NEET Test: Physics Mock Test - 2 Free Online Test 2026

The direction of oscillations of E and B fields are perpendicular to each other as well as to the direction of propagation. So, electromagnetic waves are transverse in nature. The electric and magnetic fields oscillate in same phase.

Q = It also Q = ne [e = 1.6 × 10⁻¹⁹C]
∴ ne = It or
= 6.25 × 10¹⁸ electrons s⁻¹

The maximum kinetic energy of the emitted electrons is given by
K_max​=hυ−ϕ₀​=h(4υ)−h(υ)=3hυ

If drops and bubbles do not collapse under the effect of gravity, it indicates that the pressure inside the drop is greater than the pressure outside. The greater inner pressure prevents the drop from collapsing.

The dipole moment is defined as the product of a charge and distance. The dimension of charge (current*time) is [I T] and the dimension of distance is [L]. Therefore the dimension of dipole moment is [L T I]. Its unit in the CGS and the SI system are esu*cm and C*m respectively.

Total emf = 3 x 4 = 12 V
Total resistance = 6 + 3r
Current in the circuit = 2 A
Using ohm's law
2 = 12/ (6+3r)
2(6+3r)=12
12+6r=12
6r = 12-12 = 0
r = 0/6 = 0

D is the derived expression.

The speed of revolving electron in nth state of hydrogen atom is:
v=​e²/2nhϵ₀ ​
For n=1,
v= (1.6×10⁻¹⁹)^2​/2(1)(6.6×10⁻³⁴)(8.85×10⁻¹²)
v=2.56×10^−38​/116.82×10⁻⁴⁶
 
v=0.0219×10⁸ms⁻¹
The speed of light is 3×10⁸
Hence, 
v/c​=0.0219×10^8​/3×10⁸
v/c​=1/137
 

Given, n = 1, B = 10^–2 T,
A = π(0.3)2m2, R = π²
f= (200/60) and ω = 2π(200/60)
Substituting these values and solving, we
get
I₀ = 6 × 10^–3 A = 6mA

Energy stored per unit volume

stress (stress / Young' s modulus)

 (stress)² /( Young ' s modulus)

From diagram, force on q₁(= q) at A,


where  is the unit vector along BC 
Force on 
(where   is the unit vector along AC)
Force on q₃(= -q) at C,

where  unit vector along the direction bisecting ∠BCA

Given, i₁​=i₂​=i
∴F=μ_0​i²l​/2πb
Hence, force per unit length is F=μ₀​i²​/2πb

As the electron is moving along the direction of the magnetic field, it will experience no magnetic force, but due to an electric force acting on it opposite to the direction of electric field (as it is a negatively charged particle) the velocity of the electron will decreases.

The restoring force in a simple harmonic motion is maximum in magnitude when the particle is instantaneously at rest because in SHM object’s tendency is to return to mean position and here particle is instantaneously at rest after that instant restoring force will be max to bring particle to mean position.

Sound waves can not be polarised as they are longitudinal. Light waves can be polarised as they are transverse.

As we know, f=v/2l
Now, it will act as one end and one end closed.
So, f₀=v/2l’=v/4½=v/2l=f

To conserve the momentum each time a single bullet is fired,
the reverse speed gained by the gun from one bullet is 
V = 400 X .035 / 20
= 0.7 m/s
Thus total speed gained in a second is = 0.7 X 400 = 280 m/s
As total speed is gained in one second only the acceleration produced = 280 m/s²
Thus total force applied on the gun by the bullets = 20 x 280 
= 5600 N

The dielectric constant of metal is infinite as the net electric field inside the metal is zero.

• The dielectric constant is defined as the ratio of the permittivity of a substance to the permittivity of free space.
• As the electric flux density increases, the dielectric constant increases.

Focal length in air = 20 cm
Refractive index of air-water n₁= 1.33
Refractive index of air - glass n₂= 1.5
For focal length in air,
Using formula of lens
1/f_air={(n₂/n₁)-1}(1/R₁)-(1/R₂)
Put the value into the formula
1/20={(1.5/1)-1}{(1/R₁)-(1/R₂)}
1/20=0.5{(1/R₁)-(1/R₂)}…1
We need to calculate the focal length in water
Using formula of lens
1/f_water={(1.5/1.33)-1}{(1/R₁)-(1/R₂)}
1/f_water=0.128{(1/R₁)-(1/R₂)}….2
f_water/20=0.5/0.128
f_water=78.125cm

When Young's double-slit set up for interference is shifted from air to within water then the fringe width decreases because the refractive index of water is more than that of the air.
Originally the fringe width is given by:
β1​=λD/2d​
The new fringe width within water will be given by 
β2​= λD​/2nd
So, β2​= β1/n​​
Here, n is the refractive index of medium.
 

Speed = 10 m/s; radius r = 10 m
Angle made by the wire with the vertical is:

We can observe from the table that Yˉ=A.B⇒Y=AB
So, the table represents a NAND gate.

The graph given is a velocity-time graph for the first 4 seconds of motion. To find the distance covered during this time interval, we need to calculate the area under the velocity-time graph.

The graph shows a straight line from (0, 0) to (4, 15), which forms a right-angled triangle with:

• Base = 4 seconds (time)
• Height = 15 m/s (velocity)

Distance covered = Area under the curve = Area of the triangle

Area of triangle = (1/2) × base × height

= (1/2) × 4 × 15

= 2 × 15 = 30 meters

Therefore, the distance covered in the first 4 seconds is 30 meters.

The correct answer is Option A - 0.36 R

For a uniform sphere, gravitational acceleration varies linearly with distance from the centre; at a distance r the value is g_r = g (r/R), where g is the surface acceleration and R is the radius.

At a depth d below the surface the distance from the centre is r = R - d.

Hence the acceleration at that depth is g_d = g ((R - d)/R).

Given g_d = 0.64 g, we get (R - d)/R = 0.64.

So R - d = 0.64 R, which gives d = 0.36 R.

Thus the depth is 0.36 R, matching Option A.