IIT-JEE Advanced Examination Paper-2 21-05-2017 Physics · 2018. 1. 19. · CAREER POINT Kota H.O....

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Paper-2 [ CODE – 5 ]

JEE Advanced Exam 2017 (Paper & Solution)

PAPER-2

PART-I (PHYSICS)

SECTION – 1 (Maximum Marks : 21)

• This section contains SEVEN questions

• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

• For each question, darken the bubble corresponding to the correct option in the ORS.

• For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases.

Q.1 A photoelectric material having work-function φ0 is illuminated with light of wavelength λ ⎟⎟⎠

⎞⎜⎜⎝

⎛φ

<λ0

hc .

The fastest photoelectron has a de Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then the ratio Δλd/Δλ is proportional to

(A) λ3d/λ (B) λd/λ (C) λ2

d/λ2 (D) λ3d/λ2

Ans. [D]

Sol. K.Emax = 0hc

φ−λ

λd = maxE.mK2

h

λd = ⎟⎠⎞

⎜⎝⎛ φ−

λ 0hcm2

h

λ2d =

⎟⎠⎞

⎜⎝⎛ φ−

λ 0

2

hcm2

h

CAREER POINT

Date : 21 / 05 / 2017

CAREER POINT



λ2d ⎟

⎠⎞

⎜⎝⎛ φ−

λ 0hc =

m2h2

⎟⎠⎞

⎜⎝⎛ φ−

λ 0hc =

m2h2

. ⎟⎟⎠

⎞⎜⎜⎝

⎛

λ2d

1

⎟⎟⎠

⎞⎜⎜⎝

⎛λ

λ−

=−λλ

− d3d

2

2 d2m2

h0dhc

2

3dd

dd

λλ

=λ

λ

Q.2 Consider regular polygons with number of sides n = 3, 4, 5 ...... as shown in the figure. The centre of

mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the centre of mass for each polygon is Δ. Then Δ depends on n and h as

hhh

(A) Δ = h ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−⎟⎠⎞

⎜⎝⎛ π

1

ncos

1 (B) Δ = h tan2 ⎟⎠⎞

⎜⎝⎛ π

n2 (C) Δ = h sin ⎟

⎠⎞

⎜⎝⎛ π

n2 (D) Δ = h sin2 ⎟

⎠⎞

⎜⎝⎛ π

n

Ans. [A] Sol.

nπ

nπ

h

a

h = a cos nπ

maximum height = a Δymax = a – h

= ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−π

h

ncos

h

= ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−π

1

ncos

1h

CAREER POINT



Q.3 Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of

fractional change in density ⎟⎟⎠

⎞⎜⎜⎝

⎛ρ dt

dp1 is constant. The velocity v of any point on the surface of the

expanding sphere is proportional to

(A) R3 (B) R (C) R1 (D) R2/3

Ans. [B] Sol.

Rρ

M = ρ ⎟⎠⎞

⎜⎝⎛ π 3R

34

ρR3 = constant

logρ + 3logR = constant

0dtdR

R3

dtd1

=+ρ

ρ

dt

ddtdR

R3

ρρ

−= ⎩⎨⎧

=ρ

ρdt

d constant

Velocity any point on surface RdtdR

∝

Q.4 A person measures the depth of a well by measuring the time interval between dropping a stone and

receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms–2 and the velocity of sound is 300 ms–1. Then the fractional error in the measurement, δL/L, is closest to

(A) 5% (B) 3% (C) 0.2% (D) 1% Ans. [D] Sol.

drop

L

Sound

t = ⎟⎠⎞

⎜⎝⎛+

vg2 ll

CAREER POINT



dt = ⎟⎟⎠

⎞⎜⎜⎝

⎛+

vdd.

21

g2 l

ll

vd

2ddt

2g l

l

l+=

vd

2ddt

2g ll

l

l+⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

v2ddt

2g ll

l

l

⎟⎟⎠

⎞⎜⎜⎝

⎛+=×

30020

220d01.0

210

l

l

⎟⎠⎞

⎜⎝⎛ +

×=

1515

01.05dl

l

% %1

1515

5100d≈

⎟⎠⎞

⎜⎝⎛ +

=×l

l

Q.5 A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure.

The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the centre of the loop is

I

4a

(A) a4I0

πμ 3[ 3 – 1] (B)

a4I0

πμ 6[ 3 – 1] (C)

a4I0

πμ 3[2 – 3 ] (D)

a4I0

πμ 6[ 3 + 1]

Ans. [B] Sol.

30º 30º a

O

A D

CAREER POINT



Bat O = a4iµ0

π[sin 60º × 2 – sin 30º × 2] × 6

= [ ]613a4iµ0 −

π

Q.6 A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the

Sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 × 104 times larger than the radius of the Earth. The escape velocity from Earth’s gravitational field is ve = 11.2 km s–1. The minimum initial velocity (vs) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

(A) vs = 42 km s–1 (B) vs = 62 km s–1 (C) vs = 22 km s–1 (D) vs = 72 km s–1 Ans. [A] Sol.

Earth Sun

R

m Msun = 3×105 mv

2.5 ×104 R

ve = RGm2

21 m0vs

2 – R

Gmm0 – R105.2

mm103G4

05

×××× = 0

21 vs

2 = R

Gm + R

Gm × 5.2

30

2

v2s =

RGm [1 + 12]

vs = 11.2 13 = 42 km/s

Q.7 Three vectors →P ,

→Q and

→R are shown in the figure. Let S be any point on the vector

→R . The distance

between the points P and S is b |→R |. The general relation among vectors ,

→P ,

→Q and

→S is

CAREER POINT



Y

O X

Q

P

S

b |→R |

→P →

S →Q

→R =

→Q –

→P

(A) →S = (1 – b2)

→P + b

→Q (B)

→S = (1 – b)

→P + b

→Q

(C) →S = (1 – b)

→P + b2

→Q (D)

→S = (b – 1)

→P + b

→Q

Ans. [B]

Sol. Sr

= RbPrr

+

Sr

= )PQ(bPrrr

−+

Sr

= Qb)b1(Prr

+−


• The section contains SEVEN questions

• Each question has FOUR options (A), (B), (C) and (D) . ONE OR MORE THAN ONE of these four option(s) is(are) correct.

• For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

• For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect

option is darkened Zero Marks : 0 If none of the bubble is darkened. Negative Marks : –2 In all other cases.

• For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get –2 marks, as a wrong option is also darkened.

Q.8 The instantaneous voltages at three terminals marked X, Y and Z are given by VX = V0 sin ωt,

VY = V0 sin ⎟⎠⎞

⎜⎝⎛ π

+ω3

2t and

CAREER POINT



VZ = V0 sin ⎟⎠⎞

⎜⎝⎛ π

+ω3

4t

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading (s) of the voltmeter will be

(A) 23VV 0

rmsXY = (B)

21VV 0

rmsYZ =

(C) Independent of the choice of the two terminals (D) 0rmsXY VV =

Ans. [B,C] Sol. Vxy = Vx + Vy

V0

V0

2π/3

Vnet = V0

∴ Vxy rms = 2

V0

Vyz = Vy + Vz

V0

V0

2π/3

Vnet = V0

Vyz rms = 2

V0

Q.9 A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the

figure. Which of the following statements is/are Correct?

R

+Q

(A) Total flux through the curved and the flat surfaces is 0

Qε

(B) The circumference of the flat surface is an equipotential (C) The component of the electric field normal to the flat surface is constant over the surface

(D) The electric flux passing through the curved surface of the hemisphere is ⎟⎟⎠

⎞⎜⎜⎝

⎛ε 2

1–12Q–

0

Ans. [D]

CAREER POINT



Sol. Lets take a close hemispherical surface of radius R

R

+Q

45º R

φtotal = φcurve + φFlat

0 = φ curve + ]º45cos–1[2q

0ε

φcurve = – ⎥⎦

⎤⎢⎣

⎡ε 2

1–12q

0

Component of electric field along the curve surface is non zero so curve surface is not equipotential. Electric field normal component or flat surface not constant Q.10 Two coherent monochromatic point sources S1 and S2 of wavelength λ = 600 nm are placed

symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ. Which of the following options is/are correct?

S1 S2P2

P1

d

Δθ

(A) The angular separation between two consecutive bright spots decreases as we move from P1 to P2

along the first quadrant (B) A dark spot will be formed at the point P2

(C) At P2 the order of the fringe will be maximum (D) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000 Ans. [C,D] Sol.

S1 S2 P2

P1

d

θ θ

CAREER POINT



Δx = dcosθ for constructive interference d cosθ = nλ

n = λ

θcosd

n ∝ cosθ

when θ↓ cosθ ↑ and separation between consecutive bright spots increases For P2 Δx = 1.8 mm for dark spot

Δx = (2n–1) 2λ

(2n–1) 2λ = 1.8 mm

(2n–1) 3–9

108.1210600

×=× −

2n –1 = 9

3

10600106.3

−

−

××

2n–1 = 60001061010

60036 63

8

3=×=× +−

−

−

2n = 6001

n = 2

6001

is not a integer. So it is not dark spot for maxima at P2 nλ = Δx n 300×10–9 = 1.8×10–3

n = 6

3

9

3

9

3

1010

618

1010

60018

10300108.1

−

−

−

−

−

−

×=×=××

= 3 × 103 n = 3000 So at P it is maxima Q.11 A uniform magnetic field B exists in the region between x = 0 and x = y (region 2 in the figure) pointing

normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region 2 from region 1 at point P1(y = –R). Which of the following option(s) is/are correct ?

CAREER POINT



Region 1 Region 2 Region 3×

×××××××

×

×××××××

×

×××××××

O

+Q P1

(y = –R)

P2x

B

3R/2

y

(A) For a fixed B, particles of same charge Q and same velocity v, the distance between the point P1 and

the point of re-entry into region 1 is inversely proportional to the mass of the particle.

(B) For B = 138

QRp , the particle will enter region 3 through the point P2 on x-axis

(C) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1 and the farthest point from y-axis is p/ 2

(D) For B > 32

QRp , the particle will re-enter region1

Ans. [B,D] Sol.

O

Br

2R3

y= –R

Region1

×

qBmvr =

The distance of re-entry point from P1 into region 1 in 2r

d = qBmv2

d∝m

(b) B = qR13p8

In such case radius of circular part is r = qBp

r = p8qqR13p

×× =

8R13 = 1.6 R

Which is greater then 2R3

CAREER POINT



R 8R13

8R13

2R3

⎟⎠⎞

⎜⎝⎛ 0,

2R3

In such case equation of circle is

(x – 0)2 + 2

8R5–y ⎟

⎠⎞

⎜⎝⎛ =

2

8R13

⎟⎠⎞

⎜⎝⎛

Point P2 ⎟⎠⎞

⎜⎝⎛ 0,

2R3 satisfy the above equation

So point P2 lies on circle Q.12 A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the

figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct ?

R

S

Q

X

P

(A) If the force is applied normal to the circumference at point X then τ is constant (B) If the force is applied tangentially at point S then τ ≠ 0 but the wheel never climbs the step (C) If the force is applied at point P tangentially then τ decreases continuously as the wheel climbs (D) If the force is applied normal to the circumference at point P then τ is zero Ans. [A,D] Sol. When force apply normal to the point x

F

x

F

Force and its perpendicular distance of line of action Remain constant so torque is constant

CAREER POINT



When force applies normal to point P its line of action always passes through point O so torque is

F

Q.13 A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2 through

a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following options is/are correct ?

+ – V L1 L2

S

(A) After a long time, the current through L1will be RV

21

2

LLL+

(B) The ratio of the currents through L1 and L2 is fixed at all times (t > 0)

(C) After a long time, the current through L2will be RV

21

1

LLL+

(D) At t = 0, the current through the resistance R is RV

Ans. [A,B,C] Sol.

dtdiL 2

2−

L1 dtdiL 1

1−

i1 i2

L2

R

V

at t = 0 No current through the resistor

dtdiL

dtdiL 2

21

1 −=−

L1 di1 = L2 di2 L1 i1 = L2 i2

1

2

2

1

LL

ii

= (Const.)

After a long time current in Resistor

RVi =

RVii 21 =+

CAREER POINT



RVi

LLi 1

2

11 =+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=21

21 LL

LRVi

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=21

12 LL

LRVi

Q.14 A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown

in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which of the following statements about its motion is/are correct ?

θ

B

L

A

O

θ

(A) Instantaneous torque about the point in contact with the floor is proportional to sin θ (B) The trajectory of the point A is a parabola (C) The midpoint of the bar will fall vertically downward (D) When the bar makes an angle θ with the vertical, the displacement of the its midpoint from the

initial position is proportional to (1 – cos θ) Ans. [C,D] Sol.

A

smooth x

B

COM

Y

θ

2l

θ

XA = – θsin2l

YA = θ+θ cos2

cos2

ll

YA = l cosθ

1Y4/

X2

2A

2

2A =+

ll

ellipse

CAREER POINT



COM (mid point) of bar will fall on vertical straight-line (there is no any horizontal force)

displacement of mid point = θ− cos22ll

= )cos1(2

θ−l

∝ (1 – cos θ)

acm

θ α

l/2N

mg

θ

α

θ

θα sin2l

2lα 90 – θ

wrt to COM

N α=θ12

msin2

2ll

mg – N = macm

cmasin2

=θαl

mg – θα

=θ

α sin2

msin6

m ll

g = lα ⎟⎠⎞

⎜⎝⎛ θ

+θ 2

sinsin61

α = )sin31(

sing62 θ+

θl

About the point which is in contact with ground.

τ = Iα

CAREER POINT



τ = 3

m 2αl

τ = ⎟⎟⎠

⎞⎜⎜⎝

⎛

θ+θ

)sin31(sing6

3m

2

2

l

l

τ = )sin31(

sinmg22 θ+

θl


• This section contains TWO paragraphs

• Based on each paragraph, there are TWO questions

• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct

• For each question, darken the bubble corresponding to the correct option in the ORS

• For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 In all other cases

PARAGRAPH 1 Consider a simple RC circuit as shown in Figure 1. Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e.,

charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.

Process 2 : In a different process the voltage is first set to 3

V0 and maintained for a charging time T >> RC.

Then the voltage is raised to 3V2 0 without discharging the capacitor and again maintained for a time

T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1.

These two processes are depicted in Figure 2.

R

S

+ – V

C

Figure 1

V0/3

2V0/3

V0Process 1

Process 2

T >> RC

T 2T Figure 2

t

V

CAREER POINT



Q.15 In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by :

(A) EC = ED ln 2 (B) EC = 21 ED (C) EC = ED (D) EC = 2ED

Ans. [C] Sol.

R

+ – V0 C

U = C2

q20 = energy stored

EC = 20CV

21 .......... (1)

Work done by battery = q0V0

= 20CV

∴ heat loss = 20CV –

C2q2

0

= 20

20 CV

21CV −

ED = 20CV

21 .......(2)

∴ 11

CV21

CV21

EE

20

20

C

D ==

Q.16 In Process 2, total energy dissipated across the resistance ED is :

(A) ED = 31

⎟⎠⎞

⎜⎝⎛ 2

0CV21 (B) ED = 3 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ 20CV

21

(C) ED = 20CV

21 (D) ED = 3 2

0CV

Ans. [A] Sol.

R

+ – 3

V0

C 3CVq 0=

18CVU

20

1 =

CAREER POINT



(battery) W = ⎟⎠⎞

⎜⎝⎛

3CV

3V 00

= 9

CV 20

Energy of capacitor = 18

CV 20

Heat loss E1 = 18

CV–9

CV 20

20

E1 = 20CV

181

R

+ –

3V2 0 C 0CV

32q =

C2qU

2

f =

= 20CV

184

20f CV

92U =

ΔU = 6

CV 20

(Battery) W = ⎟⎠⎞

⎜⎝⎛

3CV

3V2 00

= 20CV

92

∴ heat loss = 6

CV–CV

92 2

020

= 20CV

543

E2 = 18

CV 20

R

+ – C 0CVq =

0CV91q =Δ

V0

Work done by battery = V0 ⎟⎠⎞

⎜⎝⎛

0CV31

= 3

CV 20

ΔU = 20CV

185

CAREER POINT



Total heat loss E3 = 3

CV 20 – 2

0CV185

E3 = 20CV

181

∴ Total heat dissipated

E = E1 + E2 + E3

= 20

20

20 CV

181CV

181CV

181

++

= ⎟⎠⎞

⎜⎝⎛= 2

02

0 CV21

31CV

61

PARAGRAPH 2

One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In

the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface

of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the

finger is in contact is r. The finger rotates with an angular velocity ω0. The rotating ring rolls without

slipping on the outside of a smaller circle described by the point where the ring and the finger is in

contact (Figure 2). The coefficient of friction between the ring and the finger is μ and the acceleration

due to gravity is g.

Figure 1

R

R

Figure 2

Q.17 The total kinetic energy of the ring is

(A) 20Mω (R – r)2 (B) 2

0Mω R2

(C) 23 2

0Mω (R – r)2 (D) 21 2

0Mω (R – r)2

Ans. [A]

CAREER POINT



Q.18 The minimum value of ω0 below which the ring will drop down is

(A) )r–R(

g2μ

(B) )r–R(2

g3μ

(C) )r–R(

gμ

(D) )r–R(2

gμ

Ans. [C]

Sol.

N

f

mg

N = mω2 (R – r)

τN = τmg

μω2(R – r) R = mg R

ω = )rR(

g−μ

CAREER POINT



PART-II (CHEMISTRY) SECTION – 1 (Maximum Marks : 21)

• This section contains SEVEN questions.




Q.19 Which of the following combination will produce H2 gas ? (A) Cu metal and conc. HNO3

(B) Au metal and NaCN (aq) in the presence of air (C) Zn metal and NaOH (aq) (D) Fe metal and conc. HNO3 Ans. [C] Sol. (A) Cu + Conc. HNO3 ⎯→ Metal nitrate + NO2 + H2O

(B) 4Au + 8NaCN ])CN(Au[Na4 2OH2O

2

2⎯→⎯ + NaOH

(C) Zn + 2NaOH ⎯⎯⎯ →⎯Aqueous Na2ZnO2 + H2

(D) Fe + conc. HNO3 surfacemetalonanyoflayerotactivePr

32OFe⎯→⎯

Q.20 The order of basicity among the following compounds is -

H3C

NH

NH2 H2N

NH2

NH N NH HN N

I II III IV

(A) I > IV > III > II (B) II > I > IV > III (C) IV > I > II > III (D) IV > II > III > I Ans. [C] Sol. Order According to value of Kb.

CAREER POINT



Q.21 The major product of the following reaction is -

NH2

OH

(i) NaNO2, HCl, 0°C (ii) aq. NaOH

(A)

OH

N=N

(B)

OH

Cl

(C)

ONa+

N2Cl

–

(D)

OH N=N

Ans. [A] Sol. Step- 1

OH

NaNO2 + HCl

OH

N=N Cl– ⊕

NH2

0°C

Step- 2

OH

NaOH

OH

N=N

Diazo Coupling

⊕

Cl – N=N

(Azo dye)

(aq.)

H

Q.22 For the following cell, Zn(s) |ZnSO4 (aq) || CuSO4 (aq) | Cu(s) When the concentration of Zn2+ is 10 times the concentration of Cu2+ , the expression for ΔG (in J mol–1)

is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V] (A) 2.303 RT + 1.1 F (B) 2.303RT – 2.2F (C) 1.1 F (D) –2.2 F Ans. [B] Sol. Zn(s) + Cu+2 (aq) ⎯→ Zn+2 (aq) + Cu(s)

]Cu[]Zn[log

n0591.0EE 2

2

10CuCu +

+° −=

CAREER POINT



]Cu[10]Zn[ 22 ++ =

]Cu[]Cu[1010log

F2RT303.21.1E 2

2

Cu +

+

−=

F2RT303.21.1ECu −=

CuEnFG −=Δ

= ⎥⎦⎤

⎢⎣⎡ −−

F2RT303.21.1F2

ΔG = 2.303 RT – 2.2 F Q.23 The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3, and H4P2O6 is - (A) H3PO3 > H3PO2 > H3PO4 > H4P2O6 (B) H3PO4 > H3PO2 > H3PO3 > H4P2O6 (C) H3PO2 > H3PO3 > H4P2O6 > H3PO4 (D) H3PO4 > H4P2O6 > H3PO3 > H3PO2 Ans. [D] Sol. Oxidation state of phosphorous Atom In – H3PO2 ; H3PO4 ; H3PO3 And H4P2O6 is

POH POH OPH POH2

23

4

33

4

624

5

43

++++>>>

H3PO2 ⇒ +3 + x – 4 = 0, x = +1

H3PO4 ⇒ +3 + x – 8 = 0, x = +5

H3PO3 ⇒ +3 + x – 6 = 0, x = +3

H4P2O6 ⇒ +4 + 2x – 12 = 0, x = +4 Q.24 Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the

freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T).

[molecular weight of ethanol is 46 g mol–1]. Among the following, the option representing change in the freezing point is -

(A)

Water

Water + Ethanol V.P

./Bar

270 273 T/K

Ice

I –

(B)

Water

Water + Ethanol

V.P

./Bar

270 273T/K

Ice I –

(C)

Water

Water + Ethanol V.P

./Bar

271 273 T/K

Ice

I –

(D)

Water

Water + Ethanol

V.P

./Bar

271 273T/K

Ice I –

CAREER POINT



Ans. [B]

Sol. mikT ff =Δ

= 50046

10005.3421×

×××

k3Tf =Δ

ΔTf = Fp of solvent – Fp of solution

3 = 273 – x

x = 270

Q.25 The standard state Gibbs free energies of formation of C (Graphite) and C (diamond) at T = 298 K are

ΔfGº [C(graphite)] = 0 kj mil–1

ΔfGº [C(diamond)] = 2.9 kj mil–1

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversing of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10–6 m3 mol–1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is

[Useful information: 1 J = 1 kg m2 s–2; 1 Pa = 1 kg m–1 s–2 ; bar = 105 Pa]

(A) 29001 bar (B) 1450 bar

(C) 14501 bar (D) 58001 bar

Ans. [C]

Sol. At equillibrium ΔG = 0

ΔG° = dp(ΔV)

1362

3 molm)102)(1P(mol

J109.2 −−×−=×

1

36

22

23

molm)102)(1P(

molskgm109.2 −

−×−=××

29

2 mskg10

29.21P ×=−

= 1.45 ×109 + 1 pa

= 14501 bar

CAREER POINT





• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.




option is darkened Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. • For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get

+4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get –2 marks, as a wrong option is also darkened.

Q.26 Among the following, the correct statement(s) is(are) (A) AlCl3 has the three-centre two-electron bonds in its dimeric structure (B) BH3 has the three-centre two-electron bonds in its dimeric structure (C) Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure (D) The Lewis acidity of BCl3 is greater than that of AlCl3

Ans. [B, C, D]

Sol. (A) AlCl3 ⎯⎯ →⎯Dimer Al2Cl6 (B) BH3 ⎯⎯ →⎯Dimer B2H6

Cl Al

Cl

Cl

Cl Al

Cl

Cl

HB

HB

H

H

H

H

3C–4e– Bond's 3C–2e– Bond's

(C) Al(CH3)3 ⎯⎯ →⎯Dimer

H3C

Al AlCH3

CH3

C

CH3C

H H

H

H HH

3C–2e– Bond

(D)

Cl—B—Cl

Cl

Weak pπ– dπ back bonding

But AlCl3 form dimer )octetComplete(

ClAl 62

Hence BCl3 is more acidic than AlCl3

CAREER POINT



Q.27 For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

(A) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive

(B) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative

(C) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases

(D) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases

Ans. [C, D] Sol. Factual Q.28 Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S

is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes halfoform reaction but not Cannizzaro reaction.

(i) P OH/Zn)ii

ClCH/O)i

2

223 ⎯⎯⎯⎯ →⎯ )OHC( 88

Q

(ii) R OH/Zn)ii

ClCH/O)i

2

223 ⎯⎯⎯⎯ →⎯ )OHC( 88

S

The option(s) with suitable combination of P and R, respectively, is(are)

(A) CH3

H3C

and

CH3

H3C

CH3

(B)

CH3

H3C

CH3

and

CH3

CH3

CH3

(C) H3C

and

CH3

(D) H3C

and

CH3

H3C

Ans. [A, C]

CAREER POINT



Sol. (A)

CH

CH3

CH CH3O3/CH2Cl2

Zn/H2OC

CH3

H

O

[+ve Cannizzaro]

C CH3

O3/CH2Cl2

Zn/H2O C CH3

O

[+ve haloform]

(P) (Q)

(R)

C CH3

CH3

(S)

& (C)

CH CH2O3/CH2Cl2

Zn/H2OC H

O

[+ve Cannizzaro]

C CH3

O3/CH2Cl2

Zn/H2OC CH3

O

[+ve haloform]

(P) (Q)

(R)

CH2

(S)

CH3 CH3

Q.29 For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions

is(are)

Br

I

Br

II

H3C—C—Br

CH3

CH3

III

Br

IV

CH3

(A) I and II follow SN2 mechanism (B) Compound IV undergoes inversion of configuration (C) The order of reactivity for I, III and IV is: IV > I > III (D) I and III follow SN1 mechanism Ans. [A, B, C, D] Sol. (A) I & II follow SN2 mechanism [Back side attack occurs] (Both are 1º halide SN2 can occur)

CAREER POINT



(B) Compound IV under goes inversion of configuration

CH3

Br NU –

NU

H

CH3

H

Walden inversion

SN2

(C) Order of reactivity is IV > I > III due to stability of intermediate in SN1 reaction. (D) I & III follow SN1 mechanism due to formation of carbocation intermediate. Q.30 In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct

option(s) among the following is(are) (A) The value of frequency factor predicted by Arrhenius equation is higher than that determined

experimentally (B) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius

equation (C) The activation energy of the reaction is unaffected by the value of the steric factor (D) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used Ans. [B, C] Sol. Factual

Q.31 The correct statement(s) about surface properties is(are) (A) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion

medium (B) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system (C) Brownian motion of colloidal particles does not depend on the size of the particles but depends on

viscosity of the solution (D) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption

of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature

Ans. [B, D] Sol. Adsorption is exothermic process ΔH = –ve In adsorption process, randomness of the molecules decreases. So entropy decreases. Q.32 The option(s) with only amphoteric oxides is(are) (A) ZnO, Al2O3, PbO, PbO2 (B) Cr2O3, BeO, SnO, SnO2 (C) NO, B2O3, PbO, SnO2 (D) Cr2O3, CrO, SnO, PbO Ans. [A, B] Sol. Amphoteric oxide's are → ZnO, Al2O3, BeO, Cr2O3, GeO PbO/PbO2 And SnO/SnO2

CAREER POINT




• Type instruction

• This section contains TWO paragraphs.

• Based on each paragraph, there are TWO questions.

• Each question has FOUR options [A], [B],[C] and [D] ONLY ONE of these four options is correct

• For each question, darken the bubble corresponding to the correct option in the ORS

• For each question, marks will be awarded in one of the following categories: Full marks : +3 If only the bubble corresponding to the option is darkened Zero Marks : 0 In all other cases

PARAGRAPH 1

Upon heating KClO3 in the presence of catalytic amount of MnO2 a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives Y and Z.

Q.33 Y and Z are, respectively (A) N2O5 and HPO3 (B) N2O4 and HPO3 (C) N2O4 and H3PO3 (D) N2O3 and H3PO4 Ans. [A]

Sol. MnO2

Catalyst 3KClO3 3O2 + 3KCl

P4O10 P4 HNO3 HPO3 + N2O5

Q.34 W and X are, respectively (A) O3 and P4O10 (B) O2 and P4O6

(C) O2 and P4O10 (D) O3 and P4O6 Ans. [C]

Sol. MnO2

Catalyst 3KClO3 3O2 + 3KCl

P4O10 P4 HNO3 HPO3 + N2O5

PARAGRAPH 2 The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by additional of H2O gives

Q. The compound Q on treatment with H2SO4 at 0°C gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O produces compound S. [Et in compound P is ethyl group]

(H3C)3C

Q R S

P

CO2Et

CAREER POINT



Q.35 The Product S is

(A)

(H3C)3C

HO3S

COCH3

CH3O (B)

(H3C)3C

COCH3 CH3

(C)

(H3C)3C

COCH3

CH3H3C

(D) (H3C)3C

H3COCCH3H3C

Ans. [C] Sol.

(H3C)3C(i) CH3MgBr

P

CO

O Et(excess)(C2H5)2O

(ii) H2O

(H3C)3C

Q

CCH3

CH3HO

(H3C)3C

Q

CCH3

CH3HO (H3C)3CH+

[Reaction intermediate]

C CH3

0°C–H2O

CH3 ⊕

Dehydration

(H3C)3CCH3

CH3

R

ArSE (Friedel craft Alkylation)

(H3C)3CCH3

CH3

R

(H3C)3CCH3

CH3

S

CO

CH3

CH3COClAlCl3

Friedel craft Acylation

Q.36 The reactions, Q to R and R to S, are (A) Friedel-Craft alkylation and Friedel-Craft acylation (B) Friedel-Craft alkylation, dehydration and Friedel-Craft acylation (C) Dehydration and Friedel-Craft acylation (D) Aromatic sulfonation and Friedel-Craft acylation Ans. [B] Sol. Q to R is Friedel-crafts and Dehydration alkylation [Ar.SE] R to S is Friedel-crafts alkylation [Ar.SE]

Frid

el-C

raft

-alk

ylat

ion

Frid

el-C

raft

- al

kyla

tion

CAREER POINT



PART-III (MATHEMATICS) SECTION – 1 (Maximum Marks : 21)

• The section contains SEVEN questions




Q.37 The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is

(A) 14x + 2y + 15z = 31 (B) 14x – 2y + 15z = 27 (C) –14x + 2y + 15z = 3 (D) 14x + 2y – 15z = 1 Ans. [A] Sol. Let the direction ratios of the normal of required plane are a,b,c then 2a + b – 2c = 0 3a – 6b – 2c = 0

⇒ λ=−

=−

=− 15

c2

b14a

(a, b, c) = (14λ, 2λ, 15λ) Hence equation of plane will be 14(x – 1) + 2(y – 1) + 15(z – 1) = 0 14x + 2y + 15z – 31 = 0 Q.38 How many 3 × 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal

entries of MT M is 5 ? (A) 162 (B) 198 (C) 126 (D) 135 Ans. [B]

Sol. Let M = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ihgfedcba

Then MTM = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ihgfedcba

ifchebgda

Sum of the diagonal elements = a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 = 5 This is possible if 7 of them are 0, 1 is 1 and 1 is 2.

CAREER POINT



Number of cases = 9C7 × 2C1 × 1C1 = 72 or 5 of them are 1, 4 of them are 0 Number of cases = 9C5 × 4C4 = 126 Total cases = 198 Q.39 Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10.

Then the probability that z is even, is

(A) 21 (B)

5536 (C)

115 (D)

116

Ans. [D] Sol. x + y + z = 10 Total cases = 3 + 10 – 1C10 = 12C10 = 66 when z is even Let z = 0 ⇒ x + y = 10 Total cases = 11

z = 2 ⇒ x + y = 8 Total cases = 9

z = 4 ⇒ x + y = 6 Total cases = 7

z = 6 ⇒ x + y = 4 Total cases = 5

z = 8 ⇒ x + y = 2 Total cases = 3

z = 10 ⇒ x + y = 0 Total cases = 1 36

Required probability = 6636 =

116

Q.40 Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OS·OROQ·OP + = OS·OQOP·OR + = OS·OPOR·OQ +

Then the triangle PQR has S as its (A) orthocenter (B) incentre (C) centroid (D) circumcentre Ans. [A]

Sol. OS·OROQ·OP + = OS·OQOP·OR +

⇒ 0)OROQ(OS)OROQ·(OP =−−−

0)OROQ(·)OSOP( =−−

0QR·PS =

Similarly 0PR·QS =

So, S is the orthocentre

CAREER POINT



Q.41 If y = y(x) satisfies the differential equation dyx9x8 ⎟⎠⎞⎜

⎝⎛ + = dxx94

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛++ , x > 0 and

y(0) = 7 , then y(256) =

(A) 3 (B) 9 (C) 16 (D) 80 Ans. [A]

Sol. ∫dy = ∫+++ x94x9x8

dx

Let x9 + = t ⇒ x92

1

+. dx

x21 = dt

y = ∫ + t42dt

y = )t4( + + c

y = x94 ++ + c

as y(0) = 7 ⇒ c = 0

∴ y = x94 ++

∴ y(256) = 3

Q.42 If f : R → R is a twice differentiable function such that f ′′(x) > 0 for all x ∈ R and f ⎟⎠⎞

⎜⎝⎛

21 =

21 , f(1) = 1,

then

(A) 21 < f ′(1) ≤ 1 (B) f ′(1) > 1 (C) 0 < f ′(1) ≤

21 (D) f ′(1) ≤ 0

Ans. [B] Sol. f ′′(x) > 0

⇒ f ′(x) is increasing function.

Applying LMVT in ⎥⎦⎤

⎢⎣⎡ 1,

21

f ′(x) =

211

21f)1(f

−

⎟⎠⎞

⎜⎝⎛−

=

21

211−

= 1

as f ′(x) is increasing

f ′(x) < f ′(1) ∀ x ∈ ⎟⎠⎞

⎜⎝⎛ 1,

21

∴ f ′(1) > 1

CAREER POINT



Q.43 Let S = {1, 2, 3, ... 9}. For k = 1, 2, .... 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 =

(A) 126 (B) 125 (C) 210 (D) 252 Ans. [A] Sol. N1 = 5C1 × 4C4 = 5 N2 = 5C2 × 4C3 = 40 N3 = 5C3 × 4C2 = 60 N4 = 5C4 × 4C1 = 20 N5 = 5C5 = 1 ∴ N1 + N2 + N3 + N4 + N5 = 126



• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct.




option is darkened Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases.

• For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get –2 marks, as a wrong option is also darkened.

Q.44 If f(x) = xcosxsinxsinxsin–xcosxcos–)x2sin()x2cos()x2cos(

, then

(A) f(x) attains its maximum at x = 0 (B) f ′(x) = 0 at exactly three points in (– π, π)

(C) f ′(x) = 0 at more than three points in (– π, π) (D) f(x) attains its minimum at x = 0 Ans. [A,C] Sol. f(x) = cos 2x + cos22x – sin22x f(x) = cos 2x + cos 4x f ′(x) = –2 sin 2x – 4 sin 4x

CAREER POINT



= – 2 sin 2x – 8 sin 2x cos 2x = – 2 sin 2x (1 + 4 cos 2x) f ′(x) = 0

sin 2x = 0 ⇒ x = 2π

− , 0, 2π

or cos 2x = 41

− ⇒ 1 – 2 sin2x = –41

⇒ sin2x = 85

⇒ x = ± sin–1

85 , π – sin–1

85 , – π + sin–1

85

So, f ′(x) = 0 at 7 points in (–π, π) f(x) will attain maximum value at x = 0

Q.45 Let f (x) = |x–1|

|)x–1|1(x–1 + cos ⎟⎠⎞

⎜⎝⎛

x–11 for x ≠ 1. Then

(A) −→1x

lim f(x) = 0

(B) +→1x

lim f(x) does not exist

(C) −→1x

lim f(x) does not exist

(D) +→1x

lim f(x) = 0

Ans. [A, B] Sol. LHL = )x(flim

1x −→

= ⎟⎠⎞

⎜⎝⎛

−−−+−

−→ x11cos

)x1())x1(1(x1lim

1x

= ⎟⎠⎞

⎜⎝⎛

−−

−→ x11cos)x1(lim

1x = 0

RHL = )x(flim1x +→

= ⎟⎠⎞

⎜⎝⎛

−−−+−

+→ x11cos

x1)1x1(x1lim

1x

= ⎟⎠⎞

⎜⎝⎛

−+

+→ x11cos)x1(lim

1x

limit does not exist because of non unique value.

CAREER POINT



Q.46 If dx)1x(x

1kI98

1k

1k

k∑∫=

+

++

= , then

(A) 5049I > (B) I < loge 99 (C) I > loge 99 (D)

5049I <

Ans. [A] Sol. x ∈ (k, k + 1) x > k x + 1 > k + 1

1x

1+

< 1k

1+

1x1k

++ < 1

∫+

++

1k

k

dx)1x(x

1k < ∫+1k

k

dxx1

∫+

++

1k

k

dx)1x(x

1k < 1kknx +

l = k

1kn +l

∑ ∫=

+

++98

1k

1k

k

dx)1x(x

1k < k

1kn98

1k

+∑=

l

< 9899n....

23n

12n lll +++

< 9899....

23.

12nl

I < ln 99 x < k + 1

)1x(x

1k+

+ > )1x)(1k(

1k++

+ k ≤ 98

)1x(x

1k+

+ > 1x

1+

> 100

1

∑ ∫=

+

++98

1k

1k

k

dx)1x(x

1k > ∑ ∫=

+98

1k

1k

k

dx100

1 ⇒ I > 100

1 (98)

I > 5049

Q.47 If ∫ −=)x2sin(

xsin

1 dt)t(sin)x(g , then

(A) π−=⎟⎠⎞

⎜⎝⎛ π 2

2'g (B) π−=⎟

⎠⎞

⎜⎝⎛ π− 2

2'g (C) π=⎟

⎠⎞

⎜⎝⎛ π 2

2'g (D) π=⎟

⎠⎞

⎜⎝⎛ π

− 22

'g

CAREER POINT



Ans. [Bonus]

Sol. g(x) = ∫ −x2sin

xsin

1 dttsin

g ′(x) = 2 cos 2x sin–1(sin 2x) – cos x sin–1(sin x)

g ′ ⎟⎠⎞

⎜⎝⎛ π

2 = 2(–1) (0) – 0 ⎟

⎠⎞

⎜⎝⎛ π

2 = 0

g ′ ⎟⎠⎞

⎜⎝⎛ π

−2

= 2(–1) (0) – 0 ⎟⎠⎞

⎜⎝⎛ π−

2 = 0

None of the given option is correct. So Answer Bonus

Q.48 If the line x = α divides the area of region R = {(x, y) ∈ R2 : x3 ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal parts,

then

(A) 0142 24 =+α−α (B) 014 24 =−α+α (C) 121

<α< (D) 210 ≤α<

Ans. [A, C]

Sol. 0

x = α

1

∫α

−0

3 dx)xx( = ∫ −1

0

3 dx)xx(21

α

⎥⎦

⎤⎢⎣

⎡−

0

42

4x

2x =

1

0

42

4x

2x

21

⎥⎦

⎤⎢⎣

⎡−

⇒ 42

42 α−

α = 81

⇒ 2α4 – 4α2 + 1 = 0

if α = 21 then ∫ −

2/1

0

3 dx)xx( = 647

which is less than 81

So, 21 < α < 1

CAREER POINT



Q.49 If f : R → R is a differentiable function such that f '(x) > 2f(x) for all x ∈ R, and f(0) = 1, then

(A) f(x) is decreasing in (0, ∞) (B) f(x) > e2x in (0, ∞)

(C) f '(x) < e2x in (0, ∞) (D) f(x) is increasing in (0, ∞) Ans. [B, D] Sol. f ′(x) > 2 f(x)

e–2x f ′(x) > 2 e–2x f(x)

dxd (f(x) . e–2x) > 0

⇒ g(x) = f(x) e–2x is increasing function

So, g(x) > g(0) ∀ x ∈ (0, ∞)

g(x) > f(0)e–0 ∀ x ∈ (0, ∞)

x2e)x(f > 1 ∀ x ∈ (0, ∞)

f(x) > e2x ∀ x ∈ (0, ∞)

f ′(x) > 2 f (x) > 2e2x > 0 ∀ x ∈ (0, ∞)

⇒ f ′(x) > 0

⇒ f(x) is increasing ∀ x ∈ (0, ∞)

Q.50 Let α and β be nonzero real numbers such that 2(cos β – cos α) + cos α cos β = 1. Then which of the following is/are true ?

(A) 02

tan32

tan =⎟⎠⎞

⎜⎝⎛ β

+⎟⎠⎞

⎜⎝⎛ α (B) 0

2tan3

2tan =⎟

⎠⎞

⎜⎝⎛ β

−⎟⎠⎞

⎜⎝⎛ α

(C) 02

tan2

tan3 =⎟⎠⎞

⎜⎝⎛ β

−⎟⎠⎞

⎜⎝⎛ α (D) 0

2tan

2tan3 =⎟

⎠⎞

⎜⎝⎛ β

+⎟⎠⎞

⎜⎝⎛ α

Ans. [A, B] Sol. 2(cos β – cos α) + cos α cos β = 1

⇒ ⎟⎠⎞

⎜⎝⎛ β

+⎟⎠⎞

⎜⎝⎛ α

+

⎟⎠⎞

⎜⎝⎛ β

−⎟⎠⎞

⎜⎝⎛ α

−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

α+

α−

−β

+

β−

2tan1

2tan1

2tan1

2tan1

1

2tan1

2tan1

2tan1

2tan1

222

22

2

2

2

2

⇒ ⎟⎠⎞

⎜⎝⎛ β

+α

=⎟⎠⎞

⎜⎝⎛ β

−α

2tan

2tan2

2tan

2tan4 2222

⇒ 2

tan32

tan 22 β=

α

⇒ 2

tan32

tan β±=

α

CAREER POINT




• This section contains TWO paragraphs.

• Based on each paragraph, there are TWO questions.



• For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 In all other cases.

PARAGRAPH # 1

Let O be the origin, and OZ,OY,OX be three unit vectors in the directions of the sides PQ,RP,QR , respectively, of a triangle PQR.

Q.51 OYOX × =

(A) sin 2R (B) sin (P + R) (C) sin (Q + R) (D) sin (P + Q) Ans. [ D] Sol.

P

RQ

OY

OXOZ

Rsin|OY||OX||OYOX| =× = sin R = sin (180 – (P + Q)) = sin (P + Q) Q.52 If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R+ P) is

(A) 23

− (B) 35 (C)

35

− (D) 23

Ans. [A] Sol. cos(P+Q) + cos(Q+R) + cos (R+P) = – cos R – cos P – cos Q = – (cos P + cos Q + cos R)

CAREER POINT



3RQP ++ π

x

(Q, cos Q)(P, cos P)

y

(R, cos R)

3

RcosQcosPcos3

RQPcos ++>⎟

⎠⎞

⎜⎝⎛ ++

RcosQcosPcos23

++>

23)RcosQcosP(cos −>++−

So minimum value = 23

−

PARAGRAPH # 2 Let p, q be integers and let α, β be the roots of the equation, x2 – x – 1 = 0, where α ≠ β. For n = 0, 1, 2, ……, let an = pαn + qβn.

FACT : If a and b are rational numbers and 05ba =+ , then a = 0 = b.

Q.53 If a4 = 28, then p + 2q = (A) 14 (B) 12 (C) 7 (D) 21 Ans. [B]

Sol. ,2

51+=α

251−

=β

⇒ ,2

532 +=α

2532 −

=β

⇒ ,2

5374 +=α

25374 −

=β

Now, a4 = 28 pα4 + qβ4 = 28

282

537q2

537p =⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟

⎟⎠

⎞⎜⎜⎝

⎛ +

56)qp(53)q7p7( =−++

∴ 7p + 7q = 56 and p – q = 0

∴ p = q = 4

∴ p + 2q = 12

CAREER POINT



Q.54 a12 = (A) a11 – a10 (B) a11 + a10 (C) a11 + 2a10 (D) 2a11 + a10 Ans. [B] Sol. an = 4(αn + βn)

a1 = 4(α + β) = 4

a2 = 4(α2 + β2) = 4[(α + β)2 – 2 αβ] = 12

a3 = 4(α3 + β3) = 4[(α + β)3 – 3αβ (α + β)] = 16 a4 = 28 (given) ∴ an = an – 1 + an – 2

∴ a12 = a11 + a10

IIT-JEE Advanced Examination Paper-2 21-05-2017 Physics · 2018. 1. 19. · CAREER POINT Kota H.O....

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Transcript of IIT-JEE Advanced Examination Paper-2 21-05-2017 Physics · 2018. 1. 19. · CAREER POINT Kota H.O....