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Answers & Solutions
forforforforfor
JEE (Advanced)-2020
Time : 3 hrs. Max. Marks: 198
Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005
Ph.: 011-47623456 Fax : 011-47623472
PAPER - 2
DATE : 27/09/2020
PART - I : PHYSICS
SECTION - 1 (Maximum Marks : 18)
This section contains SIX (06) questions.
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, BOTH INCLUSIVE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screenvirtual numeric keypad in the place designated to enter the answer.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct INTEGER value is entered.
Zero Marks : 0 If the question is unanswered;
Negative Marks : –1 in all other cases.
1. A large square container with thin transparent vertical walls and filled with water
4refractive index
3 is kept on a horizontal table. A student holds a thin straight wire vertically
inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the
wire from this corner, another student sees two images of the wire, located symmetrically on each
side of the line of sight as shown. The separation (in cm) between these images is ____________.
O
O
12 cm
Line of sight
Answer (4)
JEE (ADVANCED)-2020 (Paper-2)
2
Sol.
12 ( , 0)0
S i
(x, y)
i
y
x
45°
y co-ordinate of first image S is 'y'
4 1 3sini sin45 sini3 42
23
4 23
i i 32
= 45 – i
So slope of the line passing through (–12, 0) is m = tan(45 – i) =
1 tani
1 tani
23 3
m23 3
23 3 x 12 23 3Eq y
23 3 23 3
Other equation y = –x
Now
23 3 y 12 23 3y
23 3 23 3
12 23 32 23y
23 3 23 3
2y =
12 23 3
4.5 423
So, separation between two images is 4 cm.
2. A train with cross-sectional area St is moving with speed v
t inside a long tunnel of cross-sectional
area S0(S
0 = 4S
t). Assume that almost all the air (density ) in front of the train flows back between
its sides and the walls of the tunnel. Also, the air flow with respect to the train is steady and laminar.
Take the ambient pressure and that inside the train to be p0. If the pressure in the region between
the sides of the train and the tunnel walls is p, then 20 t7
p p v .2N
The value of N is ________.
Answer (9)
JEE (ADVANCED)-2020 (Paper-2)
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Sol. Taking the velocity w.r.t. train
v
vt
4Stvt = 3S
tv
t
4v v
3
Also ambient air will be moving with speed vt w.r.t. train
2 20 t t
1 1 16p v P v
2 2 9
2
0 t
1 7p p v
2 9
N = 9
3. Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch
as shown in the figure. Charged oil drops of density 900 kg m–3 are released through a tiny hole at
the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to
apply a voltage of 200 V across the discs. As a result, an oil drop of radius 8 × 10–7 m stops moving
vertically and floats between the discs. The number of electrons present in this oil drop is ________.
(neglect the buoyancy force, take acceleration due to gravity = 10 ms–2 and charge on an electron
(e) = 1.6 × 10–19 C)
Switch
200 V0.01 m
Answer (6)
Sol.
E 0.01 m
v 200E 100
l 1
JEE (ADVANCED)-2020 (Paper-2)
4
4E 2 10 v/m
mg
Eq
Also q = Ne (N is the number of electrons)
NeE = mg
NeE = 34
r g3
37
4 19
8 10 900 104N
3 2 10 1.6 10
N = 6
4. A hot air balloon is carrying some passengers, and a few sandbags of mass
1 kg each so that its total mass is 480 kg. Its effective volume giving the balloon its buoyancy is V.
The balloon is floating at an equilibrium height of 100 m. When N number of sandbags are thrown
out, the balloon rises to a new equilibrium height close to 150 m with its volume V remaining
unchanged. If the variation of the density of air with height h from the ground is
0h
h
0(h) eh , where
0 = 1.25 kg m–3 and h
0 = 6000 m, the value of N is _________.
Answer (4)
Sol. Variation of density (h) = 0h/h0e
Assuming the density of air inside the balloon to be negligible compared to outside density of air then
480 g = Vg(100)
(480 – N)g = Vg(150)
480 (100)
480 N (150)
0
0
100
h
0
150
h
0
e480
480 N
e
0
50
h x480 50e e x
480 N 6000
JEE (ADVANCED)-2020 (Paper-2)
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480 501
480 N 6000
N 4
5. A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole
moment �
p is now brought towards q from infinity so that the charge moves away. The final
equilibrium position of the system including the direction of the dipole, the angles and distances is
shown in the figure below. If the work done in bringing the dipole to this position is N × (mgh), where
g is the acceleration due to gravity, then the value of N is _________ . (Note that for three coplanar
forces keeping a point mass in equilibrium, F
sin is the same for all forces, where F is any one of
the forces and is the angle between the other two forces)
l
l
�
p
q h
2lsin
2
Answer (2)
Sol.
P mg90
–
2
Fd
90 – 2
T
Clearly d
Fmg T
sin 180sin 90 sin 90
2 2
Aslo d 3
0
q2pF Eq
4 2lsin2
Initial PE = (0)
Final PE =
2
0
pqmgh W
4 2lsin2
JEE (ADVANCED)-2020 (Paper-2)
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Also 22lsin sin 2lsin h2 2 2
d
3
0
F mg 2pq mgsin
sincos cos4 2lsin2 22
3
0
mg 2lsin sinpq 2
42 cos
2
3
2
mg 2lsin sin2
W mgh
2cos 2lsin2 2
mgsin 2lsin2W mgh
2cos2
mg 2sin cos lsin2 2 2W mgh
cos2
2W mg 2lsin mgh 2mgh2
N 2
6. A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two
equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus
the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure
below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K.
The partition is now released and moves without any gas leaking from one part of the vessel to the
other. When equilibrium is reached, the distance of the partition from the top (in m) will be _______
(take the acceleration due to gravity = 10 ms–2 and the universal gas constant
= 8.3 J mol–1K–1).
8 m
Answer (6)
JEE (ADVANCED)-2020 (Paper-2)
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Sol. Assuming the gas to be an ideal mono-atomic gas
V
3C R
2
and area of cross section to be A.
Initially
0
P 4A 0.1R 300 (For both)
P1
P2
1 x
Let piston shifts by x meter
and final temperature is T.
Then
2
P A(4 x) 0.1RT
2
0.1RT RTP A
4 x 10(4 x)
And 1RT
P A10(4 x)
But finally 2 1
RT 1 1P A P A 83
10 4 x 4 x
2
RT 2x83
10 16 x
V V
0.2 C 300 mg x 0.2C T
3 2 2 3R 300 83x RT2 10 10 2
900R RT83x 3
10 10
300R 83 RTx
10 3 10
283x 2x
30R 833 16 x
2
x 2x83 3 83
3 16 x
2
(9 x) (2x)1
3 16 x
JEE (ADVANCED)-2020 (Paper-2)
8
18x + 2x2 = 48 – 3x2
5x2 + 18x – 48 = 0
x 1.78 2
So, the distance from top is 6 m.
SECTION - 2 (Maximum Marks : 24)
This section contains SIX (06) questions.
Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correctanswer(s).
For each question, choose the option(s) corresponding to (all) the correct answer(s).
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.
7. A beaker of radius r is filled with water
4refractive index
3 up to a height H as shown in the figure
on the left. The beaker is kept on a horizontal table rotating with angular speed . This makes thewater surface curved so that the difference in the height of water level at the center and at the
circumference of the beaker is h (h
JEE (ADVANCED)-2020 (Paper-2)
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Answer (A, D)
Sol.
2
2h r2g
R2 = r2 + (R – h)2
2 2
2
r h g hR
2h 2
r
R (R – h)
H
Using formula for refraction of curved surface
4 41
1 3 3
v H R
1 4 1
v 3H 3R
3HRv
4R H
Apparent depth =
13H H
| v | 14 4R
=
123H H
14 4g
8. A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in
the figure) at the bottom of the ramp with speed v0 and wants to turn around over a semicircular path
xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as
shown in the figure. Assume that the energy loss is negligible and the force required for this turn at
the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)
Rz
y
x
30°
h
(A) 201
v 2gh gR2
(B) 203
v 2gh gR2
(C) the centripetal force required at points x and z is zero
(D) the centripetal force required is maximum at points x and z
Answer (A, D)
JEE (ADVANCED)-2020 (Paper-2)
10
Sol. At point y
mg sing30° =
2mv
R
2gR
v2
y
mg sin 30°xz
Using conservation of Mechanical Energy
2 2
0
1 1mv mgh mv
2 2
2
0
gRv 2gh
2
Along path xyz speed is maximum at points x & z, so maximum centripetal force will be required.
9. A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same
mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets
embedded in it. The combined system now rotates with angular speed about the pivot. The maximumangular speed
M is achieved for x = x
M. Then
L
x
v
(A) 2 23vx
L 3x(B) 2 2
12vx
L 12x
(C) ML
x3
(D) Mv
32L
Answer (A, C, D)
Sol. Using conservation of angular momentum about hinge
L, m
x
m
v
2
2 mLmvx mx
3
2 2
3vx
L 3x
M
d L0 x
dx 3
& M3v
2L
JEE (ADVANCED)-2020 (Paper-2)
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10. In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode)
at a distance d from the cathode. The target is kept at a potential V higher than the cathode resulting
in emission of continuous and characteristic X-rays. If the filament current I is decreased to I,
2 the
potential difference V is increased to 2V, and the separation distance d is reduced to d,
2 then
(A) the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will
remain the same
(B) the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain the
same
(C) the cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease
(D) the cut-off wavelength will become two times larger, and the intensity of all the X-rays will
decrease
Answer (A, C)
Sol. As energy of incident electrons is doubled, cut off wavelength will reduce to half. Wavelength of
characteristic X-ray will not be affected. Intensity will decrease due to decrease in filament current.
11. Two identical non-conducting solid spheres of same mass and charge are suspended in air from a
common point by two non-conducting, massless strings of same length. At equilibrium, the angle
between the strings is . The spheres are now immersed in a dielectric liquid of density 800 kg m–3and dielectric constant 21. If the angle between the strings remains the same after the immersion,
then
(A) electric force between the spheres remains unchanged
(B) electric force between the spheres reduces
(C) mass density of the spheres is 840 kg m–3
(D) the tension in the strings holding the spheres remains unchanged
Answer (A, C)
Sol. Net force on a sphere will decrease, but force between spheres will be same.
When in air
mg
TFe
eF
tan2 mg
& 2 2e
T F (mg)
When immersed in liquid
e2
2e
FFktan & T (mg B)
2 mg B k
b
mg kmg 1
�
b
k 21 800
k 1 21 1
�
= 840 kg/m3
JEE (ADVANCED)-2020 (Paper-2)
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12. Starting at time t = 0 from the origin with speed 1 ms–1, a particle follows a two-dimensional
trajectory in the x-y plane so that its coordinates are related by the equation 2x
y .2
The x and y
components of its acceleration are denoted by ax and a
y, respectively. Then
(A) ax = 1 ms–2 implies that when the particle is at the origin, a
y = 1 ms–2
(B) ax = 0 implies a
y = 1 ms–2 at all times
(C) at t = 0, the particle’s velocity points in the x-direction
(D) ax = 0 implies that at t = 1 s, the angle between the particle’s velocity and the x axis is 45°
Answer (A, B, C, D)
Sol.2x
y2
dy dxx
dt dt
2xy
2
22 2
2 2
d y d x dxx
dtdt dt
At t = 0, x = 0 vx = 1 m/s & v
y = 0 m/s
for ax = 1 at x = 0
ay = 1 m/s2
for ax = 0
ay = 1 m/s2 for all values of x.
At t = 1 s for ax = 0
x yˆ ˆv v i v j
�
ˆ ˆ1i 1j
SECTION - 3 (Maximum Marks : 24)
This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screenvirtual numeric keypad in the place designated to enter the answer. If the numerical value has more
than two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct numerical value is entered.
Zero Marks : 0 In all other cases.
JEE (ADVANCED)-2020 (Paper-2)
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13. A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water
pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius
becomes (R – a). For a
JEE (ADVANCED)-2020 (Paper-2)
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Sol. 4 2 23 3 3 3
R R [1 E) R [1 4 10 10 ) R [1 4 10 ]
3
R 300[1 0.04] 300 1.04 3 104 312
1
50I A
372
P S 1
50V V I 60 60
372
500
60 100
312 = R3
I1 I2
TS
P
50 V
P S
500V V volt
62
2
50 50 5I A
100 500 600 60
P T
5 50V V 100 volt
60 6
P T P S S T
50 500 500 500(V V ) (V V ) V V
6 62 60 62
S T1 1 2 1000 100
V V 500 50060 62 60 62 60 62 6 62
100 0.2688 volt372
S T
V V 0.27 volt
15. Two capacitors with capacitance values C1 = 2000 10 pF and C
2 = 3000 15 pF are connected in
series. The voltage applied across this combination is V = 5.00 0.02 V. The percentage error in the
calculation of the energy stored in this combination of capacitors is _______.
Answer (1.30)
Sol.
1 2
1 2
C C 2000 3000 6000C pf
C C 5000 5
1 22 2 2
1 2 1 2
C C1 1 1 C
C C C C C C
2 1 2
2 2
1 2
C C 9 4C C 10 15 6nF
25 25C C
22
2 2
1
C 6000 1 9
5 25C (2000)
JEE (ADVANCED)-2020 (Paper-2)
15
2 22
2
2
C 6000 1 4
5 3000 25C
21 U C 2 VU CV 100 100 1002 U C V
U 6 5 0.02100 100 2 100U 6000 5
= 0.5 + 0.8 = 1.30
16. A cubical solid aluminium (bulk modulus = dP
VdV
= 70 GPa) block has an edge length of 1 m on the
surface of the earth. It is kept on the floor of a 5 km deep ocean. Taking the average density of water
and the acceleration due to gravity to be 103 kg m–3 and 10 ms–2, respectively, the change in the edge
length of the block in mm is _____.
Answer (0.24)
Sol. dP
V BdV
dV dP
V B
3( dl) dPl B
3 34
9
l dP l gh 1 10 10 5 10l 2.4 10 0.24 mm
3 B 3 B 3 70 10
17. The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of
the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are
specified in the given circuit. After both the switches are closed simultaneously, the total work done
by the batteries against the induced EMF in the inductors by the time the currents reach their steady
state values is________ mJ.
V =
5 V
1 L =
10
mH
1
R = 5 1
I1
M =
5 m
H
R = 10 2
I2
V =
20
V2L
= 2
0 m
H2
Answer (55)
JEE (ADVANCED)-2020 (Paper-2)
16
Sol. 15
I 1 A5
2
20I 2 A
10
Total work done by battery against induced emf = Increase in mag. energy
= U1 + U
2 + U
12
2 21 1 2 2 1 21 1
W L I L I MI I2 2
3 2 2 3 31 110 10 1 20 (2) 10 5 10 1 22 2
3(5 40 10) 10
= 55 mJ
18. A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than
the surroundings. The average energy per unit time per unit area received due to the sunlight is 700
Wm–2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss
from the water to the surroundings is governed by Newton’s law of cooling, the difference (in ºC) in
the temperature of water and the surroundings after a long time will be _____________. (Ignore effect
of the container, and take constant for Newton’s law of cooling = 0.001 s–1, Heat capacity of water =
4200 J kg–1 K–1)
Answer (8.33)
Sol. In equilibrium
Heat received per sec = Heat loss per sec
E × A = S
0K ms[ ] M
( – 0) =
3E A 700 0.05
ºCK S M 10 4200 1
=
2
3 2
700 5 10 35 350 508.33ºC
4.2 42 610 42 10
JEE (ADVANCED)-2020 (Paper-2)
17
PART-II : CHEMISTRY
SECTION - 1 (Maximum Marks : 18)
This section contains EIGHT (06) questions.
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, BOTH INCLUSIVE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screenvirtual numeric keypad in the place designated to enter the answer.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct INTEGER value is entered.
Zero Marks : 0 If the question is unanswered;
Negative Marks : –1 in all other cases.
1. The 1st, 2nd, and the 3rd ionization enthalpies, I1, I
2, and I
3, of four atoms with atomic numbers n, n +
1, n + 2, and n + 3, where n < 10, are tabulated below. What is the value of n?
Atomicnumber
Ionization Enthalpy (kJ/mol)
I1
I2
I3
n
n + 1
n + 2
n + 3
1681
2081
738
496
3374
1451
3952
4562
6050
6910
6122
7733
Answer (9)
Sol. n = 9 (Fluorine)
n + 1 = 10 (Ne)
n + 2 = 11 (Na)
n + 3 = 12 (Mg)
2. Consider the following compounds in the liquid form: O2, HF, H
2O, NH
3, H
2O
2, CCl
4, CHCl
3, C
6H
6,
C6H
5Cl.
When a charged comb is brought near their flowing stream, how many of them show deflection as
per the following figure?
Answer (7)
Sol. Paramagnetic liquid and polar liquid will be attracted towards charged comb. Hence liquid showing
deflection are O2, HF, H
2O, NH
3, H
2O
2, CHCl
3, C
6H
5Cl.
3. In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution,
what is the number of moles of I2 released for 4 moles of KMnO
4 consumed?
Answer (0)
Sol. 4 2 2 3
2MnO H O I 2MnO 2OH IO
JEE (ADVANCED)-2020 (Paper-2)
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4. An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When
it was shaken after the addition of 1 mL of 3% H2O
2, a blue alcohol layer was obtained. The blue
color is due to the formation of a chromium (VI) compound ‘X’. What is the number of oxygen atoms
bonded to chromium through only single bonds in a molecule of X?
Answer (4)
Sol. 24 2 2 5 2
CrO 2H 2H O CrO 3H O
X is CrO5
Cr
O
O
O
O
O
Number of oxygen atoms bonded to chrominum through only single bond is 4.
The blue colour is attributed to the presence of chrominum pentaoxide.
5. The structure of a peptide is given below.
HO
H N2
O
H
NN
H
O
OH
O
CO H2
NH2
If the absolute values of the net charge of the peptide at pH = 2, pH = 6, and pH = 11 are |z1|, |z
2|,
and |z3|, respectively, then what is |z
1| + |z
2| + |z
3|?
Answer (5)
Sol. At pH = 2
|Z1| = 2 (All amino groups exist in the form of
3NH )
At pH = 6
|Z2| = 0 (Zwitterion form)
At pH = 11
|Z3| = |–3| = 3 (All COOH & OH exist in the form of –COO and –O )
|Z1| + |Z
2| + |Z
3| = 2 + 0 + 3 = 5
6. An organic compound (C8H
10O
2) rotates plane-polarized light. It produces pink color with neutral
FeCl3 solution. What is the total number of all the possible isomers for this compound?
Answer (6)
Sol.
OH
C* — CH3
OH
H
OH
C* — CH3
OH
HOH
HC* — CH3
OH
Phenolic group gives positive test with neutral FeCl3 solution.
JEE (ADVANCED)-2020 (Paper-2)
19
SECTION - 2 (Maximum Marks : 24)
This section contains SIX (06) questions.
Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correctanswer(s).
For each question, choose the option(s) corresponding to (all) the correct answer(s).
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.
7. In an experiment, m grams of a compound X (gas/liquid/solid) taken in a container is loaded in a
balance as shown in figure I below. In the presence of a magnetic field, the pan with X is either
deflected upwards (figure II), or deflected downwards (figure III), depending on the compound X.
Identify the correct statement(s).
magnet
Nm x
S SN
(I)Balanced;Magnetic
field absent
(II)Upward
deflection;
Magnetic field present
(III)Downwarddeflection;
Magnetic field present
(A) If X is H2O(l), deflection of the pan is upwards
(B) If X is K4[Fe(CN)
6](s), deflection of the pan is upwards.
(C) If X is O2 (g), deflection of the pan is downwards.
(D) If X is C6H
6(l), deflection of the pan is downwards.
Answer (A,B,C)
Sol. H2O(l), K
4[Fe(CN)
6](s) and C
6H
6(l) are diamagnetic compounds. These compounds will be deflected
upwards from the magnetic field.
O2(g) is paramagnetic in nature. It will be attracted towards the magnetic field (downward).
JEE (ADVANCED)-2020 (Paper-2)
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8. Which of the following plots is(are) correct for the given reaction?
([P]0 is the initial concentration of P) H C3 Br
CH3
CH3
+ NaOH H C3
OH
CH3
CH3
+ NaBr
P Q
(A)
[P]0
t1/2
(B)
[P]0
Init
ial
ra
te
(C)
time
0
[Q]
[P]
(D)
time
0
[P]ln
[P]
Answer (A)
Sol. Rate [Substrate]1 (for SN
1 reaction)
Rate = k[[P]0 – X]
Order of reaction = 1
For 1st order reaction Q = [P]0(1 – e–kt)
t1/2
Independent of initial concentration.
9. Which among the following statement(s) is(are) true for the extraction of aluminium from bauxite?
(A) Hydrated Al2O
3 precipitates, when CO
2 is bubbled through a solution of sodium aluminate
(B) Addition of Na3AlF
6 lowers the melting point of alumina
(C) CO2 is evolved at the anode during electrolysis
(D) The cathode is a steel vessel with a lining of carbon
Answer (A,B,C,D)
Sol. Al2O
3 + 2NaOH + 3H
2O 2Na[Al(OH)
4]
2Na[Al(OH)4] + 2CO
2 Al
2O
3xH
2O + 2NaHCO
3
Al2O
3xH
2O 1473K Al
2O
3(s) + xH
2O
JEE (ADVANCED)-2020 (Paper-2)
21
10. Choose the correct statement(s) among the following.
(A) SnCl2.2H
2O is a reducing agent
(B) SnO2 reacts with KOH to form K
2[Sn(OH)
6]
(C) A solution of PbCl2 in HCl contains Pb2+ and Cl– ions
(D) The reaction of Pb3O
4 with hot dilute nitric acid to give PbO
2 is a redox reaction
Answer (A,B)
Sol. Sn4+ > Sn2+ ... stability order
Sn2+ acts as a reducing agent.
SnO2 + 2KOH + 2H
2O K
2[Sn(OH)
6]
In conc. HCl, PbCl2 exists as chloroplumbous acid, H
2[PbCl
4] PbCl
2 + 2HCl H
2[PbCl
4]
Pb3O
4 + 4HNO
3 2Pb(NO
3)2 + PbO
2 + 2H
2O
(2PbO + PbO2)
11. Consider the following four compounds I, II, III, and IV.
N
H C3
CH3
NH2
NH2
NO2
O N2
NO2
N
NO2
O N2
NO2
H C3
CH3
I II
III IV
Choose the correct statement(s).
(A) The order of basicity is II > I > III > IV.
(B) The magnitude of pKb difference between I and II is more than that between III and IV.
(C) Resonance effect is more in III than in IV.
(D) Steric effect makes compound IV more basic than III.
Answer (C,D)
Sol. Basicity order: II > I > IV > III
In IV, because of steric effect–N(Me)2 goes out of plane.
That's why resonance of N with the ring decreases and its basic nature increases.
JEE (ADVANCED)-2020 (Paper-2)
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12. Consider the following transformations of a compound P.
R(Optically active)
(i) NaNH2
(ii) C H COCH36 5
(iii) H O /3
+
P(C H )
9 12
(i) X (reagent)
(ii) KMnO /H SO /4 2 4
Q(C H O )
8 12 6
(Optically active acid)
CH3
Pt/H2
Choose the correct option(s).
(A) P is (B) X is Pd-C/quinoline/H2
(C) P is (D) R is
Answer (B,C)
Sol. P
X Pd-C/Quinoline/H2
R
*
O
(P)
X
(ii) KMnO /H SO /4 2 4
COOHCOOH
* COOH
(Q)
SECTION - 3 (Maximum Marks : 24)
This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screenvirtual numeric keypad in the place designated to enter the answer. If the numerical value has more
than two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct numerical value is entered.
Zero Marks : 0 In all other cases.
JEE (ADVANCED)-2020 (Paper-2)
23
13. A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of
the solution with the volume of HA added is shown in the figure below. What is the pKb of the base?
The neutralization reaction is given by B + HA BH+ + A–.
12
10
8
6
4
2
0 2 4 6 8 10 12
pH
Volume of HA (mL)
Answer (3)
Sol. B + HA BH+ + A–
Volume of HA used till equivalence point = 6 mL (from given diagram)
At half of equivalence point, solution will be basic buffer with B and BH+.
b
[BH ]pOH pK log
[B]
∵
At half equivalence point : [BH+] = [B]
pOH = pKb = 14 – 11 = 3
bpK 3
14. Liquids A and B form ideal solution for all compositions of A and B at 25°C. Two such solutions with
0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What
is the vapor pressure of pure liquid B in bar?
Answer (0.2)
Sol. Using Raoult's law
o o
T A A B BP x P x P
o o
A B0.3 0.25 P 0.75 P ... (i)
o o
A B0.4 0.5 P 0.5 P ... (ii)
From (i) and (ii)
o
AP 0.6 bar
o
BP 0.2 bar
Ans 0.20
JEE (ADVANCED)-2020 (Paper-2)
24
15. The figure below is the plot of potential energy versus internuclear distance (d) of H2 molecule in
the electronic ground state. What is the value of the net potential energy E0 (as indicated in the
figure) in kJ mol–1, for d = d0 at which the electron-electron repulsion and the nucleus-nucleus
repulsion energies are absent? As reference, the potential energy of H atom is taken as zero when
its electron and the nucleus are infinitely far apart.
Use Avogadro constant as 6.023 × 1023 mol–1.
H Hd
d0
Internuclear distance (d)
Po
ten
tia
l e
ne
rgy
(kJ m
ol
)–1
E0
Answer (–5242.42)
Sol. Partial energy of H-atom is taken as zero when electron and nucleus are at infinite distance (given).
P.E. of a H-atom with electron in its ground state
= – 27.2 eV (from Bohr's Model)
At internuclear distance 'd0; electron-electron repulsion and nucleus-nucleus repulsion are
absent.
P.E. of two H-atom = – 2 × 27.2 eV = – 54.4 eV
=
19 2354.4 1.6 10 6.023 10kJ/mol
1000
= – 5242.42 kJ/mol
16. Consider the reaction sequence from P to Q shown below. The overall yield of the major product
Q from P is 75%. What is the amount in grams of Q obtained from 9.3 mL of P? (Use density of
P = 1.00 g mL–1; Molar mass of C = 12.0, H = 1.0, O = 16.0 and N = 14.0 g mol–1)
NH2
P
(i) NaNO + HCl/0-5°C2
OHQ
+ NaOH(ii)
(iii) CH CO H/H O3 2 2
Answer (18.6)
Sol.
NH2
N
N2Cl
(P)
(Q)
(i) NaNO + HCl/0-5°C 2
+ –
(ii) OH
,
N = N
O
CH CO H/H O3 2 2
OH
NNaOH
JEE (ADVANCED)-2020 (Paper-2)
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M.M. = 248
9.3 mL of P 9.3 g of P
Moles of P = 9.3
0.193
Moles of Q obtained = 0.1 × 0.75 = 0.075
Mass of Q obtained = 248 × 0.1 × 3
4 = 18.6 g
17. Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the
minimum temperature (in K) at which the reduction of cassiterite by coke would take place.
At 298 K: fH°(SnO
2(s)) = –581.0 kJ mol–1,
fH°(CO
2(g)) = –394.0 kJ mol–1,
S°(SnO2(s)) = 56.0 J K–1 mol–1, S°(Sn(s)) = 52.0 J K–1 mol–1,
S°(C(s)) = 6.0 J K–1 mol–1, S°(CO2(g)) = 210.0 J K–1 mol–1.
Assume that the enthalpies and the entropies are temperature independent.
Answer (935)
Sol. SnO2(s) + C(s) Sn(s) + CO
2(g)
o
rH 394 581 187 kJ/mol
o
rS 52 210 56 6 200 J/K mol
Tmin
(at which reduction will take place) =
o
r
o
r
H935K
S
18. An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molar
concentration (M) of H+ required to prevent the precipitation of ZnS?
Use Ksp
(ZnS) = 1.25 × 10–22 and overall dissociation constant of H2S, K
NET = K
1K
2= 1 × 10–21.
Answer (0.2)
Sol. ZnS(s) 2 2Zn (aq) S (aq)
[Zn2+] = 0.05 M
[S2–]max
(to prevent precipitation) = sp
2
K (ZnS)
[Zn ]
=
22211.25 10
2.5 10 M0.05
H2S 2H+ + S2–
2 2
1 2
2
[H ] [S ]K K
[H S]
2 2121 [H ] 2.5 101 10
0.1
2 1[H ]25
1[H ] 0.2 M
5
Ans 0.20
JEE (ADVANCED)-2020 (Paper-2)
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PART-III : MATHEMATICS
SECTION - 1 (Maximum Marks : 18)
This section contains SIX (06) questions.
The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 TO 9, BOTH INCLUSIVE.
For each question, enter the correct integer corresponding to the answer using the mouse and theon-screen virtual numeric keypad in the place designated to enter the answer.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct integer is entered;
Zero Marks : 0 If the question is unanswered;
Negative Marks : –1 In all other cases.
1. For a complex number z, let Re(z) denote the real part of z. Let S be the set of all complex numbers
z satisfying z4 – |z|4 = 4i z2, where i 1. Then the minimum possible value of |z1 – z2|2, where
z1, z2 S with Re(z1) > 0 and Re(z2) < 0, is _____
Answer (8)
Sol. Let z2 = a + ib
z4 – |z|4 = 4iz2
(a2 – b2) + 2abi – (a2 + b2) = 4i(a + ib)
b2 = 2b and ab = 2a
So z2 = 0 or z2 = + 2i, R
Let z = x + iy
x2 – y2 + 2xy = + 2i
xy = 1
Locus of z is a rectangular hyperbola xy = 1
If Re(z1) > 0 and Re(z2) < 0 then
Min. distance between z1 and z2 is the line segment joining two vertices of this hyperbola.
2 21 2 min
z z (1 1) (1 1) 8
21 2 minz z 8
2. The probability that a missile hits a target successfully is 0.75. In order to destroy the targetcompletely, at least three successful hits are required. Then the minimum number of missiles thathave to be fired so that the probability of completely destroying the target is NOT less than 0.95,is _____
Answer (6)
Sol. Probability to hit the target 3
H4
Probability to miss the target 1
M4
JEE (ADVANCED)-2020 (Paper-2)
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Required Probability that atleast 3 missile hit the target is
Here 1 – (P(x = 0) + P(x = 1) + P(x = 2)) 0.95
Here P(x = r) represents that r missile hits.
Here n n 1 2 n 2
n n n0 1 2
1 3 1 3 11 C C C 0.95
4 4 4 4 4
n
2 2n 3 9n(n 1) 1202 4
...(1)
By putting different value of n.
We will get least value of n is 6.
For n = 5 : L.H.S. of equation (1) is 212
0.107031252 1024
For n = 6 : L.H.S. of equation (1) is 308 0.075
0.0372 4096 2
Hence n = 6 missiles should be fired.
3. Let O be the centre of the circle x2 + y2 = r2, where 5
r .2
Suppose PQ is a chord of this circle and
the equation of the line passing through P and Q is 2x + 4y = 5. If the centre of the circumcircle ofthe triangle OPQ lies on the line x + 2y = 4, then the value of r is _____
Answer (2)
Sol. P
O
Q
R(x , y )1 1M
2x + 4y = 5
The circumcircle of OPQ is mid-point of OR.
Coordinate of O = (0, 0)Let coordinate of R be (x1, y1) then equation of PQ is
xx1 + yy1 = r2
On comparing with 2x + 4y = 5, we get :
21 1x y r
2 4 5
2 2
1 12r 4r
R (x ,y ) ,5 5
Circumcentre 2 2r 2r
M ,5 5
JEE (ADVANCED)-2020 (Paper-2)
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M lies on x + 2y = 4
2 2r 4r
45 5
r2 = 4
r = 2
4. The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matrixsuch that the trace of A is 3 and the trace of A3 is –18, then the value of the determinant of A is
Answer (5)
Sol. Let 2x y
A A 3x x yzz 3 x
Now
2 x y x yA
z 3 x z 3 x
2 2
2 2
xy 3y xy 3yx yz x yz
xz 3z xz 3zyz 9 x 6x yz 9 x 6x
2
32
3y x yx yzA
3z z 3 xyz 9 x 6x
3 2 2
2 2 2 3 2
x xyz 3yz x y y z 9y 3xy
3xz yz 9z x z 6xz 6yz 27 3x 18x xyz 9x x 6x
Trace 3 2 3 2x xyz 3yz 3yz 27 3x 18x xyz 9x x 6x 18
9yz + 9x2 – 27x + 27 = –18
yz + x2 – 3x + 3 = –2
x2 – 3x + yz = –5
3x – x2 – yz = 5
|A| = 5
5. Let the functions f : (–1, 1) and g : (–1, 1) (– 1, 1) be defined byf(x) = |2x – 1| + |2x + 1| and g(x) = x - [x],
where [x] denotes the greatest integer less than or equal to x. Let f o g: (–1, 1) be thecomposite function defined by (f o g) (x) = f(g(x)). Suppose c is the number of points in the interval(–1, 1) at which f o g is NOT continuous, and suppose d is the number of points in the interval (–1, 1)at which f o g is NOT differentiable. Then the value of c + d is _____
Answer (4)
Sol. Here f(x) = |2x – 1| + |2x + 1| (fog)(x) = |2{x} – 1| + |2{x} + 1|
12x 1 2x 3 , x 1,
2
12x 1 2x 3 , x , 0
2fog x
12x 1 2x 1, x 0,
2
12x 1 2x 1, x , 1
2
JEE (ADVANCED)-2020 (Paper-2)
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12, x 1,
2
14x 4, x , 0
2fog (x)
12, x 0,
2
14x, x , 1
2
f(g(x)) is discontinuous at x = 0.
and
10, x 1,
2
14, x ,0
2fog x
10, x 0,
2
14, x ,1
2
f(g(x)) is non-differentiable at 1 1
x , 0,2 2
c = 1 and d = 3.
c + d = 4
6. The value of the limit
x2
4 2 sin3x sinxlim
3x 5x 3x2sin2x sin cos 2 2 cos2x cos
2 2 2
is _____
Answer (8)
Sol.x
2
4 2(2sin2x cosx)lim
x 7x 5x 3xcox – cos cos – cos – 2(1 cos2x)
2 2 2 2
2
2x2
16 2 cos x sinxlim
3x 5x2cos .cosx – 2cox .cosx – 2 2 cos x
2 2
2
2x2
16 2 cos x sinxlim
x2cos x 2sin2x sin – 2 2 cos x
2
x 2
16 2 sinx 16 2lim
x 4 2 – 2 28sinx sin – 2 22
= 8
JEE (ADVANCED)-2020 (Paper-2)
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SECTION - 2 (Maximum Marks : 24)
This section contains SIX (06) questions.
Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correctanswer(s).
For each question, choose the option(s) corresponding to (all) the correct answer(s).
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, both ofwhich are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is acorrect option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –2 In all other cases.
7. Let b be a nonzero real number. Suppose f : is a differentiable function such that f(0) = 1. If thederivative f of f satisfies the equation
2 2
f(x)f (x)
b x
for all x , then which of the following statements is/are TRUE?
(A) If b > 0, then f is an increasing function
(B) If b < 0, then f is a decreasing function
(C) f(x)f(–x) = 1 for all x
(D) f(x) – f(–x) = 0 for all x
Answer (A, C)
Sol. 2 2
f x 1f x b x
2 2f (x) dx
dx cf(x) b x
ln 11 x
f(x) tan cb b
( f(0) = 1 c = 0)
11 xtan
b bf(x) e
So 11 xtan
b bf(x) e or
11 xtanb bf(x) e
but f(0) = 1, so 11 xtan
b bf(x) e
f(x) > 0 for all values of b
JEE (ADVANCED)-2020 (Paper-2)
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8. Let a and b be positive real numbers such that a > 1 and b < a. Let P be a point in the first quadrant
that lies on the hyperbola 2 2
2 2
x y– 1
a b. Suppose the tangent to the hyperbola at P passes through the
point (1, 0), and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinateaxes. Let denote the area of the triangle formed by the tangent at P, the normal at P and the x-axis.If e denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?
(A) 1 e 2 (B) 2 e 2
(C) 4a (D) 4b
Answer (A, D)
Sol. P(asec , btan )
(1, 0)A 2 2a b ,0
B
Let P (asec, btan)
x yT : sec tan 1
a b
(1, 0) lies on the tangent, so a = sec
2 2ax byN : a bsec tan
Slope of normal is –1, so a
sin 1b
b = tan
hence, a2 – b2 = 1
2 2
2 2 2
b a 1 1e 1 1 2
a a a
As a > 1, so e 1, 2
Area of PAB 2 21 a b 1 .b tan2
2 2 4 41 2 tan tan tan b2 9. Let f : and g: be functions satisfying
f(x + y) = f(x) + f(y) + f(x)f(y) and f(x) = xg(x)
for all x, y . If
x 0lim g(x) 1, then which of the following statements is/are TRUE?
(A) f is differentiable at every x (B) If g(0) = 1, then g is differentiable at every x (C) The derivative f(1) is equal to 1
(D) The derivative f(0) is equal to 1
Answer (A, B, D)
JEE (ADVANCED)-2020 (Paper-2)
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Sol.
h 0
f(x h) f(x)f (x) lim
h
h 0
f(x) f(h) f(x) f(h) f(x)f (x) lim
h
h 0
f(h)f (x) lim (1 f(x))
h
f (x) 1 f(x)
f (x)
11 f(x)
ln(1 + f(x)) = x + c
1 + f(x) = ex + c ( f(0) = 0 c = 0) f(x) = ex – 1
f(x) = ex
10. Let , , , be real numbers such that 2 + 2 + 2 0 and + = 1. Suppose the point (3, 2, –1) isthe mirror image of the point (1, 0, –1) with respect to the plane x + y + z = . Then which of thefollowing statements is/are TRUE?
(A) + = 2 (B) – = 3
(C) + = 4 (D) + + =
Answer (A, B, C)
Sol. P(3, 2, –1) and Q(1, 0, –1). If image of P is Q w.r. t plane . Then PQ is normal to the plane and mid
point of PQ lies on plane .
: 1 (x – 2) + 1 (y – 1) + 0 (z + 1) = 0
: x + y = 3
Clearly = 0, so = 1 (because + = 1)
Then = 1 and = 3
11. Let a and b be positive real numbers. Suppose ˆ ˆPQ ai bj
and ˆ ˆPS ai – bj
are adjacent sides of a a
parallelogram PQRS. Let u and v be the projection vectors of ˆ ˆW i j
along PQ and PS
,
respectively. If u v w
and if the area of the parallelogram PQRS is 8, then which of the following
statements is/are TRUE?
(A) a + b = 4
(B) a – b = 2
(C) The length of the diagonal PR of the parallelogram PQRS is 4
(D) w
is an angle bisector of the vectors PQ and PS
Answer (A, C)
Sol.
2 2 2 2
a b a – bv and u
a b a b
2 2
a b a – bv u w 2
a b
WLOG, a b
JEE (ADVANCED)-2020 (Paper-2)
33
So
2 22a a b
a = b
Also area of PQRS PQ PS 2ab 8
So a = b = 2, Also
2 2PQ PS a – b 0
So PQRS is a square of side length 2 2
12. For nonnegative integers s and r, let
s!if r s,
s r !(s – r)!
r0 if r s.
For positive integers m and n, let
m n
p 0
f(m,n,p)g(m,n)
n p
p
where for any nonnegative integer p,
p
i 0
m n i p nf(m,n,p) ,
i p p – i
Then which of the following statements is/are TRUE?
(A) g(m, n) = g(n, m) for all positive integers m, n
(B) g(m, n + 1) = g(m + 1, n) for all positive integers m, n
(C) g(2m, 2n) = 2g(m, n) for all positive integers m, n
(D) g(2m, 2n) = (g(m, n))2 for all positive integers m, n
Answer (A, B, D)
Sol.
pm
ii 0
p nn if(m,n,p) C
p n i – p n i p – i
p
mi
i 0
n p 1f m,n,p C
p p – i n – (p – i)
p
m ni p–i
i 0
n pf m,n,p C C
n p
m n n p m np p pn p
f m,n,p C C Cn p
m nm n m n
pp 0
g(m,n) C 2
JEE (ADVANCED)-2020 (Paper-2)
34
SECTION - 3 (Maximum Marks : 24)
This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screenvirtual numeric keypad in the place designated to enter the answer. If the numerical value has morethan two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct numerical value is entered.
Zero Marks : 0 In all other cases.
13. An engineer is required to visit a factory for exactly four days during the first 15 days of every monthand it is mandatory that no two visits take place on consecutive days. Then the number of all possibleways in which such visits to the factory can be made by the engineer during 1-15 June 2021 is _____
Answer (495)
Sol. Let he visit 1st time after x1 days
2nd time after x2 days
3rd time after x3 days
4th time after x4 days
and then x5 are remaining days
x1 + x2 + x3 + x4 + x5 = 11
x1, x5 0 and x2, x3, x4 1
Number of solutions = 12C4 = 12 11 10 9
55.9 4954 3 2
14. In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in sucha way that each of these rooms contains at least one person and at most two persons. Then thenumber of all possible ways in which this can be done is _____.
Answer (1080)
Sol. Groups can be only 2, 2, 1, 1
Number of ways of dividing persons in group = 22 26!
(2!) (1!) 2!
Number of ways after arranging = 46!
4! 1080(2!)
15. Two fair dice, each with faces numbered 1,2,3,4,5 and 6, are rolled together and the sum of thenumbers on the faces is observed. This process is repeated till the sum is either a prime number ora perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a primenumber. If p is the probability that this perfect square is an odd number, then the value of 14p is _____
Answer (8)
Sol. Output Ways to Occur
Perfect squares 1, 4, 9 7 (1 will never occur)
Primes 2, 3, 5, 7, 11 15
Remaining 6, 8, 10, 12 14
JEE (ADVANCED)-2020 (Paper-2)
35
Given process stopped by a perfect square = E (Let event name)
and Process stopped by a odd square = O (Let event name)
To find
O P(O E)P
E P(E)
P(O E) = Probability of occuring 9 before 1, 4, 2, 3, 5, 7, 11
24 4 14 14 4......
36 36 36 36 36
P(E) = Probability of occuring 4 or 9 before 2, 3, 5, 7, 11
27 14 7 14 7......
36 36 36 36 36
4O 436P p
7E 736
14p = 8
16. Let the function f : [0, 1] be defined by
x
x
4f(x)
4 2
.
Then the value of
1 2 3 39 1f f f ..... f f
40 40 40 40 2
is _____
Answer (19)
Sol. f(x) = x
x4
4 2
x 1 x
x 1 x4 4
f x f 1 x 14 2 4 2
1 2 3 39 1
f f f .... f f40 40 40 40 2
= 1 39 2 38 1
f f f f .... f40 40 40 40 2
= 20 1
1 1 ... 19 times f f40 2
= 19
JEE (ADVANCED)-2020 (Paper-2)
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17. Let f : be a differentiable function such that its derivative f' is continuous and f()=–6. If
F : [0, ] is defined by x
0F(x) f(t)dt , and if
0
f (x) F(x) cosx dx 2
,then the value of f(0) is __________
Answer (4)
Sol. F(x) = f(x) so F() = –6
0
f (x) F(x) cosx dx
=0 0
F (x)cosx dx F(x)cosx dx
= 0 0 0
F x cosx sinx.F x dx F x cosxdx
= 0 0 0F F 0 sinx.F x cosx.F x dx F(x)cosx dx
= 6 – F(0)
6 – F(0) = 2 F(0) = 4 = f(0)
18. Let the function f : (0, ) be defined by2 4f( ) (sin cos ) (sin cos ) ,
Suppose the function f has a local minimum at precisely when 1 r{ ,......, }, where 0 < 1 < ....
r < 1. Then the value of 1 + .... + r is _______
Answer (0.50)
Sol. f() = (sin + cos)2 + (sin – cos)4
Let (sin + cos)2 = x
f() = x + (2 – x)2 = x2 – 3x + 4
2
3 7f( ) x –
2 4
f() is least when 3
x2
3
sin cos2
3
sin4 2
2or4 3 3
5or12 12
So, 11
12 and 2
512
1 212
= 0.50
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