IIT JEE Advanced 2017 Paper 2 (Code 9)

Test Name: IIT JEE Advanced 2017 Paper 2 (Code 9)

Question 1:

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)

Option 1: vs = 22 km s–1
Option 2: vs = 72 km s–1
Option 3: vs = 42 km s–1
Option 4: vs = 62 km s–1
Correct Answer: 3

 

Solutions:Given:

ve = 11.2 Km/sec = 2GMeReFrom energy conservation12mvs – GMsmr  – GMemRe = 0where, r = distance of rocket from sun⇒vs = 2GMeRe+ 2GMsrGiven:

Ms = 3 × 105 Me , r = 2.5 × 104 Re   ⇒vs = 2GMeRe + 2G×3 × 105 Me 2.5 × 104 Re        = 2GMeRe1 + 3 × 105  2.5 × 104         = 2GMeRe×13⇒vs ≃ 42 Km/sHence, the correct answer is option C.

Question 2:

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 :

Option 1: S→=1-b P→+b2 Q→
Option 2: S→=b-1 P→+bQ→
Option 3: S→=1-b P→+bQ→
Option 4: S→=1-b2 P→+bQ→
Correct Answer: 3

 

Solutions:Let vector from point P to point S be

C→ ⇒ C→ = bR→R→R→ = bR→ = b(Q→ – P→)From triangle rule of vector additionP→ + C→ = S→P→ +  b(Q→ – P→) = S→⇒ S→ = (1 – b)P→ + bQ→Hence, the correct answer is option C.

Question 3:

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 center of the loop is :

Option 1: µ0I4πa33-1
Option 2: µ0I4πa63-1
Option 3: µ0I4πa63+1
Option 4: µ0I4πa32-3
Correct Answer: 2

 

Solutions:                           Direction of magnetic field at ‘O’ due to each segment is same. Since it is symmetric star shape, magnitude will also be same. Therefore, magnetic field due to section B1 is

B1 = kIasin (+60) – sin(30) = kI2a(3 – 1)Therefore, net magnetic field due to 12 segments at the centre ‘O’ isBnet = 12 × B1 = 6kIa(3 – 1)where,  k = μo4π⇒ Bnet = 6μoI4πa(3 – 1)Hence, the correct answer is option B.

Question 4:

A photoelectric material having work-function Ï•0 is illuminated with light of wavelength

λλ<hcϕ0. 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

Option 1: λd3/λ2
Option 2: λd3/λ
Option 3: λd2/λ2
Option 4: λd/λ
Correct Answer: 1

 

Solutions:According to photoelectric equation:

KEmax = hcλ – ϕop22m = hcλ – ϕohλ22m = hcλ  – ϕoAssuming small changes and differentiating both sides, we get

h22m-2dλdλd3 = hcλ2dλdλddλ ∝  λd3λ2Hence, the correct answer is option A.

Question 5:

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 second 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

Option 1: 0.2 %
Option 2: 5 %
Option 3: 3 %
Option 4: 1 %
Correct Answer: 4

 

Solutions: The time interval (T) between dropping a stone and receiving the sound of impact with the bottom of the well is

T = 2Lg + LcSo, .

T + δT = 2(L + δL)g +  (L + δL)c             = 2Lg1 +δLL +  Lc1 +δLLSince,  δT T is very small, hence δLL is also small, so taking binomial approximationT + δT = 2Lg 1 +12δLL + Lc1 +δLLT + δT = 2Lg +122LgδLL + Lc+ LcδLL             = 2Lg + Lc+122Lg + LcδLL             =T + 122 × 2010 + 20300δLL⇒ δT = 1 + 115δLL⇒ δLL = 1516 δT ⇒ δLL = 15161100 % error = δLL × 100 %            = 1516 % ≃ 1 %Hence, the correct answer is option D.

Question 6:

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

1ρdρdtis constant. The velocity v of any point on the surface of the expanding sphere is proportional to :

Option 1: R3
Option 2: 1R
Option 3: R
Option 4: R2/3
Correct Answer: 3

 

Solutions:Let the mass of the sphere be M. Then

M = 43πR3ρDifferentiating both sides w.r.t. t, we get0 = 43π3R2dRdtρ +R3dρdt   …..(i)Dividing (i) by ρ0 = 3R2dRdt + R31ρdρdtSince 1ρdρdt is a constant (given), therefore3R2dRdt = -R3kwhere, 1ρdρdt = k⇒ dRdt ∝ Rand we know the velocity v of any point on the surface is equal to

dRdt 

⇒ v =

dRdt or

v ∝ RHence, the correct answer is option C.

Question 7:

Consider regular polygons with number of sides n = 3, 4, 5 ….. as shown in the figure. The center 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 center of mass for each polygon is Δ. Then Δ depends on n and h as :

Option 1: ∆=h sin2 πn
Option 2: ∆=h sin 2πn
Option 3: ∆=h 1cos πn-1
Option 4: ∆=h tan2 π2n
Correct Answer: 3

 

Solutions: OA = h

OB = hcosπnInitial height of COM = hFinal height of COM =  hcosπn∴∆ = hcosπn – hor, ∆ =h 1cosπn – 1 

Hence, the correct answer is option C.

Question 8:

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 ?

Option 1: When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to (1 – cosθ)
Option 2: The midpoint of the bar will fall vertically downward
Option 3: Instantaneous torque about the point in contact with the floor is proportional to sin θ
Option 4: The trajectory of the point A is a parabola
Correct Answer: 1,2,3

 

Solutions:When the bar makes an angle θ with the vertical, the height of its Centre of Mass (COM) is

L2cosθTherefore, displacement =

L2-L2cosθ=L2(1-cosθ)Since, force on COM is only along the vertical direction, so COM is falling vertical downward. Instantaneous torque about point of contact,

τ=Mg×L2sinθ 

τ α sinθFor the trajectory of the point A :

x=L2sinθy=Lcosθx2(L/2)2+y2L2=1Path of point A is an ellipse.

Hence, the correct answers are options A,B and C.

Question 9:

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 = 1.8 mm. 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 ?

Option 1: A dark spot will be formed at the point P2
Option 2: The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant
Option 3: At P2 the order of the fringe will be maximum
Option 4: The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000
Correct Answer: 3,4

 

Solutions:At point P2:

∆x=d=1.8 mm=3000λHence a bright fringe will be formed at P2. Now,

∆x=dcosθ=nλcosθ=nλd⇒-sinθ ∆θ=(∆n)λd⇒ ∆θ=-(∆n)λdsinθ∆θ increases as θ decreasesAt P2, the order of fringe will be maximum.The total number of bright fringes d=nλ ⇒n=3000Therefore, total number of fringes between P1 and P2 are 3000.Hence, the correct answers are options C and D.

Question 10:

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?

Option 1: The ratio of the currents through L1 and L2 is fixed at all times (t > 0)
Option 2: After a long time, the current through L1 will be

VRL2L1+L2

Option 3: After a long time, the current through L2 will be

VRL1L1+L2

Option 4: At t = 0, the current through the resistance R is

VR

Correct Answer: 1,2,3

 

Solutions:Since inductors are connected in parallel.

VL1=VL2L1dI1dt=L2dI2dtL1I1=L2I2I1I2=L2L1Therefore, the ratio of currents through L1 and L2 is fixed at all times.

After a long time, inductors will start acting as short circuit, so current, I, through resistor R will be V/R.

Also,

 

I1+I2 = I⇒I1+I2 = VR⇒L2I2L1+I2=VR⇒I2=L1L1+L2VRSimilarly,I1=L2L1+L2VRAt = 0, current in the circuit is 0.

Hence, the correct answers are options are A,B and C.

Question 11:

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 ?

Option 1: If the force is applied normal to the circumference at point X then τ is constant
Option 2: If the force is applied tangentially at point S then τ ≠ 0 but the wheel never climbs the step
Option 3: If the force is applied normal to the circumference at point P then τ is zero
Option 4: If the force is applied at point P tangentially then τ decreases continuously as the wheel climbs
Correct Answer: 3

 

Solutions:

If force is applied normal to surface at point X

τ=FyRsinθTherefore,

τ depends on

θ and it is not constant. So, option A is incorrect.

If force is applied tangential at S

τ=F×R≠0but the wheel can climb step due to this non zero torque. Hence, option B is also incorrect.

If the force is applied normal to the surface at P then line of action of force will pass from Q and thus torque is zero. So, option C is correct.

If the force is applied at P tangentially, then

τ=F×2R=constant

Hence, the correct answer is option C.

Question 12:

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

VY=V0 sin ωt+2π3 and VZ=V0 sin ωt+4π3An 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:-

Option 1: VXYrms=V0
Option 2: VYZrms=V012
Option 3: Independent of the choice of the two terminals
Option 4: VXYrms=V032
Correct Answer: 3,4

 

Solutions:Potential difference between X and Y = VX – VY Potential difference between Y and Z = VY – VZ Phasors of Voltages:

∴ VX-VY=3V0VXYrms=3V02Similarly, VXYrms=3V02This voltage difference is independent of the choice of terminals.

Hence, the correct answers are options C and D.

Question 13:

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 ?

Option 1: The circumference of the flat surface is an equipotential
Option 2: The electric flux passing through the curved surface of the hemisphere is

-Q2ε01-12

Option 3: Total flux through the curved and the flat surfaces is

Qε0

Option 4: The component of the electric field normal to the flat surface is constant over the surface.
Correct Answer: 1,2

 

Solutions:(A) Every point on circumference of flat surface is at equal distance from point charge Hence circumference is equipotential.

(B) Flux passing through curved surface = – flux passing through flat surface. Flux through the ring

(dϕ)through the ring = Ecosθ.dA = 14πεoQr2+ R22Rr2+ R2.2πRdr∴∫dϕ = QR4πεo2π∫rdrr2+ R232 = Q2εo1-12∴Flux through curved surface = -Q2εo1-12Hence, the correct answers are options A and B.

Question 14:

A uniform magnetic field B exists in the region between x = 0 and

x=3R2(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 options(s) is/are correct ?

Option 1: For

B=813pQR, the particle will enter region 3 through the point P2 on x-axis

Option 2: For

B>23pQR, the particle will re-enter region 1

Option 3: For a fixed B, particle 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.
Option 4: When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the chage in its linear momentum between point P1 and the farthest point from y-axis is

p2.

Correct Answer: 1,2

 

Solutions: Radius of the particle with charge Q moving in the magnetic field of magnitude B is given as,

r=pQB        …..(i)For B=813×pQRFrom (i), we haver=p×13QR8Qp 

r=138RUsing pythagorous theorem, we get

y=5R8As,

y+R=13R8This implies that the particle will enter region 3 through point P2.

For

B>23pQRRadius of the path,

r<3pQR2Qp=32RFrom this value we can observe that the particle will not enter in region 3 and will re-enter region 1.

Charge in momentum of the particle is

2p when particle enters region 1 between entry point and leaving point.

Hence, the correct options are A and B.

Question 15:

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

v03and maintained for a charging time T >> RC. Then the voltage is raised to

2v03without 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.

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

Option 1: EC = ED
Option 2: EC = 2ED
Option 3: EC=12ED
Option 4: EC = ED ln2
Correct Answer: 1

 

Solutions: When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be q = CV

Energy stored in the capacitor is,

EC=12CV2        …..(i)Work done by the battery,

W=Vq=V(CV)=CV2Energy dissipated along the resistor of resistance R, ED = Work done by the battery – Energy stored in the capacitor

ED=CV2-12CV2=12CV2      …..(ii)From (i) and (ii), we see that EC = ED

Hence, the correct answer is option A.

Question 16:

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

v03and maintained for a charging time T >> RC. Then the voltage is raised to

2v03without 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.

In Process 2, total energy dissipated across the resistance ED is :-

Option 1: ED=1312CV02
Option 2: ED=312CV02
Option 3: ED=12CV02
Option 4: ED=3CV02
Correct Answer: 1

 

Solutions:In step 1 of process 2 the voltage is first set to 

v03, thus charge stored in the capacitor is

q1=CV03Energy stored in the capacitor,

E1=12CV032=CV0218Work done by the battery,

W1=CV03×V03=CV029Heat loss in the process,

ED1=CV029-CV0218=CV0218For step 2: Charge stored in the capacitor is

q2=2CV03Extra charge flow through the battery =

CV03Work done by the battery,

W2=CV03×2V03=2CV029Final energy stored in the capacitor after step 2,

E2=12C2V032=4CV0218Energy stored in the step 2 =

4CV0218-CV0218=3CV0218 

Heat loss in the step 2 = Work done by the battery in step 2 – Energy stored in the capacitor in step 2

ED2=2CV029-3CV0218=CV0218For step 3:  Charge stored in the capacitor is

q3=CV0Extra charge flow through the battery =

CV0-2CV03=CV03Work done by the battery,

W3=CV03×V0=CV023Final energy stored in the capacitor after step 3,

EF=12CV02Energy stored in the step 3 =

CV022-4CV0218=5CV0218 

Heat loss in the step 3 = Work done by the battery in step 3 – Energy stored in the capacitor in step 3

ED3=CV023-5CV0218=CV0218Total heat loss in process 2 is

ED=CV0218+CV0218+CV0218=CV026So, we can say that

ED=CV026=13CV022=13EFHence, the correct answer is option A.

Question 17:

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.

The total kinetic energy of the ring is :-

Option 1: Mω02R2
Option 2: Mω02 R-r2
Option 3: 12Mω02 R-r2
Option 4: 32Mω02 R-r2
Correct Answer: 2

 

Solutions: Point A is the ICOR of the system.

ωR=ω0R-r⇒ω=ω0R-rRThe total kinetic energy of the ring is

K=12Iω2⇒K=122mR2ω2⇒K=mω02R-r2Hence, the correct answer is options B.

Question 18:

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.

The minimum value of ω0 below which the ring will drop down is :-

Option 1: 3g2µR-r
Option 2: gµR-r
Option 3: 2gµR-r
Option 4: 2g2µR-r
Correct Answer: 2

 

Solutions:Normal reaction at the point of contact is

N=Mω2(R-r)         …..(i)

Friction force at the point of contact,

f=Mg    …..(ii)

The ring will not drop when,

f≥μR    …..(iii)

From (i), (ii) and (iii), we get

Mg≥μMω02R-r⇒ω0=gμR-r Mg≥μMω02R-r⇒ω0=gμR-rHence, the correct answer is options B.

Question 19:

Which of the following combination will produce H2 gas ?

Option 1: Zn metal and NaOH(aq)
Option 2: Au metal and NaCN(aq) in the presence of air
Option 3: Cu metal and conc. HNO3
Option 4: Fe metal and conc. HNO3
Correct Answer: 1

 

Solutions:Zinc metal reacts with sodium hydroxide to form sodium zincate and evolves hydrogen gas.

Zn + 2 NaOH →          Na2ZnO2 + H2Hence, the correct answer is option  (A).

Question 20:

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 represents 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 –

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 4

 

Solutions:Given: Mass of ethanol (wb) = 34.5 g Mass of water (wa) = 500 g Molar mass of ethanol, C2H5OH (Mb) = 46 g/mol Freezing point of pure water (

Tf°) = 273 K Freezing point depression constant of water (Kf) = 2 K kg mol–1 Depression in freezing point =

∆Tf = ?Freezing point of solution (

Tf) = ?

∆Tf = Kf × m = Kf × wb ×1000Mb ×Wa= 2 ×34.5 ×100046 × 500= 3 K∆Tf = Tf° – Tf3 = 273 – Tf⇒Tf = 270 KAlso, with decrease in temperatur, vapour pressure of both solvent and solution decreases. These facts can be shown by the following plot given in figure D.

Hence, the correct answer is option D.

Question 21:

The order of the oxidation state of the phosphorus atom in H3PO2 , H3PO4 , H3PO3 and H4P2O6 is

Option 1: H3PO4 > H4P2O6 > H3PO3 > H3PO2
Option 2: H3PO3 > H3PO2 > H3PO4 > H4P2O6
Option 3: H3PO2 > H3PO3 > H4P2O6 > H3PO4
Option 4: H3PO4 > H3PO2 > H3PO3 > H4P2O6
Correct Answer: 1

 

Solutions:

Oxoacid of P Structure Oxidation State of P
H3PO2 +1
H3PO4 +5
H3PO3 +3
H4P2O6 +4

The order of the oxidation state of the phosphorus atom in H3PO2 , H3PO4 , H3PO3 and H4P2O6 is H3PO4 > H4P2O6 > H3PO3 > H3PO2

Hence, the correct answer is option A.

Question 22:

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are

∆fG° [C(graphite)] = 0 kJ mol–1

∆fG° [C(diamond)] = 2.9 kJ mol–1

The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion 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; 1 bar = 105 Pa]

Option 1: 14501 bar
Option 2: 29001 bar
Option 3: 58001 bar
Option 4: 1405 bar
Correct Answer: 1

 

Solutions:The change in standard Gibbs energy for conversion of graphite to diamond is given as:

Cgraphite →           CdiamondΔG° = ΔfG°diamond – ΔfG°graphiteΔG° =2.9 – 0 = 2.9 kJ mol-1dG = VdP∫ΔG1ΔG2dΔG = ∫P1P2ΔVdPΔG2-ΔG1 = ΔVP2-P12.9×103-0 = -2×10-61-P2P2-1 = 2.9×1032×10-6P2-1 = 1.45×104 barP2 = 14501 barHence, the correct answer is option (A).

Question 23:

The major product of the following reaction is

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 3

 

Solutions:The complete reaction is given following:

Hence, the correct answer is option C.

Question 24:

The order of basicity among the following compounds is

Option 1: II > I > IV > III
Option 2: IV > II > III > I
Option 3: I > IV > III > II
Option 4: IV > I > II > III
Correct Answer: 4

 

Solutions:(I)

(II)

(III)

It is the least basic because of electron withdrawing effect of -CH = CH- group.

(IV)

Gaunidine is the most basic due to more number of equivalent resonating structures of conjugate acid hence, more stability of conjugate acid.

The basic strength of compound depends on various factors i.e. resonance, stability of conjugate acid, inductive effect, hyperconjugation effect. Due to interpaly of these effects, the order of basicity among the given compounds is IV > I > II > III.

Hence, the correct answer is option D.

Question 25:

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 ∆G (in J mol–1) is

[F is Faraday constant , R is gas constant, T is temperature , Eº(cell) = 1.1V]

Option 1: 2.303 RT + 1.1F
Option 2: 2.303 RT – 2.2F
Option 3: 1.1 F
Option 4: –2.2 F
Correct Answer: 2

 

Solutions:The cell reaction can be represented as:

Zns + Cu2+aq →          Zn2+aq + CusThe Gibbs energy change for this cell reaction is determined as:

Given: E° = 1.1 Vn = 2Zn2+ = 10Cu2+ΔG° = -nFE°ΔG° = -2×F×1.1 = -2.2 FΔG = ΔG° + 2.303RT logQΔG = -2.2F + 2.303RT logZn2+Cu2+ΔG = -2.2F + 2.303RT log10Cu2+Cu2+ΔG = -2.2F + 2.303RTHence, the correct answer is option (B).

Question 26:

In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are):

Option 1: The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally
Option 2: The activation energy of the reaction is unaffected by the value of the steric factor
Option 3: Since P = 4.5, the reaction will not proceed unless an effective catalyst is used.
Option 4: Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation.
Correct Answer: 2,4

 

Solutions:The temperature dependence of the rate of a chemical reaction is accurately explained by Arrhenius equation:

k = PZAB e-Ea/RTHere,P = Probability or steric factorZAB =Collision frequency of reactantse-Ea/RT = Fraction of molecules with energies equal to or greater than EaThe activation energy of a reaction is unaffected by the steric factor.

Now, in the given question, the value of P is 4.5, it implies that experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation.

​Hence, the correct answers are options (B) and (D).

Question 27:

Compound 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 haloform reaction but not Cannizzaro reaction.

i P →ii Zn/H2Oi O3/CH2Cl2  QC8H8Oii R →ii Zn/H2Oi O3/CH2Cl2  SC8H8OThe option(s) with suitable combination of P and R, respectively, is(are)

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 1,3

 

Solutions:

Hence, the correct answers are both options A and C.

Question 28:

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

Option 1: With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive
Option 2: With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases
Option 3: With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases
Option 4: With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system negative
Correct Answer: 2,3

 

Solutions:

The entropy change of the surroundings is given by the expression:

ΔSSurroundings = -qSurroundingsTSurroundingsIf the enthalpy change of the reaction is positive (ΔH > 0), then with the increase in temperature the value of K for endothermic reaction increases. This happens as ΔSSurrounding decreases, i.e. surrounding becomes unfavourable.

If the enthalpy change of the reaction is negative (ΔH < 0), then with the increase in temperature, the value of K for exothermic reaction decreases. This happens as ΔSSurrounding decreases, i.e. surrounding becomes favourable.

​Hence, the correct answers are options (B) and (C).

Question 29:

The option(s) with only amphoteric oxides is (are):

Option 1: Cr2O3, CrO, SnO, PbO
Option 2: NO, B2O3, PbO, SnO2
Option 3: Cr2O3, BeO, SnO, SnO2
Option 4: ZnO, Al2O3, PbO, PbO2
Correct Answer: 3,4

 

Solutions:Among the given oxides, those which react with acids, as well as bases, will be amphoterric oxides. These are the oxides of metals like Cr, Be, Zn, Al, Sn and Pb.

Hence, the correct answers are options (C) and (D).

Question 30:

Among the following, the correct statement(s) is are

Option 1: Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure
Option 2: AlCl3 has the three-centre two-electron bonds in its dimeric structure
Option 3: BH3 has the three-centre two-electron bonds in its dimeric structure
Option 4: The Lewis acidity of BCl3 is greater than that of AlCl3
Correct Answer: 1,3,4

 

Solutions:Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure

AlCl3 has the three-centre four-electron bonds in its dimeric structure

BH3 has the three-centre two-electron bonds in its dimeric structure

The Lewis acidity of BCl3 is greater than that of AlCl3 because Lewis acidic strength decreases down the group.

Hence, the correct answers are options A, C and D.

Question 31:

The correct statement(s) about surface properties is (are)

Option 1: Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
Option 2: Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system.
Option 3: Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution.
Option 4: The critical temperatures of ethane and nitrogen and 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.
Correct Answer: 2,4

 

Solutions:Emulsions are those colloids in which both the dispersion phase and the dispersion medium are liquids. Option (A) is incorrect.

Adsorption is an exothermic process which is accompained by decrease in the entropy of the system, both ΔH and ΔS are negative. Option (B) is correct.

Brownian motion of colloidal particles increases as the size and viscosity of the particles decreases. Option (C) is incorrect.

The extent of adsorption is higher for those gases which have high critical temperatures. So among ethane (critical temperature 563 K) and nitrogen (critical temperature 126 K), ethane will be adsorbed to a greater extent than nitrogen on the same amount of activated charcoal at a given temperature. Option (D) is correct.

Hence, the correct answers are option (B) and (D).

Question 32:

For the following compounds, the correct statement(s) with respect of nucleophilic substitution reactions is(are);

Option 1: I and II follow SN2 mechanism
Option 2: The order of reactivity for I, III and IV is : IV > I > III
Option 3: I and III follow SN1 mechanism
Option 4: Compound IV undergoes inversion of configuration
Correct Answer: 1,2,3,4

 

Solutions:(A) I and II follow SN2 mechanism because both are primary halide (B) Reactivity order IV > I > III due to stability of intermediate in SN1 reaction. (C) I and III follow SN1 mechanism as they form stable carbocation intermediate (D) Compound IV undergoes inversion of configuration.

Hence, the correct answers are A,B, C and D.

Question 33:

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.

W and X are, respectively

Option 1: O3 and P4O6
Option 2: O2 and P4O10
Option 3: O3 and P4O10
Option 4: O2 and P4O6
Correct Answer: 2

 

Solutions:The sequence of the chemical reactions is as follows:

2 KClO3 →MnO2   Δ   2 KCl + 3 O2WP4White Phosphorus + 5 O2Excess of W →           P4O10XW = O2 X = P4O10

Hence, the correct answer is option (B).

Question 34:

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.

Y and Z are , respectively

Option 1: N2O4 and H3PO3
Option 2: N2O4 and HPO3
Option 3: N2O5 and HPO3
Option 4: N2O3 and H3PO4
Correct Answer: 3

 

Solutions:The sequence of the chemical reactions is as follows:

2 KClO3 →MnO2   Δ   2 KCl + 3 O2WP4White Phosphorus + 5 O2Excess of W →           P4O10XP4O10X + 4 HNO3 →               2 N2O5Y + HPO34ZY = N2O5 Z = (HPO3)​4

Hence, the correct answer is  option  (C).

Question 35:

The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O produces compounds S.

[Et it compounds P is ethyl group]

The reactions, Q to R and S to S, are –

Option 1: Dehydration and Friedel-Crafts acylation
Option 2: Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation
Option 3: Aromatic sulfonation and Friedel-Crafts acylation
Option 4: Friedel-Crafts alkylation and Fridel-Crafts acylation
Correct Answer: 2

 

Solutions: The reactions, Q to R and S to S, are Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation.

Hence, the correct answer is option B.

Question 36:

The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O produces compounds S.

[Et it compounds P is ethyl group]

The product S is –

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 4

 

Solutions:

Therefore, the product S is Hence, the correct answer is opton D.

Question 37:

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

Option 1: 3655
Option 2: 611
Option 3: 511
Option 4: 12
Correct Answer: 2

 

Solutions:Suppose z = 2k, where k = 0, 1, 2, 3, 4, 5 ∴ x + y = 10 − 2k                 …..(1) Number of non-negative integral solutions of (1)

=∑k=05C2-110- 2k+2-1=∑k=05C111- 2k=∑k=0511-2k=11+9+7+5+3+1=36Total number of cases = Number of non-integral solutions of x + y + z = 10 =

C3-110+3-1=C212=66∴ Required probability =

3666=611Hence, the correct answer is option B.

Question 38:

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 =

Option 1: 125
Option 2: 252
Option 3: 210
Option 4: 126
Correct Answer: 4

 

Solutions:It is given that S = {1, 2, 3,…,9} and Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Now, N1 = Number of subsets of S containing five elements out of which exactly 1 element is odd =

C15×C44=5×1=5N1 = Number of subsets of S containing five elements out of which exactly 2 elements are odd =

C25×C34=10×4=40N1 = Number of subsets of S containing five elements out of which exactly 3 elements are odd =

C35×C24=10×6=60N1 = Number of subsets of S containing five elements out of which exactly 4 elements are odd =

C45×C34=5×4=20N1 = Number of subsets of S containing five elements out of which exactly 5 elements are odd =

C55×C04=1×1=1∴ N1 + N2 + N3 + N4 + N5 = 5 + 40 + 60 + 20 + 1 = 126

Hence, the correct answer is option D.

Question 39:

If f space colon straight real numbers rightwards arrow straight real numbers is a twice differentiable function such that ƒ”(x) > 0 for all straight x element of straight real numbers, and

f 12=12, f1=1, then

Option 1: 0<f’1≤12
Option 2: f ‘(1) ≤ 0
Option 3: f ‘(1) > 1
Option 4: 12<f’1≤1
Correct Answer: 3

 

Solutions:Using Lagranges Mean Value Theorem on f(x) for

x∈12,1, we have

∴f’c=f1-f121-12, where c∈12,1⇒f’c=1-1212⇒f’c=1, where c∈12,1It is given that f ”(x) > 0

∀x∈RSo, f ‘(x) is an increasing function

∀x∈R. ∴ f ‘(x) > 1

Hence, the correct answer is option C.

Question 40:

If y = y(x) satisfies the differential equation

8×9+x dy=4+9+x-1 dx, x > 0 and y 0=7, then y 256=

Option 1: 80
Option 2: 3
Option 3: 16
Option 4: 9
Correct Answer: 2

 

Solutions:It is given that,

8×9+xdy=4+9+x-1dx, x>0.

⇒dy=4+9+x-18×9+xdx⇒y=18∫4+9+x-1×9+xdx       =18∫1×9+x4+9+xdxPut 9+x=t ⇒dxx·9+x=4dt∴y=48∫dt4+t⇒y=4+t+C⇒yx=4+9+x⇒y256=3Hence, the correct answer is option B.

Question 41:

How many 3 × 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of MTM is 5 ?

Option 1: 198
Option 2: 126
Option 3: 135
Option 4: 162
Correct Answer: 1

 

Solutions:Let

M=abcdefghi, where entries are {0, 1, 2}.

MTM=adgbehcfiabcdefghi∴trMTM=a2+b2+c2+d2+e2+f2+g2+h2+i2=5       GivenNow, only two cases are possible for the above equation to hold true. Case 1: Five entries are 1 and other four are 0. Number of matrices =

9C5×1= 126 Case 2: One entry is 2, one is 1 and other are 0. Number of matrices =

9C2×2!= 72 Therefore, the total number of matrices are 126 + 72 = 198.

Hence, the correct answer is option A.

Question 42:

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

OP→.OQ→+OR→.OS→=OR→.OP→+OQ→.OS→=OQ→.OR→+OP→.OS→Then the triangle PQR has S as its

Option 1: incentre
Option 2: orthocenter
Option 3: circumcentre
Option 4: centroid
Correct Answer: 2

 

Solutions:Let the position vector of P, Q, R and S with respect to O are

p→, q→, r→ and s→respectively.

Now,OP→·OQ→+OR→·OS →=OR→·OP→+ OQ→·OS→⇒p→·q→+r→·s→=r→·p→+q→·s→⇒p→-s→q→-r→=0       …..(i)Also, OR→·OP→+OQ→·OS →=OQ→·OR→+ OP→·OS→⇒r→·p→+q→·s→=q→·r→+p→·s→⇒r→-s→p→-q→=0       …..(ii)Also, OP→·OQ→+OR→·OS →=OQ→·OR→+ OP→·OS→⇒p→·q→+r→·s→=q→·r→+p→·s→⇒q→-s→·p→-r→=0       …..(iii)

Therefore, triangle PQR has S as its orthocentre. Hence, the correct answer is option B.

Question 43:

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

Option 1: 14x + 2y + 15z = 31
Option 2: 14x + 2y – 15z = 1
Option 3: –14x + 2y + 15z = 3
Option 4: 14x – 2y + 15z = 27
Correct Answer: 1

 

Solutions:Let the equation of the plane passing through the point (1, 1, 1) be

ax-1+by-1+cz-1=0. Now, Direction ratio of the normal to the plane =

i^j^k^21-23-6-2=i^-2-12-j^-4+6+k^-12-3 =-14i^-2j^-15k^So,

a-14=b-2=c-15=λ⇒a=-14λ,b=-2λ,c=-15λThus, the required equation of the plane is -14λx-1-2λy-1-15λz-1=0⇒14x+2y+15z=31Hence, the correct answer is option A.

Question 44:

If

I=∑k=198∫kk+1k+1xx+1dx, then

Option 1: I<4950
Option 2: I < loge99
Option 3: I>4950
Option 4: I > loge99
Correct Answer: 2, 3

 

Solutions:

I=∑k=198∫kk+ 1k+1xx+1dx       …..(1) Put x = k + p

I=∑k=198∫01k+1k+pk+p+1dpI>∑k=198∫01k+1k+p+12dp On integrating,

I=∑k=198k+1-1k+p+101⇒I>∑k=198k+11k+1-1k+2⇒I>∑k=1981k+2⇒13+14+15+…+1100<I 

⇒98100<I⇒4950<ISo, option C is correct.

Now, from (1)

k+1xx+1<k+1xk+1       ∵k+1 is the least value of x+1⇒k+1xx+1<1x             …..2From (1) and (2), we have

I<∑k=198∫kk+11xdx⇒I<∑k=198lnk+1-lnk⇒I<ln99So, option B is correct.

Hence, the correct answers are options B and C.

Question 45:

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

Option 1: ƒ(x) > e2x in (0,∞)
Option 2: ƒ(x) is decreasing in (0,∞)
Option 3: ƒ(x) is increasing in (0,∞)
Option 4: ƒ'(x) < e2x in (0,∞)
Correct Answer: 1,3

 

Solutions:Given f‘(x) > 2f(x) for all x

R. So, f‘(x) − 2f(x) > 0 for all x

R

∴e-2xf’x-2fx>0∀x∈R⇒ddxe-2xfx>0∀x∈R Let

px=e-2xfx 

p’x>0 ∀x∈RThus, p(x) is strictly increasing

∀x∈R. p(0) = 1 so, x > 0 p(x) > p(0) = 1

e-2xfx>1∀x∈0,∞⇒fx>e2x∀x∈0,∞Hence, option A is correct.

f’x>2fx>2e2x>2 ∀x∈0,∞       …(1) So, f(x) is strictly increasing for every x

∈0,∞Hence, option (C) is correct.

f’x>2e2x for all x∈0,∞Thus, Option (D) is incorrect.

So, the correct answers are options A and C.

Question 46:

If

fx=cos 2xcos 2xsin 2x-cos xcos x-sin xsin xsin xcos x, then

Option 1: f ‘(x) = 0 at exactly three points in (–π, π)
Option 2: f(x) attains its maximum at x = 0
Option 3: f(x) attains its minimum at x = 0
Option 4: f ‘(x) = 0 at more than three points in (–π, π)
Correct Answer: 2,4

 

Solutions:

fx=cos 2xcos 2xsin 2x-cosxcosx-sinxsinxsinxcosx⇒fx=cos2xcos2x+sin2x-cos2x-cos2x+sin2x+sin2x-sinxcosx-sinxcosx⇒fx=cos2x+cos22x-sin22x 

⇒fx=cos2x+cos4x⇒fx=2cos22x+cos2x-1Let cos2x = y

∴fx=2y2+y-1 where y∈-1,1 Thus, f(x) attains its minima at

y=-14∈-1,1 

fxmin=216-14-1=-98Similarly, we can find the maxima as follows:

fxmax=2+1-1=2, when cos2x = 1 or x = 0. f’x=-4sin4x-2sin2xNow, f’x=0⇒-4sin4x-2sin2x=0⇒-2sin2x1+4cos2x=0 

⇒sin2x=0 or cos2x=-14 

sin2x=0⇒x=nπ2

cos2x=-14 Thus, there are more than three solutions.

Hence, the correct answers are options B and D.

Question 47:

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

Option 1: tan α2-3 tan β2=0
Option 2: 3 tan α2- tan β2=0
Option 3: tan α2+3 tan β2=0
Option 4: 3 tan α2+ tan β2=0
Correct Answer: 1,3

 

Solutions:It is given that 2(cosβ – cosα) + cosα cosβ = 1.

⇒2cosβ – cosα+cosα cosβ-1=0    …..1Now,cosβ=1-tan2β21+tan2β2 and cosβ=1-tan2α21+tan2α2Using these trigonometric identities in (1), we get

⇒21-tan2β21+tan2β2 – 1-tan2α21+tan2α2+1-tan2α21+tan2α21-tan2β21+tan2β2-1=0⇒21-tan2β21+tan2α2-1-tan2α21+tan2β2+1-tan2α21-tan2β2-1+tan2α21+tan2β2=0⇒21+tan2α2-tan2β2-tan2α2tan2β2-1-tan2α2+tan2β2-tan2α2tan2β2+1-tan2α2-tan2β2+tan2α2tan2β2-1+tan2α2+tan2β2+tan2α2tan2β2=0⇒tan2α2=3tan2β2⇒tanα2=±3tanβ2⇒tanα2+3tanβ2=0 or tanα2-3tanβ2=0Hence, the correct answers are options A and C.

Question 48:

If

gx=∫sin xsin2xsin-1t dt,then

Option 1: g’π2=-2π
Option 2: g’-π2=2π
Option 3: g’π2=2π
Option 4: g’-π2=-2π
Correct Answer: 0

 

Solutions:It is given that

gx=∫sin xsin2xsin-1t dt, 

⇒g’x=2sin-1sin2x ×cos2x-sin-1sinxcosx⇒g’π2=0 and g’-π2=0Disclaimer: No given option is matching with the answer.

Question 49:

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

Option 1: 12<α<1
Option 2: α4 + 4α2 – 1 = 0
Option 3: 0<α≤12
Option 4: 4 – 4α2 + 1 = 0
Correct Answer: 1,4

 

Solutions:Consider the figure:

Area between y = x3 and y = x for 0 ≤ x ≤ 1 is

A=∫01x-x3dx=14So, area of the region OPQ =

A2=18 

∴∫0αx-x3dx=18⇒2α4-4α2+1=0⇒α2-12=12⇒α2=2-12⇒α=2-12>12 

∴12<α<1Hence, the correct answers are options A and D.

Question 50:

Let

fx=1-x1+|1-x||1-x|cos11-x for x ≠1. Then

Option 1: limx→1+ f(x) does not exist
Option 2: limx→1 f(x) does not exist
Option 3: limx→1 f(x) = 0
Option 4: limx→1+ f(x) = 0
Correct Answer: 1, 3

 

Solutions:

fx=1-x1+|1-x||1-x|cos11-xfor x ≠ 1 We know,

1-x=1-x,      x<1-1-x,   x>1 

∴fx=1-xcos11-x, x<1-1+xcos11-x, x>1RHL=limx→1+fx=limh→0-2+hcos1hSo, RHL does not exist.LHL=limx→1-fx=limh→0hcos1h=0Hence, the correct answers are options A and C.

Question 51:

Paragraph 1
Let O be the origin, and OX with rightwards arrow on top comma space OY with rightwards arrow on top comma space OZ with rightwards arrow on top, be three unit vectors in the directions of the sides QR with rightwards arrow on top comma space RP with rightwards arrow on top comma space PQ with rightwards arrow on top, respectively, of a triangle PQR.

OX→×OY→=

Option 1: sin(Q + R)
Option 2: sin(P + R)
Option 3: sin 2R
Option 4: sin(P + Q)
Correct Answer: 4

 

Solutions:It is given that, OX with rightwards arrow on top comma space OY with rightwards arrow on top comma space OZ with rightwards arrow on top be three unit vectors in the directions of the sides QR with rightwards arrow on top comma space RP with rightwards arrow on top comma space PQ with rightwards arrow on top, respectively, of a triangle PQR.

OX→×OY→=OX→OY→sinR⇒OX→×OY→=sinR⇒OX→×OY→=sinπ-P+Q⇒OX→×OY→=sinP+Q=sinP+QHence, the correct answer is option D.

Question 52:

Paragraph 1

Let O be the origin, and OX with rightwards arrow on top comma space OY with rightwards arrow on top comma space OZ with rightwards arrow on top, be three unit vectors in the directions of the sides QR with rightwards arrow on top comma space RP with rightwards arrow on top comma space PQ with rightwards arrow on top, respectively, of a triangle PQR.

If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is

Option 1: 32
Option 2: -32
Option 3: 53
Option 4: -53
Correct Answer: 2

 

Solutions:

cosP+Q+cosQ+R+cosR+P=cosπ-R+cosπ-P+cosπ-Q=-cosR-cosP-cosQ=-cosP+cosQ+cosRNow, in any triangle PQR, the maximum value of cosP + cosQ + cosR =

32. So, the minimum value of −(cosP + cosQ + cosR) is

-32. Hence, the correct answer is option B.

Question 53:

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

a+b5=0, then a = 0 = b.

If a4 = 28, then p + 2q =

Option 1: 14
Option 2: 7
Option 3: 12
Option 4: 21
Correct Answer: 3

 

Solutions:It is given that α, β are the roots of the equation, x2x – 1 = 0, where αβ.

∴α2=α+1Squaring both sides, we get

α4=α+12=α2+1+2a 

∴ α4=α+1+1+2α=3α+2               α2=α+1Similarly, β4=3β+2Now,a4=28⇒pα4+qβ4=28⇒p3α+2+q3β+2=28⇒p3α+2+q3-3α+2=28             α+β=1⇒α3p-3q+2p+5q=28⇒3p-3q=0, 2p+5q=28                   α∈R-Q                     ⇒p=q, 2p+5q=28⇒p=q=4∴p+2q=12Hence, the correct answer is option C.

Question 54:

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

a+b5=0, then a = 0 = b.

a12 =

Option 1: 2a11 + a10
Option 2: a11 – a10
Option 3: a11 + a10
Option 4: a11 + 2a10
Correct Answer: 3

 

Solutions:It is given that α, β are the roots of the equation, x2x − 1 = 0, where αβ. Also, an = n + n Now,

α2=α+1Multiplying both sides by αn-2, we getαn=αn-1+αn-2Similarly, βn=βn-1+βn-2So, an=pαn+qβn⇒an=pαn-1+αn-2+qβn-1+βn-2⇒an=pαn-1+qβn-1+pαn-2+qβn-2⇒an=an-1+an-2Putting n=12, we geta12=a11+a10Hence, the correct answer is option C.

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