CUET Exam  >  CUET Tests  >  CUET Mock Test Series  >  Physics: CUET Mock Test - 9 - CUET MCQ

Physics: CUET Mock Test - 9 - CUET MCQ


Test Description

30 Questions MCQ Test CUET Mock Test Series - Physics: CUET Mock Test - 9

Physics: CUET Mock Test - 9 for CUET 2025 is part of CUET Mock Test Series preparation. The Physics: CUET Mock Test - 9 questions and answers have been prepared according to the CUET exam syllabus.The Physics: CUET Mock Test - 9 MCQs are made for CUET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Physics: CUET Mock Test - 9 below.
Solutions of Physics: CUET Mock Test - 9 questions in English are available as part of our CUET Mock Test Series for CUET & Physics: CUET Mock Test - 9 solutions in Hindi for CUET Mock Test Series course. Download more important topics, notes, lectures and mock test series for CUET Exam by signing up for free. Attempt Physics: CUET Mock Test - 9 | 50 questions in 45 minutes | Mock test for CUET preparation | Free important questions MCQ to study CUET Mock Test Series for CUET Exam | Download free PDF with solutions
Physics: CUET Mock Test - 9 - Question 1

Which of the following is a widely used variety of commercial resistor?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 1

A wire bound resistor is an electrical passive component that limits current. Wire-bound resistors are made by winding the wires of an alloy like manganin on an insulating base. They are relatively insensitive to temperature.

Physics: CUET Mock Test - 9 - Question 2

The phase difference (Δϕ) between two superimposing waves to obtain constructive interference and hence bright band, is:

Detailed Solution for Physics: CUET Mock Test - 9 - Question 2

The correct option for the phase difference (Δϕ) between two superimposing waves to obtain constructive interference and hence a bright band is:
Δϕ = np; n = 1, 2, 3, 4, 5

  • For constructive interference to occur, the waves must be in phase, i.e., their crests and troughs must coincide.
  • If two waves of the same frequency and amplitude meet at a point with a path difference of integer multiple of wavelength (nλ), then they will be in phase and add up constructively to produce a bright band.
  • The phase difference between the two waves in this case is Δϕ = 2π(nλ/λ) = 2πn, where n is an integer.
  • Since 2πn is equivalent to np, where p is the wavelength, the correct option is Δϕ = np; n = 1, 2, 3, 4, 5.

The correct answer is option "1"

Physics: CUET Mock Test - 9 - Question 3

If two sources of intensities I0 each have a randomly varying phase difference ϕ, the resultant intensity at centre of screen will be :

Detailed Solution for Physics: CUET Mock Test - 9 - Question 3

When two sources of intensities Io interfere, the resultant intensity depends on the phase difference ϕ between the two sources.

For all other values of ϕ, the resultant intensity is given by:
I = 4Io cos²(ϕ/2)

Since the phase difference ϕ is randomly varying, we need to take the average of the above equation over all possible values of ϕ. The average value of cos²(ϕ/2) over all possible values of ϕ is 1/2.

Therefore, the average value of I is:
I = 4Io × 1/2 = 2Io
we will observe an average intensity that will be given by I = 2 Io at all points.
So, the resultant intensity at the centre of the screen is 2Io
The correct answer is option "3"

Physics: CUET Mock Test - 9 - Question 4

In Young's double slit experiment, interference pattern is obtained on the screen. If one of the slits is closed, then :

Detailed Solution for Physics: CUET Mock Test - 9 - Question 4
  • In Young's double-slit experiment, when both slits are open, light passing through the two slits creates an interference pattern on the screen with bright and dark fringes. However, when one of the slits is closed, the interference pattern disappears, and only a diffraction pattern is obtained on the screen.
  • The diffraction pattern obtained when one of the slits is closed is due to the bending of light as it passes through the open slit.
  • The light spreads out, creating a wider central maximum and a series of secondary maxima and minima on either side. The intensity of the central maximum decreases because the light from only one slit is contributing to it, while the intensity of the secondary maxima and minima increases because the light from one slit is focused into a narrower region.
  • If one of the slits is closed in Young's double-slit experiment, the interference pattern disappears, and only a diffraction pattern is obtained on the screen.
  • The width of the central maximum increases, and the intensity of the central maximum decreases while the intensity of the secondary maxima and minima increases.
    ​The correct answer is option "3"
Physics: CUET Mock Test - 9 - Question 5

In Young's double-slit experiment, the separation between the slits is halved and distance between the slits and screen is doubled. The fringe width will be:

Detailed Solution for Physics: CUET Mock Test - 9 - Question 5

Concept:
In Young's double slit experiment, the fringe width (w) is given by the equation:
w = λD/d
where λ is the wavelength of light,
D is the distance between the slits and the screen, and
d is the distance between the two slits.

Explanation:
If the separation between the slits is halved, then d is also halved. Therefore, the new value of d will be d/2.
If the distance between the slits and the screen is doubled, then the new value of D will be 2D.
Substituting these values in the above equation, we get:
w' = λ(2D)/(d/2) = 4λD/d
Therefore, the new fringe width (w') will be four times i.e., it gets quadrupled the original fringe width (w).
The correct answer is option (4)

Physics: CUET Mock Test - 9 - Question 6

In Young's double slit experiment, the fringe width is 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, without changing its geometry, the new fringe width will be:

Detailed Solution for Physics: CUET Mock Test - 9 - Question 6

Concept:
In Young's double slit experiment, the fringe width (wavelength of light λ divided by the distance between the two slits d) is given by the formula:
w = λd/D
where D is the distance between the slits and the screen.
When the whole apparatus is immersed in water of refractive index n, the wavelength of light effectively decreases by a factor of n.
This is because the speed of light in water is slower than in air, and the frequency of light remains constant. Therefore, the new fringe width w' can be calculated as:
w' = λd/nD
where λ/n is the new wavelength of light in water.
Since the refractive index of water is 4/3, the new wavelength of light in water is (3/4)λ.
Substituting this into the equation above, we get:
w' =
= (3/4)w
where w is the original fringe width in the air.
Therefore, the new fringe width in water is (3/4) times the original fringe width in air.
Substituting the given value of the original fringe width w = 0.4 mm, we get:
w' = (3/4) × 0.4 mm = 0.3 mm
Therefore, the new fringe width when the whole apparatus is immersed in water of refractive index 4/3 is 0.3 mm.
The correct answer is option (3)

Physics: CUET Mock Test - 9 - Question 7

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10–11 m. The radius of the n = 3 orbit is

Detailed Solution for Physics: CUET Mock Test - 9 - Question 7
Radius of nth orbit of electron in hydrogen atom is given as : rn = r1n2, where n is the number of orbit and r1 is the innermost electron orbit of Hydrogen atom.
Given : r1 = 5.3×10-11m
rn = r1n2
∴ For n = 3 orbit,
⇒r3 = 5.3×10-11×(3)2
r3 = 47.7×10-11 m
r3 = 4.77 × 10-10 m
Hence, the correct option is (4)
Physics: CUET Mock Test - 9 - Question 8
The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is
Detailed Solution for Physics: CUET Mock Test - 9 - Question 8

CONCEPT:

  • Electromagnetic Spectrum is the orderly distribution of electromagnetic waves in accordance with their wavelength(λ) or frequency(ν) into a group of waves having widely differing properties. There are no sharply defined boundaries and these different waves do overlap each other.
  • The table below represents the frequency range and application of various electromagnetic waves

EXPLANATION:

  • The spectrum represents the arrangement in the order of frequency and wavelength.
  • The decreasing order of wavelengths is opposite to the increase in frequency.
  • The wavelength and frequency of waves are inversely related.

-------(1)

Then the waves can be arranged as

According to relationship 1, Equation 2 can be rewritten as

  • Hence Option 2 is the answer
Physics: CUET Mock Test - 9 - Question 9
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10−2 nm, the potential difference is :
Detailed Solution for Physics: CUET Mock Test - 9 - Question 9

CONCEPT:

The De Broglie wavelength of the electron is written as;

°A

here, is the wavelength in °A and V in Volt is the potential difference.

CALCULATION:

Given,

The wavelength of the electron is 1.227 × 10−2 nm

1.227 × 10−2 nm = 1.227 × 10−11 m

1.227 × 10−11 m = 1.227 × 10−11 × 1010 °A

1.227 × 10−2 nm = 12.27 × 10−2 °A

Using equation (1) we have;

°A

⇒ V = 104 volts

Hence, option 2) is the correct answer.

Physics: CUET Mock Test - 9 - Question 10
Induced EMF in a coil during the phenomenon of electromagnetic induction is directly proportional to:
Detailed Solution for Physics: CUET Mock Test - 9 - Question 10

The correct answer is rate of change in magnetic flux.Key Points

  • Induced EMF in a coil during the phenomenon of electromagnetic induction is directly proportional to the rate of change in magnetic flux.
  • Electromagnetic induction is the process of generating an electromotive force (EMF) in a conductor by changing the magnetic field around it.
  • The magnitude of the induced EMF depends on the rate of change of magnetic flux through the coil.
  • This means that the faster the magnetic field changes, the greater the induced EMF will be.

Additional Information

  • Magnetic flux is the amount of magnetic field passing through a surface.
    • It is given by the product of the magnetic field strength and the area of the surface perpendicular to the field.
    • However, the induced EMF is directly proportional to the rate of change of magnetic flux, not the magnetic flux itself.
  • Resistance of the circuit does not affect the magnitude of induced EMF.
    • It only affects the current flowing through the circuit according to Ohm's Law.
  • The phenomenon of electromagnetic induction was discovered by Michael Faraday in 1831.
  • Faraday's Law of Electromagnetic Induction states that the EMF induced in a circuit is proportional to the rate of change of magnetic flux through it.
  • The direction of induced EMF is given by Lenz's Law, which states that the direction of the induced EMF is such that it opposes the change in magnetic flux that produced it.
Physics: CUET Mock Test - 9 - Question 11

A Zener diode is used as ___________.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 11

Zener Diode is mostly used as Voltage Regulator.

CONCEPT:

Zener diode: A Semiconductor Diode blocks current in the reverse direction, but will suffer from premature breakdown or damage if the reverse voltage applied across becomes too high.

Zener diode as voltage regulator: The Zener diode is connected with its cathode terminal connected to the positive rail of the DC supply so it is reverse biased and will be operating in its breakdown condition. Resistor Rs is selected so to limit the maximum current flowing in the circuit.

EXPLANATION:

  • A Zener diode is used as a voltage regulator in reverse biased mode.
  • The breakdown voltage in the Zener diode when connected in the reverse-biased is called Zener voltage. This Zener voltage so steady and constant, it has huge applications in circuits, most importantly, voltage regulation.

So option 4 is correct.

Physics: CUET Mock Test - 9 - Question 12

What is the use of a Van de Graff generator?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 12

Van de Graff generator is used to create a large amount of static electricity. A Van de Graff generator uses static electricity and a moving belt to charge a large metal sphere to a very high voltage. As the belt moves, electrons move from the rubber belt to the silicon roller, causing the belt to become positively charged and the roller to become negatively charged. As a result, it builds up positive charge.

Physics: CUET Mock Test - 9 - Question 13

What is the order of potential difference built up by the Van de Graff generator?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 13

A Van de Graff generator, by means of a moving belt and suitable brushes, transfers charge continuously to a large spherical conducting shell. As a result, a potential difference of the order of several million volts is built up and this can be used for accelerating charged particles.

Physics: CUET Mock Test - 9 - Question 14

When was Van de Graff generator invented and by whom?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 14

Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University.

Physics: CUET Mock Test - 9 - Question 15

Find the true statement.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 15

A Van de Graff generators produces large voltage and less current. A Van de Graff generator is an electrostatic generator which creates very high electric potentials. It produces very high voltage direct current electricity at low current levels.

Physics: CUET Mock Test - 9 - Question 16

Identify the type of modulation where the frequency of the modulated wave is equal to that of the carrier wave.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 16

In Amplitude modulation, the modulating wave is superimposed on a high-frequency carrier wave such that the amplitude of the modulated wave varies as the amplitude of the modulating wave.

Physics: CUET Mock Test - 9 - Question 17

Why is AM used for broadcasting in the communication systems?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 17

Amplitude modulation (AM) is used for broadcasting because it avoids receiver complexity. Moreover, only a diode and a capacitor are sufficient to separate the audio signal from the amplitude-modulated wave.

Physics: CUET Mock Test - 9 - Question 18

A carrier of peak voltage 15 V is used to transmit a message signal. If the modulation index is 70%, then what will be the peak voltage of the modulating signal?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 18


Physics: CUET Mock Test - 9 - Question 19

‘X’ for an AM is the ratio of change in amplitude of the carrier wave to the amplitude of the original carrier wave. Identify ‘X’.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 19

The modulation index for an AM wave is defined as the ratio of change in the amplitude of the carrier wave to the amplitude of the original carrier wave.

Physics: CUET Mock Test - 9 - Question 20

 Suggest a possible communication channel for the transmission of a message signal that has a bandwidth of 200kHz.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 20

The communication channel for the transmission of a message signal that has a bandwidth of 200kHz is FM radio. AM radio is the channel required for a message signal having a bandwidth of 5 MHz. Other options are not valid.

Physics: CUET Mock Test - 9 - Question 21

Identify the type of modulation where the modulating wave is superimposed on a high-frequency carrier wave.

Detailed Solution for Physics: CUET Mock Test - 9 - Question 21

In Amplitude modulation, the modulating wave is superimposed on a high-frequency carrier wave such that the amplitude of the modulated wave varies as the amplitude of the modulating wave. The frequency of the modulated wave is equal to the frequency of the carrier wave.

Physics: CUET Mock Test - 9 - Question 22

How can the noise be reduced in AM signal?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 22

Noise is a form of amplitude variations in the transmitted signal due to several factors such as the atmosphere, industries, etc. This noise can be reduced by increasing the deviation of the frequency of the signal.

Physics: CUET Mock Test - 9 - Question 23

Of the following which is the preferred modulation scheme for digital communication?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 23

Pulse code modulation is the one preferred for digital communication. Pulse codes are used to convert an analog signal to a digital one. It is the standard form of digital audio in computers, compact discs, etc.

Physics: CUET Mock Test - 9 - Question 24

What does AGC stand for?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 24

AGC stands for automatic gain control. Automatic gain control changes (AGC) the overall gain of a receiver automatically so that the strength of the received signal remains almost constant.

Physics: CUET Mock Test - 9 - Question 25

What is a rectifier used for?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 25

A rectifier is based on the fact that a forward bias p-n junction conducts and a reverse bias p-n junction does not conduct electricity. The rectifier is used to convert alternating current voltage (ac) to direct current voltage (dc).

Physics: CUET Mock Test - 9 - Question 26

What is the ripple factor for a half-wave rectifier?

Detailed Solution for Physics: CUET Mock Test - 9 - Question 26

For a half-wave rectifier,

r =1.21

Physics: CUET Mock Test - 9 - Question 27

Flow of charges in direction of electrons is called 

Detailed Solution for Physics: CUET Mock Test - 9 - Question 27

The flow of electrons is termed electronic current. Electrons flow from the negative terminal to the positive.

Physics: CUET Mock Test - 9 - Question 28

If the length of the filament of a heater is reduced by 10% the power of the heater will

Detailed Solution for Physics: CUET Mock Test - 9 - Question 28

Power P= V2/R​
If length reduced 10% then new resistance of filament will be R′.
R′=R−10% of R
R′=0.9R
Now new power of heater is P2​
P2​= V2​​/R′ = V2/0.9R​​=1.1P
% increase power=11%

Physics: CUET Mock Test - 9 - Question 29

Potentiometer is

Detailed Solution for Physics: CUET Mock Test - 9 - Question 29

Potentiometer is a long wire of uniform cross section made of manganin.

Physics: CUET Mock Test - 9 - Question 30

The instrument for the accurate measurement of the e.m.f of a cell is

Detailed Solution for Physics: CUET Mock Test - 9 - Question 30
  • Both potentiometer and voltmeter are devices to measure potential difference.
  • EMF is the terminal p.d between the electrodes of a cell in open circuit, i.e., when no current is drawn from it.
  • Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection.
  • So, accurate measurement of p.d is done using a potentiometer.
View more questions
8 docs|148 tests
Information about Physics: CUET Mock Test - 9 Page
In this test you can find the Exam questions for Physics: CUET Mock Test - 9 solved & explained in the simplest way possible. Besides giving Questions and answers for Physics: CUET Mock Test - 9, EduRev gives you an ample number of Online tests for practice
Download as PDF