Students must start practicing the questions from RBSE 12th Physics Model Papers Set 2 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 2 with Answers in English

Time . 2 Hours 45 Min.

Max Marks: 56

General Instruction to the Examinees:

- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A

Multiple Choice Questions

Question 1.

Write the Correct Answer from multiple choice question 1 (i to ix) and write in given answer book:

(i) Total electric flux coming out of a unit positive charge put in air is: [1]

(a) ε_{0
}(b) ε^{-1}

(c) \(\frac{q}{4 \pi \varepsilon_{0} a}\)

(d) 4πε_{0}

Answer:

(b) ε^{-1}

(ii) The distance between H^{+} and Cl^{–} ions in HCl molecules is 1.38 Å. The potential due to this dipole at a distance of 10 Å on the axi s of dipole is: [1]

(a) 2.1V

(b) 1.8 V

(c) 0.2 V

(d) 1.2 V

Answer:

(c) 0.2 V

(iii) Equivalent resistance of the given network is: [1]

(a) 28

(b) 18

(c) 26

(d) 25

Answer:

(b) 18

(iv) Which of the following is not correct about cyclotron? [1]

(a) It is a machine to accelerate charged particle or ions to high energies.

(b) Cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles.

(c) The operation of the cyclotron is based on the fact that the time for one revolution of an ion is independent of its speed or radius of its orbit.

(d) The charged particles and ions in cyclotron can move on a arbitrary path.

Answer:

(d) The charged particles and ions in cyclotron can move on a arbitrary path.

(v) Lenz’s law is a consequence of the law of conservation of: [1]

(a) Charge

(b) Energy

(c) Induced emf

(d) Induced current

Answer:

(b) Energy

(vi) Which one among the following shows particle nature of light?

(a) Photoelectric effect

(b) Interference

(c) Refraction

(d) Polarization

Answer:

(a) Photoelectric effect

(vii) Following process is known as hv → e^{+} + e^{–}: [1]

(a) Pair production

(b) Photoelectric effect

(c) Compton effect

(d) Zeeman effect

Answer:

(a) Pair production

(viii) The PN unction diode is used as: [1]

(a) An amplifier

(b) A rectifier

(c) An i pcillator

(d) A modulator

Answer:

(b) A rectifier

(ix) Which one is forward bias: [1]

(d) None of these

Answer:

Question 2.

Fill in the blanks:

(i) CO_{2} is the ……………………… dielectrics. [1]

(ii) Meter Bridge is the simplest practical application of the ………………….. . [1]

(iii) A.C. generator is based on the principle of …………………. . [1]

(iv) In forward bias depletion layer is very …………………… . [1]

Answer:

(i) Non-polar

(ii) Wheatstone bridge

(iii) electromagnetic induction

(iv) thin.

Question 3.

Give the answer of the following questions in one line:

(i) When a potential difference is applied across the ends of a conductor, how is the drift velocity of the electrons related to the relaxation time? [1]

Answer:

Average drift velocity,

υ_{d} = \(\frac{e E}{m}\)τ

where, e = charge on electron,

m = mass of electron,

E = electric potential or field across conductor

and τ = relaxation time.

(ii) Mention two characteristics properties of the material suitable for making the core of a transformer. [1]

Answer:

- Low retentivity,
- High permeablility.

(iii) A metallic piece gets hot when surrounded by a coil carrying high frequency alternating current. Why? [1]

Answer:

When a metallic piece is surrounded by a coil carrying high frequency alternating current, then eddy current is developed in the metallic piece and due to presence of resistance, metallic piece becomes hot.

(iv) Define the term intensity in photon picture of electromagnetic radiation. [1]

Answer:

It is the number of photons passing through an area in a given interval of time. Its SI unit is watt/metre^{2}.

(v) For a given photosensitive material and with a source of constant frequency of incident radiation, how does the photocurrent vary with the intensity of incident light? [1]

Answer:

Threshold frequency does not depend upon the intensity of light. The intensity of light mainly depends on the number of photons for given frequency of incident radiation. Therefore, the photoelectric current increases with the intensity of incident light.

(vi) A nucleus \({ }_{92}^{238} \mathbf{U}\) undergoes α-decay and transforms to thorium. What is: [1]

(i) the mass number, and

(ii) atomic number of the nucleus produced?

Answer:

In α-decay, the mass number of parent nucleus decreases by 4 units and atomic number decreases by 2 units.

∴ ^{238}U_{92} → ^{234}Th_{90} + _{2}He^{4}

(a) Mass number of the nucleus produced = 234

(b) Atomic number of nucleus produced = 90

(vii) What is the relationship between decay constant and mean life of a radioactive nucleus? [1]

Answer:

Relationship between decay constant and mean life of a radioactive nucleus is

τ = \(\frac{1}{\lambda}\)

(viii) Identify the semiconductor diode whose V-I characteristics are as shown. [1]

Answer:

The diode having these type of V-I characteristics is photodiode.

Section – B

Question 4.

A long charged cylinder of linear charge density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders? [1½]

Answer:

In figure, A is a long charged cylinder of linear charge density λ, length l and radius a.

The charge q = λl spreads uniformly on the outer surface of A.

The electric flux through cylindrical Gaussian surface is

Φ_{E} = \(\int \vec{E} \cdot \overrightarrow{d s}=\int E d s \cos 0^{\circ}\)

\(E \int d \mathrm{~s}\) = E(2πrl)

The electric flux through the end faces of the cylindrical Gaussian

surface is zero, as \(\vec{E}\) is parallel to them.

According to Gauss’s theorem,

Φ_{E} = E(2πrl) = \(\frac{q}{\varepsilon_{0}}=\frac{\lambda l}{\varepsilon_{0}}\)

E = \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\)

Question 5.

(a) A comb run through one’s dry hair attracts small bits of paper, why?

(b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary?

(c) A bird perches on a bare high power line and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why? [1½]

Answer:

(a) This is because the comb gets charged by friction. The molecules in the paper get polarised by the charged comb, resulting in a net force of attraction.

(b) The tyres of aircraft are made by a special quality of rubber having conductivity to enable then to conduct charge (produced by friction) to ground.

(c) Current passes only when there is difference in potential.

Question 6.

Find the relation between drift velocity and relaxation time of charge carriers in a conductor. A conductor of length L is connected to DC source of emf E. If the length of the conductor is tripled by stretching it, keeping E constant, explain how its drift velocity would be affected. [1½]

Answer:

When a conductor is subjected to an electric field E, each electron experiences a force

F = – eE, and free electron acquires an acceleration,

a = F/m = – eE/m ……………… (i)

where, m = mass of electron, e = electronic charge and E = electric field.

The average time difference between two consecutive collisions is known as relaxation time of electron and

\(\bar{\tau}=\frac{\tau_{1}+\tau_{2}++\tau_{n}}{n}\) …………. (ii)

Draft Velocity

v_{d} = 0 + aτ

[ Average thermal velocity in n collisions = 0]

v_{d} = -(eE/m)τ [from Eq. (i)]

This is the required expression of drift speed of free electrons.

Drift velocity, υ_{d} ∝ E_{0} [electric field]

Thus,

\(\frac{\left(v_{d}\right)_{f}}{\left(v_{d}\right) i}=\frac{\left(E_{0}\right)_{f}}{\left(E_{0}\right) i}\)

= \(\frac{E / l_{f}}{E / l_{i}}\) = \(\frac{l_{i}}{l_{f}}=\frac{l_{o}}{3 l_{o}}=\frac{1}{3}\)

Thus, (υ_{d})_{f} =(υ_{d})_{i}/3

Thus, drift velocity decreases three times.

Question 7.

Using KirchhofPs rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit. [1½]

Answer:

Taking loops clockwise as shown in figure.

Using KVL in ABCDA,

-80 + 20I_{1} +40 (I_{1} -I_{2}) = 0

⇒ 3I_{1} – 2I_{2} = 4 ………………. (i)

Using KVL in DCFED

-40(I_{1} – I_{2}) + 10I_{2} – 40 = 0

-4I_{1} + 5I_{2} = 4 ……………….. (ii)

From Eqs. (i) and (ii), we get

I_{1} = 4 A and I_{2} = 4 A

Thus, I_{40} = I_{1} – I_{2} = 0 A

I_{20} = I_{1} = 4 A

Question 8.

What are eddy currents? Write their two applications. [1½]

Answer:

Eddy current : Eddy currents are the currents induced in the bulk pieces of conductors when the amount of magnetic flux linked with the conductor changes.

Eddy currents can be minimised by taking laminated core, consists of thin metallic sheet insulated from each other by varnish instead of a single solid mass. The plane of the sheets should be kept perpendicular to the direction of the currents. The insulation provides high resistance hence, eddy current gets minimsed. Applications:

- Electromagnetic damping
- Induction furnace.

Question 9.

A coil whose inductance is 2H and resistance is 10 Ω, is connected to a battery of 100 V and negligible internal resistance. Determine: [1½]

(i) Time constant of the circuit.

(ii) The stable current in the circuit.

(iii) The energy stored to obtain stable current in the coil.

Answer:

(i) Time constant of circuit

\(\frac{L}{R}=\frac{2}{10}\) = 0.2

(ii) Stable current in the circuit,

I_{0} = \(\frac{\varepsilon}{R}=\frac{100}{10}\) = 10 A

(iii) Stored energy in the coil,

U = \(\frac{1}{2} L I_{0}^{2}\) = \(\frac{1}{2}\) × 2 × 10 × 10= 100 J

Question 10.

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain. [1½]

Answer:

Angular magnification or magnifying power of compound microscope is defined as ratio of angle made at eye by image formed at infinity to the angle made by object, if placed at distance of distinct vision from an unaided eye.

Magnification = LD / f_{o} . f_{e}

where, L is length of the tube of microscope.

As m ∝ \(\frac{1}{f_{o}}\) and m ∝ \(\frac{1}{f_{o}}\)

∴ Both eyepiece and objective must be of smaller focal lengths, so that magnification is higher.

Question 11.

Use the mirror equation to deduce that the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. [1½]

Answer:

For mirror formula

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

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

υ = \(\frac{u f}{u-f}\)

For convex f is positive and u is negative.

υ = \(\frac{(-u) f}{-u-f}=\frac{u f}{u+f}\)

υ = \(\frac{u f}{u+f}=\frac{f}{\left(1+\frac{f}{u}\right)}\)

(1 + \(\frac{f}{u}\)) > 1

∴ υ < f

i.e., the image will always form between pole and focus.

Question 12.

Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed? [1½]

Answer:

According to question,

n_{1} = 1 [Given]

n_{2} = 1.5 ⇒ R = 20 cm

⇒ n = -100 cm

So, from surface formula

⇒ υ = \(\frac{200 \times 1.5}{3}=\frac{300}{3}\)

= 100 cm

Question 13.

At the time of sunset (or sunrise) the sun appears some what higher than its actual position. [1½]

Answer:

With altitude, the density and hence the refractive index of air-layers decreases. The light rays starting from the Sun (S) travel from rarer to denser layers. Hence, they bend more and more towards the normal.

However, an observer sees an object in the direction of the rays reaching his eyes. So, to an observer standing on the Earth, the Sun which is actually in a position S below the horizon appears in the position S above the horizon. The apparent shift in the direction of the Sun is by about 0.5°. Thus, the Sun appears to rise early by about 2 minutes and for the same reason.

Question 14.

In both β^{–} and β^{+} decay process, the mass number of nucleus remains the same, where as the atomic number Z increases by one in β^{–} decay and decrease by one in β^{+} decay. Explain giving reasons. [1½]

Answer:

In β^{–} -decay, a β-particle of zero mass and -1 charge is emitted. The decay process is shown as below:

The mass number remains the samebut here the atomic number decreases by 1 due to the loss of 1 positive charge.

Similarly, for a β^{+}-decay, a β-particle of negligibly small and +1 charge is emitted. The decay process is shown as below:

The mass number remains the same but here the atomic number decreases by 1 due to the loss of 1 positive charge.

Question 15.

Tritium has a half-life of 12.5 years undergoing beta decay. What fraction of a sample of pure tritium will remain undecayed after 25 years? [1½]

Answer:

Number of nuclei left undecayed,

N = N_{0}(\(\frac{1}{2}\))^{n}

t = 25 years

T_{1/2} = 12.5 years

but, n = \(\frac{t}{T_{y 2}}\)

∴ n = \(\frac{25}{12.5}\) = 2

⇒ \(\frac{N}{N_{0}}\) = (\(\frac{1}{2}\))^{2} = \(\frac{1}{4}\)

Therefore, (1/4)^{th} of the sample of the initial tritium will remain undcayed.

Section – C

Question 16.

Using Ampere’s circuit law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis. [3]

Or

(a) How is a toriod different from a solenoid?

(h) Use Ampere’s circuital law to obtain the magnetic field inside a toroid.

(c) Show that in an ideal toroid, the magnetic field (i) inside the toroid and (ii) outside the toroid at any point in the open space is zero. [3]

Answer:

Magnetic Field Inside an Infinitely Long Solenoid

To calculate the magnetic field inside an ideal solenoid we analyse over a rectangular amperian loop PQRS which is partially inside the solenoid and partially outside it. Side PQ is parallel to the axis of the solenoid. Hence, SR is also parallel to the magnetic field \(\overrightarrow{\mathrm{B}}\) Suppose the length of PQ is x and the number of turns in the solenoid per unit length is n. The number of turns in PQ is nx. Suppose, I current is flowing in every turn of the solenoid.

From Ampere’s Law

Because, here \(\overrightarrow{\mathrm{B}}\) is either perpendicular to \(\overrightarrow{\mathrm{d} l}\) (or inside the solenoid) or \(\overrightarrow{\mathrm{B}}\) = 0 is outside the solenoid. This is true for \(\int_{S}^{P} \vec{B} \cdot \vec{d} l\) The path selected in \(\int_{S}^{P} \vec{B} \cdot \vec{d} l\) is totally outside of magnetic field; which means that all points B = 0.

In this way

\(B \int_{P}^{Q} d l\) + 0 + 0 + 0 = μ_{0}Σi …(1)

Here, \(\int_{P}^{Q} d l\) = x

Σi = nxI

(Since, due to loop there is current i is in circuit)

Bx = μ_{0}nxI

or B = μ_{0}nI …………… (2)

The line PQ can be at any distance from the axis in a magnetic field.

Question 17.

Draw a labelled ray diagram showing image formation of a distant object by refracting telescope and describe briefly. [3]

Or

Draw a ray diagram for a convex mirror showing the image formation of an object placed anywhere in front of the mirror. [3]

Answer:

Formation of image is shown in figure by ray diagram.

The parallel rays coming from the object, situated at infinity, form the inverted, small and real image A’ B’ at the second focus F0 after refraction through objective lens. This intermediate image A’B’ acts as virtual object for eye lens. Hence eye lens is displaced forward or backward till this image comes within first focal plane of eye lens. Thus virtual, larger and erect image A”B” with respect A’B’ is formed which is the final image.

Question 18.

Why is wave theory of electromagnetic radiation not able to explain photoelectric effect? How does photon picture resolve this problem? [3]

Or

A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions giving reasons:

(i) Do the emitted photoelectrons have the same kinetic energy?

(ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?

(ii) On what factors does the number of emitted photoelectrons depend? [3]

Answer:

The wave theory of light is not able to explain the observed features of pohtoelectric current because of following reasons :

(i) The greater energy incident per unit time per unit area increases with the increase of intensity which should facilitated liiberation of photoelectron of greater kinetic energy which is in contradiction of observed feature of photoelectric effect.

(ii) Wave theory states that energy carried by wave is independent of frequency of light wave and hence wave of high intensity and low frequency (less than threshold frequency) should stimulate photoelectric emission but partically, it does not happen.

Considering the following few properties of photon, the above problem was resolved.

(i) In interaction of radiation with radiation behaves as if it is made up of particle called photon.

(ii) Energy of photon is directly proportional to the frequency of the incident light.

Section – D

Question 19.

(i) Define torque acting on a dipole of dipole moment p placed in a uniform electric field E. Express it in the vector form and point out the direction along which it acts.

(ii) What happens if the field is non-uniform?

(iii) What would happen if the external field E is increasing (a) parallel to p and (b) anti-parallel to p ? [4]

Or

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point of the configuration. Show that the equilibrium of the test charge is necessary unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign place a certain distance apart. [4]

Answer:

(i) τ = pE sinθ

In vector notation,

τ = p × E

SI unit of torque is newton-metre (N/m) and its dimensional formula is [ML^{2}T^{-2}]. Torque is always directed in plane perpendicular to the plane of dipole movement and electric field.

Case 1 : If θ = 0°, then τ = 0

The dipole is in stable equilibrium.

Case 2 : If θ = 90°, then τ = pE (maximum value)

The torque acting on dipole will be maximum.

Case 3 : If θ = 180°, then τ = 0

The dipole is in unstable equilibrium.

(ii) If the field is non-uniform, there would be a net force on the dipole in addition to the torque and the

combination of translation and rotation.

τ = p × E(r)

Net torque acts on the dipole depending on the location, where r is the position vector of the centre of the dipole.

(iii) (a) E is increasing parallel to p, then θ = 0°. So, torque becomes zero but the net force on the dipole will be in the direction of increasing electric field and hence it will have linear motion along the^ipole moment.

(b) E is increasing anti-parallel to p. So, the torque still remains zero but the net force on the dipole will be in the direction of increasing electric field which is opposite to the dipole moment, hence it will have linear motion opposite to the dipole moment.

Question 20.

(i) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction.

(ii) Name the device which is used as a light detector. Draw the necessary circuit diagram and explain its working. [4]

Or

Draw the energy band diagram of (i) n-type, and (ii) p-type semiconductor at temperature, T > OK.

In the case n-type Si semiconductor, the donor energy level is slightly below the bottom of conduction band whereas in p-type semiconductor, the acceptor energy level is slightly above the top of the valence band. Explain, what role do these energy levels paly in conduction and valence bands. [4]

Answer:

(i) When we are dealing with depletion layer formation we have to keep in mid the majority charge carriers, diffusion will always happen from high concentration to low concentration.

The two process involved in the formation of p-n junction.

(a) Diffusion

(b) Drift.

Holes and electrons diffuse from p to n and n to p respectively.

The majority charge carrier drifts under the influence of applied electric field such that

(a) holes along applied E and

(b) electron opposite to E

(ii) Photo diode is used as a light detector.

Figure shows the photo diode being used as a light detector in the circuit. In the absence of incident light due to reverse biasing, the reverse current is very small. In the presence of incident light the photons are absorbed at the junction place and generate more electron hole pair which are isolated due to the electric field at the junction and flow through the junction, due to this the value of reverse current increases.

The current is very high in comparison to the current flowing in the absence of light. If by keeping the frequency of the incident light constant theintensity of light is increased then the value of current increases more. For various intensities (Φ) the changes related in reverse current are shown in the figure. In the presence of light the reverse current can be of many hundred microampere. Diode has to be kept at voltage less than the breakdown voltage. Due to change in the intensity of light, the change generated in the reverse current is measured easily.

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