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Bihar PGT Physics Mock Test - 4 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test - Bihar PGT Physics Mock Test - 4

Bihar PGT Physics Mock Test - 4 for Bihar PGT/TGT/PRT 2024 is part of Bihar PGT/TGT/PRT preparation. The Bihar PGT Physics Mock Test - 4 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar PGT Physics Mock Test - 4 MCQs are made for Bihar PGT/TGT/PRT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Bihar PGT Physics Mock Test - 4 below.
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Bihar PGT Physics Mock Test - 4 - Question 1

"कविः नृप धनं याचते" का हिंदी में अनुवाद होगा-

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 1

"कवि: नृप धनं याचते" वाक्य का हिंदी अनुवाद है" कवि राजा से धन मांगता है"I

यहां " दुह्याच्पच्दण्ड" संज्ञा अर्थनिबंधना से "अकथितं च" सूत्र से याच् धातु के कारण नृप में कर्म संज्ञा होकर द्वितीया विभक्ति होती है ।

अकथित कारक — दुह्, याच्, पच्, दण्ड्, रुध्, प्रच्छ्, चि, ब्रू, शास्, जि, मथ्, मुष्, नी, हृ, कृष्, और वह् —16 धातुओं के साथ 'कर्म' का योग होने पर, अपादान आदि कारक अकथित हो जाते है। इस अकथित कारक की कर्म संज्ञा हो जाती है I

अत: विकल्प (A) सही है I

Bihar PGT Physics Mock Test - 4 - Question 2

नाटक को पंचम वेद किसने कहा था?

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 2

आचार्य भरत मुनि ने नाट्यशास्त्र में पंचम वेद बताया है। भरत मुनि के अनुसार ऐसा कोई ज्ञान शिल्प, विद्या, योग एवं कर्म नहीं है, जो नाटक में दिखाई न पड़े I

अत: विकल्प (A) सही है I

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Bihar PGT Physics Mock Test - 4 - Question 3

निर्देश: निम्नलिखित मुहावरे का अर्थ बताइये।

गुड़ गोबर करना

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 3

मुहावरा: गुड़ गोबर करना

अर्थ: बनाया काम बिगाड़ना

वाक्य प्रयोग: वीरू ने जरा-सा बोलकर सब गुड़-गोबर कर दिया।

Bihar PGT Physics Mock Test - 4 - Question 4

If a man has to travel to point E from point A (through these points by the shortest distance), which of the following points will he pass through first?

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 4

According to the information given in the question, we can draw the following diagram.

Thus we can see if a man has to travel to point E from point A  (through these points by the shortest distance), he has to follow a path point A → B → F → E.

Thus travelling from point A, he will pass the first point B.

Hence, the correct answer is "Point B".

Bihar PGT Physics Mock Test - 4 - Question 5
Whose birthday is celebrated as the “Good Governance Day”?
Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 5

The Correct Answer is "Atal Bihari Vajpayee".

Key Points

  • Good governance day is celebrated annually on 25th December to mark the birth anniversary of the former Prime Minister Atal Bihari Vajpayee.
  • The aim is to create awareness of accountability in government among the citizens of India.

Governance:

  • It is the process of decision-making and the process by which decisions are implemented (or not implemented).
  • Governance can be used in several contexts such as corporate governance, international governance, national governance, and local governance.

Eight Characteristics of Good Governance (as identified by the United Nations):

Important Points

Atal Bihari Vajpayee :

  • Atal Bihari Vajpayee was born on 25th December 1924 in the erstwhile princely state of Gwalior (now a part of Madhya Pradesh).
  • Atal Bihari Vajpayee had served three terms as the Prime Minister of India, first for a term of 13 days in 1996, then for a period of 13 months from 1998 to 1999, followed by a full term from 1999 to 2004.
  • He was conferred with the country's highest civilian honor, The Bharat Ratna in 2015 and the second-highest civilian honor, the Padma Vibhushan in 1994.

Jawaharlal Nehru :

  • Jawaharlal Nehru was the first Prime Minister of India.

Indira Gandhi :

  • Indira Priyadarshini Gandhi was the first and, to date, only female Prime Minister of India.
  • Indira Gandhi was the daughter of Jawaharlal Nehru, the first prime minister of India.
Bihar PGT Physics Mock Test - 4 - Question 6

Select the number which can be placed at the sign of the question mark (?) from the given alternatives.

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 6

The logic followed is:

The bottom number = sum of the square of all the first three number column-wise 

In column-wise operation:

Column 1:

69 = 72 + 42 + 22 

69 = 49 + 16 + 4

69 = 69

Column 2:

91 = 32 + 92 + 12 

91 = 9 + 81 + 1

91 = 91

Similarly,

Column 3:

? = 22 + 62 + 52 

? = 4 + 36 + 25

? = 65 

Hence, the correct answer is "65".

Additional Information

The most common pattern asked in exams are based on:

  • It could be the sum of two numbers divided by a constant
  • It could be the average of numbers.
  • It could be in the form of alphabets, where alphabets are increased by constant or increased by the square of numbers or increased by prime numbers.
  • It could be the difference of product of two diametrically opposite numbers.
  • It could be the difference in the sum of adjacent numbers.
  • The difference of the numbers could be in the patterns 1± 1, 2± 1,  and so on.
  • The difference of the numbers could be in the patterns 1± 1, 2± 1, and so on.
  • Numbers could be 12, 22, 32, and so on or 13, 23, 33, and so on.
  • The difference could be prime numbers or the difference could be the square of prime numbers.
  • The difference could be in the form of N± N or N± N.
  • The difference could be in the form of ×N + N or ×2 + 1, ×2 + 2, ×2 + 3, and so on.
  • The difference could be in the form of ×2 ± 1 alternatively.
  • Numbers could be ×1, ×2, ×3, and so on. 
Bihar PGT Physics Mock Test - 4 - Question 7

Which of the figures (1), (2), (3) and (4) can be formed from the pieces given in figure?

Problem Figure

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 7

Hence, the above image is the correct answer.

Bihar PGT Physics Mock Test - 4 - Question 8

If A is the father of B and B is the father of C, then how is C related to A?

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 8

By using the symbols in the table given below, we can draw the following family tree:

Clearly, C is the grandchild of A.

Hence, ‘Grandchild’ is the correct answer.

Bihar PGT Physics Mock Test - 4 - Question 9
Name the majority group of the Russian Social Democratic Workers Party led by Lenin?
Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 9

The correct answer is i.e. Bolsheviks.

Some facts about the Russian Socialist Democratic party(RSDP):

  • The formation was the party that happened in 1898.
  • The party followed Marxist ideas, believing the industrial working class, not peasants, would bring about revolution.
  • The party ideas were based on the philosophies of Karl Marx and Friedrich Engels, namely that despite Russia's agrarian life, the real radical potential extended with the industrial working class.
  • They played an important role in the politics of Russia at that time and they also won few seats in the elections.
  • It was mainly divided into two groups - Bolsheviks and Mensheviks.
  • The following table will help you to understand them better:
Bihar PGT Physics Mock Test - 4 - Question 10

Identify the odd one from the following -

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 10

The logic followed here is:

The place value of A is 1, D is 4, O is 15 and U is 21. 

In each of the given figure the alphabet and its place value are given except  U and 23 in figure D.

Hence, figure D is the odd one from the given figures.

Bihar PGT Physics Mock Test - 4 - Question 11
The average pattern of wind speed, temperature, rainfall etc., in place over a long period of time is called ___________.
Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 11

The correct answer is Climate.

Key Points

  • The average pattern of wind speed, temperature, rainfall, etc, in place over a long period of time, is called climate.
  • Climate is the average weather in a place over many years.
  • The weather can change in just a few hours whereas climate takes millions of years to change.
  • Planet earth has witnessed many variations in climate since the beginning.
  • There are several causes of climate change.
  • The most significant anthropogenic effect on the climate is the increasing trend in the concentration of greenhouse gases in the atmosphere.
*Multiple options can be correct
Bihar PGT Physics Mock Test - 4 - Question 12

AB and CD are smooth parallel rails, separated by a distance l, and inclined to the horizontal at an angle q. A uniform magnetic field of magnitude B, directed vertically upwards, exists in the region. EF is a conductor of mass m, carrying a current i. For EF to be in equilibrium,

               

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 12

Force on EF,
Fmag​​=i∫(dl×B)=iLBsin(90−θ)=iLBcosθ
Fmag​​ up the inclined plane.
Component of weight of EF down the inclined plane: mgsinθ
For EF to be in equilibrium, iLBcosθ=mgsinθ
∴iLB=mgtanθ
For Fmag​ to be up the inclined plane, it needs to flow from E to F.

Bihar PGT Physics Mock Test - 4 - Question 13

Two bodies with same mass “m” separated by a distance “r” exert a gravitational force of F on each other. Suppose the distance between them is doubled and the force becomes F’. The ratio of two forces is

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 13

We know that the force of gravitation is inversely proportional to square of the distance between the two bodies,

i.e. F∝ r-2

Hence, when the distance between them will be doubled, the force will be reduced by 4 times

So, the ratio will be 4:1

Bihar PGT Physics Mock Test - 4 - Question 14

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 14

In a uniformly charged hollow conducting sphere

Bihar PGT Physics Mock Test - 4 - Question 15

Relation between ray and wave front is

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 15

A wave front is a line representing all parts of a wave that are in phase and an equal number of wavelengths from the source of the wave. A ray is a line extending outward from the source and representing the direction of propagation of the wave at any point along it. Rays are perpendicular to wave fronts.

Bihar PGT Physics Mock Test - 4 - Question 16

For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 16

Moment of inertia depends on distribution of mass around axis. The more the mass near the axis lesser is the moment of inertia. Four rods forming a square have more moment of inertia because of less mass near axis.

Bihar PGT Physics Mock Test - 4 - Question 17

When a ball is thrown upwards, as it rises, the vertical component of its velocity

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 17

Explanation:if a ball is thrown upward at an angle? The ball has a vertical and a horizontal component of velocity. The force of gravity is acting on it and its acceleration is in the downward direction. The vertical component of velocity is therefore changing. As the motion and acceleration is in opposite direction so vertical component of velocity decrease.

Bihar PGT Physics Mock Test - 4 - Question 18

Under the influence of the coulomb field of charge +Q, a charge −q is moving around it in an elliptical orbit. Find out the correct statement(s). 

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 18

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

Bihar PGT Physics Mock Test - 4 - Question 19

Joule is the SI unit of:

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 19

The joule (symbol J), is a derived unit of energy in the International System of Units.

It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of its motion through a distance of one metre (1 newton metre or N⋅m).

Bihar PGT Physics Mock Test - 4 - Question 20

The main reason for preferring usage of AC voltage over DC voltage

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 20

Explanation:By using the phenomenon of mutual induction, transformers allow us to easily change voltage of AC. This is necessary to cut down poer losses while supplying electricity to our homes

Bihar PGT Physics Mock Test - 4 - Question 21

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is:      

                      [AIEEE 2011]

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 21

The maximum range of water coming out of the fountain,
Rm = v2/g
Total area around fountain,

Bihar PGT Physics Mock Test - 4 - Question 22

Comparing X-rays and Gamma rays

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 22

Gamma-ray photons have the highest energy in the EMR spectrum and their waves have the shortest wavelength. Scientists measure the energy of photons in electron volts (eV). X-ray photons have energies in the range 100 eV to 100,000 eV (or 100 keV). Gamma-ray photons generally have energies greater than 100 keV.

Bihar PGT Physics Mock Test - 4 - Question 23

What is the number of electric field lines coming out from a 1C charge?

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 23

According to Gauss’s Law, the total number of electric field lines coming out of a charge q is =where εo is the absolute permittivity of air. Its value is 8.85 * 10-12. Therefore the number of lines coming out from a 1C charge = 1/8.85 * 10-12.

Bihar PGT Physics Mock Test - 4 - Question 24

A circular loop of radius R, carrying current I, lies in x-y plane with its centre at origin. The total magnetic flux through x-y plane is 

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 24

The magnetic lines of force created due to current will be in such a way that on x – y plane these lines will be perpendicular. Further, these lines will be in circular loops. The number of lines moving downwards in x – y plane will be same in number to that coming upwards of the x – y plane. Therefore, the net flux will be zero.
One such magnetic line is shown in the figure.

Bihar PGT Physics Mock Test - 4 - Question 25

A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 25

Work required to turn the dipole is given by
W=MB[cosθi- cosθf]
Here θi is the initial angle made by a dipole with a magnetic field and   is the final angle made by a dipole with a magnetic field.
magnetic moment is normal to the field direction.
so, θi=0 and θf=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J
 
magnetic moment is opposite to the field direction.
so, θi=0 and θf=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

Bihar PGT Physics Mock Test - 4 - Question 26

A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 26

In air, all the colours of light travel with the same velocity with the same velocity, but in glass, velocities of different colours are different. Velocity of red colour is largest and velocity of violet colour is smallest. Therefore, after travelling through the glass slab, red colour will emerge first.

Bihar PGT Physics Mock Test - 4 - Question 27

The multiplication of 10.610 with 0.210 upto correct number of significant figure is:

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 27

The multiplication of the numbers:
10.610 × 0.210 = 2.2281

Since, the numbers 10.610 and 0.210 have 5 and 3 significant digits, respectively. So, the final result must also have 3 significant digits.

The final answer is 2.23 after rounding off 2.2281. 

Bihar PGT Physics Mock Test - 4 - Question 28

For a rectangular slab, refraction takes place at

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 28

The refraction takes place at both the air-glass interface and glass-air interface of a rectangular glass slab. When the light ray incident on the air-glass interface (DC) obliquely, it bends towards the normal.

*Multiple options can be correct
Bihar PGT Physics Mock Test - 4 - Question 29

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 29

There is no current inside the pipe. Therefore

Bihar PGT Physics Mock Test - 4 - Question 30

A particle of mass m moving with constant velocity v strikes another particle of same mass m but moving with the same velocity v in opposite direction stick together. The joint velocity after collision will be

Detailed Solution for Bihar PGT Physics Mock Test - 4 - Question 30

Concept:

  1. Momentum: momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
  2. The unit of momentum (P) is kg m/s.
  3. Dimension: [MLT-1]
  4. Law of conservation of Momentum: A conservation law stating that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
  5. P= P2
  6. m1 v1 = m2 v2
  7. Where, P1 = initial momentum of system, P2 = final momentum of system, m1 = mass of first object, v1 = velocity of first object, m= mass of second object and v2 = velocity of second object.

Calculation:
Given:  m1 = m kg,  m2 = m kg,  u= v m/s,  u2 ​=  -v m/s

Let the common velocity of the combined body be V m/s

Mass of combined body      M = m + m = 2m

Applying conservation of momentum:          

mv1 + m2 v2 = M V

mv + (-mv) = 2mV

0 = 2mV

V = 0 m/s
Hence the correct answer will be zero (0) m/s.

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