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


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30 Questions MCQ Test DSSSB PGT Mock Test Series 2025 - DSSSB PGT Physics Mock Test - 2

DSSSB PGT Physics Mock Test - 2 for DSSSB TGT/PGT/PRT 2025 is part of DSSSB PGT Mock Test Series 2025 preparation. The DSSSB PGT Physics Mock Test - 2 questions and answers have been prepared according to the DSSSB TGT/PGT/PRT exam syllabus.The DSSSB PGT Physics Mock Test - 2 MCQs are made for DSSSB TGT/PGT/PRT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for DSSSB PGT Physics Mock Test - 2 below.
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DSSSB PGT Physics Mock Test - 2 - Question 1

Physical quantities are:

Detailed Solution for DSSSB PGT Physics Mock Test - 2 - Question 1
  • A physical quantity is a physical property of a phenomenon, body, or substance, that can be quantified by measurement.
  • A physical quantity can be expressed as the combination of a magnitude expressed by a number – usually a real number – and a unit.
  • All these given above can be expressed as explained so these are physical quantity.
DSSSB PGT Physics Mock Test - 2 - Question 2

Vectors can be added by

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

Explanation:The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the triangle method or the parallelogram method.

DSSSB PGT Physics Mock Test - 2 - Question 3

A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (g = 10 m/s2

[AIEEE 2004]

Detailed Solution for DSSSB PGT Physics Mock Test - 2 - Question 3

Let mass of the block be m.

Frictional force in rest position F = mg sin 300

[This is static frictional force and may be less than the limiting frictional force]

DSSSB PGT Physics Mock Test - 2 - Question 4

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon (h = 6.6 × 10-34 J s).

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

For every large distance, P.E. = 0
Total energy = 2.6 + 0 = 2.6 eV
Finally, in the first excited state of H atom, total energy = -3.4 eV
Loss in total energy = 2.6 - (-3.4) = 6 eV
It is emitted as photon.

= 1.45 × 109 MHz

DSSSB PGT Physics Mock Test - 2 - Question 5

When does the potential energy of a spring increase?

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

Potential energy of a spring is proportional to the square of the difference of the springs length and its original length, hence whether it is compressed or stretched the potential the potential energy will eventually increase only.

DSSSB PGT Physics Mock Test - 2 - Question 6

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 DSSSB PGT Physics Mock Test - 2 - Question 6


On solving we get 3L/4
 

DSSSB PGT Physics Mock Test - 2 - Question 7

Two particles of equal masses have velocities    First particle has an acceleration    while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of  

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



DSSSB PGT Physics Mock Test - 2 - Question 8

A body of mass m moving at a constant velocity v hits another body of the same mass moving with a velocity v/2 but in the opposite direction and sticks to it. The common velocity after collision is 

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


Hence v = v/4
 

DSSSB PGT Physics Mock Test - 2 - Question 9

What is the force required to produce an acceleration of 9.8 m/s2 on a body of weight 9.8N? Take g = 9.8 m/s2.

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

Acceleration (a) = 9.8 m/s2

Weight of the body = 9.8 N

Since, Weight of the body = mg

⇒ Mass = Weight/g = 9.8/9.8 = 1 kg

We know that,

F = m × a

⇒ F = 1 kg × 9.8 m/s2

⇒ F = 9.8 Newton

DSSSB PGT Physics Mock Test - 2 - Question 10

What pulls a ball back to earth?

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

The gravitational force of the earth acts on every object having mass. When a ball is thrown upwards the gravitational force of the earth acts against the velocity of the object and pulls the ball downwards, i.e, towards itself.

Thus, gravity pulls a ball back to earth.

DSSSB PGT Physics Mock Test - 2 - Question 11

The earth attracts the moon with a gravitational force of 1020 N. Then the moon attracts the earth with the gravitational force of  

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

The universal law of gravitation:

  • Sir Isaac Newton put forward the universal law of gravitation in 1687 and used it to explain the observed motions of the planets and moons.
  • The law states that every particle attracts every other particle in the universe with force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
  • Formula, Gravitational force,  where, F is the gravitational force between bodies, m1, and m2 are the masses of the bodies, d is the distance between the centers of two bodies, and G is the universal gravitational constant.
  • Here, universal gravitational constant, G = 6.67 × 10-11 Nm/kg2

  • The Universal Gravitational Law can explain almost anything, right from how an apple falls from a tree to why the moon revolves around the Earth.

Newton's third law of motion

  • It states that for every action, there is an equal and opposite reaction.
  • Action and reaction always act on two different bodies.

Explanation:

By Newton's Third Law and Newton's Law of Universal gravitation, the gravitational force the Earth exerts on the Moon is exactly the same as the force the Moon exerts on the Earth. 
F(Earth) = F(Moon)
Therefore, the moon attracts the earth with a gravitational force of 1020N.

DSSSB PGT Physics Mock Test - 2 - Question 12

As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth, this would

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

Concept:

Newton's law of gravitation: The force of attraction between any objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

  • The force acts along the line joining the two bodies.
  • The gravitational force is a central force that is It acts along the line joining the centers of two bodies.
  • It is a conservative force. This means that the work done by the gravitational force in displacing a body from one point to another is only dependent on the initial and final positions of the body and is independent of the path followed.

Explanation:

  • The gravitational force on mercury due to the sun is very large than that due to the earth.
  • As observed from the earth, the sun appears to move around the earth in a circular orbit though, in reality, the earth moves around the sun due to gravitational force between them
  • All planets move around the sun due to the huge gravitational force of the sun acting on them.
  • Since the gravitational force on mercury due to the earth is smaller as compared to that on it due to the sun therefore it revolves around the sun and not around the earth.

Hence, option (3) is the correct answer.

DSSSB PGT Physics Mock Test - 2 - Question 13

A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?

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

DSSSB PGT Physics Mock Test - 2 - Question 14

If a person studies about a fluid which is at rest, what will you call his domain of study?

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

Fluid Mechanics deals with the study of fluid at rest or in motion with or without the consideration of forces, Fluid Statics is the study of fluid at rest, Fluid Kinematics is the study of fluid in motion without consideration of forces and Fluid Dynamics is the study of fluid in motion considering the application forces.

DSSSB PGT Physics Mock Test - 2 - Question 15

According to the Stefan-Boltzmann law of thermal radiation for a perfect radiator, the rate of radiant energy per unit area is proportional to

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

The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

DSSSB PGT Physics Mock Test - 2 - Question 16

Suppose we have a box filled with gas and a piston is also attached at the top of the box.What are the ways of changing the state of gas (and hence its internal energy)?

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

As change of state depends on both temperature and pressure, so, either method A or C can be used. We can understand this by ideal gas equation PV = nRT

DSSSB PGT Physics Mock Test - 2 - Question 17

Kelvin- Planck statement states that

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

It works on the principle that no machine is 100% efficient. For instance if taking an ideal machine into consideration, work can be completely converted to heat. Now if the conditions are reversed, all heat cannot be converted to equal amount of work until and unless backed up by an external force.

DSSSB PGT Physics Mock Test - 2 - Question 18

Value of gas constant, R for one mole of a gas is independent of the

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

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

DSSSB PGT Physics Mock Test - 2 - Question 19

If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be

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

Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, E=F/q​=100N/2C​=50 N/C.
Hence, the value of field intensity will be 50 N/C.

DSSSB PGT Physics Mock Test - 2 - Question 20

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 DSSSB PGT Physics Mock Test - 2 - Question 20

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.

DSSSB PGT Physics Mock Test - 2 - Question 21

Gauss Law can be used to compute which of the following?

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

Gauss law relates the electric flux density and the charge density. Thus it can be used to compute the radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

DSSSB PGT Physics Mock Test - 2 - Question 22

When number of identical cells, in parallel combination are increased, the voltage of the circuit will

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

The voltage developed by the cells in parallel connection cannot be increased by increasing the number of cells present in the circuit. It is because they do not have the same circular path. In parallel connection the connection provides power based on one cell.
The voltage will remain the same.

DSSSB PGT Physics Mock Test - 2 - Question 23

An infinitely long conductor PQR is bent to form a right angle as shown in Figure. A current I flows through PQR. The magnetic field due to this current at the point M is H1. Now, another infinitely long straight conductor QS is connected at Q so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H2. The ratio H1/H2 is given by

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

Case 1 : Magnetic field at M due to PQ and QR is

Case 2 : When wire QS is joined.
H2 = (Magnetic field at M due to PQ) + (magnetic field at M due to QR) + (Magnetic field at M due to QS)

NOTE : The magnetic field due to an infinitely long wire carrying current at a distance R from the end point is half that at a distance R from the middle point.

DSSSB PGT Physics Mock Test - 2 - Question 24

Which of the following statements about bar magnets is correct ?

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

Explanation:As magnetic monopole does not exist. If we split the bar magnet into two pieces  each part will have its own north and south pole.

DSSSB PGT Physics Mock Test - 2 - Question 25

Eddy currents do not cause:

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

Eddy current when produced it causes damping heat loss energy loss but it don't produce any type of sparking because the electrons are only at the surface of the conductor and the electrons do not jump off to enter the air and heat it. 

DSSSB PGT Physics Mock Test - 2 - Question 26

The magnifying power of the telescope can be increased by

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

Magnifying power of a telescope is f0/fe​ ​​, so as  1/​​fe increases, magnifying power increases.
Hence the correct answer is option D.

DSSSB PGT Physics Mock Test - 2 - Question 27

A lamp and a screen are set up 100 cm apart and a convex lens is placed between them. The two positions of the lens forming real images on the screen are 40 cm apart. What is the focal length of the lens ?

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

L1L2=40cm and O I=100cm
Therefore, x+40+x=100 or x=30cm
For lens at L1, we have
U=-30cm and v=+70
Thus,
(1/v)-(1/u)=1/f
Or, 1/f=(1/70) - (1/-30)
Or, f=+21cm
 

DSSSB PGT Physics Mock Test - 2 - Question 28

Shape of the wave front of light emerging out of a convex lens when a point source is placed at its focus.

Detailed Solution for DSSSB PGT Physics Mock Test - 2 - Question 28

Plane. The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

DSSSB PGT Physics Mock Test - 2 - Question 29

If the work function of a material is 2eV, then minimum frequency of light required to emit photo-electrons is 

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

Φ= hνλ => 2eV= 6.626 x 10-34 x ν

2 x 1.6 x 10-19= 6.626 x 10-34 x ν

On solving we get ν = 4.6 x 1014 Hz.

DSSSB PGT Physics Mock Test - 2 - Question 30

A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then the maximum velocity with which the car can move is

Detailed Solution for DSSSB PGT Physics Mock Test - 2 - Question 30
When a car goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the car have a tendency to leave the curved path and regain the straight line path. Force of friction between the wheels and the road opposes this tendency of the wheels. This force (of friction) therefore acts, towards the centre of the circular track and provides the necessary centripetal force.

Hence, the maximum velocity with which a vehicle can go round a level curve, without skidding is Here, r = 40 m

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