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1

Answers & Solutionsfor

JEE (Advanced)-2020

Time : 3 hrs. Max. Marks: 198

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005Ph.: 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 thewire from this corner, another student sees two images of the wire, located symmetrically on eachside 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 3

sini sin45 sini3 42

23

4 2 3

i i 32

= 45 – i

So slope of the line passing through (–12, 0) is m = tan(45 – i) =

1 tani1 tani

23 3m

23 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 2

0 t7

p p v .2N

The value of N is ________.

Answer (9)


3

Sol. Taking the velocity w.r.t. train

vvt

4Stvt = 3Stv

t4

v v3

Also ambient air will be moving with speed vt w.r.t. train

2 20 t t1 1 16

p v P v2 2 9

20 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 V 0.01 m

Answer (6)

Sol.

E 0.01 m

v 200E 100

l 1


4

4E 2 10 v/m

mg

Eq

Also q = Ne (N is the number of electrons)

NeE = mg

NeE = 34 r g

3

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

h0(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

100h

0150h

0

e480480 N

e

0

50h x480 50e e x

480 N 6000


5

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,

Fsin

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

–

2Fd

90 – 2

T

Clearly

dFmg Tsin 180

sin 90 sin 902 2

Aslo d 30

q2pF Eq

4 2lsin2

Initial PE = (0)

Final PE =

2

0

pqmgh W

4 2lsin2


6

Also 22lsin sin 2lsin h

2 2 2

d3

0

F mg 2pq mgsinsin cos cos4 2lsin2 22

3

0

mg 2lsin sinpq 2

4 2 cos2

3

2

mg 2lsin sin2

W mgh

2cos 2lsin2 2

mgsin 2lsin2W mgh

2cos2

mg 2sin cos lsin2 2 2W mghcos

2

2W mg 2lsin mgh 2 mgh2

N 2

6. A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into twoequal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thusthe partition is held initially at a distance of 4 m from the top, as shown in the schematic figurebelow. 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 theother. 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)


7

Sol. Assuming the gas to be an ideal mono-atomic gas

V3

C R2

and area of cross section to be A.

Initially

0P 4A 0.1R 300 (For both)

P1

P2

1 x

Let piston shifts by x meter

and final temperature is T.

Then

2P A(4 x) 0.1RT

2

0.1RT RTP A

4 x 10(4 x)

And 1RT

P A10(4 x)

But finally

2 1RT 1 1

P A P A 8310 4 x 4 x

2

RT 2x83

10 16 x

V V0.2 C 300 mg x 0.2C T

3 2 2 3

R 300 83x RT2 10 10 2

900R RT83x 3

10 10

300R 83 RTx

10 3 10

283x 2x

30R 833 16 x

2x 2x

83 3 833 16 x

2

(9 x) (2x)1

3 16 x


8

18x + 2x2 = 48 – 3x2

5x2 + 18x – 48 = 0

x 1.78 2

So, the distance from top is 6 m.



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).


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, andboth of which 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. 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 thecircumference of the beaker is h (h


9

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

4 411 3 3

v H R

1 4 1v 3H 3R

3HRv

4R H

Apparent depth = 1

3H H| v | 1

4 4R

=

123H H1

4 4g

8. A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown inthe figure) at the bottom of the ramp with speed v0 and wants to turn around over a semicircular pathxyz of radius R during which he/she reaches a maximum height h (at point y) from the ground asshown in the figure. Assume that the energy loss is negligible and the force required for this turn atthe 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)


10

Sol. At point y

mg sing30° = 2mv

R

2gR

v2

y

mg sin 30°xz

Using conservation of Mechanical Energy

2 20

1 1mv mgh mv

2 2

20

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 samemass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and getsembedded in it. The combined system now rotates with angular speed about the pivot. The maximumangular speed M is achieved for x = xM. Then

Lx

v

(A) 2 23vx

L 3x (B)

2 212vx

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

22 mLmvx mx

3

2 2

3vx

L 3x

Md L

0 xdx 3

& M3v

2L


11

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 willremain the same

(B) the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain thesame

(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 willdecrease

Answer (A, C)

Sol. As energy of incident electrons is doubled, cut off wavelength will reduce to half. Wavelength ofcharacteristic 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 acommon point by two non-conducting, massless strings of same length. At equilibrium, the anglebetween the strings is . The spheres are now immersed in a dielectric liquid of density 800 kg m–3

and 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

eFtan2 mg

& 2 2eT F (mg)

When immersed in liquid

e 22e

FFktan & T (mg B)

2 mg B k

b

mg k mg 1

bk 21 800k 1 21 1

= 840 kg/m3


12

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 ay, respectively. Then

(A) ax = 1 ms–2 implies that when the particle is at the origin, ay = 1 ms

–2

(B) ax = 0 implies ay = 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 & vy = 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


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.


Full Marks : +4 If ONLY the correct numerical value is entered.

Zero Marks : 0 In all other cases.


13

13. A spherical bubble inside water has radius R. Take the pressure inside the bubble and the waterpressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radiusbecomes (R – a). For a


14

Sol. 4 2 23 3 3 3R R [1 E) R [1 4 10 10 ) R [1 4 10 ]

3R 300[1 0.04] 300 1.04 3 104 312

150

I A372

P S 150

V V I 60 60372

500

60 100

312 = R 3

I1 I2TS

P

50 V

P S500

V V volt62

2

50 50 5I A

100 500 600 60

P T5 50

V V 100 volt60 6

P T P S S T50 500 500 500

(V V ) (V V ) V V6 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 TV V 0.27 volt

15. Two capacitors with capacitance values C1 = 2000 10 pF and C2 = 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 2

2 2 21 2 1 2

C C1 1 1 C

C C C C C C

2 1 22 21 2

C C 9 4C C 10 15 6nF

25 25C C

22

2 21

C 6000 1 95 25C (2000)


15

2 22

22

C 6000 1 45 3000 25C

2

1 U C 2 VU CV 100 100 100

2 U C V

U 6 5 0.02

100 100 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 waterand the acceleration due to gravity to be 103 kg m–3 and 10 ms–2, respectively, the change in the edgelength of the block in mm is _____.

Answer (0.24)

Sol. dP

V BdV

dV dPV B

3( dl) dP

l 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)


16

Sol. 15

I 1 A5

220

I 2 A10

Total work done by battery against induced emf = Increase in mag. energy

= U1 + U2 + U12

2 21 1 2 2 1 21 1

W L I L I MI I2 2

3 2 2 3 31 1

10 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 35 350 508.33º C

4.2 42 610 42 10


17

PART-II : CHEMISTRY



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.


Full Marks : +3 If ONLY the correct INTEGER value is entered.


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 I3n

n + 1

n + 2n + 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 asper the following figure?

Answer (6)

Sol. Polar liquid will be attracted towards charged comb. Hence liquid showing deflection are HF, H2O,

NH3, 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 32MnO H O I 2MnO 2OH IO


18

4. An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. Whenit was shaken after the addition of 1 mL of 3% H

2O

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 atomsbonded to chromium through only single bonds in a molecule of X?

Answer (4)

Sol. 24 2 2 5 2CrO 2H 2H O CrO 3H O

X is CrO5

CrO

O

O

O

O

Number of oxygen atoms bonded to chromium through only single bond is 4.

The blue colour is attributed to the presence of chromium pentaoxide.

5. The structure of a peptide is given below.

HO

H N2O

HN

NH

O

OH

OCO 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| + |Z2| + |Z3| = 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* — CH 3

OH

H

OH

C* — CH3

OH

HOH

HC* — CH3

OH

Phenolic group gives positive test with neutral FeCl3 solution.


19



Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correct

answer(s).





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;



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 C6H6(l), deflection of the pan is downwards.

Answer (A,B,C)

Sol. H2O(l), K4[Fe(CN)6](s) and C6H6(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).


20

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 r

ate

(C)

time

0

[Q][P]

(D)

time

0

[P]ln

[P]

Answer (A)

Sol. Rate [Substrate]1 (for SN1 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 Al2O3 precipitates, when CO2 is bubbled through a solution of sodium aluminate

(B) Addition of Na3AlF6 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. Al2O3 + 2NaOH + 3H2O 2Na[Al(OH)4]

2Na[Al(OH)4] + 2CO2 Al2O3xH2O + 2NaHCO3

Al2O3xH2O 1473K Al2O3(s) +

xH2O


21

10. Choose the correct statement(s) among the following.

(A) SnCl2.2H2O is a reducing agent

(B) SnO2 reacts with KOH to form K2[Sn(OH)6]

(C) A solution of PbCl2 in HCl contains Pb2+ and Cl– ions

(D) The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction

Answer (A,B)

Sol. Sn4+ > Sn2+ ... stability order

Sn2+ acts as a reducing agent.

SnO2 + 2KOH + 2H2O K2[Sn(OH)6]

In conc. HCl, PbCl2 exists as chloroplumbous acid, H2[PbCl4]

PbCl2 + 2HCl H2[PbCl4]

Pb3O4 + 4HNO3 2Pb(NO3)2 + PbO2 + 2H2O

(2PbO + PbO2)

11. Consider the following four compounds I, II, III, and IV.

N

H C3 CH3NH2

NH2NO2O N2

NO2

NNO2O 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.


22

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 4Q

(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*

(P)

X

(ii) KMnO /H SO /4 2 4

COOHCOOH

* COOH

(Q)



For each question, enter the correct numerical value of the answer using the mouse and the on-screen

virtual 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.





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 oT A A B BP x P x P

o oA B0.3 0.25 P 0.75 P ... (i)

o oA B0.4 0.5 P 0.5 P ... (ii)

From (i) and (ii)

oAP 0.6 bar

oBP 0.2 bar

Ans 0.20


24

15. The figure below is the plot of potential energy versus internuclear distance (d) of H2 molecule inthe electronic ground state. What is the value of the net potential energy E0 (as indicated in thefigure) in kJ mol–1, for d = d0 at which the electron-electron repulsion and the nucleus-nucleusrepulsion energies are absent? As reference, the potential energy of H atom is taken as zero whenits electron and the nucleus are infinitely far apart.

Use Avogadro constant as 6.023 × 1023 mol–1.

H Hd

d0

Internuclear distance (d)

Po

tent

ial e

nerg

y(k

J m

ol)

–1

E0

Answer (–5242.42)

Sol. Potential 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 areabsent.

P.E. of two H-atom = – 2 × 27.2 eV = – 54.4 eV

=

19 2354.4 1.6 10 6.023 10

kJ/ mol1000

= – 5242.42 kJ/mol

16. Consider the reaction sequence from P to Q shown below. The overall yield of the major productQ from P is 75%. What is the amount in grams of Q obtained from 9.3 mL of P? (Use density ofP = 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

N 2Cl

(P)

(Q)

(i) NaNO + HCl/0-5°C 2

+ –

(ii) OH

,

N = N

O

CH CO H/H O3 2 2

OH

NNaOH


25

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 × 34

= 18.6 g

17. Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine theminimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At 298 K: fH°(SnO2(s)) = –581.0 kJ mol–1, fH°(CO2(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) + CO2(g)

orH 394 581 187 kJ / mol

orS 52 210 56 6 200 J/ K mol

Tmin (at which reduction will take place) =

or

or

H935 K

S

18. An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molarconcentration (M) of H+ required to prevent the precipitation of ZnS?

Use Ksp (ZnS) = 1.25 × 10–22 and overall dissociation constant of H2S, KNET = K1K2= 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 ]

= 22

211.25 10 2.5 10 M0.05

H2S 2H+ + S2–

2 2

1 22

[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


26

PART-III : MATHEMATICS



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.


Full Marks : +3 If ONLY the correct integer is entered;



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


27

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 circumcentre of OPQ is mid-point of OR.

Coordinates of O = (0, 0)Let coordinates 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


28

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 6yz 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


29

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


30



Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correctanswer(s).





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;



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


31

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)


32

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


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


34



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.




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


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


36

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

Ans & Sol_JEE (Advanced)-2020_Physics_Paper-2Ans & Sol_JEE (Advanced)-2020_Chemistry_Paper-2Ans & Sol_JEE (Advanced)-2020_Maths_Paper-2

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