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Test: NEET 2016 Past Year Paper - NEET MCQ


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30 Questions MCQ Test - Test: NEET 2016 Past Year Paper

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Test: NEET 2016 Past Year Paper - Question 1

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 1

The question is illustrated in the figure below,

Let, the tension at point A be TA.
Using Newton's second law, we have

Energy at point C is,

At point C, using Newton's second law,

In order to complete a loop, T≥ 0
so,

From equation (i) and (ii)
Using the principle of conservation of energy,

Test: NEET 2016 Past Year Paper - Question 2

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the

angle between these vectors is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 2

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Test: NEET 2016 Past Year Paper - Question 3

At what height from the surface of earth the gravitational potential and the value of g are -5.4 x 10-7J

kg-2 and 6.0 ms-2 respectively ? Take the radius of earth as 6400 km.

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 3

-GM/r = 5.4 x 10-7

-GM/r= 6.0

dividing both the equations, r = 9000 km.

so height from the surface = 9000 - 6400 = 2600 km

Test: NEET 2016 Past Year Paper - Question 4

A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid is

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 4

Given,
Number of turns in the solenoid, N = 1000
Current, I = 4 A
Magnetic flux, straight phi subscript straight B = 4 x 10-3 Wb
Self-inductance of solenoid is given by,

Putting the value of equation in (i), we get

= 1H

Test: NEET 2016 Past Year Paper - Question 5

An inductor 20 mH, a capacitor 50 μF and a resistor. 40Ω are connected in series across a source of

emf V = 10 sin 340t. The power loss in A.C. circuit is

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 5


Test: NEET 2016 Past Year Paper - Question 6

Two identical charged spheres suspended from a common point by two mass less strings of lengths ℓ are initially at a distance d(d << ℓ) a part because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 6



Test: NEET 2016 Past Year Paper - Question 7

A capacitor of 21F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 7

Consider the figure given above.
When switch S is connected to point 1, then initial energy stored in the capacitor is given as,

When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,

Therefore, per centage loss of energy 

Test: NEET 2016 Past Year Paper - Question 8

A particle moves so that its position vector is given by r = cosωtx + sinωty. Where ω is a constant.

Which of the following is true?

 

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 8

Test: NEET 2016 Past Year Paper - Question 9

From a disc of radius R and mass M1 a circular hole of diameter R1 whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about at perpendicular axis, passing through the centre ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 9



Test: NEET 2016 Past Year Paper - Question 10

The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 10

Test: NEET 2016 Past Year Paper - Question 11

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 11

According to question, emf of the cell is directly proportional to the balancing length.
i.e., E∝ ℓ    ... (i)
Now, in the first case, cells are connected in series
That is,
Net EMF = E1 + E2
From equation (i),
E1 + E2 = 50 cm (given)  ... (ii)
Now, the cells are connected in series in the opposite direction,
Net emf = E1 - E2
From equation (i)
E1 - E2 = 10       ... (iii)
From equation (ii) and (iii),

Test: NEET 2016 Past Year Paper - Question 12

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 ms-1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is : (Take velocity of sound in air = 330 ms-1)

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 12



 

Since the observer and the wall are stationary so frequency of echo observed by the observer will also

be 838 Hz.

Test: NEET 2016 Past Year Paper - Question 13

To get output 1 for the following circuit, the correct choice for the input is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 13

The resultant boolean expression of the above logic circuit is given by,
Y = (A+B).C
Using the inputs given in the options,
If A = 0, B = 0,C = 0, we have
Y = (0+0).0
i.e., Y = 0
If A = 1, B = 1, C = 0, then we have
Y = (1+1).0
i.e., Y = 1.0 = 0
If A = 1, B = 0, C = 1, then
Y = (1+0).1
i.e., Y = 1.1 = 1
If A = 0, B = 1, C = 0, then
Y = (0+1).0
i.e., Y = 1.0
Y = 0
Therefore, output Y = 1 only when inputs A = 1, B = 0 and C = 1.

Test: NEET 2016 Past Year Paper - Question 14

In a diffraction pattern due to a single slit of width 'a' the first minimum is observed at an angle 30 deg when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 14

Given that, first minimum is observed at an angle of 30º in a diffraction pattern due to a single slit of width a.
i.e., n = 1, θ = 30º
According to the Bragg's law of diffraction,

For 1st secondary maxima,

Putting the value of a from Eqn. (i) to Eqn. (ii), we get

Test: NEET 2016 Past Year Paper - Question 15

When a metallic surface is illuminated with radiation of wavelength λ the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2:, the stopping potential is V/4. The threshold wavelength for the metallic surface is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 15

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V.
Photoelectric equation can be written as,

Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,

From equations (i) and (ii), we get

When a metallic surface is illuminated with radiation of wavelength λ stopping potential is V.
Photoelectric equation can be written as,

Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,

From equations (i) and (ii), we get

Test: NEET 2016 Past Year Paper - Question 16

When an α-particle of mass 'm' moving with velocity ' v ' bombards on a heavy nucleus of charge 'Ze' its distance of closet approach from the nucleus depends on m as :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 16

When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,


This is the required distance of closest approach to alpha particle from the nucleus.
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,


This is the required distance of closest approach to alpha particle from the nucleus.

Test: NEET 2016 Past Year Paper - Question 17

Match the corresponding entries of column-1 with column-2. [Where m is the magnification produced

by the mirror]

Column - 1                   Column - 2

(A) m = -2               (a) Convex mirror

(B) m = -1/2            (b) Concave mirror

(C) m = +2              (c) Real image

(D) m = =1/2            (d) Virtual image

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 17

(A) m = -2, so image is magnified and inverted. Which is possible only for concave mirror. since image is i inverted so it will be real.

(B) M = -1/2 , so image is inverted and diminished. since image is inverted, so it will be real, and the

mirror will be concave.

(C) M = +2, image is magnified so the mirror will be concave. Image is erect so it will be virtual.

(D) m = +1/2 , image is erect so image will be virtual. Image is virtual and diminished, so the mirror should be convex. Ans. will be (2)

Test: NEET 2016 Past Year Paper - Question 18

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10-4J by the end of the second revolution after the beginning of the motion ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 18

Given, mass of particle. m = 0.01 kg
Radius of circle along which particle is moving , r = 6.4 cm
Kinetic energy of particle, K.E. = 8 x 10-4 J

Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.
It means, initial velocity (u) is 0 m/s at this moment.
Now, using the Newton's 3rd equation of motion,

Test: NEET 2016 Past Year Paper - Question 19

A small signal voltage V(t) = V0 sinωt is applied across an ideal capacitor C :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 19

Capacitor does not consume energy effectively over full cycles

Test: NEET 2016 Past Year Paper - Question 20

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 20

Time does not depend on mass, else

K2/Ris least for sphere and hence least time is taken by sphere

Test: NEET 2016 Past Year Paper - Question 21

Coefficient of linear expansion of brass and steel rods are α1 and α2 . Lengths of brass and steel rods are ℓ1 and ℓ2 respectively. If (ℓ2 - ℓ1 ) is maintained same at all temperatures, which one of the following relations holds good ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 21

Coefficient of linear expansion of brass = α1
Coefficient of linear expansion = 
Length of brass and steel rods are l1 and l2 respectively.
Given,
Increase in length (l2'-l1' ) is same for all temperature.
So,

Test: NEET 2016 Past Year Paper - Question 22

A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 22

According to a question,
Focal length of the objective lens, Fo = +40 cm

Focal length of eyepiece, Fe = 4 cm

Object distance for objective lens (uo) = -200 cm
Applying lens formula for objective lens,


Image will be formed at the focus of a eyepiece lens.
So, for normal adjustment distance between objectives and eyepiece (length of tube) will be, v + Fe = 50 + 4 = 54 cm

Test: NEET 2016 Past Year Paper - Question 23

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. it is subjected to a torque which produces a constant angular acceleration of 2.0 rad s-2. Its net acceleration in ms-2 at the end of 2.0 s is approximately :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 23

The angular speed of disc increases with time, and hence centripetal acceleration

Test: NEET 2016 Past Year Paper - Question 24

A refrigerator works between 40 celcius and 300 celcius. It is requiredto remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is :(Take 1 cal = 4.2 Joules)

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 24

Given,
Temperature of the source, T = 30o C = 30 + 273  = 303 K
Temperature of sink, T2 = 4o C = 4 + 273 = 277 K
We know,

where, Q2 is the amount of heat drawn from the sink at T2,
W is the work done on the working substance,
Q1 is the amount of heat rejected to source at room temperature T1.
That is,

Test: NEET 2016 Past Year Paper - Question 25

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 25

Directly from graph the magnitude of work done = Area under p-v plot is larger for adiabatic compression

Test: NEET 2016 Past Year Paper - Question 26

The intensity at the maximum in Young's double slit experiment is I0. Distance between two slits is d = 5λ, where λ is the wavelength of light used is the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 26

Given, 

Maximum intensity = Io

Distance between the slits, d = 5λ

Distance of screen from the slit, D = 10d
Using the formula,
Path difference, 
Here,

D = 10d = 50λ
So,

Corresponding phase difference will be



Test: NEET 2016 Past Year Paper - Question 27

Two non-mixing liquids of densities ρ and nρ (n > 1) are put in container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 27

Test: NEET 2016 Past Year Paper - Question 28

Consider the junction diode as ideal. The value of current flowing through AB is :

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 28

For diode as ideal
I = ΔV/R = 4-(-6)/ 10= 10-2 A

Test: NEET 2016 Past Year Paper - Question 29

A car is negotiating a curved road of radius R. The road is banked at an angle .. The coefficient of friction between the tyres of the care and the road is 1s. The maximum safe velocity on this road is:

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 29

A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
The given situation is illustrated as:

In the case of vertical equilibrium,

Dividing Eqns. (i) and (ii), we get

Test: NEET 2016 Past Year Paper - Question 30

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B`, at a radial distances a/2 and 2a respectively, from the axis of the wire is:

Detailed Solution for Test: NEET 2016 Past Year Paper - Question 30

Consider two amperian loops of radius a/2 and 2a as shown in the diagram.

Applying Ampere's circuital law for these loops, we get

B1 acts at a distance a/2 from the axis of the wire.
Similarly, for bigger amperian loop,

Total current enclosed by amperian loop 2.
 at a distance 2a from the axis of the wire.

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