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JEE Main 2014 April 12 Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2014 April 12 Paper & Solutions

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JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 2

The expression of time taken by man is,

The expression for time taken by escalator is,

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JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 3

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

JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 11

The rate of change temperature is given as,

Substitute the values.

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

JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 12

The values for heat in Carnot engine is,

The expression for efficiency is,

So,

JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 13

The equation of rms velocity is,

From ideal gas equation,

Thus,

The molar mass of hydrogen is 2g.

JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 27

The radioactivity in tth time,

Substitute the values.

Take log both sides, and use property of log.

JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 29

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

Detailed Solution for JEE Main 2014 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - 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 April 12 Paper & Solutions - Question 30

The potential gradient is,

The voltage across 220 cm is,

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

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