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DSSSB PGT Physics Mock Test - 5 - DSSSB TGT/PGT/PRT MCQ


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30 Questions MCQ Test DSSSB PGT Mock Test Series 2024 - DSSSB PGT Physics Mock Test - 5

DSSSB PGT Physics Mock Test - 5 for DSSSB TGT/PGT/PRT 2024 is part of DSSSB PGT Mock Test Series 2024 preparation. The DSSSB PGT Physics Mock Test - 5 questions and answers have been prepared according to the DSSSB TGT/PGT/PRT exam syllabus.The DSSSB PGT Physics Mock Test - 5 MCQs are made for DSSSB TGT/PGT/PRT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for DSSSB PGT Physics Mock Test - 5 below.
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DSSSB PGT Physics Mock Test - 5 - Question 1

What is the dimension of k/m where k is the force constant and m is the mass of the oscillating object?

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 1

F=k x (compression in string).
X=L.
a=LT-2.
F=ma.
ma= k L.
LT-2/L=k/m.
T-2=k/m.

DSSSB PGT Physics Mock Test - 5 - Question 2

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity w. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity w' = 

[AIEEE 2006]

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

The initial angular momentum of the ring will be:
Li​=mR2ω
whereas the final angular momentum is given as:
Lf​=mR2ωnew​+2MR2ωnew
Conserving angular momentum
mR2ω=(mR2+2MR2new​
⇒ωnew​=m​ω/m+2M

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DSSSB PGT Physics Mock Test - 5 - Question 3

For the given uniform square lamina ABCD, whose centre is O

                        

[AIEEE 2007]

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 3


By symmetry we can say that
IEF = IGH
IAC = IBD
By perpendicular axis theorem, we can say that
IZ = IEF + IGH = 2IEF
Also, IZ = IAC + IBD = 2IAC
Hence, IAC = IEF

DSSSB PGT Physics Mock Test - 5 - Question 4

Two point masses of mass 4m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be

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

Let the velocity of m is v and velocity of 4m is u
Therefore, v = 4u and we know that kinetic energy is 1/2mv2 so for m:
KE = ½ x m x v2 = 1
And for 4m:
KE = ½ x 4m x 42 = 4
So, ratio of their kinetic energy is 1:4
Hence A

DSSSB PGT Physics Mock Test - 5 - Question 5

with the orbit radius of a body in circular planetary motion. Find the correct state ment about the

curves A, B and C

                       

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

The body is in circular planetary motion this implies that Potential energy will be negative (attractive)  and the total energy of the system will lie between zero and potential energy (because of circular motion).

DSSSB PGT Physics Mock Test - 5 - Question 6

A satellite which appears to be at a fixed position at a definite height to an observer is called:

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 6

As the relative velocity of the satellite with respect to the earth is zero, it appears stationary from the Earth surface and therefore it is called is geostationary satellite or geosynchronous satellite.

DSSSB PGT Physics Mock Test - 5 - Question 7

Bernoulli’s theorem is important in the field of:

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 7

Bernoulli's theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar.

DSSSB PGT Physics Mock Test - 5 - Question 8

The process by which heat flows from the region of higher temperature to the region of lower temperature by actual movement of material particles is called

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

In this process, heat is transferred in the liquid and gases from a region of higher temperature to a region of lower temperature. Convection heat transfer occurs partly due to the actual movement of molecules or due to the mass transfer

DSSSB PGT Physics Mock Test - 5 - Question 9

The second law of thermodynamics says

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 9

The second law of thermodynamics gives a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of a refrigerator. It says that the efficiency of a heat engine can never be unity or 100%, this implies that the heat released to the cold reservoir can never be made zero.
For a refrigerator the second law says that the coefficient through performance can never be infinite, this implies that the external work can never be zero.

DSSSB PGT Physics Mock Test - 5 - Question 10

The ratio of the adiabatic to isothermal elasticities of a triatomic gas is

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 10

For triatomic gas γ = 4/3

DSSSB PGT Physics Mock Test - 5 - Question 11

One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 11

Explanation:

At const volume, 

Q = 500 J

Q=nCPΔT

500=1×2.5×8.31ΔT

ΔT=24.06

W=nRΔT=1×8.31×24.06=200J

DSSSB PGT Physics Mock Test - 5 - Question 12

A particle of mass m is allowed to oscillate on a smooth parabola x2 = 4ay, a > 0, about the origin O (see figure). For small oscillations, find the angular frequency ( ω )

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 12

During its oscillations, at a particular instant let the coordinates of the particle are (x, y)  The total energy of the particle is 

As oscillations are very small, we can ignore the middle term. 

DSSSB PGT Physics Mock Test - 5 - Question 13

The pendulum of a wall clock executes

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 13

The bob of a pendulum moves in such a way that it repeats its positions several time but alternately the time gap between these positions is always equal which proves that the motion of a pendulum is oscillatory.

DSSSB PGT Physics Mock Test - 5 - Question 14

If the path difference between the interfering waves is nl, then the fringes obtained on the screen will be

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 14

B because of constructive interference.

DSSSB PGT Physics Mock Test - 5 - Question 15

A point charge causes an electric flux of −1.0×103Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge? 

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 15

Electric flux, Φ = −1.0 × 103 N m2/C

Radius of the Gaussian surface,

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103 N m2/C.

Electric flux is given by the relation,

qϕ=q∈0

Where,

q = Net charge enclosed by the spherical surface

0 = Permittivity of free space = 8.854 × 10−12 N−1 Cm−2

∴ qq=ϕ∈0

= −1.0 × 103 × 8.854 × 10−12

= −8.854 × 10−9 C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

DSSSB PGT Physics Mock Test - 5 - Question 16

Under the influence of the coulomb field of charge +Q, a charge −q is moving around it in an elliptical orbit. Find out the correct statement(s). 

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 16

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

DSSSB PGT Physics Mock Test - 5 - Question 17

A point charge causes an electric flux of −1.0×103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 17

Electric flux is given by  since amount of charge not depends on size and shape so by making radius <  double the amount of charge remain same so electric flux remain same.

DSSSB PGT Physics Mock Test - 5 - Question 18

A point charge 2nC is located at origin. What is the potential at (1,0,0)?

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 18

 V = Q/(4πεr), where r = 1m
V = (2 X 10-9)/(4πε x 1) = 18 volts.

DSSSB PGT Physics Mock Test - 5 - Question 19

Electromotive force is a

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 19

EMF is equal to potential difference between plates when the battery is not connected to any external circuit. 
Potential difference is due to the charge accumulation on the plates. So, EMF produces a potential difference between the two terminals which drives the current.

DSSSB PGT Physics Mock Test - 5 - Question 20

It is necessary to use satellites for long distance TV transmission because

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 20

TV signals being of high frequency are not reflected by the ionosphere. Therefore, to reflect these signals, satellites are needed. That is why, satellites are used for long distance TV transmission.
 Most long-distance shortwave (high frequency) radio communication—between 3 and 30 MHz—is a result of skywave propagation.
This 3-30 MHz is a range of frequencies which are used in sky waves propagation so that the ionosphere is capable of reflecting it.

DSSSB PGT Physics Mock Test - 5 - Question 21

According to Maxwell’s equations

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 21

Maxwell’s Fourth Equation
It is based on Ampere’s circuital law. To understand Maxwell’s fourth equation it is crucial to understand Ampere’s circuital law,
Consider a wire of current-carrying conductor with the current I, since there is an electric field there has to be a magnetic field vector around it. Ampere’s circuit law states that “The closed line integral of magnetic field vector is always equal to the total amount of scalar electric field enclosed within the path of any shape” which means the current flowing along the wire(which is a scalar quantity) is equal to the magnetic field vector (which is a vector quantity)

DSSSB PGT Physics Mock Test - 5 - Question 22

The Brewster’s angle for a transparent medium is 600.The angle of incidence is​

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 22

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.
So the angle of incidence would be 60°

DSSSB PGT Physics Mock Test - 5 - Question 23

Number of ejected photoelectrons increases with increase

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 23

The number of electrons ejected can be measured as a function of intensity.
Intensity is equal to energy per unit time per unit area.
Keeping frequency constant, increasing intensity will increase energy thus increase of ejected electrons.
Thus,
Number of ejected photoelectrons will increase with increase in intensity of light. 

DSSSB PGT Physics Mock Test - 5 - Question 24

Reason why there are many lines in an atomic spectrum is because

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 24

Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states.

DSSSB PGT Physics Mock Test - 5 - Question 25

α-rays are

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 25

Alpha particles, also called alpha rays or alpha radiation, consist of two protons and two neutrons bound together into a particle identical to a helium-4 nucleus. They are generally produced in the process of alpha decay, but may also be produced in other ways.

DSSSB PGT Physics Mock Test - 5 - Question 26

90% of a radioactive sample is left undisintegrated after time τ has elapsed, what percentage of initial sample will decay in a total time 2τ?

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 26

Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed 
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.
 

DSSSB PGT Physics Mock Test - 5 - Question 27

The depletion layer in the p-n junction is caused

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 27

When a p−n junction is formed, some of the free electrons in the n region diffuse across the junction and combine with holes to form negative ions. In doing so they leave behind positive ions at the donor impurity sites. Similarly, holes from p side diffuse to the n side and thus form a layer called diffusion layer at the junction.
So the depletion region is caused by the diffusion of charge carriers.
 

DSSSB PGT Physics Mock Test - 5 - Question 28

How many depletion regions does a transistor have?

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 28

There are two depletion regions in a transistor, since there are two PN junctions.

DSSSB PGT Physics Mock Test - 5 - Question 29

A fixed volume of iron is drawn into a wire of length l. The extension x produced in this wire by a constant force F is proportional to

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 29
DSSSB PGT Physics Mock Test - 5 - Question 30

The ratio of densities of nitrogen and oxygen is 14 : 16. The temperature at which the speed of sound in nitrogen will be same as that of oxygen at 55 oC is

Detailed Solution for DSSSB PGT Physics Mock Test - 5 - Question 30
  p p TN=287K=14∘C
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