
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 × 10^{5} times heavier than the Earth and is at a distance 2.5 ×10^{4 }times larger than the radius of the Earth. The escape velocity from Earth’s gravitational field is v_{e} = 11.2 km s^{–1}. The minimum initial velocity (v_{s}) 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:  v_{s} = 22 km s^{–1} 
Option 2:  v_{s} = 72 km s^{–1} 
Option 3:  v_{s} = 42 km s^{–1} 
Option 4:  v_{s} = 62 km s^{–1} 
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→=1b P→+b2 Q→ 
Option 2:  S→=b1 P→+bQ→ 
Option 3:  S→=1b P→+bQ→ 
Option 4:  S→=1b2 P→+bQ→ 
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:  µ0I4πa331 
Option 2:  µ0I4πa631 
Option 3:  µ0I4πa63+1 
Option 4:  µ0I4πa323 
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 workfunction Ï•_{0} is illuminated with light of wavelength
λλ<hcϕ0. The fastest photoelectron has a deBroglie wavelength λ_{d}. A change in wavelength of the incident light by Δλ results in a change Δλ_{d} in λ_{d}. Then the ratio Δλ_{d}/Δλ is proportional to
Option 1:  λd3/λ2 
Option 2:  λd3/λ 
Option 3:  λd2/λ2 
Option 4:  λd/λ 
Solutions:According to photoelectric equation:
KEmax = hcλ – ϕop22m = hcλ – ϕohλ22m = hcλ – ϕoAssuming small changes and differentiating both sides, we get
h22m2dλdλd3 = 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 % 
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:  R^{3} 
Option 2:  1R 
Option 3:  R 
Option 4:  R^{2/3} 
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 πn1 
Option 4:  ∆=h tan2 π2n 
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 
Solutions:When the bar makes an angle θ with the vertical, the height of its Centre of Mass (COM) is
L2cosθTherefore, displacement =
L2L2cosθ=L2(1cosθ)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 S_{1} and S_{2} 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 P_{2} 
Option 2:  The angular separation between two consecutive bright spots decreases as we move from P_{1} to P_{2} along the first quadrant 
Option 3:  At P_{2} the order of the fringe will be maximum 
Option 4:  The total number of fringes produced between P_{1} and P_{2} in the first quadrant is close to 3000 
Solutions:At point P_{2}:
∆x=d=1.8 mm=3000λHence a bright fringe will be formed at P_{2}. 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 L_{1} and L_{2} 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 L_{1} and L_{2} is fixed at all times (t > 0) 
Option 2:  After a long time, the current through L_{1} will be
VRL2L1+L2 
Option 3:  After a long time, the current through L_{2} will be
VRL1L1+L2 
Option 4:  At t = 0, the current through the resistance R is
VR 
Solutions:Since inductors are connected in parallel.
VL1=VL2L1dI1dt=L2dI2dtL1I1=L2I2I1I2=L2L1Therefore, the ratio of currents through L_{1} and L_{2} 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 
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 V_{X} = V_{0} 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 
Solutions:Potential difference between X and Y = V_{X} – V_{Y} Potential difference between Y and Z = V_{Y} – VZ Phasors of Voltages:
∴ VXVY=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
Q2ε0112 
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. 
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 = Q2εo112∴Flux through curved surface = Q2εo112Hence, 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 xaxis enters region 2 from region 1 at point P_{1}(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 P_{2} on xaxis 
Option 2:  For
B>23pQR, the particle will reenter region 1 
Option 3:  For a fixed B, particle of same charge Q and same velocity v, the distance between the point P_{1} and the point of reentry into region 1 is inversely proportional to the mass of the particle. 
Option 4:  When the particle reenters region 1 through the longest possible path in region 2, the magnitude of the chage in its linear momentum between point P_{1} and the farthest point from yaxis is
p2. 
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 P_{2}.
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 reenter 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 V_{0 }(i.e., charging continues for time T >> RC). In the process some dissipation (E_{D}) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is E_{C}. 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 V_{0} and the capacitor is charged to the same final voltage V_{0} as in Process 1. These two processes are depicted in Figure 2. 
In Process 1, the energy stored in the capacitor E_{C} and heat dissipated across resistance E_{D} are related by:
Option 1:  E_{C} = E_{D} 
Option 2:  E_{C} = 2E_{D} 
Option 3:  EC=12ED 
Option 4:  E_{C} = E_{D} ln2 
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, E_{D} = Work done by the battery – Energy stored in the capacitor
ED=CV212CV2=12CV2 …..(ii)From (i) and (ii), we see that E_{C} = E_{D}
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 V_{0 }(i.e., charging continues for time T >> RC). In the process some dissipation (E_{D}) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is E_{C}. 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 V_{0} and the capacitor is charged to the same final voltage V_{0} as in Process 1. These two processes are depicted in Figure 2. 
In Process 2, total energy dissipated across the resistance E_{D} is :
Option 1:  ED=1312CV02 
Option 2:  ED=312CV02 
Option 3:  ED=12CV02 
Option 4:  ED=3CV02 
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=CV029CV0218=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 =
4CV0218CV0218=3CV0218
Heat loss in the step 2 = Work done by the battery in step 2 – Energy stored in the capacitor in step 2
ED2=2CV0293CV0218=CV0218For step 3: Charge stored in the capacitor is
q3=CV0Extra charge flow through the battery =
CV02CV03=CV03Work done by the battery,
W3=CV03×V0=CV023Final energy stored in the capacitor after step 3,
EF=12CV02Energy stored in the step 3 =
CV0224CV0218=5CV0218
Heat loss in the step 3 = Work done by the battery in step 3 – Energy stored in the capacitor in step 3
ED3=CV0235CV0218=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 ω_{0}. 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 Rr2 
Option 3:  12Mω02 Rr2 
Option 4:  32Mω02 Rr2 
Solutions: Point A is the ICOR of the system.
ωR=ω0Rr⇒ω=ω0RrRThe total kinetic energy of the ring is
K=12Iω2⇒K=122mR2ω2⇒K=mω02Rr2Hence, 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 ω_{0}. 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 ω_{0} below which the ring will drop down is :
Option 1:  3g2µRr 
Option 2:  gµRr 
Option 3:  2gµRr 
Option 4:  2g2µRr 
Solutions:Normal reaction at the point of contact is
N=Mω2(Rr) …..(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≥μMω02Rr⇒ω0=gμRr Mg≥μMω02Rr⇒ω0=gμRrHence, the correct answer is options B.
Question 19:
Which of the following combination will produce H_{2} 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. HNO_{3} 
Option 4:  Fe metal and conc. HNO_{3} 
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: 
Solutions:Given: Mass of ethanol (w_{b}) = 34.5 g Mass of water (w_{a}) = 500 g Molar mass of ethanol, C_{2}H_{5}OH (M_{b}) = 46 g/mol Freezing point of pure water (
Tf°) = 273 K Freezing point depression constant of water (K_{f}) = 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 H_{3}PO_{2} , H_{3}PO_{4} , H_{3}PO_{3} and H_{4}P_{2}O_{6} is
Option 1:  H_{3}PO_{4} > H_{4}P_{2}O_{6} > H_{3}PO_{3} > H_{3}PO_{2} 
Option 2:  H_{3}PO_{3} > H_{3}PO_{2} > H_{3}PO_{4} > H_{4}P_{2}O_{6} 
Option 3:  H_{3}PO_{2} > H_{3}PO_{3} > H_{4}P_{2}O_{6} > H_{3}PO_{4} 
Option 4:  H_{3}PO_{4} > H_{3}PO_{2} > H_{3}PO_{3} > H_{4}P_{2}O_{6} 
Solutions:
Oxoacid of P  Structure  Oxidation State of P 
H_{3}PO_{2}  +1  
H_{3}PO_{4}  +5  
H_{3}PO_{3}  +3  
H_{4}P_{2}O_{6}  +4 
The order of the oxidation state of the phosphorus atom in H_{3}PO_{2} , H_{3}PO_{4} , H_{3}PO_{3} and H_{4}P_{2}O_{6} is H_{3}PO_{4} > H_{4}P_{2}O_{6} > H_{3}PO_{3} > H_{3}PO_{2}
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
âˆ†_{f}G° [C(graphite)] = 0 kJ mol^{–1}
âˆ†_{f}G° [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} m^{3} 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 m^{2} s^{–2} ; 1 Pa = 1 kg m^{–1} s^{–2}; 1 bar = 10^{5} Pa]
Option 1:  14501 bar 
Option 2:  29001 bar 
Option 3:  58001 bar 
Option 4:  1405 bar 
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 mol1dG = VdP∫ΔG1ΔG2dΔG = ∫P1P2ΔVdPΔG2ΔG1 = ΔVP2P12.9×1030 = 2×1061P2P21 = 2.9×1032×106P21 = 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: 
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 
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)  ZnSO_{4} (aq.)  CuSO_{4} (aq.)  Cu(s)
when the concentration of Zn^{2+} is 10 times the concentration of Cu^{2+}, the expression for âˆ†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 
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. 
Solutions:The temperature dependence of the rate of a chemical reaction is accurately explained by Arrhenius equation:
k = PZAB eEa/RTHere,P = Probability or steric factorZAB =Collision frequency of reactantseEa/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.
â€‹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 C_{8}H_{8}O. 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: 
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 
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 ΔS_{Surrounding} 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 ΔS_{Surrounding} decreases, i.e. surrounding becomes favourable.
â€‹Hence, the correct answers are options (B) and (C).
Question 29:
The option(s) with only amphoteric oxides is (are):
Option 1:  Cr_{2}O_{3}, CrO, SnO, PbO 
Option 2:  NO, B_{2}O_{3}, PbO, SnO_{2} 
Option 3:  Cr_{2}O_{3}, BeO, SnO, SnO_{2} 
Option 4:  ZnO, Al_{2}O_{3}, PbO, PbO_{2} 
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(CH_{3})_{3} has the threecentre twoelectron bonds in its dimeric structure 
Option 2:  AlCl_{3} has the threecentre twoelectron bonds in its dimeric structure 
Option 3:  BH_{3} has the threecentre twoelectron bonds in its dimeric structure 
Option 4:  The Lewis acidity of BCl_{3} is greater than that of AlCl_{3} 
Solutions:Al(CH_{3})_{3} has the threecentre twoelectron bonds in its dimeric structure
AlCl_{3} has the threecentre fourelectron bonds in its dimeric structure
BH_{3} has the threecentre twoelectron bonds in its dimeric structure
The Lewis acidity of BCl_{3} is greater than that of AlCl_{3} 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. 
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 S_{N}2 mechanism 
Option 2:  The order of reactivity for I, III and IV is : IV > I > III 
Option 3:  I and III follow S_{N}1 mechanism 
Option 4:  Compound IV undergoes inversion of configuration 
Solutions:(A) I and II follow S_{N}2 mechanism because both are primary halide (B) Reactivity order IV > I > III due to stability of intermediate in S_{N}1 reaction. (C) I and III follow S_{N}1 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 KClO_{3} in the presence of catalytic amount of MnO_{2} , a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO_{3} gives Y and Z.
W and X are, respectively
Option 1:  O_{3} and P_{4}O_{6} 
Option 2:  O_{2} and P_{4}O_{1}_{0} 
Option 3:  O_{3} and P_{4}O_{1}_{0} 
Option 4:  O_{2} and P_{4}O_{6} 
Solutions:The sequence of the chemical reactions is as follows:
2 KClO3 →MnO2 Δ 2 KCl + 3 O2WP4White Phosphorus + 5 O2Excess of W → P4O10XW = O_{2} X = P_{4}O_{10}
Hence, the correct answer is option (B).
Question 34:
Upon heating KClO_{3} in the presence of catalytic amount of MnO_{2} , a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO_{3} gives Y and Z.
Y and Z are , respectively
Option 1:  N_{2}O_{4} and H_{3}PO_{3} 
Option 2:  N_{2}O_{4} and HPO_{3} 
Option 3:  N_{2}O_{5} and HPO_{3} 
Option 4:  N_{2}O_{3} and H_{3}PO_{4} 
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 = N_{2}O_{5} Z = (HPO_{3})_{â€‹4}
Hence, the correct answer is option (C).
Question 35:
The reaction of compound P with CH_{3}MgBr (excess) in (C_{2}H_{5})_{2}O followed by addition of H_{2}O gives Q, The compound Q on treatment with H_{2}SO_{4} at 0ºC gives R. The reaction of R with CH_{3}COCl in the presence of anhydrous AlCl_{3} in CH_{2}Cl_{2} followed by treatment with H_{2}O produces compounds S.
[Et it compounds P is ethyl group]
The reactions, Q to R and S to S, are –
Option 1:  Dehydration and FriedelCrafts acylation 
Option 2:  FriedelCrafts alkylation, dehydration and FriedelCrafts acylation 
Option 3:  Aromatic sulfonation and FriedelCrafts acylation 
Option 4:  FriedelCrafts alkylation and FridelCrafts acylation 
Solutions: The reactions, Q to R and S to S, are FriedelCrafts alkylation, dehydration and FriedelCrafts acylation.
Hence, the correct answer is option B.
Question 36:
The reaction of compound P with CH_{3}MgBr (excess) in (C_{2}H_{5})_{2}O followed by addition of H_{2}O gives Q, The compound Q on treatment with H_{2}SO_{4} at 0ºC gives R. The reaction of R with CH_{3}COCl in the presence of anhydrous AlCl_{3} in CH_{2}Cl_{2} followed by treatment with H_{2}O produces compounds S.
[Et it compounds P is ethyl group]
The product S is –
Option 1:  
Option 2:  
Option 3:  
Option 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 
Solutions:Suppose z = 2k, where k = 0, 1, 2, 3, 4, 5 ∴ x + y = 10 − 2k …..(1) Number of nonnegative integral solutions of (1)
=∑k=05C2110 2k+21=∑k=05C111 2k=∑k=05112k=11+9+7+5+3+1=36Total number of cases = Number of nonintegral solutions of x + y + z = 10 =
C3110+31=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 N_{k} be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =
Option 1:  125 
Option 2:  252 
Option 3:  210 
Option 4:  126 
Solutions:It is given that S = {1, 2, 3,…,9} and N_{k} be the number of subsets of S, each containing five elements out of which exactly k are odd. Now, N_{1} = Number of subsets of S containing five elements out of which exactly 1 element is odd =
C15×C44=5×1=5N_{1} = Number of subsets of S containing five elements out of which exactly 2 elements are odd =
C25×C34=10×4=40N_{1} = Number of subsets of S containing five elements out of which exactly 3 elements are odd =
C35×C24=10×6=60N_{1} = Number of subsets of S containing five elements out of which exactly 4 elements are odd =
C45×C34=5×4=20N_{1} = Number of subsets of S containing five elements out of which exactly 5 elements are odd =
C55×C04=1×1=1∴ N_{1} + N_{2} + N_{3} + N_{4} + N_{5} = 5 + 40 + 60 + 20 + 1 = 126
Hence, the correct answer is option D.
Question 39:
If is a twice differentiable function such that ƒ”(x) > 0 for all , 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 
Solutions:Using Lagranges Mean Value Theorem on f(x) for
x∈12,1, we have
∴f’c=f1f12112, where c∈12,1⇒f’c=11212⇒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+x1 dx, x > 0 and y 0=7, then y 256=
Option 1:  80 
Option 2:  3 
Option 3:  16 
Option 4:  9 
Solutions:It is given that,
8×9+xdy=4+9+x1dx, x>0.
⇒dy=4+9+x18×9+xdx⇒y=18∫4+9+x1×9+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 M^{T}M is 5 ?
Option 1:  198 
Option 2:  126 
Option 3:  135 
Option 4:  162 
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 
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 
Solutions:Let the equation of the plane passing through the point (1, 1, 1) be
ax1+by1+cz1=0. Now, Direction ratio of the normal to the plane =
i^j^k^212362=i^212j^4+6+k^123 =14i^2j^15k^So,
a14=b2=c15=λ⇒a=14λ,b=2λ,c=15λThus, the required equation of the plane is 14λx12λy115λz1=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 < log_{e}99 
Option 3:  I>4950 
Option 4:  I > log_{e}99 
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+11k+p+101⇒I>∑k=198k+11k+11k+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+1lnk⇒I<ln99So, option B is correct.
Hence, the correct answers are options B and C.
Question 45:
If ƒ : â„ → â„ is a differentiable function such that ƒ'(x) > 2ƒ(x) for all x ∈ â„, and ƒ(0) = 1, then
Option 1:  ƒ(x) > e^{2x} in (0,∞) 
Option 2:  ƒ(x) is decreasing in (0,∞) 
Option 3:  ƒ(x) is increasing in (0,∞) 
Option 4:  ƒ'(x) < e^{2x} in (0,∞) 
Solutions:Given f‘(x) > 2f(x) for all x
∈R. So, f‘(x) − 2f(x) > 0 for all x
∈R
∴e2xf’x2fx>0∀x∈R⇒ddxe2xfx>0∀x∈R Let
px=e2xfx
p’x>0 ∀x∈RThus, p(x) is strictly increasing
∀x∈R. p(0) = 1 so, x > 0 p(x) > p(0) = 1
e2xfx>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 2xcos xcos xsin 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) 
Solutions:
fx=cos 2xcos 2xsin 2xcosxcosxsinxsinxsinxcosx⇒fx=cos2xcos2x+sin2xcos2xcos2x+sin2x+sin2xsinxcosxsinxcosx⇒fx=cos2x+cos22xsin22x
⇒fx=cos2x+cos4x⇒fx=2cos22x+cos2x1Let cos2x = y
∴fx=2y2+y1 where y∈1,1 Thus, f(x) attains its minima at
y=14∈1,1
fxmin=216141=98Similarly, we can find the maxima as follows:
fxmax=2+11=2, when cos2x = 1 or x = 0. f’x=4sin4x2sin2xNow, f’x=0⇒4sin4x2sin2x=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 α23 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 
Solutions:It is given that 2(cosβ – cosα) + cosα cosβ = 1.
⇒2cosβ – cosα+cosα cosβ1=0 …..1Now,cosβ=1tan2β21+tan2β2 and cosβ=1tan2α21+tan2α2Using these trigonometric identities in (1), we get
⇒21tan2β21+tan2β2 – 1tan2α21+tan2α2+1tan2α21+tan2α21tan2β21+tan2β21=0⇒21tan2β21+tan2α21tan2α21+tan2β2+1tan2α21tan2β21+tan2α21+tan2β2=0⇒21+tan2α2tan2β2tan2α2tan2β21tan2α2+tan2β2tan2α2tan2β2+1tan2α2tan2β2+tan2α2tan2β21+tan2α2+tan2β2+tan2α2tan2β2=0⇒tan2α2=3tan2β2⇒tanα2=±3tanβ2⇒tanα2+3tanβ2=0 or tanα23tanβ2=0Hence, the correct answers are options A and C.
Question 48:
If
gx=∫sin xsin2xsin1t dt,then
Option 1:  g’π2=2π 
Option 2:  g’π2=2π 
Option 3:  g’π2=2π 
Option 4:  g’π2=2π 
Solutions:It is given that
gx=∫sin xsin2xsin1t dt,
⇒g’x=2sin1sin2x ×cos2xsin1sinxcosx⇒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) ∈ â„^{2} : x^{3} ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal parts, then
Option 1:  12<α<1 
Option 2:  α^{4} + 4α^{2} – 1 = 0 
Option 3:  0<α≤12 
Option 4:  2α^{4} – 4α^{2} + 1 = 0 
Solutions:Consider the figure:
Area between y = x^{3} and y = x for 0 ≤ x ≤ 1 is
A=∫01xx3dx=14So, area of the region OPQ =
A2=18
∴∫0αxx3dx=18⇒2α44α2+1=0⇒α212=12⇒α2=212⇒α=212>12
∴12<α<1Hence, the correct answers are options A and D.
Question 50:
Let
fx=1x1+1x1xcos11x for x ≠1. Then
Option 1:  lim_{x→1+} f(x) does not exist 
Option 2:  lim_{x→1–} f(x) does not exist 
Option 3:  lim_{x→1–} f(x) = 0 
Option 4:  lim_{x→1+} f(x) = 0 
Solutions:
fx=1x1+1x1xcos11xfor x ≠ 1 We know,
1x=1x, x<11x, x>1
∴fx=1xcos11x, x<11+xcos11x, x>1RHL=limx→1+fx=limh→02+hcos1hSo, RHL does not exist.LHL=limx→1fx=limh→0hcos1h=0Hence, the correct answers are options A and C.
Question 51:
Paragraph 1
Let O be the origin, and , be three unit vectors in the directions of the sides , 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) 
Solutions:It is given that, be three unit vectors in the directions of the sides , respectively, of a triangle PQR.
OX→×OY→=OX→OY→sinR⇒OX→×OY→=sinR⇒OX→×OY→=sinπP+Q⇒OX→×OY→=sinP+Q=sinP+QHence, the correct answer is option D.
Question 52:
Paragraph 1
Let O be the origin, and , be three unit vectors in the directions of the sides , 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 
Solutions:
cosP+Q+cosQ+R+cosR+P=cosπR+cosπP+cosπQ=cosRcosPcosQ=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, x^{2} – x – 1 = 0, where α ≠ β. For n = 0, 1, 2,…., let a_{n} = pα^{n} + qβ^{n}. FACT : If a and b are rational numbers and a+b5=0, then a = 0 = b. 
If a_{4} = 28, then p + 2q =
Option 1:  14 
Option 2:  7 
Option 3:  12 
Option 4:  21 
Solutions:It is given that α, β are the roots of the equation, x^{2} – x – 1 = 0, where α ≠ β.
∴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+q33α+2=28 α+β=1⇒α3p3q+2p+5q=28⇒3p3q=0, 2p+5q=28 α∈RQ ⇒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, x^{2} – x – 1 = 0, where α ≠ β. For n = 0, 1, 2,…., let a_{n} = pα^{n} + qβ^{n}. FACT : If a and b are rational numbers and a+b5=0, then a = 0 = b. 
a_{12} =
Option 1:  2a_{11} + a_{10} 
Option 2:  a_{11} – a_{10} 
Option 3:  a_{11} + a_{10} 
Option 4:  a_{11} + 2a_{10} 
Solutions:It is given that α, β are the roots of the equation, x^{2} − x − 1 = 0, where α ≠ β. Also, a_{n} = pα^{n} + qβ^{n} Now,
α2=α+1Multiplying both sides by αn2, we getαn=αn1+αn2Similarly, βn=βn1+βn2So, an=pαn+qβn⇒an=pαn1+αn2+qβn1+βn2⇒an=pαn1+qβn1+pαn2+qβn2⇒an=an1+an2Putting n=12, we geta12=a11+a10Hence, the correct answer is option C.