Physics: CUET Mock Test - 2 - CUET MCQ

A scalar quantity is a quantity with magnitude only but no direction. But a vector quantity possesses both magnitude and direction. An electric field has a very specific direction (away from a positive charge or towards a negative charge). Hence electric field is a vector quantity. Moreover, we have to use a vector addition for adding two electric fields.

Calculation:

The given circuit can be rearranged as

Where the resistance of each resistors are 2 ohms and voltage applied is 2 V.
We can see that arm resistances R₁ ,R₂,R3,R4

R₁R3 = R₂R₄

the above circuit is a wheatstone bridge.

The current flowing through the middle branch would be zero.

Net resistance "R" will be = (2 + 2) II (2 + 2)

R = 2 ohm.

The current from the battery would be

I = V/R

V = 2 Volts.

So, I = 2/2 = 1 A

I = 1 A.

The correct answer is option (4).

Calculation:
We know that the voltmeter is always connected in parallel.
The voltmeter resistance is given as 150 Ω and is connected across AB.
This means 150 Ω is connected in parallel with 100 Ω .
The resistance across AB will be a parallel combination of 150 and 100 Ω resistors.
The resistance across AB = (150) II (100)
= 150*100/(150 + 100)
= 60 Ω.
The net resistance of the circuit would be series combination of 60 and 100 Ω .
R = 60 + 100
= 160 Ω.
Current I = V/R
I = 40/160
= 0.25.
The voltage across AB will be the voltmeter reading which is equal to I*R.
Voltmeter reading = 0.25*60
= 15 Volts.
The correct answer is option (3)

Calculation:

In the given potentiometer experiment, the balancing length l is 1.6 m, and the total length of the wire AB is 2 m. The resistance per unit length of the wire AB is R = 10/2 = 5 Ω/m.

Let V be the emf of the standard cell to be determined. At the balancing point, the potential difference across the wire AB is equal to the emf of the standard cell, i.e., V = IR, where I is the current flowing through the wire AB.

The total resistance in the circuit is the sum of the resistance of the standard cell (0.2 Ω), the internal resistance of the cell in the potentiometer (2 Ω), and the resistance of the wire AB up to the balancing point (5 Ω/m x 1.6 m = 8 Ω).

The current I in the circuit can be calculated using the potential difference across the circuit, which is the sum of the emf of the cell in the potentiometer and the potential drop across the wire AB up to the balancing point. Thus, I = (2.4 V)/(2 Ω + 5 Ω/m x 1.6 m) = 0.24 A.

Substituting the value of I and R into the equation V = IR gives V = 0.24 A x (5 x 1.6 + 0.2) = 1.968 V.

The correct answer is option (2) i.e., nearly equal to 1.9 V.

Concept:

• The drift velocity of electrons refers to the average velocity at which free electrons drift in a conductor under the influence of an electric field.
• When a voltage is applied to a conductor, the electric field generated accelerates free electrons and they start to move in the direction of the electric field.
• However, due to the collisions with atoms and other electrons present in the conductor, the free electrons do not move with a constant velocity, but rather move with a small, average velocity in the direction of the electric field. This average velocity is called the drift velocity.
• The drift velocity of electrons in a conductor depends on various factors such as the applied voltage, the type of material of the conductor, the density of free electrons, and the temperature.
• However, in a typical metallic conductor at room temperature, the drift velocity of electrons is typically on the order of a few millimeters per second, which is quite slow compared to the speed of light or even the speed of sound.

The drift velocity of electrons is directly proportional to the voltage applied across a conductor.

The other factors mentioned in the options, such as temperature, length of the conductor, and area of cross section of conductor, can affect the resistance of the conductor, but they do not directly affect the drift velocity of electrons.

So, the correct answer to the question is (2)

Additional Information

Drift velocity of electrons is directly proportional to the electric field applied to the conductor.

This means that as the strength of the electric field increases, the drift velocity of the electrons in the conductor also increases. Mathematically, the relationship between the drift velocity and the electric field is described by the following equation:

vd = μE

where vd is the drift velocity,

E is the electric field strength, and

μ is the electron mobility, which is a material property that describes how easily electrons can move through a material.

The higher the electron mobility, the higher the drift velocity for a given electric field strength.

Electric field intensity is defined as the force on a unit positive charge kept in an electric field. Hence we can simply consider its dimension as (the dimension of force/the dimension of charge). The dimension of force is [MLT^-2] and the dimension of charge is [IT]. Therefore the dimension of field intensity is [M L T^-3 I^-1].

Torque = NIBA sinθ.
Torque = 50 × 20 × 0.2 × 200 × 10^-4 × sin (90)
Torque = 4Nm.

Explanation:for electromagnetic induction magnet and coil both may move but relative motion between must be present

The coil of a galvanometer is wound on a light metallic frame (such as aluminium). When the coil rotates in the magnetic field of the permanent magnet, eddy currents are induced in the metallic frame. These eddy currents create magnetic fields that oppose the motion of the coil (according to Lenz's Law), thus providing electromagnetic damping. This damping helps the coil to return to zero quickly and prevents oscillations.

In a series RLC circuit there becomes a frequency point were the inductive reactance of the inductor becomes equal in value to the capacitive reactance of the capacitor. In other words, XL = XC.
Series Resonance circuits are one of the most important circuits used electrical and electronic circuits.
 

In an RLC circuit, the voltage is always used as a reference and according to the phase of the voltage, the phase of the other parameters is decided.

The speed of electromagnetic waves in free space depends on two important properties: permeability and permittivity. In this context:

• μ₀ represents the permeability of free space, which indicates how well magnetic fields can pass through a vacuum.
• ε₀ signifies the permittivity of free space, reflecting how electric fields interact with the vacuum.

These properties are fundamental to understanding how electromagnetic waves propagate in free space.

All regions have the same speed.
The speed of a wave is not dependent upon its frequency and wavelength but rather upon the properties of the medium through which it travels.
 

Maxwell predicted that an accelerated charge produces a sinusoidal time varying magnetic field, which in turn produces a sinusoidal time varying electric field. These two fields are mutually perpendicular to each other and sources of each other. They (fields) form an electromagnetic wave i.e. the wave that is propagated by simultaneous periodic variations of electric field and magnetic field intensity.

Solution : 

The correct option is Option A.

Laws of refraction;-

The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.

For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.

When a ray of light passes through a glass slab, ∠i,∠r and the normal all lie in the same plane.

When a ray of light passes from one medium to another, here from air to glass or glass to air, the ratio sini / sinr = constant.

The relation between radius of curvature, image distance and object distance is given by
n₂/v-n₁/u=n₂-n₁
un₂-vn₂/vu=n₂-n₁/R
R=vu(n2-n2)/un₂-vn₂
=uv(2)/u(2)-v(1)
=[uv/2u-v)

For diffraction, the width of the slit must be less than the wavelength of the incident ray.

Diffraction of light takes place for all types of waves transverse or longitudinal.

In the photoelectric effect, the work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal.Electrons ejected from a sodium metal surface were measured as an electric current.The minimum energy required to eject an electron from the surface is called the photoelectric work function.
This energy (work function) is a measure of how firmly a particular metal holds its electrons. The work function is important in applications involving electron emission from metals, as in photoelectric devices and cathode-ray tubes.

A photon is massless, has no electric charge, and is a stable particle. In vacuum, a photon has two possible polarization states. The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavour quantum numbers) are zero.
 

Threshold frequency: - frequency of minimum energy required to remove electron
 
work function=3.3ev
f=3.3×1.6×10⁻¹⁹J​ /6.626×10⁻³⁴J−s
=0.8×10¹⁵Hz
=8×10¹⁴Hz

Solution:

• The energy of a photon can be calculated using the formula: E = hc / λ.
• Here, h is Planck's constant, which is 6.63 × 10^-34 J·s.
• The speed of light (c) is 3 × 10⁸ m/s.
• The wavelength (λ) is given as 5500 × 10^-10 m.

Substitute these values into the formula:

• E = [6.63 × 10^-34 J·s × 3 × 10⁸ m/s] / 5500 × 10^-10 m
• Calculating gives: E = [0.36 × 10^-26] / 10^-8
• This simplifies to: E = 3.6 × 10^-19 J

Convert energy from joules to electronvolts:

• 1 electronvolt (eV) = 1.6 × 10^-19 J
• Therefore, E = 3.6 × 10^-19 J / 1.6 × 10^-19 J/eV
• The result is approximately 2.26 eV.

In Thomson’s plum pudding model, the positive charge is spread out uniformly across the atom, so α-particles passing through would experience only very small, gentle deflections, leading to a small average angle of deflection.

In contrast, Rutherford’s nuclear model postulates a dense, positively charged nucleus. While most α-particles pass through undeflected, a small number undergo large-angle scattering, which increases the overall average deflection angle.

Therefore, the average angle of deflection predicted by Thomson’s model is less than that predicted by Rutherford’s model.

Binding energy per nucleon is the ratio of the binding energy of a nucleus to the number of the nucleons.

Binding energy per nucleon = (Total binding energy) / (Number of nucleon)

Measure of stability of the nucleus: Larger the binding energy per nucleon, the greater the work that must be done to remove the nucleon from the nucleus, the more stable the nucleus.

Zener Diodes can be used to produce a stabilised voltage output with low ripple under varying load current conditions. By passing a small current through the diode from a voltage source, via a suitable current limiting resistor (RS), the zener diode will conduct sufficient current to maintain a voltage drop of V_out.

In physics and in electronic engineering, dark current is the relatively small electric current that flows through photosensitive devices such as a photomultiplier tube, photodiode, or charge-coupled device even when no photons are entering the device; it consists of the charges generated in the detector when no outside radiation is entering the detector.

It is referred to as reverse bias leakage current in non-optical devices and is present in all diodes. Physically, dark current is due to the random generation of electrons and holes within the depletion region of the device.

Efficiency of half-wave and full wave rectifier is given by
η_h​=40.6% and η_f​=81.2%
So,  η_h/η_f​​​=40.6/81.2 ​=1/2​
∴f/n ​=2η_h​
Hence, the efficiency of a full wave rectifier is double of half wave-rectifier.

This energy is emitted in the form of heat and light. The electrons dissipate energy in the form of heat for silicon and germanium diodes but in gallium arsenide phosphide (GaAsP) and gallium phosphide (GaP) semiconductors, the electrons dissipate energy by emitting photons.

The forces acting on an electron in uniform electric and magnetic fields pointing in the same direction are explained below:

• Magnetic Force: Since the electron is moving parallel to the fields, the magnetic force (v × B) is zero.
• Electric Force: The electric field exerts a force (F_e = -E) on the electron, acting opposite to its direction of motion.

As a result:

• The electron experiences a backward force due to the electric field.
• This force causes the electron to decelerate, reducing its velocity.