JEE Main Part Test - 6 - JEE MCQ

# JEE Main Part Test - 6 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Part Test - 6

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JEE Main Part Test - 6 - Question 1

### Let nr and nb be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

Detailed Solution for JEE Main Part Test - 6 - Question 1

Since, Pr​=Pb​
r for red and b for blue.
Pr​=Pb​
or, nr​× (hc/λr)  ​=nb​× (hc​/λb)​
or,  ​(nr/nb)​​= λr​​/λb
Since, the wavelength of red bulb is greater than the wavelength of blue bulb.
or,  nr​>nb

JEE Main Part Test - 6 - Question 2

### 10-3 W of 5000 Å light is directed on a photoelectric cell. If the current in the cell is 0.16 mA, the percentage of incident photons which produce photoelectrons, is

Detailed Solution for JEE Main Part Test - 6 - Question 2

Current is 0.16×10−6 Amp it means 0.16×10−6 Coulomb charge is flowing per second
So, n=0.16×10−6C /1.6×10−19C ​=1012 electrons are generated per second
Now we notice that one photon has energy E, E=hc/λ=​=(6.62×10−34Js×3×108ms-1)/(5000×10−10m) ​=3.972×10−19Joule
So, number of photon in 10−3W will be N=10−3/3.972×10−19 ​=0.25×1016 this is number of photons incident per second
So required percentage is (n/N)×100=1014/(0.25×1016) ​=0.04%

JEE Main Part Test - 6 - Question 3

### If the frequency of light in a photoelectric experiment is doubled, the stopping potential will

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

The maximum kinetic energy for the photoelectrons is
Emax​=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
eV0​=hν−ϕ ...................(1)
where, V0​ is the stopping potential and e is the electronic charge.
When, the frequency of light in a photoelectric experiment is doubled,
eV0′​=2hν−ϕ
eV0′​=2[hν−(ϕ/2​)].........................(2)
From the above two equations we can say that the K.E. in (2) is more than double of K.E in (1). Hence, when the frequency of light in a photoelectric experiment is doubled, the stopping potential becomes more than double.
So, the answer is option (C).

JEE Main Part Test - 6 - Question 4

The stopping potential for the photo electrons emitted from a metal surface of work function 1.7eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect

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

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,
KEmax​=10.4V
Now, in photoelectric effect,
Energy of incident radiation (Ein​) = work function + K.Emax​
⇒ Ein​=1.7+10.4
⇒ Ein​=12.1eV
Now, for 0 hydrogen atom,
Energy of first energy level, E1​=−13.6eV
Energy of second energy level, E2​=−3.4eV
Energy of third energy level, E3​=−1.5eV
Hence, a transition from third to first energy level will result in emission of radiation of energy = E3​−E1​=12.1eV which is same as the energy of incident radiation of above photoelectric effect.
Thus, correct answer is n=3 to 1

JEE Main Part Test - 6 - Question 5

When a photon of light collides with a metal surface, number of electrons, (if any) coming out is

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

When photon strikes with the electron it completely transfers it’s energy to the electron as during photoelectric experiment the threshold frequency required is used by the electron to eject from the atom which is also called the work function and the remaining energy in electron is kinetic energy which can be measured. Now (before collision) since photon is a particle it must have mass and thus it has energy equivalent to E=mc2 .
After collision when it completely transfers it’s energy to electron thus E=0
And therefore 0=mc2 thus i guess photon vanishes.

JEE Main Part Test - 6 - Question 6

A point source of light is used in photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential

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

When the source is moved away from the emitter, intensity of the incident radiation decreases but frequency remains the same so there will be no change in the stopping potential. Thus, it remains constant.

JEE Main Part Test - 6 - Question 7

A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

Detailed Solution for JEE Main Part Test - 6 - Question 7

Saturation current is the maximum current possible and it will be directly proportional to the number of number of electrons falling on collector plate per second which depend on number of photons
incident on the cathode as one photon contribute in one electron and the number of photons is actually
proportional to intensity which varies
As intensity I∝1/r2​, where r is the distance
So the correct graph will be decreasing with power2 of distance and it will be rapidly decreasing with a higher value of r.

JEE Main Part Test - 6 - Question 8

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential is Volts is

Detailed Solution for JEE Main Part Test - 6 - Question 8

Given, the maximum kinetic energy: Kmax​=4eV
If V0​ be the stopping potential, then Kmax​=eV0​
⇒eV0​=4eV
⇒V0​=4V

JEE Main Part Test - 6 - Question 9

Radiation of two photon energies twice and five times the work function of metal are incident sucessively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted is the two cases will be

Detailed Solution for JEE Main Part Test - 6 - Question 9

As,
Maximum K.E= incident photon energy − work function
(½)​mV12​=2W−W=W⟶(1)
and
Ve​mV22​=5W−W=4W⟶(2)
from (1) and (2)
V1/V2​ ​​=1/2​
V1​:V2​=1:2

JEE Main Part Test - 6 - Question 10

Cut off potentials for a metal in photoelectric effect for light of wavelength l1, l2 and l3 is found to be V1, V2 and V3 volts if V1, V2 and V3 are in Arithmetic Progression and l1, l2 and l3 will be

Detailed Solution for JEE Main Part Test - 6 - Question 10

We know that,
eV=(hc/λ)-w
V=(hc/eλ)-(w/e)
Arithmetic progression =>V2=(V1+V2)/2
Now,
(hc/eλ2)-w/e=1/2[(hc/eλ1)-(w/e) +(hc/eλ3) -(w/e)]
=>1/ λ2=1/2[(1/ λ1)+(1/λ3)]
=>2/ λ2=1/ λ1 + 1/λ3
Hence the correct answer is harmonic Progression.

JEE Main Part Test - 6 - Question 11

The energy released by the fission of one uranium atom is 200 MeV. The number of fissions required per second to produce 3.2 W of power is

Detailed Solution for JEE Main Part Test - 6 - Question 11

The energy released by the fission of one uranium atom, E = 200 MeV
E = 200 × 10× 1.6 × 10-19 J
The number of fissions required per second, n/t = P/E.

JEE Main Part Test - 6 - Question 12

If in a nuclear fission, a piece of uranium of mass 0.5 g is lost, then the energy obtained in kWh is

Detailed Solution for JEE Main Part Test - 6 - Question 12

E = mc2 = (0.5/1000) × (3 × 108)2 = 4.5 × 1013 Ws
1000 W = 1 kW
3600 seconds = 1 hour
So, by unit conversion, E = 1.25 × 107 kWh

JEE Main Part Test - 6 - Question 13

What is the rest mass energy of an electron?

Detailed Solution for JEE Main Part Test - 6 - Question 13

JEE Main Part Test - 6 - Question 14

Which of the following are not emitted by a radioactive substance?

Detailed Solution for JEE Main Part Test - 6 - Question 14

Neutrons are not emitted by a radioactive substance.

JEE Main Part Test - 6 - Question 15

During a negative beta decay,

Detailed Solution for JEE Main Part Test - 6 - Question 15

Negative β decay is expressed by the following equation:
stands for antineutrino.
Hence, the correct choice is (c).

JEE Main Part Test - 6 - Question 16

The half-life of Pa-218 is 3 minutes. What mass of a 16 g sample of Pa-218 will remain after 15 minutes?

Detailed Solution for JEE Main Part Test - 6 - Question 16

Since 15 minutes = 5 x 3 minutes = 5 half-lives, the number of nuclei left after 15 minutes = 1/25 = 1/32 of the original number

Therefore, the mass of 16 g sample left after 15 minutes = 16/32 = 0.5 g

Hence, the correct choice is (d).

JEE Main Part Test - 6 - Question 17

Mp denotes the mass of a proton and Mn that of a neutron. A given nucleus of binding energy B contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by

Detailed Solution for JEE Main Part Test - 6 - Question 17

We know that,
BE(B) = [ZMp + NMn - M(N, Z)]c2
Where Z is number of protons
N is number of neutrons
Mp is mass of a proton
and Mn is mass of the nucleus

JEE Main Part Test - 6 - Question 18

The circuit shown in the figure contains two diodes D1 and D2, each with a forward resistance of 50 ohms and infinite backward resistance. If the battery voltage is 6 V, the current (in amperes) through the 100 ohm resistance is

Detailed Solution for JEE Main Part Test - 6 - Question 18

In the given circuit only diode D1 will allow the current to pass through as it is forward-biased.
Hence, the current through the 100 ohm resistance is (∵ resistance of D1 = 50 Ω)

= 0.02 A

JEE Main Part Test - 6 - Question 19

I is the current passing through a circular wire. If the radius of the wire is changed to twice, then what will be the current passing through the wire?

Detailed Solution for JEE Main Part Test - 6 - Question 19

Current, I = V/R
If the radius becomes twice, the area becomes 4 times and resistance becomes R/4, as resistance is inversely proportional to the area of wire.
New current = 4V/R = 4I

JEE Main Part Test - 6 - Question 20

To obtain a p-type semiconductor germanium crystal, it must be doped with foreign atoms whose valency is

Detailed Solution for JEE Main Part Test - 6 - Question 20

To obtain a p-type semiconductor the impurity must be trivalent, i.e. it must be doped with foreign atoms whose valency is 3. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. P-type semiconductors are created by doping an intrinsic semiconductor with acceptor impurities. A common p-type dopant for silicon is boron.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 21

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x × 1015 Hz. The value of x is ____________.

(Given h = 4.25 × 10-15 eVs)

Detailed Solution for JEE Main Part Test - 6 - Question 21

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

where En is the energy of the nth level and n is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

From n = 2 to n = 1
From n = 3 to n = 1
From n = 3 to n = 2
From n = 4 to n = 1
From n = 4 to n = 2
From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

The frequency of the incident light is related to the energy difference by the equation:
ΔE = hv
where h is the Planck's constant and v is the frequency.
Now, we can solve for the frequency:

So, the value of x is 3.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 22

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 is 0.42 V. If the threshold frequency is x × 1013 /s, where x is _________ (nearest integer).

(Given, speed light = 3 × 108 m/s, Planck's constant = 6.63 × 10−34 Js)

Detailed Solution for JEE Main Part Test - 6 - Question 22

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 23

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be __________.

Detailed Solution for JEE Main Part Test - 6 - Question 23

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 24

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is 1: 21/3 Their respective speed have a ratio of n : 1. The value of n is __________.

Detailed Solution for JEE Main Part Test - 6 - Question 24

Let the masses of the two nuclear parts be m1 and m2 and their respective speeds be v1 and v2
According to the problem, the ratio of their nuclear sizes is 1 : 21/3. Since the nuclear size is proportional to the cube root of the mass, we can write:

Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:

From the problem statement, the ratio of their respective speeds is n : 1, so we can write:

Substitute the expression for v1 into the momentum conservation equation:

We know the mass ratio, so substitute that into the equation:

Divide both sides by

Now, solve for n:
n = 2
Thus, the value of n is 2.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 25

A nucleus with mass number 242 and binding energy per nucleon as 7.6 Me V breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is _________MeV.

Detailed Solution for JEE Main Part Test - 6 - Question 25

The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.

The initial total binding energy of the nucleus is 242 × 7.6 MeV.

After the break, each fragment has a total binding energy of 121 × 8.1 MeV.

Since there are two such fragments, the final total binding energy is 2 × 121 × 8.1MeV.

The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:

2 × 121 × 8.1 MeV – 242 × 7.6 MeV = 1960.2 MeV – 1839.2 MeV = 121 MeV.

JEE Main Part Test - 6 - Question 26

Choose the correct statement regarding the physical properties of carbonyl compound.

Detailed Solution for JEE Main Part Test - 6 - Question 26
• Higher aldehydes are soluble in Benzene. Therefore, statement (B) is correct.
• Lower Aldehydes are pungent but higher Aldehydes are fragrant. Therefore, statement (A) is incorrect.
• Acetone is more polar than n-Butane and weight is comparable, so acetone has higher boiling point. Therefore, statement (C) is incorrect.
• Methanal is soluble in water. Therefore, statement (D) is incorrect.
JEE Main Part Test - 6 - Question 27

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:

Detailed Solution for JEE Main Part Test - 6 - Question 27

Presence of one vinyl (or ethenyl) group gives formaldehyde as one of the product in ozonolysis.

CH2=CH- on ozonolysis will give HCHO.

• Ozonolysis is an organic reaction where the unsaturated bonds of alkenes, alkynes, or azo compounds are cleaved with ozone.
• Alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group while azo compounds form nitrosamine.

JEE Main Part Test - 6 - Question 28

The correct order of increasing acidic strength is:

Detailed Solution for JEE Main Part Test - 6 - Question 28

The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid.

• Phenol is more acidic than ethanol because in phenol, the phenoxide ion obtained on deprotonation is stabilized by resonance which is not possible in case of ethanol.
• Also carboxylic acids are more acidic than alcohols and phenols as the carboxylate ion is stabilized by resonance. Chloroacetic acid is more acidic than acetic acid due to inductive effect of chlorine atom which stabilizes the carboxylate anion.
JEE Main Part Test - 6 - Question 29

Presence of unsaturation in organic compounds can be tested with:

Detailed Solution for JEE Main Part Test - 6 - Question 29

Baeyer's reagent (alk. KMnO4) which is pink in colour decolourises due to the presence of unsaturation. Thus, it shows the presence of unsaturation in an organic compound.

JEE Main Part Test - 6 - Question 30

Which of the following will not undergo aldol condensation?

Detailed Solution for JEE Main Part Test - 6 - Question 30
• For aldol condensation to occur, α -hydrogen atom is required for carbonyl compound.
• For formaldehyde, no α -carbon and no α -hydrogen is present. Therefore, H C H O will not undergo aldol condensation.
• An alpha (α) hydrogen is the hydrogen which is bonded to alpha carbon. Alpha carbon is the first carbon which is bonded to the functional group.
JEE Main Part Test - 6 - Question 31

Which of the following gives benzoic acid on oxidation?

Detailed Solution for JEE Main Part Test - 6 - Question 31

Benzyl chloride gives benzoic acid on oxidation.

The reaction takes place as follows:

C6H5CH2Cl + 2KOH + 2[O] → C6H5COOK + KCl + H2O

JEE Main Part Test - 6 - Question 32

Toluene reacts with halogen in presence of iron(III) chloride giving ortho and para halo compounds, the reaction is?

Detailed Solution for JEE Main Part Test - 6 - Question 32

Toluene reacts with halogen in presence of iron(III) chloride giving ortho and para halo compounds, the reaction is electrophilic substitution reaction.

The H atom or aromatic nucleus is replaced with Cl atom. The electrophile in this reaction is Cl+ ion obtained by the reaction of Cl2 with AlCl3.

JEE Main Part Test - 6 - Question 33

Which of the following carboxylic acid is highly insoluble in water?

Detailed Solution for JEE Main Part Test - 6 - Question 33

Higher carboxylic acids are insoluble in water due to the increased hydrophobic interaction of the hydrocarbon part.

• Thus, decanoic acid is highly insoluble in water, as it contains longer hydrocarbon chain.
• Decanoic acid is a ten-carbon, saturated fatty acid. It is present in palm kernel, coconut fat and in milk fat.
JEE Main Part Test - 6 - Question 34

What is the name of the above reaction?

Detailed Solution for JEE Main Part Test - 6 - Question 34

The above reaction is known as Etard reaction.

• The Etard Reaction is a chemical reaction that involves the direct oxidation of an aromatic or heterocyclic bound methyl group to an aldehyde using chromyl chloride.
• When toluene is reacted with Chromyl Chloride, then a chromium complex is formed (Etard Complex) whose hydrolysis gives Benzaldehyde.
JEE Main Part Test - 6 - Question 35

Acetic acid is obtained when which of the given reaction takes place?

Detailed Solution for JEE Main Part Test - 6 - Question 35

Acetic acid is obtained when acetaldehyde is oxidised with potassium dichromate and sulphuric acid.

The reaction takes place as follows:

JEE Main Part Test - 6 - Question 36

What is the name of the following compound?
CH3CH2COCI

Detailed Solution for JEE Main Part Test - 6 - Question 36

Given:
CH3CH2COCI
Numbering the carbon atoms as:

• This is an acid chloride.
• It consists of three carbon atoms in the parent chain and chloroformyl group (-COCI)
• In the IUPAC name, the suffix "oyl chloride" is added. Therefore, the IUPAC name will be propanoyl chloride.
JEE Main Part Test - 6 - Question 37

Identify the product for the following reaction.
CH ≡ CH + HOCI →

Detailed Solution for JEE Main Part Test - 6 - Question 37

HOCI → OH- + CI+
CI+ shall attack acetylene as an electrophile and OH- will end the addition with a nucleophilic addition. Therefore, the final reaction will be:
CH ≡ CH + 2 HOCI → CH (OH)2 - CH CI2 CH CI2 - C H O
Unstable

JEE Main Part Test - 6 - Question 38

Which of the following names is correct for

Detailed Solution for JEE Main Part Test - 6 - Question 38

Given compound:

Numbering the carbon atoms, we get:

• Since there are three carbon atoms in long chain and it contains 3 -CHO groups (one on each carbon atom).
• IUPAC suffix "aldehyde" will added due to presence of functional group -CHO.

Therefore, the correct IUPAC name will be propane-1, 2, 3-tricarbaldehyde.

JEE Main Part Test - 6 - Question 39

Propanal on treatment with dilute sodium hydroxide gives:

Detailed Solution for JEE Main Part Test - 6 - Question 39

An alpha (α) hydrogen is the hydrogen which is bonded to alpha carbon. Alpha carbon is the first carbon which is bonded to the functional group. As propanal contains alpha hydrogen, it undergoes self aldol condensation to give beta hydroxy (aldol) compound. The reaction is as follows:

JEE Main Part Test - 6 - Question 40

Which of the following organic compounds does not undergo diazotisation?

Detailed Solution for JEE Main Part Test - 6 - Question 40

Benzylamine, on reaction with nitrous acid, gives unstable diazonium salt which in turn gives alcohol with a gas, that is nitrogen as shown below:

JEE Main Part Test - 6 - Question 41

The coupling of diazonium salt of 4-amino benzene sulphonic acid with N, N-dimethyl benzamine produces

Detailed Solution for JEE Main Part Test - 6 - Question 41

The coupling of diazonium salt of 4-aminobenzene sulphonic acid with N, N-dimethyl benzamine produces methyl orange.

JEE Main Part Test - 6 - Question 42

Which among the following statements is incorrect?

Detailed Solution for JEE Main Part Test - 6 - Question 42

Tertiary amines do not have any hydrogen atom attached directly to the nitrogen and hence, do not exhibit hydrogen bonding. Hence, the option (c) is incorrect.

JEE Main Part Test - 6 - Question 43

Which of the following compounds give carbylamine reaction?

Detailed Solution for JEE Main Part Test - 6 - Question 43

The carbylamine reaction (also known as the Hoffmann Isocyanide Synthesis) is the synthesis of an isocyanide by the reaction of a primary amine, chloroform, and base.
CH3NH+ 3KOH + CHCl3 →CH3NC + 3KCl + 3H2O

JEE Main Part Test - 6 - Question 44

Which of the following indicates open chain structure of glucose?

Detailed Solution for JEE Main Part Test - 6 - Question 44

Pentacetyl formation of glucose indicates that the latter contain 5–OH groups, indicating the open chain structure of glucose.

JEE Main Part Test - 6 - Question 45

The numbers of stereo centres present in linear and cyclic structures of glucose are respectively:

Detailed Solution for JEE Main Part Test - 6 - Question 45

The number of stereo centres in the linear structure of glucose is 4 and in the cyclic structure is 5.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 46

A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is ______. (to the nearest integer)
(Atomic mass: C = 12; H = 1; O = 16)

Detailed Solution for JEE Main Part Test - 6 - Question 46

Molecular weight of C7H6O2 = 122

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 47

How many of the transformations given below would result in aromatic amines?

Detailed Solution for JEE Main Part Test - 6 - Question 47

1, 3, 4 will give Aniline.
Gabriel phthalimide synthesis cannot be used to prepare Aniline.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 48

The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is ___________.

Detailed Solution for JEE Main Part Test - 6 - Question 48

C4H5N

3 double bond equivalent are present in compound

Only 1 sp3 hybridised carbon is there
(Keeping compound as acyclic)

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 49

The mass of NH3 produced when 131.8 kg of cyclohexanecarbaldehyde undergoes Tollen's test is ________ kg. (Nearest Integer)

Molar Mass of
C = 12g/mol
N = 14g/mol
O = 16g/mol

Detailed Solution for JEE Main Part Test - 6 - Question 49

RCHO + 2 [Ag(NH3)2] OH → RCOONH4 + 2Ag + 3NH3 + H2O

1 mole of aldehyde produces 3 moles of NH3
1.176 x 103 mol aldehyde will produce
= 3 × 1.176 × 103 moles of NH3
Mass of NH3 produced = 3 x 1.176 × 103 × 17
= 59.97 × 103
≈ 60 kg

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 50

Number of isomeric compounds with molecular formula C9H10O which (i) do not dissolve in NaOH (ii) do not dissolve in HCI. (iii) do not give orange precipitate with 2,4-DNP (iv) on hydrogenation give identical compound with molecular formula C9H12O is __________.

Detailed Solution for JEE Main Part Test - 6 - Question 50

2 possibilities

JEE Main Part Test - 6 - Question 51

The number of arbitrary constants in the solution of a differential equation of degree 2 and order 3 is

Detailed Solution for JEE Main Part Test - 6 - Question 51

Since, the number of arbitrary constants in a solution of a differential equation of order n is equal to its order.

So, number of arbitrary constants = Order of differential equation = 3

JEE Main Part Test - 6 - Question 52

If y = y(x) is the solution of the differential equation e such that y(0) = 0, then y(1) is equal to:

Detailed Solution for JEE Main Part Test - 6 - Question 52

(ey) = xex + exc
At y(0) = 0,
c = 1
y = x + ln(x + 1)
y(1) = 1 + ln2

JEE Main Part Test - 6 - Question 53

The solution of  satisfying y(1) = 0, is

Detailed Solution for JEE Main Part Test - 6 - Question 53

2xydy = (x+ 1)dx + y2dx

Integrating, we get

⇒ y2 = x- 1 + cx
At x = 1, y = 0
⇒ c = 0
⇒ x2 - y2 = 1

JEE Main Part Test - 6 - Question 54

The solution of the equation (y log y) dx +(x - logy) dy = 0, is

Detailed Solution for JEE Main Part Test - 6 - Question 54

The given equation is:

Its solution is:

JEE Main Part Test - 6 - Question 55

The integrating factor of dy/dx - 3y/x = x3 sin x is

Detailed Solution for JEE Main Part Test - 6 - Question 55

This is a linear differential equation of the form dy/dx + Px = Q.
P = -3/x
Integrating Factor (I.F.) =

JEE Main Part Test - 6 - Question 56

The distance of the point having position vector from the straight line passing through the point (2, 3, - 4) and parallel to the vector, is:

Detailed Solution for JEE Main Part Test - 6 - Question 56

JEE Main Part Test - 6 - Question 57

The value of k for which the points A (1, 0, 3), B (– 1, 3, 4), C (1, 2, 1) and D (k, 2, 5) are coplanar, is

Detailed Solution for JEE Main Part Test - 6 - Question 57

JEE Main Part Test - 6 - Question 58

Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a', b', c' from the origin, then

Detailed Solution for JEE Main Part Test - 6 - Question 58

JEE Main Part Test - 6 - Question 59

Equation of a plane bisecting the angle between the planes 2x - y + 2z + 3 = 0 and 3x - 2y + 6z + 8 = 0 is

Detailed Solution for JEE Main Part Test - 6 - Question 59

Equations of the planes bisecting the angle between the given planes are

Hence, option (2) is correct.

JEE Main Part Test - 6 - Question 60

Let a = i + j + k, b = I – j + 2k and c = xi + (x – 2)j – k. If the vector c lies in the plane of a and b, then x equals

Detailed Solution for JEE Main Part Test - 6 - Question 60

JEE Main Part Test - 6 - Question 61

A line passes through the points (6, –7, –1) and (2, –3, 1). If the angle which the line makes with the positive direction of the x-axis is acute, then the direction cosines of the line are

Detailed Solution for JEE Main Part Test - 6 - Question 61

JEE Main Part Test - 6 - Question 62

The equation of the line through the point (2, 1, 1) and the intersection of the lines 2x – y – 4 = 0 = y + 2z and x + 3z – 4 = 0 = 2x + 5z – 8 is

Detailed Solution for JEE Main Part Test - 6 - Question 62

JEE Main Part Test - 6 - Question 63

The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:

Detailed Solution for JEE Main Part Test - 6 - Question 63

The required plane is:
(2x - y - 4) + λ(y + 2z - 4) = 0 ... (1)
It passes through (1, 1, 0).
⇒(2−1−4) +λ(1 − 4) = 0
⇒ −3 − 3λ = 0 ⇒ λ = −1
Substitute the value of λ=−1 in equation (1).
The required equation of plane is:
x−y−z=0

JEE Main Part Test - 6 - Question 64

Four persons can hit a target correctly with probabilities 1/2, 1/3, 1/4 and 1/8. If all hit at the target independently, then the probability that the target would be hit is

Detailed Solution for JEE Main Part Test - 6 - Question 64

Let person be A, B, C, D.
P(Hit) = 1 - P(None of them hits)

JEE Main Part Test - 6 - Question 65

If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is

Detailed Solution for JEE Main Part Test - 6 - Question 65

Only two equilateral triangles are possible. They are A1A3A5 and A2A5A6.
So, required probability

JEE Main Part Test - 6 - Question 66

Out of 11 consecutive natural numbers, if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:

Detailed Solution for JEE Main Part Test - 6 - Question 66

Out of 11 consecutive natural numbers, there are either 6 even and 5 odd numbers or 5 even and 6 odd numbers.
When 3 numbers are selected at random, then total cases = 11C3
Since these 3 numbers are in A.P., let the numbers be a, b, c.
2b ⇒ even number

So, favourable cases = 6C2 + 5C2
= 15 + 10 = 25
P(3 numbers are in A.P.) = 25/(11C3) = 25/165 = 5/33

JEE Main Part Test - 6 - Question 67

Let A and B be two non-null events such that A ⊂ B. Then, which of the following statements is always correct?

Detailed Solution for JEE Main Part Test - 6 - Question 67

JEE Main Part Test - 6 - Question 68

A random variable has the following probability distribution:

The value of k is

Detailed Solution for JEE Main Part Test - 6 - Question 68

Random variate

P(x) = 1
0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
10k2 + 9k = 1
10k2 + 10k - k - 1 = 0
10k(k + 1) - 1(k + 1) = 0
(k + 1)(10k - 1) = 0
k = -1, k = 1/10
k ≠ -1
The probability of a variable cannot be negative, so
k = 0.1.

JEE Main Part Test - 6 - Question 69

If (1 + 3p)/3, (1 – p)/4 and (1 – 2p)/2 are the probabilities of three mutually exclusive events, then the set of all values of p is

Detailed Solution for JEE Main Part Test - 6 - Question 69

JEE Main Part Test - 6 - Question 70

The mean and variance of a binomial distribution are 4 and 2, respectively. What is the probability of two successes?

Detailed Solution for JEE Main Part Test - 6 - Question 70

Hence, P(X = 2) = 28/256 = 7/64

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 71

The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4, is ______ (in integer).

Detailed Solution for JEE Main Part Test - 6 - Question 71

We have,
1 – (Probability of all shots result in failure) > 1/4
Using binomial distribution,

So the least number of shots required is 3.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 72

Directions: The answer to the following question is a single-digit integer, ranging from 0 to 9.

Three tangents are drawn at random to a given circle. If the odds against the circle being inscribed in the triangle formed by them are K : 1, find K.

Detailed Solution for JEE Main Part Test - 6 - Question 72

Let us construct a circle and draw three tangents to it at random.

Now, construct the parallel lines to each of the tangents such that these three lines are also tangents to the same circle.

Now, we count the total triangles formed.

There are 6 small triangles on the end of the stars and two big triangles inverted to each other that form the star. These two triangles are the ones that contain the circle inside them.
Hence, these are the required triangles.

The odds of getting this triangle are given as follows:

6/2 = 3/1

(since probability of odds in against = failure/success)
Hence, the odds are 3 to 1 against the circle being inscribed in the triangle formed by the three tangents.
Thus, K = 3

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 73

A fair n (n > 1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is n/9, then n is equal to ____________.

Detailed Solution for JEE Main Part Test - 6 - Question 73

In this case, the success is defined as getting a number less than n when rolling an n-sided die. The probability of success, p, in each roll is then (n - 1)/n, and the probability of failure, q = 1 - p, is 1/n

where

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 74

Let the probability of getting head for a biased coin be 1/4. It is tossed repeatedly until a head appears. Let N be the number of tosses required. If the probability that the equation 64x2 + 5Nx + 1 = 0 has no real root is p/q, where p and q are coprime, then q - p is equal to ________.

Detailed Solution for JEE Main Part Test - 6 - Question 74

We have the quadratic equation 64x2 + 5 Nx + 1 = 0. For it to have no real roots, the discriminant (b2 - 4ac) should be less than 0. Here, a = 64, b = 5N, and c = 1.

This gives us :

Since N must be an integer (as it represents the number of tosses), the possible values of N are 1, 2, or 3.

The probability of getting the first head on the n- th toss (given the probability of getting a head is 1/4) is given by the geometric distribution formula, (1 - p)n-1 × p.

So, the probability for our specific values of N is:

Therefore, the total probability (p/q) is :

Therefore, q - p is equal to 27.

*Answer can only contain numeric values
JEE Main Part Test - 6 - Question 75

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 - 6 - Question 75

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