Physics Class 12 Mini Mock - 2 - Free MCQ Practice Test with solutions,

As 
According to the question, v‘ = 2v and B' = B/2
∴ 

A small negatively charged body placed at X would be pushed to the right

The negative charge feels a force opposite to the direction of field.

 Other cases are not possible from properties of electric lines of force.

The energy stored in the capacitor

This energy will be converted into heat in the resistor

Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

Magnetic declination is the angle on the horizontal plane between magnetic north (the direction the north end of a compass needle points, corresponding to the direction of the Earth's magnetic field lines) and true north (the direction along a meridian towards the geographic North Pole).
 

From conservation of angular momentum,

from conservation of energy,

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.
 

In electromagnetism the magnetic susceptibility (Latin: susceptibilis, “receptive”; denoted X) is a measure of how much a material will become magnetized in an applied magnetic field. Mathematically, it is the ratio of magnetization M(magnetic moment per unit volume) to  the applied magnetizing field intensity H.

μ = μ₀μ_r. μ is the permeability of medium, μ₀ is permeability of free space, and μr is relative permeability of the medium.

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.
 

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.

When the object is placed at principal focus (F) of a concave mirror, a highly enlarged image is formed at infinity.

Energy stored in capacitor

From Ampere circuital law

The direction of field at the given point will be vertical up determined by the screw rule or right hand rule.

Step 1: Calculate the energy of the incident photon (h * nu)

The energy of a photon is given by:

where:

• h = Planck's constant = 6.63 x 10⁻³⁴ Js

• c = speed of light = 3 x 10⁸ m/s

Substitute the values:
E = (6.63 x 10⁻³⁴ * 3 x 10⁸) / (3 x 10⁻⁷) = 6.63 x 10⁻¹⁹ J

Convert this energy to electron volts (eV) using 1 eV = 1.6 x 10⁻¹⁹ J:
E = (6.63 x 10⁻¹⁹) / (1.6 x 10⁻¹⁹) = 4.14 eV

Step 2: Calculate the kinetic energy of the photoelectron

Using the photoelectric equation:
KE = h * nu - phi = 4.14 eV - 1 eV = 3.14 eV

Convert KE to joules:
KE = 3.14 x 1.6 x 10⁻¹⁹ = 5.024 x 10⁻¹⁹ J

Step 3: Determine the velocity of the photoelectron

The kinetic energy is related to velocity (v) by:
KE = (1/2) * m * v²

where:

• m = mass of an electron = 9.11 x 10⁻³¹ kg

Rearrange to solve for v:
v = sqrt((2 * KE) / m)

Substitute the values:
v = sqrt((2 * 5.024 x 10⁻¹⁹) / (9.11 x 10⁻³¹))
v = sqrt(1.1 x 10¹²) ≈ 1.05 x 10⁶ m/s

Final Answer:

The velocity of the emitted photoelectrons is approximately 1 x 10⁶ m/s, which corresponds to option:

d) 1 x 10⁶ m/s

Huygen's construction of wavefront does not apply to origin of spectra which is explained by quantum theory

By law of conservation of momentum 

–ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths 

When a magnet is moved towards the coil quickly, the rate of change of flux is larger than that if the magnetic field is moved slowly, thus larger emf is induced due to quick movement of the coil.

Given that, A charge q is placed at the centre of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0

Magnifying power of a compound- microscope is given by m=− (l​D​/f₀​f_e​)
where, l is length of tube
D is least distance of clear vision
f₀​ is focal length of objective
f_e​ is focal length of eyepiece
So, clearly, it can be seen that focal length of objective and eyepiece needs to be decreased so that magnifying power increases.
 

When a transistor is used in the common emitter mode as an amplifier, the base-emitter junction acts as input junction and collector-emitter junction acts as output junction. The input junction is made forward biased by applying forward voltage and output junction is made reverse biased by applying reverse voltage. The input signal is connected in series with the voltage applied to forward bias the base-emitter junction.

The resistance of 6Ω and 3Ω are in parallel in the given circuit, their equivalent resistance is

For maximum speed of the photo electrons (½)​mv2=Ephoton​−ϕ
where ϕ=0.5 eV is the work function of the metal.
Energy of photon E1​=1eV
∴  (½)​mv12​=1−0.5
We get  (½) ​mv12​=0.5 eV        ....(1)
Energy of photon E₂​=2.5eV
∴  (½) ​mv₂²​=2.5−0.5
We get  (½)​mv₂²​=2 eV        ....(2)
Ratio of maximum speeds ​v12​​/  v22=0.5/2​=1/4
We get   v₁/v₂​ ​​=1/2

An inductor stores energy in its magnetic field.

The polarization angle (also called Brewster's angle)is defined as the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.
Brewster's Law states that the tangent of the angle of polarization is numerically equal to the refractive index of the medium.
μ = tan i ...................... (1)
From Snell's law:
μ = sinr sini​ ....................(2)
Comparing (1) and (2)
cos i = sin r = cos (2π​ - r)
i.e. i = (2π​ - r)
So, i + r = 2π​
So, as seen from the diagram, if i + r = 90^o Then the angle between reflected ray and refracted ray will be 180^o - 90^o = 90^o