# 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

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

Dividing both side by equation (1), we get

[[VIDEO:13867]]

#### Question 2:

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

, 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

$\frac{{I}_{\mathrm{B}}}{{I}_{\mathrm{A}}}=\frac{n}{10}$

, the value of n is

Solutions:

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

$\delta x$

.  Its volume (dV) will be

dV = 4πx2

$\delta x$

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

dm = 4πx2dx ρ(

$\delta x$

)

Moment of inertia (dI) is given by

$\delta I=\frac{2}{3}\left(\delta m\right){x}^{2}\phantom{\rule{0ex}{0ex}}⇒\delta I=\frac{2}{3}4\mathrm{\pi \rho }\left(x\right).{x}^{4}\mathit{.}\delta x\phantom{\rule{0ex}{0ex}}$

[[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

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

${A}_{r}^{2}={A}^{2}+{A}^{2}+2AA\mathrm{cos}60°\phantom{\rule{0ex}{0ex}}⇒{A}_{r}^{2}=3{A}^{2}$

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

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

, 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

$\frac{{R}_{\mathrm{P}}}{{R}_{\mathrm{Q}}}=\frac{\mathit{n}}{\mathit{e}}$

, then the value of n is

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

${A}_{\mathrm{P}}={\lambda }_{\mathrm{P}}{N}_{P}{e}^{–{\lambda }_{P}t}$

The activity of source Q (AQ) is given by

${A}_{\mathrm{Q}}={\lambda }_{Q}{N}_{Q}{e}^{–{\lambda }_{Q}t}$

We have given

AP = AQ at t = 0

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

the value of θ is 60° and

$\frac{d\theta }{dn}=m$

. The value of m is

Solutions:

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

Thus, the value of m is 1.

#### Question 6:

In the following circuit, the current through the resistor

is I Amperes. The value of I is

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

#### 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

Solutions:According to the Bohr's postulates

Radius of nth shell is given by

From the de-Broglie's hypothesis

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

$m=k\left(\frac{M}{288}\right)·$

The value of k is

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

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

Comparing equation (1) and (2) we have

$\frac{GM}{9{l}^{2}}–\frac{Gm}{{l}^{2}}=\frac{GM}{16{l}^{2}}+\frac{Gm}{{l}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{M}{9}–\frac{M}{16}=m+m\phantom{\rule{0ex}{0ex}}⇒\frac{7M}{144}=2m\phantom{\rule{0ex}{0ex}}⇒m=\frac{7M}{288}=k\left(\frac{M}{288}\right)\phantom{\rule{0ex}{0ex}}⇒k=7$

#### 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

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

$\frac{{C}_{2}}{{C}_{1}}$

is

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

Solutions:We know that,

${C}_{1}=\frac{{\mathrm{\epsilon }}_{0}A}{d}$

From figure, we have

The equivalent capacitance C2 is thus given by

${C}_{2}=\frac{CC\text{'}}{C+C\text{'}}+C\text{'}\text{'}\phantom{\rule{0ex}{0ex}}⇒{C}_{2}=\frac{4{\epsilon }_{0}s}{3d}+\frac{{\epsilon }_{0}s}{d}=\frac{7{\epsilon }_{0}s}{3d}\phantom{\rule{0ex}{0ex}}\therefore \frac{{C}_{2}}{{C}_{1}}=\frac{7}{3}$

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 $\frac{1}{4}{P}_{1}{V}_{1}$ 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 $\frac{7}{3}{P}_{1}{V}_{1}$ Option 4: If V2 = 3 V1 and T2 = 4T1, then the heat supplied to the gas is $\frac{17}{6}{P}_{1}{V}_{1}$

Solutions:

#### Question 11:

A fission reaction is given by

${}_{92}{}^{236}U\to {}_{54}{}^{140}Xe+{}_{38}{}^{94}Sr+\mathrm{x}+\mathrm{y},$

where x and y are two particles. Considering

${}_{92}{}^{236}U$

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

and

${}_{38}{}^{94}Sr$

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

Solutions:As x and y share equal energy = 4 MeV;
â€‹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,

$219–4–4$

= 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

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,

$\mathrm{Strain}=\frac{1}{\gamma }×\mathrm{stress}$

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: Option 3: $\frac{P\left(r=3R/5}{P\left(r=2R/5}=\frac{16}{21}$ Option 4: $\frac{P\left(r=R/2\right)}{P\left(r=R/3\right)}=\frac{20}{27}$

Solutions:

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

We will find,

and

$\frac{P\left(r=3R/5\right)}{P\left(r=2R/5\right)}=\frac{16}{21}$

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

${\stackrel{\to }{V}}_{p}$

and Q alone in L1 has terminal velocity

${\stackrel{\to }{V}}_{Q}$

, then

 Option 1: $\frac{|{\stackrel{\to }{V}}_{P}|}{{\stackrel{\to }{|V}}_{Q}|}=\frac{{\eta }_{1}}{{\eta }_{2}}\phantom{\rule{0ex}{0ex}}$ Option 2: Option 3: Option 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

We know that, the terminal velocity is given by

$V=\frac{2}{9}\frac{{r}^{2}\left(\sigma –\rho \right)g}{\eta }$

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 Æ0, permeability µ0 and speed of light c, the dimensionally correct equation(s) is(are)

 Option 1: µ0I2 = Æ0V2 Option 2: Æ0I = µ0V Option 3: I= Æ0cV Option 4: µ0cI = Æ0V

Solutions:Dimensions of I = T-1Q
â€‹Dimensions of V = MLT-2Q-1
Dimensions of µ= MLQ-2
Dimensions ofâ€‹ Æ= M-1L-3T2Q2
Dimensions ofâ€‹ â€‹c = LT-2
â€‹
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

then which of the following is the correct statement?

 Option 1: $\stackrel{\to }{\mathrm{E}}$ is uniform, its magnitude is independent of R2 but its direction depends on $\stackrel{\to }{\mathrm{r}}$ . Option 2: $\stackrel{\to }{\mathrm{E}}$ is uniform, its magnitude depends on R2 and its direction depends on $\stackrel{\to }{\mathrm{r}}$ . Option 3: $\stackrel{\to }{\mathrm{E}}$ is uniform, its magnitude is independent of $\mathrm{\alpha }$ but its direction depends on $\stackrel{\to }{\mathrm{\alpha }}$ . Option 4: $\stackrel{\to }{\mathrm{E}}$ is uniform and both its magnitude and direction depend on $\stackrel{\to }{\mathrm{\alpha }}$ .

Solutions:

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

${\mathrm{n}}_{1}=\sqrt{45}/4$

and

${\mathrm{n}}_{2}=3/2,$

and S2 with

${\mathrm{n}}_{1}=8/5$

and

${\mathrm{n}}_{2}=7/5$

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 $\frac{16}{3\sqrt{15}}.$ Option 2: NA of S1 immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ 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 $\frac{4}{\sqrt{15}}.$ Option 4: NA of S1 placed in air is the same as that of S2 placed in water.

Solutions:

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: $\frac{{\mathrm{NA}}_{1}{\mathrm{NA}}_{2}}{{\mathrm{NA}}_{1}+{\mathrm{NA}}_{2}}$ Option 2: NA1 + NA2 Option 3: NA1 Option 4: NA2

Solutions:The numerical aperture of the combination is given by

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 $\stackrel{\to }{\mathrm{B}}$ 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

Solutions:Force experienced by charge carriers will beâ€‹ given by

When w1 = w2 and d1 =2d2

When when w1 = 2w2 and d1 = d2

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 $\stackrel{\to }{\mathrm{B}}$ 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

Solutions:

Similarly,

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

#### Question 21:

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

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

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

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

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:

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

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

Solutions:Since,

We can say that

Also,

For HX,

${\lambda }_{m}\left(\mathrm{HX}\right)={\lambda }_{m}^{°}{\alpha }_{\mathrm{HX}}$

For HY,

${\lambda }_{m}\left(\mathrm{HY}\right)={\lambda }_{m}^{°}{\alpha }_{\mathrm{HY}}$

#### Question 26:

A closed vessel with rigid walls contains 1 mol of

${}_{92}{}^{238}\mathrm{U}$

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

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

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

Decomposition of uranium-238 is shown below:

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.

#### Question 27:

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

${\mathrm{MnO}}_{4}^{–}$

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

$\left[{\mathrm{MnO}}_{4}^{–}\right]$

is

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

${\mathrm{MnO}}_{4}^{–}$

is given as:

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

${\mathrm{MnO}}_{4}^{–}$

can be calculated as follows:

#### Question 28:

The number of hydroxyl group(s) in Q is

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

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

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

${\mathrm{\pi }}^{*}2\mathrm{p}$

orbital. Thus, occupancy of

${\mathrm{\pi }}^{*}2\mathrm{p}$

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:

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:

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:

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:

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)

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+

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 Option 2: 10 Option 3: 24.5 Option 4: 51.4

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 =

$\frac{–5600}{0.1}$

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 Option 4: 7

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

= −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:

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:

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

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

[[VIDEO:13801]]

#### Question 42:

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

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)}

$\to$

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

$\to$

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

$\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{5}=1$

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

$\left(\frac{1}{{\mathrm{m}}_{1}^{2}}+{\mathrm{m}}_{2}^{2}\right)$

is

Solutions:Equation of the given ellipse is :

Hence, the value of

is 4.

#### Question 44:

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

$\underset{\mathrm{\alpha }\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{e}}^{\mathrm{cos}\left({\mathrm{\alpha }}^{\mathrm{n}}\right)}–\mathrm{e}}{{\mathrm{\alpha }}^{\mathrm{m}}}\right)=–\left(\frac{\mathrm{e}}{2}\right)$

then the value of

$\frac{\mathrm{m}}{\mathrm{n}}$

is

Solutions:

Therefore, the value of

$\frac{m}{n}$

is 2.
[[VIDEO:13804]]

#### Question 45:

If

$\alpha =\underset{0}{\overset{1}{\int }}\left({e}^{9x+3{\mathrm{tan}}^{–1}}x\right)\left(\frac{12+9{x}^{2}}{1+{x}^{2}}\right)$

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

$\left({\mathrm{log}}_{e}\left|1+\alpha \right|–\frac{3\mathrm{\pi }}{4}\right)$

is

Solutions:

Therefore, the value of

$\left({\mathrm{log}}_{e}\left|1+\alpha \right|–\frac{3\mathrm{\pi }}{4}\right)$

is 9.
[[VIDEO:13805]]

#### Question 46:

Let f :

$\mathrm{ℝ}\to \mathrm{ℝ}$

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

$\frac{1}{2}$

. Suppose that F (x) =

${\int }_{–1}^{x}$

f (t)dt for all x

$\in$

[−1, 2] and G (x) =

${\int }_{–1}^{x}$

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

$\in$

[−1, 2]. If

$\underset{x\to 1}{\mathrm{lim}}$

then the value of f

$\left(\frac{1}{2}\right)$

is

Solutions:Given that F(x) =

${\int }_{–1}^{x}$

f(t)dt â€‹  for all x

$\in$

[−1, 2]
and  G(x) =

${\int }_{–1}^{x}$

t|f[f(t)]|dt â€‹ for all x

$\in$

[−1, 2]â€‹

Also,

$\underset{x\to 1}{\mathrm{lim}}$

$\underset{x\to 1}{⇒\mathrm{Lim}}$

(using L'Hospital's rule)

Value of

$f\left(\frac{1}{2}\right)$

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

Therefore, the value of

$f\left(\frac{1}{2}\right)$

is 7.
[[VIDEO:13806]]

#### Question 47:

Suppose that

$\stackrel{\to }{\mathrm{p}}$

,

$\stackrel{\to }{\mathrm{q}}$

and

$\stackrel{\to }{\mathrm{r}}$

are three non–coplanar vectors in â„3. Let the components of a vector

$\stackrel{\to }{\mathrm{s}}$

along

$\stackrel{\to }{\mathrm{p}}$

$\stackrel{\to }{\mathrm{q}}$

and

$\stackrel{\to }{\mathrm{r}}$

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

$\stackrel{\to }{\mathrm{s}}$

along

and

$\left(–\stackrel{\to }{\mathrm{p}}–\stackrel{\to }{\mathrm{q}}+\stackrel{\to }{\mathrm{r}}\right)$

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

Solutions:

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

#### Question 48:

For any integer k, let

where

$\mathrm{i}=\sqrt{–1}$

.  The value of the expression

$\frac{\sum _{\mathrm{k}=1}^{12}\left|{\mathrm{\alpha }}_{\mathrm{k}+1}–{\mathrm{\alpha }}_{\mathrm{k}}\right|}{\sum _{\mathrm{k}=1}^{3}\left|{\mathrm{\alpha }}_{4\mathrm{k}–1}–{\mathrm{\alpha }}_{4\mathrm{k}–2}\right|}$

is

Solutions:

Therefore, the value of the expression

$\frac{\sum _{k=1}^{12}\left|{\mathrm{\alpha }}_{k+1}–{\mathrm{\alpha }}_{k}\right|}{\sum _{k=1}^{3}\left|{\mathrm{\alpha }}_{4k–1}–{\mathrm{\alpha }}_{4k–2}\right|}$

is 4.

#### Question 49:

Let f, g : [−1, 2] → â„ 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)

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 â€ˆ∈

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

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

Solutions:

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

#### Question 51:

Let

$f\text{'}\left(x\right)=\frac{192{x}^{3}}{2+{\mathrm{sin}}^{4}\mathrm{\pi }x}$

for all x ∈ â„ with

, then the possible values of m and M are

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

Solutions:Given that

Now,

#### 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: Option 2: Option 3: Option 4:

Solutions:We have

$\alpha$

x2x

$\alpha$

= 0

Since the roots of the quadratic equation are real, so

D > 0

⇒ 1 − 4.

$\alpha$

.

$\alpha$

> 0

Hence, the correct options are A and D.

#### Question 53:

If

, 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

Solutions:

∴ 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

$=\frac{2\sqrt{2}}{3}$

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

 Option 1: ${e}_{1}^{2}+{e}_{2}^{2}=\frac{43}{40}$ Option 2: Option 3: $\left|{e}_{1}^{2}–{e}_{2}^{2}\right|=\frac{5}{8}$ Option 4:

Solutions:

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 (â„“, m) is the centroid of the triangle âˆ†PMN, then the correct expression(s) is(are)

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

Solutions:

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)

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

Solutions:We have

$L=\frac{\underset{0}{\overset{4\mathrm{\pi }}{\int }}{e}^{t}\left[{\mathrm{sin}}^{6}\left(at\right)+{\mathrm{cos}}^{4}\left(at\right)\right]dt}{\underset{0}{\overset{\mathrm{\pi }}{\int }}{\mathrm{e}}^{t}\left[{\mathrm{sin}}^{6}\left(at\right)+{\mathrm{cos}}^{4}\left(at\right)\right]dt}$

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)

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)

Solutions:

f(x) = xF(x)

⇒ f '(x) = F(x) + xF '(x)

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

Now,

f(2) = 2F(2) < 0         (âˆµ F(2) < 0)

Also,

f '(x) = F(x) + xF '(x) < 0 ∀ x ∈ (1, 3)                 [âˆµ 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

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

 Option 1: 9f'(3) + f'(1) − 32 = 0 Option 2: Option 3: 9f'(3) + f'(1) + 32 = 0 Option 4:

Solutions:

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

$\frac{1}{3}$

, 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

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,

$\mathrm{P}\left({B}_{1}\right)=\mathrm{P}\left({B}_{2}\right)=\frac{1}{2}$

Also,

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

$\frac{1}{3},\phantom{\rule{0ex}{0ex}}$

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

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 =

$\frac{{n}_{1}}{{n}_{1}+{n}_{2}}$

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 =

$\frac{{n}_{1}–1}{{n}_{1}+{n}_{2}–1}$

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

$\frac{{n}_{1}}{{n}_{1}+{n}_{2}}$ $\frac{{n}_{1}–1}{{n}_{1}+{n}_{2}–1}$

Probability of transferring a black ball =

$\frac{{n}_{2}}{{n}_{1}+{n}_{2}}$

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 =

$\frac{{n}_{1}}{{n}_{1}+{n}_{2}–1}$

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

$\frac{{n}_{2}}{{n}_{1}+{n}_{2}}$ $\frac{{n}_{1}}{{n}_{1}+{n}_{2}–1}$

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