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Bihar PGT Physics Mock Test - 8 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test - Bihar PGT Physics Mock Test - 8

Bihar PGT Physics Mock Test - 8 for Bihar PGT/TGT/PRT 2024 is part of Bihar PGT/TGT/PRT preparation. The Bihar PGT Physics Mock Test - 8 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar PGT Physics Mock Test - 8 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 - 8 below.
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Bihar PGT Physics Mock Test - 8 - Question 1

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

Shock

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 1

Shock is to surprise someone in an unexpected or unwanted way. It represents chaos whereas calm means to stay composed and stay cool.
Thus option 2 is the correct option.
Example- After a cyclone, there is always a period of calm that follows.

Bihar PGT Physics Mock Test - 8 - Question 2

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

Filter

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 2

Filter is to purify something whereas pollute means to make the things dirty. Thus, pollute is the best antonym to filter.
Example- The smoke from old diesel vehicles pollute the air the most.

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Bihar PGT Physics Mock Test - 8 - Question 3

'अखरावट' के रचयिता हैं:

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 3

मलिक मुहम्मद जायसी हिन्दी साहित्य के भक्ति काल की निर्गुण प्रेमाश्रयी धारा के कवि थे। वे अत्यंत उच्चकोटि के सरल और उदार सूफ़ी महात्मा थे। जायसी मलिक वंश के थे। उनकी प्रमुख कृतियों में पद्मावत, अखरावट, आख़िरी कलाम, कहरनामा, चित्ररेखा आदि प्रमुख हैं।

Bihar PGT Physics Mock Test - 8 - Question 4

'पदलालित्य' के विषय में कौन प्रसिद्ध है?

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

संस्कृत सहित्य में प्रत्येक कवियों को उनकी रचनाओं में पाये जाने वाली विशेषताओं के अनुसार कुछ पङ्क्तियों में गुणगान किया गया है-

विद्वानों में मान्यता है कि दण्डी कृत् ‘दशकुमारचरित’ आदि काव्यों में नित्य नूतन शब्दों का प्रयोग हुआ है इसलिए 'दण्डिनः पदलालित्यं' कहा जाता है।

अतः स्पष्ट है कि 'पदलालित्य' के विषय में महाकवि दण्डी प्रसिद्ध है।

Bihar PGT Physics Mock Test - 8 - Question 5

‘लाल किले पर तिरंगा फहराया गया।’ इसके रेखांकित शब्द में कौन-सा समास होगा? 

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 5

दिए गए विकल्पों में सही उत्तर विकल्प A ‘द्विगु समास’ होगा। 

वह समास जिसका पहला पद संख्यावाचक विशेषण होता है तथा समस्तपद किसी समूह या फिर किसी समाहार का बोध करता है तो वह द्विगु समास कहलाता है।

तिरंगा : तीन रंगों का समूह

Bihar PGT Physics Mock Test - 8 - Question 6

'जय-पराजय' में कौन-सा समास है?

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 6

जिस समास में दोनों पद अथवा सभी पदों की प्रधानता होती है। जैसे - 

जय-पराजय, जीत-हार आदि।

Bihar PGT Physics Mock Test - 8 - Question 7

सन्देश रासक किस कवि की रचना है?

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

"संदेश रासक", "अब्दुल रहमान" की रचना है।

  • 223 छंदों का छोटा सा विरह काव्य है।
  • अब्दुल रहमान का संदेश रासक पहला धर्मेतर रास  ग्रंथ है।
  • रचना 1000-1100 ई के आसपास है।
  • यह ग्रन्थ उस अपभ्रंश में है जिससे लहन्दा, पंजाबी और सिन्धी आदि पश्चिमी भारतीय भाषाएँ जन्मी हैं।
  • इस ग्रंथ की पाण्डुलिपियाँ मुनि जिनविजय ने जैन पुस्तकालयों से प्राप्त की थी।
  • यह काव्य कालिदास द्वरा रचित मेघदूतम् से प्रेरित है।
  • अपभ्रंश में किसी मुसलमान द्वारा रचित यह प्रथम और एकमात्र ग्रंथ है।
  • इसकी हस्तलिखित प्रति पाटण के जैन भंडार में मिली है।

Bihar PGT Physics Mock Test - 8 - Question 8

'भाषा ठीक करने से पहले मैं मनुष्यों को ठीक करना चाहता हूँ, समझे'।

प्रस्तुत संवाद 'चन्द्रगुप्त' नाटक के किस पात्र का है?

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

यह संवाद चन्द्रगुप्त नाटक में चाणक्य के द्वारा कहा गया है। 

नाटक कार - जयशंकर प्रसाद 
प्रकाशन वर्ष - 1931
प्रमुख पात्र - चन्द्रगुप्त, चाणक्य, शकटार, आंभीक, सेल्यूकस, कार्नेलिया, कल्याणी, मालविका आदि 

Bihar PGT Physics Mock Test - 8 - Question 9

प्रश्न के शब्द-युग्म के सही अर्थ-भेद का चयन कीजिए।

अपेक्षा-उपेक्षा

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 9

‘अपेक्षा’ का अर्थ है ‘आशा’ तथा ‘उपेक्षा’ का अर्थ है ‘तिरस्कार’। इस आधार पर शब्द युग्म 'अपेक्षा-उपेक्षा' का अर्थ है 'आशा-तिरस्कार'। अतः सही विकल्प आशा-तिरस्कार है।

Bihar PGT Physics Mock Test - 8 - Question 10

किस पुस्तक का अंत एलोरा की गुफाओ में इतिहास खोजने की कोशिश से हुआ, पंक्तियाँ है-

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 10

"अरे यायावर याद रहेगा" का अंत एलोरा की गुफाओं में इतिहास खोजने की कोशिश से हुआ। 

रचनाकार: अज्ञेय जी
रचना वर्ष: 1953
विधा: यात्रा साहित्य
"एक बूंद सहसा उछली" अज्ञेय जी का अन्य यात्रा वृतांत है।

Bihar PGT Physics Mock Test - 8 - Question 11

'उच्चारण' शब्द का संधि विग्रह है-

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 11

‘उच्चारण’ का संधि विच्छेद करने पर ‘उत्+चारण’ होगा तथा यहाँ व्यंजन संधि है।

व्यंजन संधि में यदि त् + च या छ आये तो त् का च् हो जाएगा। जैसे– उत् + चारण = उच्चारण, महत् + छत्र = महच्छत्र।

स्वर संधि-स्वर वर्ण के साथ स्वर वर्ण के मेल से विकार उत्पन्न होता है, जैसे– विद्या + अर्थी = विद्यार्थी, महा + ईश = महेश।

व्यंजन संधि-एक व्यंजन से दूसरे व्यंजन या स्वर के मेल से विकार उत्पन्न होता है, जैसे - अहम् + कार = अहंकार, उत् + लास = उल्लास।

विसर्ग संधि-विसर्ग के साथ स्वर या व्यंजन के मेल से विकार उत्पन्न होता है, जैसे– दुः + आत्मा =दुरात्मा, निः + कपट =निष्कपट।

Bihar PGT Physics Mock Test - 8 - Question 12

'खुद से कहता हूँ' वाक्य में किस कारक का प्रयोग हुआ है?

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 12

भाववाचक संजा से क्रिया विशेषण बनाते समय करण कारक का प्रयोग होता है, जैसे -

नम्रता से बात करो।

खुद से कहता हूँ।

Bihar PGT Physics Mock Test - 8 - Question 13

What is the molecular formula of Butane?

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 13

The correct answer is option 2 i.e C4H10.

Concept:

Hydrocarbons

  • Compounds made of carbon and hydrogen atoms only are called hydrocarbons.
  • Saturated hydrocarbons:
    • The hydrocarbons in which carbon atoms are singly bonded are called saturated hydrocarbons.
    • Saturated hydrocarbons are also called alkanes or paraffin.
      • The general formula of alkane– CnH2n+2.
  • Unsaturated hydrocarbons:
    • The hydrocarbons in which carbon atoms are either doubly or triply bonded are called unsaturated hydrocarbons.
    • Doubly bonded (carbon atoms) hydrocarbons are called alkenes.
    • The general formula of alkene is CnH2n.
  • The formula for alkane should be mentioned CnH2n+2

Explanation:


Bihar PGT Physics Mock Test - 8 - Question 14
The Pyrimidine bases in a DNA are :
Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 14

The correct answer is Thymine and Cytosine.

Key Points

  • The pyrimidines in DNA are C & T. In RNA, U replaces T.
  • Cytosine (C) formula is C4H5N3O and Thymine (T) formula is C5H6N2O2.

Additional Information

  • Deoxyribonucleic acid, commonly known as DNA, is the hereditary complex molecule present in humans and almost all other organisms.
  • The information in DNA (double helix structure) is stored as a code made up of four chemical bases: adenine (A), guanine (G), cytosine (C), and thymine (T).
    • DNA has two base pairs, A-T and G-C, attached to a sugar-phosphate backbone.
  • DNA is a long molecule that contains our unique genetic code.
    • It holds the instructions for making all the proteins in our bodies and the function of living things.
  • In the cell nucleus, the DNA (deoxyribonucleic acid) molecule is packaged into thread-like structures called chromosomes.
    • A chromosome (microscopic structure) is made up of DNA tightly coiled many times around proteins called histones that support its structure.
    • A chromosome has two arms - a p arm (short arm) and a q arm (long arm), joined in the location of the centromere.

Important Points

  • Watson and Crick Proposed the Double Helix structure of DNA with the two strands connected by hydrogen bonds.
Bihar PGT Physics Mock Test - 8 - Question 15

From the given alternatives, select the word which CANNOT be formed using the letters of the given word.

FEARLESS

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 15

1) GRASS – FEARLESS (Cannot be formed because G is missing)

2) RESEAL – FEARLESS (Can be formed)

3) LESSER – FEARLESS (Can be formed)

4) ERASE – FEARLESS (Can be formed)

Hence, the correct answer is “GRASS”.

Bihar PGT Physics Mock Test - 8 - Question 16

When the interference occurs with coherent sources, _______ of the superimposingwaves add like _______.

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 16

The answer provided in the forum is wrong, The correct answer would be option A.
Solution:
When the interference occurs with non-coherent sources, intensities of the superimposing waves add like scalars, I = I1 + I2

Bihar PGT Physics Mock Test - 8 - Question 17

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s about an axis passing through the centre point O. Other end of the metallic rod slides on a Metallic ring . A constant and uniform magnetic field of 2 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 17

l = 1 m

= 314V

Bihar PGT Physics Mock Test - 8 - Question 18

Three identical cells, each of 2V and internal resistance of 0.2 ohm are connected in series to an external resistance of 7.4 ohm. The current in the circuit will be​

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 18

From the Question,

  • Emf (E) = 2 V

  • Internal Resistance (r) = 0.2 Ω

  • Resistance (R) = 7.4 Ω

Three cells of 2V and 0.2 each are connected in series.

Consider two cells connected to each other.

The total potential difference would be :

Total Potential Difference in a cell is given as :

V=E-IR

Thus,

⇒E-Ir=E1-Ir1+E2+Ir2

⇒Eeq=E1+E2 and req=r1+r2

Therefore,

For three cells each of emf 2V,

E=2+2+2=6V

For three cells with internal resistance 0.2 Ω,

R=.02+.02+.02=0.6 Ω

We know that,

I=E/r+R

Implies,

 I=6/0.6+7.4

I=6/8

 I=0.75A

Bihar PGT Physics Mock Test - 8 - Question 19

The First Law of Thermodynamics states that:

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 19

The first law of thermodynamics states that the total energy of an isolated system is constant. Energy can be transformed from one form to another, but can neither be created nor destroyed.

According to this law, some of the heat given to system is used to change the internal energy while the rest in doing work by the system. Mathematically,

ΔQ=ΔU+ΔW

where,

ΔQ =  Heat supplied to the system

ΔW= Work done by the system.

ΔU = Change in the internal energy of the system.

If Q is positive, then there is a net heat transfer into the system, if  W is positive, then there is work done by the system. So positive Q adds energy to the system and positive W takes energy from the system.

It can also be represented as  ΔU=ΔQ-ΔW

We can say that internal energy tends to increase when heat is given to the system and vice versa.

Bihar PGT Physics Mock Test - 8 - Question 20

The materials which have the same elastic properties in all directions are called __________.

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 20

Same elastic properties in all direction is called the homogenity of a material.

Bihar PGT Physics Mock Test - 8 - Question 21

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 21

We know that P=V2/R

For a given potential difference at a particular temperature

It is given that the powers of the bulbs are in the order 

100W > 60 W > 40W

Bihar PGT Physics Mock Test - 8 - Question 22

The only component that dissipates energy in ac circuit is:

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 22

The only component that dissipates energy in ac circuit is the resistor because  Pure Inductive and pure capacitive circuits have no power loss.

Bihar PGT Physics Mock Test - 8 - Question 23

The reason why cyclists bank when taking a sharp turn is

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 23

Explanation:In order to take a safe turn, the cyclist has to bend a little from his vertical position. In this case, a component of the reaction provides the required centripetal force.If q is angle made by the cyclist with the vertical then

In actual practice, the value of q is slightly less because the force of friction also contributes towards the centripetal force.

Bihar PGT Physics Mock Test - 8 - Question 24

A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is: (Take b = 0.25 m-3/2 s-1)

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 24

From the work energy theorem,
Work done by net force = ΔK.E.

w = 16 J

Bihar PGT Physics Mock Test - 8 - Question 25

The temperature range upto which Newton’s law of cooling holds good

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 25

Newton’s law of cooling is to be used at temperatures around room temperature.

Bihar PGT Physics Mock Test - 8 - Question 26

If a machine is lubricated with oil

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 26

When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.

Bihar PGT Physics Mock Test - 8 - Question 27

The acceleration at any instant is the slope of the tangent of the ________ curve at that instant:

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 27

This can be verified graphically

ObtaIning instantaneous acceleration from graph

Bihar PGT Physics Mock Test - 8 - Question 28

An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is at a distance u > f from the mirror. Its image will have a length

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 28

At infinity, image of B will form at focus since it is at infinity. Image of A will be at A which can be calculated by mirror formula.
1/v​+(1/−u)​=(1/−f)
⟹v= fu /(f−u)​=−( fu/(u−f)​)
Image length = v−f=(fu/(u−f)​)−(f)

= f2​/(u−f)
(We take the absolute values of the distances to calculate the rod length)

Bihar PGT Physics Mock Test - 8 - Question 29

A body moves along a circular path of radius 5 m. The coefficient of friction between the surface of the path and the body is 0.5. The angular velocity in rad/s with which the body should move so that it does not leave the path is (g = 10 m/s2)  

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 29

Bihar PGT Physics Mock Test - 8 - Question 30

A (nonrotating) star collapses onto itself from an initial radius Ri with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration ag on the surface of the star as a function of the radius of the star during the collapse ?

                       

Detailed Solution for Bihar PGT Physics Mock Test - 8 - Question 30

g∝1​/R2
Curve b represents this variation.
 

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