
IIT JEE Advanced 2017 Paper 1 (Code 9)
Test Name: IIT JEE Advanced 2017 Paper 1 (Code 9)
Question 1:
A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse 1) of wavelength λ_{0 }is produced at point O on the rope. The pulse takes time T_{OA} to reach point A. If the wave pulse of wavelength λ_{0} is produced at point A (Pulse 2) without disturbing the position of M it takes time T_{AO} to reach point O. Which of the following options is/are correct?
Option 1:  The time T_{AO} = T_{OA} 
Option 2:  The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope 
Option 3:  The wavelength of Pulse 1 becomes longer when it reaches point A 
Option 4:  The velocity of any pulse along the rope is independent of its frequency and wavelength. 
Solutions:(A) Speed of wave is a property of medium, so time taken to cross the string will be equal. Therefore, the time T_{AO} = T_{OA}_{ .}
(B) Speeds of the two pulses are same but their direction of propagation is different at the midpoint of the rope. Hence, velocities of the two pulses (pulse 1 and pulse 2) are different at the midpoint of the rope.
(C) Since wavelength is given as
λ= vf = 1fTμ and To > TAFrom above expression, it can be seen that wavelength of pulse become shorter on reaching A.
(D) Velocity of any pulse is given by
v = TμTherefore we can say that â€‹the velocity of any pulse along the rope is independent of its frequency and wavelength.
Hence, the correct answers are options A and D.
Question 2:
A human body has a surface area of approximately 1 m^{2}. The normal body temperature is 10 K above the surrounding room temperature T_{0}. Take the room temperature to be T_{0} = 300 K. For T_{0} = 300 K, the value of
σT04= 460 Wm^{–2} (where σ is the StefanBoltzmann constant). Which of the following options is/are correct?
Option 1:  The amount of energy radiated by the body in 1 second is close to 60 Joules 
Option 2:  If the surrounding temperature reduces by a small amount ΔT_{0} << T_{0}, then to maintain the same body temperature the same (living) human being needs to radiate ΔW =
4σT03∆T0more energy per unit time 
Option 3:  Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation 
Option 4:  If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths 
Solutions:(A) Let us assume e =1 (black body radiation)
W = σA(T4 T04)Now, Wrad = σAT4 = σ.1.( T0 + 10 )4 =σ.T04(1 + 10 T0 )4 …..(i)Putting T0 = 300 in (i), we get =σ.(300)4(1+40 300) = 460×1715 = 520 WWnet = 520 – 460 ≈60 W ⇒ Energy radiated in 1 second =60 J(B)
P = σA(T4 T04)dP = σA(0 – 4T3.dT) and dT = ∆T⇒dP = 4σAT3∆T(C) Surface area decreases
⇒ Energy radiation decreases
Hence, the correct answers are options A, B and C.
Question 3:
A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a coordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?
Option 1:  The x component of displacement of the centre of mass of the block M is :
mRM+m 
Option 2:  The position of the point mass is :
x=2mRM+m 
Option 3:  The velocity of the point mass m is :
v=2gR1+mM 
Option 4:  The velocity of the block M is :
V=mM2gR 
Solutions:(A)
Ms∆x¯cm = m1∆x¯ + m2∆x¯20 = m(+R + x¯ ) + mx¯ x¯ = mRM + m(C)
If speed of point mass is v, then using conservation of linear momentum we get the velocity of the block of mass M as
V = mvM
mgR = 12mv2 + 12M(mvM)2 ⇒ mgR = 12mv2 (1 + mM)v = 2gR1 + mMHence, the correct answers are options A and C.
Question 4:
A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field
B→points into the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω in the magnetic field. Which of the following options is/are correct?
Option 1:  The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper 
Option 2:  The net emf induced due to both the loops is proportional to cos ωt 
Option 3:  The emf induced in the loop is proportional to the sum of the areas of the two loops 
Option 4:  The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone 
Solutions:We know that,
ϕ = BAcos θHere, θ = ωtϕ = BAcos (ωt)Now, rate of change of magnetic flux is
dϕdt= ε =ddt[BAcos(ωt)]dϕdt= ε = BAωsin(ωt)⇒dϕdt ∝ sin(ωt)Therefore, rate of change of magnetic flux is maximum when
ωt = π2 i.e. when the plane of the loops is perpendicular to plane of the paper.
Also, the net emf induced due to both the loops will be the difference of emfs in both the loops because their polarities are opposite. Hence,
εnet = ε2A – εA = B(2A)ωsinωt – B(A)ωsinωt = BAωsinωtHence, the correct answers are options A and D.
Question 5:
For an isosceles prism of angle A and refractive index μ, it is found that the angle of minimum deviation δ_{m} = A. Which of the following options is/are correct ?
Option 1:  At minimum deviation, the incident angle i_{1} and the refracting angle r_{1} at the first refracting surface are related by r_{1} = (i_{1}/2) 
Option 2:  For this prism, the refractive index μ and the angle of prism A are related as
A=12cos1µ2 
Option 3:  For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is
i1=sin1sin A4cos2A21cos A 
Option 4:  For the angle of incidence i_{1} = A, the ray inside the prism is parallel to the base of the prism. 
Solutions: For minimum deviation i_{1} = e , r_{1 }= r_{2} and r_{1 }+ r_{2 }=_{ } A
⇒ r1 = A2(A)
δm = 2i1 – A = A (given)⇒ i1 = A⇒ r1 = A2 = i12
(B)
μ = sin(A)sin (A2) = 2cos(A2)⇒A = 2cos1 (μ2)(C)
μ sin(r2) = 1sin(r2) = 1μ ⇒r2 = sin1 (1μ)Also, r1 +r2 = A r1 = A – r2 = A – sin1 (1μ)Using snell’s law,
sin (i1) = μ sin (r1) …..(ii) using (i) in (ii), we geti1 = sin1 μsin A sin11μi1 = sin1 μ2 – 1 sin A cosA =sin1 4cos2A2 – 1 sin A cosAHere, μ = 2cosA2(D) Condition for minimum deviation are i_{1} = e and
r1 = r2 = A2Rays will be parallel to base.
Hence, the correct answers are options A, C and D.
Question 6:
In the circuit shown, L = 1 μH, C = 1 μF and R = 1 kΩ. They are connected in series with an a.c. source V = V_{0} sin ωt as shown. Which of the following options is/are correct ?
Option 1:  The frequency at which the current will be in phase with the voltage is independent of R. 
Option 2:  At ω ~ 0 the current flowing through the circuit becomes nearly zero 
Option 3:  At ω >> 10^{6} rad.s^{–1}, the circuit behaves like a capacitor. 
Option 4:  The current will be in phase with the voltage if ω = 10^{4} rad.s^{–1}. 
Solutions:(A) The current in the given RLC circuit will be in phase with the voltage if
XL = XC⇒ ωL = 1ωC⇒ ω = 1LCor f = 12π1LCTherefore, the frequency at which the current will be in phase with the voltage is independent of R.
(B) Current in the circuit can be given as
i = V0sinωtR2 + (ωL)2 1ωC2At ω ~ 0i ~ 0(C) At
ω = 106 rad s1
XL = ωL =106 × 106 = 1 ΩFor ω>>106 , XL≫1 ΩandXc = 1 ωC =1106 × 106 = 1 ΩFor ω>>106 , Xc≪1 ΩR is independent of
ωTherefore, at ω >> 10^{6} rad s^{–1}, the circuit behaves like a inductor.
(D) ω = 1LC ω = 1106 × 106 = 106 rad s1Hence, the correct answers are options A and B.
Question 7:
A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true?
Option 1:  The resistive force experienced by the plate is proportional to v 
Option 2:  The pressure difference between the leading and trailing faces of the plate is proportional to uv. 
Option 3:  The plate will continue to move with constant nonzero acceleration, at all times 
Option 4:  At a later time the external force F balances the resistive force. 
Solutions:
Let,
u’=αuForce due to gas molecules hitting the plate from front side per second,
FLeading=2ρAu’+v2Force due to gas molecules hitting the plate from back side per second,
FTrailing=2ρAu’v2Net force on the plate due to gas moleculesâ€‹
F’=FLeadingFTrailing=2ρAu’v=8ρAu’vPressure difference=FLeadingFTrailingArea=8ρAu’vA⇒Pressure difference=8ρu’v=8ραuvNet force on the plate,
Fnet=F8ρAαuv=mdvdtAfter a long time the velocity will be sufficent so force (F) become
8ρAαuv. After that velocity of the plate become constant and achieve terminal velocity.
Hence, the correct answers are options A, B and D.
Question 8:
A drop of liquid of radius R = 10^{–2} m having surface tension
0.14πNm1divides itself into K identical drops. In this process the total change in the surface energy ΔU = 10^{–3} J. If K = 10^{α} then the value of α is
Solutions:From the conservation of mass we have
ρ×43πR3=K×ρ×43πr3⇒R=K13rThe total change in the surface energy is given as,
∆U=S∆A
⇒∆U=S×K×4πr24πR2⇒∆U=S×K×4πR2K2/34πR2⇒∆U=4πR2×S×K131Substituting given values, we get
⇒103=4π×104×0.14π×K131⇒100=K131⇒K13≈100=102Given, K=10α∴10α3=102⇒α3=2⇒α=6The correct value of
α is 6.
Question 9:
^{131}I is an isotope of Iodine that β decays to an isotope of Xenon with a halflife of 8 days. A small amount of a serum labelled with ^{131}I is injected into the blood of a person. The activity of the amount of ^{131}I injected was 2.4 × 10^{5} Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person’s body, and gives an activity of 115 Bq. The total volume of blood in the person’s body, in liters is approximately (you may use e^{x} ≈ 1 + x for x << 1 and ln 2 ≈ 0.7).
Solutions:The half life of â€‹^{131}I is an isotope of Iodine,
t1/2=8 daysInital activity of â€‹^{131}I is an isotope of Iodine,
A0=2.4×105 Bq=λN0Let the volume of the toal boold is the person’s body be V. At, t = 0
A0=λN0At, t = 11.5 hrs
A=λNUsing the relation
115=λNV×2.5⇒115=λN0eλtV×2.5 As,N=N0eλt⇒115=A0eln28×24×11.5V×2.5 A0=λN0⇒115=2.4×105×e124V×2.5 ⇒V=2.4×105115×2.51124=5 litersThe total volume of blood in the person’s body, in liters is approximately 5 liters.
Question 10:
An electron in a hydrogen atom undergoes a transition from an orbit with quantum number n_{i} to another with quantum number n_{f}. V_{i} and V_{f} are respectively the initial and final potential energies of the electron. If
ViVf=6.25, then the smallest possible n_{f} is.
Solutions:The potential energy of the electron in the nth orbit of hydrogen atom is â€‹given as
Vn=Z2n2V0Vn=1n2V0 (For hydrogen atom, Z=1)Accoding to the problem
ViVf=nf2ni2=6.25 …..(i)Take n_{i }= 1 , we get n_{f} = 5
The smallest possible value of n_{f} is 5.
Question 11:
A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle θ = 30°. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as n_{m} = n – mΔn, where n_{m }is the refractive index of the m^{th} slab and Δn = 0.1 (see the figure). The ray is refracted out parallel to the interface between the (m – 1)^{th} and m^{th} slabs from the right side of the stack. What is the value of m ?
Solutions:Using snell’s law, we have
nsini=n’sinr⇒1.6sinθ=nm∆nsinπ2⇒1.6sin30=nm∆n×1⇒1.6×12=nm∆n⇒nm∆n=0.8⇒1.6m×0.1=0.8⇒m=8The correct value of m is 8.
Question 12:
A stationary source emits sound of frequency f_{0} = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms^{–1}. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms^{–1} and the car reflects the sound at the frequency it has received).
Solutions:Frequency of sound as received by large car approaching the source.
f1=v+vovfo⇒f1=330+2330×492⇒f1=494.98 HzNow, this large car acts as a source of sound for reflected sound wave.
∴fr=494.98 HzThe frequency of sound received by the source is
f2=vvv0×fr⇒f2=3303302×494.98⇒f2=3303302×494.98⇒f2=497.99≈498The beat frequency of the resulting signal is
=f0f2=492498=6 HzThe correct answer is 6.
Question 13:
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0)
with a given initial velocity
v→. A uniform electric field E→and a uniform magnetic field B→exist everywhere. The velocity v→, electric field E→and magnetic field B→are given in column 1, 2 and 3, respectively. The quantities E_{0}, B_{0} are positive in magnitude. 

Column1  Column2  Column3 
(I) Electron with
v→=2E0B0x^ 
(i)
E→=E0z^ 
(P)
B→=B0x^ 
(II) Electron with
v→=E0B0y^ 
(ii)
E→=E0y^ 
(Q)
B→=B0x^ 
(III) Proton with
v→=0 
(iii)
E→=E0x^ 
(R)
B→=B0y^ 
(IV) Proton with
v→=2E0B0x^ 
(iv)
E→=E0x^ 
(S)
B→=B0z^ 
In which case will the particle move in a straight line with constant velocity ?
Option 1:  (II) (iii) (S) 
Option 2:  (IV) (i) (S) 
Option 3:  (III) (ii) (R) 
Option 4:  (III) (iii) (P) 
Solutions:The particle moves in a straight line with constant velocity when the net force on the particle is zero.
F→net=F→E+F→B=0⇒F→net=qE→+qv→×B→=0⇒qE→=qv→×B→ …..(i)From (i), we conclude that the velocity, electric field and magnetic field are mutually perpendicular to each other. And they are given by
v→=E0B0y^,
E→=E0x^ and
B→=B0z^.
Hence, the correct answer is options A.
Question 14:
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0)
with a given initial velocity
v→. A uniform electric field E→and a uniform magnetic field B→exist everywhere. The velocity v→, electric field E→and magnetic field B→are given in column 1, 2 and 3, respectively. The quantities E_{0}, B_{0} are positive in magnitude. 

Column1  Column2  Column3 
(I) Electron with
v→=2E0B0x^ 
(i)
E→=E0z^ 
(P)
B→=B0x^ 
(II) Electron with
v→=E0B0y^ 
(ii)
E→=E0y^ 
(Q)
B→=B0x^ 
(III) Proton with
v→=0 
(iii)
E→=E0x^ 
(R)
B→=B0y^ 
(IV) Proton with
v→=2E0B0x^ 
(iv)
E→=E0x^ 
(S)
B→=B0z^ 
In which case will the particle describe a helical path with axis along the positive zdirection?
Option 1:  (II) (ii) (R) 
Option 2:  (IV) (ii) (R) 
Option 3:  (IV) (i) (S) 
Option 4:  (III) (iii) (P) 
Solutions:The particle will describe a helical path with axis along the positive zdirection if the particle experiences a centripetal acceleration in xy plane. For the given set of options only option C satisfy the condition. Path is helical with increasing pitch.
Hence, the correct answer is options C.
Question 15:
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0)
with a given initial velocity
v→. A uniform electric field E→and a uniform magnetic field B→exist everywhere. The velocity v→, electric field E→and magnetic field B→are given in column 1, 2 and 3, respectively. The quantities E_{0}, B_{0} are positive in magnitude. 

Column1  Column2  Column3 
(I) Electron with
v→=2E0B0x^ 
(i)
E→=E0z^ 
(P)
B→=B0x^ 
(II) Electron with
v→=E0B0y^ 
(ii)
E→=E0y^ 
(Q)
B→=B0x^ 
(III) Proton with
v→=0 
(iii)
E→=E0x^ 
(R)
B→=B0y^ 
(IV) Proton with
v→=2E0B0x^ 
(iv)
E→=E0x^ 
(S)
B→=B0z^ 
In which case would the particle move in a straight line along the negative direction of yaxis (i.e., move along
y^) ?
Option 1:  (IV) (ii) (S) 
Option 2:  (III) (ii) (P) 
Option 3:  (II) (iii) (Q) 
Option 4:  (III) (ii) (R) 
Solutions:For a charged particle to move in a straight line along the negative direction of yaxis there are two conditions â€‹(i) its velocity is along negative ydirection (if its initial velocity is non zero) and net force acting on the particle is zero (ii) it must experiences a net force in the negative ydirectionâ€‹ (if its initial velocity is zero)
For the given set of options option D satisfy the second condition. Hence, the correct answer is options D.
Question 16:
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.  
Column1  Column2  Column3 
(I)
W1→2=1γ1P2V2P1V1 
(i) Isothermal  
(II)
W1→2=PV2+PV1 
(ii) Isochoric  
(III)
W1→2=0 
(iii) Isobaric  
(IV)
W1→2=nRT InV2V1 
(iv) Adiabatic 
Which of the following options is the only correct representation of a process in which ΔU = ΔQ – PΔV?
Option 1:  (II) (iv) (R) 
Option 2:  (II) (iii) (P) 
Option 3:  (II) (iii) (S) 
Option 4:  (III) (iii) (P) 
Solutions:
W1→2 =PV1 + PV2 ⇒ IsobaricHence, the correct answer is option B.
Question 17:
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.  
Column1  Column2  Column3 
(I)
W1→2=1γ1P2V2P1V1 
(i) Isothermal  
(II)
W1→2=PV2+PV1 
(ii) Isochoric  
(III)
W1→2=0 
(iii) Isobaric  
(IV)
W1→2=nRT InV2V1 
(iv) Adiabatic 
Which one of the following options is the correct combination?
Option 1:  (III) (ii) (S) 
Option 2:  (II) (iv) (R) 
Option 3:  (II) (iv) (P) 
Option 4:  (IV) (ii) (S) 
Solutions:
W1 → 2 = 0 ⇒ Isochoric processHence, the correct answer is option A.
Question 18:
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.  
Column1  Column2  Column3 
(I)
W1→2=1γ1P2V2P1V1 
(i) Isothermal  
(II)
W1→2=PV2+PV1 
(ii) Isochoric  
(III)
W1→2=0 
(iii) Isobaric  
(IV)
W1→2=nRT InV2V1 
(iv) Adiabatic 
Which one of the following options correctly represents a thermodynamics process that is used as a correction in the determination of the speed of sound in an ideal gas?
Option 1:  (III) (iv) (R) 
Option 2:  (I) (ii) (Q) 
Option 3:  (IV) (ii) (R) 
Option 4:  (I) (iv) (Q) 
Solutions:Adiabatic process is used as a correction in the determination of the speed of sound in an ideal gas. Also, for adaibatic process,
W1→2=1γ1P2V2P1V1Hence, the correct answer is option D.
Question 19:
The IUPAC name(s) of the following compound is(are)
Option 1:  4methylchlorobenzene 
Option 2:  4chlorotoluene 
Option 3:  1chloro4methylbenzene 
Option 4:  1methyl4chlorobenzene 
Solutions:The name ‘toluene’ (a monocyclic substituted aromatic hydrocarbons) is retained by IUPAC as PIN (Preferred IUPAC Name). Also, for IUPAC nomenclature of substituted benzene compounds, the substituent is placed as prefix to the word benzene.
On the basis of these facts, the IUPAC names of the given compound may be given as:
(1) 4Chlorotoluene (2) 1chloro4methylbenzene
Hence, the correct answers are both options B and C.
Question 20:
The correct statement(s) for the following addition reactions is(are)
(i)
(ii)
Option 1:  (M and O) and (N and P) are two pairs of diastereomers 
Option 2:  Bromination proceeds through transaddition in both the reactions 
Option 3:  O and P are identical molecules 
Option 4:  (M and O) and (N and P) are two pairs of enantiomers 
Solutions:There are different possibilities for products resulting from the same substrate with different stereoconformations. First we will consider the cisconfiguration as a starting material.
Two different products (O and P) are observed. However, they are formed in a 1 : 1 ratio, so 50% of the products are of the (R,R) configuration and the other 50% are of the (S,S) configuration. So, (O) and (P) are enantiomers.
Now we will consider the transconfiguration as a starting material. We can draw two possible products (M and N) for this reaction.
Initially it might appear as if two different products are formed. However they can be proved as identical. Let us focus on the stereochemistry of one of our above products. If we rotate the molecule of product N so that carbon 2 is now on top and carbon 1 is on the bottom. The new stereochemical configuration is shown below.
Furthermore, if we take that configuration and rotate it so the methyl groups are in the same position as our original products from the bromination reaction, product M is observed.
So, it can be concluded that
 (M) and (O) are diastereomers. Similarly, (N) and (P) are diastereomers of each other.
 Addition of Br_{2} on alkene follows anti or trans addition.
 (M) and (N) are identical while (O) and (P) are enantiomers.
Hence, the correct answers are options A and B.
Question 21:
Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl_{2} . 6H_{2}O (X) and NH_{4}Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. Among the following options, which statements is(are) correct ?
Option 1:  The hybridization of the central metal ion in Y is d^{2}sp^{3} 
Option 2:  Z is tetrahedral complex 
Option 3:  Addition of silver nitrate to Y gives only two equivalents of silver chloride 
Option 4:  When X and Z are in equilibrium at 0°C, the colour of the solution is pink 
Solutions:Using the spin only magnetic moment, the central metal ion is Co^{2+} (d^{7} configuration). The pinkcoloured complex is [Co(H_{2}O)_{6}] Cl _{2} . The reactions are given below:
CoH2O6Cl2X →O2(Air) Excess NH4OH/NH4Cl CoNH36Cl3YCoH2O62+X + 4 Cl → CoCl42ZIn the conversion of complex X to Y, the oxidation state of cobalt changes from +2 to +3 (d^{6} configuration). Since, NH_{3} is a stongfield ligand, therefore, it causes pairing of electrons. The hybridisation of cobalt in complex Y is d ^{2} sp ^{3} . Option (A) is correct.
In complex Z, since Cl^{−} ion is a weak field ion, hence pairing of electrons does not occur. The hybridisation of cobalt is sp^{3} giving rise to tetrahedral shape of the complex Z. Option (B) is also correct.
In complex Y, three chlorine atoms are present outside the coordination entity, hence addition of silver nitrate solution to complex Y will give three equivalents of silver chloride. Option (C) is incorrect.
The conversion of complex X to Z is an exothermic process, so at equilibrium, if ice is added to bring the temperature at 0 ^{o}C, then it shifts towards complex X and the colour of the solution is pink. Option (D) is also correct.
Hence, the correct answer is option (A), (B) and (D).
Question 22:
For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here x_{L} and x_{M} represent mole fractions of L and M, respectively, in the solution. the correct statement(s) applicable to this system is(are) –
â€‹
Option 1:  Attractive intramolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L–M when mixed in solution 
Option 2:  The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when x_{L} → 0 
Option 3:  The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when x_{L} → 1 
Option 4:  The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from x_{L} = 0 to x_{L} = 1 
Solutions: From the graph, it can be implied that the solution shows positive deviation from the ideal behaviour . The increasing order of the forces of attraction in the solution is as follows: F _{L−L} , F_{M−M} > F_{L−M} Option (A) is correct.
Point Z in the graph represents the vapour pressure of pure liquid L (denoted on the xaxis). Option (C) is correct.
Hence, the correct answers are options (A) and (C).
Question 23:
An ideal gas is expanded from (p_{1} , V_{1} , T_{1}) to (p_{2} , V_{2} , T_{2}) under different conditions. The correct statement(s) among the following is(are)
Option 1:  The work done on the gas is maximum when it is compressed irreversibly from (p2 , V_{2}) to (p_{1} , V_{1}) against constant pressure p1 
Option 2:  The work done on the gas is less when it is expanded reversibly from V_{1} to V_{2} under adiabatic conditions as compared to that when expanded reversibly from V_{1} to V_{2} under isothermal conditions. 
Option 3:  The change in internal energy of the gas (i) zero, if it is expanded reversibly with T_{1} = T_{2} , and (ii) positive, if it is expanded reversibly under adiabatic conditions with T_{1} ≠ T_{2} 
Option 4:  If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic 
Solutions:Isothermal expansion of the gas is represented in the graph below: The work done by the gas in the process of irreversible expansion is more than that in the reverible expansion. Option (A) is correct.
The isothermal expansion and adiabatic expansion is represented in the graph below: The area under the curve AB is more than that under AC ( a gas cools in the adiabatic expansion), so the work done in isothermal expansion is more than in the adiabatic expansion. Option (B) is correct.
(i) If a gas is expanded reversibly at a constant temperature, it is known as isothermal expansion. The change in internal energy of the gas is zero, We know that change in internal energy is given as:
ΔU = nCvΔTΔT = 0 (Isothermal process)ΔU = 0(ii) If a gas expands reversibly in an adiabatic process, then the change in internal energy of the gas is negative. We know that during adiabatic expansion, a gas cools down (T_{2} < T_{1}), the change in internal energy is given as:
ΔU = nCvΔTΔT = ()ΔU = nCvΔTOption (C) is incorrect.
When a gas undergoes free expansion, then q = 0 w = 0 This makes the change in internal energy equal to zero too.
ΔU = nCvΔT = 0ΔT = 0Hence, the process is isothermal. Option (D) is correct.
Hence, the correct answers are options (A), (B) and (D).
Question 24:
The correct statements(s) about the oxoacids, HClO_{4} and HClO, is (are) –
Option 1:  HClO_{4} is more acidic than HClO because of the resonance stabilization of its anion 
Option 2:  HClO_{4} is formed in the reaction between Cl_{2} and H_{2}O 
Option 3:  The central atom in both HClO_{4} and HClO is sp^{3} hybridized 
Option 4:  The conjugate base of HClO_{4} is weaker base than H_{2}O 
Solutions:HClO_{4} is more acidic than HClO because of the resonance stabilisation of its anion.
HClO4 ⇌H+ + ClO4HClO ⇌H+ + ClOClO4 is resonance stabilised.
HClO is formed in the reaction between Cl_{2} and H_{2}O as shown following:
Cl2 + H2O →HClO + HClThe central atom in both HClO_{4} and HClO is sp^{3} hybridised.
Acidic: HClO4 > H3O+Basic: ClO4 <H2O Hence, the correct answers are options A, C and D.
Question 25:
The colour of the X_{2} molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to –
Option 1:  the physical state of X_{2} at room temperature changes from gas to solid down the group 
Option 2:  decrease in HOMOLUMO gap down the group 
Option 3:  decrease in π*σ* down the group 
Option 4:  decrease in ionization energy down the group 
Solutions:
The colour of the halogen molecules intensifies on moving down the group due to decreasing energy gap between higher occupied molecular orbital (HOMO) and lower unoccupied molecular orbital (LUMO). This makes the transition of electron easy from HOMO to LUMO.
The molecular orbital electronic configuration of the halogen molecules is as follows:
Halogen Molecule  Molecular Orbital Electronic Configuration 
Flourine (F_{2})  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < σ2pz2 <π2px2 = π2py2 <π*2px2 = π*2py2 < σ*2pz 
It can be seen that the electron transitions from π^{*} to σ^{*}.
Hence, the correct answers are options (B) and (C).
Question 26:
Among H_{2},
He2+, Li_{2}, Be_{2}, B_{2} ,C_{2}, N_{2},
O2, and F_{2} , the number of diamagnetic species is – (Atomic number) : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8 , f = 9)
Solutions:The molecular orbital electronic configuration of the halogen molecules is as follows:
Species  Molecular Orbital Electronic Configuration  Magnetic Nature 
H_{2}  σ1s2  Diamagnetic 
He2+  σ1s2 < σ*1s1  Paramagnetic 
Li_{2}  σ1s2 < σ*1s2 < σ2s2  Diamagnetic 
Be_{2}  σ1s2 < σ*1s2 < σ2s2 < σ*2s2  Diamagnetic 
B_{2}  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < π2px1 = π2py1  Paramagnetic 
C_{2}  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < π2px2 = π2py2  Diamagnetic 
N_{2}  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < π2px2 = π2py2 < σ2pz2  Diamagnetic 
O2  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < σ2pz2 <π2px2 = π2py2 <π*2px2 = π*2py1  Paramagnetic 
F_{2}  σ1s2 < σ*1s2 < σ2s2 < σ*2s2 < σ2pz2 <π2px2 = π2py2 <π*2px2 = π*2py2  Diamagnetic 
From the molecular orbital electronic configuration of the given species, it can be seen that there are 6 diamagnetic species (including highly unstable Be_{2}). However, if the existence of Be_{2} species is neglected then there are 5 diamagnetic species.
Hence, the correct answer is 5 or 6.
Question 27:
Among the following, the number of aromatic compound (s) is
Solutions:Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems whether or not having benzene ring, possessing following characteristics. (i) Planarity (ii) Complete delocalisation of the π electrons in the ring (iii) Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2, . . .). This is often referred to as Hückel Rule.
On these facts the following five compounds are aromatic. , , , and
Question 28:
The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm^{2}. The conductance of this solution was found to be 5 × 10^{–7}S. The pH of the solution is 4. The value of limiting molar conductivity
Λm0of this weak monobasic acid in aqueous solution is Z × 10^{2}S cm^{–1}mol^{–1}. The value of Z is.
Solutions:
pH=4logH+ = 4H+ = 104 mol/LH+ = cα =cΛmΛm∞=cκ×1000cΛm∞=κ×1000Λm∞=G×la×1000Λm∞H+ = G×la×1000Λm∞Λm∞ =G×la×1000H+Z×102 = 5×107×1201×1000104Z = 1102×5×107×1201×1000104 =6Hence, the correct answer is 6.
Question 29:
The sum of the number of lone pairs of electrons on each central atom in the following species is.
[TeBr_{6}]^{2–}, [BrF_{2}]^{+} , SNF_{3} and [XeF_{3}]^{–}
[Atomic number : N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54]
Solutions:
Species  Hybridisation of Central Atom  Number of Bond Pairs on Central Atom  Number of Lone Pairs on Central Atom 
[TeBr_{6}]^{2−}  sp^{3}d^{2}  6  1 
[BrF_{2}]^{+}  sp^{3}  2  2 
SNF_{3}  sp^{3}  4  0 
[XeF_{3}]^{−}  sp^{3}d^{2}  3  3 
Total  15  6 
Hence, the correct answer is 6.
Question 30:
A crystalline solid of a pure substance has a facecentred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8g cm^{–3}, then the number of atoms present in 256g of the crystal is N × 10^{2}^{4}. The value of N is
Solutions:
Given:Structure of the crystal solid is facecentred cubic. Z = 4a = 400 pm = 4×108 cmd = 8 g cm3w = 256 gNumber of atoms in 256 g = N×1024To find:N=?Number of atoms = wM×NAd = Z×MNA×a3M = d×NA×a3ZN×1024=w×Zd×a3N=256×48×1024×4×1083=2Hence, the correct answer is 2.
Question 31:
The wave function
ψn,l,m1is a mathematical function whose value depends upon spherical polar coordinates (r, θ, Ï•) of the electron and characterized by the quantum numbers n, l and m_{1}. Here r is distance from nucleus, θ is colatitude and Ï• is azimuth. In the mathematical functions given in the Table, Z is atomic number a_{0} is Bohr radius.
Column1  Column2  Column3 
(I) 1s orbital  (i)
ψn,l,m1∝Za032eZrae 

(II) 2s orbital  (ii) One radial node  (Q) Probability density at nucleus
∝1a03 
(III) 2p_{z} orbital  (iii)
ψn,l,m1∝Za052reZr2ae cosθ 
(R) Probability density is maximum at nucleus 
(IV)
3dz2orbital 
(iv) xy – plane is a nodal plane  (S) Energy needed to excite electron from n = 2 state to n = 4 state is
27 32times the energy needed to excite electron from n = 2 state to n = 6 state 
For the given orbital in column 1, the only CORRECT combination for any hydrogen – like species is :
Option 1:  (IV) (iv) (R) 
Option 2:  (II) (ii) (P) 
Option 3:  (III) (iii) (P) 
Option 4:  (I) (ii) (S) 
Solutions:The number of radial nodes for an orbital = (n−l−1) Radial nodes in the following orbitals are: 1s = 1−0−1 = 0 2s = 2−0−1 = 1 2p = 2−1−1 = 0 3d = 3−2−1 = 0 For the 2s orbital, the plot of wave function against the distance from the nucleus is correctly represented as:
Hence, the correct answer is option (B).
Question 32:
The wave function
ψn,l,m1is a mathematical function whose value depends upon spherical polar coordinates (r, θ, Ï•) of the electron and characterized by the quantum numbers n, l and m_{1}. Here r is distance from nucleus, θ is colatitude and Ï• is azimuth. In the mathematical functions given in the Table, Z is atomic number a_{0} is Bohr radius.
Column1  Column2  Column3 
(I) 1s orbital  (i)
ψn,l,m1∝Za032eZrae 

(II) 2s orbital  (ii) One radial node  (Q) Probability density at nucleus
∝1a03 
(III) 2p_{z} orbital  (iii)
ψn,l,m1∝Za052reZr2ae cosθ 
(R) Probability density is maximum at nucleus 
(IV)
3dz2orbital 
(iv) xy – plane is a nodal plane  (S) Energy needed to excite electron from n = 2 state to n = 4 state is
27 32times the energy needed to excite electron from n = 2 state to n = 6 state 
For He^{+} ion, the only INCORRECT combination is
Option 1:  (II) (ii) (Q) 
Option 2:  (I) (i) (S) 
Option 3:  (I) (i) (R) 
Option 4:  (I) (iii) (R) 
Solutions:In the combination given in option (D) is incorrect, as the wave function for 1s orbital cannot have angular component. However, for all sorbitals the probability of finding an electron is highest at the nucleus.
Hence, the correct answer is option (D).
Question 33:
The wave function
ψn,l,m1is a mathematical function whose value depends upon spherical polar coordinates (r, θ, Ï•) of the electron and characterized by the quantum numbers n, l and m_{1}. Here r is distance from nucleus, θ is colatitude and Ï• is azimuth. In the mathematical functions given in the Table, Z is atomic number a_{0} is Bohr radius.
Column1  Column2  Column3 
(I) 1s orbital  (i)
ψn,l,m1∝Za032eZrae 

(II) 2s orbital  (ii) One radial node  (Q) Probability density at nucleus
∝1a03 
(III) 2p_{z} orbital  (iii)
ψn,l,m1∝Za052reZr2ae cosθ 
(R) Probability density is maximum at nucleus 
(IV)
3dz2orbital 
(iv) xy – plane is a nodal plane  (S) Energy needed to excite electron from n = 2 state to n = 4 state is
27 32times the energy needed to excite electron from n = 2 state to n = 6 state 
For hydrogen atom, the only CORRECT combination is
Option 1:  (I) (iv) (R) 
Option 2:  (I) (i) (P) 
Option 3:  (II) (i) (Q) 
Option 4:  (I) (i) (S) 
Solutions:
The wave function for 1s orbital has no angular dependance. The orbital has spherical symmetry.Ψ∝Za032eZra0Energy needed to excite an electron from energy level n1 to n2ΔE = 13.6Z2×1n221n12 eVFor excitation from n1=2 to n2 =4:ΔE1 =13.6Z2×316 eVFor excitation from n1=2 to n2 =6:ΔE2 =13.6Z2×836 eVΔE1ΔE2=13.6Z2×316 eV13.6Z2×836 eV=2732ΔE1 = 2737ΔE2Hence, the correct answer is option (D).
Question 34:
Columns 1, 2 and 3 contains starting materials, reaction conditions, and type of reactions, respectively.  
Column1  Column2  Column3 
(I) Toluene  (i) NaOH/Br_{2}  (P) Condensation 
(II) Acetophenone  (ii) Br_{2} / hv  (Q) Carboxylation 
(III) Banzaldehyde  (iii) (CH_{3}CO)_{2}O/CH_{3}COOK  (R) Substitution 
(IV) Phenol  (iv) NaOH/CO_{2}  (S) Haloform 
For the synthesis of benzoic acid, the only CORRECT combination is
Option 1:  (III) (iv) (R) 
Option 2:  (IV) (ii) (P) 
Option 3:  (I) (iv) (Q) 
Option 4:  (II) (i) (S) 
Solutions:Ketone or aldehyde having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform (bromoform). Thus, haloform reaction of acetophenone yields benzoic acid.
Thus, the correct match is [(II) (i) (S)]
Hence, the correct answer is option D.
Question 35:
Columns 1, 2 and 3 contains starting materials, reaction conditions, and type of reactions, respectively.  
Column1  Column2  Column3 
(I) Toluene  (i) NaOH/Br_{2}  (P) Condensation 
(II) Acetophenone  (ii) Br_{2} / hv  (Q) Carboxylation 
(III) Banzaldehyde  (iii) (CH_{3}CO)_{2}O/CH_{3}COOK  (R) Substitution 
(IV) Phenol  (iv) NaOH/CO_{2}  (S) Haloform 
The only CORRECT combination in which the reaction proceeds through radical mechanism is
Option 1:  (I) (ii) (R) 
Option 2:  (II) (iii) (R) 
Option 3:  (III) (ii) (P) 
Option 4:  (IV) (i) (Q) 
Solutions:Hydrogen atom of toluene can be replaced by halogens. Halogenation takes place either at higher temperature (573773 K) or in the presence of diffused sunlight or ultraviolet light. These reactions are known as substitution reactions.
So, the correct match is [(I)(ii)(R)].
Hence, the correct answer is option A.
Question 36:
Columns 1, 2 and 3 contains starting materials, reaction conditions, and type of reactions, respectively.  
Column1  Column2  Column3 
(I) Toluene  (i) NaOH/Br_{2}  (P) Condensation 
(II) Acetophenone  (ii) Br_{2} / hv  (Q) Carboxylation 
(III) Banzaldehyde  (iii) (CH_{3}CO)_{2}O/CH_{3}COOK  (R) Substitution 
(IV) Phenol  (iv) NaOH/CO_{2}  (S) Haloform 
The only CORRECT combination that gives two different carboxylic acids is
Option 1:  (IV) (iii) (Q) 
Option 2:  (III) (iii) (P) 
Option 3:  (II) (iv) (R) 
Option 4:  (I) (i) (S) 
Question 37:
Which of the following is(are) NOT the square of a 3 × 3 matrix with real entries?
Option 1:  100010001 
Option 2:  100010001 
Option 3:  100010001 
Option 4:  100010001 
Solutions:For any two matrices, if A = B^{2} , then
⇒A=B2=+veConsider the matrix
100010001.
100010001=11=1 negativeMatrix B can not be possible.
Consider the matrix
100010001.
100010001=1 negativeMatrix B can not be possible.
Consider the matrix
100010001.
100010001=1 positiveMatrix B can be possible.
Consider the matrix
100010001.
100010001=110=1 positiveMatrix B can be possible.
Hence, the correct answers are options A and B.
Question 38:
If a chord, which is not a tangent, of the parabola y^{2} = 16x has the equation 2x + y = p, and midpoint (h, k), then which of the following is(are) possible value(s) of p, h and k?
Option 1:  p = 5, h = 4, k = −3 
Option 2:  p = −1, h = 1, k = −3 
Option 3:  p = −2, h = 2, k = −4 
Option 4:  p = 2, h = 3, k = −4 
Solutions:The equation of a chord with mid point (h, k) is
ky16x+h2=k216h⇒8xky+k28h=0Comparing with 2x+yp=0, we getk=4; 2hp=4Only option D satisfies the given relation. Hence the correct answer is option D.
Question 39:
Let a, b, x and y be real numbers such that a – b = 1 and y ≠ 0. If the complex number z = x + iy satisfies
Im az+bz+1=y, then which of the following is(are) possible value(s) of x?
Option 1:  11y2 
Option 2:  1+1+y2 
Option 3:  11+y2 
Option 4:  1+1y2 
Solutions:It is given that complex number
z=x+iysatisfies
Imaz+bz+1=y.
∴ Imax+iy+bx+iy+1=y⇒Imax+iay+bx+1iyx+12+y2=y⇒yax+b+ayx+1=yx+12+y2⇒aby=yx+12+y2∵y≠0 and ab=1⇒x+12+y2=1⇒x=1±1y2Hence, the correct answers are options A and D.
Question 40:
Let X and Y be two events such that
PX=13, PXY=12 and PYX=25. Then
Option 1:  PX’Y=12 
Option 2:  PX∩Y=15 
Option 3:  PX∪Y=25 
Option 4:  PY=415 
Solutions:It is given that,
Px=13, PY∩XPY=12,PY∩XPX=25. By the above information, we get
PX∩Y=215, PY=415∴PX∪Y=13+415215=715PX¯/Y=PX¯∩YPY=PYPX∩YPY⇒PX¯/Y=1215415=12Hence the correct answer is options A and D.
Question 41:
Let [x] be the greatest integer less than or equal to x. Then, at which of the following point(s) the function ƒ(x) = x cos (π(x + [x])) is discontinuous?
Option 1:  x = –1 
Option 2:  x = 0 
Option 3:  x = 2 
Option 4:  x = 1 
Solutions:The given function is
fx=xcosπx+x
⇒fx=1xxcosπxCheck the continuity by putting x = n.
fn=ncos2nπ=nfn+=ncos2nπ=nfn=ncos2n1π=nThus, it will be discontinuous at all integers except 0. Hence, the correct answers are options A, C and D.
Question 42:
If 2x – y + 1 = 0 is tangent to the hyperbola
x2a2y216=1, then which of the following CANNOT be sides of a right angled triangle?
Option 1:  2a, 4, 1 
Option 2:  2a, 8, 1 
Option 3:  a, 4, 1 
Option 4:  a, 4, 2 
Solutions:Let the equation of the tangent be
y=mx+c. The line
y=mx+c will be a tangent to the hyperbola
x2a2y2b2=1 if
c2=a2m2b2. Thus, for the given hyperbola,
x2a2y216=1
12=4a216⇒4a2=12+16⇒4a2=17⇒a2=174
a=172Now for option (A): 2a, 4, 1, the sides will be
17,4,1. This represents the sides of a right angled triangle.
For option (B): 2a, 8, 1, the sides will be
17,8,1. In this case the triangle is not possible.
For option (C): a, 4, 1, the sides are
172,4,1. The triangle is not possible in this case.
For option (D): a, 4, 2, the sides are
172,4,2. In this case triangle exists but it will not be a right angled triangle as sum of the squares of two sides is not equal to the square of the third side.
Hence, the correct answers are options B, C and D.
Question 43:
Let
f:ℝ→0,1be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval (0, 1)?
Option 1:  ex∫0xft sin tdt 
Option 2:  x^{9} – f(x) 
Option 3:  fx+∫0π2ft sin tdt 
Option 4:  x∫0π2xft cos tdt 
Solutions:In option (A), let h(x) =
ex∫0xftsintdtNow,
ex∈1,e for x∈0,1 and
0<∫0xftsintdt<1 in 0,1So,
h’x=exfxsinx>0∀x∈0,1Thus, h(x) is strictly increasing function. Also, h(0) = 1.
⇒hx>1∀x∈0,1Therefore, option (A) is not possible.
In option (B), let
gx=x9fx
So, g0=f0<0 ∵f∈0,1and g1=1f1>0 ∵f∈0,1 Thus, g0· g1<0So, option (B) is correct.
In option (C),
let kx=fx+∫0π2ftsintdtAs f∈0,1⇒kx>0∀x∈0,1So, option (C) is not possible.
In option (D), let
px=x∫0π2xftcostdtNow, for x = 0 we have
p0=0∫0π2ftcostdt<0Also, for x = 1,
p1=1∫0π21ftcostdt>0So,
p0·p1<0Hence, option (D) is correct.
Therefore, the correct answers are options (B) and option (D).
Question 44:
The sides of the right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side?
Solutions:Let the sides of the right angled triangle be
a,ad and a+d. a > 0 and d > 0. Length of the smallest side will be
adSince it is a right angled triangle, so applying Pythagoras theorem
a+d2=a2+ad2⇒2ad=a22ad⇒a24ad=0⇒aa4d=0Thus we have,
a=0 or a=4dBut
a≠0, so a = 4d.
Now area of the given right triangle will be A=12×a×ad=24⇒48=aad Thus, a = 8 and d = 2. The length of the smallest side will be
ad=82=6 units.
Question 45:
For how many values of p, the circle x^{2} + y^{2} + 2x + 4y – p = 0 and the coordinate axes have exactly three common points?
Solutions:Here we have three cases. Case I. When circle passes through origin. So, p = 0 Now, the equation of circle becomes
x2+y2+2x+4y=0. Case II. When circle touches yaxis and cuts xaxis. In that case,
g2c>0 and f2c=0⇒1–p>0 and 4–p=0⇒p>1 and p=4This is not possible. Case III. When circle touches xaxis and cuts yaxis.
g2c=0 and f2c>0⇒1–p=0 and 4–p>0⇒p=1 and p>4⇒p=1So we got the two possible values of p i.e., 0 and −1. Thus, there are two values of p for which the given circle and the coordinate axes have exactly three common points.
Question 46:
For a real number α, if the system
1αα2α1αα2 α1 xyz=111of linear equations, has infinitely many solutions, then 1 + α + α^{2} =
Solutions:If the system of linear equations has infinitely many solutions, then
1αα2α1αα2α1=0⇒11α2ααα3+α2α2α2=0⇒1α2α2+α4=0⇒α212=0⇒α21=0⇒α=±1For α = 1, the system of linear equations has no solution. For α = −1,
1+α+α2=1Hence, the value of
1+α+α2is 1.
Question 47:
Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then,
y9x=
Solutions:If x is the number of such words where no letter is repeated, then
x=10!And, if y is the number of such words where exactly one letter is repeated twice and no other letter is repeated, then
y=C110.C8910!2! =10!1!×9!.9!1!×8!.10!2! =10!×5×9
∴y9x=10!×5×99×10!=5Hence, the value of
y9xis 5.
Question 48:
Let f : R → R be a differentiable function such that f(0) = 0, f
π2= 3 and f'(0) = 1. If
gx=∫xπ2f’t cosec tcot t cosec t ft dtfor
x∈0,π2, then limx→0 gx=
Solutions:Here,
gx=∫xπ2f’tcosectcott.cosect ftdt =∫xπ2ft.cosect’dt =fπ2cosecπ2fxsinx =3fxsinx∴limx→0gx=3limx→0fxsinx =31 f’0=1 =2Hence, the value of
limx→0gxis 2.
Question 49:
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P)
am2,2a2 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii)
y=mx+am2+1 
(Q)
mam2+1,am2+1 
(III) y^{2} = 4ax  (iii)
y=mx+a2m21 
(R)
a2ma2m2+1,1a2m2+1 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv)
y=mx+a2m2+1 
(S)
a2ma2m21,1a2m21 
The tangent to a suitable conic (Column 1) at
3,12is found to be
3x+2y=4, then which of the following options is the only CORRECT combination?
Option 1:  (II) (iii) (R) 
Option 2:  (IV) (iv) (S) 
Option 3:  (IV) (iii) (S) 
Option 4:  (II) (iv) (R) 
Solutions:Let us consider the option A. Consider the conic
x2+a2y2=a2. If
3,12lies on it, then
a2=32+a2122⇒a2=3+a24⇒3a24=3⇒a2=4So, the equation of the conic for a^{2} = 4 is
x2+4y2=4.
x2+4y2=4⇒x24+y21=1Here, a^{2} = 4 and b^{2} = 1. We know that the equation of the tangent to the ellipse
x2a2+y2b2=1is
y=mx±a2m2+b2and the points of contact are
±a2ma2m2+b2,∓b2a2m2+b2. For b^{2} = 1, the equation of tangent to ellipse
x2+4y2=4in terms of m is
y=mx±a2m2+1and the points of contact are
±a2ma2m2+1,∓1a2m2+1. So, option A is not correct.
Now, one of the equation of tangent to the ellipse
x2+4y2=4in terms of m is
y=mx+a2m2+1and the points of contact are
a2ma2m2+1,1a2m2+1. Also, the equation of the tangent to
x24+y21=1at
3,12is
3×4+12y=1⇒3x+2y=4This is same as the equation of tangent given in the question.
Hence, the correct answer is option D.
Question 50:
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P)
am2,2a2 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii)
y=mx+am2+1 
(Q)
mam2+1,am2+1 
(III) y^{2} = 4ax  (iii)
y=mx+a2m21 
(R)
a2ma2m2+1,1a2m2+1 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv)
y=mx+a2m2+1 
(S)
a2ma2m21,1a2m21 
If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8,16), then which of the following options is the only CORRECT combination ?
Option 1:  (III) (i) (P) 
Option 2:  (III) (ii) (Q) 
Option 3:  (II) (iv) (R) 
Option 4:  (I) (ii) (Q) 
Solutions:Let us consider the option A. Consider the conic
y2=4ax. If (8, 16) are the points of contact of the tangent to it, then
256=32a⇒a=8So, the equation of conic for a = 8 is
y2=32x. We know that the equation of the tangent to the parabola
y2=4axis
y=mx+am(or my = m^{2}x + a) and the points of contact are
am2,2am. For a = 8, the point of contact are
8m2,16mComparing
8m2,16mwith (8, 16), we get m = 1 So, the equation of tangent to the parabola
y2=32xat (8, 16) is
y=x+81Or y=x+8This is same as the equation of tangent given in the question.
Hence, the correct answer is option A.
Question 51:
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P)
am2,2a2 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii)
y=mx+am2+1 
(Q)
mam2+1,am2+1 
(III) y^{2} = 4ax  (iii)
y=mx+a2m21 
(R)
a2ma2m2+1,1a2m2+1 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv)
y=mx+a2m2+1 
(S)
a2ma2m21,1a2m21 
For
a=2, if a tangent is drawn to a suitable conic (Column 1) at the point of contact (–1,1), then which of the following options is the only CORRECT combination for obtaining its equation ?
Option 1:  (II) (ii) (Q) 
Option 2:  (III) (i) (P) 
Option 3:  (I) (i) (P) 
Option 4:  (I) (ii) (Q) 
Solutions:Let us consider the option A. Consider the conic
x2+a2y2=a2. For
a=2and the point of contact of the tangent at
1,1, we get
12+22×12=22Or 3 = 2, which is not true. So, option A is not correct.
Let us now consider option B. Consider the conic
y2=4ax. For
a=2and the point of contact of the tangent at
1,1, we get
12=421⇒1=42, which is not true. So, option B is not correct.
Let us consider option C. Consider the conic
x2+y2=a2. For
a=2and the point of contact of the tangent at
1,1, we get
12+12=22Or 2 = 2, which is true. We know that the equation of the tangent to the circle
x2+y2=a2in terms of m is
y=mx±am2+1and the points of contact are
±amm2+1,∓am2+1. Now, one of the equation of tangent to the circle
x2+y2=a2in terms of m is
y=mx+am2+1and the points of contact are
mam2+1,am2+1. So, option C is incorrect. Thus, option D is correct.
Hence, the correct answer is option D.
Question 52:
* Column 1 contains information about zeros of ƒ(x), ƒ'(x) and ƒ”(x). * Column 2 contains information about the limiting behavior of ƒ(x), ƒ'(x) and ƒ”(x) at infinity. * Column 3 contains information about increasing/decreasing nature of ƒ(x) and ƒ'(x).  
Column1  Column2  Column3 
(I) ƒ(x) = 0 for some x ∈ (1,e^{2})  (i) lim_{x→∞} ƒ(x) = 0  (P) ƒ is increasing in (0,1) 
(II) ƒ'(x) = 0 for some x ∈ (1,e)  (ii) lim_{x→∞} ƒ(x) = –∞  (Q) ƒ is decreasing in (e,e^{2}) 
(III) ƒ'(x) = 0 for some x ∈ (0,1)  (iii) lim_{x→∞} ƒ'(x) = –∞  (R) ƒ’ is increasing in (0,1) 
(IV) ƒ”(x) = 0 for some x ∈ (1,e)  (iv) lim_{x→∞} ƒ”(x) = 0  (S) ƒ’ is decreasing in (e,e^{2}) 
Which of the following options is the only CORRECT combination ?
Option 1:  (IV) (i) (S) 
Option 2:  (I) (ii) (R) 
Option 3:  (III) (iv) (P) 
Option 4:  (II) (iii) (S) 
Solutions:The given function is
fx=x+logxxlogx, x∈0,∞.
fx=x+logxxlogx, ∀x>0⇒f’x=1+1xx×1x+logx×1=1xlogx⇒f”x=1×21xFor x ∈ (1, e^{2}),
f1fe2=1+00e2+22e2=2e2<0So, f(x) = 0 for some x ∈ (1, e^{2}). For x ∈ (1, e),
f’1f’e=101e1<0So, f ‘(x) = 0 for some x ∈ (1, e). For x ∈ (0, 1),
f’0f’1=∞>0So, f ‘(x) ≠ 0 for some x ∈ (0, 1). For x ∈ (1, e),
f”1f”e=111e21e>0So, f ”(x) ≠ 0 for some x ∈ (1, e). Now,
limx→∞fx=limx→∞x+logxxlogx =limx→∞x1+logxxlogx =∞ limx→∞logxx→0, limx→∞logx→∞
limx→∞f’x=limx→∞1xlogx=∞
limx→∞f”x=limx→∞1×21x=0Now,
f”x=1×21x<0 ∀x∈0,∞∴ f ‘(x) is strictly decreasing function for x ∈ (0, ∞). Also, f ‘(x) > 0 for ∈ (0, 1), so f is increasing in (0, 1). f ‘(x) < 0 for x ∈ (e, e^{2}), so f is decreasing in (e, e^{2}).
Hence, the correct answer is option D.
Question 53:
* Column 1 contains information about zeros of ƒ(x), ƒ'(x) and ƒ”(x). * Column 2 contains information about the limiting behavior of ƒ(x), ƒ'(x) and ƒ”(x) at infinity. * Column 3 contains information about increasing/decreasing nature of ƒ(x) and ƒ'(x).  
Column1  Column2  Column3 
(I) ƒ(x) = 0 for some x ∈ (1,e^{2})  (i) lim_{x→∞} ƒ(x) = 0  (P) ƒ is increasing in (0,1) 
(II) ƒ'(x) = 0 for some x ∈ (1,e)  (ii) lim_{x→∞} ƒ(x) = –∞  (Q) ƒ is decreasing in (e,e^{2}) 
(III) ƒ'(x) = 0 for some x ∈ (0,1)  (iii) lim_{x→∞} ƒ'(x) = –∞  (R) ƒ’ is increasing in (0,1) 
(IV) ƒ”(x) = 0 for some x ∈ (1,e)  (iv) lim_{x→∞} ƒ”(x) = 0  (S) ƒ’ is decreasing in (e,e^{2}) 
Which of the following options is the only CORRECT combination ?
Option 1:  (III) (iii) (R) 
Option 2:  (I) (i) (P) 
Option 3:  (IV) (iv) (S) 
Option 4:  (II) (ii) (Q) 
Solutions:The given function is
fx=x+logxxlogx, x∈0,∞.
fx=x+logxxlogx, ∀x>0⇒f’x=1+1xx×1x+logx×1=1xlogx⇒f”x=1×21xFor x ∈ (1, e^{2}),
f1fe2=1+00e2+22e2=2e2<0So, f(x) = 0 for some x ∈ (1, e^{2}). For x ∈ (1, e),
f’1f’e=101e1<0So, f ‘(x) = 0 for some x ∈ (1, e). For x ∈ (0, 1),
f’0f’1=∞>0So, f ‘(x) ≠ 0 for some x ∈ (0, 1). For x ∈ (1, e),
f”1f”e=111e21e>0So, f ”(x) ≠ 0 for some x ∈ (1, e). Now,
limx→∞fx=limx→∞x+logxxlogx =limx→∞x1+logxxlogx =∞ limx→∞logxx→0, limx→∞logx→∞
limx→∞f’x=limx→∞1xlogx=∞
limx→∞f”x=limx→∞1×21x=0Now,
f”x=1×21x<0 ∀x∈0,∞∴ f ‘(x) is strictly decreasing function for x ∈ (0, ∞). Also, f ‘(x) > 0 for ∈ (0, 1), so f is increasing in (0, 1). f ‘(x) < 0 for x ∈ (e, e^{2}), so f is decreasing in (e, e^{2}).
Hence, the correct answer is option D.
Question 54:
* Column 1 contains information about zeros of ƒ(x), ƒ'(x) and ƒ”(x). * Column 2 contains information about the limiting behavior of ƒ(x), ƒ'(x) and ƒ”(x) at infinity. * Column 3 contains information about increasing/decreasing nature of ƒ(x) and ƒ'(x).  
Column1  Column2  Column3 
(I) ƒ(x) = 0 for some x ∈ (1,e^{2})  (i) lim_{x→∞} ƒ(x) = 0  (P) ƒ is increasing in (0,1) 
(II) ƒ'(x) = 0 for some x ∈ (1,e)  (ii) lim_{x→∞} ƒ(x) = –∞  (Q) ƒ is decreasing in (e,e^{2}) 
(III) ƒ'(x) = 0 for some x ∈ (0,1)  (iii) lim_{x→∞} ƒ'(x) = –∞  (R) ƒ’ is increasing in (0,1) 
(IV) ƒ”(x) = 0 for some x ∈ (1,e)  (iv) lim_{x→∞} ƒ”(x) = 0  (S) ƒ’ is decreasing in (e,e^{2}) 
Which of the following options is the only INCORRECT combination ?
Option 1:  (II) (iii) (P) 
Option 2:  (II) (iv) (Q) 
Option 3:  (I) (iii) (P) 
Option 4:  (III) (i) (R) 
Solutions:The given function is
fx=x+logxxlogx, x∈0,∞.
fx=x+logxxlogx, ∀x>0⇒f’x=1+1xx×1x+logx×1=1xlogx⇒f”x=1×21xFor x ∈ (1, e^{2}),
f1fe2=1+00e2+22e2=2e2<0So, f(x) = 0 for some x ∈ (1, e^{2}). For x ∈ (1, e),
f’1f’e=101e1<0So, f ‘(x) = 0 for some x ∈ (1, e). For x ∈ (0, 1),
f’0f’1=∞>0So, f ‘(x) ≠ 0 for some x ∈ (0, 1). For x ∈ (1, e),
f”1f”e=111e21e>0So, f ”(x) ≠ 0 for some x ∈ (1, e). Now,
limx→∞fx=limx→∞x+logxxlogx =limx→∞x1+logxxlogx =∞ limx→∞logxx→0, limx→∞logx→∞
limx→∞f’x=limx→∞1xlogx=∞
limx→∞f”x=limx→∞1×21x=0Now,
f”x=1×21x<0 ∀x∈0,∞∴ f ‘(x) is strictly decreasing function for x ∈ (0, ∞). Also, f ‘(x) > 0 for ∈ (0, 1), so f is increasing in (0, 1). f ‘(x) < 0 for x ∈ (e, e^{2}), so f is decreasing in (e, e^{2}).
Hence, the correct answer is option D.