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KVS PGT Physics Mock Test - 2 - KVS PGT/TGT/PRT MCQ


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30 Questions MCQ Test KVS PGT Exam Mock Test Series 2024 - KVS PGT Physics Mock Test - 2

KVS PGT Physics Mock Test - 2 for KVS PGT/TGT/PRT 2024 is part of KVS PGT Exam Mock Test Series 2024 preparation. The KVS PGT Physics Mock Test - 2 questions and answers have been prepared according to the KVS PGT/TGT/PRT exam syllabus.The KVS PGT Physics Mock Test - 2 MCQs are made for KVS PGT/TGT/PRT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for KVS PGT Physics Mock Test - 2 below.
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KVS PGT Physics Mock Test - 2 - Question 1

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

Castigate

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 1

The correct answer is Option 2, i.e., ‘spare’.
The word ‘castigate’ means ‘to subject to severe punishment.’ The word ‘spare’ has an opposite meaning of ‘to choose not to punish.’ Hence, the correct option is Option 2.

KVS PGT Physics Mock Test - 2 - Question 2

Directions: In the following question, one part of the sentence may have error(s). Find out the part of the sentence having an error and select the appropriate option. If a sentence is free from error, select 'No error' as your answer. 

Q.The stolen property (1)/ was find (2)/ at the dacoit's hiding place. (3) /No error (4)

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 2

The error is in part 2 of the sentence. The sentence is in past tense and therefore the verb 'find' is incorrectly used. The correct form of the verb 'find' would be 'found'.

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KVS PGT Physics Mock Test - 2 - Question 3

‘मत्स्य’ का पर्यायवाची है-

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 3
सही विकल्प मीन है।
  • 'मत्स्य' का पर्यायवाची है - मीन
  • मत्स्य शब्द के अन्य पर्यायवाची शब्द - जलजीवन, शफरी, मकर, मछली
Key Points
अन्य विकल्प-

Additional Informationपर्यायवाची शब्द के अन्य उदाहरण -

KVS PGT Physics Mock Test - 2 - Question 4
मदनश्लाका किसका पर्यायवाची है?
Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 4
मदनश्लाका कोयल का पर्यायवाची है। Key Points Additional Informationकबूतर
  • बहुवचन - कबूतर
  • समानार्थी शब्द - कपोत
  • लिंग - पुल्लिंग
  • संज्ञा के प्रकार - जातिवाचक

चिड़िया

  • बहुवचन - चिड़ियाँ
  • समानार्थी शब्द - पंछी , पक्षी , परिंदा
  • लिंग - स्त्रीलिंग
  • संज्ञा के प्रकार - जातिवाचक

कोयल

  • बहुवचन - कोयलें
  • समानार्थी शब्द - कोकिल , कोकिला
  • लिंग - स्त्रीलिंग
  • संज्ञा के प्रकार - जातिवाचक

कौआ

  • समानार्थी शब्द - काग , काक , कौवा
  • विलोम शब्द - कौवी , कौवी
  • लिंग - पुल्लिंग
KVS PGT Physics Mock Test - 2 - Question 5

The Tropic of Cancer does NOT pass through which of the following state?

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 5

The correct answer is Odisha.

Key Points

  • The Tropic of Cancer is an imaginary line north from the Equator at an angle of 23.50 degrees.
  • Tropic of Cancer passes through 16 countries, 3 continents, and 6 water bodies.
  • It passes through 8 Indian states.

Additional Information

KVS PGT Physics Mock Test - 2 - Question 6

A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)

[AIEEE 2002]

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 6

As the plane is frictionless, thus there will be only slipping and no rolling. So, for all of them acceleration will be ‘gsinθ’

KVS PGT Physics Mock Test - 2 - Question 7

Mary bought a tank whose dimensions are 5.6cm, 8.2cm and 12.8 cm. The volume of water (in cm³) that can be stored in tank (correct up to four significant figures) is:

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

Volume of water = Volume of the tank = Length * breadth * height = 5.6 * 8.2 * 12.8 = 587.776

Volume, correct up to 4 significant figure = 587.8

*Multiple options can be correct
KVS PGT Physics Mock Test - 2 - Question 8

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10–2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3 . The value of the magnetic field at its centre is

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

KVS PGT Physics Mock Test - 2 - Question 9

Three identical blocks of masses m = 2 kg are drawn by a force F = 10.2N with an acceleration of 0.6 ms-2 on a frictionless surface, then what is the tension (in N) in the string between that blocks B and C ? 

[AIEEE 2002]

                      

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 9

Let the tension between B and C be T

M=2kg

F−T=ma

10.2−T=ma

10.2−T=2×0.6

10.2−T=1.2

10.2−1.2=T

T=9N

KVS PGT Physics Mock Test - 2 - Question 10

Select the correct choice(s) :

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 10

If the TE = 0 that means KE + PE = 0
KE = - PE which is enough for the body to leave the system (here, Earth)

KVS PGT Physics Mock Test - 2 - Question 11

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is      

[AIEEE 2003]

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 11

Let R be the normal contact force by wall on the block.

R = 10 N
fL = w and f = μR
∴ μR = w
or w = 0.2 x 10 = 2N

KVS PGT Physics Mock Test - 2 - Question 12

Phasor diagrams show

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 12

Explanation:By convention phasor diagrams are made to show the rotation of phasor in counterclockwise direction with constant speed

KVS PGT Physics Mock Test - 2 - Question 13

A wheel of moment of inertia 0.40 kg-m2 is making 2100 rotations per minute. The wheel is brought to a complete stop in 2 seconds. The angular acceleration of the wheel is

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 13

CONCEPT:

  • Angular velocity (ω): How fast the angular position or orientation of an object changes with time or How fast an object rotates or revolves relative to another point is known as angular velocity.
    • Angular velocity in terms of RPM (revolutions per minute) 

ω = 2 π f / 60

where ω is in terms of radian per second and f is in revolution per minute.

  • Angular acceleration (α): The rate of change of angular velocity is known as angular acceleration.

The relation between α and changing ω is given by: 

ω1 = ω0 + α t

where ω1 is the final angular velocity, ω0 is the initial angular velocity α is the angular acceleration.

CALCULATION:

Given that f0 = 2100 rpm; f1 = 0 rpm; t = 2 sec.

ω0 = 2 π f0 / 60 = 2 π 2100 / 60 = 70π 

ω1 = 0

ω1 = ω0 + α t

0 = 70π + α 2

α = -35π rad/sec2

So the correct answer is option 1.

KVS PGT Physics Mock Test - 2 - Question 14

The gravitational force on a body of mass m at a distance r from the centre of the Earth for r < R, where R is the radius of Earth, is proportional to:

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 14

We know, the gravitational force 
,
This is valid when R is the centre to centre distance between the two masses M and m.
At a distance r < R, the mass of Earth cannot be taken as M.

The new mass (M'):
The mass of earth will be contained for a radius r. Let us assume that the mean density of the earth is ρ and that earth is a sphere with a radius equal to the distance r. We know, mass = density × volume

Therefore the gravitational force, F = 

Hence the gravitational force, F ∝ r

KVS PGT Physics Mock Test - 2 - Question 15

Which of the following is the definition for magnetic meridian of Earth?

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 15

A vertical plane passing through the axis of a freely suspended or pivoted magnet is known as magnetic meridian of Earth. The vertical plane passing through the geographical North Pole and South Pole at a given place is known as the geographical meridian of that place.

KVS PGT Physics Mock Test - 2 - Question 16

 A ball is released from the top of a tower of height h meter. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?   

[AIEEE 2004]

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 16

Second law of motion gives:

KVS PGT Physics Mock Test - 2 - Question 17

The number of significant digits in 2,076 is:

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 17

There are three rules for determining how many significant figures are in a number: 

  • Non-zero digits are always significant.
  • Any zeros between two significant digits are significant.
  • A final zero or trailing zeros in the decimal portion ONLY are significant.

Keeping these rules in mind, we can say that there are 4 significant digits.

KVS PGT Physics Mock Test - 2 - Question 18

What is the angular spread of the first minimum and the central maximum when diffraction is observed with 540 nm light through a slit of width 1mm?​

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 18

The First minimum is located at the place where the central maximum ends. And the first minimum spreads further upto half the fringe width.

KVS PGT Physics Mock Test - 2 - Question 19

The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 19

We know that PE + KE = TE = constant
Hence at the extremis TE = PE = ½ kz2  
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE =  ½ kz2
Hence we get  ½ kx2 = ¾  ½ kz2
We get x/z = √3/4 
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.

KVS PGT Physics Mock Test - 2 - Question 20

A force time graph for the motion of a body is as shown in figure. Change in linear momentum between 0 and 6 s is 

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 20

Area under F-t curve = change in momentum = 2 x (-2) + 1 x 4 = -4 + 4 = 0  

KVS PGT Physics Mock Test - 2 - Question 21

If the coefficient of friction of a surface is μ = √3, then the angle of inclination of a plane to make a block on it to just start sliding is

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 21

KVS PGT Physics Mock Test - 2 - Question 22

Two wooden blocks are moving on a smooth horizontal surface such that the mass m remains stationary with respect to block of mass M as shown in figure. The magnitude of force P is 

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 22

The difterent torces acting on mass m are shown in the adjoining figure. Acceleration of the system

KVS PGT Physics Mock Test - 2 - Question 23

Referring to the Young’s double slit experiment, Phase difference corresponding to a Path Difference of λ /3 is

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 23

We know that,
Phase difference=K× path difference. K=2π/ λ.
Therefore, phase difference=(2π/ λ) ×( λ/3)=2π/3=120 degree.

KVS PGT Physics Mock Test - 2 - Question 24

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure.  The least count of the scale used in the metre bridge is 1mm. The unknown resistance is

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 24

In case of a meter bridge

For finding the value of R

For finding the value of ΔR

Therefore, R = (60 ± 0.25)Ω

KVS PGT Physics Mock Test - 2 - Question 25

According to Atomic Hypothesis:

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 25

Explanation:atomic hypothesis that all things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another.

KVS PGT Physics Mock Test - 2 - Question 26

The relative error in a physical quantity raised to the power k is:

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 26

The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.

Suppose Z = A2,
Then,
ΔZ/Z = (ΔA/A) + (ΔA/A) = 2 (ΔA/A).
Hence, the relative error in A2 is two times the error in A.

KVS PGT Physics Mock Test - 2 - Question 27

Measurement of a physical quantity is essentially the:

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 27

The Measurement of a given quantity is essentially an act or result of comparison between a quantity whose magnitude (amount) is unknown, with a similar quantity whose magnitude (amount) is known, the latter quantity being called a Standard.

KVS PGT Physics Mock Test - 2 - Question 28

An automobile travelling at a speed of 60 km/h, can apply the brake to stop within a distance of 20 m. If the car is going twice as fast, i.e. 120 km/h, the stopping distance will be: 

[AIEEE 2004]

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 28
  • We know that with retardation of a car with an initial speed of 60 kmph stops after 20m.
  • We know that v2 = 2as  (third equation of motion). Thus we get that for some constant retardation, stopping distance is proportional to the square of the initial speed.
  • Hence when the initial speed is 120kmph we get,
    s = 20 x (120/60)2
    s = 80 m
KVS PGT Physics Mock Test - 2 - Question 29

Which of the following statements false?

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 29

Explanation:A displacement is vector that is the shortest path length from the initial to the final position of the body. It is not always eqaul to path length. It can be zero and -ve also.

*Multiple options can be correct
KVS PGT Physics Mock Test - 2 - Question 30

Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle θ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1? 

Detailed Solution for KVS PGT Physics Mock Test - 2 - Question 30

Magnetic field due to current in wire 1 at point P distant r from the wire is

 (directed perpendicular to the plane  of paper, inwards)

The force exerted due to this magnetic field on current element i2 dl is

dF = i2 dl B sin 90°

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