Students must start practicing the questions from RBSE 12th Physics Model Papers Board Model Paper 2022 with Answers in English Medium provided here.

## RBSE Class 12 Physics Board Model Paper 2022 with Answers in English

Time: 2:45 Hours

Maximum Marks: 56

General Instruction to the Examinees

- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulasory.
- 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) The unit of permittivity of free space is: [1]

(a) C^{2}N^{1}m^{-2}

(b) C^{2}N^{-1}m^{-2}

(c) C^{-2}N^{-2}m^{-2}

(d) C^{2}N^{-1}m^{2}

Answer:

(b) C^{2}N^{-1}m^{-2}

(ii) Dielectric materials are: [1]

(a) conductors

(b) non-conductors

(c) semi-conductors

(d) none of these

Answer:

(b) non-conductors

(iii) Resistance of the carbon resistor shown in figure in ohm will be: [1]

(a) 420 ± 10%

(b) 240 ± 10%

(c) 420 ± 5%

(d) 240 + 5%

Yellow Golden

Answer:

(c) 420 ± 5%

(iv) An electric charge q is moving with uniform velocity v in the direction of magentic field B. Magnetic force acting on the charge will be: [1]

(a) qυB

(b) zero

(c) \(\frac{q v}{B}\)

(d) \(\frac{v B}{q}\)

Answer:

(b) zero

(v) “The magnitude of the induced emf in a circuit is equal to the time rate of change of

magnetic flux through the circuit. ” This law is given by: [1]

(a) Lenz

(b) Ampere

(c) Faraday

(d) Henery

Answer:

(c) Faraday

(vi) Photo electric equation of Einstein is: [1]

(a) hυ = \(\frac{1}{2} m v_{m}^{2}\) – Φ_{0}

(b) hυ = \(\frac{1}{2} m v_{m}^{2}\) + Φ_{0}

(c) \(\frac{1}{2} m v_{m}^{2}\) = eV_{0}

(c) Φ_{0} = hυ_{0}

Answer:

(b) hυ = \(\frac{1}{2} m v_{m}^{2}\) + Φ_{0}

(vii) The atomic masses of two different nuclei are 3 and 81, then the ratio of their radius are: [1]

(a) 3 : 81

(b) 81 : 3

(c) 1 : 3

(d) 3 : 1

Answer:

(c) 1 : 3

(viii) The acceptor impurity in following is: [1]

(a) arsenic

(b) indium

(c) antinomy

(d) phosphorus

Answer:

(b) indium

(ix) The universal gate in following is: [1]

(a) AND

(b) OR

(c) NOR

(d) NOT

Answer:

(c) NOR

Question 2.

Fill in the blanks:

(i) The capacitance of the parallel plate capacitor in vacuum is C_{0}. A material of dielectric

constant K is filled completely between the plates, then the new value of capacitance is …………………….. . [1]

Answer:

KC_{0}

(ii) The resistivity of semiconductor is …………………….. with increasing of the temperature. [1]

Answer:

decrease

(iii) Lenz’s law obey the ……………………. conservation law. [1]

Answer:

energy

(iv) ………………………… diode is used as a voltage regulation. [1]

Answer:

Zener

Question 3.

Give the answer of the following questions in one line:

(i) Draw circuit diagram for comparison of emf of two primary cells with the help of potentiometer. [1]

Answer:

(ii) How will be a galvanometer converted into a voltmeter? [1]

Answer:

If in a pivotet coil galvanometer a very high resistance known as ‘series resistance’ is connected in series then the galvanometer is converted to voltmeter.

(iii) A bar magnet NS is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. In which coil the direction of current will look like anti-clockwise if viewed from left side? [1]

Answer:

According to Lenz’s law the coil AB develops a south pole at the end B and the coil CD also develops a south pole at the end C. Hence, induced current flows along anticlockwise in coil AB.

(iv) Write the name of experiment supporting the wave nature of particle. [1]

Answer:

Davisson and Germer Experiment.

(v) An alpha particle and a proton have same kinetic energies. Which one of these particles has lowest de-Broglie wavelength? [1]

Answer:

For a particle dc-Brogue wavelength, λ = h/P

Kinetic energy, K = p^{2}/2M

Then = λ = \(\frac{h}{\sqrt{2 m K}}\)

m_{α} > m_{α}

∴ λ_{α} < m_{p}

(vi) Write the difference between nuclear fission and nuclear fusion. [1]

Answer:

Both nuclear fission and fussion are nuclear reactions that produce energy, but the processes are very different. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, the fusion is the process where two light nuclei combine together.

(vii) Draw a graph between the binding energy per nucleOn and mass number. [1]

Answer:

(viii) Write the names of the logié gates related to figure P and Table Q. [1]

Answer:

Figure P represents symbols of OR gate.

Figure Q represents truth table of NAND gate.

Question 4.

Derive an expression for electric potential due to a point charge at a distance r from the’charge. [1½]

Answer:

Suppose a point, charge +q is placed at point O and potential due to this charge at point P situated at distance r from the charge, is to be determined. For this purpose, let us consider another point P’ situated at distance x from O beyond P and + q_{0} positive test charge is placed at P’, the electrostatic force acting on + q_{0},

Question 5.

Calculate the equivalent capacity of the combination of capacitors between A and Bin given diagram. [1½]

Question 5.

Calculate the equivalent capacity of the combination of capacitors between A and Bin given diagram. [1½]

Answer:

Two capacitor of 5 μF are connected in parallel. So net capacitance C’ = 5 + 5 = 10 μF C and 10 μF are in series.

So, equivalent capacitance of circuit is

\(\frac{1}{C_{e q}}=\frac{1}{C^{\prime}}+\frac{1}{10}\)

\(\frac{1}{C_{e q}}=\frac{1}{10}+\frac{1}{10}\)

C_{eq} = 5μF

Question 6.

Write two differences between the terminal voltage (V) and emf (e) of the cell. [1½]

Answer:

- Emf is an open circuit voltage. It is the potential difference between the two terminals of a battery or cell in a closed circuit. Terminal voltage is a closed-circuit voltage.
- Emf is independent of the resistance of the electrical circuit but is dependent upon the internal resistance of the circuit. Terminal voltage is dependent of the resistance.

Question 7.

In the given circuit diagram, find the value of unknown resistance S, in the balancing condition of meter bridge. [1½]

Answer:

For balancing meter bridges:

\(\frac{R}{S}\) = \(\frac{l}{100-l}\), where l = 40 cm

and R = 4 Ω

S = \(\frac{\left(100^{\circ}-40\right)}{40}\) × 4

S = 6 Ω

Question 8.

A conducting rod of length T is moving with constant linear speed υ in a uniform magnetic field B. This arrangement is mutually perpendicular. Obtain the expression of motional electromotive force. [1½]

Answer:

Magnetic force acts on these electrons,

\(\overrightarrow{\mathbf{F}_{m}}\) = q(\((\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)) ………………. (1)

If the developed electric field is E, then the force that acts on the electron of charge q,

\(\overrightarrow{\mathrm{F}_{e}}=q \overrightarrow{\mathrm{E}}\)

In the equilibrium state, q\(\overrightarrow{\mathrm{E}}\) + q(\(\vec{v} \times \vec{B}\)) = 0

or, q\(\overrightarrow{\mathrm{E}}\) = -q(\(\vec{v} \times \vec{B}\)) or, \(\overrightarrow{\mathrm{E}}\) = -(\(\vec{v} \times \vec{B}\))

i.e., \(\overrightarrow{\mathrm{E}}\) is directed opposite to the direction of (\(\vec{v} \times \vec{B}\)) or is from end a to 6 in the conductor.

\(|\vec{E}|=-|(\vec{v} \times \vec{B})|\) = υB.sin90°

or E = υB

The magnitude of electric field, E = VB

ε = El thus, ε = VBl ………………. (3)

Here, l is distance from the negative terminal to the positive terminal.

Question 9.

Current in a circuit falls from 5 Amp to zero in 0.1 second. If an average emf of 100 volt is induced, then calculate self-inductance of inductor in the circuit. [1½]

Answer:

Here; \(\frac{d I}{d t}\) = \(\frac{5 A-0 A}{0.1 \mathrm{~s}}\) = 50A/s,ε = 100V

As ε = \(\frac{L d I}{d t}\) L = \(\frac{\varepsilon}{d I / d t}\) = \(\frac{100 \mathrm{~V}}{50 \mathrm{~A} / \mathrm{s}}\) = 2H

Question 10.

Define total internal reflection. Write the name of any two phenomenons based on it. [1½]

When the angle of incidence in denser medium increases after the critical angle, the light reflects to the same medium in place of refraction to rarer medium. This phenomenon is known as the “T.otal internal reflection.”

Two phenomenons based on total internal reflection are:

- Mirage in desert.
- Sparkling of diamond.

Question 11.

Write any three differences between primary and secondary rainbow. [1½]

Answer:

Differences between primary and secondary rainbow:

Primary Rainbow | Secondary Rainbow |

1. Its upper end is red. | Its upper end is violet. |

2. Its inner end is violet. The violet ray makes an angle of 41° with Sim rays. | Its inner end is red and the violet ray makes an angle of 54° with Sun rays. |

3. The angle between the Sun rays and the rays of red colour in position of minimum deviation is 43°. | Rays of red colours make an angle of 51° with Sun rays. |

Question 12.

Establish the relation between focal length (f) and radius of curvature (R) for a spherical mirror. [1½]

Answer:

Relation between focal length (f) and radius of curvature (R).

According to law of reflection;

∠i = ∠r = θ (let)

∴ ∠OAF = ∠i + ∠r = θ + θ = 2θ

and ∠OAF ∠AFP = 2θ (because both are alternate angles)

Similarly, ∠OAC = ∠ACP θ(these are also alternate angles) ‘

In right angled ΔANC,

tanθ = \(\frac{A N}{C N}\)

If θ is very small, then

(i) tanθ ≈ θ and (ii) N and P are very close to each other and considered to be same point.

∴ θ = \(\frac{A N}{C P}=\frac{A N}{R}\) ………………. (1)

Similarly from right angled Δ ANF

tan 2θ = \(\frac{A N}{F N}\)

If 2θ is very small, then

(i) tan 2θ ≈ 2θ and (ii) N and P are very close to each other and considered to be same point.

∴ 2θ = \(\frac{A N}{F P}=\frac{A N}{f}\) ………………….. (2)

From equations (1) and (2), we get

2θ = \(\frac{2 A N}{R}=\frac{A N}{f}\)

or \(\frac{2}{R}=\frac{1}{f}\)

Question 13.

The focal lengths of an objective lens and eyepiece are 192 cm and 8 cm respectively in a telescope. Calculate its magnifying power and separation between the two lens. [1½]

Answer:

Given, f_{0} = 192 cm

and f_{e} = 8 cm

Magnifying power of telescope

m = \(\frac{f_{0}}{f_{e}}\) = – \(\frac{192}{8}\) = -24

Separation between the two lens

l = |f_{0}| + |f_{e}|

= 198 + 8 = 200 cm = 2 m

Question 14.

Write the function of moderator, coolant and control rods in the nuclear reactor. [1½]

Answer:

Function of Moderator: The generated neutrons in the range of 1 – 100 eV have lesser possibility to absorb the U^{238} nuclei. The core contains a moderator (water, heavy water or graphite) to slow down the velocity of neutrons.

Function of Coolent: The heat taken from reactor by the coolent is used to convert water into steam.

Function of control Rods: Control rods which are of cadmium are used to control fission.

Question 15.

In a radioactive sample the numbers of active nuclei remains 6.25% of its initial value in 6 hours. Find the half life of the radioactive sample. [1½]

Answer:

According to question, \(\frac{N}{N_{0}}\) × 100 = 6.25

\(\frac{N}{N_{0}}\) = \(\frac{6.25}{100}=\frac{1}{16}\)

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

n = 4

Half life t_{1/2} = \(\frac{t}{n}\) = \(\frac{6}{4}\) = 1.5 hours

Section – C

Question 16.

Obtain an expression for magnetic Reid on the axis of a current carrying circular loop (coil) with the help of Bio-savart law. Draw necessary diagram. [2 + 1 = 3]

Or

Obtain an expression for magnetic field on the axis of a current carrying toroid with the help of Ampere’s circuital law. Draw necessary diagram. [2 + 1 = 3]

Answer:

Let us consider a circular loop of radius ‘α’ with a current I as shown in figure. P is a point along the axis of the coil at a distance x from the centre O of the coil.

According to Biot-Savart law, the magnetic field at P due to the element dl is

dB = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{I d l \sin \theta}{r^{2}}\) where θ is the angle between Ml and r.

Here θ = 90°

∴ dB = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{I d l}{r^{2}}\)

The direction of dB is perpendicular to the current element Ml and CP.

Considering the diametrically opposite element A’ B’, the magnitude of dB at P due to this element is the same as that for AB but its direction as along PM. Let the angle between the axis of the coil and the line joining the element (dl) and the point P be a. dB is resolved into two components : dB sin a along OP and dB cos a perpendicular to OP. dB cos a components due to two opposite elements cancel each other whereas dB sina components get added up. So, the total magnetic field at P due to the entire coil is

Or

Magnetic Field on the Axis of Toroid:

Suppose there are n turns per unit length of toroid and / is the current flowing through it. Due to flow of current, a magnetic field is produced inside the turns of toroid. Inside the toroid the magnetic field lines are in the form of concentric circles and at each point of symmetry, there is magnetic field which along the tangential direction.

Magnetic field inside the toroid : Suppose, there is a circular path of radius r inside the toroid. Applying Ampere’s circuital law for this loop,

\(\oint \vec{B} \cdot \vec{d} l\) = μ_{0}ΣI …………….. (1)

Total current flowing through closed loop,

ΣI = n × 2πr × I

= 2πrnI

Therefore from equation (1)

\(\oint \vec{B} \cdot \vec{d} l\) = μ_{0}2πrnI ……………… (2)

Here \(\vec{B}\)and \(\vec{d} l\) are in same direction.

∴ \(\oint \vec{B} \cdot \overrightarrow{d l}=\oint B d l \cos 0^{\circ}=\oint B d l\)

= \(B \oint d l\) = B 2πr

Therefore from equation (2)

B2πr = μ_{0}In 2πr

or B = μ_{0}nI

Magnetic field in space outside the toroid :

In space covered by the toroid

Suppose, there is a circular path of radius R_{1}(r_{1} < r) which is in the space covered by the toroid (fig. 4.41). No current is enclosed by this loop i. e

Question 17.

For refraction at a spherical surface derive the relation \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\) object distance (u), image distance (v) refractive index of media (n_{1}, n_{2}) and radius of curvature (R). Draw necessary ray diagram. [3]

Or

Draw a ray diagram for the formation of image by compound microscope. Derive an expression of the angular magnification of it when the final image is formed at the closest comfortable distance (D) for viewing the image. [3]

Answer:

Formula for refraction at concave surface : Suppose AB is concave spherical surface on the left side .of which, there is rarer medium and on the right side, the medium is denser. P is pole and C the centre of curvature of concave surface. O is point object on the principal axis and I is its image. Now from snell’s law.

_{1}n_{2} = \(\frac{\sin i}{\sin r}\)

n = \(\frac{\sin i}{\sin r}\)(using temporarily μ for _{1}μ_{2}

If i and r are small, then

sini = i and sinr = r = r

or n = \(\frac{i}{r}\) or i = nr …………….. (1)

In triangle, external angle is equal to sum of facing internal angles.

∴ From Δ MOC, α + i = γ ⇒i = (γ – α)

Similarly from Δ MIC, r + β = γ ⇒ r = (γ – β)

∴ From equation (1), (γ – α) = μ (γ – β) ……………… (2)

If point M is not for away from principle axis, then

(i) P and P’ will coincide each other.

(ii) Angles α,β and γ will bee small

Adjustment and Formation of Image: In adjustment process, eye lens oi eye pieces is adjusted first. For this, eye piece is moved forward or backward upto such extent that cross wire should be seen clearly. Now the object is placed infrqnt of field lens and it is moved to adjust in such a way that the clear image of the object would be seen. In this situation inverted, larger and virtual image, of the object forms on cross-wire. The ray diagram of formation of image is given in figure below.

AB is a small object and its inverted, real and larger image A’ B’ is formed by objective lens. This image acts as virtual object for eye lens, therefore it is moved forward or backward such that the image A’ B’ may lie within the focal length f_{e} of the eye piece. The image A’B’ acts as an object for the eye piece which essentially acts like a simple microscope. The eye piece forms a virtual and magnified final image A”B” of the object AB. Clearly, the final image A “B ” is inverted with respect to the object A.

If final image forms at least distance of distinct vision : Then,

Question 18.

(i) Define photo electric effect.

(ii) Plot a graph of variation of photo electric current with collector plate potential for two incident radiations of same intensity and different frequencies.

(iii) The work function for cesium metal is 3.31 × 1.6 × 10^{-19} Jule. Determine its threshold frequency for it. [1 + 1 + 1 = 3]

Or

(i) Define stopping voltage.

(ii) Plot a graph of variation of photo electric current with collector plate potential for two incident radiations of same frequency and different intensities.

(iii) Determine the de-Broglie wave length of an electron accelerated by potential difference of 100 volt. [1 + 1 + 1 = 3]

Answer:

(i) Photoelectric effect : The phenomenon of emission of electron from a metal surface, when electromagnetic radiations of sufficiently high frequency are incident on it, is called photoelectric effect.

(ii)

Variation of photielectnc current with collector plate potential for different frequencies of iriddent radiation

(iii) Work function Φ_{0} = hυ_{0}

υ_{0} = \(\frac{\phi_{0}}{h}\)

Where Φ_{0} = 3.31 × 16 × 10^{-19} Joule

υ_{0} = \(\frac{3.31 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}\)

υ_{0} = 8 × 10^{14} Hz

Or

(1) Stopping potential : It is the minimum v.alue of the negative potential that must be

applied to the anode of photocell to make the photoelectric current zero. It depends upon the

frequency of incident light.

(ii)

Variation of photocurrent with collector plate potential for different intensities of incident radiation

(iii) de-Brogue wavelength of an electron acelerated.

λ = \(\frac{12.27}{\sqrt{V}}\) Å

V= 100 Volts

λ = \(\frac{12.27}{\sqrt{100}}\) = 1.227 Å

Section – D

Question 19.

Define electric dipole. Derive an expression for electric field intensity due to an electric dipole at a point on the equatorial plane of the electric dipole. Draw necessary diagram. [1 + 2 + 1 = 4]

Or

Define electric field intensity. Derive an expression for electric field intensity at a point due to an infinitely long uniformly charge straight wire with the help of Gauss law. Draw necessary diagram. [1 + 2 + 1 = 4]

Answer:

Electric Dipole: “When two equal and opposite charges are situated at a small distance, then they form electric dipole.”

Electric field of a dipole at a point on the equitorial plane of the dipole :

The electric field at any general point P is obtained by adding the electric field F_{-q} due to the charge – q and E_{+q} due to the charge + q, by the parallelogram law of vector.

Or

Electric Field Intensity : The electric field intensity at a point is the force experienced by a unit positive charge placed at that point.

Electric field intensity is a vector quantity. It is denoted by ‘E formula.

Applications of Gauss’s Law

Electric Field Intensity due to an Infinite Long Line Charge

Let the linear charge density is X and the intensity of electric field at point P is to be determined. For this purpose let us consider two small elements of length dl at A and B at equal distance from O. The intensities of electric field due to these elements at P, dE_{1} and dE_{2} are equal in magnitude. If the electric fields are resolved in two normal components, then the components along OP provide the electric field \(\vec{E}\) along OP and normal components dE_{1} sinθ and dE_{2} sinθ cancel out each other. Similar will be the result for other pairs of length elements considered.

Thus the direction of electric field at P will be in OP direction i. e., normal to linear charge.

Calculation of Electric Field: For application of Gauss’s law we require a Gaussian surface. For this purpose, we draw a cylindrical Gaussian surface of radius r and axis coinciding the linear charge AB. Point P lies on the curved surface of the cylinder. The enclosed charge

Σq = λl

where l is the length of the cylinder. Therefore from Gauss’s law,

Φ = \(\frac{\Sigma q}{\varepsilon_{0}}\) or Φ = \(\frac{\lambda l}{\varepsilon_{0}}\)

From the definition of the flux,

where r̂ = unit vector in direction OP i.e., in perpendicular to linear charge.

From equation (3), E ∝ \(\frac{1}{r}\)

Question 20.

What do you mean by rectification? Draw circuit diagram and explain the working of full wave rectifier. Represent the wave form of input and output voltages also. [4]

Or

What do you mean by reverse biasing of P-N junction diodes? Draw circuit diagram for reverse biasing of P-N junction diode and explain its working process. Draw V-I characteristic curve for reverse biasing. [4]

Answer:

Rectification : Rectification is the process by which alternating current is changed to direct current. The device used for such work is called rectifier. Full Wave Rectifier In a full wave rectifier circuit the output voltage current is obtained for both the half cycles of input alternating voltage. Here two junction diodes are so used that one diode represents the positive half cycles of input ac and the other represents the negative half cycles. Figures shows a circuit of full wave rectifier.

Here, the P-side of the diodes are connected to the ends of the secondary of the transformer. The N-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer. So for a full wave rectifier the secondary of the transformer is provided with a center tapping end so it is called center tap transformer. Suppose, the input voltage to A with respect to the centre tap at any instant is positive. It is clear, at that instant, voltage at B being out of phase will be negative.

So diode D_{1} gets forward biased and conducts (while D_{2} being reverse biased is not conducting). Hence, during the positive half cycle we get an output current (and a output voltage across the load resistor R_{L}). In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D_{1} would not conduct but diode D_{2} would giving an output current and output voltage (across R) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the half wave rectifier.

Or

Reverse biasing of PN junction diode: In reverse biasing of junction of p-terminal is at power potential than N-terminal.

Reverse Bias Characteristics: For the V-I characteristics of a reverse biasing P-N junction diode the experimental circuit arrangement is shown in the figure. Here with the help of a potential divider arrangement the P and N ends of the diode are connected to the negative and positive ends of the battery. Due to the reverse current being very small a microammeter is used in place of milliammeter in this circuit. At different reverse potential the related different reverse current value are noted and a curve is drawn as shown in the figure.

As explain in previous part in reverse biasing, the current is very small as it is due to minority- charge-carriers and remains constant till the time the reverse potential reaches the value of breakdown voltage (V_{B}). The reverse current increases very rapidly with a slight increase in the breakdown voltage.

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