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

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JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 1

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 1

A dimensionless and unitless term will have same value in all system of units. Option (2) is a dimensionless term.
Thus, this term would have same value in different systems of units.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 2

A person climbs up a stalled escalator in 60 s. If standing on the same but escalator running with constant velocity he takes 40 s. How much time is taken by the person to walk up the moving escalator ?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 2

The expression of time taken by man is,

The expression for time taken by escalator is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 3

Three masses m, 2m and 3m are moving in x-y plane with speed 3u, 2u, and u respectively as shown in figure. The three masses collide at the same point P and stick together. The velocity of resulting mass will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 3

According to the law of conservation of momentum,
Left momentum Right momentum

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 4

A 4 g bullet is fired horizontally with a speed of 300 m/s into 0.8 kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3, how far will the block slide approximately?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 4

According to the law of conservation of momentum,
mv = Mv'
0.004 x 300 = 0.8v'
v' = 1.5 m/s
From kinetics, write the expression for the velocity,
v2 = v'2 + 2as
0 = 1.52 + 2 x 3 x s
s ≈ 0.379 m

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 5

A spring of unstitched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity v is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 5

The kinetic energy in elemental form is,

The velocity in elemental form is,

The mass in elemental form is,

The kinetic energy in integral form is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 6

A particle is moving in a circular path of radius a, with a constant velocity v as shown in the figure. The center of circle is marked by 'C. The angular momentum from the origin O can be written as:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 6

Consider the diagram shown below,

The expression of the component of angular velocity through OC is,
OC = v × a ……(1)
By using the property of triangle,

The expression for the angular velocity for an arbitrary point N from the figure is,
ON = OC + CN ……(2)
The expression of the component of angular velocity through CN is,

Substitute the values into equation (2)
Angular momentum = va +vacos2θ
= va (1 + cos2θ)

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 7

Two hypothetical planets of masses m1 and m2 are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is 'd' ? (Speed of m1 is v1 and that of m2 is v2):

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 7

Let reference point is at infinity. The total energy at infinity will be 0.
So, initial energy of the system is 0.
The expression for the final energy of the system will be,

From law of conservation of energy,

.....(1)
As the momentum is conserved for the system, the kinetic energy will be inversely proportional to the mass.
The expression of the kinetic energy for mass m1 is,

Substitute above expression in equation (1),

Similarly for mass m2 KE is,

Solve the above expression.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 8

Steel ruptures when a shear of 3.5 ×108 Nm–2 is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 8

The expression for stress is,

The expression for area is,
Area 2πrt

The expression for the force is,
Force A x stress

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

A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is :

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

If the liquid is at x height from orifice at time t then, the volume coming from the orifice is given by the expression,
V = vA …… (1)
The expression of the velocity of water is,

Write expression for rate of volume.

Compare equation (2) with the above expression.

Solve the above expression.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 10

Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S the change in total surface area, T is the surface tension and P atmospheric pressure, which of the following relation is correct?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 10

Let the radius of the soap bubble is r and the radius of the larger bubble is R.
So, the surface tension is given by,
By solving the above expression,

Rearrange the above expression to form,
3PV + 4TS = 0

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 11

Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is :​

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 11

The rate of change temperature is given as,

Substitute the values.

....(1)
Similarly,
...(2)
Thus,
T = 10 °C

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

A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be :

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

The values for heat in Carnot engine is,

The expression for efficiency is,

So,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 13

At room temperature a diatomic gas is found to have an r.m.s. speed of 1930 ms–1. The gas is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 13

The equation of rms velocity is,

From ideal gas equation,

Thus,

The molar mass of hydrogen is 2g.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 14

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 14

Only linear equations represent SHM along a straight line; and the degree of x as 1 is present only in option (4) so it is a linear equation. Thus option (4) is correct.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 15

A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives waves of frequency 2150 Hz. The source A receives waves, reflected from ground, of frequency nearly : (Speed of sound =343 m/s)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 15

By using Doppler effect,

The velocity from the above expression is,

v = 55.8 m/s
The reflected frequency is,

= 2500 Hz

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 16

A spherically symmetric charge distribution is characterised by a charge density having the following variation:

Where r is the distance from the centre of the charge distribution and ρ0 is a constant. The electric field at an internal point (r <R ) is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 16

The expression of charge for a symmetrically charged spherical body,
dq = ρ × 4πr2dr
The equation of Charge density is,

So, the total charge is given by,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 17

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
K(x) = Ko + λx (λ = a constant)
The capacitance C, of this capacitor, would be related to its ‘vacuum’ capacitance Co as per the relation :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 17

The equation of potential is,

Let the charge density is σ and the surface area is s.
The capacitance is given by,

Solve above expression by multiply the denominator and numerator by d in the right hand side.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 18

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 Ω, 9 Ω and 9 Ω and a capacitor 5.0 µF.

How much is the current I in the circuit in steady state?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 18

The current would not flow through the branch with capacitor in the steady state.
Therefore, net potential is 8 V and equivalent resistance is 12 Ω.
So,
I = 8/12
= 0.67 A

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 19

A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time ∆t so that the charge reverses its direction crossing the y axis at a distance d. Then :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 19

Write expression for the energy balance for the charge,

...(1)

The equation of the total time for charge particles is,
...(2)
The equation of the time for reversal of direction is,

Solve equation (1) and (2),

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 20

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field B1, at its centre when a current (I) passes through it. The second wire is bent into a coil with three identical loops adjacent to each other and produces magnetic field B2 at the centre of the loops when current 1/3 passes through it. The ratio B1 : B2 is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 20

Consider the following diagram,

The equation of magnetic field for loop wire is,

The equation of magnetic field for coiled wire is,

Taking ratio,

=1/3

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 21

A sinusoidal voltage V(t) = 100 sin (500t) is applied across a pure inductance of L = 0.02 H. The current through the coil is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 21

The inductive reactance is,
XL = ωL
Substitute the values.
XL 500 x 0.2
= 10 Ω
The current lags behind voltage by π/2 because the circuit is
pure inductive. So,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 22

A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 22

The intensity of electric field is,
...(1)
Also,

From equation (1),

The electric field intensity will be half of the total intensity and thus,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 23

The refractive index of the material of a concave lens is µ. It is immersed in a medium of refractive index µ1. A parallel beam of light is incident on the lens. The path of the emergent rays when µ1 > µ is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 23

The light bends towards the normal as it passes through rarer to denser medium and as it goes through lens to medium.
Thus, option (1) is correct.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 24

Interference pattern is observed at 'P' due to superimposition of two rays coming out from a source 'S' as shown in the figure. The value ‘l’ for which maxima is obtained at 'P' is (R is perfect reflecting surface) :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 24

From the geometry of the given figure,

The expression of total path difference by light is,

For maxima, the path difference should be nλ,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 25

In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is 6600 Å , then wavelength of first maximum will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 25

Write the expression for the first minima of red.
...(1)
Write the expression for maxima for a wavelength λ2.
...(2)
As the light coincides, So compare (1) and (2).

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 26

A beam of light has two wavelengths 4972 Å and 6216 Å with a total intensity of 3.6 ×10–3 Wm−2 equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 26

Write the expression for the number of photons per second by a monochromatic light.

Substitute the values,

The number of photons ejected in 2s is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 27

A piece of bone of an animal from a ruin is found to have 14C activity of 12 disintegrations per minute per gm of its carbon content. The 14C activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die ? (Given half life of 14C is t1/2 = 5760 years):

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 27

The radioactivity in tth time,

Substitute the values.

Take log both sides, and use property of log.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 28

For LED's to emit light in visible region of electromagnetic light, it should have energy band gap in the range of:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 28

λ should lie between 4000 × 10−10 to 7600 ×10−10 m for emitting light. The equation of minimum energy is,

Substitute the values in above expression,

Similarly,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 29

For sky wave propagation, the radio waves must have a frequency range in between:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 29

The range of frequency for radio wave is 5 MHz to 25 MHz for sky wave propagation. Thus, option (2) is correct.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 30

In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potentiometer wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of voltmeter will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 30

The potential gradient is,

The voltage across 220 cm is,

The error in reading is,
0.5 − 0.55= −0.05 V

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 31

If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 31

The total energy of a revolving electron in the hydrogen atom is calculated by adding the potential energy and kinetic energy of the electron.

The value of Z for hydrogen atom is 1.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 32

The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity 100 ms–1 is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 32

The momentum of the particle is,
p = m·v

The de Broglie wavelength of the given particle,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 33

What happens when an inert gas is added to an equilibrium keeping volume unchanged?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 33

The molar concentration of reactants and products does not change by the addition of inert gas at constant volume, and due to which the state of equilibrium remains unchanged. Thus, the addition of an inert gas at constant volume does not affect the equilibrium.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 34

The amount of BaS04 formed upon mixing 100 mL of 20.8% BaCl2 solution with 50 mL of 9.8% H2SO4 solution will be:
(Ba = l37, Cl = 35.5, S=32, H = l and O = 16)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 34

The number of moles of BaCl2 is,

= 0.01mol
The molarity of the given BaCl2 solution is,

= 0.1M
The number of moles of H2SO4 is,

= 0.01mol
The molarity of the given H2SO4 solution is,

= 0.2 M
The number of moles of BaSO4 formed from BaCl2 and H2SO4 solution is 0.1 mol because the number of moles of barium is limited. The mass of BaSO4 formed in the given system is calculated by multiplying its moles with the molar mass.
Mass of BaSO4 = 0.1 mol x (137 g/mol + 32 g/mol 4 x 16 g/mol)
= 23.3 g

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 35

The rate coefficient (k) for a particular reaction is 1.3 ×10−4 M−1s−1 at 100 C, and 1.3 ×10−3 M−1s−1 at 150 C. What is the energy of activation (EA) (in kJ) for this reaction? (R = molar gas constant = 8.314 JK−1mol−1)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 35

The activation energy of the given reaction is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 36

How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate?
(Atomic mass of copper = 63.5 u, NA = Avogadro's constant) :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 36

The number of molecules of copper formed in the given electrolysis process is,

Since one copper molecule is formed by 2 electrons.
So, the number of electrons is double the number of molecules of copper.
Hence, the number of electrons required in the given system is calculated as,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 37

The entropy (Sº) of the following substances are :
CH4 (g) 186.2 J K–1 mol–1
O2 (g) 205.0 J K–1 mol–1
CO2 (g) 213.6 J K–1 mol–1
H2O (l) 69.9 J K–1 mol–1
The entropy change (∆Sº) for the reaction
CH4 (g) + 2O2 → CO2 (g) + 2H2O (l) is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 37

The standard entropy change of the given reaction is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 38

The conjugate base of hydrazoic acid is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 38

The expression of the dissociation of hydrazoic acid is,

Thus, the conjugate base of hydrazoic acid is

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 39

In a monoclinic unit cell, the relation of sides and angles are respectively:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 39

In a monoclinic unit cell, the relation between sides is a ≠ b ≠c, and the relationship between angles is γ = β = 90° and α ≠ 90°.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 40

The standard enthalpy of formation (∆f298) for methane, CH4 is– 74.9 kJ mol–1. In order to calculate the average energy given out in the formation of a C–H bond from this it is necessary to know which one of the following?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 40

Write the expression of the balanced chemical equation for the formation of methane.

If the values of enthalpy of sublimation of carbon and bond dissociation energy of H2 are known, the average energy released in the given reaction can be calculated.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 41

Which of the following xenon-OXO compounds may not be obtained by hydrolysis of xenon fluorides ?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 41

The chemical reactions of xenon fluoride with water are,

The oxidation state of xenon in XeO4 is +8 .

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 42

Excited hydrogen atom emits light in the ultraviolet region at 2.47 ×1015 Hz . With this frequency, the energy of a single photon is:
(h = 6.63×10–34 Js)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 42

The energy of a photon is,
E=hν

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 43

Which one of the following exhibits the largest number of oxidation states ?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 43

The number of oxidation states exhibited by an element depends upon the number of unpaired electrons present in the d orbital. Since, these are five unpaired electrons in the d orbital of manganese that is larger than that of titanium, vanadium, and chromium. Thus, manganese exhibits the largest number of oxidation states among the given compounds.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 44

Copper becomes green when exposed to moist air for a long period. This is due to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 44

The copper metal in presence of air for a longer time reacts with carbon dioxide, water, and oxygen to form copper carbonate and copper hydroxide. The color of copper changes to green due to the formation of copper carbonate (green color).
The chemical equation involved in the reaction of copper with moist air is given as,
2Cu + H2O + CO2 + O2→ CuCO3 + Cu (OH)2

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 45

Among the following species the one which causes the highest CFSE, ∆o as a ligand is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 45

The crystal field splitting energy of ligand depends upon the strength of ligands. The splitting of molecular orbitals occurs due to the higher strength of ligands, cause results in the high crystal field splitting energy. Since the strength of CO ligand is higher than that of the CN, NH3 and F. Therefore, CO causes highest crystal field splitting energy out of the given ligands.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 46

Similarity in chemical properties of the atoms of elements in a group of the Periodic table is most closely related to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 46

According to the modern periodic law, the similarity in the chemical properties is shown by the elements having an equal number of valence electrons in their valence shell. Thus, the similarity in the chemical properties of elements in a group is related to the number of valence electrons as all elements present in a group have an equal number of valence electrons in their outermost shell.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 47

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species O2–, S2–, N3–, P3– ?:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 47

The atomic radii of O2−,N3− and S2− is smaller than that of P3− due to the atomic radius of elements decreases along the period. The atomic radius of O2− is smaller than that of N3− because both the ions have an equal number of electrons but the number of protons is greater in O2− as compared to N3−. Thus, the increasing order of atomic radii of the given elements is O2− < N3− <S2−< P3−.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 48

Global warming is due to increase of:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 48

The increase in the amount of methane and carbon dioxide gas in the atmosphere causes global warming. The increase in the amount of these gases warms the earth’s atmosphere which results in the global warming.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 49

Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases H2O2 acts as a reducing agent in acid medium ?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 49

The reducing agent in the reaction with permanganate ion in the acidic medium is hydrogen peroxide as shown in the chemical equation,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 50

Which one of the following complexes will most likely absorb visible light?
(At nos. Sc = 21, Ti = 22, V = 23, Zn = 30)

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 50

The electronic configuration of central metal ions in the given complexes is,

The central metal ion present in the complex  has its valence electrons in the 4s orbital that can easily excite to the 3d orbital and can absorb visible light, but in case of other complexes, the central metal ion has fully filled electronic configuration. Thus, the complex  is most likely to absorb visible light among the given complexes.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 51

on mercuration-demercuration produces the major product:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 51

The mechanism involved in the mercuration-demercuration of the given compound is,

The mercuration-demercuration reaction of the given compound involves the formation of cyclic carbocation intermediate. This cyclic carbocation intermediate cannot undergo rearrangement because it is a type of non-classical carbocation.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 52

In the Victor-Meyer's test, the colour given by 1°, 2° and 3 alcohols are respectively:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 52

In the Victor Meyer’s test, primary alcohols give blood red color solution of nitrolic acid in sodium hydroxide as shown below.

In the Victor Meyer's test, secondary alcohols give a blue color solution of pseudonitrol in sodium hydroxide as shown below.

At the end of Victor Meyer’s test, tertiary alcohols form a colorless solution as shown below.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 53

Conversion of benzene diazonium chloride to chloro benzene is an example of which of the following reactions?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 53

Sandmeyer reaction involves the reaction of benzene diazonium salts with the solution of copper halide, and hydrochloric acid leads to the formation of halogen substituted benzene. The conversion of benzene diazonium salt to chlorobenzene is shown below.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 54

In the presence of peroxide, HCl and HI do not give antiMarkownikoff s addition to alkenes because :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 54

The steps involved in the antimarkovnikov addition of hydrogen chloride and hydrogen iodide on alkenes in the presence of peroxide are endothermic in nature due to which the antimarkovnikov addition of hydrogen chloride and hydrogen iodide is not favorable.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 55

The major product obtained in the photo catalysed bromination of 2-methylbutane is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 55

The photocatalyzed bromination of 2 − methylbutane proceeds via free radical mechanism. Since the stability of tertiary free radical is greater than that of primary and secondary radical, the major product obtained in the given reaction corresponds to that of tertiary free radical.

Thus, the major product formed in the given reaction is 2 − bromo − 2− methylbutane.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 56

Which of the following molecules has two sigma (σ) and two pi(π)bonds :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 56

The structure of HCN is shown below.

In the structure of HCN, there are two sigma bonds and two pi bonds.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 57

Which one of the following acids does not exhibit optical isomerism?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 57

The structure of maleic acid is shown below.

Maleic acid exhibits only geometrical isomerism because only cis and trans arrangement is possible in its structure due to the presence of two groups at each carbon atom. So, the compound which does not exhibit optical isomerism is maleic acid.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 58

Aminoglycosides are usually used as:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 58

Aminoglycosides are a class of antibiotics due to which they are very useful as antibiotics. Thus, the aminoglycosides are usually used as antibiotics.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 59

Which of the following will not show mutarotation?

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 59

Sucrose does not show mutarotation because the bond present between glucose and fructose is between the alpha hydroxyl group of both glucose and fructose due to which it cannot exists in alpha and beta form. Thus, sucrose does not undergo mutarotation.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 60

Phthalic acid reacts with resorcinol in the presence of concentrated H2SO4 to give :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 60

The formation of fluorescein takes place due to the reaction of phthalic acid with resorcinol as shown below.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 61

A relation on the set A = {x : |x| < 3, x∈ Z} , where Z is the set of integers is defined by R = {( x, y ) : y = x , x ≠ 1}. Then the number of elements in the power set of R is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 61

For |x| <3 , the set A is,
A = {−2,− 1, 0, 1, 2}
For y = |x| and x ≠ −1, the set R is,
R = {( −2, 2) (0, 0) (1,1) (1, 2)}
The power set of R is,
24 = 16

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 62

Let z ≠ −i be any complex number such that  is a purely imaginary number. Then  is

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 62

Let’s consider z = x + iy
The expression is,

This term cannot be purely imaginary. Thus,  is any non-
zero real number.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 63

The sum of the roots of the equation, x2 + |2x − 3| − 4 = 0, is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 63

Write the equation when
x2 + |2x-3| - 4 = 0
x2 + 2x - 7 = 0
Write the equation when
x2 + |2x-3| - 4 = 0
x2 - 2x + 3 - 4 = 0
x2 - 2x -1 = 0
Compute the roots of the quadratic equations when

Compute the roots of the quadratic equations when

Compute the sum of the roots as,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 64

then k  is equal to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 64

By gauss elimination, apply the first row transformation
R2 → R2 − R1 and R1 → R1 − R3.

By gauss elimination, apply the second row transformation
R3 → R3 + R1 and R1 → R2 - 1/2R2.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 65

If  and  be such that  then:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 65

The multiplication of AB is,

Solve the above matrix for first equation.
y + 2x + x = 6
3x + y = 6
Solve the above matrix for second equation.
3y - x + 2 = 8
3y - x = 6
Solve the above equations.
3x + y = 3y - x
4x = 2y
y = 2x

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 66

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such number in which the odd digits do not occupy odd places, is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 66

The expression for the ways to select 3 odd places out of 4 odd places is,

= 120

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 67

is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 67

The expression for the general term from Binomial theorem is,
Let’s consider the coefficients of Tr +1 =Tr+2.

r = 7
Thus, the first term is given as,

= 8th
And, the second term is given as,

=9th

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 68

Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of  then a : b can be:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 68

Write expression for the geometric mean of numbers a and b.

The value of b is ar.
Thus, the expression becomes,

Calculate the arithmetic mean of numbers

The ratio of 1/M : G is,

Solve above equation,

4r2 + 4 + 8r = 25r
4r2 - 17r + 4 = 0
The roots of the above equation are,

Thus, the value of a:b is,
a = br
a/b r
= 1/4

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 69

The least positive integer n such that  is,

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 69

Write the general binomial expression.

Solve the above expression.

Further solve the above expression,

n - 1 = 5
n = 6

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 70

Let f, g : R → R be two functions defined by
Statement I : f is a continuous function at x = 0.
Statement II : g is a differentiable function at x = 0.

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 70

Write the given function.

The value of the function f (x) at x → 0.

= f (0)
So, the f (x) is continuous function at x = 0.
The value of the function g (x) at x → 0.
g(0) = 0

The expression for the left hand limit of the function g (x) is,

The expression for the right hand limit of the function g (x) is,

Since, the left hand limit is equal to the right hand limit. Thus, the function g(x) at x = 0 is differentiable.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 71

If f(x) x2 - x + 5, x > 1/2, and g (x) is its inverse function, then g′ (7) is equals :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 71

Write the given function.
f(x) = x2 − x+ 5
Compute the value of x at which the value of f(x) is 7 as,
7 = x2 - x + 5
2 = x(x-1)
x = 2 and 3
The value of the function g (x) is,
g(f(x)) = x
Differentiate the above expression.

The value of the function g′ (7) is,

= 1/3

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 72

Let f and g be two differentiable functions on R such that f'(x) > 0 and g'(x) < 0, for all x ∈ R. Then for all x:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 72

The term f′(x) > 0 shows that it is an increasing function.
The term g′(x) < 0 shows that it is an decreasing function.
In case (1),
x > x - 1
g(x) > g(x-1)
f(g(x)) < f(g(x-1))
In case (2),

In case (3),

In case (4),

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 73

If 1 + x4 + x5 for all x in R, then a2 is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 73

Write the given equation.

Compare the coefficients of x5.
a5 = 1
Compare the coefficients of x4

Compare the coefficients of x3.

Compare the coefficients of x2.

a2 = -4.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 74

The integral is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 74

Write the given integral.

Let, the term tan3 x + 1 is,

The integral is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 75

If [ ] denote the greatest integer function, then the integral  is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 75

Write the given integral.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 76

If for a continuous function for all t ≥− π, then  is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 76

Write the given continuous function.

Write the expression of the function.

Thus, the value of f (x) is,

f(x) = −3x

= π

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 77

The general solution of the differential equation,

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 77

Write the given differential equation.

Compute the integration factor of the above equation as,

The expression for the general solution of the equation is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 78

If a line intercepted between the coordinate axes is trisected at a point A (4, 3), which is nearer to x-axis, then its equation is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 78

The below figure represents the diagram of the line,

Write the value of the abscissa at point B.
a/3 = 4
a = 12
Write the value of the ordinate at point C.
2b/3 = 3
b = 9/2
Write the equation of the line.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 79

If the three distinct lines x + 2ay + a = 0, x +3by +b = 0 and x + 4ay + a = 0 are concurrent, then the point (a, b) lies on a :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 79

Write the matrix form of the given equations.

By gauss elimination, apply the first row transformation
R2 → R2− R1 and R3 → R3 −R1.

Compute the determinant of the above matrix as,

a = b
The locus of (a, b) lies at x = 0 or y = x. Hence, the point (a, b) will lie on a straight line.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 80

For the two circles x2 + y= 16 and x2 + y − 2y = 0, there is/are:​

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 80

The distance between their centers is 1 unit and sum of their radii is 3 from the two given equations of the circle. Thus, one of them lie completely inside the other. So, there will be no common tangent.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 81

Two tangents are drawn from a point (–2,–1) to curve y2 = 4x. If α is the angle between them, then tan α is equal to :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 81

Write the equation of the tangent.

Write the condition of tangency.
C = a/m

Compute the roots of the above equation as,

Compute the value of the tan α is,

=3

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 82

The minimum area of a triangle formed by any tangent to the ellipse  and the coordinate axes is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 82

Let the point on an ellipse is M having coordinates (4 cosθ,9 sinθ ) as shown in figure.

Write the equation of the tangent.

Let the points P and Q intersect the tangent at M with coordinate axis.
Write the coordinates of point P.

Write the coordinates of point Q.

Compute the area of the triangle.

Compute the minimum area of the triangle as,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 83

A symmetrical form of the line of intersection of the planes x = ay + b and z = cy + d is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 83

Write the first equation.

Write the second equation.

Equate the both equations.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 84

If the distance between planes, 4x – 2y– 4z + 1 = 0 and 4x – 2y– 4z + d = 0 is 7, then d is :

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 84

The expression for the distance between the two lines is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 85

If  are three unit vectors in three-dimensional space, then the minimum value of  is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 85

Write the given expression.

The minimum value of given expression lies at

The minimum value of the given expression is,

= 6 - 3
= 3

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 86

Let  and M.D. be the mean and the mean deviation about  of n observation xi, i = 1, 2, ...., n. If each of the observation is increased by 5, then the new mean and the mean deviation about the new mean respectively, are:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 86

The expression for the mean of the n observation is,

The new mean after the increment 5 in each observation is given as,

The expression for the mean deviation of the n observation is,

The new mean deviation after the increment 5 in each observation is given as,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 87

A number x is chosen at random from the set {1, 2, 3, 4, ....., 100}. Define the event: A = the chosen number x satisfies  Then P(A) is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 87

Write the given expression.

Here, the range of x lies
x ∈ {10, 11, 12, 13K 29} ∪ {50, 51, 52 .....100} .
The total value of x is 71.
Thus, the value of P (A) is,

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 88

​Statement I : The equation  has a solution for all
Statement II : For any x ∈ R,  and

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 88

Write the given equation.

Solve the above equation.

Since, the value of cos−1 x lies,

Thus, statement I is false.
For statement II, the given equation is not applicable for any x ∈ R. So, statement II is false.

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 89

and A and B are respectively the maximum and the minimum values of f(θ), then (A, B) is equal to:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 89

The determinant of the given matrix is calculated as,

The value of sin 2θ + cos 2θ lies between
The maximum and minimum value of sin 2θ + cos 2θ is,

The maximum value of A is  and the minimum value of B is

JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 90

Let p, q, r denote arbitrary statements. Then the logically equivalent of the statement p ⇒ (q∨r) is:

Detailed Solution for JEE Main 2014 Question Paper With Solutions (12th-April-2014) - Question 90

Simplify the given statement as,
p ⇒ (q∨r)
~ pv(qvr)
(~pvq) v (~pvr)
(p⇒q)∨(p⇒r)

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