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Physics | Class 12th

CBSE Board Paper 2018

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CBSE Board Paper 2018 Set - 1

General Instructions:

1. All questions are compulsory. There are 26 questions in all. 2. This question paper has five sections: Section A, Section B,

Section C, Section D and Section E. 3. Section A contains five questions of one mark each, Section B

contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.

4. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary: c = 3 × 108 m/s

h = 6.63 × 10–34 Js

e = 1.6 × 10–19 C

μo = 4π × 10–7 T m A–1

ε0 = 8.854 × 10–12 C2 N-1 m-2

o

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4= 9 × 109 N m2 C–2

me = 9.1 × 10–31 kg

mass of neutron = 1.675 × 10–27 kg

mass of proton = 1.673 × 10–27 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10–23 JK–1

Time allowed: 3 Hours Max Marks: 70

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1. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?

Section A

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2. Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery.

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3. 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.

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4. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two- the parent or the daughter nucleus - would have higher binding energy per nucleon?

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5. Which mode of propagation is used by short wave broadcast services?

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Section B

6. Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.

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7. A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit.

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OR

In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

8. (a) Why are infra-red waves often called heat waves?

Explain.

(b) What do you understand by the statement, ‘‘Electromagnetic waves transport momentum’’?

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9. If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?

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10. A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.

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11. Four-point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the

(a) resultant electric force on a charge Q, and

(b) potential energy of this system.

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OR

(a) 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.

(b) Find out the amount of the work done to separate the charges at infinite distance.

Section C

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13. A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

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12. (a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.

(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

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14. (a) An iron ring of relative permeability μr has windings of insulated copper wire of n turns per meter. When the current in the windings is I, find the expression for the magnetic field in the ring.

(b) The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

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15. (a) Show using a proper diagram how unpolarised light can be linearly polarised by reflection from a transparent glass surface.

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16. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.

(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?

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(b) The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index 3/2 , placed in water of refractive index 4/3 . Will this ray suffer total internal reflection on striking the face AC ?

17. A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.

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18. (a) State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ?

(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

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19. (a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.

(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ?

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20. (a) A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw the labelled circuit diagram she would use and explain how it works.

(b) Give the truth table and circuit symbol for NAND gate.

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21. Draw the typical input and output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine (a) the input resistance (ri), and (b) current amplification factor (β).

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22. (a) Give three reasons why modulation of a message signal is necessary for long distance transmission.

(b) Show graphically an audio signal, a carrier wave and an amplitude modulated wave.

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23. The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.

(a) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.

(b) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.

(c) Write two values each shown by the teachers and Geeta.

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Section D

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24. (a) 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 center 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 ‘d’ from

the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

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OR

(a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.

(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.

(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1).

Section E

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25. (a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ω in a magnetic field B directed perpendicular to the axis of rotation.

(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. 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 × 10–4 T and the angle of dip is 30°.

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OR

A device X is connected across an ac source of voltage V = V0 sin(ωt). The current through X is given as

(a) Identify the device X and write the expression for its reactance.

(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.

(c) How does the reactance of the device X vary with frequency of the AC? Show this variation graphically.

(d) Draw the phasor diagram for the device X.

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26. (a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.

(b) Obtain the mirror formula and write the expression for the linear magnification.

(c) Explain two advantages of a reflecting telescope over a refracting telescope.

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OR

(a) Define a wave front. Using Huygens’ principle, verify the laws of reflection at a plane surface.

(b) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the obstacle. Explain why.

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1. The frequency of revolution of charge in a perpendicular magnetic field is given as

Where ν = frequency of revolution of charge

q = charge of particle

B = magnetic field

m = mass of the particle

As we can see from formula that frequency is inversely proportional to the mass of the particle.

We know that the mass of an electron is less than the mass of

a proton. Therefore, the electron will move in a circular path with higher frequency.

2. a) Electromagnetic radiation used for water purification is UV radiations.

b) Electromagnetic radiation used for eye surgery is a laser.

Solutions (Set-1)

Section A (Solutions)

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3.

I2>I1

4. When two light nuclei fuse to form a larger nucleus, energy is released, since the large nucleus is more tightly bound. Therefore, the daughter nucleus will be stable than the parent nucleus.

Thus, the binding energy of the daughter nucleus will be more than the parent nucleus as the daughter nucleus is more tightly bound and stable.

5. Sky waves are used for short wave broadcast services.

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6. Let, Resistance, power and current of bulb P = RP, PP, IP

Resistance, power and current of bulb Q = RQ , PQ, IQ

It is given in the question that,

We know that power dissipated across any resistance is given as

P = I2R

Where P = power dissipated

I = current through the resistance

According to the question, both the resistance is in series. Therefore, the current through both the resistance will be the same.

∴ IP = IQ …….. (ii)

From equation (ii),

From equation (i),

∴ the ratio of power dissipation in bulbs P and Q is 1:2.

7. Let the current in the circuit be i.

Applying Kirchhoff’s law in loop ABCDA

10 + 38i = 200

Section B (Solutions)

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38i = 200-10 = 190

∴ the current in circuit is 5A.

Given

The balance point of the cell in open circuit = l1 = 350 cm.

the balance point of the cell with resistance = l2 = 300cm.

External resistance R = 9Ω.

We know that internal resistance of a primary cell by a potentiometer is given as,

∴ the internal resistance of the cell is 1.5Ω.

8. a) Infrared rays are readily absorbed by the (water) molecules in most of the substances and hence increases their thermal motion. Due to the increase in thermal motion, the temperature of the substance also increases and heat is emitted from the substance. Thus, infra-red waves are often called heat waves.

b) Electromagnetic waves can set (and sustain) charges in motion. Hence, they are said to transport momentum. Electromagnetic waves are also known to exert “radiation pressure‟. This pressure is due to the force associated with the rate of change of momentum. Hence, EM waves transport momentum.

OR

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9. Given

Wavelength of incident light λ = 412.5nm = 412.5× 10-9m

The energy of photon of light of a given frequency is given as

E = hν

Where h = Planck’s constant = 6.63× 10-34Js

ν = frequency of light

We know that

Where c = speed of light = 3× 108ms

We know that 1eV = 1.60× 10-19J.

Substituting the value of λ, h and c in the above equation we get,

Photoelectric emission will only take place if the energy of the photon of the incident radiation is more than the work function of the given metal.

From the table given in the question, we can see that the work function of Na (1.92eV) and K (2.15eV) is less than the energy of a photon of incident radiation. Therefore, metals Na and K will show photoelectric emission.

10. Given

The peak voltage of carrier peak Ac = 15V

Modulation index μ = 60% = 0.6

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Let peak voltage of modulating signal = Am

We know that

Am = μ × Ac

= 0.6 × 15 = 9V

∴ the peak voltage of the modulating signal to have a modulation index of 60% is 9V.

11. a) As seen from the diagram, there will be three forces on the charge Q at one of the corners of the square.

Now we know that from Coulomb’s law, the force between two charges is given as

Where q1 and q2 are the charges separated by a distance r.

Applying coulomb’s law for charge Q (at C) and q (at B)

Section C (Solutions)

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Applying coulomb’s law for charge Q (at C) and q (at D)

Applying coulomb’s law for charge Q(at C) and Q (at A)

Now from the diagram, it can be seen that forces F1 and F2 are perpendicular to each other. So, the magnitude of their resultant will be

Since cos90° is zero and F1 = F2, therefore,

Now applying the superposition principle, we get

F = F’ + F3

b) Potential energy is between two charges is given as

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Where q1 and q2 are the charges separated by a distance r.

Due to the symmetry of the system, potential energy between charges that are adjacent to each other will be same.

Potential energy between diagonally opposite charge q

Potential energy between diagonally opposite charge Q

Therefore, the total potential energy of the system will be

U = U1 + U2 + U3

a) We know that from coulomb’s law, the force between two charges is given as

Where q1 and q2 are the charges separated by a distance r.

Applying coulomb’s law between charge -4q and q

Applying coulomb’s law for charge 2q and q

OR

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From the diagram, we can see that the angle between F1 and F2 is 120°. So, the magnitude of their resultant force F will be

Since F1 = 2F2 and cos120° = -1/2

F = √3𝐹22

F = √3F2

So, the net electric force on charge q is

b) The amount of work done in separating the charges to infinity will be equal to potential energy.

Potential energy is between two charges is given as

Where q1 and q2 are the charges separated by a distance r.

The total potential energy of the system

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12. a) Conductivity of a wire is defined as the reciprocal of resistivity ρ. Its S.I units are mho/m.

b) As we know that resistance R of a conductor is defined as

Where ρ = resistivity of the conductor

l = length of conductor

A = area of cross-section of the conductor

According to ohm’s law V = IR

Where V = voltage across a conductor and I = current through the conductor.

Substituting the value of R in ohm’s law

Now

So, the voltage becomes

V = j ρ l …. (i)

We know that electric field E is given as

Therefore V = E l.

So, equation (i) becomes

E l = j ρ l

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E = j ρ

Now we know that

Therefore j = σ E

The average drift velocity of electron is given as

Where e = charge of electron

E = electric field

m = mass of electron

τ = relaxation time

If n is the number of free electrons per unit volume, the current I is given by

I = neA |Vd|

Now

13. a) Given

Magnetic moment m = 6 J/T

Magnetic field B = 0.44T

Initial angle θ1 = 60°

Now work done in turning the magnet is given as

W = -mB (cosθ2 -cosθ1)

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i) For first case θ2 is 90°

Therefore

W = -mB (cos90 -cos60)

As Cos90° = 0 and cos60° = 1/2

We get,

W = -6× 0.44× (0-1/2)

W = 3× 0.44 = 1.32J

ii) For the second case θ2 is 180°

Therefore

W = -mB (cos180 -cos60)

Cos 180° = -1 and cos60° = 1/2

b) Torque is given as

τ = m× B = mBsinθ

where θ is the angle between m and B.

In the second orientation, magnetic moment and magnetic field are anti-parallel i.e. θ = 180°.

Now sin180° = 0.

Therefore, torque will be zero.

14. a) According to ampere’s law, the line integral of magnetic field B around any closed circuit is equal to μo times the total current passing through this closed circuit. Mathematically,

∫B.dl = μoμrIenclosed

Now ∫B.dl = B∫dl = BL , if B is constant and directed along the tangent to every point on the perimeter L of a closed curve.

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For the field inside the ring, we can write

∫B.dl = B∫dl = B. 2πr

(r = radius of the ring)

Also, Ienclosed = (2πrn)I using equation (i)

Where n = number of turns per meter

∴ B. 2πr = μoμr (n. 2πr)I

∴ B = μoμrnI

b) The susceptibility of this material is positive and lies between 0 and 1. So it is a paramagnetic material. When a paramagnetic substance is placed in a uniform magnetic field, the unpaired electron aligns itself with the external field.

The modified magnetic field when a paramagnetic material is placed in a uniform magnetic field is shown below.

15. a)

b) The angle of incidence, of the ray, on striking the face AC is

i = 60° (as from figure).

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Total internal reflection will only take place when the angle of incidence i is greater than the critical angle ic.

We know that the condition for total internal reflection is

Where μ is the relative refractive index of glass, with respect to the surrounding water. Therefore,

Therefore

Since ic is greater than the incident angle i, the total internal reflection will not take place.

16. a) We know that intensity I of the light wave is directly proportional to square amplitude A of a light wave.

I∝ A2

It is given that intensity is reduced by 50%.

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The ratio of maximum and minimum intensity is given as

b) The central fringe remains white.

No clear fringe pattern is seen after a few (coloured) fringes on either side of the central fringe.

17. Given

Focal length of convex lens f1 = y

Refractive index of lens μ1 = 1.5

Let the focal length of liquid be f2

And the refractive index of the liquid is μ2.

Let the focal length of system (lens + liquid) be f = x(given)

Now equivalent focal length is given as

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Let radius of curvature of lens be R.

Then for a convex lens R1 = R and R2 = -R

From lens maker formula

Substituting the value of f1 and μ1, to find R

Since liquid acts as a plane mirror, therefore, R1 = R = y cm and R2 = ∞.

Applying lens maker formula for liquid, we get

18. a) Bohr’s postulate, for stable orbits, states

“The electron, in an atom, revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of h/2π”, where h = Planck’s constant.

De Broglie wavelength of a particle is given as

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Where λ = wavelength of the moving particle

h = Planck’s constant = 6.63× 10-34 Js

p = momentum of moving the particle

We know that momentum is given as

p = m× v

where m = mass of the particle

v = velocity of the particle

For a stable orbit, we must have a circumference of the

orbit = nλ (n = 1,2,3, .........)

Thus, De Broglie showed that formation of stationary pattern for

Integral “n‟ gives rise to the stability of the atom.

b) We know that for hydrogen atom the energy of nth level is given as

According to question, a hydrogen atom is excited from the ground state (n = 1) to n = 4 level. So, the energy of the photon will be

E4-E1 = hν

Where h = Planck’s constant

ν = frequency of the photon absorbed.

Therefore, we can write

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∴ the frequency of the absorbed photon is 3.1× 1015 Hz.

19. a) The plot of binding energy per nucleon (BE/A) versus the mass number A is given below:

From the plot, we note that

i) During nuclear fission

A heavy nucleus in the larger mass region ( A>200) breaks into two middle-level nuclei, increasing B.E/ nucleon. This results in the release of energy.

Example:

92U235 + n1→ 56Ba144 + 36Kr89 + 30n1 + energy

ii) During nuclear fusion

Light nuclei in the lower mass region (A<20) fuse to form a nucleus having higher B.E / nucleon. Hence Energy gets released.

Example:

1H2 + 1H2→ 2He4 + energy

b) Let the initial activity be Ao.

Let the time taken for the activity to reduce = t.

Half live given T1/2 = 10years

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If we take Ao = 100, then activity after time t will be

At = 3.125 (given)

We know the law of radioactivity decay is given as

At = Ao e-λt …. (i)

Where λ = decay constant.

Therefore, substituting the values of At Ao and λ in equation (i), 3.125 = 100e-0.0693t

Taking natural logarithm on both sides of the above equation

∴ the time taken for the activity to reduce to 3.125% is 50 years.

20. a) To convert alternating current into direct current, we need a full-wave rectifier. The circuit diagram for a full-wave rectifier is given below:

The working of this circuit is as follows:

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i) During one-half cycle (of the input ac) diode D1 alone gets forward biased and conducts. During the other half cycle, it is diode D2 (alone) that conducts.

ii) In the centre tapped transformer the current though the load flows in the same direction in both the half cycles.

Hence we get a unidirectional/ direct current through the load when the input is alternating current.

b) The circuit symbol and truth table for the NAND gate is given below:

21. The input and output characteristics of n-p-n transistor in CE configuration is given below:

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Input resistance is defined as the ratio of change in the base-emitter voltage (ΔVBE) to the change in base current (ΔIB) at a constant collector-emitter voltage.

Current amplification factor (β) is defined as the ratio of the change in the collector current to the change in the base current at a constant collector-emitter voltage when the transistor is in active state.

22. a) The three reasons necessary for long-distance transmission are:

i) A reasonable length of the transmission antenna.

ii) Increase in effective power radiated by the antenna.

iii) Reduction in the possibility of “mix-up‟ of different signals.

b) In Amplitude modulation, the transmission of a wave signal is done by modulating the amplitude. At the source, the information signal is modulated and sent to the receiver, once it is received, it is then demodulated and filtered to remove any disturbance in the original signal that may have occurred during the transmission. The amplitude modulation is the process of superimposing the information signal with a carrier generated signal.

The information signal is the data message signal stored on a network that has to be transmitted from a source to a receiver.

A carrier wave is an electromagnetic wave of fixed amplitude and of a constant frequency that is modulated with an input message signal for transmission.

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23. a) The device used to change the alternating voltage to a higher or lower value is step up or step-down transformer. It works on the principle of Faraday's law of induction by converting one value of electrical energy into another.

One of the reasons for power dissipation in this device is the production of eddy currents in core. The changing magnetic field induces an EMF in the secondary winding. However, due to this changing magnetic field through the core, an EMF is induced in the core itself as the core is made of conducting material and thus current is generated in the core. These

Section D (Solutions)

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circulating currents are in the form of swirling eddies, called eddy currents.

b) The relation between the voltage in the secondary coil with a voltage in the primary coil is given as

And the relation between current in the secondary coil with a

current in the primary coil is given as

In a step-up transformer, if the voltage is getting stepped up,

then current in the secondary coil is getting reduced.

Hence the power loss (P = I2R) is reduced considerably while such stepping up is not possible for direct current.

c) The teacher is well informed and willing to share her knowledge with Geeta that reflects her concerns for her. She wants her students to learn the practical aspects of the theoretical learning that is done in the class.

Geeta, on the other hand, manifest the value of a good listener who is interested in learning the practical application of the concepts that has been taught in the class. It reflects her curious nature and the sheer willingness to learn something new.

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24. a) The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area. It is a scalar quantity.

Electric flux ∅E = ∫E.dS

Where E = electric field

dS = area element

We can imagine the square as the face of a cube with edge d and with charge q placed at its centre, as shown in the figure below.

According to Gauss’s theorem, the total flux of the electric field

E through a closed surface S is equal to 1/ϵo times the charge q enclosed by the surface S.

The symmetry of six faces of a cube about its centre ensures

that the flux ∅s through each square face is same when the charge is placed at the centre.

∴ Total flux,

Section E (Solutions)

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Hence the flux through the square of side d with charge q at a

distance d/2 directly above the head is q/6ϵo.

b) If the charge is moved to a distance and the side of the square is doubled the cube will be constructed to have a side 2d, but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.

a) Electric field (E) due to a straight uniformly charged an infinite line of charge density λ C/m.

Consider a thin long straight wire having a uniform linear

charge density λ C/m. By symmetry, the field E of the line is directed radially outwards, and its magnitude is the same at all points equidistant from the line charge. To determine the field at a distance r from the line charge, we choose a cylindrical gaussian surface of radius r, length l and with its axis along with the line charge. As shown in the figure, it has a curved surface S1 and a flat circular surface S2 and S3. From figure,

dS1||E

OR

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dS2and dS3 are perpendicular to E.

so, only the curved surface contributes towards the total flux.

As cos90° = 0 and cos0° = 1, total flux will be

∅E = E∫dS1 = E× area of the curved surface

∅E = E× 2πrl

Charge enclosed by Gaussian surface, q = λl.

Using gauss theorem, we get

b) Graph the variation of E with perpendicular distance r from the line of charge is shown below:

c) Work done in moving the charge q, through a small displacement dr

dW = F.dr

Now electric F is given as

F = qE

Where E = electric field

∴ dW = qEdr cos0° = qE dr

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Substituting the value of E from part (a) we get

Work done in moving the given charge from 𝑟1 to 𝑟2 (𝑟2> 𝑟1)

25. a) Diagram of the ac generator is given below.

Working 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 current is set up in it.

The expression for induced emf:

Let N = number of turns in the coil

A = face area of each turn

B = magnitude of the magnetic field

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θ = angle which normal to the coil makes with filed B

ω = the angular frequency with which coil rotates

Then the magnetic flux linked with the coil at any instant t will be

∅ = NBA cos(θ) = NBA cos(ωt)

By faraday’s flux rule, the induced emf is given by

ϵ = ϵo sin(ωt)

where ϵo = NBAω

b) Potential difference developed between the ends of the wings e = Bl𝜈

Given Velocity v = 900km/hour = 250m/s

The horizontal component of the Earth’s magnetic field = 5× 10-4T

Wing span (l) = 20 m

The vertical component of Earth’s magnetic field

BV = BH tan δ

= 5×10−4 (tan 30°) T

∴ The potential difference, e

= 5×10−4 (tan 30° ) × 20 × 250

a) The device X is a capacitor since clearly from the equation

𝐼 = 𝐼𝑜sin(𝜔𝑡 +𝜋

2), it can be inferred that the current leads

OR

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voltage by a phase angle of 𝜋

2. The reactance of the capacitor

Xc is given as

b) Variation of voltage and current with time over one cycle of ac, for a capacitor.

c) The reactance of the capacitor is given by the expression,

The reactance of the capacitor from the equation is inversely proportional to the frequency

Therefore, the graph of capacitor reactance against the frequency will be as following,

d) Phasor diagram for capacitor

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The phasor diagram of the capacitor represents A phasor diagram is used to show the phase relationships between two or more sine waves having the same frequency. In this case the voltage and the current equation.

26. a)

b)

From the figure given above, we can see that the two right-angled triangles A’B’F and MPF are similar. For paraxial rays,

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MP can be considered to be a straight line perpendicular to CP. Therefore,

Since ∠ APB = ∠ A’PB’, the right-angled triangles A’B’P and ABP are also similar. Therefore,

Comparing equations (i) and (ii), we get

The light travels from the object to the mirror MPN. Hence this is taken as the positive direction. To reach the object AB, image A’B’ as well as the focus F from the pole P, we have to travel opposite to the direction of the incident light. Hence, all three will have negative signs. Thus,

B’P = -v, FP = -f ,BP = -u

Using these in equation (iii), we get

This relation is known as mirror equation where v = image distance, u = object distance and f = focal length.

Linear magnification m for a mirror is given as

Where h’ = image height

h = object height

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c) Advantages of reflecting telescope over refracting telescope are:

1. In reflecting type telescope, there is no chromatic aberration as the objective is a mirror.

2. Image is brighter compared to the one formed in refracting type telescope.

a) Wavefront: A locus of the points which oscillate in the same phase is called a wavefront. Thus, a wavefront is defined as a surface of constant phase.

Let the speed of the wave in the medium be ‘v’

Let the time taken by the wavefront, to advance from point B to point C is τ

Hence BC = vτ Let CE represent the reflected wavefront

Distance AE = vτ = BC

Δ AEC and Δ ABC are congruent ∴ ∠ BAC = ∠ ECA

⇒∠ i = ∠ r.

b) The size of the central maxima is given as

Where λ = wavelength of light

D = distance between slit and screen

OR

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d = width of slit

If d⇒2d then β ⇒β/2.

∴ Size of central maxima reduces to half.

Intensity will increase. This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle because of diffraction of the light source and in case of diffraction the central maxima is always bright.

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