CBSE Sample Papers for Class 12 Physics Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Physics with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Physics Set 3 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70 Marks

General Instructions:

1. There are 35 questions in all. All questions are compulsory.
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
3. Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study-based questions of 4 marks each.
4. There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.
5. Use of calculators is not allowed.

Section – A (18 Marks)

Question 1.
Which of the following has the maximum mutual inductance. [1]

Answer:
(a)

Explanation:
Coeffcient of coupling is maximum in option (a) and hence the mutual inductance between the coils is maximum.

Related Theory
Mutual inductance between two coils depends upon the manner in which the two coils are oriented relative to each other.
K = \(\sqrt{\frac{M}{L_1 L_2}}\)

Question 2.
Which of the following will be most effective for emission of electrons from a metallic surface? [1]
(a) Microwaves
(b) Ultraviolet rays
(c) Infraredrays
(d) All of these
Answer:
(b) Ultraviolet rays

Explanation:
Ultraviolet rays are most effective for photoelectric emission because they have highest frequency which makes them the most energetic electromagnetic waves.

Question 3.
The energy gap between the valence and conduction bands of a substance is 6 eV. The substance is a: [1]
(a) conductor
(b) semiconductor
(c) insulator
(d) superconductor
Answer:
(c) insulator

Explanation:
Insulators have a significant energy gap (approximately 6eV) between the valence and conduction bands, while semiconductors have a smaller one and conductors have the lowest.

Question 4.
Considering the following figure, where V₁ > V₂, which of the following statement is correct? [1]

(a) The lines of force are tangent to the equipotential surface.
(b) The lines of force are parallel to the equipotential surface
(c) The lines of force are parallel, where potential is less and perpendicular, where potential is higher.
(d) The lines of forces are everywhere perpendicular to the equipotential surface.
Answer:
(d) The lines of forces are everywhere perpendicular to the equipotential surface.

Explanation:
The potential difference between the two surfaces has nothing to do with the formation of lines of forces.

Related Theory
The lines offerees are everywhere perpendicular to the equipotential surface and directed in the direction of increasing potential.

Question 5.
Which of the following statement is incorrect? [1]
(a) As the slit width decreases, the width of central maximum with increase.
(b) Distance between the slit and the screen is increased, the width of central maximum with increases.
(c) Distance between the slit and the screen is increased, the width of central maximum with decreases.
(d) As the light of smaller wavelength is used, the width of central maximum with decrease.
Answer:
(c) Distance between the slit and the screen is increased, the width of Central maximum with decreases.

Explanation:
Width of central maximum is inversely proportional to the slit width and directly proportional to the distance between the slit the screen and wavelength of the light
β₀ = \(\frac{2 \mathrm{D} \lambda}{a}\)
Considering the above formula, we can conclude that option (c) is incorrect.

Question 6.
Which of the following statement is incorrect? [1]
(a) The resistivity of semiconductor increases with increase in temperature.
(b) The conductors have overlapped valence and conduction bands.
(c) Substances with energy gap of order of lOeV are insulators.
(d) The conductivity of a semiconductor increases with increase in temperature.
Answer:
(a) The resistivity of semiconductor increases with increase in temperature.

Explanation:
The resistivity of a semiconductor decreases with temperature. This is because of increasing temperature, the electrons in the valence band gain sufficient thermal energies to jump to the conduction band.

Question 7.
Three capacitors of equal capacitance are combined in different arrangement as given below. Which arrangement will have the largest effective capacitance? [1]

Answer:
(a)

Explanation: Effective Capacitance for series combination is given by,
\(\frac{1}{C_{e f f}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots+\frac{1}{C_n}\)
Effective Capacitance for parallel combination is given by,
C_eff = C₁ + C₂ + C₃ + ………… + C_n
Effective capacitance in (a) will be = 3C
Effective capacitance in (b) Will be = \(\frac{3 C}{2}\)
Effective capacitance in (c) will be = \(\frac{2 C}{3}\)
Effective capacitance in (d) will be = \(\frac{C}{3}\)
Taking an arbitrary value of C. we obtain the largest value of effective capacitance in option (a).

Question 8.
Rays enter in a circular body of refractive index p and forms the image at the edge of the mirror as shown in the given figure. [1]

What will be the value of refractive index (µ)?
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(a) 2

Explanation:
\(\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\)
\(\frac{\mu}{2 R}-\frac{1}{-\infty}=\frac{\mu-1}{R}\)
\(\frac{\mu}{2 R}=\frac{\mu-1}{R}\)
µ = 2µ – 2
µ = 2

Question 9.
Which of the following statement is true for a transformer? [1]
(a) Due to resistance of the windings of primary and secondary coils, some electrical energy is always converted into heat energy.
(b) The core of the transformer stops the vibration and the Humming sound due to passage of alternating current.
(c) By eliminating the energy loss, we can produce a real-life transformer with 100% efficiency.
(d) During each cycle of magnetisation, no energy is lost due to hysteresis.
Answer:
(A) Due to resistance of the windings of primary and secondary coils, some electrical energy is always converted into heat energy.

Explanation:
In practice, a number of types of energy losses occur in transformer making the efficiency of a practical transformer less than hundred percent Due to the resistance of copper windings, electrical energy is always converted into heat energy. Due to the passage of alternating current the core of a transformer starts vibrating and produces humming noise.

Question 10.
In respect to electromagnetic spectrum, which of the following is incorrect? [1]
(a) Speed of electromagnetic waves is same in vacuum.
(b) Waves interact with matter through their electric and magnetic fields.
(c) Infrared waves incident on a substance set into oscillation all its electrons, atoms and molecules.
(d) Our eyes early sensitive to solar the spectrum.
Answer:
(d) Our eyes early sensitive to solar the spectrum.

Explanation:
Our eyes are most sensitive to the most intense wavelengths of the Solar spectrum. That is the centre of the sensitivity of our eyes consides with the centre of wavelength distribution of the solar radiation.

Question 11.
The graph between the stopping potential and frequency of the incident radiation is shown below. [1]

The work function will be
(a) 0.400eV
(b) 0.580eV
(c) 0.158eV
(d) 0.414eV
Answer:
(d) 0.414eV

Explanation:
Work function,
Φ₀ = hν₀
hν₀ = 6.63 × 10^-34 × 1 × 10¹⁴
= 6.63 × 10^-20 J
= \(\frac{6.63 \times 10^{-20}}{1.6 \times 10^{-19}}\) = 0.41 eV

Question 12.
The given figure shows emission lines of three different wavelengths due to transition between three energy levels. [1]

Then the relation between the energies in terms of wavelength will be:
(a) \(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2}+\frac{h c}{\lambda_3}\)
(b) \(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2 \lambda_3}\)
(c) \(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2} \times \frac{h c}{\lambda_3}\)
(d) \(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2}-\frac{h c}{\lambda_3}\)
Answer:
(a) \(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2}+\frac{h c}{\lambda_3}\)

Explanation:
E₃ – E₁ = (E₃ – E₂) + (E₂ – E₁)
E₃ – E₁ = \(\frac{h c}{\lambda_1}\)
E₃ – E₂ = \(\frac{h c}{\lambda_2}\)
E₂ – E₁ = \(\frac{h c}{\lambda_3}\)
Substituting the vaLues in the above equation,
we get,
\(\frac{h c}{\lambda_1}=\frac{h c}{\lambda_2}+\frac{h c}{\lambda_3}\)
Above equation can also be written os,
\(\frac{1}{\lambda_1}=\frac{1}{\lambda_2}+\frac{1}{\lambda_3}\)

Question 13.
The figure shows a prism with equal perpendicular and base. [1]

What will be the refractive index of the material which is used to make the prism?
(a) µ = 1.5
(b) µ = 2
(c) µ = \(\sqrt{2}\)
(d) µ = \(\sqrt{\frac{3}{2}}\)
Answer:
(c) µ = \(\sqrt{2}\)

Explanation:
The ray of light is incident on the face BC of the prism normally. It passes as such into a prism and after undergoing total internal reflection at the faces AB and AC, it falls on the face BC and emerges out of the prism. Angle of incidence on the face AB or AC is 45°, which is the value of critical angle for the material of the prism.
µ = \(\frac{1}{\sin C}=\frac{1}{\sin 45^{\circ}}\)
= \(\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}\)

Question 14.
An electron has a speed of 6.6 × 10⁴m sec^-2 within accuracy of 0.01%. Calculate the uncertainty in the position of the electron. [1]
(a) 8.8 × 10^-6 m
(b) 0
(c) 2.4 × 10^-6 m
(d) 3
Answer:
(a) 8.8 × 10^-6 m

Explanation:
The momentum of electron,
p = mv
= (9 × 10^-31) × (6.6 × 10⁴)
= 59.4 × 10^-27 kg m sec^-1
The uncertainty in the momentum is 0.01% of 59.4 × 10^-27 kg m sec^-1.
∆p = \(\frac{0.01}{100}\) × 59.4 × 10^-27 kg m sec^-1
= 59.4 × 10^-31 kg m sec^-1
From Heisenberg uncertainty relation,
∆x∆p ≥ \(\frac{h}{4 \pi}\)
∆x ≥ \(\frac{h}{4 \pi \Delta p}\)
= \(\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 59.4 \times 10^{-31}}\)
∆x = 8.8 × 10^-6 m

Question 15.
Particle with rest mass ma is moving with speed c. What will be the de-Broglie wavelength associated with it? [1]
(a) ω
(b) – 1
(c) 0
(d) 1
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.
Answer:
(c) 0

Explanation:
λ = \(\frac{h}{m v}\)
In the given case,
λ = \(\frac{h}{m c}\)
m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Putting v = c
We get, m = ∞
de-Brogbe wavelength associated with it, will be:
λ = \(\frac{h}{\infty}\) = 0

Question 16.
Assertion (A): Magnetic field lines can be entirely confined within the core of a toroid but not a straight solenoid. [1]
Reason (R): In a straight solenoid, the magnetic field lines cannot farm closed loops within the solenoid.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
The magnetic field lines always form closed loops. As the turns of the wire in a toroidal solenoid are wounded over its core in circular form, the field lines are confined within the core of the toroid.

Question 17.
Assertion (A): It is not possible to define ac ampere in terms of forces between two parallel wires carrying AC currents. [1]
Reason (R): AC current defined in terms of Joule heating which is independent of direction of current.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Alternating current changes direction with the source frequency and the net force would add up to zero. RMS value of alternating current in a circuit is 1 ampere of the current that produces the same average heating effect as 1 ampere of direct current would produce under the same conditions.

Question 18.
Assertion (A): Current carrying loop on a smooth horizontal plane cannot turn around itself in a uniform magnetic field. [1]
Reason (R): Area vector always acts perpendicular to the plane of the loop.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
The magnetic field cannot be set up to turn the loop around itself For turning a loop around itself, top should be along the vertical. Sense the area vector acts along the perpendicular to the plane of loop, i.e., along the vertical, torque cannot act along the vertical.

Section -B (14 Marks)

Question 19.
A circular loop consists a capacitor. If two magnets are directed towards each other through the loop facing North and South pole respectively what will be the polarity of the capacitor? [2]
OR
How does the threshold frequency depend on the nature of photo metals?
Answer:
When two magnets of opposite poles are moved to a circular loop consisting a capacitor, both magnets will induce current in clockwise direction when seen from the right-hand side of the loop. The first plate will attain positive charge and the second plate will attend negative charge.

OR

Maintaining the same intensity and frequency of incident light, different photo metals are exposed to light. Plotting the graph between stopping potential and frequency for different photo metals, straight line curves are obtained, it is clear that all graphs have the same slope but they intersect the frequency axis at different points indicating that different photo metals have different threshold frequency.

Question 20.
During diffraction of a single slit, the slit stone from position AB to A’B’ as shown below. How will it affect the central maximum? [2]

Answer:
The path difference between the rays reaching the centre of the screen will remain zero as the extra path AA’ covered by the light to travel from the source of life to the top edge of the Slate is compensated by the extra path BB1, of the light travels from bottom edge of the slit to the screen. As a result, the central maximum does not move.

Question 21.
Why are microwave best suited for long distance transmission of signals? What evidence is there that sound are not electromagnetic in nature? [2]
Answer:
Microwaves have wavelength of the order of a few millimetres. Due to their short wavelength, this is not affected too much by objects of normal dimensions. So, they can be transmitted as a propagation medium in a particular direction. Sound waves require an inertial medium for their propagation which shows that sound waves are mechanical in nature and not electromagnetic.

Question 22.
State the two factors on which the resistivity of the material of a conductor depends on. [2]
OR
In what ways an electric field is different from a magnetic field?
Answer:
The two factors on which the resistivity of the material of a conductor depends are:

1. It is inversely proportional to the number of free electrons per unit volume of the conductor. Since the value depends upon nature of the material, the resistivity of a conductor depends upon the nature of the material.
2. It is inversely proportional to the average relaxation time of free electrons in the conductor. The resistivity of conductor depends upon its temperature and it increases with increase in temperature of the conductor.

OR

1. An electric field is caused by charges both at rest and in motion, whereas a magnetic field is caused by a magnet, current flowing through a conductor, or a moving change.
2. When a permeable medium is placed at a spot, the intensity of the electric field diminishes while the strength of the magnetic field increases.
3. Electric lines of force, which represent the electric field, do not form closed courses, but magnetic lines of force do.

Question 23.
Explain the temperature dependence ofelectron and hole concentration. [2]
Answer:
As we have that at 0 K, the intrinsic semiconductor behaves as an insulator. On increasing the temperature of the intrinsic semiconductor, covalent bonds are broken due to thermal energy and electron hole pairs are generated. So, an increase in temperature of a semiconductor increases the number of free electrons and holes and also conductivity of the material.

In an intrinsic semiconductor, the number of electrons is equal to number of holes. In equilibrium, the concentration of electrons is equal to concentration of holes and is known as intrinsic carrier concentration.
i.e., n = p = n_i.

Question 24.
An n-type semiconductor crystal has more free electrons than holes. Is it negatively charged? [2]
Answer:
An n-type semiconductor has free electrons as charge carriers. These are donated by pentavalent impurity atoms which becomes positively charged. Although there are some thermally generated electron-hole pairs, but the number of these holes is negligibly small in comparison to the total number of electrons.

Thus, n-type semiconductor mainly consists of negatively charged free electrons and nearly equal number of positively charged donor ions. Hence, the material as a whole is electrically neutral.

Question 25.
Draw a labelled diagram of refraction type telescope in normal adjustment and state two shortcomings over reflection type telescope. [2]
Answer:
Rays from top of a distant object

The shortcomings of reflection type telescope are:

1. The large of defective lens used is very heavy, which makes it difficult for manufacturing and support by its edges.
2. It is difficult and expensive to make large size lenses free from chromatic aberration and distortions,

Section – C (15 Marks)

Question 26.
Draw magnetic field lines when a diamagnetic material is placed inside a magnetic field, when a paramagnetic material is placed inside a magnetic field and when a ferromagnetic material is placed inside a magnetic field. [3]
OR
Compare the set-up for Young’s double slit experiment, when sources a line at the centre line between the slits, and when it’s shifted slightly above. How will is affect the fringe formation?
Answer:
When a diamagnetic material is placed inside a magnetic field,

When a paramagnetic material is placed inside a magnetic field,

When a ferromagnetic material is placed inside a magnetic field,

OR
In the normal. position of the source with respect to double slits, the central. bright fringe is formed at the point O.

When the source of light is not exactly on the centre line, but moves slightly upwards, the central bright fringe will move downwards.

Question 27.
State any three important advantages of a reflecting type telescope over a refracting type telescope. [3]
Answer:
(A) Concave mirror of large aperture has less gathering power and absorbs very less amount of light then the lenses of large apertures. The final image formed in the reflecting type telescope is very bright. So, even very distant or faint stars can be easily viewed.

(B) The mirror requires grinding and polishing of one surface only. So, it costs much less to construct a reflecting telescope then a refracting telescope of equivalent optical quality. The use of paraboloidal mirrors reduces spherical aberration.

(C) A lens of large aperture tends to be very heavy and therefore, difficult to meet and support by its edges. On the other hand, a mirror of equivalent optical quality weighs less and can be supported over its entire back surface.

Question 28.
Draw a circuit diagram of full wave bridge rectifier with its input, output characteristics. [3]
Answer:

Question 29.
Explain the temperature dependence of the resistivity of a semiconductor and an insulator. [3]
OR
State the junction rule and the loop rule in electrical measurements.
Answer:
The vibration of resistivity with temperature in case of a semiconductor or an insulator is different from a conductor. In case of a conductor, the resistivity increases linearly with the temperature of the conductor. The resistivity of a semiconductor and insulator decreases exponentially with temperature. The resistivity at temperature T of a semiconductor and an insulator is given by,
ρ = ρ₀ e^E_g/kT

OR

The junction rule is also known as the Kirchhoffs first law. It states that the algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Σ I = 0
Kirchhoff’s second law states that in any closed path of an electrical circuit, the algebraic sum of the EMFs is equal to the algebraic sum of the product of the resistance and current flowing through them. It is also known as loop rule. Around any closed loop of the network, the algebraic sum of changes in potential must be zero.
Σ ∆V = 0
or Σ e = ΣIR

Question 30.
A rectangular loop of area 6 × 10^-2m² is placed in a magnetic field 0.6T with its plane. [3]
(a) normal to the field
(b) inclined 30° to the field
(c) parallel to the field.
What will be the flux linked in each case.
Answer:
Area of the loop = 6 × 10^-2 m²
Magnetic field = 0.6T
Let θ be the angle formed between the plane of the coil and magnetic field B.
(A) Here θ = 90° – 90° = 0
Φ = BA cos θ
= 0.6 × 6 × 10^-2 x cos 0°
Φ =3.6 × 10^-2 Wb

(B) Here θ = 90° – 30° = 60°
Φ = BA cos θ
= 0.6 × 6 × 10^-2 × cos 60°
Φ = 1.8 × 10^-2 Wb

(C) Here θ = 90°
Φ = BA cos θ
= 0.6 × 6 × 10^-2 × cos 90°
Φ = Zero

Section – D (15 Marks)

Question 31.
Charge Q coulomb is uniformly distributed over a sphere volume of radius R meters. Obtain an expression for the energy of the system. [5]
OR
Two dielectric slab of dielectric constants happen port in between the plates of capacitor. What should be the capacitance of the capacitor? Show the variation between capacitance and total energy stored in the capacitor.
Answer:
Consider a sphere of radius R. The charge Q is uniformly distributed over its volume.

If ρ is volume charge density, then,
ρ = \(\frac{Q}{\frac{4}{3} \pi R^3}\)
The sphere of volume charge density may be assumed to be built by putting a spherical layer of increasing radii around the initial spherical core, till the sphere of radius R is formed.

Suppose at any instant, there is a spherical core of radius x,
Then, charge on the spherical core,
q = \(\frac{4}{3}\) πx³ρ
Suppose a spherical layer of infinitely small thickness dx having charge density p is put around the spherical core. The small amount of charge on the spherical layer,
dq = volume of the layer × ρ
= (4π x³ dx)ρ
The system of the spherical core having charge q and the spherical layer having charge dq processes electrostatic potential energy, which is given by,
du = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q d q}{x}\)
= \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{4}{3} \pi x^3 \rho \times\left(4 \pi x^2 d x\right) \rho}{x}\)
= \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{16}{3} \pi^2 \rho^2 x^4 d x\)
The electrostatic energy of the sphere of radius R can be obtained by integrating the above between the limits x = 0 and x = R.
= \(\int_0^R d U=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{16}{3} \pi^2 \rho^2 \int_0^R x^4 d x\)

OR

The arrangement is equivalent to a series combination of two capacitors, each of plate area A, plate separation \(\frac{d}{2}\), such that one has medium of dielectric constant K₁ and other of dielectric constant K₂. C₁ and C₂ are the capacitance of two capacitors,

Total energy stored in a capacitor is given by,
U = \(\frac{q^2}{2 C}\)
U ∝ \(\frac{1}{C}\)
Therefore,

Question 32.
Explain the conditions for interference of light. [5]
OR
What is a rainbow and how does a primary rainbow differ from a secondary rainbow? Explain the phenomenon with the help of a diagram.
Answer:
If we have two or more sources of light sending wave simultaneously in any region, then according to definition, we must always have phenomenon of interference. But in practice this interference pattern is not good or visible.

To obtain a well-defined observable interference pattern, the intensity at the points corresponding constructive and destructive interference must be maintained maximum and minimum respectively. For obtaining a visible interference following conditions must be fulfilled.

That two interfering waves should be coherent i.e:, the Waves must be in the same phase or must have a constant phase difference. The phase difference is not constant and is changing with time, and then at any point the constructive interference may exist and very short time later at this point destructive interference may be. Hence, the same point will appear sometimes bright and sometimes dark or Baton will not be sustained. So, this condition is known as condition for sustained interference.

The two waves must have same wavelength and frequency i.e., sources should be monochromatic. If the sources are chromatic (have many frequencies), then overlapping of various coloured fringes will produce a confused pattern of fringes. Also, if wavelength is different, the phase difference between the two fringes will continuously vary. Hence, the intensity of light at the same point will vary with time.

In that two waves must have the same attitude. If it is not so, then the minimum amplitude will not be zero and good contrast between bright and dark fringes will not be observed. So, this condition is known as condition for good contrast.

It may also be noted that the fringe width is given by,
β = \(\frac{\mathrm{D} \lambda}{2 d}\)
So, in order to get clear what is available fringes,
The separation between the light sources should be very small.

The distance between sources and screen must be kept quiet large compared to distance between the sources.

Here, it is important to note that according to the theories of light, every source emerging the light, emits a continuous wave only. After that it emits a fresh wave having different initial phase. So, two independent sources can never be considered to behave as coherent sources and thus, to obtain sustained interference pattern we take two sources originated by the same source in young’s double slit experiment.

OR

The rainbow is nature’s most spectacular display of the spectrum of light produced by refraction, dispersion and internal reflection of sunlight by spherical rain drops. It is observed when the sun shines on rain drops, during or after a shower.

An observer standing with his back towards the sun observes it in the form of concentric circular arcs (bows) of different colours in the horizon. The inner brighter rainbow is called the primary rainbow and the outer fainter rainbow is called the secondary rainbow.

The primary rainbow is formed by rays which undergo one internal reflection and two refractions and finally emerge from the raindrops at minimum deviation. The red rays emerge from the water drops at one angle of 43° and the violet rays emerge at another angle of 41°. The parallel beam of sunlight getting dispersed at these angles produces a cone of rays at the observer’s eye. Thus, the rainbow is seen as a colourful arc, with its inner edge in violet and outer edge in red colour.

The secondary rainbow is formed by the rays which undergo two internal reflections and two refractions before emerging from the water drops at minimum deviation. Due to two internal reflections, the sequence of colour in secondary rainbow is opposite to that in the primary rays emerge from the water drops at an angle of 51° and the outer violet rays emerge at an angle of 54°.

Question 33.
Consider a long solenoid of length l, area of cross section A and number of turns per unit length n. Derive an expression for the energy stored in a solenoid? [5]
OR
Define the root mean square value of an alternating EMF. Derive the relation between root mean square value and peak value of an alternating EMF.
Answer:
By the given description, self-inductance of the solenoid is given by,
L = µ_o n²IA
Suppose that the current I it is passed through the solenoid. Then, the magnetic field inside the solenoid is,
B = µ_o n I
I = \(\frac{\mathrm{B}}{\mu_0 n}\)
When current flows through an inductor having self-inductance L, the energy stored is given by,
U = \(\frac{1}{2}\) L I²
U = \(\frac{1}{2}\) (µ_o n²IA) × \(\left(\frac{\mathrm{B}}{\mu_0 n}\right)\)
U = \(\frac{1}{2 \mu_0}\) B² A I
The energy resides in the solenoid in form of magnetic field, the magnetic field stored per unit volume in the solenoid.
u = \(\frac{U}{V}\)
u = \(\frac{B^2}{2 \mu_0}\)

OR

Root mean square value of an alternating EMF is defined as that value of a steady voltage that produces the same amount of heat in a given resistance as is produced by the given alternating EMF when applied to the same resistance for the same time. It is also called virtual or effective value of the alternating EMF.

Suppose an alternating emf e applied to a resistance R is given by,
ε = ε₀ sin ωt
Heat produced in a small time dt will be,
dH = \(\frac{\varepsilon^2}{R}\) dt
= \(\frac{\varepsilon_0^2}{R}\) sin² ωt dt
Let T be the time period of the alternating emf.
Then heat produced in time T will be,

If ε_rms is the root mean square value of the alternating emf, then the amount of heat produced by it in the same resistance R in the time T will be,

Section – E (8 Marks)

Question 34.
The current voltage graph of junction diode is non-linear, which means it does not obey Ohm’s law. The resistance of the junction diode varies with applied voltage. In such cases, it is used for to define A quantity known as dynamic or AC resistance of the voltage. It is the ratio of small change in applied voltage to the corresponding change in current. Above the threshold voltage, the diode characteristic is linear.

(A) Using a sensitive voltmeter across the terminals, is it possible to measure the potential difference of p-n junction? [1]
(B) If two p-type and n-type semiconductors have equal level of doping, then why do they have unequal connectivity? [1]
(C) Small current flowing through a reverse biased junction diode is called reverse saturation current. Why?
OR
Why does an extrinsic semiconductor behaves as an intrinsic semiconductor at high temperature? [2]
Answer:
(A) There are no free charge carriers in the depletion region. It offers infinite resistance in the absence of any forward biasing. Therefore, it would be impossible to measure the potential difference using a sensitive voltmeter,

(B) Conductivity of n-type semiconductor is greater than that of p-type semiconductor given electric field, free electrons have higher mobility than holes.

(C) The reverse current is due to the thermally generated minority carriers. We cannot increase the number of these minority Carriers by applying an increase in the reverse voltage. So, it is termed as saturation current.

OR

At high temperature concentration of thermally generated carriers (electrons and holes) will be much larger than the concentration of free electrons contributed by the donors. At this situation the hole and electron concentration will be nearly equal.

Question 35.
When a thin film is seen with a monochromatic light we find alternate bright and dark fringes. But with white light, brilliant colours are seen. This is because the path difference between two successive rays depends on refractive index, thickness of the Mm and angle of refraction. Sometimes t and r are fixed but the wavelength of light varies, show the condition for maxima and minima for different
constituents occur at different points of the thin film. As a result, the refracted light shows various colours.

(A) What is optical retardation? [1]
(B) How are the colours emerging from a prism different from colours emerging from a soap film? [1]
(C) During an interference experiment, two sources are infinitely close to each other. How will it affect the interference pattern? What if two sources are far apart from each other?
OR
What is lateral shift of fringes? [2]
Answer:
(A) When attend transparent plate is introduced in the path of one of the two interfering beams, then the optical path of that being is retarded and it is known as optical path retardation.

(B) In the prism, colours are produced due to Dispersion of light The colours of a soap film are due to interference of light

(C) When two coherent sources are placed infinitely close to each other, the fringe width becomes very large. Even a single fringe may occupy the entire screen. Due to this, the interference pattern may not be observable. As the distance between the two sources is increased, the fringe width goes on increasing. At a very large separation, it becomes too smalt to be detected. The interference pattern again cannot be observed.

OR

When a thin transparent plate is introduced in the path of one of two interfering beams, the entire fringe is shifted towards the beam in the part of which plate is introduced. This shift of entire fringe pattern is known as lateral shift of fringes.