Description

This mock test of JEE Main 2014 Question Paper With Solutions (9th-April-2014) for JEE helps you for every JEE entrance exam.
This contains 90 Multiple Choice Questions for JEE JEE Main 2014 Question Paper With Solutions (9th-April-2014) (mcq) to study with solutions a complete question bank.
The solved questions answers in this JEE Main 2014 Question Paper With Solutions (9th-April-2014) quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this JEE Main 2014 Question Paper With Solutions (9th-April-2014) exercise for a better result in the exam. You can find other JEE Main 2014 Question Paper With Solutions (9th-April-2014) extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

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 :

Solution:

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,

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}).

Solution:

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.

QUESTION: 3

Water is flowing at a speed of 1.5 ms^{-1 }through a horizontal tube of cross-sectional area 10^{-2} m^{2} 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=10^{3} kgm^{-3}).

Solution:

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

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 :

Solution:

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/s^{2}

The maximum force on block B can be calculated by the equation,

F = ma_{max}

= 9 x 3

= 27 N

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})

Solution:

The net compressive force of the spring is ,

The magnitude of maximum force exerted on the floor,

= Mg + mω^{2}A + mg

Substituting the values of parameters

F_{max} =

= 40 + 10 + 10

= 60 N

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:

Solution:

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.

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 a_{e} = 1.5×10^{11 }m, that of Mar's orbit a_{m} = 2.28×10^{11} m, taken Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to :

Solution:

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

= 1.89 x 10^{11} m

By the Kapler’s rule,

Substituting the value in the above equation,

T_{y} ≈ 518 days

Therefore, the time required for Mangalyan to reach Mars is given as

≈ 260 days

QUESTION: 8

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

Solution:

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

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).

Solution:

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 .

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:

Solution:

The pressures are equal

P_{1} = P_{2}

g_{1}x = g_{2}y

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

By equating the two equations we’ll get,

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).

Solution:

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,

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 T_{o }and P_{o} respectively. The work done by the gas when its temperature doubles isobarically will be:

Solution:

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,

T_{f} = 2T_{0}

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.

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^{-1}mol^{-1}, 1 atm = 10^{5} Pa and N_{Avogadro} = 6 ×10^{23} mol^{−1} , the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Solution:

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

QUESTION: 14

A particle which is simultaneously subjected to two perpendicular simple harmonic motions t and ω represented by ;

x = a_{1}cosωt and y = a_{2}cos 2ωt traces a curve given by:

Solution:

The displacement of particle which is under simple harmonic motion is given by,

x = a_{1} cosωt

cosωt = x/a_{1}

The given equation is

y = a_{2}cos2ωt

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

QUESTION: 15

A transverse wave is represented by :

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

Solution:

The displacement of transverse wave is given as,

The given condition is

Equation for calculation of velocity is,

λ = 40 cm

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 :

Solution:

Equation for electric field is,

The net surface charge is given by,

By substituting the values we’ll get,

QUESTION: 17

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

Solution:

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 (c) is correct.

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 :

Solution:

The minimum value of R can be calculated by,

= 220/20

= 11 Ω

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:

Solution:

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

QUESTION: 20

The magnetic field of earth at the equator is approximately 4×10^{–5} T. The radius of earth is 6.4×10^{6} m. Then the dipole moment of the earth will be nearly of the order of:

Solution:

The equation of magnetic field of earth is

Substitute the values, for dipole moment of earth,

QUESTION: 21

When the rms voltages V_{L}, V_{C} and V_{R} 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 V_{L} : V_{C} : V_{R} = 1 : 2 : 3. If the rms voltage of the AC source is 100 V, then V_{R} is close to:

Solution:

The given ratio of voltage is,

V_{L} :V_{C} :V_{R} =1:2:3

The given ratio of reactance is,

X_{L} :X_{C} :X_{R} = 1:2:3

x: 2x: 3x

The equation for current is,

Equation for calculation of voltage across the resistor is,

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.

Solution:

Vibrations of atoms and molecules - (700 nm to 1 mm)

Inner shell electrons in atoms moving from - (1 nm to 4nm)

Radioactive decay of the nucleus

Magnetron valve generates wavelength of 1 mm to 0.1 m.

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 :

Solution:

The relation for θ is

= 3/4

Equation for calculation of radius of circle is,

tan θ = r/h

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:

Solution:

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

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 :

Solution:

The equation for magnifying power is,

QUESTION: 26

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

Solution:

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

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:

Solution:

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

QUESTION: 28

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

Solution:

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.

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?

Solution:

The two towers are represented in the figure below,

The maximum distance between the towers is given by,

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 :

Solution:

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.

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 :

Solution:

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 A_{2}B_{5}

QUESTION: 32

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

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

Solution:

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 .

QUESTION: 33

the standard electrode potentials ( E^{0}_{M+ /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 :

Solution:

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

QUESTION: 34

At a certain temperature, only 50% HI is dissociated into H_{2} and I_{2} at equilibrium. The equilibrium constant is:

Solution:

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.

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:

Solution:

The molarity of the given solution is

= 2M

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

QUESTION: 36

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

Solution:

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.

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)

Solution:

Work done, W =

Total number of moles =

n factor =

= + 3

The value of X is +3 .

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 Li^{2+} is :

Solution:

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

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)

Solution:

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

is root mean square velocity.

QUESTION: 40

The standard enthalpy of formation of NH_{3} is –46.0 kJ/mol. If the enthalpy of formation of H_{2} from its atoms is –436 kJ/mol and that of N_{2} is –712 kJ/mol, the average bond enthalpy of N— H bond in NH_{3} is :

Solution:

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.

QUESTION: 41

The amount of oxygen in 3.6 moles of water is:

Solution:

There are 3.6 moles of oxygen in water. The mass in gram is,

= (3.6 moles ) x 16 g/mol

= 57.6 g

QUESTION: 42

The gas evolved on heating CaF_{2} and SiO_{2} with concentrated H_{2}SO_{4}, on hydrolysis gives a white gelatinous precipitate. The precipitate is:

Solution:

The chemical reactions of CaF_{2} , SiO_{2} and SiF_{4} in concentration of H_{2}SO_{4} are,

The precipitate is hydrofluosilicic acid.

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)

Solution:

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, VCl_{4}, vanadium has +4 oxidation state.

QUESTION: 44

An octahedral complex of Co^{3+} is diamagnetic. The hybridisation involved in the formation of the complex is :

Solution:

The electronic configuration of Co^{3+} 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.

QUESTION: 45

Which of the following is not formed when H_{2}S reacts with acidic K_{2}Cr_{2}O_{7 }solution?

Solution:

The reaction of H_{2}S with K_{2}Cr_{2}O_{7} is shows below.

QUESTION: 46

Which of the following has unpaired electron(s)?

Solution:

The molecule of O_{2}^{−} has one unpaired electron in π^{∗} orbital, as shown in figure below,

QUESTION: 47

In the following sets of reactants which two sets best exhibit the amphoteric character of Al_{2}O_{3}. xH_{2}O?

Set- 1 : Al_{2}O_{3}.xH_{2}O(s) and OH^{-}(aq)

Set- 2 : Al_{2}O_{3}.xH_{2}O(s) and H_{2}O(l)

Set- 3 : Al_{2}O_{3}.xH_{2}O(s) and H^{+}(aq)

Set- 4 : Al_{2}O_{3}.xH_{2}O(s) and NH_{3}(aq)

Solution:

Compound with amphoteric structure can act both as acid and base.

Therefore, the set -1 is Al_{2}O_{3}.xH_{2}O(s), that forms Al(OH)_{4}^{−} in the solutions.

The set-3 is of acidic nature Al_{2}O_{3}.xH_{2}O(s) that shows Al^{3+} and H_{2}O .

QUESTION: 48

The number and type of bonds in C_{2}^{2−} ion in CaC_{2} are:

Solution:

The CaC_{2} has one σ bond and one π bonds.

QUESTION: 49

The form of iron obtained from blast furnace is :

Solution:

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.

QUESTION: 50

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

Solution:

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.

QUESTION: 51

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

Solution:

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

QUESTION: 52

In the hydroboration - oxidation reaction of propene with diborane, H_{2}O_{2} and NaOH, the organic compound formed is :

Solution:

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

QUESTION: 53

The major product of the reaction

Solution:

The chemical reaction is shown below:

QUESTION: 54

For the compounds CH_{3}Cl, CH_{3}Br, CH_{3}I and CH_{3}F, the correct order of increasing C-halogen bond length is :

Solution:

The atomic radii increases from fluorine to iodine.

Thus, the bond length also increases.

Therefore the correct increasing order is

CH_{3}F < CH_{3}Cl< CH_{3}Br < CH_{3}I

QUESTION: 55

Allyl phenyl ether can be prepared by heating :

Solution:

The chemical reaction of preparation of allyl phenyl ether is

QUESTION: 56

In a nucleophilic substitution reaction :

which one of the following undergoes complete inversion of configuration?

Solution:

This is a nucleophilic substitution reaction. The compound C_{6}H_{5}CH_{2}Br will completely undergoes to inversion.

QUESTION: 57

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

Solution:

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

QUESTION: 58

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

Solution:

The reaction for polymerization of Buna-S is

QUESTION: 59

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

Solution:

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

QUESTION: 60

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

Solution:

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

QUESTION: 61

Let P be the relation defined on the set of all real numbers such that P = {(a,b):sec^{2}a − tan^{2}b=1}.Then P is:

Solution:

Since the given function is reflexive.

sec^{2}a − tan^{2}b = 1 ∀x∈ R

For being symmetric,

To prove this,

= 1

Therefore, the function is symmetric.

For transitive,

sec^{2}a − tan^{2}b=1 ............. (2)

sec^{2}b − tan^{2}c=1 .......... (3)

Therefore, point P is reflexive, symmetric and transitive.’ Hence this is a equivalence relation.

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 :

Solution:

The given equation is

............... (i)

Value of is,

.............. (ii)

Solve equation (i) and (ii).

Therefore, the correct option is (4).

QUESTION: 63

If equations ax^{2} + bx + c = 0, ( a,b,c ∈ R, a≠ 0 ) and 2x^{2} + 3x + 4 = 0 have a common root, then a : b : c equals :

Solution:

The equation is

ax^{2} + bx + c= 0 a,b,c ∈ R, a ≠ 0)

And 2x^{2} + 3x + 4= 0

For the above equation.

D ≤0

It suggests roots of the given equation imaginary and common.

QUESTION: 64

If and are the roots of the equation, ax^{2} + bx + 1= 0 (a ≠ 0, a, b ∈ R) , then the equation, x(x + b^{3}) + (a^{3 }- 3abx) = 0 has roots :

Solution:

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

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 :

Solution:

For having non-trivial solutions,

Take the values given in the equations.

ab + ac + bc = abc

QUESTION: 66

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

Solution:

=

QUESTION: 67

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

Solution:

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

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

Solution:

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

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:

Solution:

Use expression for the n^{th} term of an AP.

According to the given condition its 9^{th} term

But d has to be a integer.

Therefore,

d =4 Then,

a + 4 = 12

a = 8

The 4^{th} term of the AP is

QUESTION: 70

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

Solution:

The n^{th} term of the given series is

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

= 120/21

Hence, k = 120

QUESTION: 71

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

Solution:

Simplifying the given expression

Therefore, the required value is

= 2/9

QUESTION: 72

If , then is equal to:

Solution:

y =e^{nx }................ (1)

Upon differentiation

................ (2)

Take log both sides in equation (1).

.................... (3)

Multiply the equation (2) and (3)

QUESTION: 73

If the Rolle’s theorem holds for the function f(x) = 2x^{3 }+ ax^{2}+ bx in the interval [-1,1] for the point c= 1/2, then the value of 2a+b is :

Solution:

Substitute the values in the function.

By the rolls theorem

Differentiation of the given function is

Therefore, the required value is

QUESTION: 74

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

Solution:

By graph method

By adding both the curve

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

QUESTION: 75

dx is equal to :

Solution:

By integrating the given function,

integrating it further.

QUESTION: 76

The integral equals :

Solution:

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 I_{1}

Further simplify for I_{1}.

The integration of the required function is

QUESTION: 77

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

Solution:

The bounded curve is shown in figure below.

Solve the equations for y.

y = −2, 4

Calculating the area of the triangle,

Area =

QUESTION: 78

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

Solution:

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

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

Solution:

Equation of normal is

a = h/4 .............. (i)

b = k/4 .............. (ii)

Solve equation (1) and (2).

Therefore, the locus is.

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 :

Solution:

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.

QUESTION: 81

If the point (1, 4) lies inside the circle x^{2} +y^{2} − 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:

Solution:

The points are represented in the figure below

The distance between AB is

Hence, from the given relation

QUESTION: 82

If OB is the semi-minor axis of an ellipse, F_{1} and F_{2} are its foci and the angle between F_{1}B and F_{2}B is a right angle, then the square of the eccentricity of the ellipse is:

Solution:

From Pythagoras theorem,

a^{2}= b^{2 }+c^{2}

Substituting the given values,

From the relation of eccentricity,

c = ae

c/a = e

Hence, the value of square of eccentricity is

e^{2} = 1/2

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:

Solution:

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,

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:

Solution:

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,

QUESTION: 85

If and then equals :

Solution:

Simplifying the above equation,

Then,

Thus the value of k is 5.

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:

Solution:

Mean value of the set of 2n observations.

Substitute the values in the above equation.

Then the mean value is,

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:

Solution:

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.

QUESTION: 88

The number of values of α in [0, 2π] for which 2 sin^{3}α − 7sin^{2}α + 7sinα = 2, is:

Solution:

The trigonometric expression is

2sin^{3}α− 7 sin^{2}α+ 7sinα = 2

Simplify the above expression.

equating it to zero.

Below figure represent the sine curve.

Therefore, the number of solution here is 3.

QUESTION: 89

If cos θ = then is equal to :

Solution:

Simplify the given expression

Upon further simplification

QUESTION: 90

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

Solution:

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.”

### 2014 NEET Question Paper Solved

Doc | 33 Pages

### 2014 AIIMS Question Paper Solved

Doc | 32 Pages

### UPSC Question Paper (GS - 1) - 2014

Doc | 5 Pages

### CAT 2014: Past Year Question Paper

Doc | 13 Pages

- JEE Main 2014 Question Paper with Solutions
Test | 90 questions | 180 min

- JEE Main 2014 Question Paper With Solutions (9th-April-2014)
Test | 90 questions | 180 min

- JEE Main 2014 Question Paper With Solutions (19th-April-2014)
Test | 90 questions | 180 min

- JEE Main 2014 Question Paper With Solutions (11th-April-2014)
Test | 90 questions | 180 min

- JEE Main 2014 Question Paper With Solutions (6th-April-2014)
Test | 90 questions | 180 min