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JEE Main Part Test - 4 - JEE MCQ


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75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Part Test - 4

JEE Main Part Test - 4 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Part Test - 4 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Part Test - 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Part Test - 4 below.
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JEE Main Part Test - 4 - Question 1

Two point charges 3 × 10-6 C and 8 × 10-6 C repel each other by a force of 6 × 10-3 N. If each of them is given an additional charge of -6 × 10-6 C, then the force between them will be

Detailed Solution for JEE Main Part Test - 4 - Question 1

Key Idea: Like charges repel each other while unlike charges attract each other.

From Coulomb's law, the force of attraction/repulsion between two point charges q1 and q2 placed a distance r apart is given by

When similar charges are taken

q1 = 3 × 10-6 C, q2 = 8 × 10-6 C

When additonal charge - 6 × 106 C is given to each charge, then


Dividing Eq. (ii) by Eq. (i), we get

Negative sign indicates, force is attractive.

JEE Main Part Test - 4 - Question 2

Three charges 1 μC, 1μC and 2 μC are respectively kept at the vertices A, B and C of an equilateral triangle ABC of side 10 cm. The resultant force on the charge at C is

Detailed Solution for JEE Main Part Test - 4 - Question 2


Here, FCA = FCB = FC

Hence, the resultant force on the charge at C is

JEE Main Part Test - 4 - Question 3

If the inward flux and outward electric flux from a closed surface respectively are 8 × 103 units and 4 × 103 units, then what is the net charge inside the closed surface?

Detailed Solution for JEE Main Part Test - 4 - Question 3

The charge inside the closed surface is given by
q = net electric flux pass through the surface × ε0
= (4 × 103 - 8 × 103) ε0
Therefore, q = -4 × 103 ε0 coulomb

JEE Main Part Test - 4 - Question 4

A metallic solid sphere is placed in a uniform electric field as shown. Which path will the lines of force follow?

Detailed Solution for JEE Main Part Test - 4 - Question 4
  • The electric field is always perpendicular to the surface of a conductor.
  • On the surface of a metallic solid sphere, the electric field is perpendicular to the surface and directed towards the centre of the sphere.

Hence, the correct option is (d).

JEE Main Part Test - 4 - Question 5

An electric dipole placed in a non-uniform electric field experiences

Detailed Solution for JEE Main Part Test - 4 - Question 5

The correct option is (c). In a non-uniform electric field, a dipole experiences a force, which gives it a translational motion, and a torque, which gives it a rotational motion.

JEE Main Part Test - 4 - Question 6

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

Assertion (A): In a cavity in a conductor, the electric field is zero.

Reason (R): Charges in a conductor reside only at its surface.

Detailed Solution for JEE Main Part Test - 4 - Question 6

The charge enclosed by the Gaussian surface surrounding the cavity is zero. Hence, the electric field is also zero. So, the assertion is true. Charges in a conductor reside only at its surface. So, in the cavity there is no charge. So, the reason is also true and properly explains the assertion.

JEE Main Part Test - 4 - Question 7

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

Assertion : The Coulomb force is the dominating force in the universe.

Reason : The Coulomb force is weaker than the gravitational force.

Detailed Solution for JEE Main Part Test - 4 - Question 7

Gravitational force is the dominating force in the universe so assertion is false. Gravitational force is weaker than Coulombic force so, reason is false

JEE Main Part Test - 4 - Question 8

The potential of the electric field produced by a point charge at any point (x, y, z) is given by V = 3x2 + 5, where x, y and z are in metres and V is in volts. The intensity of the electric field at (–2, 1, 0) is

Detailed Solution for JEE Main Part Test - 4 - Question 8

Given potential,
V = 3x2 + 5
Electric field intensity,

At x = -2,
E = -6 (-2) = +12V/m

JEE Main Part Test - 4 - Question 9

A capacitor of capacitance C1 is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, is

Detailed Solution for JEE Main Part Test - 4 - Question 9

If Q is the initial charge on capacitor C1, then the initial energy is given by Ui = Q2/2C1.

As the two capacitors are connected together and the charge is distributed equally, the charge on each capacitor is Q/2.

Since the potential difference (in a parallel connection) across the two capacitors is also the same, it follows that their capacitance are equal (since C = Q/V).

Thus, C1 = C2 = C (say)

Also, Q1 = Q2 = Q/2

Therefore, the final energy stored in the two capacitors is

JEE Main Part Test - 4 - Question 10

Directions: In the following question, two statements are given. One is assertion and the other is reason. Examine the statements carefully and mark the correct answer according to the instructions given below.

Assertion: A parallel plate capacitor is charged by a battery. The battery is then disconnected. If the distance between the plates is increased, then the energy stored in the capacitor will decrease.

Reason: Work has to be done to increase the separation between the plates of a charged capacitor.

Detailed Solution for JEE Main Part Test - 4 - Question 10

The charge Q remains unchanged as the battery is disconnected. The capacitance C decreases if the separation between the plates is increased.
Now, energy stored U = Q2/2C
Since Q remains the same and C is decreased, U will increase.
Also, work has to be done to increase the separation between the plates of a charged capacitor.
Hence, option d is correct.

JEE Main Part Test - 4 - Question 11

Three point charges of 1 C, 2 C and 3 C are placed at the corners of an equilateral triangle of side 1 m. The work done (in joules) in bringing these charges to the vertices of a smaller similar triangle of side 0.5 m is

Detailed Solution for JEE Main Part Test - 4 - Question 11

JEE Main Part Test - 4 - Question 12

Three capacitors, each of capacity 4 µF, are to be connected in such a way that the effective capacitance is 6 µF. This can be done by

Detailed Solution for JEE Main Part Test - 4 - Question 12

To get equivalent capacitance 6 μF, out of the three, 4 μF capacitances, two are connected in series and the third one is connected in parallel.

JEE Main Part Test - 4 - Question 13

A 3 μF capacitor is charged to a potential of 300 V and a 2 μF capacitor is charged to a potential of 200 V. The capacitors are then connected in parallel with plates of opposite polarity joined together. What amount of charge will flow when the plates are so connected

Detailed Solution for JEE Main Part Test - 4 - Question 13

Before connections,
C1 = 3µF, V1 = 300V
Charge on C1,
Q1 = 900μC
C2 = 2µF, V1 = 200V
Charge on C2,
Q2 = 400μC
When the positive terminal of C1 is connected with the negative terminal of C2, common potential of the system is:

Substituting the values, we get
V = 100V
New charge on posiitve plate of C1,
Q'= 3 x 100 = 300µC
So, charge Q1 - Q'1 = 600 µC has flown from the positive plate of the C1 to the negative plate of C2.

JEE Main Part Test - 4 - Question 14

An insulator plate is passed through the plates of a capacitor as shown. The current

Detailed Solution for JEE Main Part Test - 4 - Question 14

As insulator plate is passed between the plates of the capacitor, its capacity increases first and then decreases as the plate slips out. As a result, positive charge on plate A increase first and then decreases, hence, current in outer circuit flows from B to A and then from A to B.

JEE Main Part Test - 4 - Question 15

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

Assertion: The electric flux of the electric field ∮ E.dA is zero. The electric field is zero everywhere on the surface.

Reason: The charge inside the surface is zero.

Detailed Solution for JEE Main Part Test - 4 - Question 15

If the flux of the electric field through a closed surface is zero, the electric field may be zero everywhere on the surface and  the charge inside the surface must be zero.

JEE Main Part Test - 4 - Question 16

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

Assertion: On bringing a positively charged rod near the uncharged conductor, the conductor gets attracted towards the rod.

Reason: The electric field lines of the charged rod are perpendicular to the surface of the conductor.

Detailed Solution for JEE Main Part Test - 4 - Question 16

Though the net charge on the conductor is still zero but due to induction the negatively charged region is nearer to the rod as compared to the positively charged region. That is why the conductor gets attracted towards the rod

JEE Main Part Test - 4 - Question 17

Two parallel, large and thin metal plates have equal surface charge densities (σ = 26.4 x 10-12 C/m2) of opposite signs. The electric field between these plates is

Detailed Solution for JEE Main Part Test - 4 - Question 17

JEE Main Part Test - 4 - Question 18

When a conductor is placed in an electric field; its free charge carriers adjust itself in order to oppose the electric field. This happen until

Detailed Solution for JEE Main Part Test - 4 - Question 18
  • When an external electric field is applied to the conductor, the free electrons in the conductor move in an opposite direction to that of the applied electric field.
  • This movement of electrons induces another electric field inside the conductor which opposes the original external electric field.
  • This continues until the induced electric field cancels out the external field. 
JEE Main Part Test - 4 - Question 19

The electric field inside a dielectric decreases, when it is placed in an external electric field. This happens due to ________

Detailed Solution for JEE Main Part Test - 4 - Question 19

The external electric field polarizes the dielectric and an electric field is produced.
The net electric field inside the dielectric decreases due to polarization.

JEE Main Part Test - 4 - Question 20

The magnetic field inside a toroidal solenoid of radius R is B. If the current through it is doubled and its radius is also doubled keeping the number of turns per unit length the same; magnetic field produced by it will be:

Detailed Solution for JEE Main Part Test - 4 - Question 20

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 21

In the given circuit, the potential difference (v) across PQ will be nearest to
(Round off up to 1 decimal place)


Detailed Solution for JEE Main Part Test - 4 - Question 21

Potential difference across PQ, i.e. potential difference across the resistance of 20Ω, is

∴ V = 0.16 × 20 = 3.2 V

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 22

If the current flowing in a coil of resistance 90Ω is to be reduced to 90%, what value of resistance (in ohms) should be connected in series with it? (In integer)


Detailed Solution for JEE Main Part Test - 4 - Question 22

Let the voltage supply be V volts.

Let the current flowing initially be i amperes.

Now, by Ohm's law, i = V/90

Let the final resistance be R ohms.

As the final current is 90% of the initial value,

Or R = 100 ohms

Let the resistance added in series be r ohms.

Now, r + 90 = 100

Or r = 10

Thus, 10 ohms resistance ought to be connected in series to the given resistor.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 23

The equivalent resistance (in ohms) across A and B is


Detailed Solution for JEE Main Part Test - 4 - Question 23


The above arrangement is a balanced wheatstone bridge, so no current will pass through the 10 ohm resistance.
Or

Or 
 ohms is the equivalent resistance.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 24

A network of four resistances is connected to 9 V battery, as shown in figure. The magnitude of voltage difference between the points A and B is ___________V.


Detailed Solution for JEE Main Part Test - 4 - Question 24


In the circuit I = 9/3 = 3A
VC - VA = 2 × 1.5 = 3 .... (I)
Vc - VB = 4 × 1.5 = 6 .....  (II)
Eqn(II) - Eqn(I)
VA - VB = 6 - 3 = 3 Volt

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 25

When a resistance of 5Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA, however when 1050Ω resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is _________ Ω.


Detailed Solution for JEE Main Part Test - 4 - Question 25

Given:

Equating the two expressions for current, i,

This equation simplifies to:
100 (5 + RG) =  1050 x 5 + RG x 5
Solving for the resistance of the galvanometer, RG:
95 RG = 4750
RG = 50Ω
So, the resistance of the galvanometer is 50Ω.

JEE Main Part Test - 4 - Question 26

For the following cell with hydrogen electrodes at two different  pressure pand p

 emf is given by

Detailed Solution for JEE Main Part Test - 4 - Question 26

For SHE E°SHE = 0.00 V
Oxidation at anode (left)

Reduction at cathode (right) 
Net

This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.

Reaction Quotient (Q) 

∵ 

JEE Main Part Test - 4 - Question 27

For the cell,

Thus (x/y) is

Detailed Solution for JEE Main Part Test - 4 - Question 27

This is a type of concentration cell using hydrogen electrode as anode and cathode.





JEE Main Part Test - 4 - Question 28

The correct order of thermal stability of the hydrides of group 16 elements is

Detailed Solution for JEE Main Part Test - 4 - Question 28



Thermal stability ∝ bond dissociation energy 
(where, E = group 16 elements)
On moving down the group, bond dissociation energy decreases due to increase in bond length. Thus, the order of bond dissociation energy or thermal stability is 
H2O > H2S > H2Se > H2Te > H2PO

*Multiple options can be correct
JEE Main Part Test - 4 - Question 29

Which of the following statements regarding ozone are correct?

Detailed Solution for JEE Main Part Test - 4 - Question 29

Ozone is thermodynamically unstable w.r.t oxygen since, its decomposition into oxygen results in liberation of heat.

JEE Main Part Test - 4 - Question 30

The solubility of CO2 in water increases with

Detailed Solution for JEE Main Part Test - 4 - Question 30

This is in accordance with Henry's law.
'At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to partial pressure of that gas in equilibrium with that liquid.'

JEE Main Part Test - 4 - Question 31

An ideal solution is formed when its components'

Detailed Solution for JEE Main Part Test - 4 - Question 31

For ideal conditions, mixing of heat and mixing of volume must be zero and there should be no interaction between the solute-solvent, solvent-solution, solvent-solvent, etc.

JEE Main Part Test - 4 - Question 32

Which of the following has the highest boiling point?

Detailed Solution for JEE Main Part Test - 4 - Question 32

As Al2(SO4)3 has maximum ions, so it has the highest boiling point.

JEE Main Part Test - 4 - Question 33

3 moles of P and 2 moles of Q are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are 80 and 60 torr respectively?

Detailed Solution for JEE Main Part Test - 4 - Question 33



Ptotal = pP + pQ = pPxP + pQxQ
= 80 x 0.6 + 60 x 0.4 = 72 torr.

JEE Main Part Test - 4 - Question 34


X, Y and Z in the given graph are

Detailed Solution for JEE Main Part Test - 4 - Question 34

Here, X = PT ​= p​+ p2 ​= Total pressure of the solution.
p1 and p2 ​is the partial pressure of component 1 and 2 respectively.
Y = x2 ​= mole fraction of component 2 in the liquid phase.
Z = y​= mole fraction of component 2 in the vapour phase.
When mole fraction of component 2 in the liquid phase is 0 then mole fraction of component 2 in the vapour phase is also 0 which means the liquid is pure of component 1 only.

JEE Main Part Test - 4 - Question 35

A cell converts:

Detailed Solution for JEE Main Part Test - 4 - Question 35

A cell converts chemical energy into electrical energy.

  • It consists of two solid electrodes and this solid electrodes are placed in an electrolyte and connected together by an electrical conductor such as wire.
  • The two electrodes should be made up of two different metals and the electrolyte solution can be an alkaline, acidic or salt solution depending on the situation.
  • These electrodes can be depicted as two half cells set up in different containers and are connected through a porous or salt solution.
JEE Main Part Test - 4 - Question 36

When water is added to an aqueous solution of an electrolyte, what is the change in specific conductivity of the electrolyte?

Detailed Solution for JEE Main Part Test - 4 - Question 36

Factors affecting electrolytic conductance are: 

Concentration of ions: The sole reason for the conductivity of electrolytes is the ions present in them. The conductivity of electrolytes increases with an increase in the concentration of ions as there will be more charge carriers if the concentration of ions is more and hence the conductivity of electrolytes will be high. 

Nature of electrolyte: Electrolytic conduction is significantly affected by the nature of electrolytes. The degree of dissociation of electrolytes determines the concentration of ions in the solution and hence the conductivity of electrolytes.

Substances such as CH3COOH, with a small degree of separation, will have less number of ions in the solution and hence their conductivity will also below, and these are called weak electrolytes. Strong electrolytes such as KNO3 have a high degree of dissociation and hence their solutions have a high concentration of ions and so they are good electrolytic conductance.

Temperature: Temperature affects the degree to which an electrolyte gets dissolved in solution. It has been seen that higher temperature enhances the solubility of electrolytes and hence the concentration of ions which results in an increased electrolytic conduction.

When water is added to an aqueous solution the number of ions per unit volume decreases i.e., the concentration of ions decreases and hence thereby conductivity gets decreased.

JEE Main Part Test - 4 - Question 37

Primary battery is a battery:

Detailed Solution for JEE Main Part Test - 4 - Question 37

Primary battery is a battery which cannot be recharged.

  • A primary cell is a battery (a galvanic cell) that is designed to be used once and discarded, and not recharged with electricity and reused like a secondary cell (rechargeable battery).
  • Primary cells cannot be recharged effectively because their reaction products are not in contact with the electrodes.
  • Primary cells are made in a range of standard sizes to power small household appliances such as flashlights and portable radios.
JEE Main Part Test - 4 - Question 38

When lead storage battery is charged:

Detailed Solution for JEE Main Part Test - 4 - Question 38

During the charging of the lead storage battery, the sulphuric acid is regenerated.

  • Sulphuric acid acts as an electrolyte in lead storage batteries.
  • Thus, lead storage batteries can be used again and again by regenerating the sulphuric acid.
JEE Main Part Test - 4 - Question 39

Which of the following methods is suitable for preventing an iron frying pan from rusting?

Detailed Solution for JEE Main Part Test - 4 - Question 39

Applying a coating of zinc methods is suitable for preventing an iron frying pan from rusting.

  • Applying paint and grease are not suitable solutions to prevent an iron frying pan from rusting as the paint can melt by flame and can be destroyed in repeated use.
  • Applying a coating of zinc is a better option as it forms a coating of corrosion-resistant zinc which prevents corrosive substances from reaching the more delicate part of the metal.
JEE Main Part Test - 4 - Question 40

Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on:

Detailed Solution for JEE Main Part Test - 4 - Question 40

Electricity is conducted through an electrolytic solution due to the movement of ions called electrolytic conduction.
Electrical conductance depends upon-

  • Nature of electrolyte: The conductance of an electrolyte depends upon the number of ions present in the solution. Therefore, the greater the number of ions in the solution, the greater is the conductance.
  • Concentration of the solution: The molar conductance of electrolytic solution varies with the concentration of the electrolyte. In general, the molar conductance of an electrolyte increases with decrease in concentration or increase in dilution.
  • Temperature: The conductivity of an electrolyte depends upon the temperature. With increase in temperature, the conductivity of an electrolyte increases.
  • Valence electrons per atom present in metals describes the electric conductance of it.
  • Nature and structure is also a significant factors which influences electrical conductance.
JEE Main Part Test - 4 - Question 41

Why do most chemical reaction rates increase rapidly as the temperature rises?

Detailed Solution for JEE Main Part Test - 4 - Question 41

As the temperature rises, more collisions start taking place which results in an increase in rate of the chemical reaction.
With rise in temperature, fraction of molecules possessing kinetic energy greater than the activation energy increases. Only such molecules are able to cause effective collisions and result in the formation of product. Thus, as the fraction of such molecules increases, rate of the reaction increases.

JEE Main Part Test - 4 - Question 42

For producing the effective collisions, the colliding molecules must posses:

Detailed Solution for JEE Main Part Test - 4 - Question 42

For producing the effective collisions, the colliding molecules must posses bimolecular reactions.

Bimolecular reaction originates from a collision between two reactants. Whether or not a collision results in a chemical reaction is determined by the energy of the reactants and their orientation.

The total energy of the two reactants must be in excess of the activation energy (Ea), and the reactants must be in a favorable orientation for the chemical reaction to occur. While there are many different orientations possible for the collisions, usually not all of them will result in a chemical reaction. For most reactions, if the orientation is not correct, the reactants will bounce off of each other without a chemical reaction.

JEE Main Part Test - 4 - Question 43

The order of increasing acidic strengths of BF3, BCl3 and BBr3 is:

Detailed Solution for JEE Main Part Test - 4 - Question 43

The order of increasing acidic strengths of BF3, BCl3 and BBr3 is: BF3 < BCl3 < BBr3.
The extent of back bonding is maximum in BF3 and minimum in BBr3.

JEE Main Part Test - 4 - Question 44

Which of the following products is not formed when excess of ammonia reacts with sodium hypochlorite?

Detailed Solution for JEE Main Part Test - 4 - Question 44

The actual reactions are:
NH3 + NaOCl → NH2Cl + NaOH
NH2Cl + NH→ NH2- NH+ HCl
NH3 + HCl → NH4Cl
NaOH + HCl → NaCl + H2O
N2O is not formed in the process.
Hence, option (d) is correct.

JEE Main Part Test - 4 - Question 45

Which of the following is the correct order of paramagnetic behavior of transition metal ions?

Detailed Solution for JEE Main Part Test - 4 - Question 45

Ni2+: d8 configuration - 2 unpaired electrons
V2+: d3 configuration - 3 unpaired electron
Fe2+: d6 configuration - 4 unpaired electrons
Mn2+: d5 configuration - 5 unpaired electrons
So, the correct order of paramagnetic behaviour is Ni2+ < V2+ < Fe2+ < Mn2+.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 46

A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?

Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.
(Round off up to 2 decimal places)


Detailed Solution for JEE Main Part Test - 4 - Question 46

PT = 0.983 atm
Pwater = 0.84 atm
Pliquid = 0.143 atm

Therefore, mass of the liquid present per gram of water in the distillate is 1.27 g.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 47

The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf is 1.86° mol–1 kg, the depression in freezing point ΔTf is ____ °C.
(Round off up to 2 decimal places)


Detailed Solution for JEE Main Part Test - 4 - Question 47

Given, mole fraction of non electrolyte = 0.07
Mole fraction of solvent = 1- Mole fraction of non electrolyte
Mole fraction of solvent = 1- 0.07 = 0.93
Kf = 1.86
ΔTf = Kf m

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 48

At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by

(Round off up to 1 decimal place)


Detailed Solution for JEE Main Part Test - 4 - Question 48

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 49

Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is _________. (Nearest integer)


Detailed Solution for JEE Main Part Test - 4 - Question 49


Ag+ forms precipitate with Cl-, i.e. AgCl.
AgClO3 is soluble.
 X is NaCl.
Y is NaClO3.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 50

Which of the following will be the correct spin magnetic moment value (B.M.) for the compound Hg[Co(SCN)4]?
(Round off up to 2 decimal places)


Detailed Solution for JEE Main Part Test - 4 - Question 50

The oxidation number of cobalt in the given complex is +2.

Co+2 in the complex is sp3 hybridised and the complex has a tetrahedral geometry.
Number of unpaired electrons (n) = 3
Spin magnetic moment (ms)

JEE Main Part Test - 4 - Question 51

The range of  is 

Detailed Solution for JEE Main Part Test - 4 - Question 51

We have , 



Therefore, range of f(x) is {-1}.

JEE Main Part Test - 4 - Question 52

Let f: R → R be a mapping such that f(x) = . Then f is

Detailed Solution for JEE Main Part Test - 4 - Question 52

Correct answer is D.

JEE Main Part Test - 4 - Question 53

The value of cos15º− sin15º is

Detailed Solution for JEE Main Part Test - 4 - Question 53

  

 

JEE Main Part Test - 4 - Question 54

The values of x which satisfy the trigonometric equation   are :

Detailed Solution for JEE Main Part Test - 4 - Question 54



JEE Main Part Test - 4 - Question 55

If cos–1 x + cos–1 y = π/2 and tan–1 x – tan–1 y = 0, then x2 + xy + y2 is equal to

Detailed Solution for JEE Main Part Test - 4 - Question 55

JEE Main Part Test - 4 - Question 56

The sum of the infinite series cot–1 2 + cot–1 8 + cot–1 18 + cot–1 32 + … is equal to

Detailed Solution for JEE Main Part Test - 4 - Question 56

JEE Main Part Test - 4 - Question 57

2 cos-1 x = cos-1 (2x2 -1) holds true if

Detailed Solution for JEE Main Part Test - 4 - Question 57

2 cos-1 x = cos-1 (2x2 -1) ⇒ cos(2 cos-1 x) = 2x2 - 1
Since,
- 1 ≤ cos(2 cos-1 x) ≤ 1
 -1 ≤ 2x2 - 1 ≤ 1
 0 ≤ 2x2 ≤ 2
 0 ≤ x ≤ 1

JEE Main Part Test - 4 - Question 58

If A = (aij)3 × 3 is a matrix satisfying the equation x3 – 3x + 1 = 0, then

Detailed Solution for JEE Main Part Test - 4 - Question 58

A3 - 3A + l = 0 ⇒ l = A(3I - A2)
Since matrix I is an identity matrix and I is invertible, A(3I - A2) is also an invertible matrix.

So, matrix A is also invertible, hence non-singular.

JEE Main Part Test - 4 - Question 59

If then J.J is

Detailed Solution for JEE Main Part Test - 4 - Question 59

JEE Main Part Test - 4 - Question 60

The rows and columns of matrix A will be linearly dependent if

Detailed Solution for JEE Main Part Test - 4 - Question 60

The rows and columns of the matrix A are linearly dependent if the determinant of the matrix A is equal to 0.

JEE Main Part Test - 4 - Question 61

If A and B are two matrices, such that, AB = B and BA = A, then A2 + B2 is equal to

Detailed Solution for JEE Main Part Test - 4 - Question 61

We have:
A2 + B2 = (BA)2 + (AB)2
= (BA)(BA) + (AB)(AB)
= B(AB)A + A(BA)B
= B(BA) + A(AB) = BA + AB = A + B

JEE Main Part Test - 4 - Question 62

If R is a relation from a non – empty set A to a non – empty set B, then

Detailed Solution for JEE Main Part Test - 4 - Question 62

Let A and B be two sets. Then a relation R from set A to set B is a subset of A × B. Thus, R is a relation from A to B ⇔ R ⊆ A × B.

JEE Main Part Test - 4 - Question 63

Let R be the relation over the set of straight lines of a plane such that l1 R l2 ⇔ l1 ⊥ l2. Then, R is

Detailed Solution for JEE Main Part Test - 4 - Question 63

To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if  l1 R l2 ⇒ l1 ⊥ l2.
⇒  l2 ⊥ l1 ⇒ l1 R l2 hence symmetric
For transitive,  if l1 R l2 and l2 R l3
⇒ l1 R l2  and l2 R l3  does not imply that l1 ⊥ l3 hence not transitive.

JEE Main Part Test - 4 - Question 64

If cos-1 x = tan-1 x, then sin(cos-1 x) is equal to

Detailed Solution for JEE Main Part Test - 4 - Question 64

JEE Main Part Test - 4 - Question 65

The order of the single matrix obtained from is

Detailed Solution for JEE Main Part Test - 4 - Question 65

JEE Main Part Test - 4 - Question 66

If ω is a complex cube root of unity, then the matrix is a

Detailed Solution for JEE Main Part Test - 4 - Question 66

JEE Main Part Test - 4 - Question 67

Let A = {X = (x, y, z)T : PX = 0 and x2 + y2 + z2 = 1}, where P = then the set A

Detailed Solution for JEE Main Part Test - 4 - Question 67

 det(P) = 0
So, the system has infinitely many solutions.
All solutions lie on the line of intersection of planes.
x + 2y + z = 0, -2x + 3y - 4z = 0 and x + 9y - z = 0

JEE Main Part Test - 4 - Question 68

In a skew symmetric matrix, the diagonal elements are all 

Detailed Solution for JEE Main Part Test - 4 - Question 68

(b) Zero

JEE Main Part Test - 4 - Question 69

The void relation (a subset of A x A) on a non empty set A is:

Detailed Solution for JEE Main Part Test - 4 - Question 69

The relation { } ⊂ A x A on a is surely not reflexive. However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetry relation on A.

JEE Main Part Test - 4 - Question 70

A relation R in a set A is called reflexive,

Detailed Solution for JEE Main Part Test - 4 - Question 70

A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore, R is reflexive.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 71

The value of is equal to


Detailed Solution for JEE Main Part Test - 4 - Question 71

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 72

Find the number of real solutions of 


Detailed Solution for JEE Main Part Test - 4 - Question 72

We know that

The number of real solutions of the given equation is 3, represented by the points of intersection of:

From the above shown graph, the number of real solutions is 3.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 73

Let A = R and A4 = [aij]. If a11 = 109, then a22 is equal to _______.


Detailed Solution for JEE Main Part Test - 4 - Question 73


Given: (x2 + 1)2 + x2 = 109
Let x2 + 1 = t
t2 + t - 1 = 109
 (t - 10)(t + 11) = 0
 t = 10 = x2 + 1 = a22

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 74

Let A = {1, 2, 3, 4} and R be a relation on the set A × A defined by
R = {((a, b), (c, d)): 2a + 3b = 4c + 5d}. Then the number of elements in R is __________.


Detailed Solution for JEE Main Part Test - 4 - Question 74

2a + 3b4c + 5d
Given A = {1, 2, 3, 4), the maximum value of 2a + 3b is 20, when (a, b) = (4, 4), and the minimum value of 4c + 5d is 9, when (c, d) = (1, 1). Therefore, the possible values for 2a + 3b = 4c + 5d are 9, 13, 14, 17, 18, and 19.

Now, let's find the combinations of (a, b), (c, d) that satisfy the given equation:


There are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d.

*Answer can only contain numeric values
JEE Main Part Test - 4 - Question 75

Let S be the set of values of λ, for which the system of equations 
6λx - 3y + 3z = 4λ2,
2x - 6λy + 4z = 1,
3x + 2y + 3λz = λ has no solution. Then is equal to ___________.


Detailed Solution for JEE Main Part Test - 4 - Question 75

Given that S be the set of values of λ for which given system of equations has no solution.

Therefore for the given set of equations

Also for each values of  we have

which implies that, for each values of λ, the given system of equations has no solution.

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