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JEE Main 2014 Question Paper With Solutions (9th-April-2014)

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JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 1

​An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 1

The time period for simple pendulum is given by the equation,

From this we get,

So, the error associated with the acceleration is given by,

Substituting the values,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 2

The position of a projectile launched from the origin at t=0 is given by m at 2=2 s. If the projectile was launched at an angle θ from the horizontal, then the θ is (take d=10 ms-1).

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 2

The horizontal distance travelled by the projectile is given by, ut = (u cos θ t) +

u cos θ = 20 .................. (i)
Vertical distance travelled by the projectile is given by.

u sin θ = 35 ............. (2)
The angle of projection of projectile.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 3

​Water is flowing at a speed of 1.5 ms-1 through a horizontal tube of cross-sectional area 10-2 m2 and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water=103 kgm-3).

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 3

As the water stops following hitting the palm, the base power that ought to be applied will be the pace of progress of force.
The minimum force exerted is given by

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 4

A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 4

The friction force between to surfaces is equal to the minimum forces exerted on block A.
The minimum acceleration of block A is given by,

= 3 m/s2
The maximum force on block B can be calculated by the equation,
F = mamax
= 9 x 3
= 27 N

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 5

Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g=10 ms-1)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 5

The net compressive force of the spring is ,

The magnitude of maximum force exerted on the floor,

= Mg + mω2A + mg
Substituting the values of parameters
Fmax =
= 40 + 10 + 10
= 60 N

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 6

A cylinder of mass Mc and sphere of mass Ms are placed at points A and B of two inclines, respectively. (See Figure). If they roll on the incline without slipping such that their accelerations are the same, then ratio is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 6

The equation for acceleration of cylinder is given as,

The equation for acceleration of sphere is given as,

The acceleration of the sphere and the cylinder is equal.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 7

​India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of Earth's orbit is ae = 1.5×1011 m, that of Mar's orbit am = 2.28×1011 m, taken Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 7

The semi-major axis of the orbit of Mangalyan is,

= 1.89 x 1011 m
By the Kapler’s rule,

Substituting the value in the above equation,

Ty ≈ 518 days
Therefore, the time required for Mangalyan to reach Mars is given as

≈ 260 days

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 8

In materials like aluminium and copper, the correct order of magnitude of various elastic modulii is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 8

The misshapening of pliable materials like copper and aluminum can without much of a stretch be conceivable under elastic pressure. In this manner, the bendable material like have low shear modulii however has high mass modulii than shear modulii and Young's moduli.
Thus, the correct order of elastic modulii for a particular material is given as, Shear modulii<Young's modulii<Bulk modulii

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 9

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in, carbon dioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 9

The damped oscillation the equation of displacement is given by,

Substitute the values for air

Substitute the values for carbon dioxide

t = 167.7 sec
Thus, the closest time is 167 s .

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 10

A capillary tube is immersed vertically in water and the height of the water column is x. When this arrangement is taken into a mine of depth d, the height of the water column is y. If R is the radius of earth, the ratio x/y is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 10

The pressures are equal
P1 = P2

g1x = g2y
The equation of change in g with depth is given by,

By equating the two equations we’ll get,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 11

Water of volume 2 L in a closed container is heated with a coil f 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27°C to 77°C? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).​

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 11

The heat absorbed by water in 1 second is,
Q = 1000 - 160
= 840 J/s
The net amount of heat required to raise the temperature of water,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 12

The equation of state for a gas is given by PV= nRT+ αV , where n is the number of moles and α is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder area To and Po respectively. The work done by the gas when its temperature doubles isobarically will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 12

The gas equation at a particular state of gas is given by,
........... (i)
The ideal gas equation is given by,
PV = nRT

The final temperature is twice of the initial temperature,
Tf = 2T0
The work done by the gas can be calculated as,

By solving the above expression we’ll get,

Integrate the equation (1)

For one mole of gas substitute n =1.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 13

​Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0×10–15 atm. at room temperature (300 K). Taking R = 8.3 JK-1mol-1, 1 atm = 105 Pa and NAvogadro = 6 ×1023 mol−1 , the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 13

Assume the average distance between the gas molecules is r. Then by ideal gas equation.

Substitute the values in the above expression,

r = 0.2 mm

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 14

A particle which is simultaneously subjected to two perpendicular simple harmonic motions t and ω represented by ;
x = a1cosωt and y = a2cos 2ωt traces a curve given by:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 14

The displacement of particle which is under simple harmonic motion is given by,
x = a1 cosωt
cosωt = x/a1
The given equation is
y = a2cos2ωt

Therefore, the curve in option (1) with square relation is correct.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 15

​A transverse wave is represented by :

For what value of the wavelength the wave velocity is twice the maximum particle velocity?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 15

The displacement of transverse wave is given as,

The given condition is

Equation for calculation of velocity is,

λ = 40 cm

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 16

​The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 16

Equation for electric field is,

The net surface charge is given by,

By substituting the values we’ll get,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 17

​Three capacitances, each of 3 µF, are provided. These cannot be combined to provide the resultant capacitance of :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 17

The capacitance of the given capacitors cannot give the given resultant capacitance. Below figure represents the possible arrangements of capacitors,

For case (a)

For case (b)

For case (c)

For case (d)

Thus, option (d) is correct.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 18

A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 Ω . The battery terminals are connected to an external resistance 'R'. The minimum value of 'R', so that a current passes through the battery to charge it is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 18

The minimum value of R can be calculated by,

= 220/20
= 11 Ω

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 19

​The mid points of two small magnetic dipoles of length d endon positions, are separated by a distance x (x >> d). The force between them is proportional to x− n where n is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 19

The force between the magnetic poles is inversely proportional to the fourth power of the distance between the magnetic poles.
................. (i)
The given equation is
..................... (ii)
Compare equation (1) with equation (2).
n =4

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 20

The magnetic field of earth at the equator is approximately 4×10–5 T. The radius of earth is 6.4×106 m. Then the dipole moment of the earth will be nearly of the order of:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 20

The equation of magnetic field of earth is

Substitute the values, for dipole moment of earth,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 21

​When the rms voltages VL, VC and VR are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio VL : VC : VR = 1 : 2 : 3. If the rms voltage of the AC source is 100 V, then VR is close to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 21

The given ratio of voltage is,
VL :VC :VR =1:2:3
The given ratio of reactance is,
XL :XC :XR = 1:2:3
x: 2x: 3x
The equation for current is,

Equation for calculation of voltage across the resistor is,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 22

​Match List - I (Wavelength range of electromagnetic spectrum) with List - II. (Method of production of these waves) and select the correct option from the options given below the lists.

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 22

Vibrations of atoms and molecules - (700 nm to 1 mm)
Inner shell electrons in atoms moving from - (1 nm to 4nm)
Magnetron valve generates wavelength of 1 mm to 0.1 m.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 23

A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is 4/3, and the diver’s eyes are 15 cm below the surface of water. Then the radius of the circle is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 23

The relation for θ is

= 3/4
Equation for calculation of radius of circle is,
tan θ = r/h

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 24

Using monochromatic light of wavelength λ, an experimentalist sets up the Young's double slit experiment in three ways as shown. If she observes that y = β′ , the wavelength of the light is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 24

The fringe width is equal to the distance from the central fringe.
y β'
dsinθ = (μ - 1)t
dθ = (μ-1)t

The distance between the slits is doubled then for calculation of wavelength of light is,

= 540 nm

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 25

​The focal lengths of objective lens and eye lens of a Gallelian Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the magnifying Power of the Gallelian telescope should be :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 25

The equation for magnifying power is,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 26

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 26

Among the given options the dust particle is difficult to verify with the de-Broglie relation.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 27

If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 27

The energy required to remove electron from first orbit of Li++ is given by the equation,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 28

Identify the gate and match A, B, Y in bracket to check.

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 28

The figure below represents the logic gate.

The value of output Y for the above logic gate is,

= AB + AB
= AB
Thus the given logic gate is an AND gate.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 29

​A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight (LOS) mode?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 29

The two towers are represented in the figure below,

The maximum distance between the towers is given by,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 30

​An n-p-n transistor has three leads A, B and C. Connecting B and C by moist fingers, A to the positive lead of an ammeter, and C to the negative lead of the ammeter, one finds large deflection. Then, A, B and C refer respectively to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 30

In n-p-n or p-n-p transistor, the emitter, collector and base are,
• Collector is moderately doped
• Emitter is heavily doped
• Base is lightly doped.
The transistors are current controlled device in which high current flows between emitter and collector during conduction state.
Hence, in the given system, A refers to Emitter, B refers to base and C refers to collector.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 31

In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 31

Effective number of atoms at corner = 8 x 1/8
Effective number of atoms at tha face =
= 5/2
The ratio of effective number of atoms

Therefore, the formula of the compound is A2B5

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 32

Vander Wall's equation for a gas is stated as,

​This equation reduces to the perfect gas equation, p = when,

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 32

In the case when temperature is high enough and pressure is low the Vander wall’s equation reduces to perfect gas equation i.e. PV = nRT .

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 33

the standard electrode potentials ( E0M+ /M) of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V and –0.76 V, respectively. The sequence of deposition of metals on applying potential is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 33

For higher value of reduction potential the rate of deposition is higher. Therefore the sequence is C,B,D,A.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 34

At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium constant is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 34

The HI is dissociated with the chemical equation,

Here, a is the concentration.
The equilibrium constant of the above chemical reaction can be calculated as follows.

= 1/4
Thus, the value of equilibrium is 0.25.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 35

Dissolving 120 g of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 35

The molarity of the given solution is

= 2M
Therefore, the molarity of the given solution is 2M.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 36

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 36

Reaction time for 50% completion is 15 min .
Then, the total number of half-life cycles is,
= 60/15
= 4
Therefore, the amount of the substance left after one hour is,

Thus, the amount left after 1 hour is 1/16 of its initial amount.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 37

​A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is : (F= 96,500 C)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 37

Work done, W =
Total number of moles =
n factor =
= + 3
The value of X is +3 .

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 38

​The energy of an electron in first Bohr orbit of H-atom is – 13.6 eV. The energy value of electron in the excited state of Li2+ is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 38

In the excited state of Lithium the energy of electron can be calculated as.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 39

​The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He =4 u, O =16 u)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 39

The root mean square velocity of oxygen atom and helium atom is equal.

is root mean square velocity.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 40

The standard enthalpy of formation of NH3 is –46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is –436 kJ/mol and that of N2 is –712 kJ/mol, the average bond enthalpy of N— H bond in NH3 is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 40

The equation for calculation of bond enthalpy.
Bond enthalpy =

= -1056 kJ/mol
The average enthalpy for three N −H bonds is,

For bond disassociation the sign would be positive.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 41

The amount of oxygen in 3.6 moles of water is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 41

There are 3.6 moles of oxygen in water. The mass in gram is,
= (3.6 moles ) x 16 g/mol
= 57.6 g

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 42

The gas evolved on heating CaF2 and SiO2 with concentrated H2SO4, on hydrolysis gives a white gelatinous precipitate. The precipitate is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 42

The chemical reactions of CaF2 , SiO2 and SiF4 in concentration of H2SO4 are,

The precipitate is hydrofluosilicic acid.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 43

Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM. This Vanadium chloride has the formula : (at. no. of V = 23)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 43

The given estimation of magnetic moment 1.73 BM is that demonstrates that vanadium simply have just unpaired electron in this valence shell. In this way, VCl4, vanadium has +4 oxidation state.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 44

An octahedral complex of Co3+ is diamagnetic. The hybridisation involved in the formation of the complex is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 44

The electronic configuration of Co3+ is is diamagnetic in nature, it implies there is no unpaired electron present in the d-orbital of cobalt. Along these lines, the hybridization engaged with the development of the complex of must be (inward orbital complex) as demonstrated as follows.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 45

Which of the following is not formed when H2S reacts with acidic K2Cr2O7 solution?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 45

The reaction of H2S with K2Cr2O7 is shows below.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 46

Which of the following has unpaired electron(s)?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 46

The molecule of O2 has one unpaired electron in π orbital, as shown in figure below,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 47

​In the following sets of reactants which two sets best exhibit the amphoteric character of Al2O3. xH2O?
Set- 1 : Al2O3.xH2O(s) and OH-(aq)
Set- 2 : Al2O3.xH2O(s) and H2O(l)
Set- 3 : Al2O3.xH2O(s) and H+(aq)
Set- 4 : Al2O3.xH2O(s) and NH3(aq)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 47

Compound with amphoteric structure can act both as acid and base.
Therefore, the set -1 is Al2O3.xH2O(s), that forms Al(OH)4 in the solutions.
The set-3 is of acidic nature Al2O3.xH2O(s) that shows Al3+ and H2O .

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 48

The number and type of bonds in C22− ion in CaC2 are:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 48

The CaC2 has one σ bond and one π bonds.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 49

The form of iron obtained from blast furnace is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 49

After processing the iron ore in blasé furnace, the form of iron is pig iron. Pig iron has 92-94% of iron and 3-5% of carbon.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 50

The correct statement about the magnetic properties of [Fe(CN)6]3– and [FeF6] 3– is : (Z = 26).

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 50

In Fluorine because of weak field ligand  the pairing of electron is not possible. This distribution of electrons in case of Fluorine is shown below,

On the contrary Cyanide is a strong field ligand in which pairing of electron is east. The distribution of electron is shown below.

There are unpaired electrons in both elements. Thus, the nature of compound is paramagnetic.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 51

Which one of the following reactions will not result in the formation of carbon-carbon bond?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 51

In the Cannizzaro reaction, the C-C bond does not form at all. The reaction is given below.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 52

In the hydroboration - oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 52

In the reaction hydroboration-oxidation of propane the product is 1-propanol.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 53

The major product of the reaction

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 53

The chemical reaction is shown below:

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 54

For the compounds CH3Cl, CH3Br, CH3I and CH3F, the correct order of increasing C-halogen bond length is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 54

The atomic radii increases from fluorine to iodine.
Thus, the bond length also increases.
Therefore the correct increasing order is
CH3F < CH3Cl< CH3Br < CH3I

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 55

Allyl phenyl ether can be prepared by heating :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 55

The chemical reaction of preparation of allyl phenyl ether is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 56

In a nucleophilic substitution reaction :

which one of the following undergoes complete inversion of configuration?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 56

This is a nucleophilic substitution reaction. The compound C6H5CH2Br will completely undergoes to inversion.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 57

In which of the following pairs A is more stable than B?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 57

Because of resonance effect the compound A is more stable compare to the compound B. The compound B does not have resonance effect.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 58

Structure of some important polymers are given. Which one represents Buna-S ?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 58

The reaction for polymerization of Buna-S is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 59

Which is the major product formed when acetone is heated with iodine and potassium hydroxide?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 59

When acetone heated with iodine and potassium hydroxide, it forms iodoform. The reaction is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 60

Which one of the following class of compounds is obtained by polymerization of acetylene?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 60

The chemical reaction of polymerization of acetylene gives poly-vne. The reaction is given below.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 61

Let P be the relation defined on the set of all real numbers such that P = {(a,b):sec2a − tan2b=1}.Then P is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 61

Since the given function is reflexive.
sec2a − tan2b = 1 ∀x∈ R
For being symmetric,

To prove this,

= 1
Therefore, the function is symmetric.
For transitive,
sec2a − tan2b=1 ............. (2)
sec2b − tan2c=1 .......... (3)
Therefore, point P is reflexive, symmetric and transitive.’ Hence this is a equivalence relation.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 62

Let w(lm w ≠0 ) be a complex number. Then the set of all complex numbers z satisfying the equation , for some real number k, is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 62

The given equation is
............... (i)
Value of is,
.............. (ii)
Solve equation (i) and (ii).

Therefore, the correct option is (4).

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 63

If equations ax2 + bx + c = 0, ( a,b,c ∈ R, a≠ 0 ) and 2x2 + 3x + 4 = 0 have a common root, then a : b : c equals : ​

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 63

The equation is
ax2 + bx + c= 0 a,b,c ∈ R, a ≠ 0)
And 2x2 + 3x + 4= 0
For the above equation.
D ≤0
It suggests roots of the given equation imaginary and common.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 64

If and  are the roots of the equation, ax2 + bx + 1= 0 (a ≠ 0, a, b ∈ R) , then the equation, x(x + b3) + (a- 3abx) = 0 has roots :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 64

The given equation is,

The product of the roots is,

And,
The sum of the roots is,

Now, Substitute the value on the equation,

Further, simplify the above equation,

Hence, the roots of the above equation is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 65

If a,b and c are non-zero real numbers and if the system of equations

has a non-trival solution, then ab+bc+ca equals :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 65

For having non-trivial solutions,

Take the values given in the equations.

ab + ac + bc = abc

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 66

​If B is a 3 × 3 matrix such that B2 = 0, then det. [(I + B)50− 50B}] is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 66

=

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 67

The number of terms in the expansion of in powers of x is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 67

Therefore, the number of terms in the expression is, 101+101 = 202 terms

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 68

The sum of the digits in the unit’s place of all the 4-digit numbers formed by using the numbers 3,4 and 6, without repetition, is

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 68

If the unit place is 3 then remaining three palaces can be filled in 3! ways.
Thus 3 appears in unit place in 3! times.
Similarly, each digit appear in unit place times.
Hence, the sum of the digits in units place is.
(6 + 5 + 4 + 3) (4 - 1)! = 18 x 3!
= 18 x 6
= 108

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 69

​Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4th term is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 69

Use expression for the nth term of an AP.

According to the given condition its 9th term

But d has to be a integer.
Therefore,
d =4 Then,
a + 4 = 12
a = 8
The 4th term of the AP is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 70

​If the sum upto 20  terms mis equal to k/21, then k is equal to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 70

The nth term of the given series is

Substitute n = 1, , 2, 3, ..., 20 in the above equation.

= 120/21
Hence, k = 120

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 71

If f (x) is continuous and then is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 71

Simplifying the given expression

Therefore, the required value is

= 2/9

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 72

If , then is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 72

y =enx ................ (1)
Upon differentiation
................ (2)
Take log both sides in equation (1).
.................... (3)
Multiply the equation (2) and (3)

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 73

If the Rolle’s theorem holds for the function f(x) = 2x3 + ax2+ bx in the interval [-1,1] for the point c= 1/2, then the value of 2a+b is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 73

Substitute the values in the function.

By the rolls theorem

Differentiation of the given function is

Therefore, the required value is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 74

​If f(x) = , x ∈ R, then the equation f(x) = 0 has:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 74

By graph method

The figure shows that there is one intersection point. Therefore, the number of solution is one.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 75

dx is equal to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 75

By integrating the given function,

integrating it further.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 76

The integral equals :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 76

Put,
2x = tan θ
Differentiate the above expression.

At x =0 , θ= 0
And at x = 1/2, θ = π/4
Substitute the values in the given function.

Solve I1

Further simplify for I1.

The integration of the required function is

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 77

​Let A = . The area (in square units) of the region A is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 77

The bounded curve is shown in figure below.

Solve the equations for y.
y = −2, 4
Calculating the area of the triangle,
Area =

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 78

If the differential equation representing the family of all circles touching x-axis at the origin is then g(x) equals :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 78

The equation of circle is
.................. (1)
Differentiate the above equation

Put value of a in equation (1).

Compare the above equation with

Therefore, the value of g (x) is
g (x) = 2x

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 79

Let a and b be any two numbers satisfying . Then, the foot of perpendicular from the origin on the variable line, , lies on

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 79

Equation of normal is

a = h/4 .............. (i)
b = k/4 .............. (ii)
Solve equation (1) and (2).

Therefore, the locus is.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 80

Given there points P, Q, R with P(5,3) and R lies on the x-axis. If equation of RQ is x − 2y = 2 and PQ is parallel to the x-axis, then the centroid of ∆ PQR lies on the line :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 80

The figure below represents the given points.

The equation of line RQ is,
x − 2y= 2
The equation of line PQ is,
y = 3
Substitute 3 for y in equation of RQ.

Therefore, the centroid of the triangle is.

Only the line 2x − 5y satisfy this point.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 81

​If the point (1, 4) lies inside the circle x2 +y2 − 6x – 10y + p= 0 and the circle does not touch or intersect the coordinate axes, then the set of all possible values of p is the interval:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 81

The points are represented in the figure below

The distance between AB is

Hence, from the given relation

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 82

If OB is the semi-minor axis of an ellipse, F1 and F2 are its foci and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 82

From Pythagoras theorem,
a2= b2 +c2
Substituting the given values,

From the relation of eccentricity,
c = ae
c/a = e

Hence, the value of square of eccentricity is
e2 = 1/2

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 83

​Equation of the plane which passes through the point of intersection of lines and  and has the largest distance from the origin is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 83

The equation of planes is.

Consider a plane with direction cosine l,m,n and distances from origin is d is.
lx + my + nz= d
Hence, the equation of the plane is,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 84

A line in the 3-dimensional space makes an angle θ with both the x and y axis. Then the set of all values of θ is the internal:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 84

The given angle θ which makes by the plane in the 3-dimensional space is,

The condition for minimum value of angle θ is:
If the line lies on x, y plane, it makes angle of 45°.
The condition for maximum value of angle θ is:
If line at z − axis, it makes an angle of 90°.
Hence, the required interval is,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 85

If and  then equals :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 85

Simplifying the above equation,

Then,

Thus the value of k is 5.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 86

In a set of 2n distinct observations, each of the observation below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then the mean of the new set of observations:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 86

Mean value of the set of 2n observations.

Substitute the values in the above equation.

Then the mean value is,

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 87

If A and B are two events such that P (A ∪ B)= P (A ∩ B), then the incorrect statement amongst the following statements is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 87

The van diagram of events is given below.

P(A ∪ B) = P(A ∩ B)
From van diagram it is clear that option (1) is incorrect.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 88

The number of values of α in [0, 2π] for which 2 sin3α − 7sin2α + 7sinα = 2, is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 88

The trigonometric expression is
2sin3α− 7 sin2α+ 7sinα = 2
Simplify the above expression.

equating it to zero.

Below figure represent the sine curve.

Therefore, the number of solution here is 3.

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 89

If cos θ = then is equal to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 89

Simplify the given expression

Upon further simplification

JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 90

The contrapositive of the statement "I go to school if it does not rain" is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (9th-April-2014) - Question 90

Consider, p is equal to the statement “if it does not rain”
And, q is equal to the statement “I go to school”
According to the Contrapositive law,

Thus,
Negation of p is “its rains”
And is “if I do not go to school”
Hence, is “If I do not go to school, it rains.”

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