JEE Exam  >  JEE Tests  >  JEE Main & Advanced Mock Test Series  >  JEE Advanced Practice Test- 17 - JEE MCQ Download as PDF

JEE Advanced Practice Test- 17 - JEE MCQ


Test Description

54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 17

JEE Advanced Practice Test- 17 for JEE 2024 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Advanced Practice Test- 17 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Practice Test- 17 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Practice Test- 17 below.
Solutions of JEE Advanced Practice Test- 17 questions in English are available as part of our JEE Main & Advanced Mock Test Series for JEE & JEE Advanced Practice Test- 17 solutions in Hindi for JEE Main & Advanced Mock Test Series course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Advanced Practice Test- 17 | 54 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study JEE Main & Advanced Mock Test Series for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 1

In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. This implies that

Detailed Solution for JEE Advanced Practice Test- 17 - Question 1
Ratio of maximum to minimum intensity of the interference pattern

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 2

A particle of charge +q and mass m moving under the influence of a uniform electric field and uniform magnetic field follows a trajectory from P to Q as shown in figure. The velocities at P and Q are Which of the following statement(s) is/are correct?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 2
Magnetic force does not do work. From work-energy theorem

∴ The correct option is (1).

At P, rate of work done by electric field

Therefore, option (2) is also correct.

Rate of work done at Q:

Of electric field = = (qE) (2v) cos 90o = 0

And, of magnetic field is always zero

Therefore, option (4) is also correct.

Note that

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 3

An incompressible and non-viscous liquid in a V-shape tube is shown in the figure. When the liquid is slightly depressed, ('A' is the area of cross-section.)

Detailed Solution for JEE Advanced Practice Test- 17 - Question 3
The liquid will execute S.H.M.

When the liquid in the tube is slightly depressed and released, a restoring force is developed due to hydrostatic pressure difference.

The tube is of uniform cross-section.

Initially, the level of liquid in two limbs is at the same height 'h' from the ground.

If the liquid is depressed by 'y' in one limb, it will rise by 'y' along the length of tube in the other limb.

Therefore, restoring force developed by hydrostatic pressure difference, F = -△p x A = (h1 + h2)gdA = -Agd(sin θ1 + sin θ2)y

As the restoring force is linear, the motion will be linear simple harmonic with force constant

The unbalanced force is F = -Agd(sin θ1 + sin θ2)y

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 4

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision,

Detailed Solution for JEE Advanced Practice Test- 17 - Question 4
We know that 0.1 × 2.0 = 2 kg m/s

Friction with ground gives no impulse.

FΔt = mΔv

= 0.1(20 - 0) = 2 N-s

Using: angular impulse = change in angular momentum, we get:

Note that we have assumed that direction of angular velocities is same before and after and since LHS of the above equation is positive, ω2 > ω1.

Thus, the ring must be slipping to the right; hence, the friction will be to the left as it will be opposite to the direction of motion.

Thus, option 3 is correct.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 5

A projectile is fired at an angle θ with the horizontal. The condition under which it lands perpendicular on the fixed inclined plane with inclination α as shown in figure.

Detailed Solution for JEE Advanced Practice Test- 17 - Question 5
Applying equation of motion perpendicular to the incline for y = 0

Along the plane of incline

v = u + at

0 = 1−2 tan α tan (θ−α)

∴ cot α = 2 tan (θ − α)

or cot (θ − α) = 2 tan α

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 6

If in a hydrogen atom, radius of nth Bohr orbit is rn, frequency of revolution of electron in nth orbit is fn, and area enclosed by the nth orbit is An , then which of the following graphs is/are correct?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 6
(i) Since in hydrogen atom rn ∝ n2, therefore graph between rn and n will be a parabola through origin and having increasing slope.

Since, rn ∝ n2, therefore rn/r1 = n2Hence, log(rn/r1) = 2 log n

(ii) It means, graph between log(rn/r1) and log n will be a straight line passing through and having positive slope (tanθ = 2)tanθ = 2.

(iii) If radius of an orbit is equal to r, then area enclosed by it will be equal to A = πr2.

Since rn ∝ n2, therefore An ∝ n4

It means, graph between log(An/A1) and log n will be a straight line passing through origin and having positive slope (tanθ = 4).

(iv) If frequency of revolution of electron is f, then its angular velocity will be equal to ω = 2π. Hence, its angular momentum will be equal to Iω = mr2. But according to Bohr's theory, it is equal to nh/2π, therefore,

It means, graph between log (fn/f1) and log n will be a straight line passing through origin and having negative slope, tan θ = −3. Hence, it will be as shown in diagram.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 7

Which of the following is/are conservative force(s) ?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 7

Clearly for forces (1) and (2) the integration do not require any information of the path taken.

Taking : x2 + y2 = t

2x dx + 2y dy = dt

Which is solvable.

Hence, (1), (2) and (3) are conservative force.

But (4) requires some more information on path. Hence non-conservative.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 8

When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then,

Detailed Solution for JEE Advanced Practice Test- 17 - Question 8
Work done = ΔKinetic energy

As the net work done is positive, the kinetic energy will increase. Due to air resistance torque, the angular momentum will decrease.

T2 ∝ r3, T will decrease.

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 9

Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.

The spread of frequency as observed by the passengers in train B is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 9

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 10

A resistance of 2Ω is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 10
Given, R = 2 Ω

From the first condition,

Now, from the second condition,

Put the value of x from equation (i) in (ii).

⇒ (100 - l)(80 - l) = l(l + 20)

⇒ 8000 - 100l - 80l + l2 = l2 + 20l

⇒ 8000 = 20l + 180l

Put this value in equation (ii).

⇒ x = 3 Ω

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 11

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s-2).

The magnitude of the normal reaction that acts on the block at the point Q is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 11

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 12

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.

Value of E ( in eV) is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 12

WA = ionization energy of electron ion 2nd orbit of hydrogen atom = 3.4 eV

WB = ionization energy of electron in 2nd orbit of He+ ion

= 13.6 eV.

Now, given that

KA = 2KB

or (E − WA) = 2(E − WB)

∴ E = 2WB − WA = 23.8 eV

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 13

When you throw a ball in air with some velocity at some angle with horizontal, vertical component of velocity at highest point is zero and horizontal component of velocity remains unchanged.

Velocity of a projectile at height 15 m from ground is Here, is in horizontal direction and vertically upwards. Then

Angle of projectile with ground is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 13
ux = vx = 20 m/s [constant]

uy = 20 m/s

θ = 45o

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 14

A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 14
When 'S' is open

After the capacitors are fully charged, current flows through resistors only.

Let the charge on each capacitor be q μC.

So, we have

Total charge at junction X = 0 μC

When 'S' is closed:

Potential drop across the 3 W resistor = 1 × 3 = 3 V

Charge on the 3 μC capacitor is q1 = 3 × 3 μC = 9 μC

Potential drop across the 6 W resistor = 6 × 1 V = 6 V

Charge on the 6 μC capacitor is q2 = 6 × 6 μC = 36 μC

Total charge at X = (36 - 9) μC = 27 μC

Total charge flowing from Y to X = 27 μC

JEE Advanced Practice Test- 17 - Question 15

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

Six point charges, each of the same magnitude q, are arranged in different manners as shown in List II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

Detailed Solution for JEE Advanced Practice Test- 17 - Question 15
List - II

1. By symmetry, E = 0, V = 0 and B = 0, μ = NIA = 0 but I = 0

Since total charge is zero, current is zero.

2. E ≠ 0, V = 0, B = 0

Since I = 0, μ = NIA = 0

3. E = 0 (By symmetry)

V ≠ 0, since distances are different.

B ≠ 0, since radii are different, μ ≠ 0

4. E = 0 (By symmetry)

V ≠ 0, since distances are different,

B ≠ 0, μ ≠ 0

5. E ≠ 0, V = 0, B = 0, since each rotating charge is equivalent to a steady current.

And μ = 0, since μ for pairs of opposite charges will cancel each other.

JEE Advanced Practice Test- 17 - Question 16

Angular frequency in SHM is given by Maximum acceleration in SHM is ω2A and maximum value of friction between two bodies in contact is μN, where N is the normal reaction between the bodies.

In the diagram shown, what can be the maximum amplitude of the system so that there is no slipping between any of the blocks ?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 16

Maximum friction between 1 kg and 2 kg blocks can be 0.6 x 1 x 10 = 6 N. Therefore, maximum acceleration of 1 kg block can be 6/1 = 6 m/s2. Maximum force of friction between 2 kg and 3 kg blocks can be 0.4 x 3 x 10 = 12 N.

Therefore, maximum acceleration of 1 kg and 2 kg blocks jointly can be 12/3 = 4 m/s2.

So, maximum acceleration of the whole system, so that there is no slipping between any of blocks is 4 m/s2.

Now, ω2Amax = 4

or Amax = 4/ω2 = 49m

JEE Advanced Practice Test- 17 - Question 17

A.DC circuit consisting of two cells of emf V and 2V having no internal resistance are connected with two capacitors of capacity C and 2C and four resistors R, R, 2R and 2R as shown in figure. The ammeter and voltmeter used in the circuit are ideal.

The reading of the ammeter as soon as the switch is closed is

Detailed Solution for JEE Advanced Practice Test- 17 - Question 17
Capacitors will offer no resistance and hence the circuit becomes

Clearly reading of ammeter will be zero.

JEE Advanced Practice Test- 17 - Question 18

An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.

The molar specific heat of the gas in this process is given by C whose value is

Detailed Solution for JEE Advanced Practice Test- 17 - Question 18
It is given that there is decrease of the internal energy while the gas expands absorbing some heat, which numerically equal to the decrease in internal energy.

Hence, dQ = −dU

Thus, dQ = dW + dU = −dU

⇒ dW = −2dU (remembering that both dW and dU are negative)

⇒ PdV = 2CdT [dU = −CdT] ...... (1)

where C is the molar specific heat of the gas

Since the gas is ideal, PV = RT ....... (2)

From (2) P = RT/v

since the gas is diatomic.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 19

For the Balmer series of hydrogen spectrum, the wave number for each line is given by: where RH is Rydberg Constant and n1 and n2 are intergers

Choose the correct statement(s) about the given information.

Detailed Solution for JEE Advanced Practice Test- 17 - Question 19
Only statements (1), (2) and (4) are correct as

(1) Beyond a certain wavelength the line spectrum become band spectrum.

(2) For Balmer series n1 = 2.

(4) For the calculation of longest wavelength, we use nearest value of n2.

Hence, for the longest wavelength in Balmer series of hydrogen spectrum is n2 and n3.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 20

3-tert-butylphenol reacts with different halogens by electrophillic substitution under appropriate conditions that are given below:

The observed pattern of electrophillic substitution can be explained by

Detailed Solution for JEE Advanced Practice Test- 17 - Question 20
(1)

Mono substituted product is obtained on iodination because of stearic hindrence of the I+ ion as well as the tertiary butyl group.

The hydrogen at the least hindered site, ortho to the -OH group is substituted.

(2)

Br+ ion is less bulky and substitution takes place at both ortho and para positions.

(3)

Chloronium ion is least bulky and a tri-substituted product is obtained.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 21

Consider the following conversions:

Among P, Q, R and S, the aromatic compound is/are

Detailed Solution for JEE Advanced Practice Test- 17 - Question 21
All P, Q, R, S are aromatic in nature as explained below

(1) This is Friedel-Crafts alkylation reaction in which cyclopropyl methyl carbocation is formed.

(2) Cyclopendiene reacts with strong base from cyclopentadienyl anion which is aromatic in nature.

(3)

(4)

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 22

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

Which of the following reactions would yield acetanilide?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 22

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 23

A freshly prepared aqueous solution of Pd(NH3)2Cl2 does not conduct electricity. It suggests that

Detailed Solution for JEE Advanced Practice Test- 17 - Question 23

Since there are no ions involved in this compound, the Cl atoms are Coordinately bonded to Pd in the coordination sphere.

No precipitate of AgCl would be formed for chlorine atoms are non-ionisable.

For non-electrolytes, the value of van't Haff's factor is 1.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 24

Which of the following statements are correct?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 24
In an isocyanide, an electrophile attacks first followed by a nucleophile at the same carbon atom bearing negative charge. on partial hydrolysis in acidic medium, will give N-methylmethanamide while on complete hydrolysis in acidic medium it gives CH3NH2 and HCOOH.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 25

One mole of an ideal gas is subjected to a two-step reversible process (A→B and B→C). The pressure at A and C is the same. Mark the correct statement(s).

Detailed Solution for JEE Advanced Practice Test- 17 - Question 25

At A

∵ VB > VA, so, the expansion of gas takes place.

VB = 200 × 0.0821 = 16.42L

Also, as pressure is same at A and C so

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 26

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 26

B+ is 1s2 2s2 (stable)

C+ is1s2 2s2 2p1 (not stable)

∴ IInd IE of B > IInd IE of C

due to the nearly same sizes of Al and Ga, their first IE are nearly same.

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 27

A glass tube AD of uniform cross-section of length 100 cm100 cm sealed at both ends contains two columns of ideal gas AB and CD separated by a column of mercury of length 20 cm20 cm. When the tube is held horizontally, AB = 20 cm cm and CD = 60 cm. When the tube is held vertically with the end A up, the mercury column moves down 10 cm10 cm. What will be the length of gas column AB when the tube is held vertically with the end D up? (Give answer upto two digits after decimal point.)


Detailed Solution for JEE Advanced Practice Test- 17 - Question 27
As we know according to Boyle's law, P1V1 = P2V2.

In horizontal position the pressure of gases in column AB and CD is equal and let it be ''P'. When the tube is held vertically with end AA up, let's consider pressure in column AB = P1 and that of CD = P2.

Volume of gas is equal to the volume of column, and volume of a cylinder is the product of its height and area of base ('A' as cross-section is uniform). As mercury column moves down by 10 cm, therefore, length of column AB becomes \30 cm and that of column CD becomes 50 cm. So,

Substituting values of P1 and P2 and P2 from (1) and (2) in (3),

Now let's consider when the tube is held vertically with end D up, then column CD moves down by ‘x’ cm, so that length of CD = 60 + x having pressure 'P2' and column AB = 20 − x having pressure P1'.

According to Boyle's law,

Substituting values of P1' and P2' from (4) and (5) in (6)

Putting value of ''P' in the above equation,

Solving for x,

Therefore, length of gas column AB = 20 − x = 20 − 6.115 = 13.885 cm

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 28

A gas is enclosed in a cylinder with a piston. Weights are added to the piston, giving a total mass of 2.20 kg. As a result, the gas is compressed and the weights are lowered 0.25 m. At the same time, 1.50 J of heat is evolved from the system. What is the change in internal energy of the system? (Take g = 9.8 m/sec2)


Detailed Solution for JEE Advanced Practice Test- 17 - Question 28

From the first law of thermodynamics,

△U = △q + w ....(i)

Given:

△q = −1.50 J (Heat evolved from the system)

Mass =2.20 kg

Work and heat are path functions. Work done on the system,

w = −Pdv = Force × displacement

w = −F.ds

F = Mass × Gravity = 2.2 × 9.8

ds = 0.25 m

w = 2.2 × 9.8 × 0.25

= 5.39 J

Work will be positive since volume of system is decreasing.

According to the first law of thermodynamics,

Internal energy can be determined as given below. Internal energy is a state function and an extensive property.

△U = △q + w

= −1.50 + 5.39

△U = 3.89 J

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 29

.0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJ

Mass of CO2 (g) produced in the reaction is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 17 - Question 29

Number of moles of CH4 (g) used = = 0.0425 mol

Number of moles of O2 (g) taken = = 0.1700 mol

Here, methane is the limiting reactant. Thus, according to the balanced equation, the number of moles of CO2 formed is same as the number of moles of methane reacted, i.e. 0.0525 mol.

Hence, mass of CO2 (g) formed = 0.0425 mol × 44 g mol-1 = 1.87 g

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 30

In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface, surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer, so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.

The number of atoms in this HCP unit cell is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 30

Atom at the face = 2

Atom at the centre = 3

Atom at corners = 12

Total number of atoms = (2 × 1/2) + (3 × 1) + (12 × 1/6) = 6

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 31

The radiolysis of water yields H2 as given below:

2H2O → H2O2 + H2

In the analysis it is found that 1 molecule of H2 is yielded per 100 eV of energy absorbed. A nuclear power reactor of 200 kW capacity has been installed based on this reaction.

The volume of H2 produced per minute in the above reactor is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 31

Energy required for one molecule = 100 eV = 100 x 1.6 x 10-19 = 1.6 x 10-17 J

Energy passed = 200 kW = 200 x 1000 W (J-1)

= 12 x 106 J min-1

1.6 x 10-17 J produces 1 molecule of H2.

12 x 106 J produces molecules of H2 min-1.

=

= 7.5 x 1023 molecules

= 1.25 mol

= 28 L [∵ 1 mol = 22.4 L]

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 32

The amount of copper in a certain alloy is determined by reacting copper with sulphuric acid as follows:

Cu(s) + 2H2SO4(aq) → CuSO4(aq) + SO2(g) + 2H2O(l)

The CuSO4 formed is then precipitated as solid CuI by reacting it with KI solution as follows:

2CuSO4(aq) + 5KI(aq) → 2CuI(s) + KI3(aq) + 2K2SO4(aq)

What mass (in g) of copper was present, if 4.86 g of CuI was recovered? (Atomic weight: Cu = 63.5, I = 127)

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 17 - Question 32
The weight of copper in the alloy can be calculated by applying the unitary method.

= 1.62 g

JEE Advanced Practice Test- 17 - Question 33

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

List - 1 includes starting materials and reagents of selected chemical reactions.

List - 2 gives the types of reactions of List - 1.

For the synthesis of ortho- and para-hydroxyacetophenone, the only correct combination is ____________.

Detailed Solution for JEE Advanced Practice Test- 17 - Question 33
The Fries rearrangement is an organic reaction used to convert a phenyl ester to an ortho- and para-hydroxy aryl ketone using a Lewis acid catalyst and Bronsted acid, such as HF, AlCl3, BF3, TiCl4 or SnCl4. The mechanism begins with the coordination of the ester to the Lewis acid, followed by a rearrangement which generates an electrophilic acylium cation.

The aromatic compound then attacks the alkyl cation (both the ortho and para attacks are allowed) via an electrophilic aromatic substitution (SEAr). The deprotonation to regenerate aromaticity and Bronsted acid work ups to regenerate the Lewis acid catalyst providing the two hydroxy aryl ketone products. The complete reaction is shown as below.

JEE Advanced Practice Test- 17 - Question 34

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

List - 1 represents the value of rate constant of various reactions (Zero order, first order, second order and third order, etc).

List - 2 represents the value of half-life period of various reactions (Zero order, first order, second order and third order, etc).

For the given graph, the only correct combination is-

Detailed Solution for JEE Advanced Practice Test- 17 - Question 34

The given graph is of a zero order reaction.

For a zero order reaction:

Rate (R) = k [a]0

Or R = k [∵ [a]0 = 1]

It is clear from the above equation that the rate of a zero order reaction is independent from the concentration of reactants.

Now, the rate of a zero order reaction is:

Or d[a] = -kdt

On integration on both sides,

[a] = - kt + c

At t = 0, [a] = [ao]

Thus, c = [ao]

t =

At t = t1/2, [a] = [ao]/2

t1/2 =

JEE Advanced Practice Test- 17 - Question 35

wo beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of naphthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.

The molar mass of solute in solution B is closest to:

Detailed Solution for JEE Advanced Practice Test- 17 - Question 35

(m = moles of unknown solute)

JEE Advanced Practice Test- 17 - Question 36

Two beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of naphthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.

Now beaker B, at equilibrium, is replaced by another beaker C which contain 31 g of same solute as in B, but in different solvent X. Also moles of X in solution C is same as mol of benzene in solution B at equilibrium. When the container containing beakers A and C is closed again and allowed to attain equilibrium, beaker C has lost some weight. Which of the following can be said with guarantee ?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 36

for beaker A,

= 0.0967 (before equilibrium)

for solvent x

= 0.0965 (before equilibrium)

Relative lowering in vapour pressure is less over beaker C i.e., vapour pressure of X is more than vapour pressure of benzene.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 37

Find is equal to

Detailed Solution for JEE Advanced Practice Test- 17 - Question 37

Let tan-1 1/7 = θ : cos2θ =

From option (2),

Let,

Options (2) and (4) are correct.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 38

If x2 - 2x + sin2 a = 0, then

Detailed Solution for JEE Advanced Practice Test- 17 - Question 38

sin2 a = 2x - x2

⇒ 0 < 2x - x2 < 1

Then, 2x - x2 ≥ 0

⇒ x2 - 2x ≤ 0

x ∈ [0, 2] .....(1)

And, 2x - x2 ≤ 1

⇒ x2 - 2x + 1 0

⇒ (x - 1)2 > 0

x ∈ R ...(ii)

Hence from (i) and (ii)

x ∈ [0, 2]

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 39

The vectors and (k being scalar) are collinear for

Detailed Solution for JEE Advanced Practice Test- 17 - Question 39
Three vectors and (k being scalar) are collinear, only if

; where x, y and z are scalar.

Then, x + y + z = 0 ... (i)

x - y - kz = 0 ... (ii)

From (i) and (ii):

The possible value of k from the options is 0, 1, or -1.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 40

For the function

f(x) = xcos,1/x, x ≥ 1,

Detailed Solution for JEE Advanced Practice Test- 17 - Question 40

We have,

f(x) = xcos,1/x, x ≥ 1,

Also

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 41

Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 − x + 3y = 0 on L1 and L2 are equal, then which of the following equation can represent L1 ?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 41

Let equation of line L1 be y = mx. Intercepts made by L1 and L2 on the circle will be equal ie, L1 and L2 are at the same distance from the centre of the circle.

Centre of the given circle is (1/2,−3/2).

Therefore,

Thus, two chords are x + 7y = 0 and y − x = 0.

Therefore, x − y = 0 and x + 7y = 0 are correct answers.

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 42

The solution of is;

Detailed Solution for JEE Advanced Practice Test- 17 - Question 42

Solving for dy/dx, we obtain

Thus, we have

Solving , we get

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 43

Let f(x) be a function defined by . Then the

Detailed Solution for JEE Advanced Practice Test- 17 - Question 43
Differentiating the given function, we

f′(x) = x(x − 1)(x − 2) =

Clearly, the function has its minimum value at x = 2 , that is

When x =1 , f(1) = 0

Where

=

Hence, the range of f(x) is [−1/4, 63/4].

*Multiple options can be correct
JEE Advanced Practice Test- 17 - Question 44

Let x1, x2, ......... be positive integers in AP, such that x1 + x2 + x3 = 12 and x4 + x6 = 14. Then, x5 is

Detailed Solution for JEE Advanced Practice Test- 17 - Question 44

Let x1 = a, Then, x2 = a + d, x3 = a + 2d,.......... xn = a + (n - 1)d

Since: x1 + x2 + x3 = 12

Then,

a + a + d + a + 2d = 12

⇒ a + d = 4 .....(i)

Also, x4 + x6 = 14

⇒ a + 3d + a + 5d = 14

⇒ a + 4d = 7 ......(ii)

By (i) and (ii), a = 3 and d = 1

x5 = a + 4d = 3 + 4 x 1 = 7

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 45

The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match, the probability that in a 5-match series, India's second win occurs in the third test, is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Practice Test- 17 - Question 45

Given: P(India wins) = p = 1/2

P(India loses) = p' = 1/2

Out of 5 matches, India's second win occurs in the third test.

⇒ India wins the third test and simultaneously it has won one match from the first two and lost the other.

Required probability = P(LWW) + P(WLW)

=

= 0.25

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 46

The order of the differential equation whose general solution is given by y = (c1 + c2) cos(x + c3) - c4ex + c5, where c1, c2, c3, c4 and c5 are arbitrary constants is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 46

The given equation is

y = (c1 + c2) cos(x + c3) - c4ex + c5

Let y = A cos(x + B) - Cex; where A = c1 + c2, B = c3 and C = c4ec5

dy/dx = -A sin(x + B) - Cex

Differentiating again,

d2y/dx2 = -A cos(x + B) - Cex

Or d2y/dx2 + y= -2 Cex

Or d3y/dx3 + dy/dx = -2Cex = [d2y/dx2] + y

Or d3y/dx3 - d2y/dx2 + dy/dx - y = 0; which is a differential equation of order 3.

Alternate Solution:

y = (c1 + c2) cos(x + c3) + (-c4 ec5)ex

Put c1 + c2 = k1, c3 = k2, and -c4 ec5 = k3

Then, y = k1 cos (x + k2) + k3ex; where k1, k2 and k3 are constants.

Now, y contains 3 independent constants; hence, the order of the differential equation is 3.

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - Question 47

If z is any complex number satisfying |z - 3 - 2i| ≤ 2, then the minimum value of |2z - 6 + 5i| is


Detailed Solution for JEE Advanced Practice Test- 17 - Question 47

|2z - 6 + 5i|

We know that

= 5/2

= 5

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - 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- 17 - 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- 17 - 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- 17 - 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

λ = 13

Hence, λ − 10 = 13 − 10 = 3

*Answer can only contain numeric values
JEE Advanced Practice Test- 17 - 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- 17 - Question 50

Let a = cos−1x, b = sin−1y

From (1)−(2), we get

a2 − b2 + b − a = 0

⇒ (a − b)(a + b − 1) =

⇒ 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- 17 - Question 51

Match the statements/expressions given in Column I with the values given in Column II.

Which of the following is the only correct option?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 51

2sin2θ + 4sin2θ cos2θ = 2

sin2θ + 2sin2θ(1 - sin2θ) = 1

3sin2θ - 2sin4θ - 1 = 0

(B) Let y = 3x / π

Then,

Now, f(y) = [2y] cos[y]

Critical points are y = 1/2, y = 1, y = 3/2, y = 3

So, points of discontinuity

(C) The matrix obtained =

Volume =

= 2π - π = π

⇒ Volume of parallelepiped = π

JEE Advanced Practice Test- 17 - Question 52

Match the statements/expressions given in Column I with the values given in Column II.

Which of the following is the only correct option?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 52

(A) f'(x) > 0, for all x belongs to (0, p/2)

F(0) < 0="" and="" f(p/2)="" /> 0

So one solution.

(B) Direction ratios obtained by sytem of planes

=

So,

k(k - 4) - 4(4 - 4) + 1(8 - 2k) = 0

k2 - 4k + 8 - 2k = 0

k2 - 6k - 8 = 0

k2 - 4k - 2k - 8 = 0

k(k - 4) - 2(k - 4) = 0

k - 2 = 0 or k - 4 = 0

k = 2, 4

(C) Let f(x) = |x + 2| + |x + 1| + |x - 1| + |x - 2|.

Then k can take value 2, 3, 4 and 5.

⇒ f(x) = 2ex - 1

⇒ f(ln2) = 3

JEE Advanced Practice Test- 17 - Question 53

Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom/atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one P = O or P → O unit and one P - OH unit. The -OH group is ionisable but H atom linked directly to P is non-ionisable. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid.

Which one is mono-basic acid?

Detailed Solution for JEE Advanced Practice Test- 17 - Question 53
A monobasic acid is an acid that has only one hydrogen ion to donate to a base in an acid-base reaction. Therefore, a monobasic molecule has only one replaceable hydrogen atom.

And the structure can be graphically represnted as follows:

JEE Advanced Practice Test- 17 - Question 54

Disaccharides are carbohydrates those contain two monosaccharides molecules, each in the hemiacetal form, joined together by the elimination of a water between two hydroxyl groups. Dehydration involves the anomeric carbon of one monosaccharide and may or may not involve the anomeric carbon of the other monosaccharide when the hemiacetal hydroxyl group on an anomeric carbon is involved in a dehydration, the resulting product is an acetal (in common) and glycoside (in carbohydrate).

Sucrose is a non-reducing sugar (while its hydrolysis products glucose and fructose are reducing sugars) because :

Detailed Solution for JEE Advanced Practice Test- 17 - Question 54
Sucrose is having structure as

In above structure there is no free carbonyl centre of an aldehydic group present ; It is a non-reducing sugar.

366 docs|219 tests
Information about JEE Advanced Practice Test- 17 Page
In this test you can find the Exam questions for JEE Advanced Practice Test- 17 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Advanced Practice Test- 17, EduRev gives you an ample number of Online tests for practice
Download as PDF
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!