Students must start practicing the questions from RBSE 12th Physics Model Papers Set 7 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 7 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

Question 1.

Multiple Choice Questions

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

(i) The electric field at a point is: [1]

(a) Always continuous

(b) Continuous if there is no charge at that point

(c) Discontinuous if there is a charge at that point

(d) Both (b) and (c) are correct.

Answer:

(d) Both (b) and (c) are correct.

(ii) The capacitance of a capacitor will decrease if we introduce a slab of: [1]

(a) copper

(b) aluminium

(c) zinc

(d) None of these

Answer:

(d) None of these

(iii) When a metal conductor connected to left gap of a meter bridge is heated, the balancing point: [1]

(a) shifts towards right

(b) shifts towards left

(c) remains unchanged

(d) remains at zero.

Answer:

(a) shifts towards right

(iv) A charge particle is moving in a cyclotron, what effect on the radius of path of this charged particle will occur when the frequency of the ratio frequency field is doubled? [1]

(a) It will also be doubled

(b) It will be halved

(c) It will be increased by four times

(d) It will remain unchanged.

Answer:

(d) It will remain unchanged.

(v) In electromagnetic induction, the induced charge is independent of: [1]

(a) change of flux

(b) time

(c) resistance of the coil

(d) none of these

Answer:

(b) time

(vi) Photoelectric effect was explained by: [1]

(a) Einstein

(b) Faraday

(c) Plank

(d) Hertz

Answer:

(a) Einstein

(vi) An electron emitted in beta radiation originates from: [1]

(a) inner orbits of atom

(b) free electrons existing in the nuclei

(c) decay of a neutron in a nuclei

(d) photon escaping from the nucleus

Answer:

(c) decay of a neutron in a nuclei

(vii) The electrical conductivity of pure germanium can be increased by: [1]

(a) increasing the temperature

(b) doping acceptor impurities

(c) doping donor impurities

(d) All of the these

Answer:

(d) All of the these

(ix) The only function of NOT gate is to: [1]

(a) stop signal

(b) invert input signal

(c) act as a universal gate

(d) None of these

Answer:

(d) None of these

Question 2.

Fill in the blanks:

(i) In …………………………….. combination, the potential difference of each capacitor is same. [1]

Answer:

parallel

(ii) Charge carriers in solids are ………………………… . [1]

Answer:

free electrons

(iii) The frequency of ac used in our houses is …………………………….. Hz. [1]

Answer:

50

(iv) Zener diode is operated in the …………………………. breakdown region. [1]

Answer:

reverse

Question 3.

Give the answer to the following questions in one line (i to viii):

(i) A low voltage supply from which one needs high currents must have very low internal resistance. Why? [1]

Answer:

When external resistance (R) is zero, I_{max} = \(\frac{\varepsilon}{r}\) To have large I_{max} with small ε (low voltage supply). r (internal resistance of the supply) should be very small.

(ii) Draw the magnetic field lines due to a current-carrying loop.[1]

Answer:

Magnetic field lines due to a current-carrying loop are given by

(iii) Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two could reach earlier and why? [1]

Answer:

The glass bob will reach earlier on ground as acceleration due to gravity is independent of mass of the

falling bodies. Glass being insulator, no induced current is developed in it due to the earth’s magnetic field.

(iv) Why is photoelectric emission not possible at all frequencies? [1]

Answer:

Photoelectric emission is not possible at all frequencies because below the threshold frequency of photosensitive surface of different atoms emission is not possible.

(v) Figure shows a plot of \(1 /\sqrt{V}\), where V is the accelerating potential versus the de-Broglie wavelength λ in the case of two particles having same charge q but different masses m_{1} and m_{2}. Which line (A or B) represents a particle of large mass? [1]

Answer:

de-Broglie wavelength,

λ = \(\frac{h}{\sqrt{2 m q V}}\) or λ = \(\frac{h}{\sqrt{2 m q}} \cdot \frac{1}{\sqrt{V}}\)

The graph of λ versus \(\frac{1}{\sqrt{V}}\) is a straight line of slope \(\frac{h}{\sqrt{2 m q}}\).

The slope of line B is small, so particle B has larger mass (charge is same).

(vi) Are the equations of nuclear reactions “balanced’ in the sense a chemical equation (e.g., 2H_{2} + O_{2} → 2H_{2}O) is? If not, in what sense are they balanced on both sides? [1]

Answer:

In a nuclear reaction, the number of atoms of each element may not be conserved, but the number of protons and that neutrons both must be separately conserved.

(vii) Define the activity of a given radioactive substance. Write its SI unit. [1]

Answer:

The activity of a sample is defined as the rate of disintegration taking place in the sample of radioactive substances. SI unit of activity is Becquerel (Bq). I Bq = 1 disintegrati on s/second

(viii) Identify the logic gate whose outs equals 1 when both of its inputs are 0 each. [1]

Answer:

NAND gate or NOR gate

Section-B

Question 4.

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 1/2 QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of factor 1/2 [1½]

Answer:

If F is the force on each plate of a parallel plate capacitor, then work done in increasing the separation between the plates by

Δx = F.Δx

This must be the increase in potential energy of the capacitor.

Now, increases in volume of capacitor =A.Δx

If u = energy density = energy stored/volume, then increase in potential energy = u.AΔx

∴ FΔx =u.AΔx

The origin on factor 1/2 in force can be explained by that fact that inside the conductor, field is zero and outside the conductor, the field is E. Therefore, the average value of the field (i.e. E /2) contributes to the force.

Question 5.

Derive the expression for the electric potential due to an electric dipole at a point on its axial line.[1½]

Answer:

Let electric potential to be obtained at point P lying on the axis of dipole at distance d from the centre of the dipole.

Potential at P due to +q charge = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{(d-l)}\)

Potential at P due to -q charge

⇒ \(\frac{1}{4 \pi \varepsilon_{0}} \frac{-q}{(d+l)}\)

Total potential at P due to dipole

= \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{(d-l)}-\frac{1}{(d+l)}\right]\)

= q × 2l / 4πε_{0} (d^{2} – l^{2})

= p /4πε_{0}(d^{2} – l^{2})

where,p =q.2l

If l< < d, then

V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{d^{2 \prime}}\)

Question 6.

Two bulbs are rated (P_{1}, V) and (P_{2}, V). If they are connected (i) in series and

(ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P_{1} and P_{2}. [1½]

Answer:

The resistance P_{1} is R_{1} = \(\frac{V^{2}}{P_{1}}\)

and that P_{2} is R_{2} = \(\frac{V^{2}}{P_{2}}\)

(i) In series, R = R_{1} + R_{2}

⇒ I = \(\frac{V}{R}=\frac{V}{R_{1}+R_{2}}\)

and P=I^{2}(R_{1}+R_{2})

= \(\frac{V^{2}}{\left(R_{1}+R_{2}\right)^{2}}\left(R_{1}+R_{2}\right) \)

(ii) In parallel, \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

⇒ \(\frac{V^{2}}{R}=\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}\)

P = P_{1} + P_{2}

Question 7.

A cell of e.m.f. 12 volts and internal resistance 5 Ω is connected with a resistance. If the current flowing in the circuit be 0.3A, then what will be the resistance of the resistor and terminal voltage of the cell in closed circuit? [1½]

Answer:

Given : E = 12 volt; r = 5Ω, i= 0.3

A, R=?;V=?

Current, i = \(\frac{E}{R+r} \)

⇒ R + r = \(\frac{E}{i} \)

∴ R + r = \(\frac{12}{0.3}\) = 40 Ω

∴ R = 40 – r = 40 – 5

or R=35Ω

Terminal voltage of the cell in closed circuit,

V=E-ir=12-0.3×5

= 12-1.5

= 10.5 volt

Question 8.

State the principle of an AC generator and explain its working with the help of a labelled diagram. [1½]

Answer:

Principle: The working of an AC. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

Working: As the armature coil rotates, the magnetic flux linked with it changes and so an induced current flows through it. Suppose, initially the coil PQRS be in the vertical position and it is rotated in the clockwise direction. The side PQ moves downward and SR moves upward. According to Fleming’s right-hand rule, the induced current flows from Q to P and from S to R.

So, during the first half rotation of the coil, the induced current flows in the direction SRQP, with brush B_{1} acting as positive terminal and brush B_{2} as negative terminals. During the second half-rotation, the side PQ moves upwards and SR moves downward. The direction of induced current is reversed, i.e. it flows along PQRS so that the brush B_{1} act as the negative terminal.

Thus, the direction of current in the external circuit is reversed after every half cycle. Hence, alternating current is produced by the generator. Such a generator which generates alternating current is called an A.C. generator or an alternator.

Question 9.

An aeroplane is flying horizontally from west to east with a velocity of 900 km/h. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the earth’s magnetic field is 5 x 10^{-4} T and the angle of dip is 30°.[1½]

Answer:

Given, u = 900 kmlh

=900 x \(\frac{5}{18} \) = 250m/s

l = distance between the ends of the

wings = 20 m

Dip angle, δ = 30°

Horizontal component of earth’s field

=B_{H} =5 x 10^{-4} T

Potential difference induces due to cutting of vertical field lines.

So, induced emf = B_{ν} .l.ν = \(\frac{B_{H}}{\cos \delta}\)

= B_{H} .tan δ.l.ν

(B_{H} = B cos δ and B_{ν} = B sin δ)

=5 x 10^{-4} x tan30° x 20 x 250

= \(\frac{2.5}{\sqrt{3}} \) ≈ 1.45 V

Question 10.

Use the mirror equation to deduce that an object placed between F and 2F of a concave mirror produces a real image beyond 2 f.[1½]

Answer:

(a) From mirror formula,

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

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

∴ υ = \(\frac{u f}{(u-f)}\)

For concave mirror u and f both are negative, therefore

Clearly, the value of υ is negative therefore the image will form in front of the mirror. As a result the image

will be ‘Real image’.

Again, from υ = \(\frac{u f}{u-f}\)

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

Question 11.

Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism. State reasons to explain these observations.[1½]

Answer:

As a refractive mix of a prism is different for different colours, therefore, different colours deviate through different angles on passing through the prism. As λ_{violet} < λ_{red} maximum deviation is of violet colour. That is why violet colour, is seen at the bottom of the spectrum when white light is dispersed by a prism.

Question 12.

Find out the relation between the refractive index (n) of the glass prism and ∠A for the case, when the angle of prism (A) is equal to the angle of minimum deviation (δ_{m}). Hence, obtain the value of the refractive index for angle of prism A = 60°.[1½]

Answer:

n= \(\frac{\sin (A+A / 2)}{\sin (A / 2)}\)

⇒ n = \(\frac{2 \sin (A / 2) \cdot \cos (A / 2)}{\sin (A / 2)}\)

⇒ n= 2cos(A/2)

This is a required relation between refractive index of the glass prism and angle of prism.

Since, ∠A = 60° (given)

⇒ n = 2 cos\(\left(\frac{60^{\circ}}{2}\right)\)

⇒ n = 2 cos 30°

⇒ n = 2 x \(\frac{\sqrt{3}}{2}\)

⇒ n = \(\sqrt{3}\)

⇒ n = 1.732

Question 13.

Danger sign is red. Why? [1½]

Answer:

Danger signals are red in colour: According to Rayleigh’s scattering law, the intensity of scattered light,

I_{s} ∝ \(\frac{1}{\lambda^{4}}\) Clearly, smaller the wavelength, more is the scattering. Since the wavelength of red colour is the most, hence its scattering will be the least. As a result, red colour is seen upto far distances. That is why the danger signals are made of red in colour.

Question 14.

Define the terms half-life and average life. Find out the relationship with the decay constant (λ).[1½]

Answer:

Half-life: We know that radioactive elements always disintegrated and as the time passes the number of united nuclei decreases. “Half-life of a radioactive sample is the time to reduce the nuclei to half of its initial value.”

Average Life of a Radioactive Substance:

“The life upto which a radioactive nucleus is exits, is called average life.” It is denoted by τ. Relation between Half-life and Decay Constant If the number of nuclei in the beginning is N0 then the number of remaining nuclei after time t

N = N_{0}e^{-λt}

When t=T then N= \(\frac{N_{0}}{2}\)

∴ \(\frac{N_{0}}{2}\) = N_{0}e^{-λt}

or \(\frac{1}{2}\) =e^{-λt}

⇒ \(\frac{1}{2}\) = \(\frac{1}{e^{\lambda T}}\)

or 2=e^{λT}

Taking the log on both sides,

1og_{e} 2 = Iog_{e} e^{λT} = λTlog_{e} e=λT

or T=log_{e}2

or T = \(\frac{\log _{e} 2}{\lambda}\) ………………………….. (2)

With the help of this equation we can find out the value of λ it value of T is known or vice-versa.

log_{e} 2 = 0.6932

∴ T = \(\frac{0.6932}{\lambda}\) ……………………………… (3)

or λ = \(\frac{0.6932}{T}\) …………………………….. (4)

Question 15.

Binding energy per nucleon of deutron (_{1}H^{2}) and a-particle _{2}He^{4} is 1.1 MeV and 7.0 MeV respectively. When two deutron make a a-particle after fusion then calculate the free energy. [1½]

Answer:

Given, binding energy per nucleon of deutron(_{1}H^{2}) = 1.1MeV

Binding enery per nucleon of α-particle(_{2}He^{4} )= 7.0 MeV

Fusion reaction

_{1}H^{2} + _{1}H^{2} → _{2}He^{4}

Binding energy before fusion

E^{i} =1.1 x 2+1.1 x 21 = 1 x 4 = 4.4MeV

Binding energy after fusion

E_{f} = 7.0 x 4=28.0 MeV

Free energy =ΔE = E_{f} – E_{i}

= 28.0 – 4.4

= 23.6 MeV

Section-C (3 Marks)

Question 16.

Explain, using a labelled diagram, the principle and working of a moving tool galvanometer. What is the function of (i) uniform radial magnetic field (ii) soft iron core? [3]

Or

Two very small identical circular loop (1) and (2) carrying equal current I are placed vertically (with respect to the plane of the paper) with their geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at the point O.[3]

Answer:

Suspended Coil Galvanometer:

Construction

It consists of a rectangular coil of a large number of turns of thin insulated copper wire wound over a light metallic frame. The coil is suspended from TH between the pole pieces of a horse-shoe magnet by a fine Phosphor-Bronze strip from a movable torsion head. The lower end of the coil is connected to a hairspring of Phosphor Bronze having only a few turns.

The other end of the spring is connected to a binding screw. A soft iron cylinder is placed symmetrically inside the coil. The hemispherical magnetic poles produce a radial magnetic field in which the plane of the coil is parallel to the magnet field in all its positions.

A small plane mirror attached to the suspension wire is used along with a lamp and scale arrangement to measure the deflection of the coil.

Principle: “Suspended coil galvanometer works on the principle that a current-carrying coil placed in a uniform magnetic field experiences a torque; whose magnitude is directly proportional to the electric current flowing in the coil.” The magnetic torque tends to rotate the coil.

If at any instant the vector area \(\overrightarrow{\mathrm{A}}\) of the coil makes an angle θ with the magnetic field \(\vec{B}\)then magnetic torque acting on the coil is;

τ_{def} = NAIB sin θ

Here, N = number of turns in the coil The magnetic field in the coil is always taken in radial direction due to which θ = 90°

∴ τ_{def} = NAIB

Due to τ_{def} the coil gets deflected from its equilibrium position and starts rotating. As a result of this, there is a twist in the Phosphor-Bronze wire and the spring, hence a counter torque starts working on the coil which is opposite to the magnetic torque. If Φ is the deflection and C is the force couple for unit radian deflection, then counter-torque is;

τR=CΦ …………………………. (i)

In equilibrium position;

τ_{def} = τ_{R}

or NAIB=CΦ

or I= \(\left[\frac{C}{N A B}\right] \phi \) ……………………. (2)

or I=kΦ ………………………………….. (3)

Here, k = \(\frac{C}{N A B}\) is a constant; which is called the “reduction factor of the galvanometer” or “the torsional constant of the spring”.

It is clear from equation (3) that current flowing in the galvanometer is directly proportional to the deflection;

I ∝ Φ

Radial Magnetic Field

Radial magnetic field is used to keep the vector area \(\overrightarrow{\mathrm{A}}\) of the coil always

perpendicular (θ =90°) to the magnetic field.

Question 17.

Draw a ray diagram to show the image formation by a convex lens—

(a) When the object is kept between its focus and optical centre

(b) When the object is kept between 2F and F (c) When the object is kept between infinite and 2 F. [3]

Or

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope? [3]

Answer:

Question 18.

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?

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

Or

What is the de-Broglie wavelength of:

(a) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.

(b) A ball of mass 0.060 kg. moving at a speed of 1.0 m/s, and

(c) A dust particle of mass 1.0 x 10^{-9} kg. Drifting with a speed of 2.2 m/s? [3]

Answer:

(i) Yes, all emitted photoelectrons have same kinetic energy as the kinetic energy of emitted photoelectrons depending upon frequency of the incident radiation for a given photosensitive surface.

(ii) No, the kinetic energy of emitted electrons does not depend on the intensity of incident radiation. If the intensity is increased, number of photons will also increase but energy of each photon remains same as the frequency is also same, The maximum kinetic energy depends on frequency not on intensity.

(iii) The number of emitted photoelectrons depends only on intensity of incident light.

Section-D

Question 19.

(i) State Gauss’s law. (ii) A thin straight infinitely long conducting wire of linear charge density λ is enclosed by a cylindrical surface of radius r and length l. Its axis coinciding with the length of the wire Obtain the expression for the electric field, indicating its direction, at a point on the surface of the cylinder. [4]

Or

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [4]

Answer:

(i) Gauss’ law states that the total flux through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times to the net charge enclosed by the closed surface.

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

Here, ε_{0} is the absolute permittivity of the free space, q is the total charge enclosed and E is the electric field at the area element dS.

(ii) electric field intensity due to an infinitely long uniformly charged wire at point P at distance r from it is obtained as follows:

Consider a thin cylindrical Gaussian surface S with charged wire on it axis and point P on its surface, then net electric flux through surface S is

Φ = \(\int_{s} \mathbf{E} \cdot d \mathbf{S} \)

Φ=O+EA +0 or Φ=E.2πrl

But by Gauss’s theorem,

Φ =q/ε_{0} = λl/ε_{0}

where q is the charge on length l of wire enclosed by cylindrical surface S and is uniform linear charge density of wire.

∴ E x 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\)

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

Thus, electric field of a line charge is inversely proportional to distance directed normal to the surface of charged wire.

Question 20.

Distinguish between n-type and p-type semiconductors on the basis of energy band diagrams. [4]

Or

(i) Explain with the help of diagram, how a depletion layer and barrier potantial are formed in a junction diode.

(ii) Draw a circuit diagram of a full-wave rectifier. Explain its working and draw input and output waveforms. [4]

Answer:

In n-type extrinsic semiconductors, the number of free electrons in conduction band is much more than the number of holes in valence band. The donor energy level lies just below the conduction band. In p-type extrinsic semiconductors, the number of holes in valence band in much more than the number of free electrons in conduction band. The acceptor energy level lies just above the valence band.

(ii)

At absolute zero temperature (0K) conduction band of semiconductor is semiconductor behaves as an insulator. At room temperature, some valence electrons acquire enough thermal energy and jump to the conduction band where they are free to conduct.

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