Concept:

Electrical energy:

• The energy that is derived from the flow of electric charge is called electrical energy.
• The SI unit of electrical energy is Joule.
• Kilowatt-hour: It is the unit of energy and is equal to the energy consumed in the circuit at the rate of 1 kilowatt for 1 hour.
• The commercial unit of electrical energy is a kilowatt-hour (kWh).

Calculation:

∵ 1 kilowatt hour = 1 kilowatt × 1 hour

∴ Writing this in SI units

⇒1 kilowatt-hour = 1000 Watt × 3600 second

⇒1 kilowatt-hour = 3.6 x 106 joule = 36 × 105 J

Additional Information

Electric Power:

• The rate at which electrical energy is consumed in a circuit is called electric power.
• SI unit is the watt.
• One Kilowatt is equal to 1000 watts.

CONCEPT:

• When highly energetic electrons are made to strike a metal target, electromagnetic radiations come out.
• A large part of this radiation has a wavelength of the order of 0.1 nm and is known as X-ray.
• These rays named X-ray as at that time its nature and properties were not known.
• Properties of X-Ray

1. X - rays travel in a straight line in a vacuum at a speed equal to that of light (3× 108).
2. X - rays are diffracted by crystals in accordance with Bragg’s law.
3. X - rays are not deflected by an electric or magnetic field as it contains no charged particles.
4. X - rays can affect a photographic plate more strongly than visible light.
5. When passed through a gas, X - rays ionize the molecules of the gas.

EXPLANATION:

• There is no charge on X-rays.

CONCEPT:

Huygen's Wave Theory:

• According to wave theory, a luminous body is a source of disturbance in a hypothetical medium ether. This medium pervades all space.
• It is assumed to be transparent and having zero inertia. The disturbance from the source is propagated in the form of waves through space.
• The waves carry energy and momentum. Huygen assumed that the waves were longitudinal.
• Further when polarization was discovered, then to explain it, light waves were, assumed to be transverse in nature by Fresnel.
• This theory explains successfully, the phenomenon of interference and diffraction apart from other properties of light.
• Huygen's theory fails to explain the photoelectric effect, Compton's effect, etc
• The wave theory introduces the concept of a wavefront.

EXPLANATION:

• The wave theory of light was proposed by Christiaan Huygens who believed that light was made up of waves vibrating up and down perpendicular to the direction of the light travels. 
• 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. Therefore option 4 is correct.

CONCEPT:

• Magnetic field (B): The region around a magnet or around a current-carrying wire in which other magnets or magnetic materials experience a magnetic force is called a magnetic field.
  • When a charged particle moves in a magnetic field it experiences a force on it at right angles to the magnetic field as well as to the velocity.
• Force on a charge moving in magnetic fields depend upon:
  F ∝ B
  F ∝ q 
  F ∝ V sin θ 
  F = BqV sin θ
  F = q ( v × B )
  where B = Uniform magnetic field; q = charge on particle; V = velocity of an object; θ = angle between uniform magnetic field and velocity
  

EXPLANATION:

• The magnetic force is independent of the nature of the charge. 
  F = BqV sin θ​
  Here q is the magnitude of charge not the nature of charge.
• The magnitude of Charge on proton 1.6 × 10^-19 C and charge on electron 1.6 × 10^-19 C 
• So, the force on proton and electron are the same.
• Hence option 3 is correct.

Concept:

• The wave theory of light was first demonstrated by Thomas Young in 1801 through Young's double-slit experiment.
• This experiment shows that the observed pattern of interference occurs due to the superposition of light waves which proves the wave nature of light.
• In this experiment, two narrow slits which are close to each other, are illuminated by a monochromatic light source.
• The two slits are responsible for the production of two different wavefronts which then superimpose on a screen forming the definite interference pattern.
• For two coherent sources s₁ and s₂, the resultant intensity at some point p is given by

\(I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}cos\phi }\)

Calculation:

Given that,

The intensity of light at a point on the screen = K units

The phase difference in terms of λ is considered to be 2π which means Phase difference corresponding to λ /3 is 2λ/3

Now using the formula for intensity,

\(I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}cos\phi }\)

In the first case, when the path difference is λ

\(K = {I_0} + {I_0} + 2{I_0}\cos 2\pi = 4{I_0}\) 

In the second case, when path difference is λ/3

\(K' = {I_0} + {I_0} + 2{I_0}\cos \frac{{2\pi }}{3} = {I_0}\)

Now,

\(\frac{{K'}}{K} = \frac{{{I_0}}}{{4{I_0}}} = \frac{1}{4} = K' = \frac{K}{4}\)

Concept:

Resistivity

• The property of a conductor that opposes the flow of electric current through them and is independent of the shape and size of the materials but depends on the nature and temperature of the materials is called resistivity.
• The unit for resistivity is the ohm-meter (Ω-m).
• The resistivity of a material depends on its nature and the temperature of the conductor and doesn't depend on its shape and size (length and area).
• Resistance: The property of any conductor that opposes the flow of electric current through it and depends on the shape and size of the materials, temperature, and nature of the materials is called resistance.
  • It is denoted by R and the SI unit is the ohm (Ω).
• The resistance is given by:

\(⇒ R=\frac{ρ l}{A}\)

ρ is resistivity, l is length, A is cross-sectional Area

• The resistivity of a conductor increases with temperature. 

Temperature coefficient of resistance (α) :

• The change in electrical resistance of a substance with respect to a per degree change in temperature.
• α depends on the type of material.

Variation of resistance with temperature:

Let ρ0  and ρ_T represent the resistivity at a temperature of T0 and T respectively. Therefore

⇒ ρT = ρ0 [1 + α (T - T0)]

Resistivity of semiconductors

• Compared to insulators, the energy band gap is very less in semi-conductors. Thus, the electrons require very little energy to jump to the conduction band.
• When the temperature is increased, most of the electrons acquire a sufficient amount of energy and jumps to the conduction band.
• Hence, the conduction increases i.e. conductivity increases.
• We know that conductivity is the reciprocal of resistivity. Thus, on increasing the temperature of a semiconductor, resistivity decreases.
• Semiconductors have a negative temperature coefficient. So, α is negative.

Explanation:

Among the given options Silicon is semiconductors and the rest options are metals and hence conductors. 

So, the resistivity of silicon will decrease with an increase in temperature. 

Additional Information

• Energy Band: The different energy levels with continuous energy variation of electrons in a material are called energy bands.
• Valance Band: The energy band which includes the energy levels of the valence electrons is called the valence band.
• Conduction Band: The energy bands above the valence bands are called the conduction bands.
• The conduction band holds those electrons that are responsible for conduction.

CONCEPT:

Resistivity:

• The property of a conductor that opposes the flow of electric current through them and independent of the shape and size of the materials but depends on the nature and temperature of the materials is called resistivity.
• The unit for resistivity is the ohm-meter (Ω-m).
• The resistivity of a material depends on its nature and the temperature of the conductor.
• The resistivity of a material doesn't depend on its shape and size (length and area).

Resistance:

• The property of any conductor that opposes the flow of electric current through it and depends on the shape and size of the materials, temperature, and nature of the materials is called resistance.
• It is denoted by R and the SI unit is the ohm (Ω).
• The resistance is given by:

\(⇒ R=\frac{ρ l}{A}\)

ρ is  resistivity, l is the length of wire, A is cross sectional area

Electrical Energy and Joules Heating:

• The heat generated by a resistive coil or device is given as

\(H = \frac{V^2}{R}t = I^2 Rt\)

H is energy, t is time, R is resistance, V is the potential difference. 

• If the conductor is pure resistive, then the whole electrical energy is converted into heat energy. 

EXPLANATION:

• Heat energy produced in the conductor depends upon the resistance. 
• The resistance further depends upon the size of the material and resistivity. 
• Resistivity depends upon material and temperature.

So, we can say the heat produced depends upon material, length and thickness.

The correct answer is the Linear curve.

• The current-voltage graph of ohmic devices is of the form linear curve.

Key Points

• Ohm’s Law
  • The electric current flowing through a conductor is directly proportional to the potential difference across its ends.
  • It is given by the formula:
  • V=IR
• The device which follows ohm's law for all voltages across it is called an ohmic device.
• The current-voltage graph of ohmic devices is a linear curve.

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.

Explanation-

• In the domestic power supply AC: 220 V and 50 Hz
• The potential value(220 V) is the rms value of voltage and 50 Hz is frequency. So option 2 is correct.

CONCEPT:

Wheatstone Bridge:

• Wheatstone bridge is an arrangement of four resistance that can be used to measure one of them in terms of rest.
• Here arms AB and BC are called ratio arms and arms AC and BD are called conjugate arms.
• The bridge is said to be balanced when deflection in galvanometer is zero i.e. no current flows through the galvanometer or in other words V_B = V_D.
• In the balanced condition,

\(\Rightarrow \frac{P}{Q}=\frac{R}{S}\)

CALCULATION:

Given P = Q = R = S = 5Ω

• From the figure, it is clear that it is a Wheatstone bridge.
• In the given diagram,

\(\Rightarrow \frac{P}{Q}=\frac{5}{5}\)

\(\Rightarrow \frac{P}{Q}=1\)     -----(1)

And

\(\Rightarrow \frac{R}{S}=\frac{5}{5}\)

\(\Rightarrow \frac{R}{S}=1\)     -----(2)

• By equation 1 and equation 2 it is clear that it is a condition of the balanced wheat stone bridge and no current will flow from the resistance connected between point B and point D.
• So the current in the 1Ω resistance is zero. Hence, option 3 is correct.

CONCEPT:

• Motional emf: The emf induced due to motion relative to a magnetic field is called the motional emf. 

​

• Consider a straight conductor PQ moving perpendicular to a uniform magnetic field B.
  • Assume the motion of the rod to be uniform with a constant velocity of (v m/s).
  • The rectangle PQRS forms a closed circuit enclosing a changing area due to the motion of the rod PQ.
• The magnetic flux ϕ enclosed by the loop PQRS can be given as:

⇒ ϕ = B × area = B × (l.x)

• Since the conductor is moving, there is a change in the rate of area moving. This causes a rate of change of flux which induces an emf.
• This induced emf is given by:

\(⇒ \epsilon =\frac{-dϕ }{dt}=\frac{-d}{dt}(Blx) =-Bl\frac{dx}{dt}=Blv\)

Where  \(-\frac{\mathrm{d} x}{\mathrm{d} t}=v\) is the speed of the conductor PQ​

CALCULATION:

Given: length of the wire (l) = 3 m, magnetic field (B) = 4.5 Wb/m², and velocity (v) = 50 m/s

• The emf induced in the wire is

⇒ ε = Blv = 4.5 × 3 × 50 = 675 V

• Hence option 2) is correct.

CONCEPT:

EXPLANATION:

• In the photoelectric experiment, light behaves as a particle.
• When light undergoes diffraction and refraction it behaves as a wave. Hence, diffraction and refraction indicate that light is a wave. Therefore, option 1 is correct. 

CONCEPT:

• The Current-Carrying Conductor produces a magnetic Field Around it. 
  • Biot Savarts Law: The direction and Magnitude of this magnetic field can be obtained from Biot Savarts Law. 

\(\overrightarrow{dB} = \frac{μ _{0}\overrightarrow{I}\times \overrightarrow{dl}}{4\pi r^{2}}\)

Where dB = Magnetic Field produced due to small wire of length dl, I = Current in wire, μ0 = Permittivity of free space, dl = Small Current Element

• The magnetic field generated due to the current-carrying circular conductor at its center is given as:

\(\Rightarrow B=\dfrac{\mu_0 i}{2r}\)

• This result has been obtained from Biot Savarts Law.

EXPLANATION:

• The Magnetic Field Produced due to the Semi-Circular wire is half of that of Circular wire.
• So, the magnetic field produced at the center of the wire 

\(\Rightarrow B = \frac{1}{2}\times \dfrac{\mu_0 i}{2r}=\dfrac{\mu_0 i}{4r}\)

• Hence option 2 is the correct option.

CONCEPT:

• LCR Circuit: The ac circuit containing the capacitor, resistor, and the inductor is called an LCR circuit.

For a series LCR circuit, Impedance (Z) of the circuit is given by:

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL = inductive reactance and XC = capacitive reactive

• Power factor (Cos Φ): The ratio of the true power to the apparent power of an a.c. the circuit is called the power factor.
  • Its value varies from 0 to 1.

The power factor (P) of a series LCR-circuit is given by:  

\(\cos {\rm{Φ }} = \frac{R}{Z} = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}\)

Where R = resistance, Z = Impedance, XL = Inductive reactance and XC = Capacitive reactance

CALCULATION:

If we have an LCR circuit and it is under resonance:

It means that XL = Xc, hence Z = R and V = VR

\(\text{Impedance }\!\!~\!\!\text{ of }\!\!~\!\!\text{ the }\!\!~\!\!\text{ circuit }\!\!~\!\!\text{ }\left( \text{Z} \right)=~\sqrt{{{R}^{2}}+~{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}\)

So Z = R

Power factor = Cosϕ = R/Z = R/R = 1

So option 4 is correct.

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.

\(⇒ E=\dfrac{F}{q_{o}}\)

Where E = electric field intensity, qo = charge on the particle

EXPLANATION:

We know that the electric field intensity at a point due to a point charge Q is given as,

\(⇒ E=\dfrac{kQ}{r^2}\)     ---(1)

⇒ E ∝ Q

\(⇒ E\propto\dfrac{1}{r^2}\)

Where k = electrostatic force constant, r = distance of the point from the charge

• The value of k depends on the medium and it is maximum for vacuum.
• Electric field intensity is directly proportional to the magnitude of the source charge.
• Electric field intensity is inversely proportional to the square of the distance of the point from the charge.
• Therefore it is clear that the electric field intensity at a point due to a point charge depends upon the magnitude of the source charges, the distance of the point from the charge, and the medium. Hence, option 4 is correct.

• Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q.
• This is because F is proportional to q, so the ratio F/q does not depend on q.
• The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. 

CONCEPT:

• Bar Magnet: A bar magnet consists of two equal and opposite magnetic pole separated by a small distance. Poles are not exactly at the ends.
• The shortest distance between two poles is called effective length (Le) and is less than its geometric length (Lg) for bar magnet.
• Magnetic field and magnetic lines of force: It is the space around a magnetic pole or magnet or current-carrying wire within which it's magnetic effect can be experienced is defined as a magnetic field.

• The magnetic field can be represented with the help of a set of lines or curves called magnetic lines of force.

EXPLANATION:

Properties of magnetic field line:

1. The magnetic field line is directed from the north pole to the south pole.
2. Magnetic field lines are closed and continuous.
3. Magnetic field lines are more crowded near poles.
4. Magnetic field lines never intersect with each other.

• Magnetic field lines originate from the north pole and terminate at the south pole while forming continuous closed paths. Therefore option 3 only satisfies this property.

Lenz’s law:

When a voltage is generated by a change in magnetic flux according to Faraday's law, the polarity of the induced voltage is such that it produces a current whose magnetic field opposes the change which produces it.

The induced emf is given by the rate of change of magnetic flux linked with the circuit i.e.

\(e=-\frac{d\text{ }\!\!\Phi\!\!\text{ }}{dt}\)

Where dΦ = change in magnetic flux and e = induced e.m.f.

EXPLANATION:

• Lenz's law is the consequence of the law of conservation of energy. So option 3 is correct.

• The magnitude of the induced emf is given by Faraday's Laws of Electromagnetic Induction. 
• Faraday's Laws of Electromagnetic Induction: Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change of flux.

The correct answer is They are positively charged.

Key Points

• Electromagnetic Wave does not have any Charge as they result from Changing electric and Magnetic field and carries No Charge.
• Electromagnetic Waves travel across space at the speed of light, which is about 300 million metres per second (3.0 × 10⁸).
• All electromagnetic waves travel at the same speed through space
• A wave with a shorter wavelength must have a higher frequency, and vice versa.
• Electromagnetic Waves consist of both electric and magnetic waves.
• The creation of all electromagnetic waves begins with an oscillating charged particle, which creates oscillating electric and magnetic fields.
• Electromagnetic waves are ubiquitous in nature.
  • This means that an electric field that oscillates as a function of time will produce a magnetic field, and a magnetic field that changes as a function of time will produce an electric field.

​Additional Information

• Examples of Electromagnetic Waves are microwaves, Radio waves, Infrared Waves, X-rays, gamma rays, RADAR, Communication devices, Military Equipment and Cancer therapy etc.

CONCEPT:

Eddy current:

• When a metal moves in a magnetic field or when the magnetic field through a stationary metal is altered, an induced current is produced in the metal.
  • This induced current flows in the metal in the form of closed loops resembling ‘eddies’ or whirlpool.
  • Hence this current is called eddy current.
  • The direction of the eddy current is given by Lenz’s law.

EXPLANATION:

• When a changing magnetic flux is applied to a metal piece of conducting material then circulating currents is called eddy currents are induced in the material. Therefore option is 1 correct.
• When a metal is kept in a steady magnetic field, then the number of magnetic fields lines passing through the metal surface does not change and, hence no eddy current is induced in it. 
• When a circular coil is placed in a magnetic field and it is not moving in the magnetic field, therefore the number of magnetic field lines passing through the circular coil does not change and, hence no eddy current is induced in it.

CONCEPT:

• Electric charge: A physical property of matter due to which the body to experience a force when it is placed in the electromagnetic field is known as Electric charge.
• There are two types of electric charge

1. positive
2. negative

• When an electron is given to a body or one electron is taken out of the body it has 1 electron charge.
• When n electron is given or taken out of a body, the charge it has given by the formula

q = n × e

where q is the charge on a body, n is no. of electron and e is the charge on one electron. 

and e = 1.6 × 10-19 C (charge on one electron)

• Formula q = ne thing represents quantization of charge.
• It tells us that charge is quantized ( in the form of electrons)
• Every charge has to be an integral multiple of ‘e’.

​CALCULATION:

given that charge on the body q = 1 nC = 10^-9 C

e = 1.6 × 10-19 C

⇒ q = n × e

⇒ n = q / e = 10-9 / (1.6 × 10^-19) = 6.25 × 109

• So the correct answer will option 2.

CONCEPT:

• According to Ohm’s law: The potential difference across a resistor in a circuit is directly proportional to the current flowing in it.

V = R I

Where V is potential difference, R is resistance and I is current.

• Kirchhoff’s current law (KCL): It states that, the total current entering a circuit’s junction is exactly equal to the total current leaving the same junction.

EXPLANATION:

Let potential at O is V’ and all the current is originating from the point O.

According to the KCL, the sum of all the current originating from the point O will be equal to zero.

Use Ohm’s law to final total current originating from O:

Current in OA + Current in OB + Current in OC = 0

(V – 8)/2 + (V – 4)/4 + (V – 2)/2 = 0

(2 V – 16 + V – 4 + 2 V – 4)/ 4= 0

5 V – 24 = 0

CONCEPT:

Rutherford’s Model of the Atom:

• He bombarded the beam of alpha particles on a very thin gold foil.
• While bombarding he observes the number of scattering of particles –
  • Most of the particles passed either un-deviated or with a very small deviation.
  • Some deviated by large angles
  • 1 out of 8000 was deflected by more than 90°
• The conclusion to this deflection:
  • Most of the space of an atom is empty.
  • At the center of an atom having a tiny positively charged particle called a nucleus.
  • The center nucleus has all the mass of an atom.
  • The amount of positive charge at the nucleus is equal to the total amount of negative charges on all the electrons of the atom.
  • All the electrons revolve around the nucleus and coulomb force provide the centripetal force

​

EXPLANATION:

• In an atom, there is maximum empty space and the electrons revolve around the nucleus in the same way as the planets revolve around the sun, and a number of protons and electrons inside an atom are equal under normal conditions (non-ionized). Therefore option 1 is correct.
• The nucleus is positively charged and its diameter (size) is of the order of 10-15 m ≈ 1 Fermi. Therefore option 2 is correct.
• The concept of electrons move in a circular path of fixed energy called orbits was put forward by Bohr and not derived from Rutherford's scattering experiment. Therefore option 3 is incorrect.
• Most of the mass (at least 99.95%) and all of the charge of an atom concentrated in a very small region is called the atomic nucleus. Therefore option 4 is correct.

Concept:

Total Internal Reflection (TIR):

• When a ray of light goes from denser to the rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in the rarer medium also increases and at a certain angle, angle of refraction becomes 90°, this angle of incidence is called critical angle (C).
• When Angle of incidence exceeds the critical angle than light ray comes back into the same medium after reflection from the interface. This phenomenon is called Total internal reflection (TIR).
• Hence, in the case of total internal reflection, the entire light is reflected back into the first medium.  

Conditions for TIR: 

1. The ray must travel from denser medium to rarer medium.
2. The angle of incidence i must be greater than the critical angle C.

​Explanation:

• Optical fiber works on the principle of total internal reflection.
• The light comes in from one end of the fiber and after thousands of successive internal reflections, the light reaches the opposite end of the fiber with almost zero loss. 
• The difference between the apparent and real depth of the pond is due to the refraction of light. 
• Mirage on hot summer days is due to total internal reflection. 

• The brilliance of a diamond is due to total internal reflection. 

CONCEPT:

Electric field lines:

• An electric field line is an imaginary line along which a positive test charge will move if left free.
• Electric field lines are drawn to represent the electric field.

Properties of electric field lines:

1. Electric field lines start from positive charges and end at negative charges. If there is a single positive charge then electric field lines start from positive charge and end at infinity. Similarly, if there is a single negative charge then electric field lines start from infinity and end at a negative charge.
2. In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
3. The tangent at any point on the electric field line gives the direction of the electric field at that point.
4. Electric field lines due to a point charge never intersect each other.
5. The electric field lines never form a closed loop.
6. The density of the electric field lines at a point indicates the strength of the electric field at that point.

EXPLANATION:

• From the above, it is clear that electric field lines start from positive charges and end at negative charges. 
• So, the electric field lines never form a closed loop. Hence, option 2 is correct.

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:
  1. Primary cell: This type of cell cannot be recharged.
  2. Secondary cell: This type of cell can be recharged.
• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

CALCULATION:

Given emf = E, internal resistance = r, and external resistance of circuit = r

For a cell, we know that,

⇒ E - V = Ir        -----(1)

By Ohm's law, we know that,

⇒ V = Ir       -----(2)

By equation 1 and equation 2,

⇒ E - Ir = Ir

⇒ 2Ir = E

\(\Rightarrow I=\frac{E}{2r}\)        -----(3)

So the potential difference is given as,

\(\Rightarrow V=\frac{E}{2r}r\)

\(\Rightarrow V=\frac{E}{2}\)

• Hence, option 1 is correct.

CONCEPT:

• Simple microscope: It is an optical instrument used to see very small objects.

It’s magnifying power is given by \(Magnifying\;power\;\left( m \right) = \frac{{Angle\;subtended\;by\;the\;image\;at\;the\;least\;distance\;of\;distant\;vision\;\left( \beta \right)}}{{Angle\;subtended\;by\;the\;object\;at\;the\;least\;distance\;of\;distant\;vision\;\left( \alpha \right)}}\)

\(m = 1 + \frac{D}{f}\)

Where D = Least distance of distant vision and f = focal length of a convex lens

EXPLANATION:

The magnifying power of a simple microscope is given

\(m = 1 + \frac{D}{f}\)

CONCEPT:

Mutual Induction: 

• Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighboring coil or circuit will also change.
• Hence an emf will be induced in the neighboring coil or circuit. This phenomenon is called ‘mutual induction’.
• Mutual induction between the two coils of area A, number of turns N1 and N2 with the length of secondary or primary l is given by –

\(M = \frac{{{\mu _{o\;{N_1}{N_2}A}}}}{l}\)

EXPLANATION:

• Mutual induction between the two coils of area A, number of turns N1 and N2 with the length of secondary or primary l is given by –

\(M = \frac{{{\mu _{o\;}}{N_1}{N_2}A}}{l}\)

Dependence of mutual inductance –

• The number of turns (N₁, N₂) of both coils.
• Area of the cross-section of coils.
• The magnetic permeability of medium between the coils (μ_o) or nature of the material on which two coils are wound.
• Length of the coil (l).
• From the above formula, mutual inductance of two coils can be increased by increasing the number of turns in the coils.

Concept:

• Conductors, semiconductors and insulators can be distinguished with respect to their conduction band and valence band
• For a material to conduct electricity the electrons from the valence band must shift to the conduction band
• In the case of conductors like metals, the valence band and conduction band overlap each other and almost all the charge carriers are found in the conduction band; This is the reason they are good conductors of electricity
• In the case of semiconductors, there is a gap between the valence band and the conduction band and only a small quantity of charge carriers are found in the conduction band; Hence, they are not very good conductors of electricity
• In the case of insulators, the gap between the valence band and the conduction band is very large, and no charge carriers are found in the conduction band; Thus, they do not conduct electricity

Explanation:

From the above explanation, we can see that insulator the gab between condition and valence band is extremely large compared to conductor and semiconductor 

CONCEPT:

• Semi-Conductors: They are materials which have a conductivity between conductors (generally metals) and nonconductors or insulators (such as most ceramics).
  • Semiconductors can be pure elements, such as silicon or germanium, or compounds such as gallium arsenide or cadmium selenide.
  • In a process called doping, small amounts of impurities are added to pure semiconductors causing large changes in the conductivity of the material.

Types of semiconductors:

• Plank's constant: It is the quantum of electromagnetic action that relates a photon's energy to its frequency.
  • The Planck constant multiplied by a photon's frequency is equal to a photon's energy.
  • In S.I units the value of Plank's constant h = 6.62 × 10^-34 J.s. 

Formula :

E = h ν 

where, E = energy released, h = plank's constant, ν = frequency .

CALCULATION:

Given that, bandgap 2.6 eV.

We know that,

E = h ν 

ν = c/λ _max...( where λ _max = maximum wavelength)

Substituting it in the above equation

E = hc/ λ _max

λ _max = hc/E = 6.62 × 10^-34 × 3 × 10⁸/2.5 × 1.6 × 10^-19 = 4.965 × 10^-7 = 4965 Å

The wavelength of signal must be less than λ_max  = 4000 Å

The correct option is 4000Å.

CONCEPT:
Root mean square value of Alternating E.M.F:

• The root mean square (r.m.s.) value of alternating e.m.f.  is defined as that value of steady volatge, which would generate the same amount of heat in a given resistance in a given time, as is done by the alternating e.m.f., when passed through the same resistance for the same time.
• The r.m.s. value is also called effective value  or virtual value of alternating e.m.f. It is represented by E_rms or E_eff or E_v.
• The relation between the peak value of alternating e.m.f. value of current (E_o) and r.m.s. value of e.m.f. is given as –
  \({E_{rms}} = \frac{{{E_o}}}{{\sqrt 2 }}\)

SOLUTION:

Given – V_rms = 30 V,

\({V_{rms}} = \frac{{{V_o}}}{{\sqrt 2 }}\)

V_o = V_rms√2

V_o = 30√2

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.

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

CALCULATION:

The given figure is,

• All charges are equal in magnitude.
• The charge -3 μC is at an equal distance from both the +3 μC charges so the force on the charge -3 μC due to both the +3 μC charges will be equal in magnitude.

So the force between on the -3μC charge due to any of the +3μC charge is given as,

\(⇒ F = 9\times10^9\times\frac{{{3\times10^{-6}}\; \times \;{3\times10^{-6}}}}{{{(9\times10^{-2})^2}}}\)

⇒ F = 10 N     -----(1)

The force on the charge -3μC due to both +3μC charges is shown in the figure below,

So by the parallelogram law of vector addition, the resultant force is given as,

\(\Rightarrow F_R=\sqrt{F^2+F^2+2\times F\times F\times cos60}\)

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

\(\Rightarrow F_R=\sqrt{3}F\)

\(\Rightarrow F_R=10\sqrt{3}\,N\)

• Hence, option 3 is correct.

Concept:

In the double-slit experiment, a beam of light is aimed at a barrier with two vertical slits. After the light passes through the slits, the resulting pattern is recorded on a photographic plate. When one slit is covered, a single line of light is displayed, aligned with whichever slit is open.

Intuitively, one might hypothesize that if both slits were open, the resulting pattern would display as two lines of light, aligned with the slits. What occurs in practice, however, is that light passing through the slits and displayed on the photographic plate is entirely separated into multiple lines of lightness and darkness in varying degrees.

\({\rm{\theta '}} = \frac{{\rm{\lambda }}}{{\rm{d}}}\)

Calculation:

Here, wavelength of light used (λ) = 5303 Å.

Distance between two slit (d) = 19.44 µ-m

Width of single slit (a) = 4.05 µ-m

Here, angular width between first and second diffraction minima

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

and angular width of a fringe due to double slit is

\({\rm{\theta '}} = \frac{{\rm{\lambda }}}{{\rm{d}}}\)

∴ Number of fringes between first and second diffraction minima, \(n = \frac{\theta }{{\theta '}}\)

\(= \frac{{\frac{{\rm{\lambda }}}{{\rm{a}}}}}{{\frac{{\rm{\lambda }}}{{\rm{d}}}}} = \frac{{\rm{d}}}{{\rm{a}}} = \frac{{19.44}}{{405}} = 481\)

or n = 5

CONCEPT:

• Gauss’s Law: Total electric flux though a closed surface is 1/εo times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)
• But we know that Electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \)

\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)

Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.

EXPLANATION:

• Gauss’s Law is used to determine electric field when electric charge is continuously distributed on an object which possesses symmetrical geometry.
• The object can be a plane, cylinder, sphere, etc.
• The theorem relates electric flux associated with an electric field enclosing a symmetrical surface to the total charge enclosed by the symmetrical surface. Therefore option 1 is correct.

CONCEPT:

• The wave which is generated due to vibration between electric field and magnetic field and it does not need any medium to travel is called an electromagnetic wave.
• It can travel through a vacuum.
• The wave having shorter wavelengths than visible light is called Ultraviolet waves.
• The wave has a wavelength greater than visible light is called Infrared waves.
• The wave has a wavelength greater than infrared is called a Radio wave.
• The wave having a shorter wavelength than that of radio waves but longer than that of infrared are called microwaves.

EXPLANATION:

• The original currency notes have some strip embedded into it.
• The forged currency doesn’t have these strips.
• The Ultraviolet waves detect these strips and we can catch the forged currency. Hence option 1 is correct.

CONCEPT:

• As the current-carrying conductor experiences a force when placed in a magnetic field, each side of a current-carrying circular coil experiences a force in a magnetic field.
• Consider a circular coil of length l and breadth b carrying a current I placed in a uniform magnetic field B and θ be the angle between the plane of the circular coil and the magnetic field.

Force acting on the circular loop:

F = I L B sinθ, where F is force, I = current, L = distance from the axis, B = strength of the magnetic field, θ = angle between the plane of the circular coil and the magnetic field.

• Acting on the left and right sides are equal and opposite along the same line of action, they cancel each other. Hence net force on the loop is equal to zero. So option 4 is correct.

• Whenever a current carrying loop is placed in a uniform magnetic field then the net force on the loop is zero. But there will be some torque on the loop.

CONCEPT:

• Electric potential difference: The amount of work done to move a unit charge from a point to another point in a space where an electric field is present is called an electric potential difference.

\({\rm{Potential\;difference}}\left( {\rm{V}} \right) = {\rm{\;}}\frac{{Work\;done\left( W \right)}}{{Charge\left( Q \right)}}\)

Work done (W) = Energy gained = Q V

• Electric field: The space or region around the electric charge in which electrostatic force can be experienced by other charged particles is called an electric field by that charge.

CALCULATION:

Given that:

Charge (Q) = 1.6 coulombs

Voltage (V) = 0.8 V

Using the equation:

⇒ Potential difference = Work Done/Quantity of charge moved

\(⇒V={W\over Q}\)

⇒ Work (W) = Q × V

⇒ W = 1.6 × 0.8 = 1.28 J

• So option 1 is correct.

Explanation:

• WiFi an electronic device that allows the exchange of data wirelessly using radio waves. 
• WiFi stands for Wireless Fidelity.
• WiFi was introduced in 1998.
• Invented by: Dr. O'Sullivan and his colleagues.
• Commonly used for local area networking (LAN).
• The name WiFi was coined by the firm Interbrand.

Additional Information

Important computer abbreviations:

CONCEPT:

Van de Graaff generator

Principle:

• Static charges are accumulated on a large radius conducting shell using corona discharge.

• This is a machine that can build up high voltages of the order of a few million volts.
• The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies needed for experiments to probe the small-scale structure of matter.

Working

• Charges from a ground metal base are collected by an insulating belt.
• These charges are driven upward by the belt attached to the pulley
• The charges are collected on a small conducting sphere attached to the pulley with the help of a metal brush.
• A metallic wire is then connected between the small and the large conducting spheres.
• The charge transfers to the large sphere due to its low potential.
• The process is repeated to accumulate a high amount of static charge on the large sphere.

EXPLANATION:

• The Van de Graaff generator is switched on.
• Some charge is accumulated on the small sphere
• This charge is transferred to the larger sphere.
• The potential at the smaller sphere is less than the potential at the larger sphere.
• This potential difference is independent of the charge at the larger sphere.

\(\Rightarrow V(r)-V(R) = \frac {KQ}{R}+\frac {Kq}{r} - \frac {KQ}{R}-\frac {Kq}{R} =Kq(\frac {1}{r}-\frac {1}{R}) > 0\)

Where V(r) is the potential at the smaller sphere, V(R) is the potential at the larger sphere, Q is the charge at the larger sphere, q is the charge at the smaller sphere.

• Therefore any charge on the smaller sphere flows to the larger sphere. ​Therefore option 1 is correct

Additional Information

Mathematics

• We have a large spherical conducting shell of radius R, on which we place a charge Q.
• This charge spreads itself uniformly all over the sphere.
• The field outside the sphere is just that of a point charge Q at the center.
• The field inside the sphere is zero.
• The potential outside is that of a point charge.
• The potential inside is constant and has a value same as the value at the radius R.
• Potential inside conducting spherical shell of radius R carrying charge Q

\(\Rightarrow V = \frac {KQ}{R}\)

• A small sphere of radius r, carrying some charge q, is placed into the large one, at the center.
• The potential due to this new charge q is

\(\Rightarrow V = \frac {Kq}{r}\) at the surface of the small sphere

\(\Rightarrow V = \frac {Kq}{R}\) at the surface of the large sphere

• The net potential at a distance R from the center due to both the spheres is

\(\Rightarrow V(R) = \frac {KQ}{R}+\frac {Kq}{R} \)

• The net potential at a distance r from the center due to both the spheres is

\(\Rightarrow V(r) = \frac {KQ}{R}+\frac {Kq}{r} \)

• The potential difference between points at distance r and R from the center is

\(\Rightarrow V(r)-V(R) = \frac {KQ}{R}+\frac {Kq}{r} - \frac {KQ}{R}-\frac {Kq}{R} =Kq(\frac {1}{r}-\frac {1}{R})\)

• Thus potential at the smaller sphere surface is higher than the potential at the larger sphere surface.
• Therefore all the charge at the smaller sphere gets transferred to the larger sphere.

Concept:

Photoelectric Effect: 

• The phenomena by which rays of light falling on the metal surface make electrons ejected out of the metal surface is known as the Photoelectric Effect.
• Work Function (ϕ) =The minimum energy required to eject an electron out of metal is known as work function. 
• Threshold Frequency: The minimum frequency required so that the photoelectric effect can took place is called threshold frequency. 
• It is given as

 ϕ = h ν₀ 

where ν₀ is the threshold frquency,  the minimum frequency required for a light ray to eject an electron from that metal, h is Planck's constant. 

This can also be written as

\(\phi = \frac{hc}{λ }\)

λ is the threshold wavelength or the maximum wavelength of the light so that the photoelectric effect can happen,  c is the speed of light. 

Calculation:

Given

Work function  ϕ = 6 eV

Plancks constant h = 6.624 × 10-34 Js

c = 3 × 108 m / sec

\(6 \times 1.602 × 10 ^{-19} J= \frac{(6.624 × 10 ^{-34})(3 × 10 ^8 )}{λ }\)

(1 ev = 1. 602 × 10 ^-19 J )

\(\implies λ = \frac{(6.624 × 10 ^{-34})(3 × 10 ^8 )}{6 \times 1.602 × 10 ^{-19} J }\)

λ ≈ 2071 Å

(1 Å = 10 ^-10 m )

So, the correct option is 2071 Å. 

Additional Information

The  Einstein equation for the photoelectric effect is given as: 

K = hν - ϕ 

Where h is Planck constant, ν is the frequency of light falling K is the kinetic energy of an electron ejected.

The light is said to be made up of small packets of energy each having energy equal to h ν, where ν is the frequency of the source light. 

• The Equation can also be written as 

K = hc/ λ - ϕ

c is the speed of light, c = 3 × 108 m/s, λ is the wavelength.

CONCEPT:

Permanent magnet:

• Substances that at room temperature retain their ferromagnetic property for a long period of time are called permanent magnets.
• Permanent magnets can be made in the following ways:
  1. By holding an iron rod in the north-south direction and hammer it repeatedly.
  2. By holding a steel rod and stroke it with one end of a bar magnet a large number of times.
  3. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current.
• The hysteresis curve allows us to select suitable materials for permanent magnets.
• The hysteresis curve should be tall and wide for the material that is used to make a permanent magnet.
• The material that is used to make a permanent magnet should have high retentivity and high coercivity.
• Steel is a one-favored choice to make a permanent magnet.
• Other suitable materials for permanent magnets are alnico, cobalt steel, and ticonal.

EXPLANATION:

• From the above, it is clear that the permanent magnets can be made in the following ways:
  1. By holding an iron rod in the north-south direction and hammer it repeatedly.
  2. By holding a steel rod and stroke it with one end of a bar magnet a large number of times.
  3. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current.
• Hence, option 3 is correct.

CONCEPT:

• Biot-Savart's Law: Biot-Savart’s law is used to determine the magnetic field at any point due to a current carrying conductor.
• This law is although for infinitesimally small conductor yet it can be used for long conductors.

EXPLANATION:

• According to Biot-Savart Law, the magnetic field at point ‘ P ’ due to the current element \(i\overrightarrow {dl} \) is given by the expression,
  \(\overrightarrow {dB} = \frac{{{\mu _o}}}{{4\pi }}i\left( {\frac{{\overrightarrow {dl} \times \vec r}}{{{r^3}}}} \right)\)
  Where, μ_o = Absolute permeability of air or vacuum, \(i\overrightarrow {dl} \) = Current element and r = distance.
  Therefore option 4 is correct.

CONCEPT:

• The conductivity of an intrinsic semiconductor is very low at room temperature.
  • So we add a small amount of impurity to the pure semiconductor.
  • This increases the conductivity of the semiconductor by manifold.
• Doping: The deliberate addition of a desirable impurity is called doping.
• Dopants: The add impurity atoms are called dopants.

EXPLANATION:

• To increase the conductivity of the semiconductors, some suitable impurity is added to them. This process is called doping.
• Doping decreases the resistance of semiconductors.
• So the correct answer is option 1.

​Additional Information

• Semiconductors: Semiconductors are materials that have conductivity between conductors and that of insulators.
  • The missing electrons (holes) are the positively charged particles whereas electrons are the negatively charged particles in a semiconductor. The electrons and holes have an equal magnitude of charge but opposite polarity.
  • The conductivity of semiconductors is improved by adding suitable atoms called impurities and this process is called doping.
  1. ​A p-type semiconductor is a semiconductor doped with atoms with three valence electrons which forms a covalent bond with the semiconductor and leaves behind a hole.
  2. An n-type semiconductor is a semiconductor doped with atoms with five valence electrons which form four covalent bond with the semiconductor and leaves behind an extra electron.

Concept:

• Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
  • Convex lens: ​A lens having two spherical surfaces, bulging outwards is called a double convex lens (or simply convex lens).
  • It is thicker in the middle as compared to the edges.
  • Convex lenses converge light rays and hence, convex lenses are also called converging lenses.

• The lens formula is given by:

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

Where V = Distance of image from the optical center, U = Distance of object from the optical center

• Depending upon the situations convex lens produces real and virtual images.

Calculation:

• Assume the distance between the object and the real image formed by the convex lens be' X'
• Let the distance of the object from the lens be, U = -Y
• So the image distance from the lens will be, V = X - Y
• The thin lens equation is given by

\(⇒ \frac{1}{f} =\frac{1}{V} -\frac{1}{U}\)

\(⇒\frac{1}{f} =\frac{1}{X-Y} -\frac{1}{-Y}=\frac{1}{X-Y} +\frac{1}{Y}\)

⇒ Y2 - XY + fX = 0

Then the value of Y according to the quadratic equation is given by, 

\(\Rightarrow Y = {-X \pm \sqrt{X^2-4fX} \over 2}\)

For the real value of Y, the value \(-X + \sqrt{X^2-4fX}≥ 0\)

⇒ X2 - 4fX ≥ 0

⇒ X2 ≥  4fx

⇒ X ≥ 4f 

• The minimum distance between the and the real image in a convex lens is 4f.

The correct answer is option 1) i.e. 2C

CONCEPT:​

• Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field in it. 
  • It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  • Capacitance is the ability of a capacitor to store charge in it. The capacitance C is related to the charge Q and voltage V across them as:

\(⇒ C =\frac{Q}{V} \)​

• Equivalent capacitance of capacitors -
  • Connected in series: When n capacitors C₁, C₂, C₃, ... C_n are connected in series, the net capacitance (C_s) 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

⇒ C_p = C1 + C2  + C3 +...  Cn

CALCULATION:

• The given circuit can be simplified as follows:

• The capacitors in arms AB and EF are in series. Hence the equivalent capacitance in each of these arms 

\(⇒ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} ⇒ C_s = C/2\)

• The circuit arms AB, CD, and EF are in parallel and therefore, the equivalent capacitance in parallel is

⇒ C_P= C/2 + C + C/2 = 2C

CONCEPT:

De Broglie's Explanation of Bohr's Second Postulate of Quantisation:

• Bohr's second postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (6.6 × 10-34 J-sec).
• Thus the angular momentum (L) of the orbiting electron is quantized.
• So the angular momentum of the electron orbiting in the nth orbit around the nucleus is given as,

\(⇒ L=\frac{nh}{2π}\)

• Louis de Broglie argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle-wave.
• For an electron moving in the nth circular orbit of radius rn, the total distance is the circumference of the orbit.
• If the wavelength of the electron orbiting in the nth circular orbit is λ, then,

⇒ 2πrn = nλ

• We know that the De Broglie wavelength of the electron moving with speed vn in the nth orbit is given as,

\(\Rightarrow \lambda=\frac{h}{p}=\frac{h}{mv_n}\)

Limitations of the Bohr model:

1. The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two-electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. The formulation of the Bohr model involves electrical force between the positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms.
2. Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

EXPLANATION:

• Bohr's second postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (6.6 × 10-34 J-sec).
• Thus the angular momentum (L) of the orbiting electron is quantized.
• So the angular momentum of the electron orbiting in the nth orbit around the nucleus is given as,

\(⇒ L=\frac{nh}{2π}\)

• Hence, option 1 is correct.

CONCEPT:

• Consider an electric dipole consisting of two equal and opposite point charge -q at A and +q at point B separated by a small distance AB = 2l, having dipole moment p = q x 2a directed from -q to +q along the axis of the dipole.
• Let this dipole be placed in a uniform electric field E at an angle θ with the direction of E. Force on charge +q at A = qE, along the direction of E. Force on the charge -q at B = qE, along the direction opposite to E.
• Since the electric field (E) is uniform, therefore, the net force on the dipole is 0. However, as the forces are equal, unlike and parallel, acting at different points, therefore, they form a couple which rotates the dipole. Hence, the couple tends to align the dipole along the direction of the electric field (E).

∴ Torque (τ) = force × arms of a couple

τ = F × 2asinθ = (q E) × 2asinθ

τ = (q × 2a)E sinθ \(\vec{\tau }=\vec{p}\times ~\vec{E}\)

τ = pE sinθ

\(\vec{\tau }=\vec{p}\times ~\vec{E}\)

EXPLANATION:

∴ From the above, it is clear that the torque acting on a dipole moment p placed in an electric field E is \(\vec{\tau }=\vec{p}\times \vec{E}\)

So option 4 is correct.

CONCEPT:

• When a magnetic dipole of moment \(\vec M\) is held at an angle θ with the direction of a uniform magnetic field \(\vec B\), then it experiences a torque.
• The magnitude of the torque acting on the dipole is

⇒ τ = M B sin θ

CALCULATION:

Given - θ = 0°, magnetic moment (M) = \(50\hat i\;A{m^2}\), and magnetic field \(\vec B = \left( {0.5\hat i + 3.0\hat j} \right)\)

• The magnitude of the torque acting on the dipole is

⇒ τ = M B sin θ

\(\Rightarrow \tau=50\hat i\times(0.5\hat i+3.0\hat j)\)

\(\Rightarrow \tau=\begin{vmatrix} \hat i & \hat j & \hat k \\ 50 & 0 & 0 \\ 0.5 & 3& 0 \end{vmatrix}=(0)\hat i+(0)\hat j+(150)\hat k=(150)\hat kN.m\)

• Torque acting will be the cross product of magnetic moment and field. Hence only j component of field will come into play. The magnitude will be 150 N-m.

CONCEPT:

Myopia (short-sightedness): 

• It is a defect of the eye due to which a person can see a nearby object clearly but cannot see far away object clearly.
• In this defect, the image is formed before the retina and far point comes closer.

• In this defect focal length or radii of curvature of the lens is reduced or the power of the lens increases or the distance between the eye lens and retina increases i.e., the eyeball becomes too long. Therefore option 4 is correct.
• This defect can be removed by using a concave lens of suitable focal length. 

CALCULATION:

Given -  v =  -4 m, and u = ∝ 

• According to the lens formula 

\(\Rightarrow \frac{1}{f} = \frac{1 }{V} - \frac{1}{U}\)

\(\Rightarrow \frac{1}{f} = -\frac{ 1}{4} -\frac{1}{\propto}=-\frac{1}{4}\)

\(\Rightarrow f = -4 \,m\)

• The power of a lens is given 

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

\(\Rightarrow P =\frac{1}{-4}=-0.25\, D\)

• Hence option 4  is the answer

CONCEPT:

Spherical Mirror:

• A spherical mirror is a part of a hollow sphere, whose one side is reflecting and the other side is opaque.
• Two types of spherical mirrors are:

1. Concave mirror, whose reflecting surface is towards the center of the sphere of which the mirror is a part.
2. Convex mirror, whose reflecting surface is away from the center of the sphere of which the mirror is a part.

• Mirror formula: The expression which shows the relation between object distance (u), image distance (v), and focal length (f) is called the mirror formula.

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

Linear magnification (m):

• It is defined as the ratio of the height of the image (hi) to the height of the object (ho).

\(\Rightarrow m = \frac{{{h_i}}}{{{h_o}}}\)

• The ratio of image distance to the object distance is called linear magnification.

\(\Rightarrow m = \frac{{image\;distance\;\left( v \right)}}{{object\;distance\;\left( u \right)}} = - \frac{v}{u}\)

• A positive value of magnification means virtual an erect image.
• A negative value of magnification means a real and inverted image.

CALCULATION:

Given - m = 1/n, focal length = +f, and object distance = -u

\(\Rightarrow \frac{-v}{-u} = \frac{1}{n} \Rightarrow v = \frac{u}{n}\)

• Now from equation of mirror,

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

\(\Rightarrow \frac{u}{n} - \frac{1}{u} = \frac{1}{f}\) [As u is negative in convex mirror]

\(\Rightarrow \frac{u}{n} - \frac{1}{u} = \frac{1}{f}\)

⇒ u = (n – 1)f

Calculation:

Given that,

We have two parallel wires kept apart by = 1 m

Current through wire 1 (I₁) = 1A

Current through wire 2 (I₂₎ = 2A

The expression of force per unit length acting between two wires is given by,

\(F = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\)

By substituting the given values

\(F = \frac{{4\pi \times {{10}^{ - 7}} \times 1 \times 3}}{{2\pi \times 1}} = 6 \times {10^{ - 7}}N/m\)

Also, we know that force between two parallel current carrying conductors in the same direction attract each other and those carrying currents in the opposite direction repel each other.

As