Test Name: JEE Advanced 2013 Paper2(Code 0)
Question 1:
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation.
Option 1: 
the rate at which heat is absorbed in the range 0100 K varies linearly with temperature T. 
Option 2: 
heat absorbed in increasing the temperature from 0100 K is less than the heat required for increasing the temperature from 400500 K. 
Option 3: 
there is no change in the rate of heat absorption in the range 400500 K. 
Option 4: 
the rate of heat absorption increases in the range 200300 K. 
dQ = mCdT
Rate of absorption of heat,
In 0100 K range the specific heat increases So R is increases but not linearly.
The value of C in the range 400500 K is more than in 0100 K thus the heat absorbed in increasing temperature from 400 to 500 K is more than in 0 to 100 K.
In the range 200300 K as C increases thus R increases.
Question 2:
The radius of the orbit of an electron in Hydrogenlike atom is 4.5 a_{0}, where a_{0} is the Bohr radius. Its orbital angular momentum is It is given that h is Planck constant and R is Rydberg constant. The possible wavelengths(s), when the atom deexcites, is (are)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Angular momentum,
Radius of the orbit,
Given:
Thus we get, n =3 and z=2.
Possible transitions are 3→2, 3→1 and 2→1
Now,
Question 3:
Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°,
Option 1: 
the absolute error in d remains constant. 
Option 2: 
the absolute error in d increases. 
Option 3: 
the fractional error in d remains constant. 
Option 4: 
the fractional error in d decreases. 
This shows that with the increase in θ the absolute error and fractional error in d decreases.
Question 4:
Two nonconducting spheres of radii R_{1} and R_{2} and carrying uniform volume charge densities +ρ and −ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
Option 1: 
the electrostatic field is zero 
Option 2: 
the electrostatic potential is constant. 
Option 3: 
the electrostatic field is constant in magnitude. 
Option 4: 
the electrostatic field has same direction. 
Consider a point P inside the overlapped portion. Now find electric field at this point.
Electric field at P,
Thus the magnitude of electric field is constant inside the shaded portion and has same direction.
Question 5:
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are)
Option 1: 
In the region 0 < r < R, the magnetic field is nonzero. 
Option 2: 
In the region R < r < 2R, the magnetic field is along the common axis. 
Option 3: 
In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis. 
Option 4: 
In the region r > 2R, the magnetic field is nonzero. 
In the region 0
In the region R
In the region r>2R, the magnetic field is due to cylinder which is nonzero 0.
Question 6:
Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f_{1}. An observer in the other vehicle hears the frequency of the whistle to be f_{2}. The speed of sound in still air is V. The correct statement(s) is (are)
Option 1: 
If the wind blows from the observer to the source, f_{2} > f_{1}. 
Option 2: 
If the wind blows from the source to the observer, f_{2} > f_{1}. 
Option 3: 
If the wind blows from the observer to the source, f_{2} < f_{1}. 
Option 4: 
If the wind blows from the source to the observer, f_{2} < f_{1}. 
If wind blows from the observer to the source,
If wind blows from the source to the observer,
Question 7:
Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are)
Option 1: 
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 
Option 2: 
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 
Option 3: 
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 
Option 4: 
The energy of the mass m remains constant. 
Energy of mass m remains conserved. Now applying energy conservation we get,
Question 8:
A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u_{0}. When the speed of the particle is 0.5 u_{0}, it collides elastically with a rigid wall. After this collision,
Option 1: 
the speed of the particle when it returns to its equilibrium position is u_{0}. 
Option 2: 
the time at which the particle passes through the equilibrium position for the first time is 
Option 3: 
the time at which the maximum compression of the spring occurs is 
Option 4: 
the time at which the particle passes through the equilibrium position for the second time is 
From conservation of energy when the particle returns to its equilibrium position its speed becomes u_{0}.
For SHM the displacement, x = Asinωt
Initial velocity, u_{0} = Aω
Now before collision, speed of the particle, v = 0.5 u_{0}
Or,
So the time at which the particle passes through its equilibrium for the first time,
The time at which maximum compression takes place =
The time at which the particle passes through the equilibrium point for the second time is
Question 9:
The speed of the block when it reaches the point Q is
Option 1: 
5 ms^{−1}

Option 2: 
10 ms^{−1}

Option 3: 

Option 4: 
20 ms^{−1}

Applying energy conservation theorem,
Question 10:
The magnitude of the normal reaction that acts on the block at the point Q is
Option 1: 
7.5 N

Option 2: 
8.6 N

Option 3: 
11.5 N

Option 4: 
22.5 N

As the body slides along the circular track thus,
Question 11:
If the direct transmission method with a cable of resistance 0.4 Ω km^{−1} is used, the power dissipation (in %) during transmission is
Option 1: 
20

Option 2: 
30

Option 3: 
40

Option 4: 
50

Total resistance offered by cable in transmission over 20 km = 0.4×20 = 8 Ω
Now, P = VI
Or, I = (600×10^{3})/(4000) = 150 A
Power dissipated during transmission, P_{d }= I^{2}R = 150^{2}(8) = 180 kW
In %,
Question 12:
In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the stepup transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the stepdown transformer is
Option 1: 
200 : 1

Option 2: 
150 : 1

Option 3: 
100 : 1

Option 4: 
50 : 1

In the stepup transformer,
In stepdown transformer,
Question 13:
The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is
Option 1: 

Option 2: 

Option 3: 
BR

Option 4: 
2 BR

Question 14:
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change is
Option 1: 
−γ BQR^{2}

Option 2: 

Option 3: 

Option 4: 
γ BQR^{2}

Given magnetic dipole moment, M=γL
And , L=Iω=
Now, change in magnetic dipole, M=
Negative sign shows that the change is in opposite direction of magnetic field.
Question 15:
The correct statement is
Option 1: 
The nucleus can emit an alpha particle.

Option 2: 
The nucleus can emit a proton.

Option 3: 
Deuteron and alpha particle can undergo complete fusion.

Option 4: 
The nucleus and can undergo complete fusion.

(A) m()=6.015123 u
And, m() + m()=(2.014102 + 4.002603)=6.016705 u
Now, m() + m()>m()
Thus cannot undergo in fission.
(B) m()=209.982876 u
And, m() + m()=(1.007825 + 208.980388) u = 209.988213 u
Now, m() + m()>m()
Thus cannot undergo in fission.
(C) m() + m()=(2.014102 + 4.002603)=6.016705 u
m()=6.015123 u
Now, m() + m()>m()
Thus, a deuteron and an alpha particle can undergo complete fusion.
(D) m() + m()=(69.925325 + 81.916709)=151.842034 u
m()=151.919803 u
Now, m() + m()
The nuclei and cannot undergo complete fusion.
Question 16:
The kinetic energy (in keV) of the alpha particle, when the nucleus at rest undergoes alpha decay, is
Option 1: 
5319

Option 2: 
5422

Option 3: 
5707

Option 4: 
5818

Kinetic energy of αparticle must be less than ΔE thus the correct answer is 5319 keV
Question 17:
One mole of a monatomic ideal gas is taken along two cyclic processes E→F→G→E and E→F→H→E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.
Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

List I 

List II 
P. 
G→E 
1. 
160 P_{0}V_{0} ln2 
Q. 
G→H 
2. 
36 P_{0}V_{0} 
R. 
F→H 
3. 
24 P_{0}V_{0} 
S. 
F→G 
4. 
31 P_{0}V_{0} 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

FG is isothermal thus,
P_{0}V_{G}=32P_{0}V_{0}
Or, V_{G}=32V_{0}
FH is adiabatic thus,
P_{0}(V_{H})^{5/3}=32P_{0}(V_{0})^{5/3} (For monatomic gas, γ=5/3)
V_{H}=8V_{0}
W_{GE}=P_{0}(V_{0}32V_{0})= 31P_{0}V_{0}
W_{GH}=P_{0}(8V_{0}32V_{0})= 24P_{0}V_{0}
Question 18:
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists:

List I 

List II 
P. 
Alpha decay 
1. 

Q. 
b^{+} decay 
2. 

R. 
Fission 
3. 

S. 
Proton emission 
4. 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

Alpha decay,
β^{+} decay,
In fission parent nucleus splits into two almost equal daughter nuclei.
Proton emission,
Question 19:
A right angled prism of refractive index μ_{1} is placed in a rectangular block of refractive index μ_{2}, which is surrounded by a medium of refractive index μ_{3}, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between μ_{1}, μ_{2} and μ_{3}, it takes one of the four possible path ‘ef’, ‘eg’, ‘eh’ or ‘ei’.
Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists:

List I 

List II 
P. 
e → f 
1. 

Q. 
e → g 
2. 
m_{2} > m_{1} and m_{2} > m_{3} 
R. 
e → h 
3. 
m_{1} = m_{2} 
S. 
e → i 
4. 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

e→f, μ_{2}>μ_{1 }and μ_{3}<μ_{2}
e→g, μ_{1}=μ_{2}
e→h, μ_{2}<μ_{1} and μ_{3}<μ_{2}, as there is no total internal reflection thus
e→i,
Question 20:
Match List I with List II and select the correct answer using the codes given below the lists:

List I 

List II 
P. 
Boltzmann constant 
1. 
[ML^{2}T^{–1}] 
Q. 
Coefficient of viscosity 
2. 
[ML^{–1}T^{–1}] 
R. 
Planck constant 
3. 
[MLT^{–3} K^{–1}] 
S. 
Thermal conductivity 
4. 
[ML^{2}T^{–2} K^{–1}] 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

Question 21:
The carbonbased reduction method is NOT used for the extraction of
Option 1: 
tin from SnO_{2} 
Option 2: 
iron from Fe_{2}O_{3} 
Option 3: 
aluminium from Al_{2}O_{3} 
Option 4: 
magnesium from MgCO_{3} · CaCO_{3} 
Extraction of Aluminium, magnesium and calcium from their ores Al_{2}O_{3}, MgCO_{3} and CaCO_{3} respectively is done by electrolytic method, whereas Tin and Iron is extracted from their oxide ores SnO_{2} and Fe_{2}O_{3} respectively by reducing them with Carbon. This is because the stable Al_{2}O_{3}, MgCO_{3} and CaCO_{3} can be reduced by carbon at sufficiently high temperatures; however the reduction is not carried out due to possibility of formation of metal carbides at high temperatures.
Question 22:
The thermal dissociation equilibrium of CaCO_{3}(s) is studied under different conditions.
For this equilibrium, the correct statement(s) is (are)
Option 1: 
ΔH is dependent on T 
Option 2: 
K is independent of the initial amount of CaCO_{3} 
Option 3: 
K is dependent of the pressure of CO_{2}_{ }at a given T 
Option 4: 
ΔH is independent of the catalyst, if any 
For decomposition of CaCO_{3} as shown below:
At equilibrium:
K_{p} =
Change in enthalpy is given by the following expression:
ΔH = C_{p}ΔT
As C_{p} is a temperature dependent quantity therefore, ΔH is also temperature dependent but is independent of the catalyst. The presence of a catalyst alters only the activation energy.
Equilibrium constant of the above reaction is dependent only on temperature and is independent of the concentration of CaCO_{3}.
Question 23:
The correct statement(s) about O_{3} is (are)
Option 1: 
O–O bond lengths are equal. 
Option 2: 
Thermal decomposition of O_{3} is endothermic. 
Option 3: 
O_{3} is diamagnetic in nature. 
Option 4: 
O_{3} has a bent structure. 
The structure of ozone molecule is:
From the structure it is evident that ozone has bent shape and all the bond lengths are equal (1.28 Å). Ozone does not contain any unpaired electron and is therefore, diamagnetic in nature.
Ozone decomposes into oxygen with liberation of heat; therefore its decomposition is exothermic in nature.
Question 24:
In the nuclear transmutation
(X, Y) is (are)
Option 1: 
(γ, n) 
Option 2: 
(p, D) 
Option 3: 
(n, D) 
Option 4: 
(γ, p) 
The nuclear transmutation is
According to the above transmutation reaction, if X is a gamma radiation (γ), then Y will be a neutron ():
But if X is a protium (), then Y will be deuterium ():
Question 25:
The major product(s) of the following reaction is (are)
Option 1: 
P 
Option 2: 
Q 
Option 3: 
R 
Option 4: 
S 
Bromination of 3 − hydroxybenzenesulphonic acid in aqueous medium results in the formation of a tri − substituted product along with evolution of SO_{3} gas as shown in the following reaction:
Question 26:
After completion of the reactions (I and II), the organic compound(s) in the reaction mixture is (are)
Option 1: 
Reaction I : P and Reaction II : P 
Option 2: 
Reaction I : U, acetone and Reaction II : Q, acetone 
Option 3: 
Reaction I : T, U, acetone and Reaction II : P 
Option 4: 
Reaction I : R, acetone and Reaction II : S, acetone 
Reaction I is haloform reaction in which acetone reacts with bromine in presence of sodium hydroxide as a base to form tribromomethane as shown in the following reaction:
While in reaction II substitution of halogen takes place in presence of acetic acid as shown in the following reaction:
Question 27:
The K_{sp} of Ag_{2}CrO_{4} is 1.1 × 10^{−12} at 298 K. The solubility (in mol/L) of Ag_{2}CrO_{4} in a 0.1 M AgNO_{3} solution is
Option 1: 
1.1 × 10^{−11} 
Option 2: 
1.1 × 10^{−10} 
Option 3: 
1.1 × 10^{−12} 
Option 4: 
1.1 × 10^{−9} 
We are given that:
K_{sp} of Ag_{2}CrO_{4} at 298 K = 1.1 × 10^{−12}
Concentration of AgNO_{3} = 0.1 M
In solution,
K_{sp} = [Ag^{+}]^{2}
$\left[{\mathrm{CrO}}_{4}^{2\u2013}\right]$1.1 × 10^{−12} = (0.1)^{2} × s
s = 1.1 × 10^{−10}
Question 28:
In the following reaction, the product(s) formed is (are)
Option 1: 
P (major) 
Option 2: 
Q (minor) 
Option 3: 
R (minor) 
Option 4: 
S (major) 
Question 29:
The precipitate P contains
Option 1: 
Pb^{2+}

Option 2: 
Hg_{2}^{2+}

Option 3: 
Ag^{+}

Option 4: 
Hg^{2+}

Out of the given ions only salts containing lead ion form precipitate with dilute HCl.
Pb^{2+} + 2HCl → PbCl_{2} + 2H^{+}
Question 30:
The coloured solution S contains
Option 1: 
Fe_{2}(SO_{4})_{3}

Option 2: 
CuSO_{4}

Option 3: 
ZnSO_{4}

Option 4: 
Na_{2}CrO_{4}

Formation of a coloured solution on treatment of the filtrate with H_{2}S in ammonical medium followed by addition of H_{2}O_{2} in basic medium, confirms the presence of Chromium ion.
Question 31:
Compounds formed from P and Q are, respectively
Option 1: 
Optically active S and optically active pair (T, U) 
Option 2: 
Optically inactive S and optically inactive pair (T, U) 
Option 3: 
Optically active pair (T, U) and optically active S 
Option 4: 
Optically inactive pair (T, U) and optically inactive S 
Question 32:
In the following reaction sequences V and W are, respectively
Option 1:  
Option 2: 

Option 3:  
Option 4: 
Question 33:
The succeeding operations that enable this transformation of states are
Option 1: 
Heating, cooling, heating, cooling

Option 2: 
Cooling, heating, cooling, heating

Option 3: 
Heating, cooling, cooling, heating

Option 4: 
Cooling, heating, heating, cooling

The transformation K → L is brought about by increasing the volume at constant pressure which increases the temperature of the system and it heats up.
The transformation L → M is carried out by decreasing the pressure at constant volume, thereby temperature of the system falls indicating cooling.
The transformation M → N occurs when the volume of the system is decreased at constant pressure which again decreases the temperature and cooling is said to take place.
The final transformation N → K is performed by increasing the pressure of the system at constant volume which again increases the temperature and heating is said to occur.
Question 34:
The pair of isochoric processes among the transformation of states is
Option 1: 
K to L and L to M

Option 2: 
L to M and N to K

Option 3: 
L to M and M to N

Option 4: 
M to N and N to K

A process in which the volume of the system remains constant is known as an isochoric process. From the given graph,
It is evident that the processes L → M and N → K are carried out at constant volume conditions, hence they are isochoric processes.
Question 35:
P and Q, respectively, are the sodium salts of
Option 1: 
hypochlorus and chloric acids

Option 2: 
hypochlorus and chlous acids

Option 3: 
chloric and perchloric acids

Option 4: 
chloric and hypochlorus acids

Reaction of Cl_{2} with dilute and concentrated NaOH can be written as:
NaOCl and NaClO_{3} are salts obtained from HClO and HClO_{3} oxoacids of chlorine.
Question 36:
R, S and T, respectively, are
Option 1: 
SO_{2}Cl_{2}, PCl_{5} and H_{3}PO_{4}

Option 2: 
SO_{2}Cl_{2}, PCl_{3} and H_{3}PO_{3}

Option 3: 
SOCl_{2}, PCl_{3} and H_{3}PO_{2}

Option 4: 
SOCl_{2}, PCl_{5} and H_{3}PO_{4}

The remaining reactions can be written as:
Question 37:
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
1. 
Conductivity decreases and then increases 

Q. 
2. 
Conductivity decreases and then does not change much 

R. 
3. 
Conductivity increases and then does not change much 

S. 
4. 
Conductivity does not change much then increases 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

(P)
As acetic acid (CH_{3}COOH) is already a weak acid and do not dissociate completely, so causing less conductivity. On addition of weak base like triethyl amine causes acidbase neutralization and new ions are formed which conduct more electricity and thus conductivity increases.
(Q) KI + AgNO_{3} → AgI (ppt.) + KNO_{3}
Since precipitation reaction takes place, there is no change in conductivity of the solution and hence conductivity remains same then increases due to greater mobility of K^{+} ion.
(R) CH_{3}COOH + KOH → CH_{3}COO^{−}K^{+} + H_{2}O
As KOH is stronger base and therefore dissociate completely and has greater conductivity. As CH_{3}COOH is added to it, neutralization takes place and conductivity decreases and remains same after neutralization
(S) NaOH + HI → NaI + H_{2}O
In HI, H^{+} ion is replaced by Na^{+} ion which has lower mobility and hence low conductivity. Therefore conductivity decreases and after neutralization conductivity increases.
Question 38:
The standard reduction potential data at 25°C is given below.
Match E° of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
1. 
–0.18 V 

Q. 
2. 
–0.4 V 

R. 
3. 
–0.04 V 

S. 
4. 
–0.83 V 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

(P)
(Q)
(R)
(S)
Question 39:
The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
1. 
NO 

Q. 
2. 
I_{2} 

R. 
3. 
Warm 

S. 
4. 
Cl_{2} 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

Question 40:
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
1. 
(i) Hg(OAc)_{2}; (ii) NaBH_{4} 

Q. 
2. 
NaOEt 

R. 
3. 
Et–Br 

S. 
4. 
(i) BH_{3}; (ii) H_{2}O_{2}/NaOH 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

(P) follows βelimination reaction.
(Q) follows S_{N}2 mechanism.
(R) is OxymercurationDemercuration of Alkenes. It is an electrophilic Addition reaction. Here overall transformation is C=C to HCCOH. It follows Markovnikov’s addition.
(S) follows antiMarkovnikov's addition.
Question 41:
Let ω be a complex cube root of unity with ω ≠ 1 and P = [p_{ij}] be a n × n matrix with p_{ij} = ω^{i}^{ + j}. Then P^{2}≠ 0, when n =
Option 1: 
57 
Option 2: 
55 
Option 3: 
58 
Option 4: 
56 
Given, and .
For n = 1:
Thus, we get
.
For n = 2:
$P={\left[\begin{array}{cc}{\omega}^{2}& {\omega}^{3}\\ {\omega}^{3}& {\omega}^{4}\end{array}\right]}_{2\times 2}={\left[\begin{array}{cc}{\omega}^{2}& 1\\ 1& \omega \end{array}\right]}_{2\times 2}\left(\because {\omega}^{3}=1\right)$
Thus, we get
${P}^{2}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}{\omega}^{2}& 1\\ 1& \omega \end{array}\right]\left[\begin{array}{cc}{\omega}^{2}& 1\\ 1& \omega \end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}{\omega}^{4}+1& {\omega}^{2}+\omega \\ {\omega}^{2}+\omega & 1+{\omega}^{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}\omega +1& {\omega}^{2}+\omega \\ {\omega}^{2}+\omega & 1+{\omega}^{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}\u2013{\omega}^{2}& \u20131\\ \u20131& \u2013\omega \end{array}\right]\left(\because 1+\omega +{\omega}^{2}=0\right)\phantom{\rule{0ex}{0ex}}\ne 0$
For n = 3:
$P=\left[\begin{array}{ccc}{\omega}^{2}& {\omega}^{3}& {\omega}^{4}\\ {\omega}^{3}& {\omega}^{4}& {\omega}^{5}\\ {\omega}^{4}& {\omega}^{5}& {\omega}^{6}\end{array}\right]=\left[\begin{array}{ccc}{\omega}^{2}& 1& \omega \\ 1& \omega & {\omega}^{2}\\ \omega & {\omega}^{2}& 1\end{array}\right]\left(\because {\omega}^{3}=1\right)$
Thus, we get
We observe that, for n = 1, n = 2 and n = 3 i.e. when n is not a multiple of 3.
We also observe tha, the numbers in options B, C and D are not multiple of 3.
Hence, the correct options are B, C and D.
Question 42:
The function has a local minimum or a local maximum at x =
Option 1: 
−2 
Option 2:  
Option 3: 
2 
Option 4: 
First, we define that given function f(x) = 2x + x + 2 − x + 2 − 2x as below.
Now, we draw the graph of this function as below.
From the graph, we see that the function f(x) has local minimum at x = −2 and x = 0 and the function f(x) has local maximum at x = .
Thus, the given function has the local minimum at x = −2 and local maximum at x = .
Hence, the correct options are A and B.
Question 43:
Let
$\omega =\frac{\sqrt{3}+i}{2}$and P = {
${\omega}^{n}$: n = 1, 2, 3, …} Further
${H}_{1}=\left\{z\in \mathrm{\u2102}:\mathrm{Re}z\frac{1}{2}\right\}$and
${H}_{2}=\left\{z\in \mathrm{\u2102}:\mathrm{Re}z\u2013\frac{1}{2}\right\}$where C is the set of all complex numbers. If z_{1}∈ P ∩ H_{1}, z_{2}∈ P ∩ H_{2} and O represents the origin, then ∠z_{1}Oz_{2} =
Option 1:  
Option 2:  
Option 3:  
Option 4: 
We have .
Given, P = {ω^{n} : n = 1, 2, 3, …}.
Thus,
$P=\left\{{e}^{\frac{i\mathrm{\pi}}{6}},{e}^{\frac{i\mathrm{\pi}}{3}},{e}^{\frac{i\mathrm{\pi}}{2}},{e}^{\frac{i2\mathrm{\pi}}{3}},{e}^{\frac{i5\mathrm{\pi}}{6}},{e}^{i\mathrm{\pi}},{e}^{\frac{i7\mathrm{\pi}}{6}},{e}^{\frac{i4\mathrm{\pi}}{3}},{e}^{\frac{i3\mathrm{\pi}}{2}},{e}^{\frac{i5\mathrm{\pi}}{3}},{e}^{\frac{i11\mathrm{\pi}}{6}},{e}^{i2\mathrm{\pi}}\right\}$
The points in the set P are located as below.
From the graph, we get
$\left(\because {H}_{1}=\left\{z\in \mathrm{\u2102}:\mathrm{Re}z\frac{1}{2}\right\}\right)$
$\left(\because {H}_{2}=\left\{z\in \mathrm{\u2102}:\mathrm{Re}z\u2013\frac{1}{2}\right\}\right)$
Thus, we get
Hence, the correct options are C and D.
Question 44:
If 3^{x} = 4^{x}^{−1}, then x =
Option 1:  
Option 2:  
Option 3:  
Option 4: 
We have
On taking log to the base 3 in both sides, we get
This is choice A.
Now, dividing the numerator and the denominator by log_{3}2, we get
This is choice B.
Again, dividing the numerator and the denominator by 2, we get
This is choice C.
Hence, the correct options are A, B and C.
Question 45:
Two lines and are coplanar. Then α can take value(s)
Option 1: 
1 
Option 2: 
2 
Option 3: 
3 
Option 4: 
4 
We have
The lines L_{1} and L_{2} are coplanar. Therefore,
Thus, we get
Hence, the correct options are A and D.
Question 46:
In a triangle PQR, P is the largest angle and Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
Option 1: 
16 
Option 2: 
18 
Option 3: 
24 
Option 4: 
22 
Let the lengths of PN, QL and RM are x, x + 2, x + 4 respectively. Triangle PQR and its incircle are shown in the following figure.
From the figure, we get
PQ = x + x + 2 = 2x +2
QR = x + 2 + x + 4 = 2x + 6
PR = x + x + 4 = 2x + 4
In the triangle PQR, we have
Since, we have, therefore,
Thus, we get
PQ = 2(8) + 2 = 18
QR = 2(8) + 6 = 22
PR = 2(8) + 4 = 20
Hence, the correct options are B and D.
Question 47:
For a ∈ R (the set of all real numbers), a ≠ −1,
Then a =
Option 1: 
5 
Option 2: 
7 
Option 3:  
Option 4: 
On dividing the numerator and denominator by n^{a}, we get
On dividing the numerator and denominator by n, we get
Thus, we get
On integrating, we get
Since we have, we get
Thus, we get
Hence, the correct option is B and D.
Question 48:
Circle(s) touching x−axis at a distance 3 from the origin and having an intercept of length on y−axis is (are)
Option 1:  
Option 2:  
Option 3: 
${x}^{2}+{y}^{2}\u20136x\u20138y+9=0$ 
Option 4: 
${x}^{2}+{y}^{2}\u20136x\u20137y+9=0$ 
Let the radius of the circle be r. There are Then, we get
The centre of the circle is (3, r).
From the general equation of the circle x^{2} + y^{2} + 2gx + 2fy + c = 0, we get
Now, the radius is
Now, the intercept on yaxis is
Thus, the equation of the circle is
Hence, the correct options are A and C.
Question 49:
Which of the following is true for 0 < x < 1?
Option 1: 
0 < f (x) < ∞

Option 2: 

Option 3: 

Option 4: 
− ∞ < f (x) < 0

It is given that,
It also can be written as .
Now, let. Then, we have. Thus, the function F(x) is concave upward.
Since, f(0) = f(1) = 0, we get f(x) < 0 for x
$\in $(0, 1).
Hence, the correct option is D.
Question 50:
If the function e^{−}^{x} f (x) assumes its minimum in the interval [0, 1] at which of the following is true?
Option 1: 

Option 2: 

Option 3: 

Option 4: 

Since the function has minimum value at, the function F(x) is decreasing in and increasing in.
Thus, we get
in 0 < x <
Correct Option: (C)
Question 51:
Length of chord PQ is
Option 1: 
7a

Option 2: 
5a

Option 3: 
2a

Option 4: 
3a

Let P(at_{1}^{2}, 2at_{1}) and Q(at_{2}^{2}, 2at_{2}) be the parametric points of the parabola y^{2} = 4ax. Then, the point of intersection of the tangents at P and Q is M(at_{1}t_{2}, a(t_{1}+t_{2})) and it will intersect on the directrix x = −a.
Thus, we get
Since the point M(at_{1}t_{2}, a(t_{1}+t_{2})) lies on the line y = 2x + a, we have
Thus, the length of the chord PQ is
Hence, the correct option is B.
Question 52:
If chord PQ subtends an angle θ at vertex of y^{2} = 4ax, then tan θ =
Option 1: 

Option 2: 

Option 3: 

Option 4: 

We have the figure as below.
The slope of OP is and the slope of the OQ is .
Thus, we get
Thus, we get
Since the tangents at P and Q will intersects at right angle, the angle is an obtuse angle.
Therefore, we get
Correct Option: (D).
Question 53:
Area of S =
Option 1: 

Option 2: 

Option 3: 

Option 4: 

Let z = x + iy. Then, S_{1}: z < 4 . Thus, S_{1} is the region enclosed by
${x}^{2}+{y}^{2}=16$whose centre is (0, 0) and the radius is 4.
Now, S_{2}:
Thus, S_{2} is a line.
And, we have S_{3}: Re (z) > 0. We get x > 0.
Now, we get the figure as below.
Thus, the area of S is the area of shaded portion in the circle.
Therefore, we get
Hence, the correct option is B.
Question 54:
Option 1: 

Option 2: 

Option 3: 

Option 4: 

is the minimum distance of z from (1, −3) in S. Thus, we get the is the perpendicular length from (1, −3) to the line .
Therefore, we get
Correct Option: (C)
Question 55:
If 1 ball is drawn from each of the boxes B_{1}, B_{2} and B_{3}, the probability that all 3 drawn balls are of the same colour is
Option 1: 

Option 2: 

Option 3: 

Option 4: 

In the box B_{1}, we have
In the box B_{2}, we have
In the box B_{3}, we have
Thus, the probability of all three ball of the same color is
P(WWW) + P(RRR) + P(BBB)
Hence, the correct option is A.
Question 56:
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B_{2} is
Option 1: 

Option 2: 

Option 3: 

Option 4: 

In the box B_{1}, we have
In the box B_{2}, we have
In the box B_{3}, we have
Let A be the event of two balls are drawn and one is W(White) and one is R(Red). Then we need to find
We have
.
Thus, we get
Correct Option: (D)
Question 57:
Match List − I with List − II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
Volume of parallelepiped determined by vectors is 2. Then the volume of the parallelepiped determined by vectors 
1. 
100 
Q. 
Volume of parallelepiped determined by vectors is 5. Then the volume of the parallelepiped determined by vectors 
2. 
30 
R. 
Area of a triangle with adjacent sides determined by vectors is 20. Then the area of the triangle with adjacent sides determined by vectors and is 
3. 
24 
S. 
Area of a parallelogram with adjacent sides determined by vectors is 30. Then the area of the parallelogram with adjacent sides determined by vectors and is 
4. 
60 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

(P)
We know that if three vectorsand pass through a point then the volume V_{1} of the parallelogram determined by these three vectors is
We need to find the volume of the parallelogram determined by the vectorsand. Let this volume be V_{2}
Therefore,
Using, we get
Thus, the match is P→3
(Q)
We know that if three vectorsandpass through a point then the volume V_{1} of the parallelogram determined by these three vectors is
We need to find the volume of the parallelogram determined by the vectorsand. Let this volume be V_{2}
Therefore,
Using, we get
Thus, the match is Q→4
(R)
Area A_{1} of a triangle with adjacent sides determined vectors andis
Thus, area A_{2} of the triangle determined by the vectorsandis
Using, we get
Thus, the match is R→1
(S)
Area A_{1} of a parallelogram with adjacent sides determined by vectors andis
Thus, area A_{2} of parallelogram determined by the vectorsandis
Using, we get
Thus, the match is S→2
Hence, the correct code is (C).
Question 58:
Consider the lines and the planes Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L_{1} and L_{2}, and perpendicular to planes P_{ 1} and P_{2}.
Match List − I with List − II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
a = 
1. 
13 
Q. 
b = 
2. 
–3 
R. 
c = 
3. 
1 
S. 
d = 
4. 
–2 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

The given lines and planes are:
Normal to the plane P is
Letand
Since, L_{1} and L_{2} intersect, therefore,
2k_{1} + 1 = k_{2} + 4 … (1)
− k_{1} = k_{2} − 3 … (2)
k_{1 }− 3 = 2k_{2} − 3 … (3)
Solving equation (1) and (2), we get k_{1} = 2, k_{2} = 1
Thus, the point of intersection of the lines L_{1} and L_{2} is
Therefore, equation of the plane P is
⇒ a = 1, b = − 3, c = − 2, d = 13
Hence, the correct code is A.
Question 59:
Match List I with List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
1. 

Q. 
possible value of is 
2. 
_{} 
R. 
3. 

S. 
then possible value of x is 
4. 
1 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

Let
Therefore,
Thus, the match is P→4
(Q)
They above equations can be written as
… (1)
… (2)
Squaring (1) and (2) and adding them, we get
Thus, the match is Q→3
(R)
The above equation can be written as
Thus, the match is R→2 or R→4
(S)
Therefore,
Thus, the match is S→1
Hence, the correct code is B.
Question 60:
A line L : y mx + 3 meets y − axis at E(0, 3) and the arc of the parabola y^{2} = 16x, 0 ≤ y ≤ 6 at the point F(x_{0}, y_{0}). Then tangent to the parabola at F(x_{0}, y_{0}) intersects the yaxis at G(0, y_{1}). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists:

List I 

List II 
P. 
m = 
1. 

Q. 
Maximum area of âˆ†EFG is 
2. 
4 
R. 
y_{0} = 
3. 
2 
S. 
y_{1} = 
4. 
1 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

The given parabola is y^{2} = 16x, 0 ≤ y ≤ 6, 0 ≤ x ≤ 9/4
Tangent to the parabola at F (x_{0}, y_{0}) = (4t^{2}, 8t) is y t = x + 4t^{2}
At x = 0, y = 4t, thus G ≡ (0, 4t)
Thus, (4t^{2}, 8t) satisfies the line y = m x + 3
⇒ 8t = 4mt^{2} + 3
⇒ 4mt^{2} − 8t + 3 = 0
From the figure area A of the triangle EFG is
For maxima of A, differentiate A with respect to t
Differentiate A with respect to t again
Thus, maxima occur at t = 1/2
Therefore,
G (0, 4t) ≡ G (0, 2), y_{1} = 2, (x_{0}, y_{0}) = (4t^{2}, 8t) = (1, 4)
Putting t = 1/2 in 4mt^{2} − 8t + 3 = 0, we get m = 1
Thus, the match is (P) →4, (Q) →1, (R) → 2, and (S) →3
Hence, the correct code is A.