JEE Exam  >  JEE Tests  >  JEE Main 2014 April 6 Paper & Solutions - JEE MCQ

JEE Main 2014 April 6 Paper & Solutions - JEE MCQ


Test Description

30 Questions MCQ Test - JEE Main 2014 April 6 Paper & Solutions

JEE Main 2014 April 6 Paper & Solutions for JEE 2024 is part of JEE preparation. The JEE Main 2014 April 6 Paper & Solutions questions and answers have been prepared according to the JEE exam syllabus.The JEE Main 2014 April 6 Paper & Solutions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main 2014 April 6 Paper & Solutions below.
Solutions of JEE Main 2014 April 6 Paper & Solutions questions in English are available as part of our course for JEE & JEE Main 2014 April 6 Paper & Solutions solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main 2014 April 6 Paper & Solutions | 90 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
JEE Main 2014 April 6 Paper & Solutions - Question 1

​The current voltage relation of diode is given by I = (e1000V/T – 1) mA, where the applied V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 1

 I = (e1000 V/T – 1)mA
When I = 5 mA, e1000 V/T = 6 mA
Also, dI =
 
= 0.2 mA

JEE Main 2014 April 6 Paper & Solutions - Question 2

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 2

Time taken to reach highest point is t1 = u/g
Speed on reaching ground = 
Now, v = u + at 
                                     

1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Main 2014 April 6 Paper & Solutions - Question 3

A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 3

a = Rα
mg  – T = ma
T × R = mR2α
or T = ma
a = g/2

JEE Main 2014 April 6 Paper & Solutions - Question 4

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. f the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 4

tanθ = dy/dx = x2/2
At limiting equilibrium,
μ = tanθ

⇒ x = ±1
Now, y = 1/6

JEE Main 2014 April 6 Paper & Solutions - Question 5

When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is :

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 5

JEE Main 2014 April 6 Paper & Solutions - Question 6

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support.
The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 6

τ = mg × l sin θ. (Direction parallel to plane of rotation of particle)

as perpendicular to direction of L changes but magnitude remains same.

JEE Main 2014 April 6 Paper & Solutions - Question 7

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 7





JEE Main 2014 April 6 Paper & Solutions - Question 8

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is :
(For steel Young's modulus is 2 × 1011 Nm–2 and coefficient of thermal expansion is 1.1 × 10–5 K–1)

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 8

As length is constant,
Strain = 
Now pressure = stress = Y × strain
= 2 × 1011 × 1.1 × 10–5 × 100
= 2.2 × 108 Pa

JEE Main 2014 April 6 Paper & Solutions - Question 9

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio is

  

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 9

Equating pressure at A

(Rcosα + Rsinα)d2g = (Rcosα – Rsinα)d1g
⇒ 

JEE Main 2014 April 6 Paper & Solutions - Question 10

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is (Density of water is ρω)

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 10

When the bubble gets detached, Buoyant force = force due to surface tension

JEE Main 2014 April 6 Paper & Solutions - Question 11

​Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 11


Q = Q1 + Q2

200 – 2T = 2T + T
⇒ T = 40°C
Q =  =  4.8 cal/s

JEE Main 2014 April 6 Paper & Solutions - Question 12

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 12


For BC, ΔT = –200 K
ΔU = –500R

JEE Main 2014 April 6 Paper & Solutions - Question 13

​An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 13


P + x = P0
P = (76 – x)
8 × A × 76 = (76 – x) × A × (54 – x)
x = 38
Length of air column = 54 – 38 = 16 cm.

JEE Main 2014 April 6 Paper & Solutions - Question 14

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s, it travels 2a in same direction then

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 14

As it starts from rest, we have
x = Acosωt. At t = 0, x = A
when t = , x = A – a
when t = , x = A – 3a

Now, A – a = Acos

JEE Main 2014 April 6 Paper & Solutions - Question 15

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 15


∴ Answer is 6.

JEE Main 2014 April 6 Paper & Solutions - Question 16

Assume that an electric field exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 16

JEE Main 2014 April 6 Paper & Solutions - Question 17

A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 17


= 2.2 × 8.85 ×10–12 × 3 ×104 ≈ 6 × 10–7  C/m2

JEE Main 2014 April 6 Paper & Solutions - Question 18

​In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 18

15 × 40 + 5 × 100 + 5 × 80 + 1000
= V × I
600 + 500 + 400 + 1000 = 220 I

I = 12 A.

JEE Main 2014 April 6 Paper & Solutions - Question 19

A conductor lies along the z-axis at –1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A indirection (see figure). For a fieldT, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10–3 s. Assume parallel motion along the x-axis

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 19

Average Power = work/time


= 2.97 × 10–3 J

JEE Main 2014 April 6 Paper & Solutions - Question 20

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 103 A m–1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 20


I = 3 A.

JEE Main 2014 April 6 Paper & Solutions - Question 21

In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 21

Applying Kirchhoff's law in closed loop, –VR – VC = 0
⇒ VR/VC = –1

JEE Main 2014 April 6 Paper & Solutions - Question 22

During the propagation of electromagnetic waves in a medium

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 22

Energy is equally divided between electric and magnetic field

JEE Main 2014 April 6 Paper & Solutions - Question 23

​A thin convex lens made from crown glass has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f1 and f2 respectively. The correct relation between the focal lengths is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 23

By Lens maker's formula

⇒ f1 = 4f & f2 = –5f

JEE Main 2014 April 6 Paper & Solutions - Question 24

A green light is incident from the water to the air-water interface at the critical angle(θ). Select the correct statement

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 24

sin θ= 1/μ

For greater  wavelength (i.e. lesser frequency) μ is less
So, θc would be more. So, they will not suffer reflection and come out at angles less then 90°.

JEE Main 2014 April 6 Paper & Solutions - Question 25

Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and Irespectively, then equals

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 25

By law of Malus, I = I0cos2θ
Now, IA′ = IAcos230
IB′ = IBcos260
As IA′ = IB′

IA/IB = 1/3

JEE Main 2014 April 6 Paper & Solutions - Question 26

​The radiation corresponding to 3→2 transitions of hydrogen atoms falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 26



= 0.8 V
For transition between 3 to 2,

JEE Main 2014 April 6 Paper & Solutions - Question 27

​Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ1, λ2, λ3 and λ4 respectively then approximately which one of the following is correct?

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 27


JEE Main 2014 April 6 Paper & Solutions - Question 28

The forward biased diode connection is

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 28


For forward Bias, p-side must be at higher potential than n-side.

JEE Main 2014 April 6 Paper & Solutions - Question 29

Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists :

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 29

(a) Infrared rays are used to treat muscular strain
(b) Radiowaves are used for broadcasting
(c) X-rays are used to detect fracture of bones
(d) Ultraviolet rays are absorbed by ozone

JEE Main 2014 April 6 Paper & Solutions - Question 30

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

Detailed Solution for JEE Main 2014 April 6 Paper & Solutions - Question 30

As measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD – 1 VSD = 0.1 mm

View more questions
Information about JEE Main 2014 April 6 Paper & Solutions Page
In this test you can find the Exam questions for JEE Main 2014 April 6 Paper & Solutions solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main 2014 April 6 Paper & Solutions, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE