PAPER – 2

Transcription

PAPER – 2

JEE – Advanced 2015
Paper – 2 (Hints & Solution)
®
K. SINGH
ISION CLASSES
JEE MAIN + ADVANCED/AIPMT
JEE ADVANCED – 2015 (PAPER – 2) (CODE–7) Hints & Solution
Part – I (Physics)
SECTION - I (One Integer Value Correct Type)
This section contains 8 questions. Each questions , when worked out will result in one integer
from 0 to 9 (both inclusive).
01.
In the following circuit, the current through the resistor R(  2) is I Amperes. The value of I is
Ans. [1]
Sol. By removing 8 and 10 from the circuit because it is balanced wheat stone bridge
Req.  2 
i
02.
6  18 13

6  18
2
6.5  2
1
13
A electron in an excited state of Li2+ ion has angular momentum 3h / 2 . The de Broglie wavelength of the
electron in this state is pa0 (where a0 is the Bohr radius). The value of p is
Ans. [3]
Sol.
L  mvr 
r
3h
2
9
a0  3a0
3
K. Singh Vision Classes, Opp. Gate No.-2, Kendriya Vidyalaya, Kankarbagh, Patna-20; Vision Tower, Po st
Office Lane, Near P.N. Anglo School Naya Tola, Patna-4; Jai Kamla Plaza, Near Lalita Hotal, East boring Canal
Road, Patna-1 ;
3rd Floor, Karni Heights, 1-Club Road, Ranchi; Kabir Nagar Road, Durgakund,
Varanasi (U.P.)
1

h
 2a0
mv
 P2
03.
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). The point masses are connected by a rigid massless rod of length l and this
assembly is free to move along the line connecting them. All three masses interact only through their mutual
gravitational interaction. When the point mass nearer to M is at a distance r = 3l from M, the tension in the rod is
 M 
zero for m  k 
 . The value of k is
 288 
Ans. [7]
Sol. Appling NLM on the particle as system
GMm
2
(3 )
a

GMm
(4)2
 (2m)a
GM  25 


22  16  9 
Appling NLM on the particle nearer to sphere
GMm
(3)2
m
04.

GMm
( 2 )
 ma
7M
288
The energy of a system as a function of time t is given as E(t)  A 2 exp(t), where   0.2 s1. The measurement
of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of
E(t) at t = 5 s is
Ans. [4]
Sol.
A 1.25 t 1.5

,

A
100
t
100
E
 A  t
 2
  t t
E
 A 
=4
05.
The densities of two solid spheres A and B of the same radii R vary with radial distance r as
5
r 
r 
 A (r)  k   and B (r)  k   , respectively, where k is a constant. The moments of inertia of the individual
R
 
R
spheres about axes passing through their centre are IA and IB respectively. If
IB
n

, the value of n is
IA 10
Ans. [6]
2
K. Singh Vision Classes, Opp. Gate No.-2, Kendriya Vidyalaya, Kankarbagh, Patna-20; Vision Tower, Post Office
Lane, Near P.N. Anglo School Naya Tola, Patna-4; Jai Kamla Plaza, Near Lalita Hotal, East boring Canal Roa d,
Patna-1;
3rd Floor, Karni Heights, 1-Club Road, Ranchi; Kabir Nagar Road, Durgakund, Varanasi (U.P.)
Sol.
r 
A  K  
R 

B (r)  K

dm  4x2  dx  
IA
Kx
;
R
dI 
r5
R5
2
Kx 2
4x 2 dx
x
3
R
x
R
 dI 
0
8 K 5
x dx
3 R 
0
IA 
8 K  R6  48 KR5 4KR5

 
;
3 R  6  3
6
9
 dI 
R
8 
8K  R10 
K  x 9 dx 

;
3 R5
3R5  10 
0
dI 
2
Kx5
 4x 2 dx 5
3 
R
IB 
8KR5 4KR5 4KR5


30
15
15
 2
x


IB
n
21


9  6
IA 10 15
06.
Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, /3, 2/3 and . When
they are superposed, the intensity of the resulting wave is nI0. The value of n is
Ans. [3]
Sol. f  equal
2/3
/3
I0  equal
I 

3A

O
A
2
2
I  3  A 
07.
For a radioactive material, its activity A and rate of change of its activity R are defined as A  
dN
dA
,
and R  
dt
dt
where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life  ) and Q (mean life 2 ) have
RP n
the same activity at t = 0. Their rates of change of activities at t = 2 are RP and RQ, respectively. If R  e ,
Q
then the value of n is
Ans. [2]
Sol.


A  A 0 1  et ;

t
P
RP  A 0 P e

t ;
RQ

Q
A 0 Q e
R  A 0e
P 2

Q 1 ;

t

RP
2
RQ
Road, Patna-1 ;
Varanasi (U.P.)
3
08.
A monochromatic beam of light is incident at 60° on one face of an equilateral
prism of refractive index n and making an angle (n) with the normal (see the
d
 m. The value of m is
figure). For n  3 the value of  is 60° and
dn
Ans. [2]
Sol.
sin60  nsin 1
nsin(60  1)  1sin 
3
sin 1 
2n
 3

1
n
cos 1  sin 1   sin .
2
 2

cos 1 
  sin1
d

dn
4n2  3
2n
60°
1
60°
 3
4n2  3 1
3
  sin .
n

 
2n
2 2n 
 2


3
4n2  3  1

4 
1
1
60°
3 
4n2  3  1
16 

2

3
4


1

 8n 
 2 4n2  3

1

1
3
3  12
16

1
2
3 8  3 
 2


4  2(3) 
3 1
1
4
SECTION - II (One or More than one Option Correct Type)
This section contains 8 multiple choice questions. Each questions has 4 choices (a), (b), (c) and (d)
out of which only one & more than one is correct.
09.
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. Then the correct statement(s) is (are)
(A) P has more tensile strength than Q
(B) P is more ductile than Q
(C) P is more brittle than Q
(D) The Young’s modulus of P is more than that of Q
Ans. (A,B)
Sol.
From Theory
10.
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If
P(r) is the pressure at r(r < R), then the correct option(s) is(are)
(A) P(r = 0) = 0
(B)
P(r  3R/4) 63

P(r  2R/3) 80
(C)
P(r  3R/5) 16

P(r  2R/5) 21
(D)
P(r  R/2) 20

P(r  R/3) 27
Ans. (B,C)
4
Patna-1;
Sol.
dP  dA  E  dA  dr
P
P
r
r
  dP   Krdr
0
P+dP
dA
dr
R
P  (R2  r 2 )
Hence B & C
11.
A parallel plate capacitor having plates of area S and plate separation d, has
capacitance C1 in air. When two dielectrics of different relative permittivities
(1  2 and 2  4) are introduced between the two plates as shown in the
figure, the capacitance becomes C2. The ratio
(A) 6/5
C2
is
C1
(B) 5/3
(C) 7/6
(D) 7/3
Ans. (D)
2 0
Sol.
CA 
d
2
S
2  2C
1
S
4 0
2  4C
CB 
1
d
2
CA CB
K=2 K=4
CC
CD
K=2
K=2
CC = CA = 2C1
CD = CA = 2C1
C2 
12.
4
7
C1  C1  C1
3
3
.... They are in parallel
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure).
Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is
then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out
by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is
1
P1V1
4
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is
7
P1V1
3
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
17
P1V1
6
Ans. (A,B,C)
Road, Patna-1 ;
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5
Sol.
P2 A  P1A  Kx  (P2  P1)A  Kx  Ax 
Kx 2
P2  P1
...(1)
P1v1 P2 (v1  Ax)

T1
T2
P1v1T2  P2 v1T1
 Ax
P2 T1
...(2)
From (1) and (2)
(P v T  P2 v1T1 )
Kx 2
 1 1 2
(P2  P1)
P2 T1

1 2 (P2  P1 ) (P1v1T2  P2 v1T1 ) 1  P1   P1v1T2
 1   
 P2 v1 
Kx 
2  P2   T1
2
2P2
T1

2
1  P1   P1v1T2 P1v1 T2 

1




Energy stored in spring 2  P 
T1v 2 

2   T1
From given data correct options A,B,C
13.
 x  y, where x and y are two particles. Considering 236
92 U
to be at rest, the kinetic energies of the products are denotes by KXe, KSr, Kx(2 MeV) and Ky(2 MeV), respectively.
A fission reaction is given by
236
140
94
92 U  54 Xe  38 Sr
and 94
38 Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively..
Considering different conservation laws, the correct option(s) is(are)
Let the binding energies per nucleon of
236
140
92 U, 54 Xe
(A) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV
(B) x = p, y = e–, KSr = 129 MeV, KXe = 86 MeV
(C) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV
(D) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV
Ans. (A)
Sol.
236
140
94
92 U  54 Xe  38 Sr
(B.E)
236
92 U
= 7.5 × 236 = 1770;
(B.E)
94
38 Sr
= 8.5 × 94 = 799
Q  K Xe  K Sr  4 ;
14.
 x  y,
219  K Xe  K Sr  4 ;
(B.E) 140
54 Xe = 8.5 × 140 = 1190
K Xe  K Sr  215
Two spheres P and Q of equal radii have densities 1 and 2 , respectively. The spheres are
connected by a massless string and placed in liquids L1 and L2 of densities 1 and 2 and
viscosities 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). If sphere P alone in L2 has terminal velocity VP and

Q alone in L1 has terminal velocity VQ , then

VP
1
(A)   
VQ
2

VP
2
(B)   
VQ
1
 
(C) VP  VQ  0
 
(D) VP  VQ  0
Ans. (A,D)
6
Patna-1;
Sol.
Vt 
1
;



VP is upward and VQ is downward,
 
hence VP  VQ  0
15.
In terms of potential difference V, electric current I, permittivity 0 , permeability µ0 and speed of light c, the
dimensionally correct equation(s) is(are)
(A) µ0I2  0 V 2
(B) 0I  µ0 V
(C) I  0 cV
(D) µ0 cI  0 V
Ans. (A,C)
Sol.
V  ML2 T 3 A 1 ;
I=A
0  M1L3 T 4 A 2 ;
µ0 
1
0 C2
Check dimensional correctness in each option
16.
Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a
spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the electric field
 
inside the cavity at position r is E(r ), then the correct statement(s) is(are)

(A) E

(B) E

(C) E

(D) E
Ans. (D)
Sol.
is uniform, its magnitude is independent of R2 but its direction depends on r
is uniform, its magnitude depends on R2 and its direction depends on r

is uniform, its magnitude independent of a but its direction depends on a

is uniform and both its magnitude and direction depend on a


a
E
30
SECTION - III (Linked Comprehension Type)
This section contains 2 Paragraphs. Based upon the paragraph 2 multiple choice questions have to
be answered. Each question has 4 choices (a), (b), (c) and (d) out of which one or more than one is/
are correct.
Passage - I
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure.
The length, width and thickness of the strip are l, w and d, respectively.

A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers
experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface
PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along
the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the
electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by
electrons.
Road, Patna-1 ;
Varanasi (U.P.)
7
17.
Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, width are w1 and
w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite
faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1
and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the
correct statement(s) is(are
(A) If w1 = w2 and d1 = 2d2, then V2 = 2V1
(B) If w1 = w2 and d1 = 2d2, then V2 = V1
(C) If w1 = 2w2 and d1 = d2, then V2 = 2V1
(D) If w1 = 2w2 and d1 = d2, then V2 = V1
Ans. (A,D)
Sol.
E  e  ev dB
E  v dB
V
 v dB ;
w
18.
V
IBw IB

nAe nd
Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier
densities n1 and n2 respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2,
both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in
strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s)
is(are)
(A) If B1 = B2 and n1 = 2n2, then V2 = 2V1
(B) If B1 = B2 and n1 = 2n2, then V2 = V1
(C) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
(D) If B1 = 2B2 and n1 = n2, then V2 = V1
Ans. (A,C)
Passage - II
Light guidance in an optical fiber can be understood by considering
a structure comprising of thin solid glass cylinder of refractive index
n1 surrounded by a medium of lower refractive index n2. The light
guidance in the structure takes place due to successive total
internal reflections at the interface of teh media n1 and n2 as shown
in the figure. All rays with the angle of incidence i less than a
particular value im are confined in the medium of refractive index
n1. The numerical aperture (NA) of the structure is defined as sin
im.
19.
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 refractive index
(B) NA of S1 immersed in liquid of refractive index
6
15
16
3 15
.
is the same as that of S2 immersed in water
(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index
4
15
.
(D) NA of S1 placed in air is the same as that of S2 placed in water.
Ans. (A,C)
8
Patna-1;
1  3 
Sol. For S1   cos 

 45 
1  15 
For S2   cos  8 


(A)
(B)
4
sinim 
3
6
15
45
3

;
4
45
sinim 
16
3 15
sinim 
8
15

5
8
45
3 4
8
15

; sinim  
4
3
5
8
45
(C) 1sinim 
45
3

;
4
45
(D) 1sinim 
45
3

;
4
45
4
15
sinim 
8
15

5
8
4
8
15
sinim  
3
5
8
Calculating sin im
20.
If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2 (NA2 < NA1) are
joined longitudinally, the numerical aperture of the combined structure is
NA1 NA 2
(A) NA  NA
1
2
(B) NA1 + NA2
(C) NA1
(D) NA2
Ans. (D)
Sol. Smaller NA2 is the answer.
Part – II (Chemistry)
21.
A closed vessel with rigid walls contains 1 mol of
298
238
92 U to 82 Pb,
298
92 U
and 1 mol of air at 298 K. Considering complete decay of
the ratio of the final pressure to the initial pressure of the system at 298 K is
Ans. {9}
Sol.
pf
9
pi
Road, Patna-1 ;
Varanasi (U.P.)
9
22.
In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by MnO4 . For this reaction, the ratio
of the rate of change of H  to the rate of change of MnO4  is
Ans. {8}
2
 Fe3  4CO2  Mn2   6H2O
Sol.
8H  MnO4  Fe(H2 O)2 (C2O4 )2 
23.
The number of hydroxyl group(s) in Q is
H
H
heat
aqueous dilute KMnO (excess)
0ºC
4
 P 
Q
HO
H3C
CH3
Ans. {4}
OH
H
Sol. HO
H3C
24.
CH3
H
 H3C
heat
H3C
CH3
aqueous dilute KMnO4 (excess)


0ºC
HO
H3C
(P)
OH
OH
(Q)
Among the following, the number of reaction(s) that produce(s) benzaldehyde is
CHCl2
I.
CO, HCl


Anhydrous AlCl3 /CuCl
H O
100ºC
2


II.
COCl
CO2Me
H2


PdBaSO 4
III.
IV.
DIBAL H
Toluene 78ºC, H2O


Ans. {4}
CHO
Sol. I.
CO, HCl
Anhydrous AlCl3 /CuCl


CHO
CHCl2
H2O


100ºC
II.
COCl
III.
CHO
H2


PdBaSO 4
CO2Me
IV.
CHO
DIBAL H


Toluene 78ºC, H2O
25. In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe –C bond(s) is
Ans. {3}
26.
Among the complex ions, [Co(NH2 -CH2 -CH2 -CH2 -NH2 )2 Cl2 ] , [CrCl2 (C2O 4 )2 ]3 , [Fe(H2O)4 (OH2 )] ,
[Fe(NH3 )2 (CN)4 ] , [Co(NH2 -CH2 -CH2 -NH2 )2 (NH3 )Cl]2 and [Co(NH3 )4 (H2 O)Cl]2 , the number of complex
ion(s) that show(s) cis-trans isomerism is
Ans. {6}
10
Patna-1;
Sol. 3B2H 6  18 CH3OH  6 B(OCH3 )3  18H2
27. Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product
formed is
Ans. {6}
28. The molar conductivity of a solution of a weak acid HX(0.01M) is 10 times smaller than the molar conductivity of
0
0
a solution of a weak acid HY(0.10M). If  X   Y  , the difference in their pKa values, pK a (HX)  pK a (HY), is
(consider degree of ionization of both acids to be <<1)
Ans. {3}
Sol.
HX
1

HY 10

HHX

HHY

1
10
ka
HX 0.01
k aHY 0.1

1
10
kaHX
1

k aHY 1000
pK a (HX)  pK a (HY)  3
29.
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) Amongest (i) and (iv), the strongest acid is (i)
Ans. (B, C)
30. The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is/are
(a) Ba2 , Zn2
(b) Bi3 , Fe3 
(c) Cu2 , Pb2
(d) Hg2  , Bi3 
Ans. (C, D)
31. Under hydrolytic conditions, the compound used for preparation 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
Ans. (B)
32. When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE statement(s)
regarding this adsorption is/are
(a) O2 is physisorbed
(b) heat is released
(c) occupancy of *2p of O2 is increased
(d) bond length of O2 is increased
Ans. (B,C,D)
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Varanasi (U.P.)
11
33.
One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The relationship
of interatomic potential V(r) and interatomic distance r for the gas is given by
V(r)
V(r)
0
0
r
(a)
V(r)
r
(b)
V(r)
0
r
(c)
0
(d)
r
Ans. (C)
34. In the following reactions, the product S is
H3C
i. O
ii. Zn, H2O
NH
3
3 S

 R 
H3C
H3C
N
(a)
N
(b)
N
N
(c)
(d)
H3C
H3C
Ans. (A)
H3C
(i) O3
H3C
CH = O
(ii) Zn/H2O
CH2 – CH = O
NH3
H
C
H3C
C
H
Sol.
H3C
12
H3C
N
CH
CH = O
Intermolecular
Rxn
CH2 – CH = NH
N
Patna-1;
35.
The major product U in the following reactin is
CH CHCH , H
heat pressure, heat
radical initiator, O
2
3
2 U

T 
H
O
O
CH3
(a)
CH3
H3C
O
O
(b)
H
H
O
CH2
O
O
CH2
(c)
O
(d)
H
Ans. (b)
H3C CH CH3
CH3–CH=CH2/H
CH3
C
Radical initiator
Sol.
O–H
CH3
O2
(T)
36.
O
(U)
In the following reactions, the major product W is
OH
NH2
, NaOH
NaNO , HCl
0ºC
2

V
W
OH
(a)
N
N
OH
(b)
N
N
N
N
HO
OH
(c)
N
N
(d)
Ans. (A)
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13
N2 Cl
NH2
Sol.
NaNO2, HCl
0ºC
N
OH
, NaOH
N
OH
Diazocoupling Rxn
(V)
are correct.
Passage - I
In the following reactions
H2O
HgSO4,H2SO4
C8H8O
37.
Compound X is
O
OH
OH
(a)
CH3
(b)
CH3
(c)
CHO
(d)
Ans. (c)
38. The major compound Y is
CH3
(a)
CH3
(b)
CH3
CH2
CH3
(c)
(d)
CH3
Ans. (d)
14
Patna-1;
C
CH
H2, Pd/BaSO4
(i) B2H6
Ph – CH = CH2
C8H6
(ii) H2O, NaOH, H2O
Ph – CH2 – CH2 – OH
(X)
C8H6
H2O,HgSO4, H2SO4
C
OH
O
OH
CH2
C
Taut.
C
CH3
EtMgBr, H2O
CH3
CH2 – CH3
C8H6
CH2
H+ /heat
CH3
CH3
CH3
+
(Major)
Passage – II
When 100 ml of 1.0 M HCl was mixed with 100 ml of 1.0 M NaOH in an insulated beaker at constant pressure,
a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt 1). Because the enthalpy of
neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used
to measure the calorimeter constant.
In a second experiment (Expt. 2), 100 ml of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 ml of 1.0 M
NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 ºC was measured.
(Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g ml–1)
39. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the expt. 2 is
(a) 1.0
(b) 10.0
(c) 24.5
(d) 51.4
Ans. (a)
Sol. H  msT
ms 
5.7
 1  Calorimeter constant
5.7
Exp. 2
CH3 COOH  OH  CH3 COO   H2O ; H  56 kJ mol1
–56 = –57 + ionH
ionH  1kJ mol1
40. The pH of the solution after Expt. 2 is
(a) 2.8
(b) 4.7
Ans. (b)


 CH3 COONa  H2O
Sol. CH3 COOH  NaOH 
200 m mole 100m mole
0
0
100 m mole 0
100 m mol
pH  pK a  log
(c) 5.0
(d) 7.0
[salt]
 5  log2  4.7
[acid]
Road, Patna-1 ;
Varanasi (U.P.)
15
Part – III (Mathematics)
41.
Let f :    be a continuous odd function, which vanishes exactly at one point and f(1) 
x
F(x) 

x
f(t)dt for all x  [ 1,2] and G(x) 
1
F(x)
1
. Suppose that
2
1
 1

, then value of f  
 t f(f(t)) dt for all x  [1,2] . If xlim
1 G(x) 14
2
1
is
Ans. 7
x
 f(t)dt
1
lim
So.
x 1 x

1
14
 t | f(f(t)) | dt
1
As f(x) is odd function, so by applying newton leibnitz’s theorem to calculate limit by L’ - H rule.
f(x)
1

xf(f(x)) 14
Put x = 1
f(1)
1

f(f(1)) 14
1
2  1
 1  14
f 
2
1
1

 1  14
2f  
2
 1
f   7
2
 

 

Suppose that p,q and r are three non-cplanar vectors in 3 . Let the components of a vector s along p,q and
  
 be 4, 3 and 5, respectively. If the components of this vector  along      
( p  q  r ),(p  q  r ) and ( p  q  r ) are
s
r
x, y and z, respectively, then the value of 2x + y + z is
Ans. Data insufficient
42.
16
Patna-1;
12
 | k 1  k |
43.
 k 
 k 
For any integer k, k  cos    isin   , where i  1 . The value of the expression
 7 
 7 
k 1
3
 | 4k 1  4k 2 |
is
k 1
Ans. {4}
i
sol.
K 1  K  e
(K 1)
7
K

i
i



 e 7  1  e 7  1   cos  1  isin  2 sin
7
7
14


12

14
K 1

4
3

3.2
sin
 4K 1  K 2
14
 K 1  K
12.2 sin
K 1
44.
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. 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.P. is
Ans. {9}
Sol.
s7
6

;
s11 11
149  42d
6

22a  110d 11
a = 9d
t 7  s7  s6 
7
1
2a  6d  2a  5d
2
2
a + 6d = 15d
d=9
The coefficient of x 9 in the expansion of (1  x)(1  x 2 )(1  x3 )......(1  x100 ) is
Ans. {8}
45.
Sol. By multiplying we get cofficient of x 9 is 8
46.
Suppose that the foci of the ellipse
x2 y2

 1 are (f1,0) and (f2 ,0) where f1  0 and f2  0 . Let P1 and P2 be
9
5
two parabolas with a common vertex at (0, 0) and with foci at (f1,0) and (2f2 ,0) , respectively. Let T1 be a tangent
to P1 which passes through (2f2 ,0) and T2 be a tangent to P2 which passes through (f1,0) . If m1 is the slope
 1

2
of T1 and m2 is the slope of T2 , then the value of  2  m2  is
 m1

Ans. {4}
Sol.
x2 y2

1
9
5
e2  1 
5 2

9 3
f1  (2,0)
Road, Patna-1 ;
Varanasi (U.P.)
17
f2  (2,0)
equation of P1 : y 2  8x
equation of P2 : y2  16x
equation of tangent to P1 is ty  x  2t 2
it passes through (–4, 0)
t 2
1
m1 
2
equation of tangent ty  x  4t 2
It pass through (2, 0)
1
t

47.
2
1
m12
, m2  2
 m22  ( 2)2  ( 2)2  4
 cos( n )

e
e
e
lim
    then the value of m is
Let m and n be two positive integers greater than 1. If 0 
m

2

n


Ans. {2}
n
Sol.
ecos   e
lim
m
0
 2n 1m1  
e
 ;
2
lim
m m 1
0
cos n
ne
e
  ; lim
m
2 0
 sin n  n .n 1
e
.
.

 n  m1
2


e
2
1
48.
n
ecos  (  sin n )(nn 1 )
If    (e9x 3 tan
1
x
0
 12  9x 2
)
 1 x2


 dx where tan1 x takes only principal values, then the value of


3 

 loge | 1   |  4  is


Ans. {9}
1
Sol.
   e(9x 3 tan
0
1
3
9 
1 x2

x) 

 dx

Let 9x  3 tan1 x  t

3 
9 
 dx  dt
1 x2 

x0
18
t0
Patna-1;
x 1
9

t 9
3
4

0
3
4
 9 3

et dt   e 4  1




(  1)  e
9
In |   1| 
3
4
3
9
4
49.
Consider the hyperbola H : x2  y2  1 and a circle S with centre N(x2 ,0) . Suppose that H and S touch each
other at a point P(x1, y1 ) with x1  1 and y1  1 . The common tangent to H and S at P intersects the x-axis at
point M. If (,m) is the centroid of the triangle PMN , then the correct expression(s) is (are)
d
1
(A) dx  1  2 for x1  1
3x1
1
x1
dm

for x1  1
(B) dx1 3  x2  1 
1


d
1
(C) dx  1  2 for x1  1
3x1
1
dm 1
(D) dy  3 for y1  1
1
Ans. (A, B, D)
Sol.
x12  y12  1
Equation of tangent at P is xx1  yy1  1  0
1 
 M   ,0 
 x1 
Equation of circle is
x 2  y2  4xx1  (x12  y12  2)  0
x 2  2x1

1
1
1
1 2
x1  1
 3x1   ,m  y1 
3
x1 
3
3
Road, Patna-1 ;
Varanasi (U.P.)
19
4
t
6
t
6
 e (sin
50.
The option(s) with the values of a and L that satisfy the following equation is(are)
0

at  cos4 at)dt
 L?
 e (sin
4
at  cos at)dt
0
e4   1
(A) a  2,L 
(B) a  2,L 
e  1
e4   1
(c) a  4,L 
e  1
e4   1
e  1
(D) a  4,L 
e4   1
e  1
ans. (A, C)
(n 1)
Sol.
In 

et (sin6 t  cos 4 t)dt
n
Let t  n  x
dt = dx


0
0
n
In   e(n x) (sin6 (n  x)  cos 4 (n  x)dx  en . et (sin6 t  cos4 t)dt  e .I0
Given ratio 
51.
I0  e .I0  e2 .I0  e3  .I0
e4   1
 1  e   e2   e3  
I0
e  1
Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the x-axis
and the y-axis, respectively. Let S be the circle x 2  (y  1)2  2 . The straight line x + y = 3 touches the curves
S, E1 and E2 at P, Q and R, respectively. Suppose that PQ  PR 
2 2
. If e1 and e2 are the eccentricities of
3
E1 and E2 , respectively, then the correct expression(s) is (are)
2
2
(A) e1  e2 
43
40
(B) e1  e2 
7
2 10
2
2
(C) e1  e2 
5
8
(D) e1e2 
3
4
Ans. {A, B}
Sol. Coordinate of P  (1,2)
Coordinate of Q & R an given by
x 1 y  2
2 2


1
1
3

2
2
If
P  (acos ,b sin )
cos   3a
3a2 
sin   3b
2
1
5
4 b
4
,3b2  ,     1  e12 , e1 
5
3
3 a
5
Q  (A cos ,Bsin  )
20
Patna-1;
3A 2 
52.
1
8 A2 1
1
,3B2  ,
  1  e22  ;
2
3
3 B
8
8
e2 
7
2 2
  
Let f(x)  7 tan8 x  7 tan6 x  3 tan4 x  3 tan2 x for all x    ,  . Then the correct expression(s) is (are)
 2 2
 /4
(a)
xf(x)dx 

0
 /4
1
12

(B)
 /4
xf(x)dx  0

(C)
0
xf(x)dx 
0
1
6
 /4
(D)

xf(x)dx  1
0
Ans. {A, B}
Sol.
f(x)  7 tan6 x  3 tan2 x  sec 2 x


g(x)   f(x)dx  tan7 x  tan3 x  sec 2 x(tan5 x  tan3 x)
 g(x)dx 
53.
Let f '(x) 
tan6 x tan4 x

;
6
4
 /4

 /4
f(x)dx  0 ;
0

xf(x)dx 
0
1
12
1
192x3
 1
for all x   with f    0 . If m   f(x)dx  M , then the possible values of m and M are
4
2
2  sin x
1/2
(B) m 
(A) m = 13, M = 24
1
1
,M 
4
2
(C) m = -11, M = 0
(D) m = 1, M = 12
ans. {D}
Sol.
64x3  f '(x)  96x3
By integrating from
1
to x.
2
16x 4  1  f(x)  24x 4 
26

10
54.
1

1/2
f(x)dx 
3
2
39
10
Let f, g : [ 1,2]   be continuous functions which are twice differntiable on the interval (-1, 2). Let the values of
f and g at the points -1, 0 and 2 be as given in the following table:
x  1 x  0 x  2
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. Then the correct statements(s)
is(are)
(A) f '(x)  3g'(x)  0 has exactly three solutions in (1,0)  (0,2)
(B) f '(x)  3g'(x)  0 has exactly one solution in (-1, 0)
Road, Patna-1 ;
Varanasi (U.P.)
21
(C) f '(x)  3g'(x)  0 has exactly one solution is (0, 2)
(D) f '(x)  3g'(x)  0 has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)
Ans. {B, C}
Sol. Let h(x) = f(x) - 3g(x)
 h( 1)  3
h(0) = 3
h(2) = 3
 By rolles theorem there exist at least one   (1,0) &   (0,2) such that h'( )  0 and h'()  0
But h''(x)  0 x  (1,0)  (0,2)
 x = 0 is point of inflextion
55.
1  6 
1  4 
If   3 sin   and   3 cos   , where the inverse trigonometric functions take only the principal values,
 11 
9
than the correct option(s) is(are)
(A) cos   0
(B) cos   0
(C) cos(  )  0
(D) cos   0
Ans. {B, C, D}
Sol.
 6    3 
  3 sin1     , 
 11   2 4 
4
  3cos 1    
9
  3 sin1
3
 6
6
6  
   sin1  3.  4.    
11
 11
 11   2

cos(  )  0
56.
Let S be the set of all non-zero real numbers a such that the quadratic equation ax 2  x    0 has two distinct
real roots x1 and x2 satisfying the inequality | x1  x 2 | 1 . Which of the following intervals is(are) a subset(s) of
S?
 1
1 

(a)   2 , 
5

 1 
,0 
(B)  
5 

 1 

(C)  0,
5

 1 1
, 
(D) 
 5 2
Ans. {A, D}
Sol.
D
1
|a|
D
2
1
1  4 2
2
22
1
Patna-1;
1
5
2 

1
5
or   
1
5
are correct.
Passage - I
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and
black balls, respectively, in box II.
57.
One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box.
The ball was found to be red. If the probability that this red ball was drawn from box II is
1
, then the correct
3
option(s) with the possible values of n1,n2 ,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
Ans. {A, B}
Sol.
n3
2(n3  n4 )
1

n1
n3
3

2(n1  n2 ) 2(n3  n4 )
2n3
n1

n3  n4 n1  n2
58.
A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after
this transfer, is
1
, then the correct option(s) with the possible values of n1 and n2 is (are)
3
(A) n1  4 and n2  6
(B) n1  2 and n2  3
(C) n1  10 and n2  20
(D) n1  3 and n2  6
Ans. {C. D}
Sol.
n1
(n1  1)
n2
n1
1




n1  n2 (n1  n2  1) (n1  n2 ) (n1  n2  1) 3
n2  2n12  n1n2  n22  2n1
Passage - II
Let F :    be a thrice differentiable function. Suppose that F(1) = 0, F(3) = –4 and F(x)  0 for all x  (1/ 2, 3).
Let f(x)  xF(x) fpr all x  .
Road, Patna-1 ;
Varanasi (U.P.)
23
59.
The correct statement(s) is(are)
(A) f '(1)  0
(B) f(2)  0
(C) f '(x)  0 for any x  (1,3)
(D) f '(x)  0 for some x  (1,3)
Ans. {A, B, C}
1 
Sol. F(1) = 0; F(3) = -1; F’(x) < 0; x   ,3 
2 
f(x) = x F(x)
f(2) = 2F(2)
f(2) < 0
f’(x) = F(x) + xf’(x)
f’(1) = F(1) + F’(1)= 0 + F’(1) < 0
f’(x) = F(x) + xF’(x)
for x  (1,3)
F(x) < 0 & F’(x) < 0
3
60.
If
x
3
2
F '(x)dx  12 and  x3F''(x)dx  40 , then the correct expression(s) is(are)
1
1
3
(A) 9f '(3)  f '(1)  32  0 (B)  f(x)dx  12
3
(C) 9f '(3)  f '(1)  32  0 (D)
1
 f(x)dx  12
1
Ans. {C, D}
Sol.
3
3
2
 x F '(x)dx  12 ;
3
 x F''(x)dx  40 ;
1
1
3
3
 x3F '(x)  3 x 2F'(x)  40


1
1
27F '(3)  F'(1)  36  40 ; 27F’(3) - F’(1) = 4
f’(3) = F(3) + 3F’(3) = -4 + 3F’(3); f’(1) = F(1) + F’(1) = F’(1)
F'(3) 
f '(3)  4
3
F’(1) = f’(1)

9f '(3)  36  f '(1)  4
9f '(3)  f '(1)  32  0
3
3
 x2

1 2
9
1
1

F(x)

f(x)dx

xF(x)dx


 x F'(x)  2 ( 4)  2  0  2 ( 12) = -18 + 6 = -12


2
2



1
1
1
1
3
24
3
Patna-1;

PAPER – 2

Transcription

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