IIT JEE Advanced 2015 Paper 2

Test Name: IIT JEE Advanced 2015 Paper 2

Question 1:

The energy of a system as a function of time t is given as E(t) = A2 exp(−αt), where α = 0.2 s1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is

Correct Answer: 4

Solutions:We have given,
E(t) = A2 exp(−αt)     …..(1)

Taking derivative we getδE=2AδA exp αt+A2δexp αtδE=2A exp αt·δA+A2·exp αt δαtδ(E)=2A exp αt·δA+A2exp αt αδt+t δα

Dividing both side by equation (1), we get

δEE=2δAA+αδt+tδαδEE=2δAA+αδt           δα=0δEE=2×0.0125+0.2×0.015×5      δEE=0.0250+0.015=0.040              % error=δEE×100=4%

[[VIDEO:13867]]

Question 2:

The densities of two solid spheres A and B of the same radii R vary with radial distance r as

ρAr=krR and ρBr=krR5

, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IA and IB, respectively. If

IBIA=n10

, the value of n is

Correct Answer: 1

Solutions:


To calculate the moment of inertia of solid sphere, consider a thin hollow sphere has surface area 4πx2 and thickness

δx

.  Its volume (dV) will be

dV = 4πx2

δx

Now, mass of the thin hollow sphere (dm) is

dm = 4πx2dx ρ(

δx

)

Moment of inertia (dI) is given by

δI=23δmx2δI=234πρx.x4.δx

 It is clear from the above equationdI0Rx4ρx·dxAs  both spheres have same radii, so  the above equation is valid for both the sphere. IB=1R50Rx9.dx     .....1and IA=1R0Rx5.dx     .....2Dividing 1 by 2 we getIBIA=1R50Rx9·dx1R0Rx5·dx=110·R516R5=610

[[VIDEO:13868]]

Question 3:

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,
π/3, 2π/3 and π. When they are superposed, the intensity of the resulting wave is nI0. The
value of n is

Correct Answer: 3

Solutions:We have given that the harmonic waves have same intensity. So, they will also have same amplitude.

Let the amplitude of individual wave is A.

Now, using the phaser method.


Waves with phase angles 0 and π will cancel each other.

Amplitude of resultant wave is

Ar2=A2+A2+2AAcos60°Ar2=3A2

Since, the intensity is directly proportional to the square of amplitude.

 I=3I0

Thus, the value of n is 3.
[[VIDEO:13869]]

Question 4:

For a radioactive material, its activity A and rate of change of its activity R are defined as

A=dNdt and R=dAdt

, where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life τ) and Q (mean life 2τ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are RP and RQ, respectively. If 

RPRQ=ne

, then the value of n is

Correct Answer: 2

Solutions:Let NP and NQ be the number of nuclei present in the radioactive sample at t = 0. Let λP and λQ be the disintegration constant of source P and Q respectively.

The activity of source P (AP) is given by

AP=λPNPeλPt

The activity of source Q (AQ) is given by

AQ=λQNQeλQt

We have given

AP = AQ at t = 0

λPNP=λQNQNPNQ=λQλP     .....1Disintegration constant λP is given byλP=1τ     .....2   Disintegration constant λQ is given by λQ=12τ     .....3Dividing equation 3 by 2, we getλQλP=12     .....4     Rate of change of activity R isR=dAdt=d2Ndt2=λ.dNdt=λ2NRPRQ=λP2λQ2·NPeλP2τNQeλQ2τRPRQ=λP2λQ2NPNQ1e     .....5Using equation 1, 4 and 5, we getRPRQ=2e

Thus, we get n = 2.

Question 5:

A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For

n=3 

the value of θ is 60° and

dθdn=m

. The value of m is

Correct Answer: 1

Solutions:

Applying snell law's of refraction for the light incidenting on the prism, we get,

sin60°=3sinr1r1=30°Applying snell law for the light refracting out the prism, we getnsinr2=sin θsin θ=n2Differentiating both side with respect to n, we getcosθdθdn=12dθdn=12cos60°=1

Thus, the value of m is 1.

Question 6:

In the following circuit, the current through the resistor

R =2Ω

is I Amperes. The value of I is

Correct Answer: 1

Solutions:The given circuit can be simplified step by step as shown:


Net resistance of the circuit = 6.5 Ω
The current flowing through the circuit is thus given by

I=6.56.5= 1 A

Question 7:

An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

Correct Answer: 2

Solutions:According to the Bohr's postulates

Angular momentum, L=nh2πmvr=nh2π=3h2πWe get, n=3Also, mv=3h2πrLinear mometntum, P=3h2πr    P=mv     .....1

Radius of nth shell is given by

r=n2Za0For Li, Z=3r=323a0    n=3r=3a0     .....2

From the de-Broglie's hypothesis

λ=hP=h3h2πr using1λ=2πr3=2π3a03 using 2λ=2πa0=pπa0 p=2

Hence, the correct option is B.

Question 8:

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l from M, the tension in the rod is zero for

m=kM288·

The value of k is

Correct Answer: 7

Solutions:

Due to gravitational interaction between the masses, the two point masses will have some acceleration.

Forces on the mass m nearer to the large mass M at r = 3l

GMm3l2Gm2l2=ma     .....1

Force on the the mass m away from the large mass M

GMm3l2+Gm2l2=ma     .....2

Comparing equation (1) and (2) we have

GM9l2Gml2=GM16l2+Gml2M9M16=m+m7M144=2mm=7M288=kM288k=7

Question 9:

A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative permittivities 

ε1=2 and ε2=4

are introduced between the two plates as shown in the figure, the capacitance becomes C2. the ratio 

C2C1

is

Option 1: 6/5
Option 2: 5/3
Option 3: 7/5
Option 4: 7/3
Correct Answer: 4

Solutions:We know that,

C1=ε0Ad

From figure, we have

C=2ε0s2d2=2ε0sd C'=4ε0s2d2=4ε0sd andC''=2ε0s2d=ε0sd

The equivalent capacitance C2 is thus given by

C2=CC'C+C'+C''C2=4ε0s3d+ε0sd=7ε0s3dC2C1=73

Hence, the correct option is (D).
[[VIDEO:13871]]

Question 10:

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)

Option 1: If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is

14P1V1
Option 2: If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1 V1
Option 3: If V2 = 3V1 and T2 = 4T1, then the work done by the gas is

73P1V1
Option 4: If V2 = 3 V1 and T2 = 4T1, then the heat supplied to the gas is

176P1V1
Correct Answer: 2

Solutions:
 

Change in internal energy U is given byU=32nRT2T1=32P2V2P1V1 From the relationP1V1T1=P22V13T1P2=32P1U=3232P1.2V1P1V1=3P1V1  Hence, the correct option is B.  

Question 11:

A fission reaction is given by

U92236X54140e+S3894r+x+y,

where x and y are two particles. Considering

U92236

to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per nucleon of

U92236, X54140e

  and

S3894r

be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option (s) is (are)

Option 1: x = n, y = n, KSr = 129 MeV, KXe = 86 MeV
Option 2: x = p, y = e, KSr = 129 MeV, KXe = 86 MeV
Option 3: x = p, y = n, KSr = 129 MeV, KXe = 86 MeV
Option 4: x = n, y = n, KSr = 86 MeV, KXe = 129 MeV
Correct Answer: 1

Solutions:As x and y share equal energy = 4 MeV;
​x = y = n

For the reaction, Q value will be 
Q = 94 × 8.5 + 140 × 8.5 − 236 × 7.5
Q = 219 MeV

x and y share 4 MeV together
Rest of the energy,

21944

= 215 MeV will be shared between Xe and Sr nucleus.
Heavier particle will have less kinetic energy therefore, among the option, (A) and (D) , (A) will be correct as energy of KSr >  KXe 

Therefore, KSr = 129 MeV and KXe = 86 MeV                          

Hence, the correct option is (A)

Question 12:

In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are)

Option 1: P has more tensile strength than Q
Option 2: P is more ductile than Q
Option 3: P is more brittle than Q
Option 4: The Young's modulus of P is more than that of Q
Correct Answer: 1, 2

Solutions:Breaking stress of P is more than Q so P is more ductile

Therefore,  Young's module of Q > Young's modulus of P.
Also,

Strain=1γ×stress

P has more tensile strength than Q

Hence, the correct options are (A) and (B)

Question 13:

A spherical body of radius R consists of a fluid of constant density and is in equilibrium
under its own gravity. If P(r) is the pressure at r (r < R), then the correct option(s) is(are)

Option 1: P(r = 0) = 0
Option 2: P(r=3R / 4) P(r=2R / 3) =6380
Option 3: P(r=3R/5P(r=2R/5=1621
Option 4: P(r=R/2)P(r=R/3)=2027
Correct Answer: 2, 3

Solutions:

Let p be the density of the fluid. Pressure inside the spherical body at any point r can be written as:

dPdr=pg(r)where g(r)=Gm(r)r2and m(r)=43πpr3 g(r)=43πpGrdPdr=43πp2Gror 0PdP=Rr43πp2GrP=46πp2Gr2R2P=23πp2GR2r2or P= KR2r2for, r=3R4;P1= KR29R162=K7R216for, r=2R3;P2= KR24R92=K5R29for, r=3R5;P3= KR29R252=K16R225for, r=2R5;P4= KR24R252=K21R225for, r=R;P5= KR2RG2=K3R29for, r=R3;P6= KR2R92=K8R29

We will find,
 

P(r=3R / 4) P(r=2R / 3) =6380

and
 

P(r=3R/5)P(r=2R/5)=1621

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

Question 14:

Two spheres P and Q of equal radii have densities p1 and p2, respectively. The spheres are
connected by a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and
viscosities È 1 and È 2, respectively. They float in equilibrium with the sphere P in L1 and
sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal
velocity

Vp

and Q alone in L1 has terminal velocity

VQ

, then

Option 1: |VP||VQ|=η1η2

 

Option 2:  |VP||VQ|=η2η1
Option 3:  VP . VQ>0
Option 4:  VP . VQ<0
Correct Answer: 1, 4

Solutions:For floating, gravitational pull on the spheres must be equal to the upthrust experienced by them
⇒ (σ1 + σ2)Vg= (p1p2)Vg
⇒ (σ1 + σ2) = (p1p2)
σ1

p2 = p1

 σ2
Since the string is taut, we have

p1<σ1 and p2>σ2

        
We know that, the terminal velocity is given by

V=29r2σρgη VP(p1σ2)η2        VQ(p2σ1)η1|VP||VQ|=η1η2 p1σ2p2σ1                           

          

|VP||VQ|=η1η2                           

Vp & V1 have opposite signs hence their product is < zero.

Hence, the correct options are (A) and (D).

Question 15:

In terms of potential difference V, electric current I, permittivity Ɛ0, permeability µ0 and speed of light c, the dimensionally correct equation(s) is(are)

Option 1: µ0I2 = Æ0V2
Option 2: Ɛ0I = µ0V
Option 3: I= Ɛ0cV
Option 4: µ0cI = Ɛ0V
Correct Answer: 1, 3

Solutions:Dimensions of I = T-1Q
​Dimensions of V = MLT-2Q-1
Dimensions of µ= MLQ-2
Dimensions of​ Æ= M-1L-3T2Q2
Dimensions of​ â€‹c = LT-2
​
on checking all the options dimesnionally; relations µ0I2 = Æ0V2 and  I= Æ0cV are dimensionally correct. 
Hence, the correct options are (A) and (C).

Question 16:

Consider a uniform spherical charge distribution of radius R1, centred at the origin O. In this distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1R2 (see figure), is made. If the electric field inside the cavity at position

r is Er,

then which of the following is the correct statement?

Option 1: E

is uniform, its magnitude is independent of R2 but its direction depends on

r

.

Option 2: E

is uniform, its magnitude depends on R2 and its direction depends on

r

.

Option 3: E

is uniform, its magnitude is independent of

α

but its direction depends on

α

.

Option 4: E

is uniform and both its magnitude and direction depend on

α

.

Correct Answer: 4

Solutions:

Electric field inside the cavity is given by  E=ρ30OPOP=OA+APOA+AP=aE=ρ30aρ=Charge density

Hence, the correct answer is option D. 

Question 17:

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.


 

For two structures namely S1 with

n1=45/4

and

n2=3/2,

and S2 with

n1=8/5

and

n2=7/5

and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)

Option 1: NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index

16315.
Option 2: NA of S1 immersed in liquid of refractive index

615

is the same as that of S2 immersed in water.

Option 3: NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index

415.
Option 4: NA of S1 placed in air is the same as that of S2 placed in water.
Correct Answer: 1, 3

Solutions:

NA=sinim=n1 sin 90csinc=n2n1NA=sinim=n1ns1n22n21NA=1nsn21n22

Option A:    NA1=45163443=91643=916NA2=6425492516315=15516315=15×316×5=916NA1=NA2Option (A) is correct.Option (B):NA1=34615=3×154×6=158NA2=15543=31520 Option (B) is incorrect.Option (C):NA1=341=34NA2=155415=34 Option (C) is correct.Option (D): Is incorrect as (C) is correct.

Hence, the correct option is (A) and (C).

Question 18:

PARAGRAPH 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

If two structures of same cross−sectional area, but different numerical apertures NA1 and NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is

Option 1: NA1NA2NA1+NA2
Option 2: NA1 + NA2
Option 3: NA1
Option 4: NA2
Correct Answer: 4

Solutions:The numerical aperture of the combination is given by

NA=1nsn21n22as NA2<NA1The numerical aperture is limited by second slab.NA of combination will be smalles of the two apertures. Therefore, NA will be NA2. 

Hence, the correct option is â€‹(D).

Question 19:

PARAGRAPH 2

In a thin rectangular metallic strip a constant current I flows along the positive x–direction, as shown in the figure. The length, width and thickness of the strip are â„“, w and d, respectively.
A uniform magnetic field

B

is applied on the strip along the positive y–direction. Due to this, the charge carriers experience a net deflection along the z–direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z–direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x–y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is (are)

Option 1: If w1 = w2 and d1 = 2d2, then V2 = 2V1
Option 2:

If w1 = w2 and d1 = 2d2, then V2 = V1

Option 3: If w1 = 2w2 and d1 = d2, then V2 = 2V1
Option 4: If w1 = 2w2 and d1 = d2, then V2 = V1
Correct Answer: 1, 4

Solutions:Force experienced by charge carriers will be​ given by

qvB=qVmVkwVMVK=wVBhere v is the velocity of the electronsThe current is given byI=neAV=newdvwV=InedVMVK=IBned

When w1 = w2 and d1 =2d2   

 

w1V1=Ine2d2w2V2=Ined2On dividing we get, V2V1=2

When when w1 = 2w2 and d1 = d2   

2w2V1=Ined2w2V2=Ined2on dividing we get, V2V1=1

Thus, V2V1

Hence, the correct options are (A) and (D).

Question 20:

PARAGRAPH 2

In a thin rectangular metallic strip a constant current I flows along the positive x–direction, as shown in the figure. The length, width and thickness of the strip are â„“, w and d, respectively.
A uniform magnetic field

B

is applied on the strip along the positive y–direction. Due to this, the charge carriers experience a net deflection along the z–direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z–direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Consider two different metallic strips (1 and 2) of same dimensions (length â„“, width w and thickness d) with carrier densities n1 and n2 respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y–directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s) is (are)

Option 1:

If B1 = B2 and n1 = 2n2, then V2 = 2V1

Option 2: If B1 = B2 and n1 = 2n2, then V2 = V1
Option 3:

If B1 = 2B2 and n1 = n2, then V2 = 0.5V1

Option 4: If B1 = 2B2 and n1 = n2, then V2 = V1
Correct Answer: 1, 3

Solutions:

We know thatV1=IB1n1de V2=IB2n2dewhen n1=2n2 and B1=B2V1=IB22n2de V2=IB2n2deOn dividing V2V1=12=0.5 

Similarly, 

We know that,V1=IB1n1deV2=IB2n2dewhen n1=n2 and B1=2B2V1=I2B2n2de V2=IB2n2deOn dividing V2V1=2 

Hence, the correct options are (A) and (C).

Question 21:

Among the following, the number of reaction(s) that produce(s) benzaldehyde is

Correct Answer: 4

Solutions:



So, all 4 reactions produce benzaldehyde.
[[VIDEO:13904]]

Question 22:

In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe−C bond(s) is

Correct Answer: 3

Solutions:The number of Fe−C bonds in the given complex is 3.

                   

Question 23:

Among the complex ions, [Co(NH2−CH2−CH2−NH2)2Cl2]+,  [CrCl2(C2O4)2]3− , [Fe(H2O)4 (OH)2]+ ,   [Fe(NH3)2 (CN)4] , [Co(NH2−CH2−CH2−NH2)2 (NH3)CL]2+ and [Co(NH3)4 (H2O)Cl]2+ , the number of complex ion(s) that show(s) cis-trans isomerism is

Correct Answer: 6

Solutions:





As all the given complexes show cis-trans isomerism, thus, the number of complexes showing geometrical isomerism is 6.

Question 24:

Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is

Correct Answer: 6

Solutions:One mole of B2H6 reacts with six moles of methanol to produce two moles of boron containing product as shown in the reaction below:

                                                

B2H6 + 6 CH3OH  2 B(OCH3)3 + 6 H2

Thus, 3 moles of B2H6 on reacting with 18 moles of methanol will produce 6 moles of boron containing product.
[[VIDEO:13905]]

Question 25:

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If

λx0  λy0

, the difference in their pKa values, pKa(HX) − pKa(HY), is (consider degree of ionization of both acids to be << 1)

Correct Answer: 3

Solutions:Since, 

λX°  λY°

λH+° + λX° λH+° + λY°

We can say that

λHX ° λHY°

Also,

λmλm° = α

For HX,  

λm(HX)=λm°αHX

For HY,

λm(HY)=λm°αHY

 

Given, λm(HY)=10 λm(HX)λm°αHY = 10 λm°αHX             αHY = 10 αHX

 

ka=cα2(1α)  Since, α<<1ka = cα2 ka(HX)= 0.01 (αHX)2 ka(HY)= 0.1 (αHY)2           = 0.1 (10 (αHX)2

     

ka(HX)ka(HY)=0.01 (αHX)20.1 (10 αHX)2             =11000Taking log on both sideslog ka(HX)  log ka(HY) = 3i.e. pka(HX)  pka(HY) = 3

Question 26:

A closed vessel with rigid walls contains 1 mol of

U92238

and 1 mol of air at 298 K. Considering complete decay of

U92238 to Pb82206

, the ratio of the final pressure to the initial pressure of the system at 298 K is

Correct Answer: 9

Solutions:Initial moles of gas (air) = 1 mol

Decomposition of uranium-238 is shown below:

U92238  Pb82206 + 8 He24 + 6 e01

Final moles of gases (air and helium) = 1 + 8 = 9 mol

Ratio of final pressure to initial pressure of gases is directly proportional to the ratio of final moles of gases to the initial moles of gases.

Final moles of gasesInitial moles of gases=91= 9   

Question 27:

In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by

MnO4

. For this reaction, the ratio of the rate of change of [H+] of the rate of change of

MnO4

is

Correct Answer: 8

Solutions:Oxidation of Fe[(C2O4)2(H2O)2]2− by

MnO4

is given as:

FeC2O42H2O22 + MnO4 + 8 H+Mn2+ + Fe3+ + 4 CO2 + 6 H2O

The ratio of rate of change of H+ to the rate of change of 

MnO4

can be calculated as follows:

Rate of change of H+Rate of change of MnO4=81                                      = 8

Question 28:

The number of hydroxyl group(s) in Q is

Correct Answer: 4

Solutions:The given reaction scheme can be completely represented as

Thus, the number of hydroxyl groups in compound Q is 4.

Question 29:

Under hydrolytic condition, the compounds used for preparation of linear polymer and for chain termination, respectively, are

Option 1: CH3SiCl3 and Si(CH3)4
Option 2: (CH3)2SiCl2 and (CH3)3SiCl
Option 3: (CH3)2SiCl and CH3SiCl3
Option 4: SiCl4 and (CH3)3 SiCl
Correct Answer: 2

Solutions:Among the given sets of alternatives, the compound used for the preparation of linear polymers is (CH3)2SiCl2. On the other hand, the chain length of the polymer can be controlled by adding (CH3)3SiCl.

Hence, the correct option is (B).

Question 30:

When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE statement(s) regarding this adsorption is(are)

Option 1: O2 is physisorbed
Option 2: heat is released
Option 3: occupancy of π2p of O2 is increased
Option 4: bond length of O2 is increased
Correct Answer: 2, 3, 4

Solutions:Since, electronic transfer takes place when O2 is adsorbed on a metallic surface, thus, it is a process of chemisorption.

Since, chemisorption is exothermic, so heat is released in this process.

O2 accepts the electron from the metal into its vacant

π*2p

orbital. Thus, occupancy of

π*2p

orbital is increased.

Since, the incoming electron enters into an antibonding molecular orbital, as a result the bond order of O2 decreases and bond length increases.

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

Question 31:

One mole of a monoatomic real gas satisfies the equation p(V − b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Option 1:           
Option 2:

Option 3:

Option 4:
Correct Answer: 3

Solutions:A monoatomic real gas follows the equation
p(Vb) = RT

Thus, the van der Waal's constant, a for the gas is 0.

This implies that only repulsive forces are present, which operate only at short distances. Thus, the interatomic potential is never negative and increases abruptly.

Hence, the correct answer is option C.
 

Question 32:

In the following reactions, the product S is

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

Solutions:The given reaction scheme can be completely represented as

Hence, the correct option is (A).
[[VIDEO:13907]]

Question 33:

The major product U in the following reactions is

Option 1:

Option 2:

Option 3:

Option 4:
Correct Answer: 2

Solutions:The given reaction scheme can be completely represented as

Hence, the correct option is (B).
[[VIDEO:13908]]

Question 34:

In the following reactions, the major product W is

Option 1:

Option 2:

Option 3:

Option 4:
Correct Answer: 1

Solutions:The given reaction scheme can be completely represented as

Hence, the correct option is (A).

Question 35:

The correct statement(S) regarding, (i) HClO, (ii) HClO2

The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is
(are)

Option 1: The number of Cl = O bonds in (ii) and (iii) together is two
Option 2: The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
Option 3: The hybridization of Cl in (iv) is sp3
Option 4: Amongst (i) to (iv), the strongest acid is (i)
Correct Answer: 2, 3

Solutions:
 

On the basis of given diagrams, the number of lone pairs of electrons on Cl in (ii) and (iii) together is three and the hybridization of Cl in (iv) is sp3.

Thus, only (B) and (C) are correct.
[[VIDEO:13911]]

Question 36:

The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is(are)

Option 1: Ba2+, Zn2+
Option 2: Bi3+, Fe3+
Option 3: Cu2+, Pb2+
Option 4: Hg2+, Bi3+
Correct Answer: 3, 4

Solutions:Group-II ions:
Hg+2, Pb+2, Bi+3, Cu+2, Cd+2, As+3, Sb+3, Sn+2, and Sn+4

Thus, the pairs of ions given in options (C) and (D) are precipitated upon passing H2S gas in presence of dilute HCl.
[[VIDEO:13909]]

Question 37:

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0 kJ mol−1), this experiment could be used to measure the calorimeter constant.
In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10−5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured.
(Consider heat capacity of all solutions as 4.2 J g−1 K−1 and density of all solutions as 1.0 g mL−1)

Enthalpy of dissociation (in kJ mol−1) of acetic acid obtained from the Expt. 2 is

Option 1: 1.0
Option 2: 10.0
Option 3: 24.5
Option 4: 51.4
Correct Answer: 1

Solutions:

Let b = heat capacity of insulated beaker
Mass of solution in both the experiments = 200 mL × 1 g/mL = 200 g
Total heat capacity=[(200 × 4.2) + b] J/K

In Exp-1,
Moles of acid and base neutralised = 0.1 mol
⇒ Heat released for 0.1 mol = 0.1 × 57 kJ = 5700 J
Hence,

T[(200 × 4.2) + b] = 5700
⇒ 5.7[(200 × 4.2) + b] = 5700
⇒ [(200 × 4.2) + b] = 1000  ……(1)

In Exp-2,
Total heat capacity = [(200 × 4.2) + b] = 1000
Heat released = 5.6 kJ = 5600 J (for 0.1 mol)
So,

Hneutralisation =

56000.1

J/mol = 56 kJ/mol

Hdissociation = (57 − 56) kJ/mol = 1 kJ/mol

Hence, the correct option is A.
[[VIDEO:13910]]

Question 38:

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0 kJ mol−1), this experiment could be used to measure the calorimeter constant.
In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10−5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured.
(Consider heat capacity of all solutions as 4.2 J g−1 K−1 and density of all solutions as 1.0 g mL−1)

The pH of the solution after Expt. 2 is

Option 1: 2.8
Option 2: 4.7
Option 3: 5.0
Option 4: 7.0
Correct Answer: 2

Solutions:

After neutralisation, we have 0.1 mol each of CH3COOH and CH3COONa in the solution.
So, it is a buffer solution.
Hence, pH = pKa + log

Conjugate BaseAcid

               = −log(2 × 10−5) + log1
               = 5 − log2 + 0
               = 4.7
Hence, the correct option is B.
[[VIDEO:13912]]

Question 39:

In the following reactions

Compound X is

Option 1:
Option 2:

Option 3:
Option 4:
Correct Answer: 3

Solutions:

Hence, the correct option is (C).

Question 40:

In the following reactions

The major compound Y is

Option 1:
Option 2:

Option 3:

Option 4:
Correct Answer: 4

Solutions:

Hence, the correct option is D.

Question 41:

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

Correct Answer: 9

Solutions:Let the first term of the given A.P is a and common difference is d.

S7S11=611722a+6d1122a+10d=61172a+6d112a+10d=6112a+6d2a+10d=6714a+42d=12a+60d2a=18da=9dNow, given that seventh term lies between 130 and 140.Therefore, we can denote it as: 130<a7<140130<a+6d<140130<15d<14013015<d<14015d=9. Therefore, the common difference of this A.P. is 9.

[[VIDEO:13801]]

Question 42:

The coefficient of x9 in the expansion of (1+x) (1+x2) (1+x3) …. (1+x100) is

Correct Answer: 8

Solutions:The coefficient of x9 will be same as the number of ways in which the sum of powers of x is 9.

Following are the ways in which the sum of powers of x is 9:

{(0, 9), (1, 8) (2, 7) (3, 6) (4, 5)} 

5 ways
{(1, 2, 6) (1, 3, 5) (2, 3, 4)}

3 ways

Therefore, the coefficient of x9 in the expansion of (1 + x)(1 + x2)(1 + x3) … (1 + x100) is 8.
[[VIDEO:13802]]

Question 43:

Suppose that the foci of the ellipse

x29+y25=1

are (f1, 0) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). If m1 is the slope of T1 and m2 is the slope of T2, then the value of

1m12+m22

is

Correct Answer: 4

Solutions:Equation of the given ellipse is :
 

x29+y25=1     Eccentricity, e = 1b2a2  Therefore, e=159=23

Focii of ellipse given by ± ae, 0=± 3×23,0=± 2, 0    So, F1=2, 0, F2=2, 0

Parabola P1 = y2 = 4ax P1=y2=4×2x y2=8x             ... 1Parabola P2=y2=42f2x = 4(4)xy2=16x         ... 2

Now, Tangent at P1 i.e y2=8x is y=m1 x+2m1Given it passes 4, 0, 0=4 m1+2m1 m12=12Tangent P2 i.e y2=16x is y=m2x+4m2Given it passess 2, 0 0=2m2+4m2m22=2So, 1m12+m22=2+2=4

Hence, the value of 

1m12 + m22

is 4.

Question 44:

Let m and n be two positive integers greater than 1. If

limα0ecosαneαm=e2

then the value of

mn

is

Correct Answer: 2

Solutions:

Given that limα0ecosαneαm=e2 limα0eecosαn11αm=e2Multiply and divide by cosαn1limα0eecosαn11αm.cosαn1.cosαn1=e2limα0eecosαn11cosαn1.2sin2αn2.1αm=e2limα0eecosαn11cosαn1.2sin2αn2αn22.αn22αm=e2 =e2 limα0α2nm =e2Now, limα0α2nm must be equal to 1.2nm=0mn=2

Therefore, the value of 

mn

 is 2.
[[VIDEO:13804]]

Question 45:

If

α=01e9x+3tan1x12+9x21+x2

dx where tan−1x takes only principal values, then the value of

loge1+α3π4

is

Correct Answer: 9

Solutions:

Let t=9x+3 tan1x     dt=9+31+x2    dt=12+9x21+x2 α =09+3π4et dt α=et09+3π4 = e9+3π41 Now, loge1+α3π4 = loge1+e9+3π413π4= 9+3π43π4=9

Therefore, the value of 

loge1+α3π4

is 9.
[[VIDEO:13805]]

Question 46:

Let f :

be a continuous odd function, which vanishes exactly at one point and f (1) =

12

. Suppose that F (x) =

1x

f (t)dt for all x

[−1, 2] and G (x) =

1x

t|f (f(t))|dt for all x

[−1, 2]. If

limx1 F(x)G(x) = 114,

then the value of f

12

is

Correct Answer: 7

Solutions:Given that F(x) = 

1x

f(t)dt â€‹  for all x 

[−1, 2] 
and  G(x) = 

1x

 t|f[f(t)]|dt â€‹ for all x 

[−1, 2]​

Also,  

limx1

 

F(x)G(x) = 114 Limx1 1 xf(t)dt1 xt|ff(t)|dt = f(x)xf(f(x)) = 114

         (using L'Hospital's rule)
                 
Value of 

f12

 is calculated by putting the value of x = 1 in the above equation.
 

f(x)xffx=114121f12=11412f12 = 114   f12 = 7

Therefore, the value of 

f12

 is 7.
[[VIDEO:13806]]

Question 47:

Suppose that

p

,

q

and

r

are three non–coplanar vectors in ℝ3. Let the components of a vector

s

along

p

q

and

r

be 4, 3 and 5, respectively. If the components of this vector

s

along 

p+q+r, pq+r

and 

pq+r

are x, y and z, respectively, then the value of 2x + y + z is

Correct Answer: 9

Solutions:

Given:s=4p+3q+5r                                                 .....   (i)If the components of this vector s along p+q+r, pq+r and pq+r are x, y and z, respectively, thens=p+q+rx+pq+ry+pq+rzs=x+yz p+xyz q+x+y+z r      .....   (ii)Comparing (i) and (ii)x+yz=4                                                      .....  (iii)xyz=3                                                          ..... (iv)x+y+z=5                                                         .....   (v)Solving (iii), (iv) and (v)x=4, y=92, z=72 2x+y+z=24+9272=9

Therefore, the value of 2x + y + z is 9.
[[VIDEO:13807]]

Question 48:

For any integer k, let

ak=cos 7+i sin 7,

where

i=1

.  The value of the expression 

k=112αk+1αkk=13α4k1α4k2

is

Correct Answer: 4

Solutions:

We haveak=coskπ7+isinkπ7=eikπ7 k=112ak+1akk=13a4k1a4k2=k=112eik+1π7eikπ7k=13ei4k1π7ei4k2π7                               =k=112eikπ7·eiπ71k=13ei4k1π7·eiπ71                               =k=112eikπ7k=13ei4k1π7             eiθ=1                               =123=4

Therefore, the value of the expression 

k=112αk+1αkk=13α4k1α4k2

is 4.

Question 49:

Let f, g : [−1, 2] → ℝ be continuous functions which are twice differentiable on the interval (−1, 2). Let the values of f and g at the points −1, 0 and 2 be as given in the following table :

  x = −1 x = 0 x = 2
f(x) 3 6 0
g(x) 0 1 −1

In each of the intervals (−1, 0 and (0, 2) the function (f − 3g)" never vanishes. Then the correct statement(s) is (are)

Option 1: f'(x) − 3g'(x) = 0 has exactly three solutions in (−1, 0) ∪ (0, 2)
Option 2: f'(x) − 3g'(x) = 0 has exactly one solution in (−1, 0)
Option 3: f'(x) − 3g'(x) = 0 has exactly one solution in (0,2)
Option 4: f'(x) − 3g'(x) = 0 has exactly two solutions in (−1, 0) and exactly two solutions in (0, 2)
Correct Answer: 2, 3

Solutions:Let h(x) = f(x) − 3g(x)

h(−1) = 3, h(0) = 3 and h(2) = 3

⇒h'(x) = f'(x) − 3g'(x) = 0 has at least one root in (–1, 0) and at least one root in (0, 2).

But its given that h''(x) = 0 has no root in (–1, 0) and (0, 2).

Therefore, h'(x) = 0 has exactly one root in (–1, 0) and exactly one root in (0, 2).

Hence, the correct options are B and C.
[[VIDEO:13808]]

Question 50:

Let f(x) = 7tan8x + 7tan6x − 3tan4x − 3tan2x for all x  ∈

π2, π2

. Then the correct expression (s) is (are)

Option 1:  

0π/4 x f x dx 112
Option 2:  

0π/4 f x dx = 0
Option 3:  

0π/4 x f x dx 16
Option 4: 0π/4f x dx = 1
Correct Answer: 1, 2

Solutions:

f(x)=7tan8x+7tan6x3tan4x3tan2xf(x)=7tan6x1+tan2x3tan2x1+tan2xf(x)=7tan6x sec2x3tan2x sec2x 0π4fxdx =0π47tan6x3tan2xsec2x dx0π4fxdx= tan7xtan3x0π4=0                      Taking tanx=t 0π4xfxdx=0π4xsec2x7tan6x3tan2x dx                            =xtan7xtan3x0π4 0π4tan7xtan3x1 dx                =xtan7xtan3x0π4+0π4tan3x1tan4x dx                =xtan7xtan3x0π4+ 0π4tan3x1tan2xsec2x dx                =xtan7xtan3x0π4+ 0π4tan3xtan5xsec2x dx                =xtan7xtan3x0π4+tan4x4tan6x6 0π4                =0+1416=112

Hence, the correct options are A and B.
[[VIDEO:13809]]

Question 51:

Let

f'x=192x32+sin4πx

for all x ∈ ℝ with

f12=0. If mt/21 fx dxM

, then the possible values of m and M are

Option 1: m = 13, M = 24
Option 2: m=14, M=12
Option 3: m = −11, M = 0
Option 4: m = 1, M = 12
Correct Answer: 4

Solutions:Given that 

f'x=192x32+sin4πx  xR

 
Now, 

f'x=192x32+sin4πx>0For x = 12f'12 = 192 1232 + sin4(π×12) = 242+1=8and for x = 1f'1 = 192 132 + sin4(π×1) =1922 + 0 = 96So, as x increase from 12to 1,f'x increase from 8 to 96.8<f1f1/2112<964<f1<4812×12×41/21fxdx12×12×4811/21fx dx12 m=1 and M=12Hence, the correct option is D.

Question 52:

Let S be the set of all non-zero real number α such that the quadratic equation ax2x + α = 0 has two distinct real roots x1 and x2 satisfying the inequality |x1x2| < 1. Which of the following intervals is (are) a subset(s) of S?

Option 1: 12, 15
Option 2: 15, 0
Option 3: 0, 15
Option 4: 15, 12
Correct Answer: 1, 4

Solutions:We have

α

x2x

α

= 0

Since the roots of the quadratic equation are real, so

D > 0

⇒ 1 − 4.

α

.

α

> 0

14α2>0α214<012<α<12            .....1Sum of roots, x1+x2=1αProduct of roots, x1 x2=1 x1x2=x1+x224 x1 x2           =1α24 x1x2<11α24<11α24<11α2<5 15<α2α2152>0α<15 or α>15                 .....2From 1 and 2, we getα 12, 1515, 12

Hence, the correct options are A and D.

Question 53:

If

α=3 sin1 611 and β=3 cos1 49

, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)

Option 1: cos β > 0
Option 2: sin β > 0
Option 3: cos (α + β) > 0
Option 4: cos α < 0
Correct Answer: 2, 3, 4

Solutions:

We haveα=3sin1611  12<611π6<sin1611π2<3sin1611π2<α Also,β=3cos149  49<12cos149>cos112cos149>π33cos149>πβ>π

∴ cosβ < 0,  sinβ < 0, cosα < 0,  cosβ < 0 and sinα > 0

As, cos(α + β) = cosαcosβ − sinαsinβ

∴ cos(α + β) > 0

Hence, the correct options are B, C and D.
[[VIDEO:13810]]

Question 54:

Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y − 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ = PR

=223

. If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct expression(s) is (are)

Option 1: e12+e22=4340
Option 2: e1 e2=7210
Option 3: e12e22=58
Option 4: e1 e2=34
Correct Answer: 1, 2

Solutions:

Let the equation of the ellipse E1 be x2a2+y2b2=1             a>b Let the equation of the ellipse E2 bex2a2+y2b2=1              a>bEquation of the circle, S: x2+y12=2Tangent to the circle and ellipse E1 and E2, x+y=3                    ..... 1Normal to the circle S: xy=KThe tangent passes through 0, 1.K=1Normal to circle: xy=k                                                     ..... 2For point Q and R,x112=y212=±223  The cordinate of Q is 53, 43 and the cordinate of R is 13, 83Also, the normal to the equation y=x+3 and x2a2+y2b2=1 isc2=a2 m2+b29=a2+b2                                                          ..... 3As point Q53, 43 lies E1: x2a2+y2b2=1 259a2+169b2=1 a2=5 From 3, b2=4e1=1b2a2=145=15Similarly,e2=118=722 e1 e2=7210e12+e22=15+78=8+3540=4340e12e22=1578=83540=2740

Hence, the correct options are A and B.

Question 55:

Consider the hyperbola H : x2y2 = 1 and a circle X with center N(x2, 0). Suppose that H and S touch each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the x-axis at point M. If (ℓ, m) is the centroid of the triangle ∆PMN, then the correct expression(s) is(are)

Option 1: ddx1=113x12for x1>1
Option 2: dmdx1=x13x121for x1>1
Option 3: ddx1=1+13x12for x1>1
Option 4: dmdy1=13 for y1>0
Correct Answer: 1, 2, 4

Solutions:

x2y2=1Differentiating both sides with respect to x, we get2x2yy'=0y'=xy                                .....1x1y1×y1x1x2=1 x2=2x1                          .....2Equation of PM isyy1xx1=x1y1Putting y=0, we getx=x12y12x1=1x1 The coordinates of M are 1x1, 0.        The coordinates of N are 2x1, 0).

=3x1+1x13=x1+13x1ddx1=113x12m=y13dmdy1=13Also, m=13y1=13x121dmdx1=x13x121

Hence, the correct options are A, B and D.
[[VIDEO:13811]]

Question 56:

The option(s) with the values of a and L that satisfy the following equation is(are)

04πet sin6 at+cos4 at dt0πet sin6 at+cos4 at dt=L?

Option 1: a=2, L=e4π1eπ1
Option 2: a=2, L=e4π+1eπ+1
Option 3: a=4, L=e4π1eπ1
Option 4: a=4, L=e4π+1eπ+1
Correct Answer: 1, 3

Solutions:We have

L=04πetsin6at+cos4atdt0πetsin6at+cos4atdt

Let f(t) = et[sin6(at) + cos6(at)]

f(kπ + t) = ekπ+t{sin6[a(kπ + t)] + cos6[a(kπ + t))]}

⇒ f(kπ + t) = ekπf(t)             (for even values of a)

 L=04πftdt0πftdt=1+eπ+e2π+e3π0πftdt0πftdt=e4π1eπ1

Hence, the correct options are A and C.
[[VIDEO:13812]]

Question 57:

The Correct statement(s) is(are)

Option 1: f'(1) < 0
Option 2:

f(2) < 0

Option 3: f'(x) ≠ 0 for any x &epsis; (1, 3)
Option 4: f'(x) ≠ 0 for some x &epsis; (1, 3)
Correct Answer: 1, 2, 3

Solutions:

f(x) = xF(x)

⇒ f '(x) = F(x) + xF '(x)

f '(1) = F(1) + F '(1) = F '(1) < 0                         

F'x<0  x12,3

Now,

f(2) = 2F(2) < 0         (∵ F(2) < 0)

Also,

f '(x) = F(x) + xF '(x) < 0 ∀ x ∈ (1, 3)                 [∵ F(x) < 0 and xF '(x) < 0 ∀ x ∈ (1, 3)]
 
f '(x) ≠ 0 for any x ∈ (1, 3)
 

Hence, the correct options are A, B and C.
[[VIDEO:13813]]

Question 58:

If

31x2F'x dx =  12 and 31x3F''x dx = 40

, then the correct expression(s) is (are)

Option 1: 9f'(3) + f'(1) − 32 = 0
Option 2: 31fx dx =12 
Option 3: 9f'(3) + f'(1) + 32 = 0
Option 4: 31fx dx = 12 
Correct Answer: 3, 4

Solutions:

13fxdx=13xFxdx=Fxx221313x22F'xdx=92 401212=1213x3F''x=40

x3F'x13133x2F'x=40 27F'3F'1312=4027F'3F'1=4Now,  f'3=F3+3F'3 andf'1=F'1Also,9 f'3f'1+32=9F3+27F'3F'1+32=27F'3F'14=0                        F3=4

Hence, the correct options are C and D.

Question 59:

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is

13

, then the correct option(s) with the possible values of n1, n2, n3 and n4 is (are)

Option 1: n1 = 3, n2 = 3, n3 = 5, n4 = 15
Option 2: n1 = 3, n2 = 6, n3 = 10, n4 = 50
Option 3: n1 = 8, n2 = 6, n3 = 5, n4 = 20
Option 4: n1 = 6, n2 = 12, n3 = 5, n4 = 20
Correct Answer: 1, 2

Solutions:Let B1 be the event of selecting the bag I, B2 be the event of selecting the bag II and R be the event of drawing a red ball.

Then,

PB1=PB2=12

Also,

PRB1=n1n1+n2 and  PRB2=n3n3+n4  PB2R=1312×n3n3+n412n3n3+n4 +12n1n1+n2=13n3n1+n22n3n1+n2n3+n1n4=13

The values of options A and B are satisfying the equation.

Hence, the correct options are A and B.
[[VIDEO:13814]]

Question 60:

A ball is drawn at random from box I and transferred to box. II. If the probability of drawing a red ball from box I, after this transfer, is

13,

then the correct option(s) with the possible values of n1 and n2 is (are)

Option 1: n1 = 4 and n2 = 6
Option 2: n1 = 2 and n2 = 3
Option 3: n1 = 10 and n2 = 20
Option 4: n1 = 3 and n2 = 6
Correct Answer: 3, 4

Solutions:Let n1 and n2  be the number of red and black balls respectively in the box I.
Let n3 and n4  be the number of red and black balls respectively in the box II.

There are 2 possibilities of transferring balls. Either red ball is transferred or black ball is transferred.

Probability of transferring red ball =

n1n1+n2

When a red ball is transferred then the number of remaining red balls in box I = n1 − 1

Probability of drawing red ball after a red ball has been transferred =

n11n1+n21

Probability of transferring red ball and then drawing a red ball= 

n1n1+n2 n11n1+n21

Probability of transferring a black ball =

n2n1+n2

When a red ball is transferred then the number of remaining black balls in box II = n2 − 1

Probability of drawing red ball when a black ball has been transferred =

n1n1+n21

Probability of transferring a black ball and then drawing a red ball= 

n2n1+n2 n1n1+n21 n1n1+n2 n1  1n1+n2  1 + n2n1+n2n1n1+n2  1 = 13

Among the given options, (C) and (D) satisfies the equation.
Hence, the correct options are C and D.
[[VIDEO:13815]]

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