Test: NEET 2016 Past Year Paper - NEET MCQ

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_c ≥ 0
so,

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

-GM/r = 5.4 x 10^-7

-GM/r² = 6.0

dividing both the equations, r = 9000 km.

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

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

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 

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),

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

be 838 Hz.

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.

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

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

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.

(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)

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,

Capacitor does not consume energy effectively over full cycles

Time does not depend on mass, else

K²/R² is least for sphere and hence least time is taken by sphere

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,

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

Focal length of eyepiece, F_e = 4 cm

Object distance for objective lens (u_o) = -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 + F_e = 50 + 4 = 54 cm

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

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

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

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

For diode as ideal
I = ΔV/R = 4-(-6)/ 10³ = 10^-2 A

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

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.