Physics Mock Test - 5

The correct option is 'a current device'.

Key Points

• A junction transistor is formed by sandwiching a thin layer of P-type semiconductor between two N-type semiconductors or by sandwiching a thin layer of n-type semiconductor between two P-type semiconductors.
• NPN transistor: It is formed by sandwiching a thin layer of P-type semiconductor between two N-type semiconductors

• PNP transistor: It is formed by sandwiching a thin layer of N-type semiconductor between two P-type semiconductor

Explanation:

The transistors are used as a current amplifier. 

• Current amplification factor (β): The ratio of the change in the value of the transistor's collector current to the change in the value of the transistor's base current in a transistor is called the current amplification factor.

\(β = \frac{{{\Delta I_c}}}{{{\Delta I_b}}}\)

Additional Information

Transistor consists of three main regions i.e. Emitter, Base and Collector.

Emitter (E) -

• It provides majority charge carriers by which current flows in the transistor.
• Therefore the emitter semiconductor is heavily doped. Hence option 2 is correct.

Base (B) –

• The based region is thin and lightly doped.
• It provides proper interaction between emitter and collector

Collector (C) –

• The size of the collector region is larger than the two other regions and it is moderately dopped.
• The main purpose of the collector is to collect the majority of charge carriers from the emitter.

CONCEPT:

Rayleigh's law of scattering:

• According to Rayleigh's law of scattering, the intensity of light of wavelength λ present in the scattered light is inversely proportional to the fourth power of λ, provided the size of the scattering particles are much smaller than λ. Mathematically,

\(I \propto \frac{1}{\lambda }\)

• Thus the scattered intensity is maximum for shorter wavelength.

EXPLANATION:

• The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light.
• These are more effective in scattering light of shorter wavelengths at the blue end than the light of longer wavelengths at the red end.
• The red light has a wavelength of about 1.8 times greater than blue light. Thus, when sunlight passes through the atmosphere, the fine particles in the air scatter the blue colour (shorter wavelengths) more strongly than red. Therefore option 1 is correct. 
• The sky appears blue due to the scattering of light.

Important Points

• If the earth had no atmosphere, there would not have been any scattering.
• Then, the sky would have looked dark. The sky appears dark to passengers flying at very high altitudes, as scattering is not prominent at such heights.

CONCEPT:

• Young's Double Slit Experiment (YDSE): Monochromatic light (single wavelength) falls on two narrow slits S1 and S2 which are very close together act as two coherent sources when waves coming from two coherent sources, (S1, S2) superimposes on each other, an interference pattern is obtained on the screen.
  • In Young's Double Slit Experiment alternate bright and dark bands obtained on the screen. These bands are called Fringes.

EXPLANATION:

• In Young's double-slit experiment, if two independent sources are used then a permanent or sustained pattern is not obtained and the minimum and maximum intensities are not fixed. So option 3 is correct.
• This is because the two independent sources do not emit continuous waves that have the same phase or constant phase difference.

CONCEPT:

The force between multiple charges:

• Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. 
• The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
• The principle of superposition says that in a system of charges q1, q2, q3,..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges  q2, q3,..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n.

\(⇒ \vec{F_1}=\vec{F_{12}}+\vec{F_{13}}+...+\vec{F_{1n}}\)

EXPLANATION:

• When four similar charges are placed on the vertex of a square then all the charges will repel each other.
• The force on any of the charge due to the other three charges are shown in the figure,

Figure: 1

Where F_BA = force on the charge at B due to charge at A, FBC = force on the charge at B due to charge at C, and FBD = force on the charge at B due to charge at D

Let the side of the square is a.

So the force F_BA and F_BC is given as,

\(\Rightarrow F_{BA}=F_{BC}=\frac{k\times q\times q}{a^2}\)

\(\Rightarrow F_{BA}=F_{BC}=\frac{kq^2}{a^2}=F\)       ----(1)

The force FBA and FBC are perpendicular to each other so the resultant force of these two charges is given as:

\(\Rightarrow F_1=\sqrt{F_{BA}^2+F_{BC}^2}\)

\(\Rightarrow F_1=\sqrt{F^2+F^2}\)

\(\Rightarrow F_1=\sqrt{2}F\)        ----(2)

Figure: 2

The diagonal of the square is given as,

\(\Rightarrow BD=\sqrt{a^2+a^2}\)

\(\Rightarrow BD=\sqrt{2}a\)       ----(3)

So the force F_BD is given as,

\(\Rightarrow F_{BD}=\frac{k\times q\times q}{(\sqrt2a)^2}\)

\(\Rightarrow F_{BD}=\frac{kq^2}{2a^2}\)       ----(4)

By equation 1 and equation 4,

\(\Rightarrow F_{BD}=\frac{F}{2}\)      -----(5)

• From figure 2 it is clear that F₁ and F_BD both forces are acting diagonally and opposite to each other.
• Since the force F₁ is more than F_BD, so the resultant force on the charge will be in the direction of F₁.
• The direction of the force F₁ is diagonally inward so the charge at point B will tend to move diagonally inward. Hence, option 3 is correct.

CONCEPT:

Photon:

• According to Einstein's theory, the light propagates in the bundles of energy, each bundle is being called a photon.
• ​The rest mass of the photon is zero. But its effective mass is given as,
• Photon exerts pressure on the surface.
• The energy of a photon is given as,

⇒​ \( E=hν=\frac{hc}{λ}\)

Where h = 6.63 × 10-34 J-sec = Planck's constant, c = 3 × 108 m/sec = speed of light, ν = frequency, and λ = wavelength

Particle nature of light: The Photon

• The photoelectric effect gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy.
• Einstein arrived at the important result, that the light quantum can also be associated with momentum.
• The momentum of the light quantum is given as,

⇒ \( P=\frac{hν}{c}=\frac{h}{λ}\)

• A definite value of energy, as well as momentum, is a strong sign that the light quantum can be associated with a particle. That particle was later named photon.
• The particle-like behavior of light was further confirmed, in 1924, by the experiment of A.H. Compton (1892-1962) on the scattering of X-rays from electrons.
• We can summarise the photon picture of electromagnetic radiation as follows:
  1. In the interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
  2. Each photon has energy E (= hν) and momentum p (= hν/c), and speed c, the speed of light.
  3. All photons of light of a particular frequency or wavelength have the same energy and momentum, whatever the intensity of the radiation may be. By increasing the intensity of light of a given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of the intensity of radiation.
  4. Photons are electrically neutral and are not deflected by electric and magnetic fields.
  5. In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.

CALCULATION:

Given λ_A : λB = 1 : 2

⇒ \(\frac{\lambda_A}{\lambda_B}=\frac{1}{2}\)     -----(1)

• We know that the momentum of a photon is given as,

⇒ \(P=\frac{h}{λ}\)     -----(2)

Where h = 6.63 × 10-34 J-sec = Planck's constant, and λ = wavelength

• By equation 1 the momentum of the monochromatic light A is given as,

⇒ \(P_A=\frac{h}{λ_A}\)     -----(3)

• By equation 1 the momentum of the monochromatic light B is given as,

⇒ \( P_B=\frac{h}{λ_B}\)     -----(4)

By equation 3 and equation 4,

⇒ \(\frac{P_A}{P_B}=\frac{h}{λ_A}\times\frac{\lambda_B}{h}\)

⇒ \(\frac{P_A}{P_B}=\frac{\lambda_B}{λ_A}\)

⇒ \(\frac{P_A}{P_B}=\frac{2}{1}\)

• Hence, option 1 is correct.

CONCEPT:

• The material which is not a good conductor or a good insulator is called as a semiconductor.
• For example: Silicon
• The charge carriers which are present in more quantity in a semiconductor compared to other particles are called the majority charge carrier.

• The impurity atoms added are called dopants and semiconductors doped with the impurity atoms are called extrinsic or doped semiconductors.
• The pure form of semiconductor is called intrinsic semiconductor.

EXPLANATION:

• When an electron is promoted from the valence band to the conduction band of an intrinsic semiconductor then there is one hole created in the valence band. So number of electrons is equal to number of holes. So option 1 is correct.

CONCEPT: 

• Drift velocity: In a material, The average velocity attained by charged particles due to an electric field is called drift velocity.

Drift velocity of the electrons is calculated by:

\(v_d=\frac{I}{neA}\)

where v is the drift velocity, I is the current in the wire, n is the number density of free electrons in the wire, A is the cross-sectional area of the wire, and e is the charge on one electron.

• Mobility: the drift velocity per unit electric field is called mobility.

μ = vd/E

where vd is the drift velocity, E is the electric field and μ is mobility.

• If q is the electric charge, m is the mass of the charge, and τ is the relaxation time then mobility

μ = qτ/m

• From the above three equations, the drift velocity v_d is directly proportional in relaxation time τ. (V_d α τ)

EXPLANATION:

• Thermal velocity will increase: It is given that the temperatures of metal wire increases. So, the KE of electron also increases, so the thermal velocity will increase.
• The drift velocity will decrease: Due to the increase in the number of collisions, the relaxation time τ will decrease.
  • Since we know that drift velocity vd is directly proportional in relaxation time τ. So drift velocity will also decrease.
• So the correct answer is option 4.

The correct option is 4.

CONCEPT:

• Refraction of Light: The bending of the ray of light passing from one medium to the other medium is called refraction.

• The refraction of light takes place on going from one medium to another because the speed of light is different in the two media.
• The greater the difference in the speeds of light in the two media, the greater will be the amount of refraction.
• A medium in which the speed of light is more is known as an optically rarer medium and a medium in which the speed of light is less is known as an optically denser medium.

EXPLANATION:

\({\rm{Refractive\;index}}\left( \mu \right) = {\rm{}}\frac{{Speed\;of\;light\;in\;vaccum\left( C \right)}}{{Speed\;of\;light\;in\;a\;medium\left( v \right)}}\)

• As the refractive index of the medium for light rays depends on the speed of light in the medium.
• As the speed of light in glass is less than that in air. So the refractive index of glass is more than that of air.
• When light rays travel from air to glass then they will get refracted due to this difference in the refractive index of the medium. Hence option 4 is correct.
• The refraction of light has no relation with the density of the material. So option 1 is wrong.
• Options 2 and 3 are not correct because there is refraction of light not the reflection or absorption of light rays.

The correct answer is option 4) i.e. All of the above

CONCEPT:

• Electric charge is a fundamental property of matter that causes the matter to experience a force when kept in electric and magnetic fields.
  • There are two types of charges possessed by matter - negative charge and positive charge.
  • The matter that is neither positive nor negative is said to be neutral.
  • The electric charge of the matter comes from the charge carried by electrons and protons at the atomic level.
  • The SI unit of electric charge is coulomb (C).
• Coulomb's law: It states that the magnitude of the electrostatic force F between two point charges q₁ and q₂ is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance r between them. 

It is represented mathematically by the equation:

\(F = \frac{kq_1 q_2}{r^2}\)

EXPLANATION:

• It is a known fact the opposite polarities attract and like polarities repel. So, unlike charges attract and like charges repel each other.
• A charge can neither be created nor be destroyed, it can only be transferred from one system to another. Thus, the charge is conserved.
• The interaction between charges is derived from Coulomb's law and hence it is governed by this law.

Therefore, all the given statements are correct.

CONCEPT:

• Equipotential surface: Any surface that has the same electric potential at every point on it is called an equipotential surface.
• Relationship between the electric field (E), an electric potential (V) and distance (r) is given by -

\(dE = - \frac{{dV}}{{dr}}\)

• The electric field is a derivative of potential difference.
• The negative sign shows that the direction of E is opposite to the direction of dv i.e., dv decrease along the direction of E.

EXPLANATION:

• Spherical equipotential surfaces are formed when the source is a field is a point charge. Therefore it is incorrect to say that equipotential surface is always spherical. Hence, option 2 is incorrect. 
• The electric field is always normal to the equipotential surface at every point.
• If the field were not normal to the equipotential surface, it would have a non-zero component along the surface.
• So to move a test charge against this component, work would have to be done.
• But there is no potential difference between any two points on an equipotential surface and consequently, no work is required to move a test charge on the surface. Therefore option 4 is correct.
• Hence the electric field must be normal to the equipotential surface at every point. Therefore option 3 is correct

CONCEPT:

• When the monochromatic light ray fall on single slit then it gets diffracted from the slit and form bright and dark band on the screen.
• The bright pattern is also called maxima and dark band is called minima.
• At maxima the intensity is maximum and at minima the intensity of light is minimum.

The width of the maxima is given by:

\(Sin\theta = \frac{{n\lambda }}{a}\)

Where λ is wavelength of the light, n is integer value, a is slit width and D is distance of screen from the slit.

Here θ is very small.

So Sinθ is approximately equal to θ.

CALCULATION:

Given that:

Wavelength (λ) = 600 nm = 600 × 10^-9 m

Slit-width (a) = 0.2 mm = 0.2 × 10^-3 m

\(Sin\theta = \frac{{n\lambda }}{a}\)

For principal/central maxima, n = 1

\(Sin\theta = \theta = \frac{\lambda }{a}\)

CONCEPT:

Kirchoff’s first law:

• This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.

​

• In a circuit, at any junction, the sum of the currents entering the junction must be equal the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
• This law is simply a statement of “conservation of charge” as if the current reaching a junction is not equal to the current leaving the junction, the charge will not be conserved.

Kirchoff’s second law:

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in a complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
• If there are n meshes in a circuit, the number of independent equations in accordance with the loop rule will be (n - 1).

CALCULATION:

From the given figure it is clear that

⇒ I₁ + I₂ = 6A       ------(1)

Applying Kirchhoff’s rule to closed loop PQRP

⇒ -2I₁ - 2I₁ + 2I₂ = 0

⇒ -4I₁ + 2I₂ = 0

⇒ 2I₁ - I₂ = 0 -----(2)

On adding Equations (i) and (ii), we get,

⇒ 3I₁ = 6A

⇒ I₁ = 2A

On subturting I1 = 2A in equation 1, we get

⇒ I₂ = 4A

• Then option 1 is the right answer

CONCEPT:

Mutual-inductance

• When two coils are brought in proximity with each other the magnetic field in one of the coils tends to link with the other. If this magnetic field of the first coil is changed then the magnetic flux associated with the second coil changes and this leads to the generation of voltage in the second coil.
• This property of a coil that affects or changes the current and voltage in a secondary coil is called mutual inductance.
• The SI unit of the mutual inductance is Henry.
• Let two coils of the number of turns N1 and N2 are placed close to each other. Then the mutual inductance of coil 1 with respect to coil 2 is given as,

\(⇒ M_{12}=N_{1}\frac{ϕ_{1}}{I_{2}}\)

Where N1 = number of turns in coil 1, ϕ1 = flux linked with coil 1, and I2 current in coil 2

Mutual inductance for two co-axial solenoids:

• Let two long solenoids S1 and S2 of equal length are placed co-axially as shown in the figure.
• The solenoid S1 is placed inside the solenoid S2.
• The mutual inductance of both the solenoids will be equal and it is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)

For medium of relative permeability μr,

\(⇒ M_{12} = M_{21} = \mu_rμ_on_1n_2πr_{1}^{2}l\)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

EXPLANATION:

• We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.
• The mutual inductance of both the solenoids is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)     -----(1)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

When the radius of the inner solenoid is halved, (\(r_1^{'}=\frac{r_1}{2}\)), the mutual inductance is given as,

\(⇒ M_{12}^{'} = M_{21}^{'} = μ_on_1n_2π\left ( \frac{r_{1}}{2} \right )^{2}l\)

\(⇒ M_{12}^{'} = M_{21}^{'} = \frac{μ_on_1n_2πr_{1}^{2}l}{4}\)     -----(2)

By equation 1 and equation 2,

\(⇒ M_{12}^{'} = \frac{M_{12}}{4}\)

\(⇒ M_{21}^{'} = \frac{M_{21}}{4}\)

• Hence, option 2 is correct.

CONCEPT:

• The root mean square current /voltage (rms) of an AC circuit is the effective current/voltage of that circuit.
• The maximum value of the potential in an Ac circuit is called the peak value of voltage.
• The average value of the potential and current of an AC circuit for a cycle is called average potential and average current.

\(Average\;current\;over\;half\;cycle\;\left( {{I_{avg}}} \right) = \frac{{2\;I}}{\pi }\)

Average current over complete cycle = 0

\(rms\;current\;\left( {{I_{rms}}} \right) = \frac{I}{{\sqrt 2 }}\)

Where I is maximum/peak current in the circuit.

CALCULATION:

Given that:

Rms current (I_rms) = 6 A

Use the formula:

\(rms\;current\;\left( {{I_{rms}}} \right) = \frac{I}{{\sqrt 2 }}=6\)

Peak current (I) = 6√2 A

\(Average\;current\;over\;half\;cycle\;\left( {{I_{avg}}} \right) = \frac{{2\;I}}{\pi } = \frac{{12\sqrt 2 }}{\pi }\;A\)

Average current over complete cycle = 0

CONCEPT:

Amplitude modulation:

• In amplitude modulation, the amplitude of the carrier is varied in accordance with the information signal.
• Here we explain the amplitude modulation process using a sinusoidal signal as the modulating signal.
• Let c(t) = Acsin(ωct) represent carrier wave and m(t) = Amsin(ωmt) represent the message or the modulating signal.
• So the modulated signal can be represented as,

⇒ cm(t) = [Ac + Am.sin(ωmt)].sin(ωct)

⇒ cm(t) = [Ac + μAc.sin(ωmt)].sin(ωct)

Where Ac = amplitude of the carrier wave, Am = amplitude of the modulating signal, ωc = angular frequency of the carrier wave, ωm = angular frequency of the modulating signal, and μ = modulation index

• The modulation index is given as,

\(⇒μ=\frac{A_m}{A_c}\)

• In practice, μ is kept ≤ q1 to avoid distortion.
• Using the trignomatric relation 2sinA.sinB = cos(A - B) - cos(A + B), we can represent the modulated signal as,

\(⇒ c_m(t)=A_csin(ω_ct)+\frac{μ A_c}{2}cos(ω_c-ω_m)t-\frac{μ A_c}{2}cos(ω_c+ω_m)t\)

• Here (ωc - ωm) and (ωc + ωm) are respectively called the lower side and upper side frequencies.
• The modulated signal now consists of the carrier wave of frequency ωc plus two sinusoidal waves, each slightly different from ωc, known as sidebands.

• As long as the broadcast frequencies (carrier waves) are sufficiently spaced out so that sidebands do not overlap, different stations can operate without interfering with each other.

CALCULATION:

Given fc + fm = 1050 kHz and fc - fm = 950 kHz

Where fc = frequency of the carrier wave and fm = frequency of the modulating signal

• We know that that (fc + fm) and (fc - fm) are the sidebands of a modulated signal.
• Therefore by adding the two sidebands,

⇒ (fc + fm) - (fc - fm) = (1050 - 950) kHz

⇒ 2fm = 100 kHz

⇒ fm = 50 kHz

• Hence, option 2 is correct.

According to the question,

the Zener diode with a resistance ‘R_s’ across the supply ’30 V’ is given below.

Here given reference voltage is given as 10 V

 ∴ V_z = 10 V

Apply KVL in the first loop,

30 – I R_s – 10 = 0

\({R_s} = \frac{{30 - 10}}{{20 \times {{10}^{ - 3}}}} = 1\;k{\rm{\Omega }}\)

Now when supply drops to 20 V from 30 V, then Zener current (I) will be:

∴ By applying KVL, we get

20 – IR_s – 10 = 0

 IR_s = 10

\( I = \frac{{10}}{{1000}} = 10\;mA\)

CONCEPT:

• Power of Lens: The inverse of the focal length is known as the power of the lens.
  • It shows the bending strength for the light ray of the lens.
  • The unit of power of a lens is Dioptre when the focal length of the lens is taken in meter (m).

\(P = \frac{1}{f}\)

where P is the power of the lens and f is the focal length of the lens.

• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • The focal length of the concave lens is negative.
  • It has a virtual focus from the diverging rays of light that seem to converge.
• Convex lens: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.
  • The focal length of a convex lens is positive.

CALCULATION:

Given that P = - 2 D 

Focal length (f) = 1/P = 1/(-20) = - 1/2 = - 0.5 m = - 50 cm

The focal length of a concave lens is negative. So it is a concave lens.

So the correct answer is option 4.

CONCEPT:

Electric dipole

• When two equal and opposite charges are placed at a very small distance to each other then this arrangement is called an electric dipole. 
• The electric dipole moment is defined as the product of the magnitude of one charge and the distance between the charges in an electric dipole.

⇒ P = q × 2r    

Where 2r = distance between the two charges

• Work done required to rotate an electric dipole in an external uniform electric field from the angle θ1 to θ2 is given as,

⇒ W = pE(cosθ1 - cosθ2)

Where θ1 and θ2 = initial and the final angle between the dipole and the electric field

EXPLANATION:

• From the above, it is clear that work done required to rotate an electric dipole in an external uniform electric field from the angle θ1 to θ2 is given as,

⇒ W = pE(cosθ1 - cosθ2)

• Hence, option 2 is correct.

The Correct Answer is Tyndall Effect.

• The scattering of light by very small particles suspended in a gas or liquid is known as the 'Tyndall Effect.
• Some of the examples of the Tyndall Effect are - the blue colour of smoke coming out from the engine, the blue colour of the sky, etc. 

CONCEPT:

Self-Induction:

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –

\(L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, the resistance of the coil (R) = 4Ω, and A = Area of the solenoid.

Induced e.m.f can be given as,

\(⇒ e =- L\frac{{di}}{{dt}}\)

CALCULATION:

Given -

di/dt = -2 Amp/sec and L = 5 H (Since the current decreases at a rate of 2 Amp/sec).

The emf induced across the inductor is,

\(⇒ e =- L\frac{{di}}{{dt}}=-(5\times (-2) )=10\, V\)

The correct answer is option 3) i.e. same

Concept:

Electric charge is a property of particles by which they have a tendency to attract or repel each other without touching. 

Electric force: Particles possessing opposite charges attract each other and particles possessing like charges repel each other. This force of attraction or repulsion is called electric force.

The relationship between the electric charges and electric force is given by Coulomb's law.

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the distance between the two objects.

\(F = \frac{kq_1 q_2}{r^2}\)

​Where F is the electric force acting between the two charges

q1 and q2 are the two charges

r is the centre to centre distance between the two objects and

k is the proportionality constant known as the Coulomb's law constant and is equal to 9 \(\times \)109 Nm2/C2.

Conclusion:

According to Coulomb's law, the force acting between two charges is only dependent on these two charges and is independent of any charge that is near it.

In the given question, F₁₂ is the force due to q₁ on q₂.

This would remain the same always as \(F = \frac{kq_1 q_2}{r^2}\)

When a third charge is brought near them, the net force on q₁ and q₂ will change as it will now depend on the force exerted by q₃.

Common collector transistor configurations have the highest input impedance

The given circuit diagram is a common-collector amplifier, i.e.

The various characteristics of different configuration are shown 

Concept:

The magnitude of the internal polarization (J) divided by the strength of the external field (B) is called Magnetic susceptibility. 

It is denoted by the Greek letter chi (χ).

Based on magnetic property materials are of 4 types:

1. Diamagnetic
2. Paramagnetic
3. Superparamagnetic
4. Ferromagnetic

Explanation:

The susceptibility of ferromagnetic material is Large and positive. Ferromagnetic materials follow the Curie – Weiss law. When ferromagnetic substance heated above Curie temperature (T_c) 

\(i.e.\;χ \propto \frac{1}{{T - {T_C}}}\)

Below table shows χ for different materials:

So, the answer will be option 1.

CONCEPT:

Photoelectric Effect:

• When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.​
• Einstein’s photoelectric equation:

⇒ KEmax = hν - ϕo

Where h = Planck's constant, ν = frequency of incident radiation, ϕo = work function, and KE = maximum kinetic energy of electrons.​

EXPLANATION:

CONCEPT:

• Magnetic force on a current-carrying conductor placed in a field is given by

\(F = I (\vec{L}\times \vec{B})\)

where i is the current in the conductor, L is the length of the conductor, B is the magnetic field in which the conductor is placed.

• The direction of this magnetic force can be found using the vector formula.

EXPLANATION:

• Magnetic force on a current-carrying conductor placed in a field is given by

\(F = I (\vec{L}\times \vec{B})\)

So F = BiL

Hence the correct answer is option 1.

Concept:

Connected in series: When n capacitors C1, C2, C3, ... Cn are connected in series, the net capacitance (Cs) is given by:

\(⇒ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}+ \frac{1}{C_3} + ... \frac{1}{C_n}\)

Connected in parallel: When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by:

⇒ Cp = C1 + C2  + C3 +...  Cn

Calculation:

C₇, C₉ and C₈ are in series, 

12 μF, 12 μF, 12 μF 

Equivalent capacitance = (1/12) + (1/12) + (1/12) = 4 μF

4 μF is in parallel with 8 μF 

Equivalent capacitance = (4 + 8) μF = 12 μF

C₄, C₅ are in parallel with 12 μF

Equivalent = 4 μF

C₃ is in series with 8 μF

Total equivalent = 12 μF

C₁ and C₂ are parallel to 12 μF

Hence equivalent = 4 μF

• A TV can be operated using a remote control device.
• Infra-red rays are emitted by the photodiode in the remote control to transmit instructions.
• Working principle of TV remote is based on Infrared waves, as a medium of communication.
• The wavelength of infrared waves ranges from 700 nm to 1 mm.
• Changing a television channel by infrared remote control is an example of point - to - point connection. 

The correct answer is 2.

Key Points

• Beta Decay is a type of radioactive decay in which a proton is transformed into a neutron or vice versa inside the nucleus of the radioactive sample.
• While doing so, the nucleus emits a beta particle which can either be an electron or positron. Hence, option 2 is correct.
• The beta decay occurs via the weak interaction.
• Beta particle (β) are high energy, high-speed electrons or positrons that are ejected from the nucleus during beta-decay.

Additional Information

• Alpha Decay – The radioactive decay in which the ejected particle is an alpha particle.
• Gamma Decay – A radioactive nucleus first decays by the emission of an α or β particle.

• The correct answer is option 4) i.e. perpendicular to the direction of polarization and parallel to the direction of propagation

CONCEPT:

• A wave is an oscillation that carries energy from one place to another without transporting matter.
• Light is a combination of electric fields and magnetic fields that are perpendicular to each other.
  • So, the two perpendicular planes are occupied by these fields. The electric and magnetic vibrations can simultaneously occur in a number of perpendicular planes. 
  • Therefore, light is an electromagnetic wave. 
  • A wave that oscillates in many numbers of planes is called an unpolarized wave.
  • Using devices called polarizers, light can be made to vibrate along a single plane. Such light waves are called polarized light.
• Linear polarization is when the oscillation of a wave is constrained to a single plane. This is the plane of polarization.
• Circular polarization is when two linear components of a wave oscillate perpendicular to each other such that their amplitudes are equal.
• Elliptical polarization is the same as circular polarization except that the amplitudes and phase differences are not the same.

EXPLANATION:

• The plane in which the wave oscillates is known as the plane of vibration.
• The plane which is perpendicular to the plane of vibration and which is along the direction of propagation of the polarised light is known as the plane of polarisation.
• Therefore, in linear polarization, the plane of polarization is perpendicular to the direction of polarization and parallel to the direction of propagation.

CONCEPT:

Polarizing angle: The angle of incidence at which a beam of unpolarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarizing or Brewster angle. It is denoted by ip.

Brewster's law: It states that when a ray is passed through some transparent medium having refractive index μ at any particular angle of incidence, the reflected ray is completely polarized; and the angle between reflected and refracted ray is 900.

​

 μ = tan θB 

Where μ = refractive index and θB is Brewster's angle or polarizing angle (ip).​

​CALCULATION:

Given that:

The angle of incidence (i_p) = 60°

From Brewster’s law, the polarization angle,

\({i_p} = {\tan ^{ - 1}}μ \)

\({\tan ^{ - 1}}1.732\; = 60^\circ\)

So refractive index (μ) = 1.73

Hence option 3 is correct.

CONCEPT:

• Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ0 times the magnitude of charge enclosed.

\(i.e.~{{\mathbf{\Phi }}_{net}}=\frac{\left( {{Q}_{in}} \right)}{{{\epsilon }_{0}}}\)

\(i.e.~\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Where, Φ = electric flux, Qin = charge enclosed the sphere, ϵ0 = permittivity of space (8.85 × 10-12 C2/Nm2), dS = surface area

EXPLANATION:

• If 'φ' is Electric flux through a closed surface 'S', 'q' is total charge enclosed by 'S' and є_o is the permittivity of free space, then these three are related by Gauss' Law formula, φ = q/єo

So option 2 is correct.

CONCEPT:

• Electromagnetic spectrum: It is a collection of a range of different waves in sequential order from radio to gamma electromagnetic waves.
  • Frequency (ν) = speed of light (c)/wavelength (λ)
  • Radio waves: The lowest frequency portion comes in radio waves generally, have wavelengths range between 1 mm to 100 km or frequencies between 300 GHz to 3 kHz.
  • Microwaves: They comes in the range of 1 mm to 1 meter and of the frequency range between 300 MHz to 300 GHz.
    • Microwaves are generally used in Radars, Transmission towers, Microwave Ovens etc.
  • Visible Light: A portion in the spectrum of electromagnetic waves that is visible to the human eye, ranging roughly between 390 to 750 nm.
  • X–rays are electromagnetic waves not mechanical wave because they are diffracted by the crystal.

According to the above table the order of wavelength will be:

EXPLANATION:

CONCEPT:

• Convex Mirror: Convex mirror is that mirror whose reflecting surface is away from the center of the curvature.
• A convex mirror is also known as a diverging mirror.

• In a convex mirror, object distance (u) is on the left side of the mirror. Therefore object distance (u) is negative.
• Image distance (v) and focal length (f) are on the right side of the convex mirror. Therefore image distance (v) and focal length (f) will be positive. 

EXPLANATION:

Given - Object distance (u)= -2f

• By using the mirror formula 

\(\Rightarrow \frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)

\(\Rightarrow \frac{1}{v} = \frac{1}{f}- (-\frac{1}{2f} )=\frac{3}{2f}\)

\(\Rightarrow v =\frac{2f}{3}\)

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite signs and separated by an insulating medium.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.

​Capacitance:

• The property to store energy in form of charge is called capacitance.
• The unit of capacitance is Farad.
• The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q =  CV

Where C = capacitance

Energy stored in the capacitor:

• If a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,

​\(\Rightarrow U=\frac{1}{2}CV^2=\frac{Q^2}{2C}\)

CALCULATION:

Given V_A = 120 V, VB = 100 V, and U_A = U_B

• We know that if a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,

​\(\Rightarrow U=\frac{1}{2}CV^2\)       ----(1)

By equation 1 for the capacitor A,

\(\Rightarrow U_A=\frac{1}{2}C_AV_A^2\)       ----(2)

By equation 1 for the capacitor B,

\(\Rightarrow U_B=\frac{1}{2}C_BV_B^2\)       ----(3)

By equation 2 and equation 3,

∵ U_A = U_B

\(\Rightarrow \frac{1}{2}C_AV_A^2=\frac{1}{2}C_BV_B^2\)

\(\Rightarrow \frac{C_A}{C_B}=\frac{V_B^2}{V_A^2}\)

\(\Rightarrow \frac{C_A}{C_B}=\frac{100^2}{120^2}\)

\(\Rightarrow \frac{C_A}{C_B}=\frac{25}{36}\)

Hence option 2 is correct.

Concept:

Rectifier

It is a circuit that is used to convert AC supply voltage to and Pulsating DC output voltage.

There are two types of rectifier circuits.

1) Half wave rectifier

• It is a circuit that allows the one-half cycle of the AC supply voltage blocking the other half cycle.
• This operation can be achieved by using a diode at the output port of the transformer.

Positive half wave rectifier

• In a positive half-wave rectifier diode D will be forward biased for a positive half cycle of an input AC supply voltage.
• Diode D will be off during the negative half cycle for an input AC supply voltage.
• As we can see from the above figure the answer will be a positive half-wave rectifier.

Important Points

Negative half wave rectifier

• In a negative half-wave rectifier diode D will be forward biased for a negative half cycle of an input AC supply voltage.
• Diode D will be off during the positive half cycle for an input AC supply voltage.

2) Full-wave rectifier

​Center tape type

• In a Center tape type full-wave rectifier, the input is fed through the center-tapped transformer, and output is taken from the load resistance R_L.
• The diode D₁ will be forward biased during a positive half cycle of input.
• The diode D₂ will be forward biased during a negative half cycle of input.
• This will produce a  continuous pulsating DC output voltage.
• PIV of the center tape type is 2V_peak.

Bridge type

• In a bridge rectifier, we use four diodes to increase the PIV rating of the circuit.
• The diode D₁ and D₃  are forward biased during a positive half cycle of input.
• The diodes D₂ and D₄  are forward biased during a negative half cycle of input.
• This will produce a  continuous pulsating DC output voltage.
• PIV of the center tape type is V_peak.

CONCEPT:

Meter Bridge:

• A meter bridge also called a slide wire bridge is an instrument that works on the principle of a Wheatstone bridge.
  • A meter bridge is used in finding the unknown resistance of a conductor as that of in a Wheatstone bridge.
• In a meter bridge, a wire having a length of a uniform cross-section of about 1m is used.
• One known resistance and an unknown resistance are connected as shown in the figure.
• The one part of the galvanometer is connected in between the known and unknown resistances, whereas the other part of the wire is used to find the null point where the galvanometer is not showing any deflection.
• At this point, the bridge is said to be balanced.

​

EXPLANATION:

• To obtain maximum sensitivity and accuracy in a meter bridge, it is advised to obtain a null point at the middle of the wire.
  • In the middle of the wire ratio arms are nearly equal in resistance. 
• Wheatstone bridge is most sensitive when the resistance of its four-arm is of the same order. Since the Meter bridge works on the principle of Wheatstone bridge, balance point or null point is obtained at the middle of the Meter bridge wire.
• The end resistance of the copper strips also produces the least effect. Hence option 3) is correct.

Additional Information

Wheatstone bridge principle:

• It states that if four resistances R1, R2, R3, and Rx are arranged in form of a bridge with a cell of EMF E connected between A and C and a galvanometer between B and D and on closing the key if the galvanometer shows no deflection, then the bridge is balanced.
• In that case: R1R2=R3Rx" id="MathJax-Element-3-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">R1R2=R3Rx$\frac{R_1}{R{}_2}=\frac{R{}_3}{R_x}$

The correct answer is option 2 i.e Ohm’s law.

• Ohm's Law states that the electric current flowing through a metallic wire is directly proportional to the potential difference ‘V’ across its ends provided its temperature remains the same.
  • \(I\ (Current) \propto V\ (Potential\ difference)\).
  • Ohm's law was named after Georg Ohm.

The correct answer is Controlled chain reaction.

• A nuclear reactor is a device in which nuclear reactions are generated, and the chain reaction is controlled to release a large amount of steady heat, thereby producing energy.

Key Points

• Nuclear fission is the process in which the nucleus of an atom is split, forming nuclei of lighter atoms and neutrons.
• The mass of these products is less than the original mass.
• Nuclear reactors are the heart of a nuclear power plant.
  • They contain and control nuclear chain reactions that produce heat through a physical process called fission.
  • Heat is used to make steam that spins a turbine to create electricity.

Additional Information

• With more than 450 commercial reactors worldwide, including 95 in the United States, nuclear power continues to be one of the largest sources of reliable carbon-free electricity available.
• Reactors use uranium for nuclear fuel.
  • The uranium is processed into small ceramic pellets and stacked together into sealed metal tubes called fuel rods.
  • Typically more than 200 of these rods are bundled together to form a fuel assembly.
  • A reactor core is typically made up of a couple of hundred assemblies, depending on power level. 
• The main job of a reactor is to house and control nuclear fission - a process where atoms split and release energy.
•  All commercial nuclear reactors in the United States are light-water reactors.
• This means they use normal water as both a coolant and neutron moderator.

CONCEPT:

• When a circular loop is associated with the current I, it starts to act as a magnet and its magnetic moment is find as given below.

• Magnetic moment (μ): The magnetic strength and orientation of a magnet or other object that produces a magnetic field.
  • It is a vector quantity associated with the magnetic properties of electric current loops.
  • It is equal to the amount of current flowing through the loop multiplied by the area encompassed by the loop.

μ = N i A 

where μ is the magnetic moment, A is the area of the coil, N is no. of turns and I is current in the coil.

• Its direction is established by the right-hand rule for rotations.

EXPLANATION:

The Magnetic moment μ = N i A is independent of the magnetic field in which it is lying. 

• As there is no magnetic field component in the formula.
• So the correct answer is option 4.

Concept:

• In an electrical circuit, the condition that exists when the inductive reactance and the capacitive reactance are of equal magnitude, causing electrical energy to oscillate between the magnetic field of the inductor and the electric field of the capacitor, is termed as resonance.
• In a series RLC circuit at resonance, inductive reactance (XL) is equal to capacitive reactance (XC). The circuit becomes purely resistive, i.e. at resonance: XL = XC

          The impedance at resonance becomes:

​          \(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} = R\) 

• The impedance - frequency curve is shown below:

Explanation:

So, from the above graph, we can say that for a series LRC circuit, \(X_{L}\) vs frequency graph is a straight line with positive slope, and nature of curve is straight line.

CONCEPT:

Diamagnetic Substance:

• Diamagnetic substances are those which develop feeble magnetization in the opposite direction of the magnetizing field.
• Such substances are feebly repelled by magnets and tend to move from stronger to weaker parts of a magnetic field.
• Magnetic susceptibility is small and negative i.e. -1 ≤  χ ≤ 0.
• Examples: Bismuth, copper, lead, zinc, etc.

Paramagnetic substances:

• Paramagnetic substances are those which develop feeble magnetization in the direction of the magnetizing field.
• Such substances are feebly attracted by magnets and tend to move from weaker to stronger parts of a magnetic field. Therefore option 2 is incorrect.
• Magnetic susceptibility is small and positive i.e. 0<  χ.
• Example: Manganese, aluminum, chromium, platinum, etc.

Ferromagnetic substances:

• Ferromagnetic substances are those which develop strong magnetization in the direction of the magnetizing field.
• They are strongly attracted by a magnet and tend to move from weaker to the stronger part of a magnetic field.
• Magnetic susceptibility is very large and positive i.e. χ > 1000.
• Example: Iron, cobalt, nickel, gadolinium, and alloys like alnico.

EXPLANATION:

• Diamagnetic materials are slightly repelled by a magnetic field. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Therefore option 2 is correct.
• Paramagnetic materials are slightly attracted by a magnetic field. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Therefore option 1 and 4 is incorrect.
• Ferromagnetic materials exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. Therefore option 3 is incorrect. 

The correct answer is Gamma rays.

Key Points

CONCEPT:

• Electromagnetic spectrum: It is a collection of a range of different waves in sequential order from radio to gamma electromagnetic waves.

Frequency (ν) = speed of light (c)/wavelength (λ)

• Characteristic of a Gamma-ray: It is penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei.
  • It consists of the shortest wavelength electromagnetic waves and so imparts the highest photon energy.
• Characteristic of an X-ray: X-rays are produced when an element is bombarded with high-energy particles, which can be photons, electrons or ions (such as protons). 
  • When the incident particle strikes a bound electron (the target electron) in an atom, the target electron is ejected from the inner shell of the atom
• Ultra-Violet: UV light is electromagnetic radiation with a wavelength shorter than that visible light in the range of 10 nm to 400 nm.
  • UV can have effects on biological matter causing cancers and a lot more.
  • Most of the UV rays are absorbed by the ozone (O3) layer in the upper atmosphere (15-30 km above).

EXPLANATION:

• According to the above electromagnetic spectrum, gamma rays have the lowest wavelength.
• Hence option 3 is correct.

Concept:

• Neils Bohr proposed Bohr's postulate which explained that a nucleus (positively charged) is surrounded by negatively charged electrons.
• According to Bhor, Electron which is moving in an orbital does not lose or radiate energy and atom came as a modification to Rutherford’s model of an atom.
• Postulates of Bohr’s model of an atom are:
  • In an atom, electrons move in a fixed circular path around a positively charged nucleus. This path is known as orbits or shells
  • These orbits or shells have fixed energy.
  • The energy levels are represented by an integer (n = 1, 2, 3, 4, …) known as a principal quantum number.
  • If an electron moves from lower energy level to higher energy level, then it will do so by gaining energy and if it moves from higher energy level to lower energy level then it will do so by losing energy.

Explanation:

• Quantum number are those numbers that designate and distinguish various atomic orbitals and electrons present in an atom.
• There are four types of quantum number:
• The angular quantum number determines the three-dimensional shape of the orbital quantum number.
  • Denoted by the symbol ‘l’ is also known as orbital angular momentum or subsidiary quantum number.
  • It defines the three-dimensional shape of the orbital.
• Principal quantum number
  • ​Denoted by the symbol ‘n’.
  • Determines the size and to a large extent the energy of the orbital.
• Magnetic orbital quantum number
  • ​Denoted by the symbol ‘ml’.
  • Gives information about the spatial orientation of the orbital concerning standard set of co-ordinate axis.
• Electron Spin quantum number
  • ​Denoted by the symbol (ms) refers to the orientation of the spin of the electron.

Concept:

Optical fibers are cylindrical solid glass material acting as waveguides made of two concentric layers of very pure glass.

The core (the interior layer) with refractive index n₁ serves as the medium for light propagation, while the cladding (the exterior layer) has a lower refractive index n₂ where n1 > n2 assuring that light rays are reflected to the core.

Since the cladding does not absorb any light from the core, the light wave can travel great distances.

Explanation:

• The working principle of optical fibers is Total Internal Reflection.

Optical fiber mostly used for communication purposes with negligible loss of energy.

• The “Total Internal Reflection” of light is the boundary between transparent media of two different refractive indices.
• At present, Optical fiber cables are used for communication like sending images, voice messages, etc.
• The designing of this cable is done with Plastic or glass so that data can be transmitted effectively and quickly than other modes of communications.

CONCEPT:

• Wavefront: The locus of all particles in a medium, vibrating in the same phase is called waveFront.
• The direction of propagation of light (ray of light) is perpendicular to the waveFront.
• Every point on the given wavefront acts as a source of a new disturbance called secondary wavelets which travel in all directions with the velocity of light in the medium.
• A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wavefront at that instant. This is called a secondary wavefront.

EXPLANATION:

• From above it is clear that, a wavefront is an imaginary surface where all particles lying on this vibrate in the same phase. Therefore option 2 is correct.

Explanation: 

De Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well. The wavelength of material waves is also known as the De Broglie wavelength.

De Broglie wavelength (λ) of electrons can be calculated from Planks constant h divided by the momentum of the particle.

λ = h/p

Where λ = De Broglie wavelength, h = Plank's const, p = Momentum = mv

v = Velocity

In terms of Energy, de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

Where m = Mass of an electron, E = Energy of the electron.

Calculation:

Given:

v = 3 v_e, λ = λ_e ×  1.813 × 10-4 , me = 9.11 × 10-31 kg

Using λ = h/p

h = λp = λmv

As h is constant

∴ λmv = λ_em_ev_e

m = \(m_e \times \frac {\lambda _e v_e}{\lambda v}\)

m = \(9.11 \times 10^{-31} \times \frac {1}{1.813\times 10^{-4}\ \times \ 3}\)

m = 1.67 × 10-27 kg

CONCEPT:

• A rectangular coil always acts as a magnetic dipole of dipole moment M.
• The dipole moment of a rectangular coil is given by:
• The dipole moment of the coil (M) = N I A
  Where N is the number of turns, I is current in the coil and A is the area of the rectangular coil

• The torque of the magnetic field on any magnetic dipole moment is given by;

⇒ T = M × B = MB Sinθ
Where M × B = cross product of Area vector A and magnetic field vector B and θ = the angle between area vector A and magnetic field B.
EXPLANATION:

• The dipole moment of the coil is

⇒ M = N I A

Torque on rectangular coil (T) = M × B = MB Sinθ = N I A B Sinθ

• According to the above formula of the torque on a rectangular coil in a magnetic field, torque will be large when the number of turns is large. So option 1 is correct and option 2 is wrong.
• If the magnetic field is perpendicular to the plane of the coil then it will be along the area vector. So the angle between the area vector and the magnetic field will be zero and finally, the torque will be zero. So option 3 is wrong.
• If the area will be small then torques will also be small. So option 4 is wrong.

CONCEPT:

• A dielectric develops a net dipole moment in the presence of an external field E. The dipole moment per unit volume is called polarization and is denoted by P.

\(⇒ P= ε_0 χ_e E\)

where χe is a constant. It is the characteristic of the dielectric and is known as the electric susceptibility of the dielectric medium.

Electric susceptibility

• Electric susceptibility can also be expressed in terms of dielectric constant K as

\(⇒ χ_e= K-1\)

EXPLANATION:

• When a dielectric is placed in an external electric field, charges are induced, and a net dipole moment is formed.
• The dipole moment per unit volume is defined as polarisation.
• ​Linear isotropic dielectrics are defined as dielectrics for which
  • The direction of polarisation is the same as that of the external electric field. 
  • The strength of polarization is proportional to the external electric field. 
• Therefore option 1 is correct.

CONCEPT:

• Wheat-stone bridge: When 5 resistors are connected in such a way that the ratio of two resistances in one side to other side is equal then the potential difference across the middle resistance is zero and we can remove that middle resistor (R) from the circuit.

\(\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{R}_{3}}}{{{R}_{4}}}\)

When resistances are in series then the equivalent resistance is given by:

R = R₁ + R­₂

When resistances are in parallel, the equivalent resistance is given by:

\(\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\)

Where R₁ and R₂ are two resistances

CALCULATION:

After rearranging the resistances in the given diagram we get,

Here,

\(\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{10}{10}=\frac{{{R}_{3}}}{{{R}_{4}}}=\frac{10}{10}=1\) So given circuit is a wheat-stone bridge, so remove the resistance between PQ.

Here both upper and lower resistances of each 10 Ω are in series with each other,

So equivalent resistance (R) = 10 + 10 = 20 Ω

Here both 20 Ω are in parallel combination.

\(\frac{1}{R'}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10}\)

Option 4 is correct

CONCEPT:

• Electric potential (V): The amount of work done to move a unit charge from a reference point (or infinity) to a specific point in an electric field without producing an acceleration is called electric potential at that point.

\({\rm{Electric\;potential\;}}\left( {\rm{V}} \right) = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}\)

• Electrostatic Potential Energy: The amount of work done to move a charged particle from infinity to a point in an electric field is known as the potential energy of that charged particle.

CALCULATION:

Given that:

Electric charge (q) = 5 C

Potential difference (V) = 16 V

Work done (W) = charge (q) × potential difference (V)

Work done (W) = 5 × 16 = 80 J