Students must start practicing the questions from RBSE 12th Physics Model Papers Set 8 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 8 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) Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in the figure.

The force on a charge of 1 µC placed at the centre of square is: [1]

(a) 7N

(b) 8N

(c) 2 N

(d) zero

Answer:

(d) zero

(ii) For equipotential surface, which statement is not correct? [1]

(a) are closer in regions of large electric fields compared to regions of lower electric fields.

(b) will always be equally spaced.

(c) will be more crowded near sharp edges of a conductor.

(d) will be more crowded near regions of large charge densities.

Answer:

(b) will always be equally spaced.

(iii) In parallel combination of n cells, we obtain: [1]

(a) more voltage

(b) more current

(c) less voltage

(d) less current

Answer:

(b) more current.

(iv) A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes: [1]

(a) 8 times

(b) 4 times

(c) 2 times

(d) 16 times

Answer:

(b) 4 times

(v) In the figure, galvanometer G gives maximum deflection when: [1]

(a) magnet is pushed into the coil

(b) magnet is rotated into the coil

(c) magnet is stationary at the centre of the coil

(d) number of turns is the coil is reduced.

Answer:

(a) magnet is pushed into the coil

(vi) Photoelectron’s stopping potential depends on: [1]

(a) frequency of incident light and nature of the cathode material.

(b) the intensity of the incident light

(c) the frequency of the incident light

(d) nature of cathode material.

Answer:

(a) frequency of incident light and nature of the cathode material.

(vii) The half life of a radioactive nucleus is 3 hours. Its activity will be reduced to a factor of:[1]

(a) \(\frac{1}{9}\)

(b) \(\frac{1}{27} \)

(c) \(\frac{1}{6}\)

(d) \(\frac{1}{8}\)

Answer:

(d) \(\frac{1}{8}\)

(viii) Energy bands in solids are a consequence of: [1]

(a) Ohm’s law

(b) Pauli’s exclusion principle

(c) Bohr’s theory

(d) Heisenberg’s uncertainty principle.

Answer:

(b) Pauli’s exclusion principle

(ix) An OR gate has 4 inputs. One input is high and the other is low. The output is: [1]

(a) low

(b) high

(c) alternately high and low

(d) may be high or low depending on relative magnitude of inputs.

Answer:

(b) high

Question 2.

Fill in the blanks:

(i) The dimensional formula of capacitance is ……………………….. . [1]

Answer:

[M^{-1}L^{-2}T^{4}A^{2}]

(ii) The drift velocity …………………….. with rise in temperature. [1]

Answer:

increases

(iii) For a closed surface, total magnetic flux is always ………………….. .[1]

Answer:

zero

(iv) Extrinsic semiconductors are of ………………………….. types [1]

Answer:

two.

Question 3.

Give the answer of the following questions in one line :

(i) Draw a plot showing the variation of terminal voltage (V) versus the current (I) drawn from the cell. [1]

Answer:

Following plot is showing variation of terminal voltage versus the current

(ii) Write the expression in a vector form for the Lorentz magnetic force F due to a charge moving with velocity v in a magnetic field B. What is the direction of the magnetic force? [1]

Answer:

The expression in vector form is given by

F=q(u x B)

where q is the charged particle.

The direction of the magnetic force is in the direction of u x B, i.e., perpendicular to the plane containing u and B.

(iii) Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing? [1]

Answer:

Using Lenz’s law we can predict the direction of induced current opposes the cause of increasing of magnetic flux. So, induced current will be clockwise in ring 1 and anticlockwise in ring 2.

(iv) Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. [1]

Answer:

Variation of photoelectric current versus potential for different intensities and equal frequencies.

(v) Define the term stopping potential in relation to photoelectric effect.[1]

Answer:

In experimental set up of photoelectric effect, the value of negative potential of anode at which photoelectric current in the circuit reduces to zero is called stopping potential or cut-off potential for the given frequency of the incident radiation.

(vi) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction?[1]

Answer:

The difference between the total mass of the nuclei and the sum of masses of the nucleons gets converted into energy (or vice-versa) in a nuclear reaction.

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

Answer:

T_{1/2} = \(\frac{\ln 2}{\lambda}\) =τln2

where T_{1/2} is half-life and t is mean life.

(viii) Why should a photodiode be operated at a reverse bias? [1]

Answer:

Fractional change in minority charge carriers is more than the fractional change in majority charge carriers, the variation in reverse saturation current is more prominent.

Section-B

Question 4.

Figure shows as charge array known as an ‘electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r / a >> 1, and contrast your results with that due to an electric dipole and an electric monopole (i. e., a single charge). [1½]

Answer:

As is clear from figure an electric quadrupole may be regarded as system of three charges, +q, -2qand +q at A, B and C respectively.

Let AC = 2α. We have to calculate electric potential at any point P where BP = r. Using superposition principle, potential at P is given by

V = \(\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{r+a}-\frac{2 q}{r}+\frac{q}{r-a}\right]\)

= \(\frac{q \cdot 2 a^{2}}{4 \pi \varepsilon_{0} r^{3}\left(1-a^{2} / r^{2}\right)}\)

when \(\frac{r}{a}\) >>1, r>>a or a<<r

∴ \(\frac{q^{2}}{r^{2}}\) negligibly small. V = \(\frac{q .2 a^{2}}{4 \pi \varepsilon_{0} r^{3}}\)

Clearly, V ∝ \(\frac{1}{r^{3}}\)

In case of an electc dile, V ∝ \(\frac{1}{r^{2}}\) and in case of an electric monopole

(i.e, single charge), V ∝ \(\frac{1}{r}\)

Question 5.

Three-point charges q, – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side l as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q. [1½ ]

Answer:

(a) Force acting on the charge q placed at A, is due to the charges placed at points B and C.

From the given figure, magnitude of force on charge A due to charge at point C is given as

F_{Ac} = \(\frac{k(q)(2 q)}{l^{2}}\) (say) = F

Similarly, magnitude of force on charge at point A, due charge at point B is

F_{AB} = \(\frac{k(4 q) q}{l^{2}}\) say = 2F ( F_{AB} = 2F_{AC})

∴ F_{res} = \(\sqrt{F^{2}+\left(2 F^{\prime}\right)^{2}+2\left(F^{\prime}\right)(2 F) \cos 120^{\circ}}\)

= \(\sqrt{F^{2}+4 F^{2}+4 F^{2}\left(-\frac{1}{2}\right)}\)

(cos 120° = \(-\frac{1}{2}\))

= \(\sqrt{F^{2}+2 F^{2}}=\sqrt{3} F\)

∴ F = \(\sqrt{3} \times \frac{2 k q^{2}}{l^{2}}\) newton

Question 6.

A silver wire has a resistance of 2.1 Ω at 27.5°C and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver. [1½ ]

Answer:

Here; R_{1} = 2.1 , t_{1} = 27:5°C,

and R_{2} = 2.7 Ω, t_{2} = 100°C

We know that R_{2} = R_{1} [(1+ α(t_{2} – t_{1})]

Thus, α = \(\frac{R_{2}-R_{1}}{R_{1}\left(t_{2}-t_{1}\right)}\)

= \(\frac{2.7-2.1}{2.1(100-27.5)}=\frac{0.6}{2.1 \times 72.5}\)

Or α = 0.0039/°C

Question 7.

A bulb of 110 V – 50 W is connected with a sources of 55 V. Calculate the power consumed by the bulb.

[1½]

Answer:

P=50W; V=110V; V’=55V P’,=?

Resistance of the bulb,

R = \(\frac{V^{2}}{P}=\frac{110 \times 110}{50}\)

= 242 Ω

Therefore the power consumed on connecting with source of V’,

P’ = \(\frac{V^{\prime 2}}{R}=\frac{55 \times 55}{242}\)

or

P’= 12.5 watt.

Question 8.

Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other.[1½]

Answer:

Mutual Inductance Between Two co-axial Solenoids Length of both solenoids is l and number of turns per unit length are n_{1 }and n_{2}.S_{1} solenoid is wounded over S_{2} solenoid, cross-section area of solenoid is A.

Let, in solenoid S_{1} current flow is I_{1}, due to which magnetic flux induced which is linked with the coil S_{2}.

The magnetic flux (Φ_{21}) is proportional to the current I_{1}, then:

Φ_{21} ∝ I_{1} …………………. (1)

∴ Φ_{21} =M_{21}I_{1} ……………..(2)

Where M_{21} = coefficient of mutual inductance Magnetic field due to current in solenoid S_{1}

B_{1} =µ_{0} n_{1}I_{1} = \( \frac{\mu_{0} N_{1}}{l} I_{1}\) …………………….. (3)

Since S_{1} solenoid is wounded on S_{2} solenoid, therefore magnetic flux linked with each turns of solenoid ‘S_{2} ‘

= B_{1} x Area of cross-section

Φ_{21} = \(\frac{\mu_{0} N_{1} I_{1}}{l}\) A …………………….. (4)

Total flux linked with the solenoid’S_{2}

Φ_{21} = Φ_{21} x Total number of turns in’ S_{2}

= \(\frac{\mu_{0} N_{1} I_{1} A N_{2}}{l}=\frac{\mu_{0} N_{1} N_{2} I_{1} A}{l}\) ………………… (5)

From the equation (2) and (5),

M_{2}I_{1} = \(\frac{\mu_{0} N_{1} N_{2} I_{1} A}{l}\)

M_{21} = µ_{0} \(\frac{N_{1} N_{2}}{l}\) A ……………………… (6)

Similarly due to current ‘I_{2} in solenoid ‘S_{2}’ then the coefficient of mutual inductance

M_{21} = \(\frac{\mu_{0} N_{1} N_{2} A}{l}\) ……………………… (7)

From the equation (6) and (7),

M= M_{12}= M_{21} = \(\frac{\mu_{0} N_{1} N_{2}}{l}\) A

Question 9.

A metallic rod of length L is rotated with angular frequency of ω with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. [1½]

Answer:

Angular velocity of rod ω = \(\frac{2 \pi}{T}\)

where T = time period.

∴ Charge in flux in one revolution

= BA = B (πL^{2})

According to Faraday’s law of EMI, magnitude of induced emf

e= \(\frac{\Delta \phi}{\Delta T}=\frac{B \pi L^{2}}{T}\)

e = \(\frac{B \pi L^{2}}{\left(\frac{2 \pi}{w}\right)}\)

e = \(\frac { 1 }{ 2 }\)BωL^{2}

Question 10.

Does short-sightedness (Myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision? [1½ ]

Answer:

No, it does not imply necessarily that the eye lens has lost its ability of accommodation. A person may have normal ability of accommodation of the eye lens and yet may be myopic or hyperopic. Myopia arises when the eye ball from front to back gets too elongated, hypermetropia arises when it gets too shortedness. In practice, in addition the eye lens may also lose some of its ability of accommodation. When the eye ball has normal length but the eye lens loses partially its ability of accommodation (as happens with increasing age for normal eye), the defect is called ‘presbyopia’ and is corrected in the same manner as hypermetropia.

Question 11.

Why does a bird flying high in the air appear to be higher than in reality? [1½]

Answer:

When the bird is flying high in the air, it is in a rare medium whereas the air near the ground is a denser medium. When we see the bird, the light rays from the bird’s body travel from a rarer to denser medium, hence, the light rays bend towards the normal. This refraction of light causes the bird to appear flying higher in the air than in reality.

Question 12.

A biconvex lens has a focal length 2/3 times the radius of curvature of either surface.Calculate the refractive index of lens material. [1½]

Answer:

Given,

f= \(\frac{2}{3}\) R,R_{1} = +R, R_{2} =-R

Using Lens maker’s formula,

\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

\(\frac{\frac{1}{2 R}}{3}=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right) \)

\(\frac{3}{2 R}=(n-1)\left(\frac{2}{R}\right)\)

⇒ n-1 = \(\frac{3}{4}\)

n = 1+ \(\frac{3}{4}\)

⇒ n = \(\frac{7}{4}\) = 1.75.

Question 13.

Explain the cause of dispersion of light? [1½ ]

Answer:

Wavelengths of different colours of light of white light are different. Now from Cauchy’s formula we know. that µ = A+ B/λ^{2}+………………………. refractive index of a medium will be also different for different colours of light of white light. From formula δ_{m} = (µ -1) A deviation of different light rays will be different so they disperse.

Question 14.

Define the Q-value of a nuclear process. When can a nuclear process not proceed spontaneously? [1½]

Answer:

The Q-value of a nuclear process refers the energy release in the nuclear process which can be determined using Einstein’s mass-energy relation, E = mc^{2}. The Q-value is equal to the difference of mass of products and reactant nuclei multiplied by square of velocity of light. Q = [m_{x} – mxubπ – m_{He}]c^{2} in a-decay The nuclear process does not proceed spontaneously when Q-value of a process is negative or sum of masses of product is greater than sum of masses of reactant.

Question 15.

Obtain the binding energy (in MeV) of a nitrogen nucleus (_{714}N), given m (_{714} N) = 14.00307u. [1½]

Answer:

Mass of nitrogen,

m(_{714}N) = 14.00307u

Mass of proton, m_{p} = 1.007825 u

Mass of neutron, m_{n} = 1.008665 u

B.E. of _{714}N = [Zm_{p} +(A- Z)m_{n} – m (_{714}N)] c^{2}

= [7+1.007825+(14- 7) x 1.0086655 -14. 0037] c^{2
}= 0.112360 x 931.5MeV

∴ B.E. = 104.7 MeV

Section-C

Question 16.

Derive the expression for the torque acting on a current-carrying loop placed in a magnetic Geld. [3]

Or

How is moving coil galvanometer converted into a voltmeter? Explain giving the necessary circuit diagram and the required mathematical relation used. [3]

Answer:

Forces \(\overrightarrow{\mathrm{F}_{3}}\) and \(\vec{F}_{4}\) are equal and opposite. The torque of the couple force acting on the rectangular loop is;

τ = magnitude of one force of the couple force x perpendicular distance between the forces From the figure, perpendicular distance between the forces b sinθ

or τ=(IlB)(bsinθ)

= I (lb)Bsinθ

τ= IABsinθ …………………. (1)

Here, A = lb = area of the loop If on the place of a rectangular loop a rectangular coil is taken of N turns then torque on it;

τ= NIABsinθ ……………………. (2)

or τ=MBsinθ ……………………… (3)

Here, \(\overrightarrow{\mathrm{M}}\) = NI \(\vec{A}\) = Multiplication of the effective area of the current and coil, which is called the magnetic torque magnetic moment of the coil.

Question 17.

Draw a schematic diagram of a reflecting telescope (Cassegrain). Write two important advantages that the reflecting telescope has over a refracting type.[3]

Or

With the help of a suitable ray diagram, derive a relation between the object distance (u), image distance (υ) and radius of curvature R for the convex spherical surface when a ray of light travels from a rarer to denser medium. [3]

Answer:

A ‘Cassegrainian type telescope’ is shown in figure. In this, a concave mirror of aperture about 200 inches is used as objective. There is a small hole at the centre of the mirror in which eye piece is fitted. The parallel rays to principal axis of objective coming from distant object (viz, from distant star) want to centralised at the focus of the objective but before centralising these rays incident on a secondary convex mirror B and after reflection they enter the eyepiece.

Thus, the final image is seen. One thing should be kept in mind that the seen image is inverted with respect to object. But for astronomical purposes, it has no meaning because mostly the astronomical objects are spherical in shape.

For normal adjustment (i.e., final image forms at infinity) the magnifying power of the telescope,

M= \(\frac{f_{o}}{f_{e}}=\frac{R / 2}{f_{e}} \)

Where R_{e} be the radius curvature of objective mirror.

Advantages:

Advantages of reflecting telescope are as under:

- Since the objective lens is a mirror, hence there is no chromatic aberration.
- As objective is parabolic mirror, so there is no spherical aberration also.

Question 18.

(i) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.

(ii) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based. [3]

Or

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon leaser. The power emitted is 9.42 mW.

(a) Find the energy and momentum of each photon in the light beam.

(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and

(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ? [3]

Answer:

(i) The photoelectric effect cannot be explained on the basis of wave nature of light because wave nature of radiations cannot explain the following:

- The increment ejection of photoelectrons.
- The existence of threshold frequency for a metal surface.
- The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

(ii) Photon picture of electromagnetic radiation on which based on particle feature of light. Its basic features are given as below:

(a) In interaction of radiation with matter. Radiation behaves as if it is made up of particles called photons.

(b) Each photon has energy

E(= hv = \(\frac{h c}{\lambda}\) ) and momentum

P(=hv=\(\frac{h}{\lambda}\)) where,cis the speed of light, h is Planck’s constant, v and λ are frequency and wavelength of radiation, respectively.

(c) All photons of light ofa particular frequency y or wavelength have the same energy E (= hv =\(\frac{h c}{\lambda}\)) and momentum P ( = hv = \(\frac{h}{\lambda}\) ) whatever the intensity of radiation may be.

Section – D

Question 19.

Define electric flux. Is it a scalar or a vector quantity? A point charge q Is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

(b) If the point charge is now moved to a distance from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected. [4]

Or

Using Gauss’ law, obtain the expression for the electric field due to uniformly charged spherical shell of radius R at a point outside the shell. Draw a graph showing the var!ation of electric field with r, for r> Rand r < R. [4]

Answer:

Electric Flux: “Number of electric field lines of force passing normally through the surface placed in the electric field is known as the electric flux linked with the surface.” It is denoted by Φ_{E} and it is a scalar quantity.

From the given problem qis the point charge at a distance of \(\frac{d}{2}\) directly above the centre of the square side. Now, construct a Gaussian surface in form of a cube of side d to evaluate the amount of electric flux.

∴We can calculate the amount of electric flux for six surfaces by using Gauss’s law,

Φ_{E} = \(\int_{s} \mathbf{E} \cdot d \mathbf{S}=\frac{q}{\varepsilon_{0}}\)

∴For one surface of the cube, amount of electric flux is given as Φ_{E’}= \(\frac{q}{6 \varepsilon_{0}}\)

(b) Even if the point charge is moved to a distance from the centre of the square and side of the square is

doubled, but amount of charge enclosed into the Gaussian surface does not changes.

∴The amount of electric flux remains same.

Question 20.

Why is a Zener diode considered as a special purpose semiconductor diode? Draw the I-V characteristics of Zener diode and explain briefly, how reverse curent suddenly increase at the breakdown voltage? Describe briefly with the help of a circuit diagram, how a Zener diode works to obtain a constant DC Voltage from the unreg ulatey DC output of a rectifier. [4]

Or

Draw the circuit diagram of full-wave rectifier using p-n junction diode. Explain its working and show the output input waveforms. [4]

Answer:

Zener diode works only in reverse breakdown region that is why it is considered as a special purpose semiconductor.

I-V characteristics of Zener diode is given below reverse current is due to the flow of electrons from n → p and holes from p → n. As, the reverse-biased voltage increase the electric field across the junction, increases significantly and when reverse bias voltage V = V_{z}, then the electric field strength is high enough to pull the electrons from p-side and accelerated it to n-side.

These electrons are responsible for the high current at the breakdown. In reverse biasing arrangement of P-N junction diode we saw that in the state of breakdown voltage the reverse current of the diode is not less but increases rapidly, but the potential at the ends of the diode remains approximately equal to the breakdown voltage meaning remains constant. For any diode; breakdown voltage depends on the quantity of impurities mixed.

Therefore, according to the quantity of the impure element any special breakdown voltage diode’s fabrication can be done. When any diode is so formed for any special breakdown voltage that in reverse biasing also at breakdown voltage it works without being damaged; then such type of diode is called Zener thode. With the help of various quantities of impurity mixed today the diodes are available of less than 1 volt and in hundreds volt also. The sign of Zener diode is shown in the figure.

When the current flowing through the breakdown point from the Zener diode is increased highly still the potential difference at both its ends is constant. This property makes its use for voltage stabilization and voltage regulation. We know that in an ordinary dc power supply with the use of rectifier and equalizer filter, ac is changed to dc. In such type of power supplies when there is change in load current, the value of dc voltage also changes.

Therefore, in such situations where current is to be obtained at constant dc voltage such power supply cannot be used. With the help of a Zener diode, the output voltage of a power supply can be kept constant.

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