FIITJEE
ALL INDIA TEST SERIES
JEE (Advanced)-2022
FULL TEST – III
PAPER –1
TEST DATE: 18-01-2022
Time Allotted: 3 Hours
Maximum Marks: 180
General Instructions:



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The test consists of total 57 questions.
Each subject (PCM) has 19 questions.
This question paper contains Three Parts.
Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
Each Part is further divided into Three Sections: Section-A, Section-B & Section-C.
Section – A (01 – 04, 20 – 23, 39 – 42): This section contains TWELVE (12) questions. Each question has
FOUR options. ONLY ONE of these four options is the correct answer.
Section – A (05 –10, 24 – 29, 43 – 48): This section contains EIGHTEEN (18) questions. Each question has
FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).
Section – B (11 – 13, 30 – 32, 49 – 51): This section contains NINE (09) questions. The answer to each
question is a NON-NEGATIVE INTEGER.
Section – C (14 – 19, 33 – 38, 52 – 57): This section contains NINE (09) question stems. There are TWO
(02) questions corresponding to each question stem. The answer to each question is a NUMERICAL
VALUE. If the numerical value has more than two decimal places, truncate/round-off the value to TWO
decimal places.
MARKING SCHEME
Section – A (Single Correct): Answer to each question will be evaluated according to the following marking scheme:
Full Marks
:
+3
If ONLY the correct option is chosen.
Zero Marks
:
0
If none of the options is chosen (i.e. the question is unanswered);
Negative Marks :
–1
In all other cases.
Section – A (One or More than One Correct): Answer to each question will be evaluated according to the following
marking scheme:
Full Marks
:
+4
If only (all) the correct option(s) is (are) chosen;
Partial Marks
:
+3
If all the four options are correct but ONLY three options are chosen;
Partial marks
:
+2
if three or more options are correct but ONLY two options are chosen and both
of which are correct;
Partial Marks
:
+1
If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks
:
0
If none of the options is chosen (i.e. the question is unanswered);
Negative Marks :
–2
In all other cases.
Section – B: Answer to each question will be evaluated according to the following marking scheme:
Full Marks
:
+4
If ONLY the correct integer is entered;
Zero Marks
:
0
In all other cases.
Section – C: Answer to each question will be evaluated according to the following marking scheme:
Full Marks
:
+2
If ONLY the correct numerical value is entered at the designated place;
Zero Marks
:
0
In all other cases.
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Physics
2
PART – I
Section – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.
1.
Ans.
A small block of mass m lies above a thin disk of total
mass M, constant surface density  , and radius R. A
hole of radius R/2 is cut into the center of the disk,
allowing the mass to pass though. The mass starts at a
position along the central axis of the disk but is displaced
a height h from the plane of the disk, where h << R.
There are no other external forces at work on this
system; i.e., the disk and block are out in deep space.
Assume m << M. For the first order in h the block
undergoes simple harmonic motion with respect to the
centre of the plane of the disk, compute the period of
oscillation of block.
(A)
T
2R
G
(B)
T
R
G
(C)
T
R
2G
(D)
None of these
m
h
M
R/2
R
A
m
Sol.

x
g  2G 1  cos 

x
g  2G 1 
2
R  x2


 x
  2G 1  
 R

m

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
3
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22

x 
x 
g  2G  1     1 

R
R
/ 2 




2G
g
x
R
T
2.
Ans.
2R
G
Consider a metal can placed coaxially inside a long solenoid. The metal can be modelled as a
thin cylindrical shell of radius r. The magnetic field due to the solenoid varies with time t as kt
where k is a positive constant. The resistivity of the can is  and its breaking stress is  . Ignore
the self-inductance of the can and any currents induced in the flat surfaces of the can. Assume
that the magnetic field produced by the induced currents is negligible. Compute the time at which
the can breaks.
(A)
t

r 2K 2
(B)
t
2
r 2K 2
(C)
t

2r 2K 2
(D)
t

3r 2K 2
B
Sol.
Can
Solenoid
Emf is included in the can
dB 2
Emf 
r  Kr 2
dt
  2r
R
A
r
i
AK
2
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
4
A
A
i2rB
From the diagram
i  2rB  2A
2
t 2 2
r K
3.
Ans.
A motor is installed at the top of a pole rigidly fixed on a
platform. A light rod of length r = 1 m is rigidly attached
to the motor shaft at its one end and at the other end a
small ball of mass m is attached. The rod can rotated in
a vertical plane with the help of the motor. Total mass of
the platform, the pole and the motor is   4 times the
mass of the ball. The motor rotates the rod at a constant
angular velocity. The platform is placed on a horizontal
surface, where coefficient of friction is   1/ 3. At
which minimum angular velocity 0 of the rod, will the
platform start sliding?
(A)
0  2rad / s
(B)
0  3rad / s
(C)
0  5rad / s
(D)
None of these

C
Sol.

r



F on ball+m g  mw 2 r
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m
5


AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22

F on platform=m g mw 2 r
5mg
m 2r 
0  5rad / s
1  2
4.
A ring of mean radius r and cross-sectional area A is made of a
perfectly conducting wire. Inductance L of the ring is so small that
inertia of free electrons cannot be neglected in the current building
process. The free electron density in the conductor is n, mass of an
electron is m and modulus of charge on an electron is e. Initially the
ring is placed ina uniform magnetic field with its plane parallel to
induction vector B of the field as shown in the figure. Find the current
ring after it is turned through an angle 90°.
(A)
i
(B)
i
(C)
i
(D)
i
B
Br 2
m  r
L
ne2 A
B r
m  r
L
ne2 A
Br 2
m  2r
L
ne2 A
B  2 r
m  r 2
L
ne2 A
Ans.
C
Sol.
For perfect conductor R = 0
Enet = 0 in loop net  constant
net  self  k  ext
 
 
=Li  Lki  B A  Lneti  B A
ml
(Where m = mass of one electron, l = loop length)
ne 2 A
As  is constant i   f
LK 


Initially i  0 & B  A
net  0
Finally   BA ' Lneti
i
Br 2

Lnet
Br 2
m  2r
L
ne2 A
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6
Section – A (Maximum Marks: 24)
This section contains SIX (06) questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR
MOER THAN ONE of these four option(s) is (are) correct answer(s).
5.
L1
L2
For the given figure consider their mutual inductance is M then
(A)
Leq 
L1L 2  M2
When mutual inductance is not opposing
L1  L 2  2M
(B)
Leq 
L1L 2  M2
When mutual inductance is not opposing.
L1  L 2  2M
(C)
Leq 
L1L 2  M2
When mutual inductance is opposing.
L1  L 2  2M
(D)
Leq 
L1L 2  M2
When mutual inductance is opposing.
2L1  3L 2  2M
Ans.
AC
Sol.
Leq 
6.
In a photoelectric effect experiment, a point source of light of power 40W emits mono-energetic
photons of wavelength 1 that can just exit photoelectrons from an isolated metallic sphere of
radius 1 cm, placed at a distance of 1 km from the light source. Now, three other sources of
wavelength  2 , 3 and  4 which are emitting the same number of photons as that of 1 brought
L1L 2  M2
L1  L 2  2M
&
Leq 
L1L 2  M2
(Opposing)
L1  L 2  2M
-6
o
o
near the source of 1. Assume photo efficiency of 10 (Take 1  4960 A,  2  4133.33 A,
o
o
o
3  5000 A,  4  7200 A and hc=12.400 eV  A )
Ans.
(A)
Number of photoelectrons emitted from the sphere per second is 2.5  109
(B)
The potential of the sphere when emission of photoelectron stops is 0.8 V.
(C)
The potential of the sphere when emission of photoelectrons stops is 0.5 V.
(D)
The time after which the emission of photoelectrons will stop is 1.39  103 sec .
ACD
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7
Sol.
I
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
P
4R2
R=1M
P
Energy incident  Ir 2
P  r 2 

4R2 hc
For 1 no. of photon incident/sec 2.5  1015
Photo electric effect for 1 &  2 only.
No. of Photons incident/sec 
h    K max
K max  0 For 1 

12400
12400
 3ev =
 2.5ev
4133.33
4960
Vs  0.5volt
E2 
KQ
 Vs  Q2  ne
r

7.
A dog running with a constant speed v is chasing a cat that is running with a constant velocity u .
During the chase, the dog always heads towards the cat. At an instant, direction of motion of the
dog makes angle  with that of the cat and the distance between them is r.
uv sin 
.
r
(A)
The magnitude of acceleration of dog at this instant is
(B)
Radius of curvature of path followed by dog at this instant is
(C)
The magnitude of acceleration of dog at this instant is
(D)
Radius of curvature of path followed by dog at this instant
Ans.
AB
Sol.
Acceleration of dog = Normal acceleration.
vr
.
usin 
uv cos 
.
r
ur
.
v cos 
C
an
D




A  A f  A i  A i   A .
dv vd
v sin 
a

 v 
dt
dt
r
2
v
vr
R

.
an usin 
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
8.
8
A small bead of mass m can slide on a frictionless fixed ring of radius r.
With the help of two identical strings of force constant k, the bead is
connected to two nails A and B each on a diameter at a distance 0.5 r
from centre O of the ring. Relaxed length of each string is negligible as
compared to the radius of the ring. The bead is given a small velocity at
point C. What can you predict about subsequent motion of the bead
before any of the string strikes a nail? [ignore gravity]
(A)
It will keep moving with its initial speed.
(B)
It will oscillate simple harmonically about the point C.
(C)
Its angular momentum about centre O of the ring is conserved.
(D)
Total elastic potential energy stored in both the strings is 1.25kr2.
Ans.
AC
Sol.
F  2 k r toward 0.

A
O
C
B


0  0
P.t 
9.
1  2 r2 
k  r    1.25kr 2
2 
4
A solid paraboloid of base radius R0 and height h0 is having uniform volume mass density of s is
Inverted and placed just above the surface as shown, of a liquid having volume mass density of
L . When paraboloid is released from rest it has been observed that when it becomes completely
submerged in the liquid the paraboloid comes to the rest again. There exist uniform gravity of g.
Assume that the liquid body is very large so that liquid level is not changing as the paraboloid
enters into the liquid.
R0
s h 0
L
(A)
The phenomenon is true for all paraboloids irrespective of their dimensions as long as
density ratio is maintained.
(B)
The phenomenon is not true for all paraboloids irrespective of their dimensions as long as
density ratio is maintained.
(C)
The ratio of
L
is 2.
s
(D)
The ratio of
L
is 3.
s
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9
Ans.
AD
Sol.
Let Y  ax 2 ( A parabola)
Y
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
H 2
x ( from given data)
R2
Fup
x
y
y
a
R
H
mg
x
dy
v
dt
1
dv
mg  L  x 2 y  g  m 
2
dt
x2 ygL
L x 2 yg
dv
 g
g

g


dt
R2H
s R 2 H
2s
2
L 2
dv
v
 g  gy
dy
s
0
H
 vdv   (g 
0
0
o  gH 
L 2
gy ) dy
s
gL
H
3s
L
3
s
10.
A ideal gas having f degrees of freedom is kept inside a thermally insulated vessel, the vessel
has two large pistons each with charge of +Q and -Q. Assume charges are uniformly distributed.
Pistons can move without friction inside the vessel. At an instant the charge of pistons is
increased k times with the help of an external agent. Then which of the following statements are
correct
-Q
+Q
Vacuum
Vacuum
Gas
(A)
The Ratio of final to initial temperature of gas in equilibrium is
K2  f
.
2f
(B)
The Ratio of final to initial temperature of gas in equilibrium is
2K 2  f
.
2f
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
10
(C)
Work done by the agent in the process is (K 2  1)nRT1 where T1 is initial temperature.
(D)
Work done by the agent in the process is  K 2  1
Q2L1
where L1 is initial separation
A 0
between pistons.
Ans.
BCD
Sol.
Stage 1
-Q
+Q
Vacuum
Vacuum
T1
Gas
L1
P1AL1  nRT1
nRT1
Q2
 P1A 
2A 0
L1
(Force Balancing)
Stage 2
Charge is suddenly changed
-KQ
KQ
Gas
T1
L1
System is not in equilibrium
Stage 3
-KQ
KQ
T2
L 2 CL1
Gas
L2
Finally system in equilibrium
nRT2
K 2 Q2
P2 AL 2  nRT2
 P2 A 
2A0
L2
Energy of the system is conserved between stage 2 and 3.
1
f
1
f
 0E22 AL1  nRT1  0E22 AL2  nRT2
2
2
2
2
On solving all the above equation
T2 2K 2  f
Q2L1

w ext  K 2  1
 K 2  1 nRT1
T1
2f
2A0
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11
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Section – B (Maximum Marks: 12)
This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.
11.
A conducting sphere with inner radius R and outer radius 3R
has some alloys mixed in it due to which its resistivity
changes with radial distance r according to the function
r  Kr, where K is a +ve constant. Inner surface of the
sphere is grounded while outer surface of sphere in
maintained at potential V0 with the help of a battery of emf V0.
Battery is ideal. The potential of the sphere at radial distance
n a
of 2R is V0
. Find (a + b).
n b
Ans.
5
Sol.
V0  I Req
R
3R
+
-
V0
dr
3R
Rr
Req   dR
r dr
4r 2
3R
Kdr
K
 
 Req 
n(3)
4

r
4

R
Req  
Req
K r 
ln  
4  R 
V
V(r)  IR(r)  0  R(r)
Req
R(r) 
V  2R   V0
12.
n2
 n3
In the circuit shown in the figure the electromotive force of the
battery is 9 V and its internal resistance is 15  .The two
identical voltmeters can be considered ideal. Let
st
V1 and V1' reading of 1 voltmeter when switch is open and
V1
V2
20
10
nd
closed respectively. Similarly, V2 and V2' be the reading of 2
voltmeter when switch is open and closed respectively. Then
V '  V2
find the value of 2
.
V1  V1'
Ans.
1
Sol.
When switch is open
9 V
15
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
12
Current in circuit
9
I
A  0.2A
45
V1  V2  6V and because in series same current passes and voltmeters are identical so
V1  V2  3V
When switch closed
V1'  2V;V2'  4V
V2'  V2 4  3

1
V1  V1' 3  2
13.
The given thin lenses have
R1  R2  10cm and 1  1, 2  2 and 3  4. If final
image formed at object and value of ‘d’ is x  10 cm .
Find ’x’.
R1
R2
R  100cm
1  2
3
O
30cm
Ans.
 d cm 
7
Sol.
3
2
1
3 1 3  1 3   2



v
u
R1
R2
4
1
2 1 4  2



v 30
10
10
v  30
2
1
3
O
30cm
 d cm 
d  30  100
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13
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Section – C (Maximum Marks: 12)
This section contains THREE (03) questions stems. There are TWO (02) questions corresponding to
each question stem. The answer to each question is a NUMERICAL VALUE. If the numerical value has
more than two decimal places, truncate/round-off the value to TWO decimal places.
Question Stem for Question Nos. 14 and 15
Question Stem
There is a hollow prism with hexagonal base of side L0 and height H0 as
shown. There are identical non conducting uniformly charged solid spheres
are placed at each corner and two identical spheres at the centre of flat
hexagonal faces. The charges spheres have uniform density of d0 and
radius of R0 . What will be the flux passing through the shaded region and
average charge density    hexagonal prism. It is given that the flux passing
through the hexagonal flat surface is k times the total flux passing through
the prism.
R 3d
Take 0 0  1 SI units
0
&
doR0 3
 1 SI units
L20H0
14.
The value of s is _________
Ans.
00000.35
15.
The value of d is _________
Ans.
00004.83
Sol.
1
s 
1
2
2  KT
61  22  total
61  2K total   total
1 
1  2K  total
6
  total 
qin
0
Top view
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14
q0 / 6
qin  12.
q0
q
2 0
6
2
4
R03
3
1  2k  4R03 d0
1 

6
0
 3q0  qin  3d0
s 
1  2k  R0 3d0
3
0
 0.35
Average charge density 
d
qin
Vprism
8 d0R03
 4.83
2
3 3 L 0 W0
Question Stem for Question Nos. 16 and 17
Question Stem
One side of a thin metal plate is illuminated by the sun. When the air temperature is T0, the temperature
of the illuminated side is T1 and that of the opposite side is T2. The temperature of each sides be T1' & T2' ,
if another plate of double thickness is used.  Take T0  300 K, T1 =330 K, T2  320 K 
16.
The value of T1'  K  is ___________
Ans.
00333.33
17.
The value of T2' K  is ___________
Ans.
00316.67
Sol.
Left face
Power from sun = P
T1
To
T2
l
d
  T1  T0 
dt
  4eAT03
d
   T1  T0 
dt
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P    T1  T0  
kA  T1  T2 
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
....(i) ( For left surface)

KA  T1  T2 
   T2  T0 
(For right surface)

When thickness is doubled
kA T1'  T2'
P    T1 ' T0  
.....(iii)
2
KA T1'  T2'
  T1'  T0
......(iv)
2
on solving
2T 2  T1T2  T0  3T1  T2   T22
T1'  1
 333.33k.
2  T1  T0 


T2' 




T1T2  T0  T1  3T2   T22
2  T1  T0 
 316.67k.
Question Stem for Question Nos. 18 and 19
Question Stem
A boat is travelling in a river with a speed of 10 m/s along the stream flowing with a speed of 2 m/s. From
this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the
sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in
water and air is negligible. The frequency detected by a receiver kept inside the river downstream is f 0
and the transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/sec in
the direction opposite the river stream the frequency of the sound detected by the receiver is f 1.
(Temperature of the air and water = 20°C; Density of river water = 103 kg/m3; Bulk modulus of the water =
2.088  109 Pa; Gas constant R = 8.31J / mol  K; Mean molecular mass of air = 28.8  103 kg / mol;
Cp/Cv for air = 1.4)
Note: Boat velocity is with respect to ground & receiver is stationary w.r.t ground
18.
The value of f0 (KHZ) is__________
Ans.
00100.69
19.
The value of f1 (KHZ) is__________
Ans.
00103.04
Sol.
Use doppler effect.
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Chemistry
16
PART – II
Section – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.
20.
The compound (B) would be
O
(1) OH
(B)
(2) H
O
OH
(A)
(B)
O
HO
O
C
O
C
O
(C)
(D)
OH
COOH
OH
OH
Ans.
C
Sol.
O
O
OH
O
OH
COOH
H
C
O
O
OH
O
COOH
OH
O
21.
How many methoxy group present in the molecule of a compound (A) would be present by the
result of a Zeisel method. Treatment of compound (A) C20H11O 4N with hot conc. HI gives CH3I
indicating presence of methoxy group. When 4.24 mg of a compound (A) is treated with hot conc.
HI and CH3I thus produced is digested with alc. AgNO 3 , 11.62 mg of AgI is obtained.
(A)
1
(B)
2
(C)
3
(D)
4
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17
And.
D
Sol.
AgNO3
C20H11O4N  HI 
 CH3I 
 AgI
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
1.24
11.62
 .013
 0.05
329
235
No. of mole of AgI  4  mole of C20H11ON
22.
The blue colour produced on adding H2 O2 to acidified K 2 Cr2 O7 solution and shaking with ether is
due to the formation of
(A)
Cr2O3
(B)
CrO42
(C)
CrO3
(D)
CrO5
Ans.
D
Sol.
ether
K 2Cr2 O7  H2 O2  H2SO 4 
 CrO5
(Blue)
23.
Ans.
Sol.
A 0.001 molal solution of a complex [MA8] in water has the freezing point 0.0054o C . Assuming
100% ionization of the complex salt and Kf for H2O = 1.86 K-Kgmol–1, write the correct
representation of the complex.
(A)
MA 8 
(B)
MA 7  A
(C)
MA 6  A 2
(D)
MA 5  A 3
C
w  1000
 i Kfm
M.W  W
.0054  i 1.86  .001
Tf  i K f
i
.0054
 2.90
1.86  .001
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18
Section – A (Maximum Marks: 24)
This section contains SIX (06) questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR
MOER THAN ONE of these four option(s) is (are) correct answer(s).
24.
A cation which form am amine complex ion with excess of NH3 is
(A)
Ag
(B)
Cu2
(C)
Co2
(D)
Al3
Ans.
ABC
Sol.
Ag  2NH3   Ag  NH3 2 

Cu2  4NH3  Cu NH3  4 
2
Co 2  6NH3  CO NH3 6 
25.
2
Which of the following statements regarding P4O10 is correct?
(A)
It is an anhydride of metaphosphoric acid
(B)
It is a strong dehydrating agent
(C)
In P2O5, each phosphorous is attached to three oxygen atoms through normal covalent
bond and also with another oxygen atom through a coordinate bond.
(D)
The dehydrated product of HNO3 obtained by treating it with P4O10 is NO2.
Ans.
ABC
Sol.
HNO3 on dehydration gives N2O5
26.
Choose the correct option(s) for the following set of reactions
(i) MeMgBr
C6H10O
Conc. HCl
Q
(ii) H2O
S
(Major)
20%, H3PO4, 360 K
(i) H3/Ni
T
(ii) B3/hv
(Major)
CH3
(A)
Cl
(S)
R
(Major)
HBr, benzoyl peroxide
U
s
(Major)
(B)
H3C
Br
(T)
H3C
Br
(U)
CH3
Cl
(S)
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19
CH3
(C)
H3C
Cl
Sol.
(D)
CH3
Br
(S)
Ans.
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Br
H3C
(U)
(U)
Br
(T)
C
O
H 3C
H3C
OH
(i) MeMgBr
Cl
Conc. HCl
H2O
(S)
(Q)
20%, H3 PO4 , 360K
CH3
Br
CH3
CH3
H3/Ni
Benzoyl peroxide
Br3/hv
(T)
27.
Br
HBr
(R)
(U)
The reaction(s) leading to the formation of 1, 3, 5 trimethyl benzene is/are
(A)
O
(B)
Me
Conc. H2SO4
CH
heated iron tube
373 K
Δ
H3 C
(C)
CHO
(D)
CH3
Zn/Hg(HCl)
(i) Br2/NaOH
(ii) H3 O
(iii) sodalime
O
Ans.
OHC
CHO
O
ABD
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20
CH3
Sol.
O
conc. H2SO4
CH 3
H 3C
Δ
H 3C
CH3
CH 3
heated iron tube
Me
C
CH
873 K
H 3C
CH 3
CHO
CH 3
Zn/Hg(HCl)
CHO
28.
CHO
H 3C
CH 3
Which of the following statement(s) is/are correct?
(A)
If temperature coefficient is greater than zero, cell reaction is endothermic.
(B)
If temperature coefficient is less than zero, cell reaction is endothermic.
(C)
If temperature coefficient is less than zero, cell reaction is exothermic.
(D)
If Ecell is negative then G is negative and cell reaction is spontaneous.
Ans.
AC
Sol.
 dE  E 2  E1
, According to Nernst equation, on  in temp. E.M.F  es.


 dT  T2  T1
29.
In which of the following pair of solution will the values of Vant Hoff factor be same?
(Assuming same dissociation)
(A)
0.05 M K 4 Fe  CN 6  and 0.10M FeSO 4
(B)
0.10 M K 4 Fe  CN 6  and 0.05 M FeSO 4 . NH4  2 SO 4 .6H2O
(C)
0.20 M NaCl and 0.10 BaCl2
(D)
0.05 M FeSO 4 . NH4 2 SO4 .6H2 O and 0.02M KCl.MgCl2 .6H2 O
Ans.
BD
Sol.
K 4 Fe  CN6   4K   Fe  CN 6 
i  5 No. of ions
FeSO 4 . NH4  2 SO 4 .6H2O  Fe 2  2NH4  2SO42
4
i = 5 = No. of ions
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21
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Section – B (Maximum Marks: 12)
This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.
+
30.
Find the quantum numbers of the excited state of electrons (Let n1 and n2) in He ion which on
transition to ground state emit two photons of wavelength 30.4 nm and 108.5 nm respectively.
Find the sum of  n1  n2   ?
Ans.
7
Sol.
1
1
2  1
 RH  z   2  2 

 n1 n2 
for
n1  1
n2  ?
1 1
1
 1.09678  107  4  2  2 
9
30.4  10
1 n2 
n2  2
for
n1  2 (first excited state), n 2  ?
1
1
1
 1.09678  107  4  2  2 
9
108.5  10
n2 
2
n2  5
sum of n1 & n2  2  5  7
31.
A compound AB forms NaCl type crystals and another compound XY forms CsCl type crystals.
The formula. Mass of XY is three times that of AB but the cubic edge length of unit cell of AB
crystals is twice that of XY. Calculate the ratio of density of XY to that of AB.
Ans.
6
Sol.
Density of AB =
4  M1
Nar  13
...(i)
Let molar mass of AB be M1 gm/mol and edge length of unit cell be  1(m) .
1 3M1
8  3  M1
Density of XY 
…(ii)

3
Nar  13
 1 
Nar   
2
(ii) /(i) = 6
32.
Ans.
Size of ring in product formed C is?
R
RCO3H
H
(B)
R
O
(C)
5
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
R
Sol.
R
H
R
22
R
O
R
O
+
R
OH
H
R
RCO3H
R
R
O
Baeyer villager
O
R
R
OH
R
O
Section – C (Maximum Marks: 12)
This section contains THREE (03) questions stems. There are TWO (02) questions corresponding to
each question stem. The answer to each question is a NUMERICAL VALUE. If the numerical value has
more than two decimal places, truncate/round-off the value to TWO decimal places.
Question Stem for Question Nos. 33 and 34
Question Stem
For the following reaction scheme
i) O3
i) NaOCH3CH 3, etoh
CH 3OH
A
C
B
ii) HOOH
H3 O
100o
H3 O
E
D
ii) CH 3CH2CH2CH 2I
33.
The molecular wight of B is ……………
Ans.
00132.00
34.
The molecular weight of E is………….
Ans.
00116.00
Sol.
(for Q. 33-34)
COOH
H2C
COOCH3
H2C
COOH
COOCH3
HC
COOCH3
COOCH3
(B)
C5H8O4
(M.W=132)
(A)
H2C
CH2
CH2 CH3
(C)
H3O
COOH
Δ
H3C
CH2
CH2 CH2 CH2
H3C CH2 CH2 CH2 HC
COOH
( CO2 )
(E)
C6H12 O2 (M.W = 116)
COOH
(D)
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23
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Question Stem for Question Nos. 35 and 36
Question Stem
A 10–3 molal aqueous solution of a chromium complex having same number of ammonia molecule and Cl
ions freezes at 0.00558o C . If all the ammonia molecules are acting as ligands and the percent of
hydrogen in the complex is 4.6 (Given K f (H2 O)  1.86K Kgmol1 ) and assume 100% ionization self.
35.
Find the number of Cl atom is complex is___.
Ans.
00004.00
36.
The value of Vant Hoff factor is______.
Ans.
00003.00
Sol.
(for Q. 35-36)
Let the molecular formula of the complex is Cr NH3  x Clx
 % H is in the complex
3x
 100  4.6
52  17x  35.5x
x  4.08  4
Now, Tf  i k f m
i
.00558
3
1.86  10 3
Question Stem for Question Nos. 37 and 38
Question Stem
The isoelectric point (PI) is the PH at which the amino acid exists only as a dipolar ion with net charge


 
 
zero. Structure of lysine and aspartic acid are COOCH  NH3   CH2  4 NH2 and CO OCH  NH3  CH2COOH




respectively. The PKa1 ,PKa2 and PKa3 of the dictation lysine are 2.18, 8.95 and 10.53 respectively. The
PKa1 , PKa2 and PKa3 of the cation of aspartic acid are 1.88, 3.65 and 9.60 respectively.
37.
Isoelectric point lysine is__________.
Ans.
00009.74
Sol.
The dissociation of cationic form of lysine can be represented.
COOH
P 
COO
Ka1
HC
NH 3
(CH 2)3
CH 2NH 3
OH
H
HC
NH 3
P 
Ka2
OH
(CH 2)3
H2C
NH 3
Net change (+2)
(+1)
PKa2  PKa3 8.95  10.53
PI 

 9.74
2
2
H
COO
HC
NH 3
(CH 2)3
CH 2NH 2
(0)
PKa3
OH
H
COO
HC
NH2
(CH 2)3
CH 2NH 2
(–1)
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24
38.
Isoelectric point of Aspartic acid is __________.
Ans.
00002.77
Sol.
The dissociation of cationic form of Aspartic acid can be shown as
COOH
COO
COO
PKa1
P

HC
CH 2
NH 3
OH
H

HC
NH 3

Ka2
OH
CH2
COOH
COOH
Net change (+1)
(0)
PKa2  PKa2 1.88  3.65
PI 

 2.77
2
2
H

HC
NH 3
CH2
COO
(–1)
PKa3
OH
H
COO
HC
CH2
COO
(–2)
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NH2
25
Mathematics
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
PART – III
Section – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.
39.
 1
 1
 f(x)  f   . If f (10) = 1001, then

x
x
Polynomial function f (x) satisfying the condition f(x)f 
f(20) is
Ans.
Sol.
(A)
7001
(B)
8001
(C)
8000
(D)
None of these
B
 1
 1
f(x)f    f(x)  f  
x
 
x
n
let f(x)  an x  an 1x n 1  an  2 x n  2  .....  aa x  a0
a
1
n
 1 a
f    nn  nn11  an 2  n2  ..........   a0
x
x
x
x x
 1
f(x)  f    a0 an xn   an1a 0  an a1  xn1  xn2  a0 an2  a1an1  a2 an 
x
 1
f(x)  f    an x n  an1xn1  xn2 an2  ........  a1x  a0
x
a
a
a
a
 nn  nn 11  nn  22  .......  1  a0
x
x
x
x
an  1 a0  1 ; a1  a 2  a3  ......  an1  0
f(x)  x n  1
f(10)  10n  1  103  1
n=3
f(x)  x3  1
3
f(20)   20   1  8001
40.
e
x

sgn x
 e x sgn x 

  and   denotes the fractional and
integral part functions, respectively. Also h  x   log  f  x    log  g  x   then for real x, h  x  is
Let f  x   e
and g  x   e 
(A)
An odd function
(B)
An even function
, x  R where
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
(C)
Neither odd nor an even function
(D)
Both odd as well as even function
26
Ans.
A
Sol.
y  y
x
h  x   log  f  x  . g  x    log e     y   y   e sgn x
 ex , x  0

x
 h  x   e sgn x   0, x  0
 e x , x  0

x
e , x  0

 h   x    0, x  0  h  x   h   x   0 for all x.
 e x x  0

41.
The length and foot of the perpendicular from the point (2, -1, 5) to the line
x  11 y  2 z  8


10
4
11
are
(A)
14, 1,2, 3 
(B)
14, 1, 2,3 
(C)
14, 1,2,3 
(D)
None of these
Ans.
C
Sol.
Let co-ordinates of foot of perpendicular are (a, b, c)
a  11 b  2 c  8
Since


k
10
4
11
Then a  10k  11,b  4k  2,c  11k  8
Here all alternative give a = 1, then k = -1. So foot is (1,2,3) and perpendicular distance between
(1,2,3) and (2,-1,5) is 14 .
42.
If x = 9 is the chord of contact of the hyperbola x  y  9 , then the equation of the
corresponding pair of tangent is
2
(A)
9x 2  8y 2  18x  9  0
(B)
9x 2  8y 2  18x  9  0
(C)
9x 2  8y 2  18x  9  0
(D)
9x 2  8y 2  18x  9  0
2
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27
Ans.
B
Sol.
Equation of chord of contact
xh  yk – 9 = 0
….(1)
x–9=0
…..(2)
h = 1 and k = 0
pair of tangent
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
SS '  T 2
x
2
2

 y 2  9  8    x  9  9x 2  8y 2  18x  9  0
Section – A (Maximum Marks: 24)
This section contains SIX (06) questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR
MOER THAN ONE of these four option(s) is (are) correct answer(s).
43.
Which of the following options are correct?
2
(A)
1 n
3
n

n
n
n
n
n
( C0  C3  C6  .....)  2 ( C1  C2  C4  C5  .....)   4



n
C1 n C2 n C 4 n C5  .....
(B)
If a and b are two positive numbers such that a5 b2  4 then the maximum value of
log21/5  a 2   log21/2  b2  is equal to 4
(C)
Constant team in ((((((x  2)2  2)2  2)2  2)2  2)2 .......2)2 is equal to 2
(D)
 25 C

The coefficient of x 24 in  25 1  x   x  22
 C0

25

C25 
2
 x  25 25
 is equal to 2925
C24 

25
C2  
2
 x  3
25
C1  
Ans.
ABD
Sol.
(A) 1  x  n C0  n C1x  n C2 x 2  n C3 x 3  ......  n Cn x n
25
C3  
2
x  4
25
C2  
25
25

2
1
C4 
 .....
C3 
n
Put x   where   
1 i 3

2
2

n
C0  n C3  n C6  .....  ........ 


n
C0 n C3 n C6  ..... 
 
n

n

C1  n C 4  n C 7  ......  

n
 1 i 3
C1 n C4 n C7  .....   

 2
2 


1 n
i 3
C1 n C2  n C4  n C5  ...... 
2
2
(B) a5 b2  4  5log2  a   2log2  b   2

n


C0 n C3 n C6  .... 


C 2  n C 5  n C8  ...... 2  1  

n

n
 1 i 3
C 2  n C 5  n C 8  .....   

 2
2 


 
C1 n C 2  n C 4 n C 5  .....  2
y  10log2 a  4log2 b  40log2 a  log2 b
A.M.  G.M. 
log2 a  log2 b 
5log2  a   2log2 b 
2
 10log2 a  log2 b
1
 y  4,y  4 when 5log2 a  2log2 b  a5  b2  2
10
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n

n
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
28
(C) For constant term put x = 0 and get constant term = 4
(D) Coefficient of x 24  1.25  2.24  3.23  ......  25.1  2925
44.
a1,a2 ,a3 ,...... are distinct terms of an A.P. We call (p,q,r) an increasing triad if ap ,aq ,ar are in G.P.
where p,q,r  N such that p  q  r. If (5, 9, 16) is an increasing triad, then which of the following
options is/are correct
(A)
If a1 is a multiple of 4 then every term of the A.P. is an integer
(B)
 85,149,261 is an increasing triad
(C)
If the common difference of the A.P. is
(D)
ratio of (4k + 1)th term and 4k term can be 4
1
1
, then its first term is
4
3
th
Ans.
ABC
Sol.
Let R be the common ratio of the G.P. and D be the common difference of A.P.
a5  a5 ,a9  Ra5 ,a16  a5R2
a9  a5  4D   R  1 a5  4D
a16  a9  7D  R  R  1 a5  7D
From equation (1)/(2), we get
1 4
7
 R 
R 7
4
2
From equation (2) – (1), we get  R  1 a5  3D 
9a5
 3D
16
3
3
3
4D
 a1  4D   D  a1   1   D  a1 
16
16
4
3

45.
If
 x
e x 1
2
 5x  4

2x dx  AF  x  1  BF  x  4   C and F  x   
(A)
A  2 / 3
(B)
B   4 / 3  e3
(C)
A 2/3
(D)
B   8 / 3  e3
Ans.
AD
Sol.
2x
C
D


 x  1 x  4  x  1 x  4
ex
dx, then
x
2x  C  x  4   D  x  1

C  2 / 3, D  8 / 3
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e x 1
8/3 
x 1  2 / 3
  x  1 x  4  2x dx   e  x  1  x  4  dx

2
8
  F  x  1  e3F  x  4   C
3
3
A  2 / 3, B  8 / 3 e3

46.
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
The vertices of a triangle ABC are A  (2,0,2), B  1,1,1 and C  1, 2,4  . The points D and E
divide the sides AB and CA in the ratio 1:2 respectively. Another point F is taken in space such
that the perpendicular drawn from F to the plane containing ABC , meets the plan at the point of
intersection of the line segment CD and BE. If the distance of F from the plane of triangle ABC is
2 units, then
(A)
the volume of the tetrahedron ABCF is
7
cubic units
3
(B)
the volume of the tetrahedron ABCF is
7
cubic units
6
(C)
one of the equation of the line AF is r 2iˆ  2kˆ   2kˆ  ˆi    R 
(D)
one of the equation of the line AF is r  2iˆ  2kˆ   ˆi  7kˆ


Ans.

 

 


AC
Sol.
A  2iˆ  2kˆ 
F
 3iˆ  ˆj  5kˆ 

3


D 
 4iˆ  4 ˆj  10kˆ 


3


E
P

C ˆi  2 ˆj  4kˆ


B  ˆi  ˆj  kˆ

CD : r  ˆi  2ˆj  4kˆ  7 ˆj  7kˆ
3


BE : r  ˆi  ˆj  kˆ  7iˆ  7ˆj  7kˆ
3
ˆ
P  (iˆ  ˆj  3k)




 




Area of tetrahedron ABCF
1
7
  Area of base triangle   height  cubic unit
3
3



ˆ PF  PF  2 units
AB AC  7ˆj  7k,
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30

 7 ˆj  7kˆ 
PF  2 
 ˆj  kˆ  P.V. of F-P.V. of P
 49  49 


ˆ ˆ
P.V. of F=i+4k


 
Vectors equation of AF is r 2 ˆi  ˆj    ˆi  2kˆ
47.

Let the equation of a straight line L in complex form be az  az  b  0, where a is a complex
number and b is a real number, then
z  c iz  c 

 0 makes an angle of 45° with L and passes through a
a
a
point c (where c is a complex number)
(A)
the straight line
(B)
the straight line
(C)
the complex slope of the line L is 
(D)
the complex slope of the line L is
z  c iz  c 

makes an angle of 45° with L and passes through a point
a
a
c (where c is a complex number)
Ans.
ABC
Sol.
Let P(z) be any point on the required line.
a
a
a
a

Then,
CP

i.e.
CP
z c
is a unit vector parallel to it
zc
*P(z)
A(z1)
*
C(c) P(z)
45°
45°
B(z2)
Let A(z1) and B(z2) be two points on
z  z1
az  az  b  0 then 2
is a unit vector parallel to the line
z 2  z1
az  az  b  0
 
z  z1  i 4 
zc
 2
e
zc
z 2  z1
2
2

i
z  c
 z 2  z1 

e 2
 z  c  z  c   z2  z1  z2  z1 
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
 z  z1 
zc
 i  2

zc
 z2  z1 
 A(z1 ) and B  z 2  are on the line az  az  b  0 therefore az1  az1  b  0
az2  az  b  0

a z 2  z1

a z2  z1
From equation (1) and (2) we get
48.
The function f  x   x
1/3
z  c iz  c 

0
a
a
 x  1
(A)
has two inflection points
(B)
has one point of extremum
(C)
is non – differentiable
(D)
Range of f  x  is  3  28/3 , 
Ans.
ABCD
Sol.
f  x   x1/3  x  1
df  x 

4 1/ 3 1 1
1
x  . 2/3  2/ 3  4x  1
dx
3
3 x
3x
 f  x  changes sign from –ve to +ve, x  1 / 4 , which is point of minima.


Also f  x  does not exist at x  0 as f  x  has vertical tangent at x  0 .
4 1
1 2 1
2 
1
2  2x  1
. 2/3  . . 5/3  2/3 2    2/3 
9 x
3 3 x
9x 
x  9x  x 
1
 f "  x   0 at x   which is the point of inflection at x  0, f "  x  does not exists but f "  x 
2
changes sign, hence x  0 is also the point of inflection.
From the above information the graph of y  f  x  is as shown.
f " x 
Section – B (Maximum Marks: 12)
This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.
49.
A, B, C and D cut a pack of 52 cards successively in the order given. If the person who cuts a
 
spade first receives Rs. 350 and if the expectation of A is  then   is equal to (where [.]
 64 
denotes greatest integer function)_________
Ans.
2
Sol.
E = even of any one cutting a space in one cut
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
32
n  E  13 C1
n  S  52 C1
 
P(E)  1/ 4  p,P E  q
Probability of a winning  p  qqqp  qqqqqqp  .............

P
64

 128
3
175
1 q
 
 64   2
 
x 2n1  ax 2  bx
. If f  x  is continuous for all x  R , then the value of a  8b is
n
x 2n  1
50.
Let f  x   lim
Ans.
8
Sol.
ax 2  bx
for  1  x  1

a  b 1
x  1

2
f  x  a  b  1

x 1
2


1
for x  1 or x  1


x
for continuity at x  1 we have a  b 
a b 1
2
hence, a  b  1
for continuity at x  1
(1)
a  b  1
(2)
a  b  1
hence a  0 and b  1
51.
A and B are two independent events. The probability that both A and B occur is 1/6 and the
probability that at least one of them occurs is 2/3. Find 8P  A   9P B  if P  A   P B  .
Ans.
7
Sol.
P  A  B   P  A  P B   1/ 6  P  A  P B 
P  A  B   P  A   P B   P  A  B 
 2 / 3  PA  
1
1

6P  A  6
2
 6  P  A    5P  A   1  0  P  A   1/ 2,P B   1/ 3
So, 8P  A   9P  B   4  3  7
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
Section – C (Maximum Marks: 12)
This section contains THREE (03) questions stems. There are TWO (02) questions corresponding to
each question stem. The answer to each question is a NUMERICAL VALUE. If the numerical value has
more than two decimal places, truncate/round-off the value to TWO decimal places.
Question Stem for Question Nos. 52 and 53
Question Stem
Consider the locus of the complex number z in the Argand plane given by Re  z   2  z  7  2i .
Let P z1 .Q  z2  . Be two complex numbers satisfying the given locus and also satisfying
 z   2  i  
arg  1
   R .
 z   2  i   2 
 2

52.
The minimum value of PR.QR where R represents the point (7, -2) is
(A)
25
(B)
12
(C)
10
(D)
50
Ans.
00025.00
Sol.
 x  2   x  7
2
  y  2
2
It is parabola
P
R
(7,-2)
2a
Q
x=2
PR RQ   4a
2
5
=4   
 2
=25
2
53.
 z   7  2i  
arg  1
equals
 z   7  2i  
 1

Ans.
00001.57
Sol.
Re z  2  z  7  2i
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AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
34
Let P  z1  ,Q  z 2 
 z   2  i   
arg  1
  R 
 z   2  i   2 
 2

z
z1
z2
Question Stem for Question Nos. 54 and 55
Question Stem
T is the region of the plane x + y + z = 1 with x, y, z >0. S is the set of points (a, b, c) in T such that just
1
1
1
two of the following three inequalities hold: a  ,b  ,c  .
2
3
6
54.
Area of the region T is
Ans.
00000.86
Sol.
T is an equilateral triangle with the vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1)
3
 area of the region T is
.
2
55.
Area of the region S is
Ans.
00000.34
Sol.
 1 1 1
Take a point P  , ,  on the plane x + y + z = 1
2 3 6
A(0, 0,1)
(5/6, 0, 1/6)
(0, 1/3, 2/3)
(1/2, 0, 1/2)
(0, 1/2, 1/2)
(1, 0, 0)B
C(0, 1, 0)
(1/3, 2/3, 0)
(5/6, 1/6, 0)
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35
AITS-FT-III (Paper-1)-PCM-JEE(Advanced)/22
The region S is shown as shaded region.
3
3 2
Area of S 

 a  b2  c 2 , where a, b, c are sides of the small equilateral triangles
2
4
7 3

.
36
Question Stem for Question Nos. 56 and 57
Question Stem
Consider the line
x 1 y
z 1


and the point C  1, 1, 2  . Let D be the image of C in the line.
2
1
2
56.
The distance C from the line is
Ans.
00003.73
57.
The distance of the origin of the plane through the line and the point C is
Ans.
00000.45
Sol.
x2 y
z 1


z
1
2
C(-1, 1, 2)

P
Q
(1, 0, -1)
CQ  CP sin 


CP L
=

L
i
ˆj
kˆ
2 1 3
2 1 2
=
4  1 4

125
3
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