# JEE Advanced 2013 Paper-2(Code 0)

### Test Name: JEE Advanced 2013 Paper-2(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 0-100 K varies linearly with temperature T. Option 2: heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K. Option 3: there is no change in the rate of heat absorption in the range 400-500 K. Option 4: the rate of heat absorption increases in the range 200-300 K.

Solutions:

dQ = mCdT

Rate of absorption of heat,

In 0-100 K range the specific heat increases So R is increases but not linearly.

The value of C in the range 400-500 K is more than in 0-100 K thus the heat absorbed in increasing temperature from 400 to 500 K is more than in 0 to 100 K.

In the range 200-300 K as C increases thus R increases.

#### Question 2:

The radius of the orbit of an electron in Hydrogen-like atom is 4.5 a0, where a0 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 de-excites, is (are)

 Option 1: Option 2: Option 3: Option 4:

Solutions:

Angular momentum,

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.

Solutions:

This shows that with the increase in θ the absolute error and fractional error in d decreases.

#### Question 4:

Two non-conducting spheres of radii R1 and R2 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.

Solutions:

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 non-zero. 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 non-zero.

Solutions:

In the region 00nI ≠ 0.

In the region R

In the region r>2R, the magnetic field is due to cylinder which is non-zero 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 f1. An observer in the other vehicle hears the frequency of the whistle to be f2. 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, f2 > f1. Option 2: If the wind blows from the source to the observer, f2 > f1. Option 3: If the wind blows from the observer to the source, f2 < f1. Option 4: If the wind blows from the source to the observer, f2 < f1.

Solutions:

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.

Solutions:

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 mass-less 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 u0. When the speed of the particle is 0.5 u0, 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 u0. 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

Solutions:

From conservation of energy when the particle returns to its equilibrium position its speed becomes u0.

For SHM the displacement, x = Asinωt

Initial velocity, u0 = Aω

Now before collision, speed of the particle, v = 0.5 u0

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

Solutions:

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

Solutions:

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

Solutions:

Total resistance offered by cable in transmission over 20 km = 0.4×20 = 8 Ω

Now, P = VI

Or, I = (600×103)/(4000) = 150 A

Power dissipated during transmission, Pd = I2R = 1502(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 step-up 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 step-down transformer is

 Option 1: 200 : 1 Option 2: 150 : 1 Option 3: 100 : 1 Option 4: 50 : 1

Solutions:

In the step-up transformer,

In step-down 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

Solutions:

#### 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 BQR2 Option 2: Option 3: Option 4: γ BQR2

Solutions:

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.

Solutions:

(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

Solutions:

Kinetic energy of α-particle 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 P0V0 ln2 Q. G→H 2. 36 P0V0 R. F→H 3. 24 P0V0 S. F→G 4. 31 P0V0
Option 1:
 P Q R S 4 3 2 1

Option 2:
 P Q R S 4 3 1 2

Option 3:
 P Q R S 3 1 2 4

Option 4:
 P Q R S 1 3 2 4

Solutions:

FG is isothermal thus,

P0VG=32P0V0

Or, VG=32V0

P0(VH)5/3=32P0(V0)5/3 (For monatomic gas, γ=5/3)

VH=8V0

WGE=P0(V0-32V0)= -31P0V0

WGH=P0(8V0-32V0)= -24P0V0

#### 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:
 P Q R S 4 2 1 3

Option 2:
 P Q R S 1 3 2 4

Option 3:
 P Q R S 2 1 4 3

Option 4:
 P Q R S 4 3 2 1

Solutions:

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 μ1 is placed in a rectangular block of refractive index μ2, which is surrounded by a medium of refractive index μ3, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between μ1, μ2 and μ3, 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. m2 > m1 and m2 > m3 R. e → h 3. m1 = m2 S. e → i 4.
Option 1:
 P Q R S 2 3 1 4

Option 2:
 P Q R S 1 2 4 3

Option 3:
 P Q R S 4 1 2 3

Option 4:
 P Q R S 2 3 4 1

Solutions:

e→f, μ21 and μ32

e→g, μ12

e→h, μ21 and μ32, 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. [ML2T–1] Q. Coefficient of viscosity 2. [ML–1T–1] R. Planck constant 3. [MLT–3 K–1] S. Thermal conductivity 4. [ML2T–2 K–1]
Option 1:
 P Q R S 3 1 2 4

Option 2:
 P Q R S 3 2 1 4

Option 3:
 P Q R S 4 2 1 3

Option 4:
 P Q R S 4 1 2 3

Solutions:

#### Question 21:

The carbon-based reduction method is NOT used for the extraction of

 Option 1: tin from SnO2 Option 2: iron from Fe2O3 Option 3: aluminium from Al2O3 Option 4: magnesium from MgCO3 · CaCO3

Solutions:

Extraction of Aluminium, magnesium and calcium from their ores Al2O3, MgCO3 and CaCO3 respectively is done by electrolytic method, whereas Tin and Iron is extracted from their oxide ores SnO2 and Fe2O3 respectively by reducing them with Carbon. This is because the stable Al2O3, MgCO3 and CaCO3 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 CaCO3(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 CaCO3 Option 3: K is dependent of the pressure of CO2 at a given T Option 4: ΔH is independent of the catalyst, if any

Solutions:

For decomposition of CaCO3 as shown below:

At equilibrium:

Kp =

Change in enthalpy is given by the following expression:

ΔH = CpΔT

As Cp 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 CaCO3.

#### Question 23:

The correct statement(s) about O3 is (are)

 Option 1: O–O bond lengths are equal. Option 2: Thermal decomposition of O3 is endothermic. Option 3: O3 is diamagnetic in nature. Option 4: O3 has a bent structure.

Solutions:

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)

Solutions:

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

Solutions:

Bromination of 3 − hydroxybenzenesulphonic acid in aqueous medium results in the formation of a tri − substituted product along with evolution of SO3 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

Solutions:

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 Ksp of Ag2CrO4 is 1.1 × 10−12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 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

Solutions:

We are given that:

Ksp of Ag2CrO4 at 298 K = 1.1 × 10−12

Concentration of AgNO3 = 0.1 M

In solution,

Ksp = [Ag+]2

$\left[{\mathrm{CrO}}_{4}^{2–}\right]$

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)

Solutions:

#### Question 29:

The precipitate P contains

 Option 1: Pb2+ Option 2: Hg22+ Option 3: Ag+ Option 4: Hg2+

Solutions:

Out of the given ions only salts containing lead ion form precipitate with dilute HCl.

Pb2+ + 2HCl → PbCl2 + 2H+

#### Question 30:

The coloured solution S contains

 Option 1: Fe2(SO4)3 Option 2: CuSO4 Option 3: ZnSO4 Option 4: Na2CrO4

Solutions:

Formation of a coloured solution on treatment of the filtrate with H2S in ammonical medium followed by addition of H2O2 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

Solutions:

#### Question 32:

In the following reaction sequences V and W are, respectively

 Option 1: Option 2: Option 3: Option 4:

Solutions:

#### 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

Solutions:

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

Solutions:

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

Solutions:

Reaction of Cl2 with dilute and concentrated NaOH can be written as:

NaOCl and NaClO3 are salts obtained from HClO and HClO3 oxoacids of chlorine.

#### Question 36:

R, S and T, respectively, are

 Option 1: SO2Cl2, PCl5 and H3PO4 Option 2: SO2Cl2, PCl3 and H3PO3 Option 3: SOCl2, PCl3 and H3PO2 Option 4: SOCl2, PCl5 and H3PO4

Solutions:

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:
 P Q R S 3 4 2 1
Option 2:
 P Q R S 4 3 2 1
Option 3:
 P Q R S 2 3 4 1
Option 4:
 P Q R S 1 4 3 2

Solutions:

(P)

As acetic acid (CH3­COOH) is already a weak acid and do not dissociate completely, so causing less conductivity. On addition of weak base like triethyl amine causes acid-base neutralization and new ions are formed which conduct more electricity and thus conductivity increases.

(Q) KI + AgNO3 → AgI (ppt.) + KNO3

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) CH3COOH + KOH → CH3COOK+ + H2O

As KOH is stronger base and therefore dissociate completely and has greater conductivity. As CH3COOH is added to it, neutralization takes place and conductivity decreases and remains same after neutralization

(S) NaOH + HI → NaI + H2O

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:
 P Q R S 4 1 2 3
Option 2:
 P Q R S 2 3 4 1
Option 3:
 P Q R S 1 2 3 4
Option 4:
 P Q R S 3 4 1 2

Solutions:

(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. I2 R. 3. Warm S. 4. Cl2
Option 1:
 P Q R S 4 2 3 1
Option 2:
 P Q R S 3 2 1 4
Option 3:
 P Q R S 1 4 2 3
Option 4:
 P Q R S 3 4 2 1

Solutions:

#### 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) NaBH4 Q. 2. NaOEt R. 3. Et–Br S. 4. (i) BH3; (ii) H2O2/NaOH
Option 1:
 P Q R S 2 3 1 4
Option 2:
 P Q R S 3 2 1 4
Option 3:
 P Q R S 2 3 4 1
Option 4:
 P Q R S 3 2 4 1

Solutions:

(P) follows β-elimination reaction.

(Q) follows SN2 mechanism.

(R) is Oxymercuration-Demercuration of Alkenes. It is an electrophilic Addition reaction. Here overall transformation is  C=C to H-C-C-OH. It follows Markovnikov’s addition.

#### Question 41:

Let ω be a complex cube root of unity with ω ≠ 1 and P = [pij] be a n × n matrix with pij = ωi + j. Then P2≠ 0, when n =

 Option 1: 57 Option 2: 55 Option 3: 58 Option 4: 56

Solutions:

Given, and .

For n = 1:

Thus, we get

.

For n = 2:

Thus, we get

For n = 3:

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:

Solutions:

First, we define that given function f(x) = 2|x| + |x + 2| − ||x + 2| − 2|x|| 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

and

where C is the set of all complex numbers. If z1PH1, z2PH2 and O represents the origin, then ∠z1Oz2 =

 Option 1: Option 2: Option 3: Option 4:

Solutions:

We have .

Given, P = {ωn : n = 1, 2, 3, …}.

Thus,

The points in the set P are located as below.

From the graph, we get

Thus, we get

Hence, the correct options are C and D.

#### Question 44:

If 3x = 4x−1, then x =

 Option 1: Option 2: Option 3: Option 4:

Solutions:

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 log32, 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

Solutions:

We have

The lines L1 and L2 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

Solutions:

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 aR (the set of all real numbers), a ≠ −1,

Then a =

 Option 1: 5 Option 2: 7 Option 3: Option 4:

Solutions:

On dividing the numerator and denominator by na, 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: ${x}^{2}+{y}^{2}–6x–8y+9=0$ Option 4: ${x}^{2}+{y}^{2}–6x–7y+9=0$

Solutions:

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 x2 + y2 + 2gx + 2fy + c = 0, we get

Now, the intercept on y-axis 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

Solutions:

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 ex 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:

Solutions:

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

Solutions:

Let P(at12, 2at1) and Q(at22, 2at2) be the parametric points of the parabola y2 = 4ax. Then, the point of intersection of the tangents at P and Q is M(at1t2, a(t1+t2)) and it will intersect on the directrix x = −a.

Thus, we get

Since the point M(at1t2, a(t1+t2)) 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 y2 = 4ax, then tan θ =

 Option 1: Option 2: Option 3: Option 4:

Solutions:

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:

Solutions:

Let z = x + iy. Then, S1: |z| < 4 . Thus, S1 is the region enclosed by

${x}^{2}+{y}^{2}=16$

whose centre is (0, 0) and the radius is 4.

Now, S2:

Thus, S2 is a line.

And, we have S3: 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:

Solutions:

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 B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is

 Option 1: Option 2: Option 3: Option 4:

Solutions:

In the box B1, we have

In the box B2, we have

In the box B3, 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 B2 is

 Option 1: Option 2: Option 3: Option 4:

Solutions:

In the box B1, we have

In the box B2, we have

In the box B3, 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:
 P Q R S 4 2 3 1

Option 2:
 P Q R S 2 3 1 4

Option 3:
 P Q R S 3 4 1 2

Option 4:
 P Q R S 1 4 3 2

Solutions:

(P)

We know that if three vectorsand pass through a point then the volume V1 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 V2

Therefore,

Using, we get

Thus, the match is P→3

(Q)

We know that if three vectorsandpass through a point then the volume V1 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 V2

Therefore,

Using, we get

Thus, the match is Q→4

(R)

Area A1 of a triangle with adjacent sides determined vectors andis

Thus, area A2 of the triangle determined by the vectorsandis

Using, we get

Thus, the match is R→1

(S)

Area A1 of a parallelogram with adjacent sides determined by vectors andis

Thus, area A2 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 L1 and L2, and perpendicular to planes P 1 and P2.

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:
 P Q R S 3 2 4 1

Option 2:
 P Q R S 1 3 4 2

Option 3:
 P Q R S 3 2 1 4

Option 4:
 P Q R S 2 4 1 3

Solutions:

The given lines and planes are:

Normal to the plane P is

Letand

Since, L1 and L2 intersect, therefore,

2k1 + 1 = k2 + 4       … (1)

k1 = k2 − 3           … (2)

k1 − 3 = 2k2 − 3      … (3)

Solving equation (1) and (2), we get k1 = 2, k2 = 1

Thus, the point of intersection of the lines L1 and L2 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:
 P Q R S 4 3 1 2

Option 2:
 P Q R S 4 3 2 1

Option 3:
 P Q R S 3 4 2 1

Option 4:
 P Q R S 3 4 1 2

Solutions:

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 y2 = 16x, 0 ≤ y ≤ 6 at the point F(x0, y0). Then tangent to the parabola at F(x0, y0) intersects the y-axis at G(0, y1). 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. y0 = 3. 2 S. y1 = 4. 1
Option 1:
 P Q R S 4 1 2 3

Option 2:
 P Q R S 3 4 1 2

Option 3:
 P Q R S 1 3 2 4

Option 4:
 P Q R S 1 3 4 2

Solutions:

The given parabola is y2 = 16x, 0 ≤ y ≤ 6, 0 ≤ x ≤ 9/4

Tangent to the parabola at F (x0, y0) = (4t2, 8t) is y t = x + 4t2

At x = 0, y = 4t, thus G ≡ (0, 4t)

Thus, (4t2, 8t) satisfies the line y = m x + 3

⇒ 8t = 4mt2 + 3

⇒ 4mt2 − 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), y1 = 2, (x0, y0) = (4t2, 8t) = (1, 4)

Putting t = 1/2 in 4mt2 − 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.