JEE Main Practice Test- 5 - JEE MCQ

# JEE Main Practice Test- 5 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Practice Test- 5

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JEE Main Practice Test- 5 - Question 1

### A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each having negligible resistance. There is a 2Ω resistor and 10V battery as shown in figure. There is a uniform magnetic field directed into the plane of the paper of magnitude 0.5 T. A force of 0.5N to the left is required to keep the wire PQ moving with constant speed to the right. With what speed is the wire PQ moving? (Neglect self-inductance of the loop)

Detailed Solution for JEE Main Practice Test- 5 - Question 1

Solving V = 16 m/s.

JEE Main Practice Test- 5 - Question 2

### In the given figure a ring of mass m is kept on a horizontal surface while a body of equal mass 'm' attached through a string, which is wounded on the ring. When the system is released the ring rolls without slipping. Consider the following statements and choose the correct option. (i) acceleration of the centre of mass of ring is g/3 (ii) acceleration of the hanging particle is 2g/3 (iii) frictional force (on the ring) acts along forward direction (iv) frictional force (on the ring) acts along backward direction

Detailed Solution for JEE Main Practice Test- 5 - Question 2

Free body diagram of ring and mass is

Since the ring is in pure rolling so its above point acceleration will be 2a and similarly acceleration of mass will be 2a.
The equation of motion of the system is
mg – T = m(2a)      ......(1)
T = ma        .......(2)
Substitute the value of T in equation (1)
mg - ma = 2ma
3ma = mg
a = g/3
So the acceleration of the center of the ring is, g/3
And acceleration of the hanging mass is 2a = 2g/3

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JEE Main Practice Test- 5 - Question 3

### A concave mirror gives a real image magnified 6 times when the object is moved 6 cm the magnification of the real image is 4 times. What will be the focal length of the mirror?

Detailed Solution for JEE Main Practice Test- 5 - Question 3

As we know that the magnification

And in the second case,

JEE Main Practice Test- 5 - Question 4

A spring mass system with unequal masses is placed at rest on a smooth horizontal surface. The spring is initially kept compressed with a thread. When the spring is cut, the mass m moves so as to enter in a vertical circular loop of radius r. The minimum compression x in the spring so that the mass m may negotiate the vertical loop is

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

Let x be the compression in the spring, v and V be the velocities of m and M respectively, when the spring released.
According to the law of conservation of momentum and energy
MV = mv ... (1)

From (2) and (1)

To negotiate a loop, minimum velocity required

JEE Main Practice Test- 5 - Question 5

A plane electromagnetic wave of angular frequency ω propagates in a poorly conducting medium of conductivity σ and relative permittivity ε. Find the ratio of conduction current density and displacement current density in the medium.

Detailed Solution for JEE Main Practice Test- 5 - Question 5

Jc = σE0 sin (ωt – kx)

= ∈ ∈0 × A E0ω cos(ωt–kx)
Jd = ∈ ∈0E0ω

JEE Main Practice Test- 5 - Question 6

A uniform electric field of magnitude 245 V/m is directed in the negative y direction as shown in figure. The coordinates of point P and Q are (- 0.5m, - 0.8m) and (0.3m, 0.7m) respectively. Calculate the potential difference VQ - VP along the path shown in the figure,

Detailed Solution for JEE Main Practice Test- 5 - Question 6

Potential difference between points P and Q is

JEE Main Practice Test- 5 - Question 7

A cylinder of radius r = 1 m and height 3 m filled with a liquid upto 2 m high and is rotated about its vertical axis as shown in the Figure.
The speed of rotation of the cylinder when the point at the centre of the base is just exposed is

Detailed Solution for JEE Main Practice Test- 5 - Question 7

v0 = 0: H = 3 m; r = 1 m: g = 10 ms− 2

JEE Main Practice Test- 5 - Question 8

A charge particle q of mass m0 is projected along the y-axis at t = 0 from origin with a velocity V0. If a uniform electric field E0 also exists along the x-axis, then the time at which the de-Broglie wavelength of the particle becomes half of the initial value is:

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

Force on charged particle is =qE0 in x direction.
At any time t velocity in y direction will be same as at t=0 as there is no force in that direction.
At time t velocity in x direction is given by
vx =axt
vx =(qE0/m)t
so speed of particle at time t is-
debroglie wavelength is inversly proportional to speed of particle and given by-
(Where R is a constant)
So for half the wavelength speed should be doubled.

JEE Main Practice Test- 5 - Question 9

The beat frequency produced when the following two waves x1 = 12 sin (484πt − 7πx) and x2 = 12 sin (480πt − 7πx) are sounded together is

Detailed Solution for JEE Main Practice Test- 5 - Question 9

x1 = 12 sin (484 πt − 7 π x)

∴ frequency n1 = 242 Hz
x2 = 12 sin (480 πt − 7 πx)
∴ frequency n2 = 240 Hz
Beat frequency = n1 − n2 = 2

JEE Main Practice Test- 5 - Question 10

An 80 kg man standing on ice throws a 4 kg body horizontally at 6 ms−1. The frictional coefficient between the ice and his feet is 0.02. The distance through which the man slips is (g = 10 ms−2)

Detailed Solution for JEE Main Practice Test- 5 - Question 10

While throwing the body, velocity gained by the man is

JEE Main Practice Test- 5 - Question 11

Liquid cools from 50ºC to 45 ºC in 5 minutes and from 45ºC to 41.5ºC in the next 5 minutes. The temperature of the surrounding is :- [Assume newton's law of cooling is applicable]

Detailed Solution for JEE Main Practice Test- 5 - Question 11

From Newton's law of cooling

in second case

From (i) and (ii)
θ0 = 33.3oC

JEE Main Practice Test- 5 - Question 12

A long solenoid of self-inductance L and area of cross section A not carrying any current is placed in a uniform magnetic field of strength B with its axis parallel to the field direction. The total magnetic flux linked with the solenoid is φ. The energy of the magnetic field stored in the solenoid is

Detailed Solution for JEE Main Practice Test- 5 - Question 12

Flux linked with each turn of the solenoid = BA
Total number of turns = nℓwhereℓ is the length of the solenoid
Energy per unit volume
∴total energy of the magnetic field stored in the solenoid

JEE Main Practice Test- 5 - Question 13

All wires have same resistance and equivalent resistance between A and B is R. Now keys are closed, then the equivalent resistance will become:-

Detailed Solution for JEE Main Practice Test- 5 - Question 13

As R0 = R/3
Req= 7R/9

JEE Main Practice Test- 5 - Question 14

Steam at 100°C is passed into 1.4 kg of water kept in a calorimeter of water equivalent 0.03 kg at 20 °C till the temperature of the calorimeter and contents reach 80°C. Latent heat of steam is 2.26 × 106 J kg − 1 and sp.heat capacity of water is 4200 J /kg/°C. The mass of steam condensed in kilogram is nearly equal to

Detailed Solution for JEE Main Practice Test- 5 - Question 14

(M + Mw) C(80-20) = mL + mC(100 − 80)
(1.4 + 0.03) 4200 (80 − 20)= m × 2.26 × 106 + m × 4200 × 20
1.43 × 4200 × 60= m (2.26 × 106 + 4200 × 20)

= 0.1537 kg ≈ 0.15 kg

JEE Main Practice Test- 5 - Question 15

A satellite orbiting around the earth of radius R is shifted to an orbit of radius 2R. The time taken for one revolution will increase nearly by

Detailed Solution for JEE Main Practice Test- 5 - Question 15

The time period of a satellite orbiting around the earth is

When the radius of the orbit is increased to 2R

T’=  2√2 T
T' = 2 × 1.414 T
T' = 2.828 T ≈ 2.8 T

JEE Main Practice Test- 5 - Question 16

Screw gauge shown in Figure has 50 divisions and in one complete rotation of circular scale the main scale moves 0.5 mm.

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

As shown in first figure the zero error of the screw gauge is:

ZE = 5×0.5 / 50 = 0.05mm

Least count of circular scale = 0.5 / 50 = 0.01mm

Actual measurement is = MSR+CSR − ZE

1+0.25−0.05=1.20mm

JEE Main Practice Test- 5 - Question 17

The given graph shows the extension (Δl) of a wire of length 1.0 m suspended from the top of a roof at one end and loaded at the other end. If the cross sectional area of the wire is 10− 6m2, the Young’s modulus Y of the material of the wire is

Detailed Solution for JEE Main Practice Test- 5 - Question 17

JEE Main Practice Test- 5 - Question 18

In Young’s double slit experiment, a monochromatic light of wavelength 5500 Å is used. The slits are 2 mm apart. The fringes are formed on a screen placed 20 cm away from the slits. It is found that the interference pattern shifts by 16 mm when a transparent plate of thickness 0.4 mm is introduced in the path of one of the slits. The refractive index of the transparent plate is

Detailed Solution for JEE Main Practice Test- 5 - Question 18

Δy = 16 mm = 16 × 10− 3 m
d = 2 mm = 2 × 10− 3 m λ = 5500 Å
D = 20 cm = 0.2 m
t = 0.4 mm = 4 × 10− 4 m

∴ μ = 1 + 0.4 = 1.4

JEE Main Practice Test- 5 - Question 19

The average degree of freedom per molecule of a gas is 6. The gas performs 25 J work, while expanding at constant pressure. The heat absorbed by the gas is:-

Detailed Solution for JEE Main Practice Test- 5 - Question 19

Change in the internal energy of gas is

For a constant pressure process. Work done (W) = nRΔT
So heat absorbed by the gas is

JEE Main Practice Test- 5 - Question 20

The radioactive sources A and B have half lives of 2hr and 4hr respectively, initially contain the same number of radioactive atoms. At the end of 2 hours, their rates of distintegration are in the ratio:-

Detailed Solution for JEE Main Practice Test- 5 - Question 20

Given that the half-life of radioactive source A is, t1/2 = 2 hr
half-life of radioactive source B is, t1/2 = 4 hr
Decay constant of radioactive source A is,

Decay constant of radioactive source B is,

The rate of disintegration (A) of source A at the end 2 hours is,

The rate of disintegration (A) of source B at the end 2 hours is,

The ratio of rate of distintegration of source A to B is

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 21

A Young's double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and interference pattern is observed on the screen at a distance 1.33 m from plane of slits. The wavelength of light used in air is 6300A°. When one of the slit is covered by a glass sheet of thickness   and refractive index 1.53, the position of maxima and minima get interchanged. Find the maximum possible value of n.

Detailed Solution for JEE Main Practice Test- 5 - Question 21

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JEE Main Practice Test- 5 - Question 22

An oscillator of frequency 425 Hz drives two speakers. The speaker are fixed on a vertical pole at a distance 2.4m from each other. A person whose height (of ears) is same as that of lower speaker runs horizontally away from the two speakers. Find the maximum distance (in m) of person from the pole where he hears no sound, (velocity of sound in air 340 m/s)

Detailed Solution for JEE Main Practice Test- 5 - Question 22

Let x be maximum distance of person from speaker | when he hears no sound (i.e, minimal)

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JEE Main Practice Test- 5 - Question 23

A particle of mass m = 1 kg and having charge Q = 2C is projected on a rough horizontal xy plane from (4, 0, 0) m with velocity   and in the region there exist a uniform electric field  and a uniform magnetic field  The coefficient of friction between the particle and the horizontal surface is μ = 1/3 and the particle comes to rest by moving a distance 1000/n metres. Find n.

Detailed Solution for JEE Main Practice Test- 5 - Question 23

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JEE Main Practice Test- 5 - Question 24

A wedge of inclination 45° is moved towards right with a constant acceleration of 2√6 m/s2 as shown in figure. Find magnitude of acceleration of the ball placed between an inclined wall and the wedge.

Detailed Solution for JEE Main Practice Test- 5 - Question 24

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JEE Main Practice Test- 5 - Question 25

In the situation shown in figure, a particle having charge q = 2C and mass = 1 kg is projected from bottom of an inclined plane of inclination 45° at an angle 45° with horizontal, so that the particle strike the inclined plane normally due to a uniform horizontal electric field E. Find the value of E (in NC-1)

Detailed Solution for JEE Main Practice Test- 5 - Question 25

JEE Main Practice Test- 5 - Question 26

Ratio of Boyle temperature and critical temperature of gas is:

Detailed Solution for JEE Main Practice Test- 5 - Question 26

Boyle Temperature Tb= a/Rb
Critical Temperature Tc= 8a/27Rb
Dividing both
Tb/Tc =27/8
Hence option (b) is correct.

JEE Main Practice Test- 5 - Question 27

Which of the following is reducing sugar?

Detailed Solution for JEE Main Practice Test- 5 - Question 27

A reducing sugar is one which is capable of acting as reducing agent.
All monosaccharaides are reducing sugars because all monosaccharaides have an aldehyde group (if they are aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses) and can themselves get oxidized hence can act as reducing agent.

Galactose
All of them are monosaccharides.
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 28

Which of the following complex square having planner complex can exhibit geometrical isomerism?

Detailed Solution for JEE Main Practice Test- 5 - Question 28

Trans-isomer

Cis –isomer
Square planar complexes having symmetric bidentate ligands carrying one or more substituents can form geometrical isomers. Because, such ligand show kind of asymmetry. Here pn (propylenediamine) shows this kind of nature hence leading to geometrical isomerism.
All the rest options are of symmetrical in nature around metal ion so no geometrical isomerism is possible in these.
Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 29

An element has a bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm–3. Number of atoms present in 208 g of the element ?

Detailed Solution for JEE Main Practice Test- 5 - Question 29

Volume of unit cell = (288 pm)3 = (288 × 10–12 m)3
= (288 × 10–10 cm)3
=2.39 × 10–23 cm3
For bcc, zeff = 2

∴ Aw = 7.2 × 3 × 1023 × 2.39 × 10–23 g
∴ 7.2 × 3 × 2.39 g of element contains = NA atoms = 6 × 1023 atoms.
∴ 208 g of the element contains
= 24.17 × 1023 atoms
Alternatively
Volume of 208 g of the element
=28.88 cm3
Number of unit cells in this volume

= 12.08 × 1023 unit cells
Since each bcc cubic unit contains 2 atoms, therefore the total number of atoms in 208 g
= 2 (atoms/unit cell) × 12.08 × 1023 unit cells
= 24.16 × 1023 atoms
Hence option (b) is correct.

JEE Main Practice Test- 5 - Question 30

The most acidic oxyacid of halogen among the given option is

Detailed Solution for JEE Main Practice Test- 5 - Question 30

The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H5IO6 has an oxidation state of +7 and hence is the strongest acid.

JEE Main Practice Test- 5 - Question 31

The Clark cell Zn|Zn2+// Hg2SO4|Hg is often employed as a standard cell since its emf is known exactly as a function of temperature. The cell emf is 1.423 V at 298 K and its temperature coefficient of voltage is -1.2 × 10-4 V K-1. What is ΔScell at 298 K?

Detailed Solution for JEE Main Practice Test- 5 - Question 31

Before we commence, we note that the spontaneous cell reaction is

so the cell reaction is a two-electron process.
Gcell = -nF × emf .
Inserting values for the cell at 298 K gives

Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 32

Which of the following is incorrect statement?

Detailed Solution for JEE Main Practice Test- 5 - Question 32

a) H2O →OH- + H+
NH3→NH2- +H+
Water being more polar than ammonia dissociates more and gives plenty amount of H+ and hence H2 gas is released as result. In ammonia sufficient amount of H+ is not obtained due to less polar nature (so less dissociation) hence H2 gas is not evolved and then ammoniation of electron happens
b) As we dilute further dissociation of ammonia increases (because degree of dissociation is inversely proportional to concentration) and we get more H+ and NH2- ion. H+ combines with ammoniated electron and H2 gas is evolved and paramagnetism is gone.
c) NH3→NH2- +H+
M →M+ +e-
M+ +xNH2-→[M (NH2)x] type of complex
On adding d block metal ion we see that metal ion combines with NH2- and forms metal complex. Complex formation leads to ammonia dissociation even more and as a result of ammonia dissociation more H+ is released and which combines with ammoniated electron and H2 gas is evolved. Now colour and magnetic nature may or may not change. But whatever it will be it will be due to this complex.
d) Metallic bonding increases at very high concentration M2 is formed.
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 33

In which of the following, colour can be explained due to ‘Ligand to Metal Charge Transfer’?

Detailed Solution for JEE Main Practice Test- 5 - Question 33

Condition for ‘ligand to metal charge transfer’ are
i) Metal should be in high oxidation state so that it has high ionization energy. And it should also be of smaller size with vacant orbitals having low energies.
ii) Ligand should have lone pair of electrons having high energy
In KMnO4, Mn is in +7 oxidation state and have all the 3d orbitals vacant. Mn+7 ion is surrounded by four oxide ions. All oxide ions have filled 2p orbitals. There is transfer of an electron from filled 2p orbital of oxide ion to vacant d orbitals of Mn+7 ion. Due to this transfer of electron from oxides of KMnO4 to metal happens and as a result colouris intensely purple.

Trick: Almost all the metal ion of d block having d0 configuration tend to show colour due to ligand to metal charge transfer. Another example of this is K2Cr2O7
Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 34

Observe the following reaction

Which of the following statement is true regarding solvent of the above reaction

Detailed Solution for JEE Main Practice Test- 5 - Question 34

In water the product is almost all benzyl naphthol. However, in DMSO (dimethyl sulfoxide) the
major product is the ether. In water the oxyanion is heavily solvated through hydrogen bonds to
water molecules and the electrophile cannot push them aside to get close to O– (this is an entropy
effect). DMSO cannot form hydrogen bonds as it has no OH bonds and does not solvate the oxyanion, which is free to attack the electrophile.
DMSO Solvent:

Water Solvent:

Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 35

Which of the following order of aromaticity is correct

Detailed Solution for JEE Main Practice Test- 5 - Question 35

Correct order is
benzene > thiophene > pyrrole > furan.
Take a look at the structures

Electronegativity of atoms in the ring
O>N>S
In furan oxygen is in ring. It will be less reluctant to share electrons of lone pair for resonance and hence aromatic character will be least in it.
In pyrrole nitrogen is in ring. It will be more willing than furan to share electrons for resonance hence more aromatic than furan.
Thiophene has sulphur which is less electronegative than nitrogen hence more aromatic character than pyrrole.
Benzen has carbon and has 6π electrons already so most aromatic.
Hence option (c) is correct.

JEE Main Practice Test- 5 - Question 36

Consider the following reaction

A-

B-

Detailed Solution for JEE Main Practice Test- 5 - Question 36

Enthalpy is not a conclusive measurement of whether a reaction will be more spontaneous and favorable. Entropy is.
Taking a look at both reactions, in reaction A two molecules are combining and giving one molecule and in reaction B only one molecule is giving one molecule so decrease in entropy in reaction B is less. So, ∆S will be less negative in here hence more favorable.
ΔG = ΔH-TΔS
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 37

Which of the following will not produce nitrogen gas?

Detailed Solution for JEE Main Practice Test- 5 - Question 37

a) (NH4)2Cr207→ N2 + Cr2O3 +4H2O
b) 8NH3 + 3Br2→N2 + 6NH4Br
c) 2NH3 + 3CuO →N2 + 3Cu + 3H2O
d) 3Ca + 6NH3→3[Ca(NH2)6]
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 38

The hydride ion H- has the same electron configuration as helium but is much less stable. Which of the following option explains the given statement

Detailed Solution for JEE Main Practice Test- 5 - Question 38

Hydrogen 1H : 1S1
Hydride: 1H- :1S2
Helium 2He : 1S2
Helium has two proton in it and it can stabilize 2 electrons. But, in hydrogen as one more electron enters and it becomes hydride ion, now one proton has to account for two electrons, the stability is affected. In hydride one proton cannot stabilize two electrons so structural deformity arises which leads to its reactivity.
Hence option (b) is correct.

JEE Main Practice Test- 5 - Question 39

IUPAC name of the following compound is

Detailed Solution for JEE Main Practice Test- 5 - Question 39

IUPAC name of the compound will be
Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 40

XeF2 when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?

Detailed Solution for JEE Main Practice Test- 5 - Question 40

Reaction mentioned is given as below
2XeF2 +2H2O →2Xe + 4HF + O2
HF forms H-bond with F- and exists as HF2-
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 41

Which of the marked position is most susceptible to nucleophilic attack

Detailed Solution for JEE Main Practice Test- 5 - Question 41

Position a and c will be more susceptible to nucleophilic attack
Position a has bromine attached to it making it electron deficient or more electropositive and hence inviting for nucleophile
Position c has oxygen atom attached to it. High electronegativity of oxygen makes carbon highly electropositive and susceptible to nucleophilic attack.
Hence option (c) is correct.

JEE Main Practice Test- 5 - Question 42

Ester and amide are found in equilibrium which of the following is true regarding the equilibrium of following. K1 and K2 are equilibrium constants

Detailed Solution for JEE Main Practice Test- 5 - Question 42

1.As we increase the acidic conditions amine will be protonated and equilibrium will shift backward hence K1>K2

2. In basic medium phenol will be deprotonated and hence reaction will shift forward here K2>K1

Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 43

In the given reaction, the major product will be

Detailed Solution for JEE Main Practice Test- 5 - Question 43

E2 eliminations therefore take place from the anti-periplanar conformation.
Here, E2 elimination gives mainly one of two possible stereoisomers.
2-Bromobutane has two conformations with H and Br anti-periplanar, but the one that is less hindered leads to more of the product, and the E-alkene predominates.
Another way to understand will be in any alkene formation  major product is always most stable alkene. We can see alkene given in option a is trans alkene and in option b is cis alkene. Trans alkene being more stable on account of less steric repulsion will be major product.
Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 44

Arrhenius relation is described as K= A e-Ea/RT which of the following statement is correct regarding activation energy

Detailed Solution for JEE Main Practice Test- 5 - Question 44

K=
Differentiating with respect to T

Here we can see that on increasing temperature those reaction having high value of Ea will be more susceptible to change in rate.
Hence, option (b) is correct.

JEE Main Practice Test- 5 - Question 45

Which of the following is correct about acidic nature of boric acid?

Detailed Solution for JEE Main Practice Test- 5 - Question 45

a) B(OH)3 +2 H2O →B(OH)4- +H3O+
This reaction is characterized as Lewis Acidity of boric acid
b) Due to hydrogen bond present in crystalline structure it becomes difficult for it to donate H+ and hence acidic character decreases.
c) B(OH)3 +2 H2O →B(OH)4- +H3O+
Cis-diols form complex with borate ion and shift the equation forward hence ionizing the boric acid to higher extent. And acidic character increases.
Hence option (c) is correct.

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 46

The difference in the oxidation state of chromium in 'B' and 'C' is

Detailed Solution for JEE Main Practice Test- 5 - Question 46

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JEE Main Practice Test- 5 - Question 47

For the reaction NO2 + CO → CO2 + NO. The experimental rate expression is  Find the number of molecules of CO involved in the slowest step.

Detailed Solution for JEE Main Practice Test- 5 - Question 47

No CO molecules involved in rate equation

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JEE Main Practice Test- 5 - Question 48

What is the total score for the correct statement(s) from the following?

Detailed Solution for JEE Main Practice Test- 5 - Question 48

I, II, III are correct

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JEE Main Practice Test- 5 - Question 49

Give the answer of the following questions for the reaction product [X].
(P) In how many M–O bonds, the bond lengths are equal.
(Q) The oxidation state of the central metal ion.
(R) The magnetic moment of the central metal ion.
(S) The number of peroxide linkage in the product formed by the reaction of [X] with H2O2 in acidic medium.

Detailed Solution for JEE Main Practice Test- 5 - Question 49

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JEE Main Practice Test- 5 - Question 50

On moving across a period, the basic character of the oxides gradually changes first into amphoteric and finally into acidic character. The acidic character of oxides increases with increase in the oxidation states of the element that is combined with oxygen. Arrange the following oxides in order of increasing acidic character.
(1) SO3 (2) Cl2O7 (3) N2O5 (4) CO2
Fill in the boxes provided below with suitable number. Least acidic oxide will come in first box and the strongest acidic oxide in the last box.

Detailed Solution for JEE Main Practice Test- 5 - Question 50

Acidic Character of oxides increase as non-metallic character of the element that is combined with oxygen increase.
The ionisation energy is the measure of matallic and non-metallic character
AS ionisation energy of element decreases the matallic character increases and non-metallic charachter decreases and vice-versa.
Hence the correct increasing order of acidic character is increasing non-metallic character

JEE Main Practice Test- 5 - Question 51

The solution to

Detailed Solution for JEE Main Practice Test- 5 - Question 51

JEE Main Practice Test- 5 - Question 52

The area bounded by the curves x + 2|y| = 1 and x = 0 is:

Detailed Solution for JEE Main Practice Test- 5 - Question 52

JEE Main Practice Test- 5 - Question 53

The length of the common chord of the circles x2 + y2 + 2x + 3y + 1 = 0 and x2 + y2 + 4x + 2 = 0 is

Detailed Solution for JEE Main Practice Test- 5 - Question 53

Consider the circles S1
The centre of S1 is (-1,-3/2) and the radius is
and S2:
The equation of the common chord is S2 - S=0
⇒ 2x + 1 = 0

JEE Main Practice Test- 5 - Question 54

In a triangle ABC, the angle B is greater than angle A. If the values of the angle A and B satisfy the equation 3 sin x - 4 sin3 x - k = 0, 0 < k < 1, then value of C is

Detailed Solution for JEE Main Practice Test- 5 - Question 54

JEE Main Practice Test- 5 - Question 55

The value of

Detailed Solution for JEE Main Practice Test- 5 - Question 55

JEE Main Practice Test- 5 - Question 56

Suppose a, b, c are in AP and a2, b2, c2 are in GP. If a < b < c & a + b + c = 3/2, then the value of a is

Detailed Solution for JEE Main Practice Test- 5 - Question 56

⇒ ac = 1/4    …(2)
or
ac = - 1/4

From (1) & (2) a & c are roots of a quadratic
x2 -(a+c)x + ac = 0
⇒ x2 -x + (1/4) = 0

Thus, the equation has roots

Since, a<b<c, the above results are not satisfied.
If we consider ac= - 1/4, then
x2 -x - (1/4) = 0

Since, a<b<c , then

JEE Main Practice Test- 5 - Question 57

The function

Detailed Solution for JEE Main Practice Test- 5 - Question 57

JEE Main Practice Test- 5 - Question 58

If log2(4x+1 + 4)log2(4x + 1) = log2 8, then x equals

Detailed Solution for JEE Main Practice Test- 5 - Question 58

⇒ x = 0

JEE Main Practice Test- 5 - Question 59

If ‘M’ and σ2 are mean and variance of random variable X, whose distribution is given by –

Then

Detailed Solution for JEE Main Practice Test- 5 - Question 59

The mean of a probability distribution is given as
M = Σ Pixi = (¼ × 0) + (0 × 1) + (¼ × 2 ) + (0 × 3) + (½ × 4) + (0 × 5)= 5/2

σ2 = Σ Pi (M – xi)2

JEE Main Practice Test- 5 - Question 60

If a,b,c are three unit vectors such that  and b is not parallel to c, then the angle between a and c is:

Detailed Solution for JEE Main Practice Test- 5 - Question 60

Comparing on both sides of the equation

⇒ θ = π/3

JEE Main Practice Test- 5 - Question 61

The equation represents the equation of a circle with ω, ω2 as extremities of a diameter, then λ is , (where ω, ω2 are cube roots of unity)

Detailed Solution for JEE Main Practice Test- 5 - Question 61

If P(z) lies on a circle whose diametrically opposite points are A(z1) & B(z2) then

JEE Main Practice Test- 5 - Question 62

If sin–1x + sin–1y + sin–1z = π/2 then x2 + y2 + z2 + 2xyz is equal to -

Detailed Solution for JEE Main Practice Test- 5 - Question 62

Given sin–1x + sin–1y + sin–1z = π/2

taking cos on both side,
cos (sin–1x + sin–1y) = cos
⇒ cos (sin–1x) cos (sin–1y) – sin (sin–1x) sin (sin -1y) = cos

⇒ (1-x2) (1-y2) = (xy + z)2
⇒ x2 + y2 + z2 + 2xyz = 1

JEE Main Practice Test- 5 - Question 63

If b2 > 4ac for the equation 4x4 + bx2 + c = 0, then all the roots of the equation will be real if:

Detailed Solution for JEE Main Practice Test- 5 - Question 63

Substituting

Clearly, if all the roots of  are real, then the equation  must have both the roots positive. But
which is a quadratic with positive coefficient of t2. Hence for both the roots to be positive:

⇒ a & care of same sign and a & b are of opposite sign

JEE Main Practice Test- 5 - Question 64

The value of the definite integral  for 0 < α < π, equal to

Detailed Solution for JEE Main Practice Test- 5 - Question 64

JEE Main Practice Test- 5 - Question 65

Coordinates of the orthocentre of the triangle whose sides are x = 3, y = 4 and 3x + 4y = 6 will be:

Detailed Solution for JEE Main Practice Test- 5 - Question 65

Sides of a triangle ABC are given by x = 3, y = 4, 3x + 4y = 6 It forms a right angle triangle ABC with B(3, 4)as right angle.

Hence B is the orthocentre as perpendiculars drawn from A and C meet at B.

JEE Main Practice Test- 5 - Question 66

An equilateral triangle is inscribed in the parabola y= 4ax whose vertex is at the vertex of the parabola. The length of its side is

Detailed Solution for JEE Main Practice Test- 5 - Question 66

Thus, the point will satisfy the equation of the parabola

JEE Main Practice Test- 5 - Question 67

The projection of the line joining the points (3,4,5) and (4,6,3) on the line joining the points (-1,2,,4) and (1,0,5) is

Detailed Solution for JEE Main Practice Test- 5 - Question 67

The direction ratios of the line joining  are proportional to 2,-2, 1
∴ Its direction cosines are
Thus, the projection of the line joining  and  on PQ is given by

JEE Main Practice Test- 5 - Question 68

Consider the following statements regarding the events E, A, B and C .
(i) Event E can take place due to the occurrence of any of the events A, B, C.
(ii) Events A, B and C are equiprobable, mutually exclusive and exhaustive.
(iii) Probability of occurrence of event E is 5/12.

Considering above data the value of P(E/C) is:

Detailed Solution for JEE Main Practice Test- 5 - Question 68

A, B, C are equiprobable ⇒
A, B, C are mutually exclusive and exhaustive

Using the law of total probability:

JEE Main Practice Test- 5 - Question 69

(P→q) ↔ (q V ~ P) is –

Detailed Solution for JEE Main Practice Test- 5 - Question 69

Truth table

From truth table
(P→q) ↔ (q V ~ P) is always true so it is a tautology

JEE Main Practice Test- 5 - Question 70

The equation of the hyperbola whose foci are (6, 5), (- 4, 5) and eccentricity 5/4 is

Detailed Solution for JEE Main Practice Test- 5 - Question 70

Center of the hyperbola is the mid-point of the line segment joining two foci. Therefore, coordinates of the center are (1, 5). Also since the foci have same y coordinate therefore the line joining foci is parallel to x-axis. Hence the transverse axis is parallel to x-axis. Clearly the equation of the hyperbola is of form

Again the distance between the foci = 10

Hence, the equation of the hyperbola is

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 71

The number of solutions of  satisfying the equation

Detailed Solution for JEE Main Practice Test- 5 - Question 71

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 72

One mapping is selected at random from all mappings of the set S = {1, 2, 3,..., n} into itself. If the probability that the mapping is one - one is 3/32 then the value of n is

Detailed Solution for JEE Main Practice Test- 5 - Question 72

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JEE Main Practice Test- 5 - Question 73

Let ab > 0 and  then the maximum value of

Detailed Solution for JEE Main Practice Test- 5 - Question 73

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JEE Main Practice Test- 5 - Question 74

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. Find the number of ways in which we can place the balls in the boxes so that no box remains empty.

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JEE Main Practice Test- 5 - Question 75

Find the coefficient of x2009 in the expansion of (1 -x)2008 (1 +x + x2)2007

Detailed Solution for JEE Main Practice Test- 5 - Question 75

f(x)=(1−x)2008 ×(1+x+x2)2007
f(x)=(1−x)× ((1−x)(1+x+x2))2007
f(x)=(1−x)× (1+x+x2−x−x2−x3)2007
f(x)=(1−x)× (1−x3)2007
f(x)=(1−x3)2007 − x(1−x3)2007
All the terms in the expansion (1−x3)2007 is in the form x3r
And in the −x(1−x3)2007 is of the form x3p+1
So 3r = 2009 or 3p+1 = 2009
Where we do not get integer values of r,p.
So the coefficient of x2009 is 0

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