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EMRS PGT Physics Mock Test - 4 - EMRS MCQ


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30 Questions MCQ Test EMRS PGT Mock Test Series 2024 - EMRS PGT Physics Mock Test - 4

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

Among the following, the one with the highest mass is:

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

Concept:

  • The laws dealing with ideal gases naturally called ideal gas laws, and the laws determined in the seventeenth century by Boyle's observational work and in the eighteenth century by Charles.
  • The Boyles Law states that the gas pressure is inversely proportional to the gas volume for a given mass of gas stored at a constant temperature.
  • Charles Law states that the gas volume is directly proportional to the gas temperature for a given fixed mass of gas stored at a constant pressure. 
  • An ideal gas is a gas that follows the equation:

PV = nRT at all temperatures and pressure, where P = Pressure of the ideal gas. V = Volume of the ideal gas n = Amount of ideal gas measured in terms of moles 

R = Universal gas constant T = Temperature

Mole Concept -

  •  The quantity one mole of a substance signifies 6.022 × 1023 number of particles of that substance which may be atoms, molecules, or ions.
  • The quantity is a universal constant like Dozen, Gross, etc., and is known as Avogadro number, denoted by NA. after the scientist Amedeo Avogadro.
  • Examples- In one mole of H2, there are 6.022 × 1023 molecules of hydrogen, and the number of atoms is 2 × 6.022 × 1023, as one molecule of hydrogen contains two-atom each.
  • The mass of one mole of a substance is called its Molar Mass (M) or Atomic mass expressed in grams.
  • The volume occupied by a mole of gas is 22.4 L at NTP, called its Molar Volume.
  • The no. of moles (n) is calculated as =

The number of particles / Avogadro’s number.

To summarise, we can say, 

Calculation:

  • The molar mass of 1 mole of H2SO4 is =

2 × 1 + 32 + 4 × 16 = 98g

  • The molar mass of 1 mole of Silver, Ag is its molecular mass 108g.
  • The molar mass of CO2 is = 12 + 2× 16 = 44g.
  • 44g of CO2 means one mole.
  • 1 mole of Oxygen weighs 16 + 16 = 32g.
  • Hence, the highest mass is of 1 mole of Ag.
EMRS PGT Physics Mock Test - 4 - Question 2

If then, find the value of k.

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

⇒ 0.0023 + 0.024 + 0.025 + 0.26 × √k = 15.6513

⇒ 0.0513 + 0.26 × √k = 15.6513

⇒ 0.26 × √k = 15.6513 – 0.0513

⇒ 0.26 × √k = 15.6

⇒ √k = 15.6/0.26

⇒ √k = 60

⇒ k = (60)2

⇒ k = 3600
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EMRS PGT Physics Mock Test - 4 - Question 3

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

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

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. 
EMRS PGT Physics Mock Test - 4 - Question 4

Direction: Read the following information carefully and answer the questions that follow.
A blacksmith has five iron articles A, B, C, D and E each having a different weight.
I. A weight is twice as much as of B.
II. B weight is four and half times as much as of C.
III. C weight is half times as much as of D.
IV. D weight is half as much as of E.
V. E weight is less than A but more than C.

Q. Which of the following is the heaviest in weight?

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

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C

So A is the heaviest in the weight.

EMRS PGT Physics Mock Test - 4 - Question 5

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

Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 5

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.

EMRS PGT Physics Mock Test - 4 - Question 6

Which of these converts a high level programming language to machine code?

EMRS PGT Physics Mock Test - 4 - Question 7

What type of software is MS-Word?

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

Microsoft Word or MS-WORD (often called Word) is a Graphical word processing program that users can type with. It is made by the computer company Microsoft. Its purpose is to allow users to type and save documents. It's an application software

EMRS PGT Physics Mock Test - 4 - Question 8

 What will be the position of the cursor when the HOME command is given?

EMRS PGT Physics Mock Test - 4 - Question 9

 What is the most common way to discover a media type?

Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 9

A file extension tells its type. Multimedia files commonly have .png, .wav, .mp3, .mpeg extensions.

EMRS PGT Physics Mock Test - 4 - Question 10

The modern consensus regarding the sending of disciplinary cases to the principal's office is that

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

School principal fulfils many roles and dealing with students' disruptive behaviour is among the least preferred by them. So, the teacher must try to discipline the students using various means like assigning responsibility to the undisciplined, finding out the underlying cause of his behaviour, and so on. Only after she has done everything without any effect on the student should she send the student to the principal.

EMRS PGT Physics Mock Test - 4 - Question 11

The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be

Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 11

Field intensity is proportional to the inverse of the square of the distance separating the point charges:

F = k*q1*q2/d^2

Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:

K = k*q1*q2, so

F = K/d^2 (and by algebraic manipulation we have K=F*d^2).

Therefore we know that

F(at d=4) = K/4^2, and

F(at d=2) = K/2^2.

Since K is a constant, we can equate K=F*d^2 for each case:

K=F(at d=4)*4^2 = F(at d=2)*2^2

so

F(at d=2) = (F(at d=4)*4^2)/2^2

F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.

EMRS PGT Physics Mock Test - 4 - Question 12

Electric field is steepest in the direction in which the potential

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

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of the electric field.

EMRS PGT Physics Mock Test - 4 - Question 13

If a steady current I is flowing through a cylindrical element ABC. Choose the correct relationship

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

 

EMRS PGT Physics Mock Test - 4 - Question 14

The perfect gas equation is PV = nRT where n is the

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

n is the number of moles.

EMRS PGT Physics Mock Test - 4 - Question 15

A particle is moving with velocity  where k is a constant. The general equation for its path is:                                                             

[AIEEE 2010]

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

As, v = k(yi + xi)
vx = Ky → dx/dt = Ky
v= Kx → dy/dt = Kx
Dividing both equations,
► dy/dx = dy/dt divided by dx/dt = Kx / Ky = x/y
► ydy = xdx
Integrating the equation,
► y2/ 2 = x2/2 + c
► y= x+ constant

EMRS PGT Physics Mock Test - 4 - Question 16

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

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

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

EMRS PGT Physics Mock Test - 4 - Question 17

Eddy currents have negative effects. Because they produce:

Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 17
  • When a conductive material is subjected to a time-varying magnetic flux, eddy currents are generated in the conductor.
  • These eddy currents circulate inside the conductor generating a magnetic field of opposite polarity as the applied magnetic field. The interaction of the two magnetic fields causes a force that resists the change in magnetic flux.
  • However, due to the internal resistance of the conductive material, the eddy currents will be dissipated into heat and the force will die out. As the eddy currents are dissipated, energy is removed from the system, thus producing a damping effect.
*Multiple options can be correct
EMRS PGT Physics Mock Test - 4 - Question 18

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

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

There is no current inside the pipe. Therefore

EMRS PGT Physics Mock Test - 4 - Question 19

Joule is the SI unit of:

Detailed Solution for EMRS 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).

EMRS PGT Physics Mock Test - 4 - Question 20

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

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

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.

EMRS PGT Physics Mock Test - 4 - Question 21

A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected ? 

[AIEEE 2004]

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

In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.
In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.

EMRS PGT Physics Mock Test - 4 - Question 22

A sports car has a “lateral acceleration” of 0.96g = 9.4 m s−2. This is the maximum centripetal acceleration the car can sustain without skidding out of a curved path. If the car is traveling at a constant 40 m/s on level ground, what is the radius R of the tightest unbanked curve it can negotiate?

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

Explanation:

The car is in uniform circular motion because it’s moving at a constant speed along a curve that is a segment of a circle. Hence we know 

This is the minimum turning radius because arad is the maximum centripetal acceleration.

EMRS PGT Physics Mock Test - 4 - Question 23

An object has a length of 0.42 cm. Which of the following statements correct?

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

As no indication of its precision is given. In this case, all we have to go on is the number of digits contained in the data.

Thus the quantity “0.42 cm” is specified to 0.01 unit in 0.42, or one part in 42.

The implied relative uncertainty in this figure is 1/42, or about 2%.

EMRS PGT Physics Mock Test - 4 - Question 24

In order to avoid gender stereotyping in class, a teacher should:

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

In order to avoid gender stereotyping in class, a teacher should try to put both boys and girls on non-traditional roles. A teacher should take care of each child irrespective of their gender. He/she should ensure the students' maximum potential. Everyone has the right to education. A teacher should ensure that his/her students derive maximum benefit from education.
Hence, the correct answer is, 'Try to put both boys and girls in non-traditional roles.'

EMRS PGT Physics Mock Test - 4 - Question 25

If force (F) is given by F = Pt_1 + Qt, where t is time. The unit of P is same as that of

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

According to the principle of homogeneity of dimension, an equation is dimensionally correct when each term has same dimension on both sides of the equation. 

Since left side has dimension of force so the term Pt−1 will have also dimension of force. 

Thus, [P][T−1]=[F]=[MLT−2]

or [P]=[MLT−1]

We know that the dimension of momentum is [p]=[MLT−1]

EMRS PGT Physics Mock Test - 4 - Question 26

In the following question, four words are given out of which one word is incorrectly spelled. Find the incorrectly spelled word.

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

"Renaissance" is the correct word and it means the revival of something in a modified form.
Example- The renaissance was a period where great pieces of art were created.

EMRS PGT Physics Mock Test - 4 - Question 27
‘गुणहीन' का सही समास विग्रह क्‍या है?
Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 27

गुणहीन का समास विग्रह 'गुण से हीन' है। Key Points

  • गुणहीन में अपादान तत्पुरुष समास है।
  • अपादान तत्पुरुष समास- जिस समास में उत्तर पद प्रधान तथा पूर्व पद गौण हो तथा 'से' विभक्ति का लोप हो अपादान तत्पुरुष समास कहलाता है।
  • उदाहरण- भयभीत, देशनिकाला, दूरागत आदि

Additional Information

  • तत्पुरुष समास की परिभाषा- जिस समास का उत्तरपद प्रधान हो तत्पुरुष समास कहलाता है, इस समास में कारक चिह्नों का लोप होता है।
  • तत्पुरुष समास के छः भेद होते है
EMRS PGT Physics Mock Test - 4 - Question 28
कविपुंगव' का समास विग्रह है-
Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 28
कविपुंगव' का समास विग्रह है-कवि में पुंगवKey Points
  • "पुंगव" शब्द का अर्थ होता है श्रेष्ठ।
  • कविपुंगव में अधिकरण तत्पुरुष समास है।
  • अधिकरण तत्पुरुष में पहला पद प्रधान होता है।
  • वनवास – वन में वास।
  • इस समास में वास प्रधान है और ‘में’ और ‘पर’ संबंध कारक चिह्न का लोप है।

Additional Informationसमास

  • समास का शाब्दिक अर्थ छोटा रूप होता है | ​
  • दो या दो से अधिक शब्दों से मिलकर बने नए व छोटे शब्द को समास कहते है |

EMRS PGT Physics Mock Test - 4 - Question 29
निम्नांकित में अव्ययीभाव समास का उदाहरण नहीं हैः
Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 29

'आपबीती' अव्ययीभाव समास का उदाहरण नही है। अतः 'आपबीती' सही उत्तर है।
आपबीती - अपने पर बीती हुई - तत्पुरुष समास
Key Points

  • बेखटके - बिना खटके के - अव्ययीभाव समास
  • प्रतिदिन - प्रत्येक दिन - अव्ययीभाव समास
  • 'पुश्तानुपुश्त' :- पुश्तो के बाद पुश्ते - अव्ययी भाव समास
  • यथासम्भव - जैसे भी सम्भव हो - अव्ययीभाव समास

Additional Information

EMRS PGT Physics Mock Test - 4 - Question 30
निम्नलिखित मे द्वन्‍द्व समास है
Detailed Solution for EMRS PGT Physics Mock Test - 4 - Question 30

‘देशविदेश’ में 'द्वन्‍द्व' समास है।इसका समास विग्रह होगा- देश और विदेश। अन्य विकल्प गलत हैं । अतः उत्तर सही विकल्प 3 'देशविदेश​' है।

Key Points

  • 'देशविदेश' का समास विग्रह करने पर 'देश और विदेश' होगा।
  • इसमें दो प्रमुख पदों के मध्य 'और' योजक का प्रयोग होने के कारण द्वंद्व समास है।

अन्य विकल्प -

Additional Information

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