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JEE Advanced Practice Test- 16 - JEE MCQ


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54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 16

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*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 1

Three identical coils A, B and C are placed coaxially as shown with their plane parallel to one another. Coil B is fixed in position. Coil A is moved towards B with constant speed of 3 m/s and coil C is moved away from B with constant speed of 5 m/s. Then, the

Detailed Solution for JEE Advanced Practice Test- 16 - Question 1
The coil A increases the flux in the coil B therefore the direction of induced current in coil B is opposite to the direction of the current in A .The coil C decreases the flux in coil B therefore the direction of the induced current in B is same as the direction of the current in C.Rate of increase in the magnetic flux due to coil A is less compared to the rate of decrease in the magnetic flux due to the coil C. Due to the movement of the coil A and C direction of the induced current in coil B is anticlockwise.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 2

If a particle is projected at an angle 'α' with the horizontal from the foot of a plane, whose inclination to the horizontal is 'β', then it will strike the plane at right angle

Detailed Solution for JEE Advanced Practice Test- 16 - Question 2

If a projectile hits at Q at an angle of 90º to PQ, the velocity parallel to PQ is zero.

If 'T' be the time of flight from P to Q, component of the initial velocity of the projectile along the plane PQ is u cos (α - β), the component of the final velocity along the PQ plane = 0 and the component of the acceleration due to gravity = -g sin β

Hence, 0 = u cos (α - β) - (g sin β)T = 0

By definition of time of flight, we have

Hence,

or,

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 3

A cubical box of copper 6 cm on each side is floating on mercury in a beaker. The portion of the box is inside the mercury. Now, water is poured in to the beaker so that it just covers the box. Density of copper = 5.6 gm/cc, density of mercury = 13.6 gm/cc. Then,

Detailed Solution for JEE Advanced Practice Test- 16 - Question 3

As the upthrust acting on the cube = Weight of cube

Upthrust = Weight of the mercury displaced

Hence, Vin ρHgg = Vcu ρcug

Vin ρHg = Vcu ρcu

Or,

or,

, where Hin is the height of the box inside and HCu is the height of the box given

When water is poured in the beaker, then up-thrust required for the floatation of the box is provided by both mercury and water.

VCu ρCu = Vin' ρHg + (VCu - Vin') ρW

Or, HCu ρCu = Hin' ρHg + (HCu - Hin') ρW

Or,

6×5.6 = Hin'×13.6 + (6 - Hin' )×1

On solving,

We get Hin' = 2.19 cm

i.e. Now, 2.19 cm of the box will be inside mercury and the remaining portion of height 6 - 2.19 = 3.81 cm will be in water

So, height of water column poured is 3.81 cm.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 4

Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Detailed Solution for JEE Advanced Practice Test- 16 - Question 4

F = upthrust = VdFg

Equilibrium of A

At dA < />F​B dB > dFD dA + dB = 2dF​ for sphere A to be at the top in liquid, the density of sphere A should be less than the density of liquid and for sphere B to be at the bottom in liquid, the density of sphere B should be greater than the density of liquid. The FBD of the top sphere gives us VdF g = T + V dAg and of the bottom sphere gives us T + V dF g = V dB g.

Substituting T from the first equation in the second, we get

2V dF g = VdA g + V dBg or 2dF = dA + dB

Therefore, options 1, 2 and 4 are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 5

A composite block is made of slabs A,B,C,D and E of different thermal conductivities ( given in terms of a constant K ) and size ( given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state

Detailed Solution for JEE Advanced Practice Test- 16 - Question 5

Thermal resistance R = ℓ / KA

(Here w = width )

=

RA : RB : RC : RD : RE

15 : 160 : 60 : 96 : 12

So, let us write, RA = 15 R, RB = 160 R etc and draw a simple electrical circuit as shown in figure

H = Heat current = Rate of heat flow.

HA = HE = H

∴ Heat flow through A and E slabs are same, is correct.

In parallel current distributes in inverse ratio of resistance.

=

= 9 : 24 : 15

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 6

Refractive index of an equilateral prism is √2

Detailed Solution for JEE Advanced Practice Test- 16 - Question 6

From

We can see that δm = 30o for μ = √2 and A = 60o

Further, at minimum deviation

∴ sin i1 = μ sin r1

= (√2) sin 30o

= 1/√2

∴ i1 = 45o

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 7

There are three optical media, 1,2 and 3 with their refractive indices μ1 > μ2 > μ3 (TIR-total internal reflection)

Detailed Solution for JEE Advanced Practice Test- 16 - Question 7
Total internal reflection takes place when ray of light travels from denser to rarer medium

Further,

Since,

θ12 > θ13

Smaller the value of critical angle more the chance of total internal reflection.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 8

A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is

Detailed Solution for JEE Advanced Practice Test- 16 - Question 8

Least count of vernier callipers

LC = 1MSD − 1VSD

= Smallest division on main scale / Number of divisions on vernier scale

20 divisions of vernier scale = 16 divisions of main scale

∴ 1VSD = 16 / 20mm = 0.8mm

LC = 1MSD − 1VSD

= 1mm − 0.8mm = 0.2mm

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 9

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dotted lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the ratio wd/ws is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 16 - Question 9

Solid line path work done (ws) is isothermal because PV is constant (Boyle's law) and dotted line (horizontal) path work done wd is isobaric. Work done in vertical line is zero as ΔV = 0.

Total work done on solid line path (ws) = 2.303 nRT log V2/V1

= 2.303 PV log V2/V1 = 2.303 x 4 x 0.5 log 5.5/0.5 = 4.8 L atm.

Total work done on dotted line path (wd) = PΔV

= 4 x (2 - 0.5) + 1(3 - 2) + 0.5 (5.5 - 3) = 6 + 1 + 1.25 = 8.25

So, ≈ 1.72.

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 10

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nA/nB, where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]

(Round off upto 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 16 - Question 10

Excess pressure above atmospheric pressure, due to surface tension in a bubble = 4T/r

Surrounding pressure, P0 = 8 N/m2

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 11

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 mm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that by B?


Detailed Solution for JEE Advanced Practice Test- 16 - Question 11

According to Wien's displacement law,

According to Stefan Boltzmann law, rate of energy radiated by a black body is

[Here, A = 4πR2]

(Using (i))

= 9

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 12

In the figure shown below, the maximum possible unknown resistance X (in Ω), that can be measured by the post office box is given by a × 10b Ω in scientific notation, then find the value of a−b?

(in this experiment, we take out only one plug in arm AB and only one plug in arm BC, but in arm AD we can take out many plugs).


Detailed Solution for JEE Advanced Practice Test- 16 - Question 12

By Wheatstone bridge principle, we have P/Q = R/X

or X = RQ / P

Now

Xmax = 9000 × 1000 / 10

= 9 × 105

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 13

A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.


Detailed Solution for JEE Advanced Practice Test- 16 - Question 13

During heating, the process was isobaric, so, ΔW1 = PΔV

And internal energy, ΔU1 = nCvΔT = 50nCv

Thus, heat transfer, ΔQ1 = 50nCv + 55

During cooling, the process is isochoric, so, ΔW2 = 0 and ΔU2 = −50nCv

Thus, heat loss = ΔQ2 = −(ΔU2 + ΔW2) = 50nCv

∴ Heat supplied −- Heat rejected = 50nCv + 55 − 50nCv = 55 J

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 14

Consider a frictionless plane of height h = 2 m m and angle α=30°. One end of an electric string is connected to point B, the other end is passed through an orifice at point D on the plane and connected to a body of mass m = 2 kg initially stationary at point A. This body comes to rest at point C at the bottom of the inclined plane. The line drawn from point B to point D is perpendicular to side AC. Knowing that the length of the string is unstretched state is equal to the length of segment BD. Determine the elastic constant in N m−1. Take g = 10 m s−2


Detailed Solution for JEE Advanced Practice Test- 16 - Question 14
Using law of conservation of energy

JEE Advanced Practice Test- 16 - Question 15

Answer the following by appropriately matching the lists based on the information given:

List I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 - S2P1 = λ /4 and S1P2 - S2P2 = λ/3, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ (P) and the intensity by I(P). Match each situation given in List I with the statement(s) in List II valid for that situation.

Detailed Solution for JEE Advanced Practice Test- 16 - Question 15
(A) → (p, s) Intensity at P0 is maximum. It will continuously decrease from P0 toward P2.

(B) → (q) Path difference due to slap will be compensated by geometrical path difference. Hence, δ(P1) = 0.

JEE Advanced Practice Test- 16 - Question 16

Answer the following by appropriately matching the lists based on the information given:

Two transparent media of refractive indices μ1 and μ3 have a solid lens shaped transparent material of refractive index μ2 between them as shown in figures in List II. A ray traversing these media is also shown in the figures. In List I different relationship betweenμ1 , μ2 and μ3 are given. Match them to the ray diagrams shown in List II.

Detailed Solution for JEE Advanced Practice Test- 16 - Question 16
In the above equation, as sine is an increasing function in 0 to π/2 , if μ2 > μ1, i > r and if μ2 < />1, i < />

Thus light rays going from rarer medium to denser medium will bend towards the normal and vice versa.

It can be seen from the image how the light rays bend when they refract at a curved surface. (Line along the radius is the normal here as shown in the figure)

If the ray goes undeviated, then the refractive indexes are same.

So in 1, μ1 < />2 and μ2 = μ3

So in 2, μ1 > μ2​ and μ2 > μ3

So in 3, μ1 < />2 and μ2 = μ3

So in 4, μ1 > μ2​ and μ2 > μ3

So in 5, μ1 > μ2 and μ2 = μ3

JEE Advanced Practice Test- 16 - Question 17

Coefficient of mutual inductance for the given coils does not depend on

Detailed Solution for JEE Advanced Practice Test- 16 - Question 17
The mutual inductance of two coils depends on their relative orientation, shape and size and also on medium in which the coils were placed.
JEE Advanced Practice Test- 16 - Question 18

Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.

The difference in maximum kinetic energy of photoelectrons from A and from B

Detailed Solution for JEE Advanced Practice Test- 16 - Question 18

(13.6) − (3.2) = 10.2 eV

= constant

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 19

Which of the following is correct about Tetraamminedithiocyanato-s cobalt (III) tris (oxalato) cobaltate (III) ?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 19
Formula of the complex [CoIII (SCN)2(NH3)4]3[CoIII(ox)3]

Linkage isomerism is due to presence of SCN- .

Ring closure due to (ox)2- Bideutate ligand.

optical isomerism is exhibited due to presence of [Co(ox)3]-3 complex ion, which has asymmetric structure.

Geometrical isomerism is exhibited due to cationic part :

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 20

The expected product(s) in the following reaction is (are):

Detailed Solution for JEE Advanced Practice Test- 16 - Question 20

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 21

The pairs of compounds which cannot exist together in aqueous solution are:

Detailed Solution for JEE Advanced Practice Test- 16 - Question 21
  • Acid + Acid: Can exist together

  • Weak base + Weak base: Can exist together

  • Strong base + Acid: Reaction takes place

  • Slightly acidic + Acid: Reaction takes place

So, NaOH and NaH2PO4 & NaHCO3 and NaOH compounds can not exist in aqueous solution.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 22

Which of the following carbonates do not give metal oxide on heating?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 22
  • CuCO3 → CuO + CO2

  • Ag2CO3 → Ag2O + CO2

    2Ag2O → 4Ag + O2

  • Sodium carbonate is thermally stable, it decomposes only at very high temperatures.

  • MgCO3 → MgO + CO2

So, the correct options are Ag2CO3 and Na2CO3.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 23

Directions: The following question has four choices, out of which one or more is/are correct.

Which of the following statements about some of the observations during qualitative salt analysis is/are correct?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 23

Gelatenous white crystals of Al(OH)3 are soluble in NaOH(aq) because of the formation of soluble sodium aluminate.

Al(OH)3 + NaOH → Na[Al(OH)4]

However, Al(OH)3 is insoluble in aqueous NH3 because Al(OH)3 precipitates at higher pH, but dissolves when the pH is low.

Hence, option (b) is incorrect.

An aqueous solution of Co(II) thiocyanate (10% freshly prepared) and mercuric nitrate solution, when taken in equal volumes, produces deep-blue precipitate on scratching the wall of the vessel with a glass rod due to the formation of mercuric tetrathiocyanatocobaltate(II) and Hg2[Co(CNS)6] (deep blue ppt).

Green precipitate of Cr(OH)3 readily dissolves in excess of sodium hydroxide, forming a green solution.

Chromium (III) salts give green coloured borax beads in both oxidising and reducing flame.

Hence, options (a), (c) and (d) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 24

Directions: The following question has four choices, out of which one or more is/are correct.

Which of the following can result in exactly 1 mol of K4[Fe(CN)6] stoichiometrically using the given compounds as the only source of carbon?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 24
Using POAC, 1 mol of C6H12O6 and 2 mol of Al4C3 have 6 mol of carbon atoms and hence, give 1 mol of K4[Fe(CN)6].

Hence, options (b) and (d) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 25

Directions: The following question has four choices, out of which ONE or MORE are correct.

Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

Detailed Solution for JEE Advanced Practice Test- 16 - Question 25

For diatomic gas:

For monatomic gas:

Note: Cp - Cv = R (For all ideal gases)

Hence the options (a) and (c) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 26

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 26

(b) is wrong because chemical adsorption first increases and then decreases with increase in temperature.

(d) is wrong because as a result of adsorption, there is a decrease in surface energy.

(a) and (c) are correct.

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 27

The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compound is:


Detailed Solution for JEE Advanced Practice Test- 16 - Question 27
The monochlorination involves free radical substitution reaction. In order to find the maximum number of isomers, we should count the number of different hydrogens present [3°,2°,1°].

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 28

Acetic acid forms a dimer in the gas phase:

The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of the dimer. At 25°C, the equilibrium constant for the dimerisation is 1.3 × 103 (pressure in atmatm). What is ΔS for the reaction? Assume that ΔH does not vary with temperature.


Detailed Solution for JEE Advanced Practice Test- 16 - Question 28

Additional bond forms in product = ( 2 hydrogen bonds)

ΔH = −(B.E of H−bond) = −66.5 kJ

Also,

ΔG° = −2.303RT logKp (must be in atm) .....(i)

ΔG° = ∆G° = Change in standard Gibbs free energy

R = Gas constant

T=Temperature

Kp = Kp = Equilibrium constant

Putting the values in equation (i),

As ΔG° = ΔH°−TΔS°

Putting values of ΔH° and ΔG°,

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 29

A blue colour complex is obtained in the analysis of Fe+3 having formula Fe4[Fe(CN)6]3

Let a = oxidation number of iron in the coordination sphere, b = no. of secondary valencies of central iron ion. C = Effective atomic number of iron in the coordination sphere.

Then find the value of (c + a − 2b).


Detailed Solution for JEE Advanced Practice Test- 16 - Question 29

The dissociation of given complex occurs as:

Fe4[Fe(CN)6]3 → Fe3+ + [Fe(CN)6]4−

Here the coordination sphere is [Fe(CN)6]4−

The oxidation number of FeFe in the coordination sphere can be calculated as:

x − 6 = −4

x = +2

Thus, a = 2

There are 6 CN atoms around Fe ions,

Therefore,

b = Number of secondary valencies of central iron ion = 6

Effective atomic number of iron in the coordination sphere can be calculated as

EAN = Z − Oxidation no. + 2 × Coordination no.

EAN = 26 − 2 + 2 × 6

EAN = 36

c = Effective atomic number of iron = 36

Now,

The value of (c + a − 2b) can be calculated as:

(c + a − 2b) = 36 + 2 − 2 × 6 = 26

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 30

The arrangement of X ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X is 250 pm, what is the radius (in pm) of A+? (Round off up to 1 decimal place)


Detailed Solution for JEE Advanced Practice Test- 16 - Question 30

According to the given figure, A+ is present in the octahedral void of X.

The limiting radius in octahedral void is related to the radius of sphere as:

rvoid = 0.414 rsphere

= 0.414 x 250 pm = 103.5 pm

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 31

The minimum energy (MeV) of gamma ray photon to give pair production is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 16 - Question 31

For pair production,

γ → e- + e+

If e-1 and e+ are produced with zero K.E., then γ ray has minimum energy.

Rest mass energy of e-1 and e+ each

= 0.51 MeV

So, minimum energy of gamma ray photon

= 0.51 + 0.51

= 1.02 MeV

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 32

Consider the following cell reaction:

2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) + 2H2O(l); E° = 1.67 V

At [Fe2+] = 10-3 M, P(O2) = 0.1 atom and pH = 3, the cell potential (in V) at 25°C is


Detailed Solution for JEE Advanced Practice Test- 16 - Question 32

2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) + 2H2O(l); E° = 1.67 V

Ecell = 1.57 V

JEE Advanced Practice Test- 16 - Question 33

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 represents reagents.

List - 2 represents the properties of the various reagents shown in List - 1.

Q. Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 33

JEE Advanced Practice Test- 16 - Question 34

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 includes organic name reactions.

List - 2 includes appropriate reagents for the reactions shown in List - 1.

Detailed Solution for JEE Advanced Practice Test- 16 - Question 34

Reimer Tiemann Reaction:

Stephen reduction:

JEE Advanced Practice Test- 16 - Question 35

In a vessel gaseous dimethyl ether at initial pressure 'p0' atm. Dimethyl ether decomposes on heating as per the reaction by first order kinetics.

It is observed that the half-life period for this reaction is 0.2 h. After a very long time, pressure in the vessel is observed to be 1.2 atm. Assume ideal behaviour of all gases with constant V-T conditions.

The initial rate of formation of CH4 gas is

Detailed Solution for JEE Advanced Practice Test- 16 - Question 35
The initial rate of a reaction is the instantaneous rate at the start of the reaction (i.e., when t = 0). The initial rate is equal to the negative of the slope of the curve of reactant concentration versus time at t = 0.

Here,

JEE Advanced Practice Test- 16 - Question 36

Which of the following statement(s) is/are incorrect regarding chloride of iron?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 36
This problem is based on various properties of iron chlorides. To solve this problem, students must have the knowledge of solubility of iron chloride in aqueous solution, the existence of iron chloride in ether solution, oxidising properties of iron chloride.

Properties of iron chloride

(i) FeCl3 (Iron chloride) is back coloured in anhydrous conditions. But its hydrated form is yellowish-brown.

(ii) FeCl3 sublimes at 300°C3 to give a dimer.

(iii) FeCl3 is water-soluble as well as ether soluble due to solvation.

(iv) It is an oxidising agent, during oxidation. The yellow colour of aqueous Fe (III) changes to light green aqueous Fe (II).

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 37

If A and B are two independent events such that P(A) = 1/2 & P(B) = 1/5, then

Detailed Solution for JEE Advanced Practice Test- 16 - Question 37
We have

P(A ∪ B) = 1 − P(A ′∩ B′) = 1 − P(A′)(B′) [∵ A and B are independent]

=

Next, P(A∣B) = P(A) = 1/2. Because A and B are independent. Also

Lastly, since A′ ∪ B′ = (A ∩ B)′, P[(A ∩ B)|(A′ ∪ B′)] = 0.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 38

If the sides of a triangle are in geometric progression with r as the common ratio, where r ≠ 1, then which of the following can be true about r

Detailed Solution for JEE Advanced Practice Test- 16 - Question 38

If r > 1, then ar2 is the greatest term.

In case 0 < r="" />< 1,="" a="" />< ar="" +="" />2

Thus, 2sin18° < r="" />< />

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 39

Let PMPM be the perpendicular from the point (1, 2, 3) to xy-plane. If OP makes an angle θθ with the positive direction of the z-axis and OM makes an angle ϕϕ with the positive direction of x-axis, where O is the origin, θ and ϕ are acute angles, then

Detailed Solution for JEE Advanced Practice Test- 16 - Question 39

Here, P be (x, y, z)

Then, x = rsin θ ⋅ cos ϕ, y = rsin θ sin ϕ, z = rcos θ

⇒ 1 = rsin θ⋅cos ϕ, 2 = rsin θ sin ϕ, 3 = rcos θ....(1)

∴ from (1) we have

(neglecting -Ve sign assuming acute angles)

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 40

Let θ1 and θ2 be the roots of the equation sinx − 2xcosx = 0, then the value of is

Detailed Solution for JEE Advanced Practice Test- 16 - Question 40

The equation can be written as tanx = 2x

⇒ 2θ1 = tanθ1

and 2θ2 = tanθ2

Similarly

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 41

If a function y = f(x) satisfies , then

Detailed Solution for JEE Advanced Practice Test- 16 - Question 41

Given

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 42

Assume that , x > 0. If = f(k) - f(1), then the possible value(s) of k is/are

Detailed Solution for JEE Advanced Practice Test- 16 - Question 42
Consider the expression

Now, consider the given expression of the integration

f(216) - f(1) = f(k) - f(1)

Therefore

k = 216

Hence, it is the required result.

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 43

If z1 = 8 + 4i, z2 = 6 + 4i and arg , then z satisfies

Detailed Solution for JEE Advanced Practice Test- 16 - Question 43

⇒ x2 - 14x + 64 + y2 - 8y = 2y - 8

⇒ (x - 7)2 + (y2 - 10y + 25) = 2

∵ |z - 7 - 5i| = √2

*Multiple options can be correct
JEE Advanced Practice Test- 16 - Question 44

A consignment of 15 record players contains 4 defective ones. The record players are selected at random, one by one, and examined. The ones examined are not put back.

What is the probability that the 9th one examined is the last defective?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 44

The first 8 record players examined contain the remaining 3 detectives. (as 9th one examined is the last defective​)

The number of ways in which we can have 3 defectives in a total of 8 objects is C(8, 3)

= 8!/3!(8 − 3)! = 56.

The number of ways in which we can have 4 defectives in a total of 15 objects is C(15, 4)

= 15!/4!(15 − 4)!

= 1365.

The probability that the ninth one examined is the last defective = 56/1365 = 8/195

Therefore, the probability that the 9th one examined is the last defective​ is 8/195

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 45

The minimum value of 3x + 4y, subject to the condition x2y3 = 6 and x > 0, y > 0, is


Detailed Solution for JEE Advanced Practice Test- 16 - Question 45

Break 3x as 3x/2 and 3x/2 and 4y as 4y/3, 4y/3 and 4y/3.

Now, we know that AM ≥ GM.

So, applying on numbers 3x/2, 3x/2 and 4y/3, 4y/3 and 4y/3,

Putting the value of x2y3 = 6, we get

3x + 4y ≥ 10

Minimum value of 3x + 4y = 10

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 46

If 0 ≤ x ≤ , then the number of solutions of the equation is


Detailed Solution for JEE Advanced Practice Test- 16 - Question 46

Let

Hence,

y2 - 10y + 16 = 0

(y - 2)(y - 8) = 0

⇒ y = 2 and y = 8

Now,

Also,

Required number of solutions is 4.

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 47

Ten different letters of an alphabet are given. Four-lettered words are formed using these given letters. Then, the number of words which have at least one letter repeated is


Detailed Solution for JEE Advanced Practice Test- 16 - Question 47

Number of words which have at least one letter repeated = Total number of words formed - Total number of words in which no letter is repeated

= 104 - 10P4 = 10000 − 5040 = 4960

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 48

The minimum value of the expression p = |z|2 + |z − 3|2 + |z − 6i|2 is q, then the value of q/6 is :


Detailed Solution for JEE Advanced Practice Test- 16 - Question 48

= 3(x2 + y2) − 6x + 9 − 12y + 36

= 3[x2 + y2 − 2x − 4y + 15]

=3[(x − 1)2 + (y − 2)2 + 10]

For minimum value of p, x = 1, y = 2

Minimum value of p = 3(10) = 30

⇒ q / 6 = 30/6 = 5

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 49

The eccentricity of an ellipse whose foci are (2,4) & (14,9) and touches x-axis is , then the value of λ−10 is


Detailed Solution for JEE Advanced Practice Test- 16 - Question 49

Since ellipse touches x-axis, x-axis is tangent to ellipse.

Product of ⊥ distance from foci is square of semi-minor axis.

∴ b2 = 9 × 4 = 36

Distance between foci = 2ae = 13

ae = 13/2

λ = 13

Hence, λ − 10 = 13 − 10 = 3

*Answer can only contain numeric values
JEE Advanced Practice Test- 16 - Question 50

The number of ordered pairs (x,y) satisfying the system of equations (cos−1x)2 + sin−1y = 1 and cos−1x + (sin−1y)2 = 1 is (are)


Detailed Solution for JEE Advanced Practice Test- 16 - Question 50

Let a = cos−1x, b = sin−1y

⇒ a2 + b = 1 ...(1)

⇒ a + b2 = 1 ...(2)

From (1)−(2), we get

a2 − b2 + b − a = 0

⇒ (a − b)(a + b − 1) = 0

⇒ a = b ...(3)

or a + b = 1⇒ a = 1, b = 0 or a = 0, b = 1

Now, from (1) & (3), we get

a2 + a − 1 = 0

From a = 1,b = 0

⇒ (x, y) = (cos1, 0)

From a = 0,b = 1

⇒ (x, y )= (1, sin1)

JEE Advanced Practice Test- 16 - Question 51

Which of the following options correctly matches the statements in Column I with their properties in Column II?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 51

B → p,q

C → q, r

D → p, q

JEE Advanced Practice Test- 16 - Question 52

Which of the following options correctly matches the statements in Column I with their properties in Column II?

Detailed Solution for JEE Advanced Practice Test- 16 - Question 52

B → p,q

C → q, r

D → p, q

JEE Advanced Practice Test- 16 - Question 53

Let P(x, y) is a variable point such that represents hyperbola.

The eccentricity e′ of the corresponding conjugate hyperbola is

Detailed Solution for JEE Advanced Practice Test- 16 - Question 53

2a = 3

Distance between the foci (1, 2) and (5, 5) is 5.

∴ 2ae = 5

∴ e = 5/3

Now if e' is eccentricity of the corresponding conjugate hyperbola, then

⇒ e′ = 5 / 4

JEE Advanced Practice Test- 16 - Question 54

Let P(x, y) is a variable point such that represents hyperbola.

Locus of intersection of two perpendicular tangents to the hyperbola is

Detailed Solution for JEE Advanced Practice Test- 16 - Question 54

Director circle is given by

(x − h)2 + (y − k)2 = a2 − b2

where (h, k) is centre, i.e. the

Midpoint of foci ≡

Y − y = m(X − x)

= 4

Therefore, the director circle is

This does not represent any real point.

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