JEE Advanced Practice Test- 8 - JEE MCQ

# JEE Advanced Practice Test- 8 - JEE MCQ

Test Description

## 54 Questions MCQ Test - JEE Advanced Practice Test- 8

JEE Advanced Practice Test- 8 for JEE 2024 is part of JEE preparation. The JEE Advanced Practice Test- 8 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Practice Test- 8 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Practice Test- 8 below.
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JEE Advanced Practice Test- 8 - Question 1

### In the circuit shown C1 = 2C2. Switch S is closed at time t = 0. Let i1 & i2 be the current flowing through C1 and C2 at any time t, then the ratio I1/I2

Detailed Solution for JEE Advanced Practice Test- 8 - Question 1

B

JEE Advanced Practice Test- 8 - Question 2

### The capacitor of an oscillatory circuit of frequency 104Hz is enclosed in a container. When the container is evacuated, the frequency changes by 50 Hz, the dielectric constant of the gas is

Detailed Solution for JEE Advanced Practice Test- 8 - Question 2

JEE Advanced Practice Test- 8 - Question 3

### A current carrying wire is placed in the grooves of an insulating semicircular disc of radius ‘R’ as shown. The current enters atpoint A and leaves point B. Determine the magnetic field at point D.

Detailed Solution for JEE Advanced Practice Test- 8 - Question 3

B

JEE Advanced Practice Test- 8 - Question 4

The organ pipes, both closed at one end, have lengths ℓ and ( ℓ + Δ ℓ) vibrates in fundamental mode. Neglect end correction. If the velocity of sound in air is v, then the number of beats/sec is (assume Δ ℓ << ℓ)

Detailed Solution for JEE Advanced Practice Test- 8 - Question 4

C

JEE Advanced Practice Test- 8 - Question 5

Light is incident normally on face AB of a prism as shown. Aliquid of refractive index μ is placed on face AC of the prism.The prism is made of glass of refractive index 3/2. The limit of μ for which total internal reflection takes place on face AC is

Detailed Solution for JEE Advanced Practice Test- 8 - Question 5

*Multiple options can be correct
JEE Advanced Practice Test- 8 - Question 6

Figure shows variation of magnification m(produced by a thin convexlens) and distance v of image from pole of the lens. Which of the following statements are correct?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 6

A, B, C

JEE Advanced Practice Test- 8 - Question 7

A certain substance has a mass of 50g/mol. When 300J of heat is added to 25 gm of sample of this material, its temperature rises from 25 to 45oC.

Q. The specific heat capacity of the substance is

Detailed Solution for JEE Advanced Practice Test- 8 - Question 7

D
Specific heat capacity,

JEE Advanced Practice Test- 8 - Question 8

A certain substance has a mass of 50g/mol. When 300J of heat is added to 25 gm of sample of thismaterial, its temperature rises from 25 to 45oC.

Q. The molar heat capacity of the substance is

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

A
Molar heat capacity

JEE Advanced Practice Test- 8 - Question 9

Matrix – Match Type
This section contains 2 questions. Each question contains statements given in  two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example:

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:

1. Match the following

Detailed Solution for JEE Advanced Practice Test- 8 - Question 9

1.

In wrist watch spring force is not dependent on gravity.

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 10

A projectile is projected with a speed 20 m/s from the floor of a 5 m high room as shown. The maximum horizontal range of the projectile is found to be m. Find n [Take g = 10 ms-2].

Detailed Solution for JEE Advanced Practice Test- 8 - Question 10

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 11

A solid cylinder is made of radius R and height 3R having mass density ρ. Now two half spheres of radius R are removed from both ends. The moment of intertia of remaining portion about axis ZZ’ can be calculated as Find K.

Detailed Solution for JEE Advanced Practice Test- 8 - Question 11

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 12

Two particles each of mass m are connected by a string of length 1m. The particles are kept on a smooth horizontal plane. The initial separation between the particles is 0.5 m. One of the particles is given a velocity 18 m/s as shown. Find the magnitude of angular velocity of one particle with respect to the other particle just after the string becomes taut (in m/s)

Detailed Solution for JEE Advanced Practice Test- 8 - Question 12

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 13

A 10 m long nichrome wire having 80Ω resistance has current carrying capacity of 5 A. This wire can be cut into equal parts and equal parts can be connected in series or parallel. What is the maximum power which can be obtained as heat by the wire from a 200 V mains supply (in KW).

Detailed Solution for JEE Advanced Practice Test- 8 - Question 13

When the wire is divided into two equal parts and connected in parallel we get maximum power.

∴ Resistance of each part = 40Ω

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 14

In the given figure, a potential of +12 volt is given to point A, and point B is earthed. What is the potential at point P in volt?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 14

JEE Advanced Practice Test- 8 - Question 15

Which one of the following in not true about D.

Detailed Solution for JEE Advanced Practice Test- 8 - Question 15

C
Co – ordination no of Fe+2 in D is 6

(a sandwhich compound)

JEE Advanced Practice Test- 8 - Question 16

Which one of the following reaction is not possible:-

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

JEE Advanced Practice Test- 8 - Question 17

(I) Co(CO)6
(II) Co(CO)5NH3
(III) Co(CO)4(NH3)2
(IV) Co(CO)3(NH3)3
(V) Co(CO)2(NH3)4
(VI) Co(CO)(NH3)5
Which statement is correct for these compounds?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 17

A
Synergic bonding.

*Multiple options can be correct
JEE Advanced Practice Test- 8 - Question 18

Which among the following statements is/are correct?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 18

B, C
(b) conjugate base of
(c) pH of 0.1 M NaCl (aqueous solution) =

*Multiple options can be correct
JEE Advanced Practice Test- 8 - Question 19

4.5 g of Al (atomic mass = 27) is deposited at cathode from Al+3 solution by certain quantity of charge. The correct statement/s is/are:

Detailed Solution for JEE Advanced Practice Test- 8 - Question 19

A, C

Number of moles of Al produced

Number of moles of electrons used
Number of moles of H2 produced

Volume of H2 produced

JEE Advanced Practice Test- 8 - Question 20

On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic. The increasing order of stability of these compounds are is under: Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromatic follow the following conditions (according to valence bond theory)
(i) The compounds must be be cyclic in structure having (4n + 2)π e, where n = Hückel’s number = 0, 1, 2, 3 et.c
(ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.
(iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure must undergo resonance .
Compounds to be anti-aromatic, it must have 4nπe where n = 1, 2… and it must be planar and undergo resonance. Non-aromatic compounds the name itself spells that compounds must be non-planar irrespective of number of π electrons. Either it has 4nπeor (4n + 2) π electrons it does not matter. The rate of reaction of any aromatic compounds depends upon the following factors:
(i) Electron density
(ii) stability of carbocation produced
Higher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy. Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.) otherwise there will be no effect of kinetic labelling.
Q. Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towards
sulphonation?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 20

C
2nd step is r.d.s. in sulphonation reaction.

JEE Advanced Practice Test- 8 - Question 21

On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic. The increasing order of stability of these compounds are is under:
Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromatic follow the following conditions (according to valence bond theory)
(i) The compounds must be be cyclic in structure having (4n + 2)π e, where n = Hückel’s number = 0, 1,2, 3 et.c
(ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.
(iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure must undergo resonance .
Compounds to be anti-aromatic, it must have 4nπe where n = 1, 2… and it must be planar and undergo resonance. Non-aromatic compounds the name itself spells that compounds must be non-planar irrespective of number of π electrons. Either it has 4nπeor (4n + 2) π electrons it  does not matter.
The rate of reaction of any aromatic compounds depends upon the following factors:
(i) Electron density
(ii) stability of carbocation produced
Higher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy. Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.) otherwise there will be no effect of kinetic labelling.
Q. Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towards
nitrations?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 21

B
2nd step is fast step in nitration.

JEE Advanced Practice Test- 8 - Question 22

The Bertholet’s equation of state for 1 mole real gas is given as under:

Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas). One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

Q. If where Ti is the inversion temperature, then which of the following is the correct relation of the Boyle’s temperature TB obtained by the Bertholet’s equation

Detailed Solution for JEE Advanced Practice Test- 8 - Question 22

A

on converting into virial equation from, we get

to higher power of n …(i)

For Boyle’s temperature, 2nd virial co-efficient becomes zero

JEE Advanced Practice Test- 8 - Question 23

The Bertholet’s equation of state for 1 mole real gas is given as under:

Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas).  One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

Q. If the value of a and b increases by 8 times and 2 times respecively, then new Boyle’s temperature will increases by how many times and what will be effect on the ease of liquification ?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 23

A
Let Boyle’s temperature, 'TB

JEE Advanced Practice Test- 8 - Question 24

The Bertholet’s equation of state for 1 mole real gas is given as under:

Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume  having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas). One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

Q. If the third virial coefficient of 4 moles of mono-atomic as is 0.04 (lit/mol)2 at NTP, then what will be the real gas at NTP?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 24

D
b2 = 0.04
b = 0.2 lit/mol
Vreal = 4 x 22.4 + 4 x 0.2 = 90.4 lit

JEE Advanced Practice Test- 8 - Question 25

(Matrix-Match Type)
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as
illustrated in the following example:

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

1. Match the following.

Detailed Solution for JEE Advanced Practice Test- 8 - Question 25

(A) Sic covalent carbide (Not hydrolysed)

JEE Advanced Practice Test- 8 - Question 26

(Matrix-Match Type)
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as
illustrated in the following example:

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

2. Match the solution given in column I with the properties given in column II.

Detailed Solution for JEE Advanced Practice Test- 8 - Question 26

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 27

Examine the structural fromuals shown below and identify how many compounds will show coupling reaction with diazonium salt faster than anisole

Detailed Solution for JEE Advanced Practice Test- 8 - Question 27

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 28

For the first order reaction The value of x will be

Detailed Solution for JEE Advanced Practice Test- 8 - Question 28

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 29

Total no of molecules/ions which have non zero dipole moment is

Detailed Solution for JEE Advanced Practice Test- 8 - Question 29

JEE Advanced Practice Test- 8 - Question 30

Which one of following statement is true

Detailed Solution for JEE Advanced Practice Test- 8 - Question 30

JEE Advanced Practice Test- 8 - Question 31

If ‘P’ be a point on unit circle with centre at point ‘O’. If Q is another point inclined at angle  to   =  r, then the interval in which ‘r’ lies, so that

Detailed Solution for JEE Advanced Practice Test- 8 - Question 31

B
Where 'Q' is the angle between

JEE Advanced Practice Test- 8 - Question 32

If a bag contains five balls. Two balls are drawn and are found to be white. What is the probabilitythat all the balls are white?

Detailed Solution for JEE Advanced Practice Test- 8 - Question 32

C
Here two balls are drawn and the result is known that the two balls are white, therefore Bayes’s theorem should be used.
Required probability

JEE Advanced Practice Test- 8 - Question 33

If f(x) and g(x) are defined and differentiable for x ≥ x0 andthen

Detailed Solution for JEE Advanced Practice Test- 8 - Question 33

D

JEE Advanced Practice Test- 8 - Question 34

If z1 and z2 are two variable complex numbers such that, . Then themaximum value of ,is

Detailed Solution for JEE Advanced Practice Test- 8 - Question 34

JEE Advanced Practice Test- 8 - Question 35

If .........................up to ‘n’ terms &

Detailed Solution for JEE Advanced Practice Test- 8 - Question 35

JEE Advanced Practice Test- 8 - Question 36

The perpendicular from origin to the tangent at any point on a curve is equal to the obscissa ofthe point of contact. If the curve passes through point (1, 2), then it also passes through point,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 36

A

Equation of tangent at 'P' is

putting y = vx

Solving Differential equation

It passes through (1,2) ⇒ C = 5

Equation of curve ≡

JEE Advanced Practice Test- 8 - Question 37

For an ellipse  with vertices A & A’, tangent drawn at the point ‘P’ in the first quadrantmeets the y – axis in Q and the chord A’P meets the y – axis in M. If ‘O’ is the centre of ellipse,then OQ2 – MQ2 is equal to,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 37

= 4

*Multiple options can be correct
JEE Advanced Practice Test- 8 - Question 38

If . Then f(x) = 0 has,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 38

A, B, D

JEE Advanced Practice Test- 8 - Question 39

Which of the following is/are not true,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 39

*Multiple options can be correct
JEE Advanced Practice Test- 8 - Question 40

If    then

Detailed Solution for JEE Advanced Practice Test- 8 - Question 40

B, C

JEE Advanced Practice Test- 8 - Question 41

If α,β are the roots of equation, x2 - 2x a2 + 1 = 0 and γ, δ are the roots of the equation, x2 - 2(a + 1)x + x(a - 1) = 0 such that α,β ∈ γ,δ then the interval in which ‘a’ lies,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 41

A, B

JEE Advanced Practice Test- 8 - Question 42

Consider the curves,

c3 ≡ x = 0

Q. Maximum area bounded by all of these three curve is,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 42

C
The value of K1 & K2 decide the opening of parabola

JEE Advanced Practice Test- 8 - Question 43

Consider the curves,

c3 ≡ x= 0

Q. Minimum area bounded by all of these three curve is,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 43

D

JEE Advanced Practice Test- 8 - Question 44

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

Q. The area of quadrilateral BFEC is equal to,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 44

D

JEE Advanced Practice Test- 8 - Question 45

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

Q. The length of segment EF is equal to,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 45

AE = R,AF = 2RcosA
⇒ EF = R

JEE Advanced Practice Test- 8 - Question 46

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

Q. Let  , then sum of squares of three altitudes of triangle AFE is equal to,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 46

⇒ sum of square of altitudes

JEE Advanced Practice Test- 8 - Question 47

(Matrix-Match Type)
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example:

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

1. Match the following:

Detailed Solution for JEE Advanced Practice Test- 8 - Question 47

(C) 6

JEE Advanced Practice Test- 8 - Question 48

(Matrix-Match Type)
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example:

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

2. Match the following:

Detailed Solution for JEE Advanced Practice Test- 8 - Question 48

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 49

A line makes an angle of α,β,γ,δ with the four diagonals of a cube then the value of
is, (where[.]denotes the GIF)...............

Detailed Solution for JEE Advanced Practice Test- 8 - Question 49

D.C's of OP are,

Let D.C's of required line be (ℓ,m,n)

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 50

Tangents drawn from any point ‘P’ on the curve xy - 2x - y - 2 = 0 intersect the asymptotes at A & B. Then the area of triangle QAB, is 4K (where Q is centre of the curve). Then value of
K is………………

Detailed Solution for JEE Advanced Practice Test- 8 - Question 50

2
The gives equation is (x - 1) (y - 2) = 4
XY = 4
Equationof tangent is,

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 51

Consider a triangle ABC whose sides length are AB = 4, BC = 2 & CA = 3. If the bisector of the angle ‘C’ cuts AB in D and the circumcircle in E, then the value of  is…………(where, [.] denotes the GIF)

Detailed Solution for JEE Advanced Practice Test- 8 - Question 51

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 52

If      Then value of R is…………..

Detailed Solution for JEE Advanced Practice Test- 8 - Question 52

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 53

If the number of ways in which four distinct balls can be put into two identical boxes so that no box remains empty is equal to K, then K =………….

Detailed Solution for JEE Advanced Practice Test- 8 - Question 53

7
Every ball have two options
⇒ 4 balls can be put in 24 ways.
Arrangement among themselves in 2!

But, above count also includes the one case in which all the balls are put in one box,
⇒ Required number of ways = 23 – 1 = 7.

*Answer can only contain numeric values
JEE Advanced Practice Test- 8 - Question 54

The value of |K| for which the following system of equations is consistent,

Detailed Solution for JEE Advanced Practice Test- 8 - Question 54