iit-jee advanced examination paper-2 27-09-2020 p+c+m

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JEE Advanced Exam 2020 (Paper & Solution)

PAPER-2

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 integer corresponding to the answer using the mouse and the on-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.

Q.1 A large square container with thin transparent vertical walls and filed with water (refractive index 43

) 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 ...................... .

OO

12cm

Line of sight

Ans. [3] Sol.

IO

12cm

NO

LK

M

J

45º

OJ

12 = cos 45º

OJ = 12 × 12

= 122

R.D. = 122

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Date : 27 / 09 / 2020

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A.D.R.D.

= 34

A.D. = 34

× 122

= 92

OI = 122

– 92

= 32

ON = 32

× 12

= 32

So Ans. = 2 × 32

= 3 cm

Q.2 A train with cross-sectional area St is moving with speed vt inside a long tunnel of cross-sectional area

S0 (S0 = 4St). Assume that almost all the air (density ρ) 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 p0 – p = 72N

ρνt2. The value of N is ........................... .

Ans. [9] Sol.

V = ? P = ?

R

VtSt

P0

4St

SVt

Apply continuity equation at R and S 4st × Vt = V × 3 St

43

Vt = V ….(i)

Apply Bernoulli equation at R and S

Pρ

+ 2V

2 = 0P

ρ + 2tV

2

P0 – P = 2ρ (V2 – Vt

2) ….(ii)

Put the value of equation (i) in equation (ii)

P0 – P = 718

ρVt2

718

ρVt2 = 7

2N ρVt

2

N = 9

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Q.3 Two large circular discs separated by a distance of 0.01m 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 centre 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)

0.01 m 200 V

Switch

Ans. [6] Sol. qε = Mg

q = mgε

= 7 34900 (8 10 ) 10

52000.01

−× π × ×

N = qe

= 3 21

19900 4 8 10 10 0.01

200 1.6 10

−

−

× π × × × ×

× ×

N = 6 Q.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 ρ(h) = 0

hh

0e−

ρ , where ρ0 = 1.25 kg m–3 and h0 = 6000 m,

the value of N is ........................ . Ans. [4] Sol. mg = Fu 480 × 10 = ρvg

480 × 10 = 0

hh

0e−

ρ

480 × 10 = 1006000

0e−

ρ Vg ….(i)

(480 – N × 1) 10 = ρ1 vg

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(480 – N)10 = 1506000

0e−

ρ Vg ….(ii)

Divide equation (i) and equation (ii)

480480 N−

= 100 150......6000 6000e

⎛ ⎞− − −⎜ ⎟⎝ ⎠

480480 N−

= 150 100

6000e−⎛ ⎞

⎜ ⎟⎝ ⎠

480480 N−

= 50

6000e

480480 N−

= 1

120e

N = 4 Q.5 A point charge q of mass m is suspended vertically by a string of length . 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 θ

is the same for all forces, where F is any one of the forces

and θ is the angle between the other two forces).

α

p→

qh

Ans. [2] Sol.

α

T

h

R Pε

C mg

α/2 A

902α⎛ ⎞

−⎜ ⎟⎝ ⎠

w = Vf – Vi

w = mgh + 2kpr

q – 0

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w = mgh + 2kpqr

….(i)

Apply Sreli's law at point R

qsin(180 )

ε− α

= mg

sin 302α⎛ ⎞+⎜ ⎟

⎝ ⎠

q = mgε

sin

cos2

αα⎛ ⎞

⎜ ⎟⎝ ⎠

= 3mg r

2Kp 2sin cos

2 2

cos2

α α⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

α⎛ ⎞⎜ ⎟⎝ ⎠

= 3mgr

Kp sin

2α⎛ ⎞

⎜ ⎟⎝ ⎠

w = mgh + 3

2Kp mgr

r Kp sin

2α⎛ ⎞

⎜ ⎟⎝ ⎠

w = mgh + mgr sin 2α⎛ ⎞

⎜ ⎟⎝ ⎠

w = mgh + mg 2 sin2 2α⎛ ⎞

⎜ ⎟⎝ ⎠

….(ii)

∴ In ΔARC = 2

h

2sin2α⎛ ⎞

⎜ ⎟⎝ ⎠

Put this value in Eq. (ii) w = mgh + mgh = 2mgh Q.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 4m 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–1 K–1).

8m

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Ans. [6] Sol.

xA

B 8 – x

Initial Condition PA × 4 = 0.1 R × 300 ….(i) For both partition 'A' and 'B' After release For A PAA × x = 0.1 × R × 300 ….(ii) For B PBA (8 – x) = 0.1 × R × 30 ….(iii) At equilibrium PAA + mg = PBA PAA + 8.3 g = PBA ….(iv) Put the value of equation (ii) and equation (iii) in equation (iv)

0.1 8.3 300y

× × + 8.3g = 0.1 8.3 3008 y

× ×−

y2 – 2y – 24 = 0 y = 6m

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) correct

answer(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 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 questions is unanswered); Negative Marks : –2 In all other cases.

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Q.7 A beaker of radius r is filled with water (refractive index 43

) 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 ω. This makes the water surface curved so that the difference in the height of water level at the centre and at the circumference of the beaker is h (h << H, h << r), as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature R. Which of the following is/are correct? (g is the acceleration due to gravity)

H

h

ω

(A) R = 2 2h r2h+

(B) R = 23r

2h

(C) Apparent depth of the bottom of the beaker is close to 3H2

12H1

2g

−⎛ ⎞ω

+⎜ ⎟⎜ ⎟⎝ ⎠

(D) Apparent depth of the bottom of the beaker is close to 3H2

12H1

4g

−⎛ ⎞ω

+⎜ ⎟⎜ ⎟⎝ ⎠

Ans. [A,D] Sol.

h

R–h A

R

B r C

9h ΔABC (R – h)2 + r2 = R2 R2 + h2 – 2Rh + r2 = R2

h2 + r2 = 2Rh

2 2h r2h+ = R ….(i)

r >>> h

R = 2r

2h

∵ h = 2 2w r2g

R = 2

2 2r 2g

2w r

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R = 2g

w ….(ii)

1Vμ – 2

uμ = 1 1

Rμ − μ

1V

+ 43u

=

413

R

−

1V

= – 1 43R 3(H h)

⎛ ⎞+⎜ ⎟−⎝ ⎠

1V

= – 1 43R 3H

⎡ ⎤+⎢ ⎥

⎣ ⎦

Put the value of equation (ii) h << H

1V

= –24 3H w1

3H 4 39⎡ ⎤⎡ ⎤

+⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

V = – 123H w H1

4 4

−⎡ ⎤⎛ ⎞⎢ ⎥+⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Q.8 A student skates up a ramp that makes an angle 30º 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)

h

Rz

y

x

30º

(A) ν02 – 2gh = 1

2gR

(B) ν02 – 2gh = 3

2 gR

(C) the centripetal force required at points x and z is zero (D) the centripetal force required is maximum at points x and z Ans. [A,D] Sol. Apply energy conservation

12

mv02 = mgh + 1

2mv2

12

mv2 = 12

mv02 – mgh ….(i)

at top point 'y'

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mg sinθ = centripetal force = 2mv

r

2mv

R = mg sinθ

Put this value in equation (i)

mg Rsin2

θ = 12

mv02 – mgh

mgR2

12

= 12

mV02 – mgh

yR2

= V02 – 2gh

Q.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 ω about the pivot. The maximum angular speed ωM is achieved for x = xM. Then

xL

→ν

(A) ω = 2 23vx

L 3x+ (B) ω = 2 2

12vxL 12x+

(C) xM = L3

(D) ωM = v2L

3

Ans. [A,C,D] Sol. According to conservation of Angular momentum

mvx = 2

2mL mx w3

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

vx = 2

2L x w3

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

w = 2 23vx

L 3x+

For maximum

dwdx

= 0

x = L3

w = 3v2L

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Q.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 I2

, the

potential difference V is increased to 2V, and the separation distance d is reduced to d2

, 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 Ans. [A,C]

Sol. λ ∝ 1v

So when 'V' double then cut off wavelength become half. Characteristic does not depend on voltage. when I half then intensity will decrease. Q.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 α. The spheres are now immersed in a dielectric liquid of density 800 kg m–1 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 Ans. [A,C] Sol.

Tcosθ

qε

mg Tsinθ

θ T θ

T cosθ = mg ….(i) T sinθ = qε ….(ii) qε = mg tanθ ….(iii) when medium introduce

FV

qKε

mg

θ T1

θ

T1 cosθ = mg – f4

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T1 cosθ = mg s

1⎛ ⎞ρ

−⎜ ⎟⎜ ⎟ρ⎝ ⎠ ….(iv)

T1 sinθ = qKε …..(v)

Divide equation (v) and (ii)

1T

T = q

kqΣΣ

T1 = Tk

= T21

….(vi)

Divide equation (iv) and (v)

qkε = mg

s1

⎛ ⎞ρ−⎜ ⎟⎜ ⎟ρ⎝ ⎠

tanθ

qε = kmg s

1⎛ ⎞ρ

−⎜ ⎟⎜ ⎟ρ⎝ ⎠ tanθ ….(vii)

equate eqn (vii) and (iii) ρs = 840 kg/m3

Q.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 y = 2x

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 = 1s, the angle between the particle's velocity and the x axis is 45º Ans. [A,B,C,D]

Sol. y = 2x

2

dydt

= 2x2

dxdt

Vy = x Vx

ydvdt

= x xdvdt

+ dxdt

vx

ay = x ax + vx2

Put the condition of option you get the answer.


• This section contains Six (06) questions. The answer to each questions is a NUMERICAL VALUE. • For each questions, 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. trucate/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.

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Q.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 << R the magnitude of the work done in the process is given by (4πp-Ra2) X, where X is a constant and γ = Cp/Cv = 41/30. The value of X is ............................ .

Ans. [2.05] Q.14 In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are

shown in the figure below. The resistance R3 has temperature coefficient 0.0004 C–1. If the temperature of R3 is increased by 100ºC, the voltage developed between S and ? T will be .................. volt.

V

R1 = 60 Ω

R3 = 300 Ω

R2 = 100 Ω

R4 = 500 ΩT

Q

S 50V

P

Ans. [0.26] Sol.

V

60 Ω

R1

100 Ω

500 ΩTD S

50V

C B

O I2I1 312Ω

50 E

A

R1 = R0 (1 + αΔT) = 300 (1 + 0.0004 × 100) R1 = 312 Ω

I1 = 50372

A

I2 = 50600

A

Vs – 50372

× 312 + 50600

× 500 = VT

Vs – VT = 0.26 V Q.15 Two capacitors with capacitance value C1 = 2000 ± 10pF 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 ...................... .

Ans. [1.30]

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Sol. eq

1C

= 1

1C

+ 2

1C

Ceq = 2000 30005000

× = 1200 pF

eq2eq

dCC

− = 1

21

dCC

− – 222

dCC

dCeq = 6PF

ε = 12

CV2

dqε × 100 = dC 2dV 100

C V⎛ ⎞+ ×⎜ ⎟⎝ ⎠

= 6 0.0221200 5

⎛ ⎞+ ×⎜ ⎟⎝ ⎠

× 100

= 1.3 %

Q.16 A cubical solid aluminium (bulk modulus = –V dpdv

= 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 ............................... .

Ans. [0.24]

Sol. K = PVV

−Δ

dVV

= – PK

V = 3

VV

Δ = 3 Δ

3 Δ = PK−

Δ = 0.238 × 10–3 = 0.238 mm = 0.24 mm Q.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.

I1

R1 = 5Ω

L 1 =

10

mH

V1 =

5V

I2

R2 = 10Ω

V2 =

20

V

M =

5m

H

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Ans. [55.00]

Sol. e1 = L1 1dIdt

+ M 2dIdt

e2 = L2 2 2L dIdt

+ 1MdIdt

dw1 + dW2 = e1I1 dt + e2I2dt dw = L1I1 dI1 + L2I2dI2 + MI1dI2 + MI2dI1 dw = L1I1dI1 + L2I2dI2 + Md(I1I2)

w = 2

1 1L I2

+ 2

2 2L I2

+ MI1I2

I1 = 55

= 1 A ; I2 = 2010

= 2A

w = 310 10 1

2

−× × + 320 10 4

2

−× × + 5 × 10–3 × 1 × 2

w = 55 × 10–3 J w = 55 mJ Q.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–2and 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)

Ans. [8.33]

Sol. dQdt

= eσA(T4 – T04)

For small temperature change.

dQdt

= eσAT3 ΔT ….(i)

mcdTdt

= eσAT3ΔT

dTdt

= 3e AT

mCσ ΔT

3e AT

mCσ → constant for Newton law of cooling

3e AT

mCσ = 0.001

eσAT3 = mC × 0.001 = 1 × 4200 × 0.001 eσAT3 = 4.2 ….(ii)

dQdt

= 700 × 0.05 = 35 Watts ….(iii)

Put the value of equation (ii), and equation (iii) in equation (i) 35 = 4.2 ΔT

354.2

= ΔT

ΔT = 8.33

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PART-II (CHEMISTRY) SECTION – 1 (Maximum Marks : 18)

• This section contains FOUR (06) questions • This 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 the on-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.

Q.1 The 1st, 2nd, and the 3rd ionization enthalpies, I1, I2 and I3, 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 ?

Ionization enthalphy (kJ/mol) Atomic number I1 I2 I3

n 1681 3374 6050

n + 1 2081 3952 6122

n + 2 496 4562 6910

n + 3 738 1451 7733

Ans. [9]

Sol. As in case of n + 2 we can see that ΔiH, is V–less means e– is removed from 1st group have low ΔiH & then from 2nd ΔiH2 is high because e– is removed from next lower shell.

n = Q = F

(n + 1) = 10 Ne

So, n + 2 = 11 = (Na) = 1s22s22p63s1

n + 3 = 12 = Mg = 1s22s22p63s2

Q.2 Consider the following compounds in the liquid form : O2, HF, H2O, NH3, H2O2, CCl4, CHCl3, C6H6, C6H5Cl

When a charged comb is brought near their flow8ing stream, how many of them show deflection as per the following figure ?

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Ans. [6] Sol. Deflected by the charged comb if it is polar molecule ∴ Polar Hf, H2O, NH3, H2O2, CHCl3, C6H5Cl (attracted) N.P. : O2 : CCl4 ; C6H6

Q.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 KMnO4 consumed ? Ans. [6] Sol. Redox reaction : Weak Basic

7 4

– –2 4 2 2

32(2H O 1MnO 3I MnO I 4HO )2

+ +−+ + ⎯⎯→ + +

– –2 4 2 24H O 2MnO 6I 2MnO 3I 8OH−+ + ⎯⎯→ + +

4 22mole MnO 3mole I− ⎯⎯→

4 24mole MnO 6mole I− ⎯⎯→

Q.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% H2O2, a blue alcohol layer was obtained. The blue colour is due to the formation of a chromium (VI) compound 'X'. What is the number of oxygen atoms bounded to chromium through only single bonds in a molecule of X ?

Ans. [4] Sol. Preparation of CrO5 →

Blue colour solution of Cr : Cr O

O

O

O +6

–1

–1

–1

–1

O–2

∴ (4) Cr = 0

24 2 2 5 3CrO H O H CrO H O− + ++ + ⎯⎯→ +

Imm.org.solvent = butan-1-ol / amylacetates

Q.5 The structure of a peptide is given below

HO

O

OH H2N

NH

O

O

NH

CO2H

NH2

If the absolute values of the net charge of the peptide at pH = 2, pH = 6, and pH = 11 are |z1|, |z2|, and

|z3|, respectively, then what is |z1| + |z2| + |z3| ? Ans. [5]

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Sol. The structure of tripeptide will be as followed at PH = 2 (in highly acidic medium)

HO

O

OH NH3

NH

O

O

NH

CO2H

NH3

⊕

Hence |z1| = 2 at pH = 6 (in approximately neutral solution it will be

HO

O

O–

NH3NH

O

O

NH

CO2

NH3

⊕

⊕

Hence |z2| = 0 at pH = 11 (in highly basic medium) the structure will be

–O

O

O H2N

NH

O

O

NH

CO2–

NH2

Hence |z3| will be 3 Therefore |z1| + |z2| + |z3| = 2 + 0 + 3 = 5 Q.6 An organic compound (C8H10O2) rotates plane-polarized light. It produces pink colour with neutral

FeCl3 solution. What is the total number of all the possible isomers for this compounds ? Ans. [6] Sol. Since Compound C6H10O2 corresponds positively to neutral FeCl3 hence it will be a phenoli compound

the possible structure which are optically active are as followed –

(d+ )=2

OH

CH–CH3

OH

•

,

(d+ )=2

OH

CH–CH3

OH

• ,

(d+ )=2

OH

CH

HO

•

CH3

Hence total optically active isomers will be 6

CAREER POINT


Paper-2


• This section contains Six (06) questions

• Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct option(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 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 option is chosen (i.e. the question is unanswered). Negative Marks : –2 In all other cases.

Q.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) ?

m X

(I) Balanced;

Magnetic field absent

m

(II) Upward deflection;

Magnetic field present

N SMagnet

m

(III) Downward deflection;

Magnetic field present

N S

(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. Ans. [A,B,C]

Sol. mass upward = diamagnetic (repelled by B ) Eg. H2O = 18 e K4[Fe(CN)6] Benzene

Downword = Paramagnetic (Attract by B )

CAREER POINT


Paper-2

Q.8 Which of the following plots is (are) correct for the given reaction ? ([P]0 is the initial concentration of P)

H3C Br

CH3

CH3

+ NaOH H3C OH

CH3

CH3

+ NaBr

P Q

(1) t1/2

[P]0

(2)

Initi

al ra

te

[P]0

(3) [Q]

time

[P]0

(4)

[P]

time

[P]0 In

Ans. [A]

Sol. (A) 1/2t = 0.693

K

(B) Rates ∝ (conc. of reactant)

(C) K = 2.303t

log 0RR

Kt = log 0RR

– Kt = log0

RR

– mn = y (D) C = C0e–kt a0 – x = a0e–kt

a0 – a0e–kt = x a0(1 – e–Kt) = x

–Kt x1 ea

− = = 0

QP

Q

t

P

Q.9 Which among the following statement(s) is(are) true for the extraction of aluminium from bauxite ? (1) Hydrated Al2O3 precipitated, when CO2 is bubbled through a solution of sodium aluminate (2) Addition of Na3AlF6 lowers the melting point of alumina (3) CO2 is evolved at the anode during electrolysis (4) The cathode is a steel vessel with a lining of carbon Ans. [A,B,C,D]

CAREER POINT


Paper-2

Sol. (A)

2 3

2 3 2 2 4Amphotericnature ofAl O

Al O H O NaOH NaAlO Na(Al(OH ))+ + ⎯⎯→ ⎯⎯→

4 2 2 3Na(Al(OH ) CO Al O ppt+ ⎯⎯→ ↓

(B) in c-conductivity by Na3AlF6 Q.10 Choose the correct statement(s) among the following (1) SnCl2.2H2O is a reducing agent (2) SnO2 reacts with KOH to form K2[Sn(OH)6] (3) A solution of PbCl2 in HCl contains Pb2+ and Cl– ions. (4) The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction. Ans. [A,B]

Sol. 2

42SnCl Sn

++⎯⎯→ : Act as reducing agent

SnO2 + KOH ⎯⎯→ K2SnO3 ⎯⎯→ [ ]26Sn(OH) − : amphoteric nature of SnO2

PbCl2 + H+ ⎯⎯→ [PbCl4]2–

Pb3O4 + HNO3 ⎯⎯→ Pb(NO3)2 + H2O + PbO2

↓ 2PbO + PbO2

4 3

2 3PbO HNO x+ +

+ ⎯⎯→

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

NH2

I

N

II

CH3H3C

NH2

III

NO2O2N

NO2

N

IV

NO2 O2N

NO2

CH3 H3C

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 and IV (D) Steric effect makes compound IV more basic than III Ans. [C,D] Sol. The correct basic strength order is (A) K0 : IV > II > I > III: * IV is stronger base due to SIR effect * III is weakest base due to – M group of three nitro groups present at Ortho and Para positions. * II is stronger than I since III is tertiary and I primary aromatic amine (B) IV is found to be 40,000 times more basic than III while I and II differ very little in basic strength (C,D) Due to SIR effect in IV both –NO2 and Ni(CH3)2 will be out of plane hence resonance effect is

more in III than in IV

CAREER POINT


Paper-2

Q.12 Consider the following transformations of a compound P.

R(Optically active)

(i) NaNH2

(ii) C6H5COCH3

(iii) H3O+/Δ

P(C9H12)

(i) X (reagent)

(ii) KMnO4/H2SO4/ΔQ

(C8H12O6) (Optically active acid)

Pt/H2

CH3

Choose the correct option (s)

(A)

P is (B) X is Pd-C/quinoline/H2

(C) P is (D) R is

Ans. [B,C]

Sol.

H2/Pd-C KMnO4/H2 Δ

Ph

Quinoline (X)

Na/Liq.MH3( )

Ph–C–CH3

O

OH

H Δ

COOHHOOC

COOH


• This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.

• For each question, enter the correct numerical value 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.

• Answer to each question will be evaluated according to the following marking scheme.

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

Zero Marks : 0 In all other cases.

CAREER POINT


Paper-2

Q.13 A solution of 0.1 M weak base (B) is titrated with 0.1 M of 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–

0 2 4 6 8 10 12

24

68

10

12

pH

Volume of HA (mL)

Ans. [3.00]

Sol. B + HA ⎯⎯→ BH+ + A–

0.1 0 0

0.1 – n n n

B + HA ⎯⎯→ BH+ + A–

Weak base + strong acid

At mid point = buffer region.

Conc. of salt & base get same (pH = 11) = (V = 3ml) = Half neutralization (Salt form)

At equivalent point = complete neutralization

pH + pOH = 14 slope = 0

ph =11 ∴ pOH = 3 pH n + charge = Buffer region

pOH = pKb + log (salt)(base) (0.1)→

[Salt=(base) : mid point] ↓

3 = pKb here conjugate acid+base

Q.14 Liquids A and B form ideal solution for all compositions of A and B at 25ºC. 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 is bar ?

Ans. [0.20]

Sol. xA = 0.25 xA = 0.50

xB = 0.75 xB = 0.50

PT = 0.3 bar PT = 0.4 bar

0 0T A A B BP P x P x= +

0.3 = 0AP × 0.25 + 0

BP × 0.75 0.4 = 0AP × 0.50 + 0

BP × 0.50

CAREER POINT


Paper-2

0.6 = 0.50 0AP + 1.50 0

BP

0.4 = 0.50 0AP + 0.50 0

BP _____________________ 0.2 = 1 0

BP

0BP = 0.2

0AP = 0.6

Q.15 The figure below is the plot of potential energy verses 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

Internuclear distance (d) →

Pote

ntia

l Ene

rgy

(k

J mol

–1) →

E0

H H d

d0

Ans. [–5246.50]

Sol. P.E. = – 1 2

0

Kq qr

= – 9 –19 2

–109 10 (1.6 10 )

0.53 10× × ×

×

= – –199 1.6 1.6 100.53

× ××

Total potential energy for 2 hydrogen atom (per mole)

= – –19 239 1.6 1.6 10 6 100.53

× ×× × × = –5246.5 KJ mole–1

Q.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) NaNO2 + HCl / 0-5 ºC

(ii) + NaOH

OH

(iii) CH3CO2H + H2O

Q

CAREER POINT


Paper-2

Ans. [21.75] Official Answer 18.60 Sol.

NH2 NaNO2 + HCl

N2⊕ClΘ

E⊕

+OH

OH

N = N – Ph

Acidic H

NaOH OΘ

N = N – Ph

CNuΘ

(Esterification) CH3–C––O–H / H⊕

O

N = N – Ph

O–C–CH3 + H2O

O

NH2

Now

N = N – Ph

O–C–CH3

O 1 mole 1 Mole (M.M. = 93 gm/mol) (M.M. = 290 gm/mol) Given 9.3 ml Given 75% obtained

Mass = d × U ⇒ 0.1 mole

= 1 × 9.3 ml ⇒ mass = 290 × 0.1 × 0.75 = 9.3 gm = 21.75 gm

Moles = 9.393

= 0.1 mole

Q.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; ΔfH0(SnO2(s)) = – 581.0 kJ mol–1, ΔfH0(CO2(g)) = – 394.0 kJ mol–1, S0(SnO2(s)) = 56.0 JK–1mol–1, S0(Sn(s)) = 52.0 JK–1mol–1, S0(C(s)) = 6.0 JK–1mol–1, S0(CO2(g)) = 210.0 JK–1mol–1 Assume that the enthalpies and the entropies are temperature independent Ans. [935.00]

CAREER POINT


Paper-2

Sol. SnO2 + C ⎯⎯→ Sn + CO2

ΔH = ΔHf (P) – ΔHf(r) = – 394 – (581) = 187 KJ mol–1 = 187000 J mol–1

ΔS = Sp – Sr = (52 + 210) – (56 + 6) = 202 – 62 = 200

∴ ΔG = H – Shigh

Δ ΤΔ

O = 187000 – T × 200 200 T = 187000 T = 935 K Q.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, KNET = K1K2 = 1 × 10–21 Ans. [0.20]

Sol. H2S 2H+ + S2–

0.1 0 0 0.1 – x 2x x

H2S ⎯⎯→ H+ + HS– HS– ⎯⎯→ H+ + HS–

0.1 0 0 x x 0 0.1 – x x x x – y x + y y

1

2

axK0.1

= x x y

10–21 Ka1 . Ka2 = 2x y0.1

× Ka2 = xyx

= y

–22

210

x = y

Q < ksp [Zn2+] [S2–] < 1.25 × 10–22

–22

0.051.25 10×

× 10–22 = x2

0.04 = x2 0.2 = x

CAREER POINT


Paper-2

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 the on-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.

Q.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 = 4iz2, 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 __________ .

Ans. [8] Sol. z4 – |z|4 = 4iz2

⇒ z4 – (z z )2 = 4iz2

⇒ ⎟⎠

⎞⎜⎝

⎛ +2

zz ⎟⎠

⎞⎜⎝

⎛ −2

zz = 4i

⇒ xy = 1 So, x1y1 = 1 and (x2y2 = 1) z1 = x1 + iy1 z2 = x2 + iy2 x1 > 0, x2 0 then y1 > 0, y2 < 0 |z1 – z2|2 = |(x1 – x2)2 + i (y1 – y2)|2

= 21x + 2

2x – x1x2 – x1x2 + 21y + 2

2y – y1y2 – y1y2

Use A.M. ≥ G.M.

⇒ 8

yyyyyyxxxxxx 212122

212121

22

21 −−++−−+ ≥ 8

1)1(

⇒ |z1 – z2|2 ≥ 8 So, minimum value of |z1 – z2|2 = 8. Q.2 The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely,

at least three successful hits are required. Then the minimum, number of missiles that have to be fired so that the probability of completely destroying the target in NOT less than 0.95, is __________ .

Ans. [6]

CAREER POINT


Paper-2

Sol. Probability (of destroying target) ≥ 0.95

3n3

3n

41

43C

−

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ + 4n4

4n

41

43C

−

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ + .... + 0n

nn

41

43C ⎟

⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ ≥ 10095

⇒ 1 – ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−− 2n2

2n

1n1

1n

n0

0n

41

43C

41

43C

41

43C ≥

10095

⇒ 1 – 10095 ≥ nnn 4·2

)1n(n94n3

41 −

++

⇒ 201 ≥ n

2

4·2n9n9n62 −++

⇒ 20

2 1n2 + ≥ 2 – 3n + 9n2

⇒ 22n – 1 ≥ 10 – 15n + 45n2 n = 6 satisfy the condition So minimum value of n = 6.

Q.3 Let O be the centre of the circle x2 + y2 = r2, where r > 25 . 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 of the triangle OPQ lies on the line x + 2y = 4, then the value of r is ___________ .

Ans. [2] Sol.

O

P

Q

(0,0) T M

r

(4 – 2α,α)

2x + 4y = 5

OT = PT = R In ΔOPM ⇒ PM2 = r2 – OM2 ...(i) In ΔPTM ⇒ PM2 = PT2 – MT2 PM2 = OT2 – MT2 ...(ii)

OM = 205

MT = 20

5448 −α+α− = 203

OT = OM + OT = 208

From Eq. (ii), PM2 = OT2 – MT2

PM2 = 2064 –

209 =

2055

From Eq. (i), 2055 = r2 –

2025

r = 2

CAREER POINT


Paper-2

Q.4 The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matrix such that the trace of A is 3 and the trace of A3 is –18, then the value of the determinant of A is ___________

Ans. [5]

Sol. Let A = ⎥⎦

⎤⎢⎣

⎡− a3cba

|A| = 3a – a2 – bc

A2 = ⎥⎦

⎤⎢⎣

⎡− a3cba ⎥

⎦

⎤⎢⎣

⎡− a3cba = ⎥

⎦

⎤⎢⎣

⎡

−++

2

2

)a3(bcc3b3bca

A3 = ⎥⎦

⎤⎢⎣

⎡

−++

2

2

)a3(bcc3b3bca ⎥

⎦

⎤⎢⎣

⎡− a3cba

Trace of (A3) = a3 + abc + 3bc + 3bc – abc + 3(3 – a)2 – a(3 – a)2 = – 18 ⇒ 3a – a2 – bc = 5 Q.5 Let the functions ƒ : (–1, 1) → R and g : (–1, 1) → (–1, 1) be defined by ƒ(x) = |2x – 1| + |2x + 1| and g(x) = x – [x] where [x] denotes the greatest integer less than or equal to x. Let ƒ B g : (–1, 1) → R be the composite

function defined by (ƒ B g) (x) = ƒ(g(x)). Suppose c is the number of points in the interval (–1, 1) at which ƒ B g is NOT continuous, and suppose d is the number of points in the interval (–1, 1) at which ƒ B g is NOT differentiable. Then the value of c + d is __________ .

Ans. [4] Sol. ƒ [g(x)] g(x) → discontinuous at x = 0 ƒ [g(0–)] = ƒ (1 – n) = 2 (1 – n) – 1 + 2 (1 – n) + 1 ƒ [g(0+)] = ƒ(n) = – (2n – 1) + 2n + 1 = 2 1c =

ƒ(x) is not differentiable at x = 21 and x = –

21

x = 21 ⇒ ƒ ′ ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ − n21g =

0nlim

→ h21gƒn

21gƒ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −

= 0n

lim→ h

21ƒn

21ƒ

−

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −

= 0n

lim→ h

2|1n21||1n21|−

−+−+−−

= 0n

lim→ h

2n22n2−

−−+ = 0

ƒ ′ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ + n21g =

0nlim

→ h2|1n21||1n21| −+++−+ =

0nlim

→ h22n4 −+ = 4

Similarly we can check at x = – 21

So d = 0, 21 , –

21

d = 3 4dc =+

CAREER POINT


Paper-2

Q.6 The value of the limit : 2

xlim

π→ ⎟

⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ +

+

2x3cosx2cos22

2x5cos

2x3sinx2sin2

)xsinx3(sin24 is __________ .

Ans. [8]

Sol. 2

xlim

π→ ⎟

⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ +

+

2x3cosxcos22

2x5cos

2x3sinx2sin2

)xsinx3(sin24

= 2

xlim

π→ ⎟

⎠⎞

⎜⎝⎛ +−+−

2x3cosxcos22

2x5cos

2x7cos

2xcos

xcos.x2sin282

= 2

xlim

π→ xcos22

2xsin.x3sin2

2xsinxsin2

xcos.xcosxsin2.2.242−+

= 2

xlim

π→ xcos22)xcos.x2sin2(

2xsin2

xcos.xsin2162

2

−

= 2

xlim

π→ 22xsin4.

2xsin2

xsin216

− =

2224216

− = 8.


• This section contains SIX (06) questions. • Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct

option(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

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 option is chosen (i.e. the question is unanswered). Negative Marks : –2 In all other cases.

Q.7 Let b be a nonzero real number. Suppose ƒ : R → R is a differentiable function such that ƒ(0) = 1. If the derivative ƒ′ of ƒ satisfies the equation

ƒ′(x) = 22 xb)xƒ(

+

for all x ∈ R, then which of the following statements is/are true ? (A) If b > 0, then ƒ is an increasing function (B) If b < 0, then ƒ is a decreasing function (C) ƒ(x) ƒ(–x) = 1 for all x ∈ R (D) ƒ(x) – ƒ(–x) = 0 for all x ∈ R Ans. [A,C]

CAREER POINT


Paper-2

Sol. ƒ′(x) = 22 xb)xƒ(

+ ⇒

)xƒ()x(ƒ′ = 22 xb

1+

⇒ ln ƒ(x) = b1 tan–1

bx + c

∴ ƒ(0) = 1 ∴ 0 = c ⇒ c = 0

∴ ln ƒ(x) = b1 tan–1

bx

⇒ ƒ(x) = bxtan

b1 1

e−

∴ ƒ′(x) = bxtan

b1 1

e−

. 22 xb1+

> 0 for x ∈ R

⇒ ƒ(x) is increasing option (A)

ƒ(x). ƒ(–x) = bxtan

b1 1

e−

– bxtan

b1 1

e−−

≠ 1 Q.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

ax – 2

2

by = 1. 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 coordinate axes. Let Δ 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) Δ = a4 (D) D = b4 Ans. [A,D] Sol. P(a sec θ, b tan θ) A (1, 0) B135° Tangent at P

b

tanya

secx θ−

θ = 1 passes through (1, 0)

sec θ = a Now slope of AP = 1 b sec θ = 1 ⇒ b = tan θ Now b2 = a2 (e2 – 1) tan2 θ = sec2 θ (e2 – 1) ⇒ e2 – 1 sin2 θ

e2 – 1 ∈ [0, 1] ∵ θ ∈ ⎥⎦

⎤⎢⎣

⎡ π2

,0

e2 ∈ [1, 2] ⇒ 1 < e < 2 Now A (1, 0) and P(a sec θ, a tan θ) = (sec2 θ, tan2 θ)

AP = θ+θ 22 tantan ⇒ AP = 2 tan θ (∵ ΔAPB is isosceles)

Area = 21 (AP)2 =

21 × 2 tan4 θ = b4

CAREER POINT


Paper-2

Q.9 Let ƒ : R → R and g : R → R be functions satisfying : ƒ(x + y) = ƒ(x) + ƒ(y) + ƒ(x) ƒ(y) and ƒ(x) = xg(x) for all x, y ∈ R. If

0xlim

→g(x) = 1, then which of the following statements is/are true ?

(A) ƒ is differentiable at every x ∈ R (B) If g(0) = 1, then g is differentiable at every x ∈ R

(C) The derivative ƒ′(1) is equal to 1 (D) The derivative ƒ′(0) is equal to 1 Ans. [A,B,D]

Sol. ∵ Put x = y = 0 is given relation

⇒ ƒ(0) = ƒ(0) + ƒ(0) + ƒ2(0)

⇒ ƒ(0) = 0 or – 1

∵ ƒ(x + y) = ƒ(x) + ƒ(y) + ƒ(x) . ƒ(y)

⇒ y

)xƒ()yxƒ( −+ = y

))xƒ(1)(yƒ( +

⇒ 0y

lim→ ⎥

⎦

⎤⎢⎣

⎡ −+y

)xƒ()yxƒ( = 0y

lim→

(1 + ƒ(x)) .y

)yƒ(

⇒ ƒ′(x) = 1 + ƒ(x)

⇒ ƒ′(0) = 1 + ƒ(0)

⇒ ƒ′(0) = 1 + 0

⇒ ƒ′(0) = 1

∵ 0x

lim→

(x) = 0x

lim→ x

)xƒ( = 1

Again )xƒ(1

)x(ƒ+

′ = 1 ⇒ ∫ +′

)xƒ(1dx)x(ƒ dx = ∫dx

⇒ ln (1 + ƒ(x)) = x + c

⇒ ln [1 + ƒ(x)] = x

⇒ 1 + ƒ(x))\ = ex

⇒ ƒ(x) = ex – 1 ⇒ ƒ′(x) = ex s

⇒ ƒ′(1) = e′

g(x) = x

)xƒ( = x

1ex − y′(0) = 0h

lim→

gh

)0(g)h0( −+

If g(0) = 1 then

g′(0+) = 0h

lim→ h

1h

1eh−

−

= 0h

lim→ 2

h

hh1e −− =

21

g′(0–) = 0h

lim→ h

)0(g)h0(g−

−− = 0h

lim→

hh

1e h

−−

−−

0h

lim→ 2

h

hh1e +−−

= 21

g(x) is differentiable.

CAREER POINT


Paper-2

Q.10 Let α, β, γ, δ be real numbers such that α2 + β2 + γ2 ≠ 0 and α + γ = 1. Suppose the point (3, 2, –1) is the mirror image of the point (1, 0, –1) with respect to the plane αx + βy + γz = δ. Then which of the following statements is/are true ?

(A) α + β = 2 (B) δ – γ = 3 (C) δ + β = 4 (D) α + β + γ = δ Ans. [A,B,C] Sol.

A(1, 0, –1)

B

A′(3, 2, –1) a(x + βy + γz) = δ

Find point at AA′ = B(2, 1, –1)

2α + β – γ = δ ...(i)

α + γ = 1 ...(ii)

dr’s of AA′ (2, 2, 0)

2α =

2β =

0γ = λ

∴ 2λ + 0 = 1, α = 2λ, β = 2λ, γ = 0

γ = 21

∴ α = 1, β = 1, γ = 0, δ = 3

Now (A) α + β + γ = 1 + 1 + 0 = 2 ≠ d

(B) α + β + 1 + 1 = 2

(C) γ + δ = 0 + 3 = 3

(D) δ + γ = – 3 – 0 = 3

Q.11 Let a and b be positive real numbers. Suppose PQ = jbia + and PS = jbia − are adjacent sides of a

parallelogram PQRS. Let →u and

→v be the projection vectors of

→w = ji + 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

Ans. [A,C]

CAREER POINT


Paper-2

Sol. ( jbia + ) × ( jbia − ) = 0

0ba0bakji

− = 8

4ab =

|→u | + |

→v | = |

→w |

→u = {

→w .( PQ )} ( PQ )

→v = {

→w . PQ } PQ

→u = ⎟

⎟⎠

⎞⎜⎜⎝

⎛

+

+22 ba

ba 22 ba

jbia+

+ ⇒ 22 ba)ba(

++ ( jbia + )

→v = 22 ba

ba+− ( jbia − ) |

→w | = 2

22

2

ba)aba(

+− i – 22

2

ba)bab(

+− j ⇒

22 baa2+

= 2

|→u | =

22 baba

+

+ ⇒ 22 ba1+

22 )b)ba(()a)ba(( −+−

|→v | =

22 baba

+

− = 22 ba)ba(

+−

2 22 ba + = 2a ab = 4

2a2 + 2b2 = 4a2 2ba ==

b2 = a2 a + b = 0 4ba =+

0ba =−

⇒ PS + PQ = ia2 Length = 2a = 4

Q.12 For non-negative integers s and r, let ⎟⎠⎞⎜

⎝⎛rs =

⎪⎩

⎪⎨⎧

>

≤−

srif0

,srif)!rs(!r

!s. For positive integers m and n, let

g(m, n) = ∑+

= ⎟⎠⎞⎜

⎝⎛ +

nm

0p ppn

)p,n,mƒ( , where for any non-negative integer p, ƒ(m, n, p) = ∑=

⎟⎠⎞⎜

⎝⎛

−+⎟

⎠⎞⎜

⎝⎛ +⎟

⎠⎞⎜

⎝⎛

p

0iipnp

pin

im . 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 Ans. [A,B,D]

jbia +

jbia −

CAREER POINT


Paper-2

Sol. !2!)1m(!m

− × !)pin(

!)1(n−+

+ × !)1n(!)1p(!)np(

+−+

!2!)1m(!m

− × !n!p

!)np( + × !)1p(!)p1n(!n

−−+

!2!)1m(!m

− × p + nCp . nCp – i

p + nCp .∑=

−

p

0iip

n1

m C.C

p + nCp . mC0 . nCp + mC1 . mCp – 1 + mC2 . nCp – 2 + ... mCp . nC0 (1 + x)m . (1 + x)n = mC0 + mC1x + mC2x ... (nC0 + nC1x + ... + nCpxp + ... p + nCp . m + nCp)

g(m, n) = ∑+

=

+nm

0pp

nm C = 2m + n

(A) g(2m, 2n) = 22m + 2n 2g (m, n) = 2.2m + n (B) g(m, n) = 2m + nH = g(mH, n) (C) g(2m, 2n) = 22m + 2n = (2m + n)2 = (g(m, n))2 (D) g(m, n) = g(n, m)


• 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-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.

• 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 If all other cases.

Q.13 An engineer is required to visit a factory for exactly four days during the first 15 days of every month

and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1-15 June 2021 is ______________ .

Ans. [495.00] Sol. 15C4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 – 4 ⇒ 11 + 1C4 ⇒ 12C4

Gaps selection

CAREER POINT


Paper-2

Q.14 In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two person. Then the number of all possible ways in which this can be done is ______________ .

Ans. [1080.00]

Sol. 4 Rooms → 6 People 6 → 1 1 2 2

!2!2!2!2!1!)1(

!6 × 4! = 1080

Q.15 Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the

numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ______________ .

Ans. [8.00] Sol. Prime/ Perfect square Prime → 2, 3, 5, 7, 11 (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2) (6, 1), (5, 2), (4, 3), (1, 6), (2, 5), (3, 4), (5, 6), (6, 5) Square = 4, 9 → (1, 3), (3, 1), (2, 2), (6, 3), (3, 6), (5, 4), (4, 5)

P (Perfect square before prime) = ⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛67

615

67

615

367 2

+ ......

P (Perfect square is odd) = 364

3614

364

3614

364 2

⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛+ + .......

36151

6/736151

36/4

−

− =

74

⇒ 74 × 14 = 8

Q.16 Let the function ƒ : [0, 1] → R be defined by ƒ(x) = 24

4x

x

+. Then the value of

ƒ ⎟⎠

⎞⎜⎝

⎛401 + ƒ ⎟

⎠

⎞⎜⎝

⎛402 + ƒ ⎟

⎠

⎞⎜⎝

⎛403 + ........ + ƒ ⎟

⎠

⎞⎜⎝

⎛4039 – ƒ ⎟

⎠

⎞⎜⎝

⎛21 is ______________ .

Ans. [19.00]

Sol. ƒ : (0, 1) → R

ƒ(x) = 24

4x

x

+

CAREER POINT


Paper-2

ƒ ⎟⎠

⎞⎜⎝

⎛401 + ƒ ⎟

⎠

⎞⎜⎝

⎛402 + ƒ ⎟

⎠

⎞⎜⎝

⎛403 + ........ + ƒ ⎟

⎠

⎞⎜⎝

⎛4039 – ƒ ⎟

⎠

⎞⎜⎝

⎛21

ƒ(1 – x) = 24

4x1

x

+− ⇒

xx

x

4.244

44

+ = x42

2+

ƒ(x) + ƒ(1 – x) = 1

ƒ ⎟⎠

⎞⎜⎝

⎛401 + ƒ ⎟

⎠

⎞⎜⎝

⎛4039 = 1 _____________________ 19 pairs

ƒ ⎟⎠

⎞⎜⎝

⎛4020 = ƒ ⎟

⎠

⎞⎜⎝

⎛21 – ƒ ⎟

⎠

⎞⎜⎝

⎛21 = 0

19

Q.17 Let ƒ : R → R be a differentiable function such that its derivative ƒ′ is continuous and ƒ(π) = – 6. If F : [0, π]

→ R is defined by F(x) = ∫x

0)tƒ( dt, and if ∫

π

=+′0

,2dxxcos))x(F)x((ƒ then the value of ƒ(0) is

_____________ .

Ans. [4.00]

Sol. ƒ(π) = – 6 F : [0, π] → R

F(x) = ∫π

0

dx)tƒ(

)( = )′( xƒxF )xƒ()x(F′ = 1

F(x) = ∫ )xƒ(

I = ∫π

′0

xcos)x(ƒ + ∫π

0

xcos)x(F

I = cos x ∫π

′0

)x(ƒ – ∫π

−0

)xƒ(xsin + F(x) – ∫π

0

xsin)xƒ(

I = π

0)xƒ(xcos +

π

0xsin)x(F

= – ƒ(π) – ƒ(0) + 0 I = – (ƒ(π) – ƒ(0)) I = (+6 – ƒ(0)) = 2 = + ƒ(0) = – 4

40ƒ +=)(

CAREER POINT


Paper-2

Q.18 Let the function ƒ : (0, π) → R be defined by ƒ(θ) = (sin θ + cos θ)2 + (sin θ – cos θ)4. Suppose the function ƒ has a local minimum at θ precisely when θ ∈ {λ1π, ..., λrπ}, where 0 < λ1 < .... < λr < 1. The the value of λ1 + .... + λr is ______________ .

Ans. [0.50] Sol. ƒ(θ) = (sin θ + cos θ)2 + (sin θ – cos θ)4

= sin2 θ + cos2 θ + 2sin θ cos θ + (sin2 θ + cos2 θ – 2sin θ cos θ)2

⇒ (1 + sin 2θ) + (1 – sin 2θ)2 ⇒ 1 + sin 2θ + 1 + sin2 2θ – 2sin 2θ ⇒ 2 + sin2 θ – sin 2θ ƒ(θ) = sin2 θ – sin 2θ + 2

47

212sinƒ

2+⎟

⎠

⎞⎜⎝

⎛ −θ=)θ(

θ ∈ (0, π)

sin 2θ = sin 6π , sin

65π

125,

12ππ

=θ

λ1 = 121 , λ2 =

125

λ1 + λ2 = 126 =

21

iit-jee advanced examination paper-2 27-09-2020 p+c+m

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