JEE Exam  >  JEE Tests  >  Mock Tests for JEE Main and Advanced 2024  >  JEE Main Part Test - 3 - JEE MCQ

JEE Main Part Test - 3 - JEE MCQ


Test Description

75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Part Test - 3

JEE Main Part Test - 3 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Part Test - 3 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Part Test - 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Part Test - 3 below.
Solutions of JEE Main Part Test - 3 questions in English are available as part of our Mock Tests for JEE Main and Advanced 2024 for JEE & JEE Main Part Test - 3 solutions in Hindi for Mock Tests for JEE Main and Advanced 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Part Test - 3 | 75 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Tests for JEE Main and Advanced 2024 for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Main Part Test - 3 - Question 1

Kinetic theory explains the behavior

Detailed Solution for JEE Main Part Test - 3 - Question 1

Explanation:

The theory for ideal gases makes the following assumptions

1. Gases consist of particles in constant, random motion. They continue in a straight line until they collide with something—usually each other or the walls of their container.

2. Particles are point masses with no volume. The particles are so small compared to the space between them, that we do not consider their size in ideal gases.

3. No molecular forces are at work. This means that there is no attraction or repulsion between the particles.

4. Gas pressure is due to the molecules colliding with the walls of the container. All of these collisions are perfectly elastic, meaning that there is no change in energy of either the particles or the wall upon collision. No energy is lost or gained from collisions.

5. The time it takes to collide is negligible compared with the time between collisions.

6. The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas molecules have different speeds, but the temperature and kinetic energy of the gas refer to the average of these speeds.

7. The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move.

8. All gases at a given temperature have the same average kinetic energy.

9. Lighter gas molecules move faster than heavier molecules.

JEE Main Part Test - 3 - Question 2

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Detailed Solution for JEE Main Part Test - 3 - Question 2

Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011)   =   1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m
Hence 1.87 × 10–3 m

JEE Main Part Test - 3 - Question 3

Two bullets A and B are fired horizontally with speed v and 2v respectively.which of the following is true

JEE Main Part Test - 3 - Question 4

The displacement of a damped harmonic oscillator is given by:
x(t) = e-0.1t cos (10πt + φ). Hence, t is in seconds.

The time taken for its amplitude of vibration to drop to half of its initial value is close to:

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

JEE Main Part Test - 3 - Question 5

1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?

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

Explanation:

for monoatomic gas 

 

from conservation of energy

JEE Main Part Test - 3 - Question 6

The average distance a molecule can travel without colliding is called the

Detailed Solution for JEE Main Part Test - 3 - Question 6

Explanation:the mean free path is the average distance traveled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties

JEE Main Part Test - 3 - Question 7

Value of gas constant, R for one mole of a gas is independent of the

Detailed Solution for JEE Main Part Test - 3 - Question 7

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

JEE Main Part Test - 3 - Question 8

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas

Detailed Solution for JEE Main Part Test - 3 - Question 8

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas the internal energy is zero because at 0K nearly all molecular motion stops.

JEE Main Part Test - 3 - Question 9

Four moles of an ideal diatomic gas is heated at constant volume from 20° C to 30° C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol-1K-1 and the universal gas constant (R) is 8.3 Jmol-1K-1. The increase in internal energy of the gas is

Detailed Solution for JEE Main Part Test - 3 - Question 9

The value of Cp is 30.3
and as Cp-Cv = R(8.3)
hence Cv = 30.3-8.3
Cv is 22
change in internal energy = no of moles × Cv × change in temperature
hence
change in internal energy = 22 × 4 × 10
= 880j
Hence Option D is correct.
 

JEE Main Part Test - 3 - Question 10

Three moles of an ideal monoatomic gas is initially in the state A shown in the adjoining pressure-temperature graph. It is taken to state B without changing its pressure. If R is the universal gas constant, the work done by the gas in this process is 

Detailed Solution for JEE Main Part Test - 3 - Question 10

The work done by the gas in taking it from state A to state B = PΔV where ΔW is the increase in volume at constant pressure P. 

We have PV = μRT where p is the number of moles in the sample of the gas and R is the universal gas constant. 

Therefore we have PΔV = μR ΔT = 3 xR(450 - 250) = 600R 

JEE Main Part Test - 3 - Question 11

A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:

Detailed Solution for JEE Main Part Test - 3 - Question 11

 

  • Time Period, T = 2π √(l/g')where,
    l = Length of seconds pendulum 
    g’ = Apparent Gravity
  • For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
  • Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
JEE Main Part Test - 3 - Question 12

Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3 sin157t + 4 cos157t, where t is time in seconds.

Detailed Solution for JEE Main Part Test - 3 - Question 12

When the displacement of a SHM is:
y=a sin wt+ b cos wt

  • Amplitude of the SHM will be:
    A=√a2+b2

Here, a = 3, b = 4
Amplitude, A= √(32+42) = 5 cm

Hence option B is correct.

JEE Main Part Test - 3 - Question 13

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

Detailed Solution for JEE Main Part Test - 3 - Question 13

   
∴ We get, ω = √3 s-1
   T = 2π / √3

JEE Main Part Test - 3 - Question 14

What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×105 N/m and total mechanical energy of 230 J.

Detailed Solution for JEE Main Part Test - 3 - Question 14

K. Σ=1/2 K(A2-x2)
Max of mean position,
K. Σ=1/2 KA2
=1/2 x4x105x(3x10-2)2
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230-180
P.Σ=50J

JEE Main Part Test - 3 - Question 15

The velocity and acceleration amplitudes of body executing simple harmonic motion is

Detailed Solution for JEE Main Part Test - 3 - Question 15

Maximum velocity: v = ωA, where ω is uniform angular velocity and a is the radius of the circle in which a reference particle executing S.H.M.
Velocity is maximum at mean positions. The maximum value of velocity is called velocity amplitude in SHM.
Acceleration is maximum at extreme position given by A = - ω2A. The maximum value of acceleration is called acceleration amplitude in SHM.

JEE Main Part Test - 3 - Question 16

A particle of mass m is executing oscillation about the origin on X-axis. Its potential energy is V(x)=K∣x∣3. Where K is a positive constant. If the amplitude of oscillation is a, then its time period T is proportional to.

Detailed Solution for JEE Main Part Test - 3 - Question 16

JEE Main Part Test - 3 - Question 17

Energy is supplied to the damped oscillatory system at the same rate at which it is dissipating energy, then the amplitude of such oscillations would become constant. Such oscillations are called

Detailed Solution for JEE Main Part Test - 3 - Question 17

Energy is supplied to the damped oscillatory system at the same rate at which it is dissipating energy, and then the amplitude of such oscillations would become constant. Such oscillations are called maintained oscillations. By the definition of maintained oscillations.

JEE Main Part Test - 3 - Question 18

What determines the natural frequency of a body?

Detailed Solution for JEE Main Part Test - 3 - Question 18

Natural frequency is the frequency at which a body tends to oscillate in the absence of any driving or damping force.
Free vibrations of any elastic body are called natural vibration and happen at a frequency called natural frequency. Natural vibrations are different from forced vibration which happen at frequency of applied force .

JEE Main Part Test - 3 - Question 19

In the ideal case of zero damping, the amplitude of simple harmonic motion at resonance is:

Detailed Solution for JEE Main Part Test - 3 - Question 19

In an ideal environment where there is no resistance to oscillatory motion, that is, damping is zero, when we oscillate a system at its resonant frequency, since there is no opposition to oscillation, a very large value of amplitude will be recorded. Forced oscillation is when you apply an external oscillating force.

JEE Main Part Test - 3 - Question 20

The necessary condition for phenomenon of interference to occur is

Detailed Solution for JEE Main Part Test - 3 - Question 20

The necessary condition for phenomenon of interference to occur are:
1. There should be two coherent sources.
2. The frequency and amplitude of both the waves should be same.
3. The propagation of waves should be simultaneously and in same direction.
These are the conditions, no explanation.

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 21

The equation of wave is given by

Y = 10-2 sin 2π (160t – 0.5x + π/4)

where x and Y are in m and t in s. The speed of the wave is _______ km h-1.


Detailed Solution for JEE Main Part Test - 3 - Question 21

Given the wave equation:

Y = 10-2 sin 2π (160t – 0.5x + π/4)

Comparing this equation with the general form:

Y = Asin (2π(ft - kx + φ))

We can identify the wave number k = 0.5 m-1 and the frequency f = 160 Hz. 

The wave speed v can be found using the relationship between wave number, wave speed, and frequency: 

Now, we can calculate the wave speed:

Now, we need to convert the wave speed from meters per second to kilometers per hour:

So, the speed of the wave is 1152 Km/h

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 22

An organ pipe 40 cm long is open at both ends. The speed of sound in air is 360 ms-1.The frequency of the second harmonic is ___________ Hz.


Detailed Solution for JEE Main Part Test - 3 - Question 22

An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.

The frequency f of the n-th harmonic for a pipe open at both ends is given by:

where:
n is the number of the harmonic,
v is the speed of sound, and
L is the length of the pipe.

To find the frequency of the second harmonic (n = 2) we can substitute the given values into the formula:

Therefore, the frequency of the second harmonic is 900 Hz.

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 23

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be x/40 N. The value of x is ________.


Detailed Solution for JEE Main Part Test - 3 - Question 23

Given the amplitude A and the length l of the pendulum, we can find the maximum angular displacement θ0.

By conservation of energy, the following equation holds:

The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:

Substituting the conservation of energy equation, we get:

So, the value of x is 99.

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 24

The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is: ___________ cm


Detailed Solution for JEE Main Part Test - 3 - Question 24


*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 25

A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (-25x)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.


Detailed Solution for JEE Main Part Test - 3 - Question 25

JEE Main Part Test - 3 - Question 26

Which of the following correctly ranks the cycloalkanes in order of increasing ring strain per methylene group? 

Detailed Solution for JEE Main Part Test - 3 - Question 26

The correct answer is Option B.

The C-C-C bond angles in cyclopropane (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o.This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o.

JEE Main Part Test - 3 - Question 27

Which of the following cycloalkanes exhibits the greatest molar heat of combustion per —CH2 — group?

Detailed Solution for JEE Main Part Test - 3 - Question 27

The correct answer is option D
Cyclopropane is a cycloalkane molecule with the molecular formula C3H6, consisting of three carbon atoms linked to each other to form a ring, with each carbon atom bearing two hydrogen atoms resulting in D3H molecular symmetry. The small size of the ring creates substantial ring strain in the structure.

JEE Main Part Test - 3 - Question 28

The correct lUPAC name of the compound

Detailed Solution for JEE Main Part Test - 3 - Question 28

Priority of functional groups is
- COOH > - CHO >>C = O.
Hence, the lUPAC name is 4-formyl-2-oxocyclohexane carboxylicacid.

JEE Main Part Test - 3 - Question 29

How many σ and π bonds are present in HC≡C−CH=CH−CH3?

Detailed Solution for JEE Main Part Test - 3 - Question 29

JEE Main Part Test - 3 - Question 30

Arrange in increasing order of basicity. HC≡C−, CH3​CH=CH−, CH3​CH2​−

Detailed Solution for JEE Main Part Test - 3 - Question 30

JEE Main Part Test - 3 - Question 31

What are the hybridization and shapes of the following molecules?
(i) CH3F
(ii) HC ≡ N

Detailed Solution for JEE Main Part Test - 3 - Question 31

CH3F - sp3 hybridised carbon, tetrahedral shape
(ii) HC = N - sp hybridised carbon, linear shape.

*Multiple options can be correct
JEE Main Part Test - 3 - Question 32

Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. Consider the following compounds.

The correct statement regarding properties of above mentioned compounds is/are

Detailed Solution for JEE Main Part Test - 3 - Question 32
  • Both have all their C—C bonds of equal length due to conjugation.
  • I does not decolorises brown colour of bromine water solution but II does as The π bonds in Cyclooctatetraene (Compound II) react as usual for olefins, rather than as aromatic ring systems.
  • I is planar but II is not as it adopts a tub conformation.
  • Cyclooctatetraene shows various other addition reactions including Sulfonation.

Hence, Option A, B and D are correct.

*Multiple options can be correct
JEE Main Part Test - 3 - Question 33

What is true about the 1,3,5,7-cyclooctatetraene?

Detailed Solution for JEE Main Part Test - 3 - Question 33

1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.
Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:
2 K + C8H8 → K2C8H8
The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.

JEE Main Part Test - 3 - Question 34

The compound shown below evolve hydrogen gas when refluxed with potassium metal, why?

Detailed Solution for JEE Main Part Test - 3 - Question 34

Metals like Potassium tend to loose their own electrons and the excess electrons will complete the aromaticity of the system.Thus, Deprotonation of the above compound converts it into an aromatic anion witn 6 pi electrons.

JEE Main Part Test - 3 - Question 35

Which of the following behaves both as a nucleophile and as an electrophile?

Detailed Solution for JEE Main Part Test - 3 - Question 35

R − Cδ = Nδ: has electrophilic C-atom and nucleophilic N-atom.

JEE Main Part Test - 3 - Question 36

Butanone is a four-carbon compound with the functional group –

Detailed Solution for JEE Main Part Test - 3 - Question 36

In organic chemistry, the word butane shows the presence of four carbons.Thus, we can conclude that the functional group present in the compound is ketone.

JEE Main Part Test - 3 - Question 37

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because-

Detailed Solution for JEE Main Part Test - 3 - Question 37

Due to more hyperconjugation, tertiary butyl carbocation is more stable than secondary butyl carbocations.

JEE Main Part Test - 3 - Question 38

Match items of column I and II

Detailed Solution for JEE Main Part Test - 3 - Question 38
  • Water and dichloromethane can be separated by differential extraction.
  • C6H12O6 and NaCl can be separated by crystallization.
JEE Main Part Test - 3 - Question 39

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : Benzene is more stable than hypothetical cyclohexatriene

Reason R : The delocalised π electron cloud is attracted more strongly by nuclei of carbon atoms.

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

Detailed Solution for JEE Main Part Test - 3 - Question 39

Benzene is more stable than hypothetical cyclohexatriene due to the delocalization of electrons in the π electron cloud of the aromatic ring. This delocalization of electrons leads to a more stable electronic configuration in benzene, as the electrons are shared more evenly between all six carbon atoms in the ring. This is in agreement with Assertion A.

Reason R is also correct, as the delocalized π electron cloud in the aromatic ring is attracted more strongly by the nuclei of the carbon atoms. This is due to the fact that the π electrons in the ring are more spread out and are not localized to a single atom, which makes them more attracted to the positively charged nuclei of the carbon atoms. This explanation supports Assertion A.

JEE Main Part Test - 3 - Question 40

Given below are two statements.

Statement I: The compound is optically active.

Statement II: is mirror image of above compound A.

Detailed Solution for JEE Main Part Test - 3 - Question 40


Compound (A) in Statement-I and compound in Statement-II is not the mirror image of (I).

JEE Main Part Test - 3 - Question 41

 Which of the following compounds react most readily with Br(g)?

Detailed Solution for JEE Main Part Test - 3 - Question 41

The reactivity of a compound with Br(g) (Bromine gas) depends on the type of carbon-carbon bonds present in the compound. Compounds with double or triple bonds react more readily than those with single bonds because they have a higher electron density, which is more attractive to the electrophilic bromine.

Reactivity of Given Compounds:
C2H2: This compound has a triple bond between the two carbon atoms, which gives it a high electron density. However, the bond is also very strong, which makes the reaction slower.
C3H6: This compound has a double bond, which gives it a high electron density and makes it more reactive than compounds with single bonds. The double bond is weaker than the triple bond in C2H2, which makes the reaction faster.
C2H4: This compound has a double bond like C3H6. While it would also react readily with Br(g), the reaction would not be as fast as with C3H6 because C2H4 has fewer carbon atoms, and therefore less electron density.
C4H10: This compound only has single bonds, which makes it the least reactive of the four options.

In conclusion,C3Hreacts most readily with Br(g) because it has a good balance of high electron density due to its double bond and faster reaction rate due to the bond's relative weakness. This makes it more attractive to the electrophilic bromine and allows the reaction to proceed more quickly.

JEE Main Part Test - 3 - Question 42

When propene reacts with HBr in the presence of peroxide, it gives rise to

Detailed Solution for JEE Main Part Test - 3 - Question 42

Explanation of the Reaction of Propene with HBr in the Presence of Peroxide

The reaction of propene with hydrogen bromide in the presence of peroxide follows the rule of anti-Markovnikov addition. This rule states that the hydrogen (H) from HBr will add to the carbon with the most hydrogen atoms already attached, and the bromine (Br) will add to the other carbon. Peroxide promotes this anti-Markovnikov addition.

In the case of propene (CH3-CH=CH2), the hydrogen from HBr will add to the terminal carbon, which already has two hydrogens. The bromine will add to the middle carbon.

Here are the steps of the reaction:

JEE Main Part Test - 3 - Question 43

Ethylene bromide on treatment with Zn gives

Detailed Solution for JEE Main Part Test - 3 - Question 43

Reaction of Ethylene Bromide with Zinc
Ethylene bromide, also known as 1,2-dibromoethane, is a halogenated hydrocarbon. When it is treated with zinc, an alkene is formed as a result. This reaction can be detailed as follows:
The compound formed when an alcoholic solution of ethylene dibromide is  heated with granulated zinc is:(A) Ethene(B) Ethyne(C) Ethane(D) Bromoethane
Why not Alkyne or Alkane?
An alkyne would require the removal of two pairs of hydrogen and bromine atoms, which does not occur in this reaction. An alkane would not have any double bonds, and the reaction with zinc specifically creates a double bond.

In conclusion, the correct answer is B: Alkene, because the reaction of ethylene bromide with zinc results in the formation of an alkene, specifically ethene. You can learn more about organic chemistry reactions on the EduRev platform.

JEE Main Part Test - 3 - Question 44

The position of double bond in alkenes can be located by :

Detailed Solution for JEE Main Part Test - 3 - Question 44

Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon-carbon bonds have been replaced by a double bond to oxygen. The outcome of the reaction depends on the type of multiple bonds being oxidized.

Bromine water can be also used to identify the position of a double bond. In this reaction, red-brown colour of bromine gets turned into coulorless indicating that there is a double bond.

JEE Main Part Test - 3 - Question 45

The compound C3H4 has a triple bond, which is indicated by its reaction with

Detailed Solution for JEE Main Part Test - 3 - Question 45

CH3−C ≡ C−H + AgNO3 → CH3 −C ≡ C−Ag
Propyne Ammonical Siver salt of Propyne

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 46


Consider the above chemical reaction. The total number of stereoisomers possible for Product 'P' is _____________.


Detailed Solution for JEE Main Part Test - 3 - Question 46


The total number of products possible = 2

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 47

The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is _____.


Detailed Solution for JEE Main Part Test - 3 - Question 47

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

∴ No. of moles of O2 required to oxidise 1 mole of propane = 5

∴ No. of moles of O2 required to oxidise 1 mole of butane = 13/2

So, No. of moles of O2 required to oxidise 1 mole of propane and 2 moles of butane = 5 + 2 × 13/2 = 18

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 48

In the following sequence of reactions the maximum number of atoms present in molecule 'C' in one plane is _________.

(A is a lowest molecular weight alkyne)


Detailed Solution for JEE Main Part Test - 3 - Question 48

As A is a lowest molecular weight alkyne. So A is H - C ≡ C - H.

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 49

Consider the following chemical reaction.

The number of sp2 hybridized carbon atom(s) present in the product is ________.


Detailed Solution for JEE Main Part Test - 3 - Question 49


All the 7-carbon-atoms in product are sp2 hybridised.

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 50

The major product 'A' of the following given reaction has _____________ sp2 hybridized carbon atoms.


Detailed Solution for JEE Main Part Test - 3 - Question 50


Number of sp2 hybridised carbon atoms = 2

JEE Main Part Test - 3 - Question 51

In the given figure, co-ordinates of the midpoint of AB are?

Detailed Solution for JEE Main Part Test - 3 - Question 51

Here we are given two points A(0, 0) and B (6, 2).

Since the midpoint falls midway between A and B, it has x-coordinates of A and B.

x-coordinates of midpoint of AB = 0+6/2 = 3.

and y-coordinates of the midpoint of AB =  0 + 2/2 = 1.

Thus the midpoint of AB has co-ordinates (3, 1).

JEE Main Part Test - 3 - Question 52

The coordinate of any point, which lies in xy plane, is?

Detailed Solution for JEE Main Part Test - 3 - Question 52

Given that the point lies in xy plane

In xy plane, the coordinate of z will be zero

So (x,x,0) represents a point which lies in xy plane

therefore option b is correct.

JEE Main Part Test - 3 - Question 53

The point (3,0,−4) lies on the?

Detailed Solution for JEE Main Part Test - 3 - Question 53

(3,0,−4)  →   Given point

Clearly, y=0 and x and z have non-zero value.

If the point lies on x−z plane, this condition is possible.

Hence, the answer is XZ- plane.

JEE Main Part Test - 3 - Question 54

Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by?

Detailed Solution for JEE Main Part Test - 3 - Question 54

Equation of the plane through (2, –3, 1) is

a(x – 2) + b(y + 3) + c(z – 1) = 0

It passes through (–1, 1, –7) and perpendicular to the plane 

x – 2y + 5z + 1 = 0

∴ –3a+ 4b – 8c = 0 and a – 2b + 5c = 0

∴ a/4 = a/7 = c/2

Hence, equation of required plane is

4x + 7y + 2z + 11 = 0

JEE Main Part Test - 3 - Question 55

If a line OP of length r (Where 'O' is the origin) makes an angle α with x-axis and lies on the xz-plane, then what are the coordinates of P?

Detailed Solution for JEE Main Part Test - 3 - Question 55


As OP lies on XZ plane so y co-ordinate is 0.

QP = r cos α 

PR = r sin α 

Co-ordinate of P = (rcosα,0,rsinα) 

Option a is correct.

JEE Main Part Test - 3 - Question 56

If the projections of a line on the axes are 9, 12 and 8. Then the length of the line is?

Detailed Solution for JEE Main Part Test - 3 - Question 56

JEE Main Part Test - 3 - Question 57

The plane  meets the coordinate axes at A, B, C respectively. D and E are the mid-points of AB and AC respectively. Coordinates of the mid-point of DE are?

Detailed Solution for JEE Main Part Test - 3 - Question 57

We have A(a, 0, 0), B(0, 0, 0), C(0, 0, c)

For midpoint coordinates, we have

Similarly, midpoint of DE is 

Hence, the correct option is (d).

JEE Main Part Test - 3 - Question 58

The foot of the perpendicular from (0, 2, 3) to the line  is?

Detailed Solution for JEE Main Part Test - 3 - Question 58


Let P (5λ -  3, 2λ + 1, 3λ - 4) be the foot of perpendicular of A (0, 2, 3). Then AP and given line are perpendicular to each other.
∴ (5λ - 3) 5 + (2λ - 1) 2 + (3λ - 7) 3 = 0
⇒ 38λ - 38 = 0 ⇒ λ = 1
Hence, P = (2, 3, - 1)

JEE Main Part Test - 3 - Question 59

In a single throw of two dice, find the probability that neither a doublet nor a total of 10 will appear.

Detailed Solution for JEE Main Part Test - 3 - Question 59

Total outcomes = 36

Favourable outcomes for getting a doublet or a total of 10 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (4, 6), (6, 4)]

Probability = 8/36

Probability that neither a doublet nor a total of 10 will appear = 1 - 8/36 = 28/36 = 7/9

JEE Main Part Test - 3 - Question 60

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is

Detailed Solution for JEE Main Part Test - 3 - Question 60

Let E1 denote the event that the letter has come from TATANAGAR and E2 the event that the letter has come from CALCUTTA.

Let B denote the event that the two consecutive alphabets visible on the envelope are TA.

We have P (E1) =  1/2, P (E2) = 1/2, P (B | E1) = 2/8, P (B|E2) = 1/7

Therefore, by the Bayes' theorem we have

JEE Main Part Test - 3 - Question 61

Two numbers x and y are chosen at random without replacement from the first 30 natural numbers. The probability that  x2 - y2 is divisible by 3 is 

Detailed Solution for JEE Main Part Test - 3 - Question 61

Total number of ways of choosing two numbers out of 30 is 30C2 We rewrite the first 30  natural numbers in three rows as follows:

Row I: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28
Row II: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29
Row III: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

For x2 - y2 to be divisible by 3, either both x and y must be chosen from the same row or exactly one of x, y from Row I and the other from Row II.

Thus, the number of favourable ways

= 3 (10C2) + 10 x 10 = 235

∴ probability of the required event

JEE Main Part Test - 3 - Question 62

Three numbers are chosen at random from the first 20 natural numbers. The probability that the selected numbers form a G.P. is

JEE Main Part Test - 3 - Question 63

The probability that in a group of 3 people, at least two will have the same birthday is

Detailed Solution for JEE Main Part Test - 3 - Question 63

Let A be the event that at least two people have the same birthday.
Then,

JEE Main Part Test - 3 - Question 64

Past records show that on an average, 1 house out of 1000 in a certain district has a fire during a year. If there are 2000 houses in that district, the probability that 5 houses will have a fire is

JEE Main Part Test - 3 - Question 65

Alex is taking readings of an experiment. He calculated the algebraic sum of deviation of 25 observations which were measured from 40 was found out to be 5. What is the mean of the readings?

Detailed Solution for JEE Main Part Test - 3 - Question 65

Mean  

Sum of observations = 5

Number of observations = 25

Mean = 5/25 = 0.2

But he started the observations from 40

So the mean = 40 + 0.2

= 40.2

JEE Main Part Test - 3 - Question 66

A scientist was weighing each of 30 fishes. Their mean weight worked out is 30 g and standard deviation is 2 g. Later, it was found that the measuring scale was misaligned and always under-reported every fish's weight by 2 g. The correct mean and standard deviation (in g) of fishes, are respectively

Detailed Solution for JEE Main Part Test - 3 - Question 66

Since mean (X + b) = E( X + b) = E(X) + b and Var (X + b) = Var X,
We get correct mean as 30 + 2 = 32 g and (s.d.) standard deviation as 2 g.

JEE Main Part Test - 3 - Question 67

If the standard deviation of x1, x2, …, xn is 3.5, then the standard deviation of – 2x1 – 3, – 2x2 – 3, …, – 2xn – 3 is:

Detailed Solution for JEE Main Part Test - 3 - Question 67

The standard deviation of a set remains unchanged if each data is increased or decreased by a constant.
However, it changes similarly when data is multiplied or divided by a constant.
∴ The SD for the new data set will be = −2 × 3.5 = −7

JEE Main Part Test - 3 - Question 68

The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are:

Detailed Solution for JEE Main Part Test - 3 - Question 68


x5 + x6 = 21
Let other two numbers be a, (21 - a)
Now,

(Using formula for variance)

⇒ 6(10.25) + 6(6.5)2 = 94 + a2 + (21 - a)2

⇒ a2 + (21 - a)2 = 221

∴ a = 10 and (21 - a) = 21 - 10 = 11

So, remaining two observations are 10, 11.

⇒ Option (a) is correct.

JEE Main Part Test - 3 - Question 69

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

Detailed Solution for JEE Main Part Test - 3 - Question 69

JEE Main Part Test - 3 - Question 70

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is:

Detailed Solution for JEE Main Part Test - 3 - Question 70

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 71

If the variance of the terms in an increasing A.P. b1, b2, b3, ..., b11 is 90, then the common difference of this A.P. is ________.


Detailed Solution for JEE Main Part Test - 3 - Question 71


*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 72

If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to _______.


Detailed Solution for JEE Main Part Test - 3 - Question 72

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 73

If the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively, then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x - y is ______.


Detailed Solution for JEE Main Part Test - 3 - Question 73

*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 74

Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space [0, 60] is less than or equal to a. If P (A) = 11/36, then a is equal to _______.


Detailed Solution for JEE Main Part Test - 3 - Question 74



*Answer can only contain numeric values
JEE Main Part Test - 3 - Question 75

A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is q. If p: q = m : n, where m and n are coprime, then m + n is equal to:


Detailed Solution for JEE Main Part Test - 3 - Question 75

357 docs|148 tests
Information about JEE Main Part Test - 3 Page
In this test you can find the Exam questions for JEE Main Part Test - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Part Test - 3, EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!