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2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Pa
ge
-2
Paper-2 READ THE INSTRUCTIONS CAREFULLY
GENERAL
1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.
2. The paper CODE is printed on the right hand top corner of this sheet and the right hand top corner of the back cover of
this booklet.
3. Use the Optical Response Sheet (ORS) provided separately for answering the questions
4. The paper CODE is printed on the left part as well as the right part of the ORS. Ensure that both these codes are
identical and same as that on the question paper booklet. If not, contact the invigilator for change of ORS.
5. Blank spaces are provided within this booklet for rough work.
6. Write your name, roll number and sign in the space provided on the back cover of this booklet.
7. After breaking the seal of the booklet at 02:00 PM, verify that the booklet contains 36 pages and that all the 54
questions along with the options are legible. If not, contact the invigilator for replacement of the booklet.
8. You are allowed to take away the Question Paper at the end of the examination.
OPTICAL RESPONSE SHEET:
9. The ORS (top sheet) will be provided with an attached Candidate’s Sheet (bottom sheet). The Candidate’s Sheet is a
carbon-less copy of the ORS.
10. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the
corresponding place on the Candidate’s sheet.
11. The ORS will be collected by the invigilator at the end of the examination.
12. You will be allowed to take away the Candidate’s Sheet at the end of the examination.
13. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.
14. Write your name, roll number and code of the examination center, and sign with pen in the space provided for this
purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble
under each digit of your roll number.
DARKENING THE BUBBLES ON THE ORS
15. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.
16. Darken the bubble COMPLETELY.
17. The correct way of darkening a bubble is as :
18. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.
19. Darken the bubbles ONYLY IF you are sure of the answer. There is NO WAY to erase or “un-darken” a darkened
bubble.
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Pa
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-3
PART-1:PHYSICS
SECTION -1 (Maximum Marks: 21)
This section Contains SEVEN questions.
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened
Negative Marks : -1 In all other cases.
*********************************************************************************************************
Q1. Consider regular polygons with number of sides 3,4,5.....n as shown in the figure. The center of mass of all
the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without
slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for polygon is
. Then depends on n and h as
h h h
A) 2sinh
n
B)
2sinh
n
C)2tan
2h
n
D)
11
cos
h
n
Key: D
Sol:
h
d
d h
cos
hh
For regular polygon / n
( n is number of sides)
11
cos / cos /
hh h
n n
.
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Pa
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-4
Q2. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is
such that the instantaneous density remains uniform throughout the volume. The rate of fractional change in
density1 d
dt
is constant. The velocity v of any point on the surface of the expanding sphere is proportional to
A) R B)1
R C)
3R D)2/3R
Key: A
Sol: 34
3R M
3 3 11
4
MR
2
2
3 13 2
4
dR M dR
dt dt
2 1 1gives constant
1 3
dR d
dt R dt
dR
vdt
v R
Q3. A Photoelectric material having work-function 0 is illuminated with light of wavelength0
hc
. The
fastest photoelectron has a de Broglie wavelength d . A change in wavelength of the incident light by results
in a change d in d . Then the ratio /d is proportional to
A)2 2/d B) /d C)
3 /d D)3 2/d
Key: D
Sol:
2
max
1
2 d
hc hK
m
2
2 3
2
2d
d
hc h
m
3
2
d dmcx
h
3
2
d d
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Pa
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-5
Q4. Three vectors ,P Q and R are shown in the figure. Let S be any point on the vector R . The distance between the
points P and S is b R . The general relation among vectors ,P Q and S is
O
Y
P SQ
b R
R Q P
Q
X
S
A) 21S b P bQ B) 1S b P bQ
C) 1S b P bQ D) 21S b P b Q
Key: C
Sol: 1S b Q P Q
1 1 1S Q b b P
1 b P bQ
Q5. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the
sound of impact with the bottom of the well. The error in his measurement of time is 0.01T seconds and he
measures the depth of the well to be 20L meters. Take the acceleration due to gravity 210g ms and the
velocity of sound is 1300ms . Then the fractional error in the measurement, /L L , is closest to
A) 1% B) 5% C) 3% D) 0.2%
Key: A
Sol: 2L L
Tg V
2
02
dL dLdT
g VL
1
2
dLdT
g L
21
2L gT
2 2 20
210
LT s
g
1
2 22
dT dL g dL
T L LgL
2 2 0.01
0.012
dL dT
L T
1%
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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6Q. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the
Earth. The sun is 53 10 times heavier than the Earth and is at a distance
42.5 10 times larger than the radius of
the Earth. The escape velocity from Earth’s gravitational field is 111.2ev kms . The maximum initial velocity
sv required for the rocket to be able to leave the sun-Earth system is closest to
(ignore the rotation and revolution of the Earth and the presence of any other planet)
A) 172Sv kms B)
122Sv kms
C)142Sv kms D)
162Sv kms
Key: C
Sol:
x
RsRE
2
02
SE
E E
GMMmV GMM
R x R
2
0 02
SE
E E
V GMGM
R x R
2
0
2
VG
0
22 SE
E E
GMGMV
R x R
52
4
2 3 10
2.5 10
Ee
E
GMV
R
22 2
0
3 10 2
5eV V
13 10eV
11.2 3.6
1
0 40.38V kms
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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7Q. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The
distance between the diametrically opposite vertices of the star is 4a . The magnitude of the magnetic field at the
center of the loop is
4a
I
A) 0 6 3 14
I
a
B) 0 6 3 14
I
a
C) 0 3 3 14
I
a
D) 0 3 2 34
I
a
Key: A
Sol:
a
d
2
1
a
012 sin sin
4
IB
d
60 30
0 3 1
4 2 2
I
d
012
3 1
4 2
IB
d
d a
0 1212B B
0 3 112
4 2d
0 6 3 14
I
a
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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-8
SECTION -2 (Maximum Marks: 28)
This Section Contains SEVEN questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are)
Correct.
For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is
darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : -2 In all other cases.
For example, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks;
darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also
darkened.
***********************************************************************************************************
Q8. A Wheel of radius R and mass M is placed at the bottom of fixed step of a fixed step of height R as shown in the
figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without
slipping. Consider the torque about an axis normal to the plane of the paper passing through the pointQ . Which
of the following options is/are correct?
X
P
S
R
Q
A) If the force is applied normal to the circumference at point P then is zero
B) If the force is applied tangentially at point S then 0 but the wheel never climbs the step
C) If the force is applied at point P tangentially then decreases continuously as the wheel climbs
D) If the force is applied normal to the circumference at point X then is constant
Key: AD
Sol:
P
F
S
(A) net torque, when F acts at P is zero, as it intersects axis of rotation.
(B) The torque increases of the wheel can climb
Q9. Two coherent monochromatic point 1S and 2S of wavelength 600 nm are placed symmetrically on either side
of the center of the circle as shown. The sources are separate by a distance 1.8d mm. This arrangement
produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The
angular separation between two consecutive bright spots is . Which of the following options is/are correct?
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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-9
d1S
2S
2P
1P
A) The angular separation between two consecutive bright spots decreases as we move from 1p to 2p along the
first quadrant
B) A dark spot will be formed at the point 2p
C) The total number of fringes produced between 1p and 2p in the first quadrant is close to 3000
D) At 2p the order of the fringe will be maximum
Key: CD
Sol: At 2p the order of the fringe will be maximum
1S2S 2P
1P
x
0
3000n
1 maxP
2 maxP
cosn
d
sin
nd
d
increases, sin increases
d decreases
cosn
d
cosx
d
9
3
600 10cos
1.8 10n
6600 10cos
1.8n
36
1018
n
3cos 10
3
n
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
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Q10. A point charge Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure.
Which of the following statements is/are correct?
Q
R
A) The electric flux passing through the curved surface of the hemisphere is
0
11
2
Q
z
B) The component of the electric field normal to the flat surface is constant over the surface
C) Total flux through the curved and the flat surface is
0
Q
D) The circumference of the flat surface is an equipotential
Key: AD
Sol: A) 0
2 1 cos 454
q
0
11
2 2
q
45
B) as &x r are
Variable: E
E is not constant Over the surface
D) 2
kQ kQV
r R
R
r
x
E
Q
sinE E
2
kQ n
r r
y
KQx
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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Q11. A rigid uniform bar AB of length L slipping from its vertical position on a frictionless floor (as shown in the
figure). At some instant of time, the angle made by the bar with the vertical is . Which of the following
statements about its motion is/are correct?
L
A
B
O
A) Instantaneous torque about the point in contact with the floor is proportional to sin
B) The trajectory of the point A is a parabola
C) The midpoint of the bar will fall vertically downward
D) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is
proportional to 1 cos .
Key: ACD
Sol:
mg
A
CN
B
sin2
B mg A C
cos2 2
, ,
1 cos2
cX
D A C D
.
Q12. A source of constant voltage V is connected to a resistance R and two ideal inductors 1L and 2L through a switch
S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t=0, the
switch is closed and current begins to flow. Which of the following options is/are correct?
V
R
1L 2L
S
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
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A) After a long time, the current through 1L will be 2
1 2
V L
R L L
B) After a long time, the current through 2L will be 1
1 2
V L
R L L
C) The ratio of the currents through 1L and 2L is fixed at all times 0t
D) At t=0, the current through the resistance R is V
R
Key: ABC
Sol:
1L
R
2L
i1i
2i
V
Left 1 2
1 2
L L
L L
t , V
iR
1 1 2 2 1 2;Li L i i i i 2 1L i i
0, 0t v 2 21
1 2 1 2
L V Li i
L L R L L
12
1 2
V Li
R L L
Q13. A uniform magnetic field B exists in the region between x=0 and 3
2
Rx (region 2 in the figure) pointing
normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters
region 2 from region 1 at point 1P y R . Which of the following option(s) is/are correct?
O
Region 1 Region 2 Region 3
O2P x
y
1PO
+Q
y R
B
3 / 2R
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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A) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change
in its linear momentum between point 1P and the farthest point from y-axis is 2
p
B) For 8
,13
pB
QR the particle will enter region 3 through the point 2P on x-axis
C) For 2
,3
pB
QR the particle will re-enter region 1
D) For a fixed B, particle of same charge Q and same velocity , the distance between the point 1P and the point
of re-entry into region 1 is inversely proportional to the mass of the particle
Key: BC
Sol:
O
P
3
2
R2P
P
r
3
, 22
R mvr P P
qB
3
rsin2
R
P
rqB
3
sin2
P R
qB
2
sin3
PB
qR
290
3
PB
qR
d=2r 2 2p mv
qB qB
Q14. The instantaneous voltages at three terminals marked X,Y and Z are given by
0 sin ,XV V t
0
2sin
3YV V t and
0
4sin
3ZV V t
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is
connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be
A) 0
1
2YZ
rmsV V
B) 0
3
2
rms
XYV V
C) Independent of the choice of the two terminals
D) 0
rms
XYV V
Key: BC
Sol: 0
2sin sin
3XY OV V t V t
0
2sin sin
3OV t V t
2
3 3
1
0 02 cos6
V V
03V
03 sinXYV V t
032
XY YZRms Rms
VV V
SECTION -3 (Maximum Marks: 12)
This section Contains TWO paragraphs.
Based on each table, there are TWO questions
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If all other cases
***********************************************************************************************************
PARAGRAPH: 1
Consider a simple RC circuit as shown in Figure 1.
Process 1: In the circuit the switch S is closed at t=0 and the capacitor is fully charged to voltage 0V (i.e.,
charging continues for time T>>RC). In the process some dissipation DE occurs across the resistance R. The
amount of energy finally stored in the fully charged capacitor is CE .
Process 2: In a different process the voltage is first set to 0
3
V and maintained for a charging time T>>RC. Then
the voltage is raised to 02
3
V without discharging the capacitor and again maintained for a time T>>RC. The
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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-15
process is repeated one more time by raising the voltage to 0V and the capacitor is charged to the same final
voltage 0V as in Process 1.
These two processes are depicted in Figure 2.
V
R
S
C
Figure 1
0V
02
3
V
0
3
V
Process2
Process1
r 2T
T>>RC
t
Figure 2
Q15. In Process 1, the energy stored in the capacitor CE and heat dissipated across resistance DE are related by:
A) ln 2C DE E B) C DE E
C) 2C DE E D) 1
2C DE E
Key: B
Sol: 2
0
1
2cE CV
0W CV D CE W E 2
0
1
2CCV E
Q16. In Process 2, total energy dissipated across the resistance DE is:
A) 2
0
1 1
3 2DE CV
B)
2
0
13
2DE CV
C) 2
03DE CV D) 2
0
1
2DE CV
Key: A
Sol: Total energy suppliers in three stage s(Process 2
0
2
3CV
Energy storages in cap 2
0
1
2CV
Therefore, loss across R 2
0
1
6CV
2
0
1 1
3 2CV
Hence, ‘A’
PARAGRAPH: 2
One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the
process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone,
shown by the dotted line. The radius of the path traced out the point where the ring and the finger is in contact is
r. The finger rotates with an angular velocity 0 . The rotating ring rolls without slipping on the outside of a
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of
friction between the ring and the finger is and the acceleration due to gravity is g.
R
R
Figure 1 Figure 2
r
Q17. The total kinetic energy of the ring is
A) 22
0M R r B) 22
0
1
2M R r
C) 2 2
0M R D) 22
0
3
2M R r
Key: A
Sol: CMV of Ring=(R-r) 0w
For no supply
1
0 0Rw R r w rw
1
0 0Rw Rw
KE of ring 2 2 2 2
0 0
1 1
2 2M R r w mR w
Which is not matching with any answer given if r R
But we take 0 0Rw R r w
then2
KE 2
2 2
0M R r w
then A is nearest possible option
Q18. The minimum value of 0 below which the ring will drop down is
A) 2
g
R r B)
3
2
g
R r C)
g
R r D)
2g
R r
Key: C
Sol: If height of the cone h>>r
Then N mg
m 2
0R r m g
0
g
R r
PART-2:CHEMISTRY
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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SECTION -1 (Maximum Marks: 21)
This section Contains SEVEN questions.
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened
Negative Marks : -1 In all other cases.
**************************************************************************************************
Q19. The order of basicity among the following compounds is
NH
3H C2NH
I II
N NH
III
NHN
NH
IV
2NH
2H N
A) IV > I > II > III B) IV > II> III > I
C) I > IV > III > II D) II > I > IV > III
Key: A
Sol: IV > I > II > III
HN N
2Sp least basic
2 2H N C NH strongest base
NH
Q20. The major product of the following reaction is
2) , ,0
) .
i NaNO HCl C
ii aq NaOH
OH
2NH
A)
O Na
2N Cl B) N N
OH
C)
OH
Cl D)
N N OH
Key: B
2017-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
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-18
Sol:
OH
.2NaNO HCl
oO C
2N
.aq NaOH
2NH
OH
OH
N N
Coupling reaction
N N
O
Q21. Which of the following combination will produce 2H gas?
A) Au metal and NaCN(aq) in the presence of air
B) Zn metal and NaOH(aq)
C) Fe metal and conc. 3HNO
D) Cu metal and conc. 3HNO
Key: B
Sol: 2 2 22 [ ]Zn NaOH Na ZnO H
Q22. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
1graphite 0f G C kJ mol
1diamonda 2.9f G C kJ mol
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature.
The conversion of graphite [C(graphite) to diamond [C (diamond)] reduces its volume by 6 3 12 10 m mol . If
C(graphite) is converted to C(diamond) isothermally at 298T K , the pressure at which C(graphite) is in
equilibrium with C(diamond), is [Useful information: 2 21 1J kg m s ;
1 21 1Pa kg m s ; 51 10bar Pa ]
A) 14501 bar B) 29001 bar C) 1450 bar D) 58001 bar
Key: A
Sol: g DC C
( ) ( )
O O O
Rxn D gG G G
=2.9 – 0 = 2.9 KJ/mol
0G P V
62.9 / 2 10kj m P
1450 /P bar
2017-Jee-Advanced Question Paper-2_Key & Solutions
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Q23. The order of the oxidation state of the phosphorus atom in 3 2H PO , 3 4H PO , 3 3H PO and 4 2 6H PO is
A) 3 4 4 2 6 3 3 3 2H PO H PO H PO H PO B) 3 2 3 3 4 2 6 3 4H PO H PO H PO H PO
C) 3 4 3 2 3 3 4 2 6H PO H PO H PO H PO D) 3 3 3 2 3 4 4 2 6H PO H PO H PO H PO
Key: A
Sol:
3 4 4 2 6 3 3 3 2H PO H P O H PO H PO
+5 +4 +3 +1
Q24. For the following cell,
4 4| || |Zn s ZnSO aq CuSO aq Cu s
When the concentration of 2Zn
is 10 times the concentration of 2Cu
, the expression for 1G in J mol is
[F is Faraday constant; R is gas constant; T is temperature; 0 1.1E cell V ]
A) 2.303 2.2RT F B) 1.1F
C) -2.2F D) 2.303 1.1RT F
Key: A
Sol: 0 2.303 logG G RT Q
0 0
2
2
0
2 1.1 2.2
[ ]2.303 log 2.303 log
[ ]
2.303 log10 2.303
2.303 2.2
G nFE
F F
ZnRT Q RT
Cu
RT RT
G RT F
Q.25. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing
point of the solution. Use the freezing point depression constant of water as 12K kg mol . The figures shown
below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is
146 g mol ]
Among the following, the option representing change in the freezing point is
A)
water
water + Ethanol
Ice
270 273/T K
1
V.P
./bar
B)
water
water + EthanolIce
270 273/T K
1
V.P
./bar
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C)
water
water + EthanolIce
270 273/T K
1
V.P
./bar
D)
water
water + Ethanol
Ice
270 273/T K
1
V.P
./bar
Key: C
Sol:
.
34.5 1000273 2
46 500
273 3 270
f f
f
f
T K m
T
T K
SECTION -2 (Maximum Marks: 28)
This Section Contains SEVEN questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are)
Correct.
For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is
darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : -2 In all other cases.
For example, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks;
darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also
darkened.
******** ***************************************************************************************************
Q26. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. the correct option(s) among
the following is (are)
A) The activation energy of the reaction is unaffected by the value of the steric factor
B) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used
C) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally
D) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
Key: AD
Sol: aE is independent of steric factor
/. Ea RTK Ae
A Z P
Steric factor
Exp. Z value is high
If “P” high reaction is very fast
So R n need not required a catalyst
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Q27. For a reaction taking place in a container in equilibrium with its surroundings , the effect of temperature on
its equilibrium constant K in terms of change in entropy is described by
A) with increase in temperature , the value of K for exothermic reaction decreases because favourable change
in entropy of the surroundings decreases.
B) with increase in temperature , the value of K for exothermic reaction decrease because the entropy change
of the system is positive
C) with increase in temperature , the value of K for endothermic reaction increases because the entropy
change of the system is negative
D) with increase in temperature, the value of K for endothermic reaction increases because unfavourable
change in entropy of the surroundings decreases
Key: AD
Sol: For Exothermic R n
H ve
K decreases with raise in Temp
sys
q HS H ve
T T
So sys
S ve
sorr
S ve
Q28. The correct statement (s) about surface properties is (are)
A) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system
B) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
C) The critical temperatures of ethane and nitrogen are 563K and 126K ,respectively. The adsorption of ethane
will be more than that of nitrogen on same amount of activated charcoal at a given temperature
D) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of
the solution
Key: AC
Sol: Adsorption degree of freedom
Decreases so S ve
tanG ve Spon eous
S ve
H ve
G H T S
ve T ve
At low temp. favours
Cloud is not emulsion
c
xT
m as cT (critical temp )
High liquification all reactions x
m also )
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Q29. Among the following , the correct statement(s) is (are)
A) 3AlCl has the three-centre two- electron bonds in its dimeric structure
B) 3BH has the three-centre two- electron bonds in its dimeric structure
C) 3 3Al CH has the three- centre two-electron bonds in its dimeric structure
D) The lewis acidity of 3BCl is greater than that of 3AlCl
Key: BCD
Sol: option ‘D’ is not given clearly
3 3BCl AlCl in case of hard lewis base
3 3BCl AlCl in case of soft lewis base. (refer. Inorganic chemistry atkins)
Q30. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions
is(are)
Br
I
Br
II
C Br
3CH
3CH
3H C
III IV
Br
3CH
A) I and III follow 1NS mechanism
B) Compound IV undergoes inversion of configuration
C) The order of reactivity for I,III and IV is: IV>I>III
D) I and II follow 2NS mechanism
Key: ABCD
Sol: I)
Br
1NS
II)
Br
[Very slowly reactive to both the pahs]
III)
Br1NS only
IV)
Br
1NS
(A) I & III follow 1NS mechanism correct
(B) compound IV undergoes inversion of configuration
(C) the order of reactivity for I, III & IV is IV > I > III correct
(D) I & II follow 2NS Mechanism correct
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Q31. The option(s) with only amphoteric oxides is(are)
A) 2 3 2, , ,ZnO Al O PbO PbO B) 2 3, , ,Cr O CrO SnO PbO
C) 2 3, 2, ,Cr O BeO SnO SnO D) 2 3, 2, ,NO B O PbO SnO
Key: AC
Sol: 2 3 2 3 2, , , ,Cr O BeO B O PbO SnO are Amphoteric, 2PbO is not amphoteric
Q32. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is
8 8C H O . Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but
not Cannizzaro reaction.
P3 2 2) /i O CH Cl
2) /ii Zn H OQ
8 8C H O
(i)
(ii)
2) /ii Zn H O 8 8C H O
R S3 2 2) /i O CH Cl
The option(s) with suitable combination of P and R, respectively, is(are)
A) 3CH
3CH
3H C
3CH
3CH
3CH
and
B)
and3H C
3CH
C)
and
3H C
3CH
3CH
3H C
3CH
D)
3H C and3H C
3CH
Key: BC
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Sol:
3H C
(P)
3 2 2
2
( ) /
( ) /
i O CH Cl
ii Zn H O
CHO
3CH
Undergoes cannizzaro's reaction , but not haloform reaction
(Q)
8 8( )C H O
Gives haloform reaction but not cannizzaro's reaction
( ) /3 2 2
( ) /2
i O CH Cl
ii Zn H O
(R)(S)
8 8( )C H O
O
Gives cannizzaro's reaction, but not haoform reaction
( ) /3 2 2
( ) /2
i O CH Cl
ii Zn H O
8 8( )C H O
O
3CH
3CH
3CH
3CH CHO
(Q)
Gives haoform reaction, but not cannizzaro's reaction,
( ) /3 2 2
( ) /2
i O CH Cl
ii Zn H O
O
+
O
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SECTION -3 (Maximum Marks: 12)
This section Contains TWO paragraphs.
Based on each table, there are TWO questions
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If all other cases
******** ***************************************************************************************************
PARAGRAPH 1
Upon heating 3KClO in the presence of catalytic amount of 2MnO , a gas W is formed. Excess amount of W
reacts with white phosphorus to give X. The reaction of X with pure 3HNO gives Y and Z
Q33. W and X are, respectively
A) 3O and 4 6PO B) 2O and 4 6PO
C) 2O and 4 10PO D) 3O and 4 10PO
Key: C
Sol:
3 2KClO KCl O4 10 2 5P O N O
2MnO3HNO4P
3 4H PO3HPO
(X)
Q34. Y and Z are, respectively
A) 2 5 3N O and HPO B) 2 3 3 4N O and H PO
C) 2 4 3N O and HPO D) 2 4 3 3N O and H PO
Key: A
Sol:
3 2KClO KCl O4 10 2 5P O N O
2MnO3HNO4P
3 4H PO3HPO
(X)
PARAGRAPH 2
The reaction of compound P with 3 ( )CH MgBr excess in 2 5 2C H O followed by addition of 2H O gives Q. The
compound Q on treatment with 2 4H SO at 00 C gives R. The reaction of R with 3CH COCl in the presence of
anhydrous 3AlCl in 2 2CH Cl followed by treatment with 2H O produces compound S. [Et in compound P is ethyl
group]
3 3H C C
2CO Et
Q R S
P
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Q35. The reactions, Q to R and R to S, are
A) Dehydration and Friedel-Crafts acylation
B) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation
C) Friedel-Crafts alkylation and Friedel-Crafts acylation
D) Aromatic sulfonation and Friedel-Crafts acylation
Key: C
Sol:
2CO Et3
2
1. / .
2.
CH MgBr D E
excessH O
3HO C CH
3CH
2 4 ,0H SO C2OH
2H O
H
Freidel crafts alkylation
3 3
2
1. / .2.
Freidal-crafts acylation
CH COCl Anhy AlClH O
3CH R
P Q
S
Freidel-craft's alkylationQ R
Freidel-craft's acylationR S
Q36. The Product S is
A)
3 3H C C
3CHO
3COCH
3HO S
B)
3H COC
3 3H C C
3CH3H C
C)
3 3H C C
3H C3CH
3COCH D)
3 3H C C 3CH
3COCH
Key: C
Sol:
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2CO Et3
2
1. / .
2.
CH MgBr D E
excessH O
3HO C CH
3CH
2 4 ,0H SO C2OH
2H O
H
Freidel crafts alkylation
3 3
2
1. / .2.
Freidal-crafts acylation
CH COCl Anhy AlClH O
3CH R
P Q
S
2CO Et3
2
1. / .
2.
CH MgBr D E
excessH O
3HO C CH
3CH
2 4 ,0H SO C2OH
2H O
H
Freidel crafts alkylation
3 3
2
1. / .2.
Freidal-crafts acylation
CH COCl Anhy AlClH O
3CH R
P Q
S
Freidel-craft's alkylationQ R
Freidel-craft's acylationR S
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PART-3:MATHS
SECTION -1 (Maximum Marks: 21)
This section Contains SEVEN questions.
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened
Negative Marks : -1 In all other cases.
******************************************************************************************************
Q37. Let 1,2,3,...,9S For 1,2,...,5,k let kN be the number of subsets of S , each containing five elements
out of which exactly k are odd. Then 1 2 3 4 5N N N N N
A) 125 B) 210 C) 252 D) 126
Key: D
Sol: 5 4 5 4
1 4 2 3C C C C
5 4 5 5 5 2 4 4 4 2 4 3 4 4
0 1 2 0 1 2 3 41 1 ...x x C C x C x C C x C x C x C x
Coeff. Of 5x on LHS
9
4
9 8 7 6126
4 3 2C
Q38. The equation of the plane passing through the point 1,1,1 and perpendicular to the planes 2 2 5x y z
and 3 6 2 7x y z , is
A) 14 2 15 3x y z B) 14 2 15 27x y z
C) 14 2 15 1x y z D) 14 2 15 31x y z
Key: D
Sol: 1 1 1 0a x b y c z
2 2 0a b c
3 6 2 0a b c
14 2 15
a b c
: : 14 : 2 :15a b c
14 2 15 14 2 15 31x y z
Q39. Let O be the origin and Let PQR be an arbitrary triangles. The point S is such that
. . . . . .OP OQ OR OS OR OP OQOS OQOR OP OS Then the Triangle is PQR has S as it’s
A) in centre B) Circumcentre C) Orthocenter D) Centroid
Key: C
Sol: . .OP OQ OR OS OR OQ
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. . 0
. 0
. 0
OP RQ OS QR
RQ OP OS
SP RQ
SP RQ
Similarly SQ QR
& SR PQ
Orthocenter
Q40. How many 3 3 matrices M with entries from 0,1,2 are there, for which the sum of the diagonal entries of
TM M is 5?
A) 162 B) 135 C) 126 D) 198
Key: D
Sol: 2 2 2 2 2 2 2 2 2
11 21 31 12 22 32 13 23 33 5a a a a a a a a a
No of ways 9 9 8
4 1 1 198C C C
Q41. Three randomly chosen nonnegative integers ,x y and z are found to satisfy the equation 10x y z then
the probability that z is even, is
A) 6
11 B)
36
55 C)
1
2 D)
1
2
Key: A
Sol: Total Solutions = 10 3 1 12
3 1 2 66C C
10 2 1 11
2 1 10, 10z x y C C
=11
8 2 1 9
2 1 12, 8 9z x y C C
10, 0 1z x y
Sum = 1+3+5+7+9+11=36
36 6
66 11P
Q42. If :f R R is a twice differentiable function such that "( ) 0f x for all x R and 1
( )2
f = 1
2, 1 1f
then
A) 1
' 1 12
f B) 1
0 ' 12
f C) ' 1 0f D) ' 1 1f
Key: D
Sol: ' 1 1f
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1
2
1
1
21
C
As per LMVT ' 1f c for some
1,1
2c
And as " 0 'f x f x is increases
So ' 1f x for x C
As 1 , ' 1 ' 1c f f c
Q43. If y y x satisfies the differential equation 1
8 9 4 9 , 0x x dy x dx x
and
0 7y then 256y =
A) 80 B) 9 C) 16 D) 3
Key: D
Sol: 2 ,x t
dy
dx= .
dy dt
dt dx
2dx
tdt
1.
2
dy dy
dx t dt
1 1
.2 92 4 9
dy
dt tt
4 9y t
4 9y x c
0 7 0
256 4 9 16
4 5 3
y c
y
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SECTION -2 (Maximum Marks: 28)
This Section Contains SEVEN questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are)
Correct.
For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is
darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : -2 In all other cases.
For example, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks;
darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also
darkened.
************************************************************************************************************
Q44. If the line x divides the area of region 2 3, : ,0 1R x y R x y x x into two equal parts
then,
A) 1
02
B) 4 22 4 1 0
C) 4 24 1 0 D)
11
2
Key: BD
Sol: 1
3 3
0 02 x x dx x x dx
0x
1 1
1
1
2 4 1 12
2 4 2 4
4 22 4 1 0
3 1
2
;
3 1
2
1 Discarded
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Q45. If
(2 ) 2 2
,
Cos x Cos x Sin x
f x Cosx Cosx Sinx
Sinx Sinx Cosx
then
A) f x attains its maximum at 0x
B) f x attains its minimum at 0x
C) ' 0f x at more than three points in ,
D) ' 0f x at exactly three points in ,
Key: AC
Sol:
0 cos2 sin2
2cos cos sin
0 sin cos
x x
f x x x x
x x
2cos cos3 cos4 cos2f x x x x x
' 4sin4 2sin2f x x x
2sin2 1 4cos2x x
24sin cos 8cos 3x x x
' '0 ; 0f ve f is ve f x has a max at 0x
' 20 sin2 0;cos 3 / 8 4f x x x solutions
' 0 sin2 0f x x ; 1
cos24
x
2x n
,2 2
x
Q46. If
98 1
1
1
1
k
kk
kI dx
x x
, then
A) 49
50I B)
49
50I C) log 99eI D) log 99eI
Key: AC
Sol:
198
1
1
1
k
k k
kI dx
x x
= 98
1
1k
k
log 1
1
k
k
x
x
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=
298
1
11 log
2K
kk
k k
=
2 22 32 42 992log 3.log 4.log ..... 99log
1.3 2.4 3.5 98.100
=
4 6 8 10 12 196 198
3 6 8 10 196 98 99
2 .3 .4 .5 .6 98 .99log
2 3 .4 .5 98 .99 .100
=
100
99
2.99log
100
log 99e
99100
99
2 99 9999 2 99 2 0.36 99 0.72 99
100 100
Clearly it is greater than 1
Q47. Let 1 1 1
1
x xf x
x
Cos
1
1 x
for 1x .then
A) 1
0x
Lim f x B) 1 0xLim f x
C) 2 0,xf x e in D) 1 0xLim f x does not exist
Key: BC
Sol: 1x 1 2 1
.cos1 1
x x
x x
1x 1 0x 1
1 cos 01
xx
1x 1 1
os1 1
x xc
x x
1x
1 0x 1
1 cos1
xx
does not exist
Q48. If :f R R is differentiable function such that ' 2 ,f x f x for all x R and 0f =1, then
A) f x is increasing in 0, B) 2' 0,xf x c in
C) 2 0,xf x e in D) f x is decreasing in 0,
Key: AC
Sol: ' 2 0f x f x
'
2 0xf x e
2 sinxf x e isincrea g 0x
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2 00xf x e f e 2 1xf x e
2xf x e
Q49. If sin 2
1
sin
sin
x
x
g x t dt , then
A) ' 2
2g
B) ' 2
2g
C) ' 2
2g
D)
' 22
g
Key: Delete
Sol: by Leibnitz theorem
' 1 1sin sin2 2cos2 sin sin .cosg x x x x x 4 cos cosx x x x
' 2 . 1 .0 22 2
g
' 2 1 22
g
Q50. Let and be nonzero real numbers such that 2cos cos cos cos 1 . Then which of the
following is/are true?
A) tan 3 tan 02 2
B) 3 tan tan 0
2 2
C) tan 3 tan 02 2
D) 3 tan tan 0
2 2
Key: AC
Sol: 2 cos cos cos cos 1
1 2cos
cos2 cos
2
22
22
2
11 2
11tan / 2, tan / 2
112
1
x
xyx y
xy
x
2 2
2 2
1 3
1 3
y x
y x
2 23y x
3 tan tan 3 tan tan 02 2 2 2
Options A & C
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SECTION -3 (Maximum Marks: 12)
This section Contains TWO paragraphs.
Based on each table, there are TWO questions
Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct
For each question, darken the bubble corresponding to the correct option in the ORS.
For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If all other cases
******** ***************************************************************************************************
PARAGRAPH-1
Let O be the origin, and , ,OX OY OZ be three unit vectors in the directions of the sides , ,QR RP PQ ,
respectively, of a triangle PQR.
Q51. OX OY
A) sin P R B) sin2R C) sin P Q D) sin Q R
Key: C
Sol: ;QR RP
OX OYQR RP
sin sin sinQR RP
OX OY QRP R P QQR RP
Q52. If the triangle PQR varies, then the minimum value of cos cosP Q Q R cos R P
A) 3
2 B)
3
2 C)
5
3 D)
5
3
Key: A
Sol: cos cos cosP Q Q R R P
cos cos cosP Q R
3
2 as
3cos cos cos
2P Q R P Q R
PARAGRAPH-2
Let p, q be integers and let , be the roots of the equation, 2 1 0x x , where .For 0,1,2,...,n
let n n
na p q .
FACT: If a and b are rational numbers and 5 0a b , then 0a b
Q53. If 4 28a , then 2p q =
A) 12 B) 21 C) 14 D) 7
Key: A
Sol: 1 1 5 1 5
,2 2
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1 2 1
4 43 2, 3 2
4 4
4 P q
28 3 2 3 2P q
1 3 5
82 2
p q p q
4p q 2 12p q
Q54. 12a
A) 11 102a a B) 11 10a a C) 11 10a a D) 11 102a a
Key: B
Sol: 5 5
6 12 128 5 144 89; 144 89
12 12
12a p q
12 144 89a p q p q
11 10 12a a a
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