NEET Exam  >  NEET Test  >   Mock Test Series - Updated 2026 Pattern  >  Physics: Topic-wise Test- 8 - NEET MCQ

NEET Physics: Topic-wise Test- 8 Free Online Test 2026


MCQ Practice Test & Solutions: Physics: Topic-wise Test- 8 (45 Questions)

You can prepare effectively for NEET NEET Mock Test Series - Updated 2026 Pattern with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Physics: Topic-wise Test- 8". These 45 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 45

Sign up on EduRev for free to attempt this test and track your preparation progress.

Physics: Topic-wise Test- 8 - Question 1

A charge q is placed at the center of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to 

Detailed Solution: Question 1

Let q charge is situated at themidposition of the lineAB.The distance between AB is x. A and B be the positions of charges Q and Q respectively.

Physics: Topic-wise Test- 8 - Question 2

Electric field intensity varies with distance as:

Detailed Solution: Question 2

E = F/q = KQ/r2

Physics: Topic-wise Test- 8 - Question 3

If electric field lines cross each other that would mean

Detailed Solution: Question 3

Properties of Electric Field Lines. At every point on an electric field line, the tangent represents the direction of electric field. If two field lines cross each other at the point, then there would be two possible directions for electric field which is physically impossible.

Physics: Topic-wise Test- 8 - Question 4

The S.I. unit of electric field intensity is:​

Detailed Solution: Question 4

Electric Field Intensity E = F\q

Physics: Topic-wise Test- 8 - Question 5

The diagram shows the electric field lines in a region of space containing two small charged spheres, Y and Z then which statement is true?

Detailed Solution: Question 5

A small negatively charged body placed at X would be pushed to the right

The negative charge feels a force opposite to the direction of field.

 Other cases are not possible from properties of electric lines of force.

Physics: Topic-wise Test- 8 - Question 6

Which equation correctly shows the force on a charge q in an electric field E

Detailed Solution: Question 6

The force F on a charge q placed in an electric field E is given by the fundamental relation:

F = qE

Here, F and E are vector quantities, so the equation is vectorial:

→F = q →E

This means the force experienced by the charge is directly proportional to the magnitude of the charge and the electric field, and it acts in the direction of the electric field if the charge is positive (opposite direction if the charge is negative).

Explanation of options:

  • Option A: Uses capital Q instead of q, which might be a notation difference, but generally q is used for the charge on which force acts.
  • Option B: →F = q →E is the correct and standard formula for force on a charge in an electric field.
  • Option C: →E = q →F is incorrect because electric field is defined as force per unit charge, not force multiplied by charge.
  • Option D: →E = Q →F is also incorrect for the same reason as Option C.

Additional note:

The electric field E at a point is defined as the force experienced by a unit positive charge placed at that point:

E = F/q

But the question asks for the force on charge q, so the correct expression is:

F = qE

Therefore, the correct answer is option B.

Physics: Topic-wise Test- 8 - Question 7

The field lines for single negative charge are:

Detailed Solution: Question 7

Conventionally, field lines originate from a positive point charge and terminate at negative charge

Physics: Topic-wise Test- 8 - Question 8

At any point on an electric field line 

Detailed Solution: Question 8

When a tangent is drawn at any point on field line then that tangent gives the direction of electric field at that point

Physics: Topic-wise Test- 8 - Question 9

An electric field can deflect

Detailed Solution: Question 9

Only alpha rays are moving with small velocity and having charge so they will be affected by electric field.
X Rays and Gamma Rays are electromagnetic radiations. They do not carry electric charge while neutron is a charge less particle.

Physics: Topic-wise Test- 8 - Question 10

In a parallel plate capacitor, the capacity increases if

Detailed Solution: Question 10

In a parallel plate capacitor the capacity of capacitor.

The capacity of capacitor increases if area of the plate is increased.

Physics: Topic-wise Test- 8 - Question 11

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates

Detailed Solution: Question 11

Correct answer is A.

Physics: Topic-wise Test- 8 - Question 12

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0= 510 nT. Amplitude of the electric field part of the wave is

Detailed Solution: Question 12

Magnetic field part of a harmonic electromagnetic wave in vacuum
,B0​=510×10−9T
Speed of light,
C=3×108m/s
E=cBo​=153N/C

Physics: Topic-wise Test- 8 - Question 13

It is necessary to use satellites for long distance TV transmission because

Detailed Solution: Question 13

TV signals being of high frequency are not reflected by the ionosphere. Therefore, to reflect these signals, satellites are needed. That is why, satellites are used for long distance TV transmission.
 Most long-distance shortwave (high frequency) radio communication—between 3 and 30 MHz—is a result of skywave propagation.
This 3-30 MHz is a range of frequencies which are used in sky waves propagation so that the ionosphere is capable of reflecting it.

Physics: Topic-wise Test- 8 - Question 14

According to Maxwell’s equations

Detailed Solution: Question 14

Maxwell’s Fourth Equation
It is based on Ampere’s circuital law. To understand Maxwell’s fourth equation it is crucial to understand Ampere’s circuital law,
Consider a wire of current-carrying conductor with the current I, since there is an electric field there has to be a magnetic field vector around it. Ampere’s circuit law states that “The closed line integral of magnetic field vector is always equal to the total amount of scalar electric field enclosed within the path of any shape” which means the current flowing along the wire(which is a scalar quantity) is equal to the magnetic field vector (which is a vector quantity)

Physics: Topic-wise Test- 8 - Question 15

The direction of propagation of an electromagnetic plane wave is

Detailed Solution: Question 15

The direction of propagation of the electromagnetic wave is always perpendicular to the plane in which
So, the direction of the propagation of the wave, 
Hence, option B is the correct answer.

Physics: Topic-wise Test- 8 - Question 16

Infrared waves are produced by

Detailed Solution: Question 16

Infrared waves are emitted by hot bodies. They are produced due to the de-excitation of atoms.
They are called Heat waves as they produce heat falling on matter. This is because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is, they heat up and heat their surroundings.
Uses: Infra red lamps; play an important role in maintaining warmth through greenhouse effect.

Physics: Topic-wise Test- 8 - Question 17

Use the formula λm T = 0.29 cmK to obtain the characteristic temperature range for λm=5×10-7m

Detailed Solution: Question 17

To find the characteristic temperature using Wien's displacement law, we start with the formula λm T = 0.29 cmK. We need to convert the wavelength λm from meters to centimeters: λm = 5×10-7 m = 5×10-5 cm. Substitute this into the formula: T = 0.29 cmK / 5×10-5 cm, which gives T = 5800 K. Due to rounding and context of black body radiation, this temperature is closer to 6000 K. Thus, the characteristic temperature range is approximately 6000 K.

Physics: Topic-wise Test- 8 - Question 18

Optical and radio telescopes are built on the ground, but X-ray Astronomy is possible only from satellites orbiting the earth because

Detailed Solution: Question 18

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

Physics: Topic-wise Test- 8 - Question 19

Plane electromagnetic waves are

Detailed Solution: Question 19

E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation. Electromagnetic waves are transverse waves. The wave number is k = 2π/λ, where λ is the wavelength of the wave.

Physics: Topic-wise Test- 8 - Question 20

The average value or alternating current for half cycle in terms of I0 is

Detailed Solution: Question 20

For alternating current average value is taken for half cycles only,
Let, I=I0sinωt where ω=2π/T

=[(I0/ω) cosωt]oT/2 x 2/T
Iavg=2Io/ π

Physics: Topic-wise Test- 8 - Question 21

Sinusoidal peak potential is 200 volt with frequency 50Hz. It is represented by the equation

Detailed Solution: Question 21

Given peak potential is 200v
so, amplitude is 200
and frequency is 50Hz
so angular frequency is 2.πx50 = 314/sec
so the value is
E = 200sin(314t)
Hence option B is the correct answer.

Physics: Topic-wise Test- 8 - Question 22

In an L.C.R series circuit R = 1W, XL = 1000W and XC = 1000W. A source of 100 m.volt is connected in the circuit the current in the circuit is

Detailed Solution: Question 22


R=1W
XL=1000W
XC=1000W
Z= Z= √ (R2+(XL-XC)2)
  =√(12+(1000-1000)2)
Z=1W
    I=V/Z=100/1
I=1000A
 

Physics: Topic-wise Test- 8 - Question 23

A coil of inductance 0.1 H is connected to an alternating voltage generator of voltage E = 100 sin (100t) volt. The current flowing through the coil will be

Detailed Solution: Question 23

L =0.14
E=100sin(100t)
⇒XL​=wL​=100×0.1
=10Ω
⇒io​=Eo/XL​=100/10​=10A
⇒i=io​sin(100t− π/2​)
⇒i=−io​sin[(π/2)​−100]
So we get
⇒i=−10cos(100t)A
therefore, option D is correct.

Physics: Topic-wise Test- 8 - Question 24

Ampere's circuital law is given by

Detailed Solution: Question 24

The line integral of the magnetic field of induction  around any closed path in free space is equal to absolute permeability of free space μ0 times the total current flowing through area bounded by the path.
Ampere's circuital law is given by: 

Physics: Topic-wise Test- 8 - Question 25

A long straight wire in the horizontal plane carries a current of 75 A in north to south direction, magnitude and direction of field B at a point 3 m east of the wire is

Detailed Solution: Question 25

From Ampere circuital law

The direction of field at the given point will be vertical up determined by the screw rule or right hand rule.

Physics: Topic-wise Test- 8 - Question 26

If a long straight wire carries a current of 40 A, then the magnitude ol the field B at a point 15 cm away from the wire is 

Detailed Solution: Question 26

I = 40A
r = 15 cm = 15 x 10-2 m
∴  = 5.34 x 10-5 T

Physics: Topic-wise Test- 8 - Question 27

The correct plot of the magnitude of magnetic field   vs distance r from centre of the wire is, if the radius of wire is R

Detailed Solution: Question 27

The magnetic field from the centre of wire of radius R is given by
B = ((μ0I)/(2R2))r (r < R) ⇒ B ∝ r
and B = μ0I/2πr (r > R) ⇒ B ∝ 1/r
From this descriptions, we can say that the graph (b) is a correct representation.

Physics: Topic-wise Test- 8 - Question 28

Two pure inductors are connected in series, if their self inductance is L. Find the total inductance

Detailed Solution: Question 28

The total inductance of inductors connected in series is given by the following formula:
Leq = L1 + L2
Substituting the values in the equation, we get the total inductance as
Leq = L + L = 2L
Hence, the total inductance of two inductors connected in series is 2L.

Physics: Topic-wise Test- 8 - Question 29

Suppose there are two coils of length 1m with 100 and 200 turns and area of cross section of 5 x 10-3 m2. Find the mutual inductance.

Detailed Solution: Question 29

The magnetic field through the secondary of N2 turns of each other of area S is given as,
N2= Φ2=N2(BS)
     =μ0n1N2i1S
M= N2Φ2/i1
M= μ0n1N2S
Substituting the values.
M=(4πx10-7) x100x200x5x10-3
   =1287142.86x10-10
   =12.57x10-5H

Physics: Topic-wise Test- 8 - Question 30

If the current in primary coil changes from 5 A to 2 A in 0.03 sec and the induced e.m.f. is 1000 volts, find the mutual induction​

Detailed Solution: Question 30

As φ =mi
now take change in flux
dφ/dt=d(mi)/dt
we can write it as
dφ/ dt= m Δi/Δt
-e=m(2-5)/0.03
e=3m/0.03
1000=300m/3
m=10

View more questions
1 videos|19 docs|61 tests
Information about Physics: Topic-wise Test- 8 Page
In this test you can find the Exam questions for Physics: Topic-wise Test- 8 solved & explained in the simplest way possible. Besides giving Questions and answers for Physics: Topic-wise Test- 8, EduRev gives you an ample number of Online tests for practice
Download as PDF