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JEE Main 2014 April 9 Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2014 April 9 Paper & Solutions

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JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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.

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JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 8

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

Detailed Solution for JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 16

Equation for electric field is,

The net surface charge is given by,

By substituting the values we’ll get,

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 18

The minimum value of R can be calculated by,

= 220/20
= 11 Ω

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 20

The equation of magnetic field of earth is

Substitute the values, for dipole moment of earth,

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 22

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.

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 23


The relation for θ is

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

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 25

The equation for magnifying power is,

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 26

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

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 27

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

JEE Main 2014 April 9 Paper & Solutions - Question 28

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

Detailed Solution for JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - Question 29

The two towers are represented in the figure below,

The maximum distance between the towers is given by,

JEE Main 2014 April 9 Paper & Solutions - 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 April 9 Paper & Solutions - 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.

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