
35
...
...
(ii) HOClO
(iii) HOClO
O
O
(iv) HOClO
O
(A) False (B) True
(C) True
(D) False
36
...
(C) , (D)
Group2 ions are Hg+2, Pb+2, Bi+3, Cu+2, Cd+2, As+3, Sb+3, Sn+2, and Sn+4
Section 3 (Maximum Marks : 16)
This section contains TWO paragraph
...
ONE OR MORE THAN ONE
of these four option(s) is(are) correct
For each question, darken the bubble(s) corresponding to all the correct option(s) in the
ORS
Marking scheme :
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened
0 If none of the bubbles is darkened
2 In all other cases
20
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (21)
PARAGRAPH 1
When 100 mL of 1
...
0 M NaOH in an insulated beaker
at constant pressure, a temperature increase of 5
...
1)
...
0 kJ mol1), this experiment could be used to measure the calorimeter
constant
...
2), 100 mL of 2
...
0 105) was mixed
with 100 mL of 1
...
1) where a temperature rise
of 5
...
(Consider heat capacity of all solutions as 4
...
0 g mL1)
37
...
2 is
(A) 1
...
0
(C) 24
...
4
37
...
The pH of the solution after Expt
...
8
(B) 4
...
0
(D) 7
...
(B)
200 m
...
+ 100 m
...
eq
...
3010 = 4
...
Compound X is
(A)
(B)
(C)
39
...
The major compound Y is
(A)
(B)
(C)
(D)
40
...
39 & 40
C
CH
CH
Pd
...
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both
inclusive
...
Marking scheme :
+4 If the bubble corresponding to the answer is darkened
0 In all other cases
41
...
P
...
If the
ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and
the seventh term lies in between 130 and 140, then the common difference of this A
...
is
41
...
S7
6
=
S11 11
7
2a 6d
6
2
=
11
2a 10d 11
2
2a 6d
6
=
2a 10d
7
22
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (23)
14a + 42 d = 12a + 60d
2a = 18 d
a = 9d
130 < a7 < 140
130 < a + 6d < 140
130 < 15d < 140
d=9
42
...
(1 + x100) is
42
...
(1 + x100)
co-eff of x9 is number of ways sum of power of x is 9
...
x 2 y2
43
...
Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at
(f1, 0) and (2f2, 0), respectively
...
If m1 is the slope of T1 and m2 is
1
the slope of T2, then the value of 2 m2 is
2
m1
43
...
, 0 )
3
e = 1
Focus ( ae, 0) = (
= ( 2, 0)
f1 = (2, 0), f2 = (2, 0)
Parabola P1 = y2 = 4
...
Let m and n be two positive integers greater than 1
...
[2]
cos n e
e ecos 1 1
e
e
lim
= lim
=
= = 0
m
m
0
2
2
n
e ecos
= lim
0
m
n 1
e ecos 1 1 n
e
e
= = lim m
cos 1 =
2 0 cos n 1
2
n
n
n
e 2 sin 2
e 2 sin 2
n 2
n 2
e
2
2 = e
lim
= lim
= = 0
2
2
0
2
2
2
2
n
n
m
...
[9]
1
45
...
Let f :
be a continuous odd function, which vanishes exactly at one point and
x
x
1
f 1
...
If lim
1
f is
2
46
...
Suppose that p,q and r are three non-coplanar vectors in 3
...
If the components of this vector s
along p q r , p q r and p q r are x, y and z, respectively, then the value
of 2x + y + z is
47
...
For any integer k, let a k cos i sin , where i 1
...
[4]
12
3
i(K 1)
K
i
7
e
e 7
K 1
i(4K 1)
(4K 2)
i
7
e
e 7
K 1
12
e
=
i
K
7
e
i
7 1
K 1
3
i(4K 1)
e 7
K 1
i
e 7 1
12
=
K 1
3
K 1
25
i
e 7 1
i
e 7 1
=
i
12 e 7 1
3
i
e 7 1
=4
(26) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
Section 2 (Maximum Marks : 32)
This section contains EIGHT questions
...
ONE OR MORE THAN ONE
of these four option(s) is(are) correct
For each question, darken the bubble(s) corresponding to all the correct option(s) in the
ORS
Marking scheme :
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened
0 If none of the bubbles is darkened
2 In all other cases
49
...
Let the values of f and g at the points 1, 0 and 2 be as given in the
following table :
x=2
x = 1 x = 0
f(x)
3
6
0
g(x)
0
1
1
In each of the intervals (1, 0) and (0, 2) the function (f 3g) never vanishes
...
(B), (C)
Let h(x) = f(x) 3g(x)
h(1) = 3
h (0) = 3
h (2) = 3
h(x) = f(x) 3g(x)
f(x) 3 g(x) = 0 has exactly one solution is (1, 0)
& f(x) 3g(x) = 0 has exactly one solution is (0, 2)
1
0
2
50
...
Then the correct
2 2
expression(s) is (are)
4
(A)
0
4
(C)
1
x f x dx
12
1
x f x dx 6
4
(B)
f x dx 0
0
4
(D)
0
f x dx 1
0
50
...
sec2x 3 tan2 (x)
...
0
4
= x(tan xtan x) 0 (tan 7 (x)tan 3(x))
...
0
4
tan 4 (x) tan 6 (x)
1 1
1
=
= =
6 0
4 6 12
4
192x 3
51
...
If m
2
1
f (x)dxM , then the
t/2
1
1
,M=
4
2
(D) m = 1, M = 12
(A) m = 13, M = 24
(B) m =
(C) m = 11, M = 0
51
...
12
52
...
Which of the following intervals is (are) a subset(s) of S?
1
1
1
(A) ,
(B)
,0
5
5
2
1
(C) 0,
5
1 1
(D)
,
5 2
27
(28) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
52
...
> 0
1
2 0
4
1
2 0
4
1
1
2
2
x1 + x2 =
1
, x1 x2 = 1
x1 x 2
x1 x 2 =
2
4x1x 2
=
1
4
2
as x1 x 2 1
1
4 < 1
2
1
4 1
2
1
5
2
1
2
5
2
1
0
5
1
2
<
5
or >
1
5
Take intersection
1
2
,
1 1 1
,
5 5 2
6
4
53
...
(B), (C), (D)
1 6
1
6
= 3sin 1 as
11
2 11
2
6
sin 1
6
11 4
6 3
3sin 1
2
11 4
3
2
4
4 1
4
= 3cos 1 as
9 2
9
28
IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (29)
4
1
cos1 cos 1
2
9
4
cos 1
9 3
4
3cos1
9
cos < 0, sin < 0, cos < 0, cos < 0, sin > 0
cos ( + ) = cos cos sin sin > 0
54
...
The major axes of E1 and E2
lie along the x-axis and the y-axis, respectively
...
The
straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively
...
If e1 and e2 are the eccentricities of E1 and E2,
3
respectively, then the correct expression(s) is (are)
7
43
2
(A) e1 e2
(B) e1 e2
2
40
2 10
3
5
2
(C) e1 e2
(D) e1 e 2
2
4
8
54
...
(1) and (2) solving P (1, 2)
For point Q and R,
2 2
x 1
y2
=
=
1
1
3
2
2
5 4
1 8
Q , and R ,
3 3
3 3
x 2 y2
As normal y = x + 3 to 2 2 1 is
a
b
c2 = a2 m2 + b2
9 = a2 + b2
(3)
2
2
x
y
5 4
As point Q , lies E1 : 2 2 = 1
a
b
3 3
25 16
=1
9a 2 9b2
a2 = 5
29
(30) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution
From (3), b2 = 4
e1 = 1
4
b2
= 1
2
5
a
=
1
5
Similarly,
e2 = 1
1
8
=
7
2 2
7
2 10
1 7
8 35
43
2
=
=
e1 e2 =
2
5 8
40
40
1 7
8 35
27
2
2
=
| e1 e2 | = =
5 8
40
40
e1 e2 =
55
...
Suppose that H
and S touch each other at a point P(x1, y1) with x1 > 1 and y1 > 0
...
If ( , m) is the centroid of the triangle
PMN, then the correct expression(s) is(are)
x1
dm
d
1
forx1 1
(A)
(B)
1 2 forx1 1
dx1 3 x 2 1
dx1
3x1
(C)
d
1
1 2 forx1 1
dx1
3x1
(D)
dm 1
fory1 0
dy1 3
55
...
The option(s) with the values of a and L that satisfy the following equation is(are)
4
e (sin
t
6
atcos 4 at)dt
e (sin
6
atcos at)dt
0
= L?
t
4
0
e4 1
e 1
e4 1
(C) a = 4, L =
e 1
56
...
Based on each paragraph, there will be TWO questions
Each question has FOUR options (A), (B), (C) and (D)
...
Suppose that F(1) = 0, F(3) = 4 and
F(x) < 0 for all x (1/2, 3)
...
57
...
(A), (B), (C)
f (x) = x F(x)
f(x) = F(x) + x F(x)
f(1) = F(1) + F(1) = F(1) < 0
( F(x) < 0 x (1/2, 3)
f (2) = 2 F (2) < 0
( F (2) < 0)
f(x) = F(x) + x F(x) < 0 x (1, 3)
( F (x) < 0 and x F (x) < 0 for all x (1, 3))
f(x) 0 for any x (1, 3)
Hence (A), (B), (C) are correct
...
If
3
x F'(x) dx = 12 and x F"(x) dx = 40, then the correct expression(s) is(are)
2
3
1
1
3
(B) f (x)dx = 12
(A) 9f(3) + f(1) 32 = 0
1
3
(D)` f (x)dx = 12
(C) 9f(3) f(1) + 32 = 0
1
58
...
Let n3 and n4 be the
number of red and black balls, respectively, in box II
...
One of the two boxes, box I and box II, was selected at random and a ball was drawn
randomly out of this box
...
If the probability that this red ball
1
was drawn from box II is , then the correct option(s) with the possible values of n1, n2,
3
n3 and n4 is(are)
(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15
(B) n1 = 3, n2 = 6, n3 = 10, n4 = 50
(C) n1 = 8, n2 = 6, n3 = 5, n4 = 20
(D) n1 = 6, n2 = 12, n3 = 5, n4 = 20
59
...
A ball is drawn at random from box I and transferred to box II
...
(C), (D)
P (E) =
1
3
n1
n1 n1 1 n 2
1
n n n n 1 n n n n 1 = 3
2 1
2
2 1
2
1
1
Among the given options, (C) and (D) satisfies the equation