Physics: Topic-wise Test- 8 - NEET MCQ

# Physics: Topic-wise Test- 8 - NEET MCQ

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## 30 Questions MCQ Test NEET Mock Test Series 2025 - Physics: Topic-wise Test- 8

Physics: Topic-wise Test- 8 for NEET 2024 is part of NEET Mock Test Series 2025 preparation. The Physics: Topic-wise Test- 8 questions and answers have been prepared according to the NEET exam syllabus.The Physics: Topic-wise Test- 8 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Physics: Topic-wise Test- 8 below.
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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 for Physics: Topic-wise Test- 8 - Question 1

Suppose in the following figure, equilibrium of charge B is considered.

Hence for its equilibrium

Physics: Topic-wise Test- 8 - Question 2

### Electric field intensity varies with distance as:

Detailed Solution for Physics: Topic-wise Test- 8 - Question 2

E = F/q = KQ/r2

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Physics: Topic-wise Test- 8 - Question 3

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

Detailed Solution for Physics: Topic-wise Test- 8 - Question 3

Electric Field Intensity E = F\q

Physics: Topic-wise Test- 8 - Question 4

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 for Physics: Topic-wise Test- 8 - Question 4

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 5

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

Detailed Solution for Physics: Topic-wise Test- 8 - Question 5

E = F/q

Physics: Topic-wise Test- 8 - Question 6

A test charge of 3nC is placed at origin and experiences a force of 9 x 10-4 N. The electric field at origin is:​

Detailed Solution for Physics: Topic-wise Test- 8 - Question 6

E =F/q = (KQ/r2)

Physics: Topic-wise Test- 8 - Question 7

At any point on an electric field line

Detailed Solution for Physics: Topic-wise Test- 8 - Question 7

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 8

An electric field can deflect

Detailed Solution for Physics: Topic-wise Test- 8 - Question 8

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 9

In a parallel plate capacitor, the capacity increases if

Detailed Solution for Physics: Topic-wise Test- 8 - Question 9

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 10

A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations

(i) key KK is kept closed and plates of capacitors are moved apart using insulating handle
(ii) key KK is opened and plates of capacitors are moved apart using insulating handle
Which of the following statements is correct?

Detailed Solution for Physics: Topic-wise Test- 8 - Question 10

When key K is kept closed, condenser C is charged to potential V. When plates of capacitors are moved apart, its capacitance, C = ϵoA/d decreases.
As potential of condenser remains same, charge Q = CV decreases. So option is correct. Once key K is closed, condenser gets charged, Q = CV.
Now, if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains same.
As plates or capacitor are moved apart, its capacity C = ϵoA/d decreases.
Therefore, its potential, V = q/C increases

Physics: Topic-wise Test- 8 - Question 11

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 for Physics: Topic-wise Test- 8 - Question 11

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 12

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

Detailed Solution for Physics: Topic-wise Test- 8 - Question 12

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 13

The intensity of a plane electromagnetic wave is proportional to

Detailed Solution for Physics: Topic-wise Test- 8 - Question 13

Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.

Physics: Topic-wise Test- 8 - Question 14

According to Maxwell’s equations

Detailed Solution for Physics: Topic-wise Test- 8 - 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 for Physics: Topic-wise Test- 8 - 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 for Physics: Topic-wise Test- 8 - 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 for Physics: Topic-wise Test- 8 - Question 17

A body at temperature T produces a continuous spectrum of wavelengths. For a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck's law by the relation, m=0.29cmK/T. For m=106m,T=2900K.Temperatures for other wavelengths can be found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of the electromagnetic spectrum. Thus, to obtain visible radiation, say  =5×107m, the source should have a temperature of about 6000K. A lower temperature will also produce this wavelength but not the maximum intensity.

Physics: Topic-wise Test- 8 - Question 18

Without the concept of displacement current it is not possible to correctly apply Ampere’s law on a path parallel to the plates of parallel plate capacitor C in

Detailed Solution for Physics: Topic-wise Test- 8 - Question 18

Explanation:

Region between the plates:
- The region between the plates of the capacitor is where the concept of displacement current is most crucial.
- Without displacement current, the changing electric field between the plates would not be taken into account, leading to significant errors in the calculation of the magnetic field in this region.
- Displacement current ensures that Ampere's law is correctly applied in this region, considering both the current in the circuit and the changing electric field between the plates.

Physics: Topic-wise Test- 8 - Question 19

The value of alternating e.m.f. is e = 500 sin 100pt , then the frequency of this potential in Hz is

Detailed Solution for Physics: Topic-wise Test- 8 - Question 19

l=500sin(100πt)
⟹ω=100π
ω=2πf
⟹2πf=100π
f=50Hz

Physics: Topic-wise Test- 8 - Question 20

The domestic power supply is at 220 volt. The amplitude of emf will be

Detailed Solution for Physics: Topic-wise Test- 8 - Question 20

We know: Vrms​=​Vo​​/√2

Substituting values (Here RMS Voltage is 220 V)
We get: Maximum V (or amplitude) =311V

Physics: Topic-wise Test- 8 - Question 21

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

Detailed Solution for Physics: Topic-wise Test- 8 - 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

The phase difference between the alternating current and voltage represented by the following equation I = I0 sin wt, E = E0 cos (wt + p / 3), will be

Detailed Solution for Physics: Topic-wise Test- 8 - Question 22

E= εocos(ωt+π/3) –1
I=I0sin(ωt)=I0cos{(π/2)-cot)
=I0cos(ωt+(- π/2)) —2
By,1 and 2,
Phase difference between E and I,
E-VI
=(ωt+π/3)- (ωt+(- π/2))
= π/3+ π/2
=5π/6

Physics: Topic-wise Test- 8 - Question 23

The inductive reactance of a coil is 1000W. If its self inductance and frequency both are increased two times then inductive reactance will be

Detailed Solution for Physics: Topic-wise Test- 8 - Question 23

We know that,
XL= ωL
Or, XL =2πfL   [ ω can be written as, 2πf]
Initially,
XL1=2πf1L1=1000Ω
For the second case,
F  and L both are doubled, so,
XL2 =2π2f12L1 =4x2πf1L1
=4000 Ω

Physics: Topic-wise Test- 8 - Question 24

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 for Physics: Topic-wise Test- 8 - Question 24

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 25

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 for Physics: Topic-wise Test- 8 - Question 25

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 26

Alternating current lead the applied e.m.f. by p/2 when the circuit consists of

Physics: Topic-wise Test- 8 - Question 27

Ampere's circuital law is given by

Detailed Solution for Physics: Topic-wise Test- 8 - Question 27

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 28

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 for Physics: Topic-wise Test- 8 - Question 28

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 29

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 for Physics: Topic-wise Test- 8 - Question 29

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

Physics: Topic-wise Test- 8 - Question 30

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 for Physics: Topic-wise Test- 8 - Question 30

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.

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