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JEE Main Mock Test - 5 - JEE MCQ


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75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Mock Test - 5

JEE Main Mock Test - 5 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Mock Test - 5 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock Test - 5 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock Test - 5 below.
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JEE Main Mock Test - 5 - Question 1

A long infinite current-carrying wire is bent in the shape as shown in the figure. The magnetic induction at point O is:

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

An infinite wire, carrying current, is bent in the following way.
Two semi-infinite parts and a bent finite part. We have to find the induced magnetic field at an external point o, at a distance R. The corresponding figure is shown below.

Magnetic field induced by a finite current carrying conductor at a distance R is given by, 
Where, θ1 and θ2 are the angles made by the endpoints of the wire to the external point.
And using right hand thumb rule, we get direction of magnetic field,
For the 1st semi-infinite part, θ1= 0 and θ2 = 90o = π/2
Substitute the above values in equation (i), we get 
For the bent portion θ= 45o and θ2 = 135
Therefore, magnetic field will be,

For the second semi-infinite part, 
θ= π/2 θ2 = 0
Therefore, magnetic field will be, 
Total magnetic field at point o is, B = B1 + B2 + B3

JEE Main Mock Test - 5 - Question 2

Two gases A and B have equal pressure P, temperature T, and volume V. The two gases are mixed together and the resulting mixture has the same temperature T and volume V as before. The ratio of pressure exerted by the mixture to either of the two gases is:

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

Let P1 and P2 be the partial pressures of gases A and B, respectively. Under equal conditions of temperature and volume, P1 = P2
Let P be the total pressure exerted by the mixture of two gases A and B. Using Dalton's law of partial pressures, P = P1 + P2
⇒ P = 2P1 ⇒ P/P1 = 2/1 ⇒ P : P1 = 2 : 1

JEE Main Mock Test - 5 - Question 3

A current of 1.6 A is passed through CuSO4 solution. The number of Cu++ ions liberated per minute are:

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

Current I = 1.6 Amps, Time t = 1 minute or 60 seconds
Two electrons are needed to liberate one ion of Cu++, i.e. the charge required is = 2 × 1.6 × 10-19 C = 3.2 × 10-19C
The total charge flow in 1 minute is Q = It = 1.6A × 60s = 96C
Thus, the number of ions liberated is n

JEE Main Mock Test - 5 - Question 4

A book with many printing errors contains four different formulae for the displacement of a particle undergoing a certain periodic motion x. Which one is the wrong formula on dimensional grounds (A = amplitude, ω = angular velocity, T = time period of motion)?

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

The argument in the trigonometric function should be a dimensionless quantity.
The dimensions of time (t)  angular velocity (ω) and amplitude (A) are [t] = [T], [ω] = [T-1], [A] = [L]
A. The dimension of the argument in the equation:

Thus, option 'A' is not wrong on dimensional grounds.
B. In the equation: x = A / T sin (t/A), the dimension of the argument in the sine function is [t/A] = |T|/L

Thus, option 'B' is wrong on dimensional grounds.
C. The dimension of the argument in equation:

Thus, option 'C' is not wrong on dimensional grounds.
D. The dimension of the argument in equation: x = A sin (ωt) is [ωt] =  [T-1] [T] = dimensionless quantity
Thus, option 'D' is not wrong on dimensional ground. All the equations, except x = A/T sin (t/A) i.e. option 'B' have dimensionless argument.

JEE Main Mock Test - 5 - Question 5

The figure below shows four plates each of area A and separated from one another by a distance d.

What is the capacitance between P and Q?

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

The given figure is shown below:

The above arrangement is equivalent to two parallel plate capacitors connected in parallel with the same separation d and area A.
Now, the capacitance of each capacitor is 
The equivalent capacitance of C1 and C2 is C = C1 + C

JEE Main Mock Test - 5 - Question 6

A spring of spring constant 5 × 103 is stretched initially by 5 cm from the unstretched position. The work required to stretch it further by another 5 cm is

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

Spring constant, k = 5 × 103 N/m
The work required to stretch the spring by the displacement x is W = 1/2 kx
When the spring is stretched by 5 cm, work done is W1 = 1/2 × (5 × 103N/m) × (5 × 10-2m)2
W1 = 6.25J
When the spring is again stretched further by 5 cm, the net displacement of the spring from its equilibrium position is
x' = 5 cm, +5 cm = 10 cm
Thus, the work required to stretch it further by an additional 5 cm is W2 = 1/2 x (5 × 103N/m) × (10 × 10-2m)
⇒ W2 = 25J
So, the extra work required to stretch the spring by an additional 
5 cmis ΔW = W- W1 ⇒ ΔW = 25J - 6.25J ⇒ ΔW = 18.75J or 18.75 N - m

JEE Main Mock Test - 5 - Question 7

Two sources of sound A and B produce progressive waves given by y1 = 6 cos(100πt) and y2 = 4 cos(102πt) ear the ears of an observer. It will hear

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

y1 = 6 cos(100πt) ...... (i)
y2 = 4 cos(102πt) ...... (ii)
Standard wave equation is given by y = Ao cos(ωt) ....... (iii)
From equation (i) and (iii), we get angular frequency ω1 = 100π = 2πf1 ⇒ f= 50 Hz
Amplitude, A01 = 6 unit From equation (ii) and (iii), we get ω=102π = 2πf2 ⇒ f2 =51 Hz
Amplitude, A02 = 4 unit
Beat frequency = difference in frequencies of the two given waves = f2 - f1 = 51 Hz - 50 Hz = 1 Hz, i.e 1 beat per sec
Resultant amplitude A of superposition of two waves of amplitudes A01 and A02 is A =
For maximum resultant amplitude, phase 

For minimum resultant amplitude, phase 

JEE Main Mock Test - 5 - Question 8

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be:

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

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°).
We have to find the angle of refraction at the position of minimum deviation.
In the position of minimum deviation, light passes symmetrically through the prism irrespective of the light wavelength used.

In the position of minimum deviation, r + r = 2r = A[ as deviation δ = 0]
  for both the colour.

JEE Main Mock Test - 5 - Question 9

A planet of radius R = 1/10 × ( radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linear mass density 10-3 kg m-1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is: (take the radius of Earth = 6 × 106 m and the acceleration due to gravity on Earth is 10 ms-2)

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

Radius of the planet R = 1/10 x (radius of Earth)
Depth of the well = R/5 Linear mass density of the wire, μ = 10-3 kg m-1
We have to find the force applies at the top of the wire by a person holding it in.
Acceleration due to gravity on the planet will be given by g = GM/R2

(i) Acceleration due to gravity of earth  
Dividing equation (i) by (ii), we get 
Now, we will consider a mass element dm of width dx at depth x below the planet.
Mass of this element will be dm = μdx
Also, acceleration due to gravity at depth x below the planet will be 
Force on the small segment of wire 
Total force on the wire is obtained by integrating the above equation.

JEE Main Mock Test - 5 - Question 10

A stone of mass m is attached to one end of a wire of cross-sectional area A and Young's Modulus Y. The stone is revolved in a horizontal circle at speed so that the wire makes an angle θ with the vertical, the strain produced in the wire is

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

The free-body diagram of the system is shown below:

[ac is centripetal acceleration]
Centripetal acceleration is in the horizontal direction as seen in the diagram. So, no acceleration is there in a vertical direction. So, the net force in a vertical direction is zero.
Thus

If L is the original length of the string, then the increase in length is given by ΔL = (T/A) (L/Y)
where, Y: Young's modulus of the wire, A: cross-sectional area of the wire

JEE Main Mock Test - 5 - Question 11

When the temperature of a metal wire is increased from 0°C to 10°C, its length increases by 0.02%. The percentage change in its mass density will be closest to:

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

JEE Main Mock Test - 5 - Question 12

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if the process is purely adiabatic and W3 if the process is purely isobaric. Then,

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


Area under the curve is work.
W2 < W1 < W3

JEE Main Mock Test - 5 - Question 13

A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion, is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C relative to the density at T = 0°C is close to: (Answer up to 2 decimal places)

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

JEE Main Mock Test - 5 - Question 14

A monoatomic gas at pressure P and volume V is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

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

Constant entropy means the process is adiabatic.

JEE Main Mock Test - 5 - Question 15

The efficiency of a Carnot's engine, working between steam point and ice point, will be:

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

The ice point is 273 K and the steam point is 373 K.
Efficiency of the Carnot engine,

JEE Main Mock Test - 5 - Question 16

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of 'm' is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1 °C-1)

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


Heat lost by steam = heat gained by water
180 × 1 × (31 - 25) + 20 × (31-25) = m × 540 + m × 1 × (100-31)
180 × 6 +20 × 6 = 540m + 100 m - 31m
1080 + 120 = 640 m - 31m
1200 = 609m
m = 1200/609
= 1.97
m = 2

JEE Main Mock Test - 5 - Question 17

The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)

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

Here, the water will provide heat for ice to melt. Therefore,

= 0.0617 kg
= 61.7 grams
Remaining ice will remain unmelted.

JEE Main Mock Test - 5 - Question 18

7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R = 8.3 JK-1 mol-1)

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

For a quasi-static process, the change in internal energy of an ideal gas is:
ΔU = nCV ΔT

[Molar heat capacity at constant volume for monoatomic gas = 3R/2]

JEE Main Mock Test - 5 - Question 19

A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

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

JEE Main Mock Test - 5 - Question 20

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is ______. (R is the universal gas constant.

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

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

An asteroid is moving directly towards the centre of the earth. When at a distance of 10 R (R is the radius of the earth) from the earth's centre, it has speed of 12 km/s. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s)? Give your answer to the nearest integer in kilometre(s).


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

Conserving energy between (1) and (2) of Earth and asteroid system

Given v1 = 12 Km/s
v1 = 12000 m/s
Me = Mass of earth
From above equation

v2 = 16031.57 m/s
v2 = 16 km/s

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 22

A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)
[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105 Jkg-1]


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

Energy released by water = 0.3 × 25 × 4,200 = 31,500 J
Suppose m kg of ice melts.
m × 3.5 × 105 = 31,500

m = 0.09 kg = 90 gm
x = 90

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 23

A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1,200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle, is (in integers)


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


Given, W = 1,200 J

So, W = Q1 - Q2
⇒ Q2 = Q1 - W = 1,800 - 1,200 = 600 J

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 24

A block of mass 'm' (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when its speed is halved. The value of spring constant of used spring will be nE Nm-1 for n = _____________. (in integers


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

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 25

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm-1. The mass is pulled slightly downward and released, so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal is T/x. The value of x is _______. (in integers)


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

JEE Main Mock Test - 5 - Question 26

The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilisation energy (CFSE) of the complex ion, in kJ mol-1, is (1 kJ mol-1 = 83.7 cm-1)

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


CFSE = -0.4Δ0 = -0.4 × 20,300 = -8120 cm-1 

JEE Main Mock Test - 5 - Question 27

The Crystal Field Stabilisation Energy (CFSE) of [CoF3(H2O)3] (Δ0 < P) is

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

[CoF3(H2O)3] ⇒ Co3+ ⇒ d6 or t42Pg2
CFSE = [-4 × 0.4 + 2 × 0.6] Δ0 + 0
= -0.4Δ0

JEE Main Mock Test - 5 - Question 28

The incorrect statement is:

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

(1) [Fe(H2O)6]2+, Fe2+ → 3d6 → 4 unpaired electron
[Cr(H2O)6]2+, Cr2+ → 3d4 → 4 unpaired electron
(2) [Ni(NH3)4(H2O)2]2+, Ni2+ → 3d8 → 2 unpaired electron
μm = 2.83 B.M
(3) In gemstone, ruby has Cr3+ ion occupying the octahedral sites of aluminium oxide (Al2O3) normally occupied by Al3+ ion.
(4) The complementary color of violet is yellow.

JEE Main Mock Test - 5 - Question 29

Consider the hydrated ions of Ti2+, V2+, Ti3+ and Sc3+. The correct order of their spin-only magnetic moments is:

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

As we know that
μ = 
where n = number of impaired electrons, i.e. greater the number of impaired electrons more will be the spin-only magnetic moment.
Ti2+: [Ar] 3d2n = 2
Ti3+: [Ar] 3d1 n = 1
V2+: 3d3 n = 3
Sc3+: 3d0 n = 0
The correct order of spin only magnetic moments is:
V2+ > Ti2+ > Ti3+ > Sc3+
So, option B is correct.

JEE Main Mock Test - 5 - Question 30

The coordination numbers of Co and Al, in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively are
(en = ethane-1, 2-diamine)

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

[CoCl(en)2]Cl
en is a bidentate ligand.
Cl- is a monodentate ligand.
Since the central metal ion(Co+3) is involved in 5 coordination bonds, the co-ordination number of Co is 5.
K3[Al(C2O4)3]
C2O42- is a bidentate ligand.
Since the central metal ion(Al+3) is involved in 6 coordination bonds, the co-ordination number of Al is 6.

JEE Main Mock Test - 5 - Question 31

Assertion A: Enol form of acetone [CH3COCH3] exists in < 0.1% quantity. However, the enol form of acetyl acetone [CH3COCH2OCCH3] exists in approximately 15% quantity.
Reason R: Enol form of acetyl acetone is stabilised by intramolecular hydrogen bonding, which is not possible in enol form of acetone.

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


Enol from of acetone is very less (< 0.1%).

JEE Main Mock Test - 5 - Question 32

For the reaction given below:

The compound which is not formed as a product in the reaction is a

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

JEE Main Mock Test - 5 - Question 33

Which one of the following reactions will not yield propionic acid?

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

All except option 1 give propionic acid as the product, but option 1 yields butanoic as the product.

JEE Main Mock Test - 5 - Question 34

Two statements are given below:

Statement I: The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series.

Statement II: The solubility of monocarboxylic acids in water decreases with increase in molar mass.

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

Statement I is correct as monocarboxylic acids with even number of carbon atoms show better packing efficiency in solid state. Statement II is also correct as the solubility of carboxylic acids decreases with increase in molar mass

JEE Main Mock Test - 5 - Question 35


The correct order of their reactivity towards hydrolysis at room temperature is:

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

The reactivity order of hydrolysis is:
Acid halide > anhydride > ester > amide

JEE Main Mock Test - 5 - Question 36

Arrange the following coordination compounds in the increasing order of magnetic moments.
(Atomic numbers: Mn = 25; Fe = 26)

(A) [FeF6]3-
(B) [Fe(CN)6]3-
(C) [MnCl6]3- (high spin)
(D) [Mn(CN)6]3-

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

A) [FeF6]3-
Fe+3 →3d5 4s0
n = 5
(B) [Fe(CN)6]3-  

Fe+3 → 3d5 4s0
n = 1
(C) [MnCl6]3-

Mn+3 → 3d4 4s0
n = 4
(D) [Mn(CN)6]3- 

Mn+3 → 3d4 4s0
n = 2

The more the number of unpaired electrons, the more the magnetic moment.
μ 
⇒ A > C > D > B

JEE Main Mock Test - 5 - Question 37

Given below are two statements:

Statement I: Potassium permanganate on heating at 573 K forms potassium manganate.
Statement II: Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature.

In the light of the above statements, choose the most appropriate answer from the options given below:

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

JEE Main Mock Test - 5 - Question 38

Fe3+ cation gives a Prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

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

JEE Main Mock Test - 5 - Question 39

In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?

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

Cupric salts give green flame with blue centre.
Colour of other salts:
Sr2+ - Crimson red
Ca2+ - Brick red
Ba2+ - Green

JEE Main Mock Test - 5 - Question 40

Given below are two statements:

Statement I: The E° value of Ce4+/Ce3+ is +1.74 V.

Statement II: Ce is more stable in Ce4+ state than Ce3+ state.

In the light of the above statements, choose the most appropriate answer from the options given below:

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

The E° value for Ce4+/Ce3+ is +1.74 V because the most stable oxidation state of lanthanide series elements is +3.
It means that Ce3+ is more stable than Ce4+.
So, option D is the correct answer.

JEE Main Mock Test - 5 - Question 41

The reaction of H2O2 with potassium permanganate in acidic medium leads to the formation of mainly:

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

The reaction of KMnO4 with H2O2 in acidic medium is as 2KMnO4 + 3H2SO4 + 5H2O2 → K2'SO4 + 2MnSO4 + 8H2O + 5O
∴ Mn2+ will be formed as the product.

JEE Main Mock Test - 5 - Question 42

Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature?
(Given: Z for Nd = 60, Yb = 70, La = 57, Ce = 58)

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

Yb (70) = 4f14 6s2
Yb+2 = 4f14 6s0
Ytterbium shows +2 oxidation state with diamagnetic nature.

JEE Main Mock Test - 5 - Question 43

The functional groups that are responsible for the ion-exchange property of cation and anion exchange resins, respectively, are:

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

Cation exchanger contains -SO3H or -COOH groups while anion exchanger contains basic group like -NH2.

JEE Main Mock Test - 5 - Question 44

Which one of the following is an example of artificial sweetener?

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

Alitame is a second generation dipeptide sweetener, that is 200 times sweeter than sucrose.

JEE Main Mock Test - 5 - Question 45

Which of the following is not an essential amino acid?

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

Essential amino acids cannot be made by the body. The nine essential amino acids are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
Tyrosine in not an essential amino acid.

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

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 Mock Test - 5 - Question 46

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)

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 47

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 Mock Test - 5 - Question 47

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 48

The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr at pH 12.5 will be _________. (Integer answer)


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


Total negative charge produced = 4

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 49

The work function of sodium metal is 4.41 × 10-19 J. If photons of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be (h = 6.63 × 10-34 J s; c = 3 × 108 m/s) ________ × 10-21 J. (Nearest integer)


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

w, work function of sodium metal = 4.41 × 10-19 J
λ, wavelength of the incident light = 300 nm = 3 × 10-7 m
According to photoelectric effect,

6.63 × 10-19 = 4.41 × 10-19 + KE
KE = 2.22 × 10-19 J = 222 × 10-21 J

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 50

If the solubility product of AB2 is 3.20 × 10-11 M3, then the solubility of AB2 in pure water is _____ × 10-4 mol L-1. [Assuming that neither kind of ion reacts with water]
[Answer in 2 significant digits]


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


KSP = S1 × (2s)2 = 4s3
3.2 × 10-11 = 4 × S3
S = 2 × 10-4 M/L

JEE Main Mock Test - 5 - Question 51

If f(x) is a quadratic expression such that f(1) + f(2) = 0, and -1 is a root of f(x) = 0, then the other root of f(x) = 0 is:

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

JEE Main Mock Test - 5 - Question 52

The function f :  defined as f(x) = 

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


∴ From the above diagram of f(x), f(x) is surjective but not injective.

JEE Main Mock Test - 5 - Question 53

The number of four-lettered words that can be formed using the letters of the word BARRACK is:

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

B's - 1
A's - 2
R's - 2
C's - 1
K's - 1

The number of four-lettered words can be formed from (2 identical, 2 identical) or (2 identical, 2 different) or 4 different letters.

So, number of ways

JEE Main Mock Test - 5 - Question 54

The number of ways of arrangements of 10 persons in four chairs is -

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

Given:

The number of ways of arrangements of 10 persons in four chairs  

Formula used:

nPr = n!/(n – r)!

Where, n = Number of persons

r = Number of chairs

Calculation:

According to the question

nPr = n!/(n – r)!

⇒ 10!/(10 – 4)!

⇒ 10!/6!

⇒ (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1)

⇒ (10 × 9 × 8 × 7)

⇒ 5040

∴ The required value is 5040

JEE Main Mock Test - 5 - Question 55

If tan (π/9), x, tan (7π/18) are in arithmetic progression and tan (π/9), y, tan (5π/18) are also in arithmetic progression, then |x - 2y| is equal to:

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

JEE Main Mock Test - 5 - Question 56

Let Sn denote the sum of first n terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 - S6 is equal to:

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

S10 = 530 ⇒ 10/2 {2a + 9d} = 530
⇒ 2a + 9d = 106 ...(1)
and S5 = 140 ⇒ 5/2 {2a + 4d} = 140
⇒ 2a + 4 d = 56 ...(2)
⇒ 5 d = 50 ⇒ d = 10 ⇒ a = 8
Now, S20 - S6 = 20/2 {2a + 19 d} - 6/2 {2a + 5d}
= 14a + 175 d
= (14 × 8) + (175 × 10)
= 1862

JEE Main Mock Test - 5 - Question 57

Let a1, a2, ... an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + ... + an. If a1 = 1, an = 300 and 15 ≤ n ≤ 50, then the ordered pair (Sn - 4, an - 4) is equal to

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

a1 = 1 and an = 300; d∈Z
300 = 1 + (n - 1)d

∴ n - 1 = 13 or 23 (as d is an integer)
⇒ n = 14 or 24
⇒ n = 24 and d = 13
a20 = 1 + 19 × 13 = 248

= 2490

JEE Main Mock Test - 5 - Question 58

Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 - S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:

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

JEE Main Mock Test - 5 - Question 59

If  then the remainder when K is divided by 6 is

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


Now, 310 - 210 = (35 - 25)(35 + 25)
= (211)(275)
= (35 × 6 + 1)(45 × 6 + 5)
= 6λ + 5
Hence, the remainder is 5.

JEE Main Mock Test - 5 - Question 60

The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + ... + 492 + 49 + 1, is

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

1 + 49 + 492 + ........ 49125 

Hence, k = 63

JEE Main Mock Test - 5 - Question 61

Let α be a root of the equation x2 + x + 1 = 0 and the matrix A  then the matrix A31 is equal to:

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

Given: x2 + x + 1 = 0, and α is a root of the given equation.
So, α = ω, ω

A4 = I
 A31 = A28 . A3 = A3

JEE Main Mock Test - 5 - Question 62

Let A = and B = then the value of A'BA is:

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



JEE Main Mock Test - 5 - Question 63

Let A and B be any two 3 × 3 symmetric and skew-symmetric matrices, respectively. Then which of the following is NOT true?

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

Given: AT = A, BT = -B
From option 1:
Let C = A4 - B4
CT = (A4 - B4) = (A4)T - (B4)T = A4 - B4 = C
From option 2:
Let C = AB - BA
CT = (AB - BA)T = (AB)T - (BA)T
= BTAT - ATBT = -BA + AB = C
From option 3:
Let C = B5 - A5
CT = (B5 - A5)T = (B5)T - (A5)T = -B5 - A5
From option 4:
Let C = AB + BA
CT = (AB + BA)T = (AB)T + (BA)T
= -BA - AB = -C
∴ Option 3 is not true.

JEE Main Mock Test - 5 - Question 64

Let A = [aij] be a square matrix of order 3 such that aij = 2j-i, for all i, j = 1, 2, 3. Then, the matrix A2 + A3 + ... + A10 is equal to:

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

JEE Main Mock Test - 5 - Question 65

Let A = [aij] be a real matrix of order 3 x 3, such that ai1 + ai2 + ai3 = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A3 is equal to:

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


⇒ AX = X
Replace X by AX
A2X = AX = X
Replace X by AX
A3X = AX = X

JEE Main Mock Test - 5 - Question 66

If for some α and β in R, the intersection of the following three planes
x + 4y - 2z = 1
x + 7y - 5z = β
x + 5y + α z = 5
is a line in R3, then α + β is equal to

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

∵ Three equations have infinitely many solutions, so

Putting the value of α in equation (3),
x + 4y - 2z = 1 and x + 5y - 3z = 5
Solving, y = z + 4 and x = -2z - 15
Substituting these values in equation (2),
x + 7y - 5z = β = -2z - 15 + 7z + 28 - 5z

JEE Main Mock Test - 5 - Question 67

The ordered pair (a, b), for which the system of linear equations

3x - 2y + z = b
5x - 8y + 9z = 3
2x + y + az = -1

has no solution, is:

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


Therefore, the required ordered pair is 

JEE Main Mock Test - 5 - Question 68

Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then

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

AB = 0
⇒ |AB| = 0

If |A| ≠ 0, B = 0 (not possible)
If |B|  0, A = 0 (not possible)
Hence, |A| = |B| = 0
 AX = 0 has infinitely many solutions.

JEE Main Mock Test - 5 - Question 69

Let A be a 3 × 3 invertible matrix. If |adj(24A)| = adj(3adj(2A))|, then |A|2 is equal to:

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

JEE Main Mock Test - 5 - Question 70

Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is

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

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

If A =  and M = A + A2 + A3 + ... + A20, then the sum of all the elements of the matrix M is equal to __________. (in integers)


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

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

Let I be an identity matrix of order 2 × 2 and P = Then the value of n ∈ N for which Pn = 5I - 8P is equal to _______.(in integers)


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

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 73

Let the function f(x) = 2x2 - logex, x > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y2 = 4ax at a point P on it passes through the point (8a, 8a - 1) but does not pass through the point  If the equation of the normal at P is x/α + y/β = 1, then α + β is equal to____. (in integer)


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


*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 74

Let AD and BC be two vertical poles at A and B, respectively, on a horizontal ground. If AD = 8 m, BC = 11 m and AB = 10 m, then the distance (in metres) of a point M on AB from the point A such that MD2 + MC2 is minimum is ________. (in integer)


Detailed Solution for JEE Main Mock Test - 5 - Question 74

(MD)2 + (MC)2 = h2 + 64 + (h - 10)2 + 121
= 2h2 - 20h + 64 + 100 + 121
= 2(h2 - 10h) + 285
= 2(h - 5)2 + 235
It is minimum if h = 5.

*Answer can only contain numeric values
JEE Main Mock Test - 5 - Question 75

Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ R If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)


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

f(x) = |(x - 1)(x + 1)(x - 3)| + (x - 3)

f'(3+) > 0 f'(3-) < 0 → Minimum
f'(1+) > 0 f'(1-) < 0 → Minimum
x ∈ (1, 3) f'(x) = 0 at one point → Maximum
x ∈ (3, 4) f'(x)  0
x ∈ (0, 1) f'(x)  0
So, there are 2 points of minima and 1 point of maxima.
Hence, m + M = 3

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