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Bihar PGT Physics Mock Test - 7 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test Bihar PGT Exam Mock Test Series 2024 - Bihar PGT Physics Mock Test - 7

Bihar PGT Physics Mock Test - 7 for Bihar PGT/TGT/PRT 2024 is part of Bihar PGT Exam Mock Test Series 2024 preparation. The Bihar PGT Physics Mock Test - 7 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar PGT Physics Mock Test - 7 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 - 7 below.
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Bihar PGT Physics Mock Test - 7 - Question 1

In the following question, out of the four alternatives, select the word similar in meaning to the word given.

Daze

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 1

Daze means a situation that is slow and causes a sort of numb. Thus, trance is the correct synonym for daze. 
Example- After the accident, Mohan went into a state of trance.

Bihar PGT Physics Mock Test - 7 - Question 2

The following sentence has been broken into four parts with an error in one part. Identify that part and mark it as your answer. If there are no errors in any of the given parts, mark option 4 or ‘No error’ as your answer.

Kajal was sitting within (1)/ between Prachi and Isha (2)/ at the party that night. (3)/ No error (4). 

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 2

The word 'between' is used when two things or persons are being compared or depicted.
Example-I had to select between the two destinations for the vacation.
Thus, the statement should read as 'Kajal was sitting between Prachi and Isha at the party that night.'

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Bihar PGT Physics Mock Test - 7 - Question 3

'कविवचन सुधा' के सम्पादक कौन थे?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 3

'कविवचन सुधा' पत्रिका के संपादक "भारतेन्दु हरिश्चंद्र' थे। काशी से प्रकाशित 'कविवचन सुधा' पत्रिका का आरंभ 1868 में हुआ था। इसके प्रकाशक भारतेन्दु हरिश्चंद्र थे।

Bihar PGT Physics Mock Test - 7 - Question 4

रचनाकाल के आधार पर निम्नलिखित रचनाओं का सही अनुक्रम है:

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

रचनाकाल के आधार पर निम्नलिखित रचनाओं का सही अनुक्रम है: चंदायन (1379 ई.), मृगावती ( 1501 ई. ), मधुमालती ( 1545 ई. ), चित्रावली ( 1613 ई.) चंदायन, मृगावती, मधुमालती और चित्रावली हिंदी के प्रमुख सूफ़ी काव्य हैं। भारत में सूफ़ी धर्म का प्रचार - प्रसार 12 वीं शताब्दी में चिश्ती ने किया था।

Bihar PGT Physics Mock Test - 7 - Question 5
The manure also considered as bio-fertilizer is:
Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 5
  • Is a substance that contains living microorganisms.
  • These, when applied to seeds, plant surfaces, or soil, colonize the rhizosphere or the interior of the plant and promotes growth by increasing the supply or availability of primary nutrients to the host plant. 
  • Biofertilizers add nutrients through the natural processes of Nitrogen fixation.
  • Examples: Rhizobium, Azotobacter, Blue-green algae, Azospirilium.

Key PointsBlue green algae:

  • Blue-green algae (BGA) is a photoautotrophic, prokaryotic algae.
  • They are free-living creatures and also known as Cyanobacteria.
  • It fixes the atmospheric nitrogen in moist soils.
  • So BGA has been recommended as a biofertilizer.
  • The algae include unicellular as well as filamentous species.
  • Some of the filamentous forms have specialized cells known as heterocysts.

Important Points

Biofertilizers:

Thus, the manure also considered as bio-fertilizer is Blue green algae.

Additional Information

Compost: 

  • Composting is a process in which both aerobic and anaerobic micro-organisms decompose organic matter.
  • Farm compost is a mass of rotted organic matter made from farm waste like sugarcane trash, paddy straw, weeds and other plants.
  • While town compost is a mass of rotted organic matter made from town refuses like night soil, street sweepings and dustbin refuse.

​Green Manure:

  • Green manures are the crops that are grown primarily for building and maintaining soil fertility and structure.
  • Green manure crops should be such that they can be grown quickly.
  • Green manure crops include legumes such as vetch, clover, beans and peas; grasses such as annual ryegrass, oats, rapeseed, winter wheat and winter rye and buckwheat.
  • Green manure crops provide nitrogen as the major nutrient.

Farmyard Manure:

  • It is the decomposed mixture of dung and urine of farm animals.
  • It also contains litter and leftover materials from roughages or fodder fed to cattle.
Bihar PGT Physics Mock Test - 7 - Question 6
Each water molecule, contains _______ hydrogen atoms and one Oxygen atom.
Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 6

The Correct Answer is Two.

 Key Points

  • The water molecule is composed of two hydrogen atoms, each linked by a single chemical bond to an oxygen atom.
  • Most hydrogen atoms have a nucleus consisting solely of a proton. 
  • Water is a tasteless and odorless liquid at room temperature.
  • It has the important ability to dissolve much other substance.

 Additional Information

  • Two isotopic forms, deuterium and tritium, in which the atomic nuclei also contain one and two neutrons, respectively, are found to a small degree in water.
  • Deuterium oxide (D2O) is called heavy water
  • It is used as a neutron moderator in some nuclear reactors.
Bihar PGT Physics Mock Test - 7 - Question 7

Manav born on 03 December 1999, is in final year of Engineering of a Regular Course, with an aggregate of 75% till his 5th semester, he has permanent body tattoo on his inner face of forearm.

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

Hence, the person does not satisfy all the conditions.

Bihar PGT Physics Mock Test - 7 - Question 8

What will be the next figure in the series given below?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 8

The logic followed is:

In each next step 1 line segment is deleting from the given letter.

In figure A

The given shape is the letter "E".

In figure B

One line segment get deleted from the right end

Thus we get 

In figure C

Again the one-line segment gets deleted from the right end.

Thus we get 

Similarly,

In figure D

Again one line segment will get deleted from the previous image to obtained figure D.

Thus only one figure is satisfying the pattern.

Hence, the correct answer is "Option 2".

Bihar PGT Physics Mock Test - 7 - Question 9

Which place in India has the highest rainfall?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 9

Mawsynram, Meghalaya:

  • Mawsynram place in India has the highest average rainfall recorded.
  • Mawsynram has a subtropical highland climate.
  • Mawsynram receives nearly 10,000 millimetres of rain in an average year.
  • Mawsynram is located in the East Khasi Hills district of Meghalaya.
  • It is on the windward side of the hills and is surrounded by hills on three sides.
  • It receives rain from southwest monsoon winds (Bay of Bengal branch), as the winds cannot move further striking the hills and due to condensation receives heavy rainfall.

Thus, Mawsynram in India has the highest rainfall.

Important Point

  • The average rainfall in Bengaluru is 776.6 mm
  • The average rainfall in Mysore is 500 mm.
  • The average rainfall in Srinagar is 720 mm.

Additional Information

Bihar PGT Physics Mock Test - 7 - Question 10

Select the Venn diagram that best represents the relationship between the following classes.

Crockery, Plate, Bowl

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 10

The Venn diagram that best represents the relationship between Crockery, Plate, and Bowl is shown below:

Bowl and Plate are name of Crockery items.

Hence, ‘option 3’ is the correct answer.

Bihar PGT Physics Mock Test - 7 - Question 11
Which of the following is NOT a root?
Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 11

The correct answer is Potato.

Modifications of Root

  1. For storage of food
    • Roots are modified in some plants for storing reserve food materials.
    • These modified roots usually are swollen and assume different forms such as spindle-shaped, e.g., radish; top-shaped, e.g., beet, turnip; cone-like, e.g., carrot; indefinite shape, e.g., sweet potatoes.
    • Dahlia, Asparagus, Portulaca are some other examples of plants with modified roots for food storage.
  2. Nodulated roots
    • The roots of pea and other leguminous plants have numerous swollen nodules on fine branches of roots.
    • These nodules are formed due to the symbiotic association of Rhizobium (bacterium) that live inside the root cortical cells of the roots.
    • They fix nitrogen and an active nodule is pink in colour.
  3. For mechanical support
    • Roots are modified to provide mechanical support as seen in a banyan tree which has roots growing vertically/obliquely downwards (prop roots); sugarcane/maize in which roots arise from the nodes in the cluster at the base of the stem (stilt roots) and betel/black pepper in which nodes and internodes bear roots which help in climbing.
  4. For gaseous exchange
    • Pneumatophores or breathing roots are found in plants growing in mangroves or swamps with saline water for exchange of gases.
    • They are erect peg-like structures with numerous pores through which air circulates e.g., Rhizophora mangle.
  • Modifications of Stem
  1. For storage of food
    • Stems get modified into underground structures for storage of food as seen in potato (tuber), ginger (rhizome), garlic (bulb), yam (corm).
    • Presence of an eye (node) in potato, distinct nodes with internodes and scaly leaves in ginger/yam, a cluster of roots at the base of the reduced stem in garlic/ onion, all indicate that these underground plant parts are modified stem.
  2. For vegetative propagation
    • Plants besides reproducing sexually also propagate through vegetative parts.
    • For this purpose, stems may be modified into a runner (Cyanodon dactylon, Oxalis). Runners are slender prostrate branches arising from axillary buds;
    • stolon (e.g., mint, strawberry) which is a slender lateral branch arising from the base of the stem grows upward and then down to develop new daughter plants;
    • offset having a single long horizontal internode growing up to some distance and producing a tuft of leaves above and a cluster of roots below at the apex (Eichornia, Pistia) and
    • sucker, which arises from the underground part of the stem, grow obliquely and gives rise to a new shoot. (Chrysanthemum, Banana, Pineapple).
  3. For protection
    • Some modified stem provides protection as thorns which are hard, pointed structures each representing a branch that arises from the axil of a leaf.
    • Thorns are found in plants like Duranta, Pomegranate, Acacia, Ber, Prosopis, Bougainvillea, Citrus, etc.
  4. For support
    • Tendrils are modifications of the stem to provide support to plants, e.g., Vitis, passionflower, Bignonia etc.
  5. For photosynthesis
    • Stems are also modified into Phylloclade, to facilitate photosynthesis.
    • Phylloclades are flattened/cylindrical stem or branches of unlimited growth (Cactus).
Bihar PGT Physics Mock Test - 7 - Question 12

What roles do human play in a food chain?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 12

Concept:

  • Food chain:​ The sequence of living organisms in a community in which one organism consumes another organism to transfer food energy is called a food chain.
  • In simple words, A list of organisms showing 'who eats whom' is called the food chain.
  • The food can be transferred from one organism to the other through the food chain.
  • Each step in the food chain is represented by a trophic level.
  • Producers occupy the base of the food chain, producers are followed by Consumers → Secondary consumers → Tertiary consumers → Decomposers.

Explanation:

  • Humans occupy the position of consumers in a food chain.
  • The human can be divided into two types of consumers - Primary consumer & Secondary consumer.
  • Primary Consumer - When humans derive their nutrition directly from the producers (green plants) in the form of fruits & vegetables.
  • Secondary Consumer - When humans derive the nutrition indirectly by eating primary consumers like goat, cow & buffalo.
Bihar PGT Physics Mock Test - 7 - Question 13

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential is Volts is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 13

Given, the maximum kinetic energy: Kmax​=4eV
If V0​ be the stopping potential, then Kmax​=eV0​
⇒eV0​=4eV 
⇒V0​=4V

Bihar PGT Physics Mock Test - 7 - Question 14

An object is placed at a distance of 10 cm from a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30 cm, what is the nature and focal length of lens A ?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 14

Bihar PGT Physics Mock Test - 7 - Question 15

Frictional force acting on the block is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 15

Net force = 50 - 20 = 30 N
Limiting friction on the body = u (static) × m × g = 0.6 × 10 × 10 = 60 N.
F = 30 N is less than the limiting friction so the body is static. So, a = 0.
Force of friction acting on the body is static friction, f = driving force = 30 N.

Bihar PGT Physics Mock Test - 7 - Question 16

Two identical thin uniform rods of length L each are joined to form T shape as shown in the figure. The distance of centre of mass from D is  

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 16


On solving we get 3L/4
 

Bihar PGT Physics Mock Test - 7 - Question 17

A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 17

Due to verifying electric field, it experiences an verifying force :-
F=QE=QE0​sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax​=QE0​
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
2A=QE0​
⇒A= QE0/mω2

Bihar PGT Physics Mock Test - 7 - Question 18

In Ferromagnetic materials

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 18

A ferromagnetic substance contains permanent atomic magnetic dipoles that are spontaneously oriented parallel to one another even in the absence of an external field. The magnetic repulsion between two dipoles aligned side by side with their moments in the same direction makes it difficult to understand the phenomenon of ferromagnetism. It is known that within a ferromagnetic material, there is a spontaneous alignment of atoms in large clusters. A new type of interaction, a quantum mechanical effect known as the exchange interaction, is involved. A highly simplified description of how the exchange interaction aligns electrons in ferromagnetic materials is given here.

Bihar PGT Physics Mock Test - 7 - Question 19

On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 19

Potential difference between two points is given by

Va - Vb = W/q0

Work, W = 2 J

Charge, q0 = 20 C

Potential difference = 2/20 = 0.1 V

The correct option is C.

Bihar PGT Physics Mock Test - 7 - Question 20

Which of the given relation is true for Newton’s law of cooling?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 20

Newton's law of cooling states that the heat released by a body with respect to time (or) the rate of heat released is directly proportional to the difference between the body's temperature and the surrounding temperature. 
dH/dt = k(T – Ts) where t = surrounding's temperature and T = temperature of the body 
Consider two bodies A and B, of equal surface areas, such that A's temperature is more that B's temperature and the surrounding temperature is less than both A and B. Then according to Newton's law of cooling A loses more heat to the surroundings when compared B during the same time interval. So, A will cool faster than B.

Bihar PGT Physics Mock Test - 7 - Question 21

A battle ship simultaneously fires two shells at enemy ships. Both are fired with the same speed but with different directions as shown. If the shells follow the parabolic trajectories shown, which ship gets hit first?

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 21

Ship 2 gets hit first. The time of flight only depends on the y-component of motion, not the x-component.  The higher you throw something up in the air, the more time it spends in the air.  It can also be shown from equation-

y=v0y t – ½ g t2=0  

    v0y for projectile μ1 is greater than for μ2.

Bihar PGT Physics Mock Test - 7 - 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 Bihar PGT Physics Mock Test - 7 - 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

Bihar PGT Physics Mock Test - 7 - Question 23

The mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference is known as

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 23

The mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference is known as conduction. This is the definition of conduction.

Bihar PGT Physics Mock Test - 7 - Question 24

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, change of flux linkage with the other coil is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 24

Bihar PGT Physics Mock Test - 7 - Question 25

In general the term diffraction is used

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 25

The process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the waveforms produced.

Bihar PGT Physics Mock Test - 7 - Question 26

The potential energy of a body at height h is mgh. Then its kinetic energy just before hitting the ground is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 26

From the law of conservation of mechanical energy, the potential energy of a body at height h is converted into kinetic energy when it falls down. So, the magnitude of kinetic energy just before hitting the ground is equal to the potential energy at height h which is mgh.

Bihar PGT Physics Mock Test - 7 - Question 27

What is the relationship between Em and E0

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 27

peak value Em=2E0
so 2/π value is 0.637
therefore,
Em=-0.637 E0

Bihar PGT Physics Mock Test - 7 - Question 28

A hollow spherical shell is compressed to half its radius. The gravitational potential at the centre

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 28

Gravitational Potential V = -GM/R for hollow spherical shell at the centre. If we replace R by R/2 then we get V = -2GM/R. Therefore it decreases.

Bihar PGT Physics Mock Test - 7 - Question 29

The pressure inside a soap bubble of radius R and surface tension S is

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 29

If R is the radius of a soap bubble and S its surface tension, then the excess pressure inside is 4S/R.

Bihar PGT Physics Mock Test - 7 - Question 30

The equation for two waves obtained by two light sources are as given below :

y1= A1 sin 3wt, y2 = A2 cos (3wt + p/6). What will be the value of phase difference at the time t _

Detailed Solution for Bihar PGT Physics Mock Test - 7 - Question 30

y1=Asin3 ωt
y2=A2cos(3ωt+p/6)
y2=A2cosd2
y1=A sin(3ωt+p/2- p/2)
   =A sin(p/2+3ωt- p/2)
y1=A cos(3ωt- p/2)  [sin(p/2+θ)=Cosθ]
y1=A1cosd1
d2-d1=(3ωt+p/6)- (3ωt- p/2)
=p/6+p/2=2p/3

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