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CBSE 12th Physics 2017 Guess Paper
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CBSE 12th Physics 2017 Guess Paper By 4ono.com
TIME - 3HR. | QUESTIONS - 26
THE MARKS ARE MENTIONED ON EACH QUESTION _________________________________________________________________________
SECTION β A
Q.1. Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? 1 mark
Ans. Since, the magnetic field induction outside the toroid is zero.
Q.2. A plot of magnetic flux (π) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of
self-inductance? 1 mark
Ans. Since π = LI β L = π
1 = slope of π β I graph β΄ slope of inductor A = slope of
inductor B. Hence the inductor A has larger value of self-inductance.
Q.3. Show graphically, the variation of the de-Broglie wavelength (Ξ») with the potential (V) through which an electron is accelerated from rest. 1 mark
Ans.
Q.4. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 Ξ© as shown in the figure. Find the value of the current in circuit. 1 mark
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Ans. Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by
= 200 -10 = 190v.
Hence, the current in the circuit is given by πΌ =Ξ΅
π .
πΌ =190
38= 5π΄
Q.5. Define electric dipole moment. Write its S.I. unit. 1 mark
Ans. Electric dipole moment: Dipole moment ( ) is a measure of strength of electric dipole. It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and the distance between them. Si unit of dipole moment is coulomb-meter (cm).
SECTION - B
Q.6. How does the resistivity of a conductor depend upon temperature electrical conductivity? 2 marks
Ans. (i) The resistivity of a conductor increases with increase with increase in temperature
β΄ ππ = π0[1 + πΌ(π β π0)]
(ii) The resistivity of a conductor is the reciprocal of electrical conductivity
β΄ π =1
π
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Q.7. In the meter bridge experiment, balance point was observed at J with
AJ = l. 2 mark
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected? 2 mark
Ans. (i) the value of R and X were doubled and then interchanged. Hence the new position of balance point will 100 β πΌ.
(ii) π΄π½ = π.
From the principle of Wheat Stones Bridge,
π
π=
π
100 β π
π = π (100 β π
π)
Hence, the galvanometer and tell are interchanged, the condition for a balance bridge is still satisfied. Therefore, the galvanometer will not show any deflection.
Q. 8. State Kirchhoff's rules. Explain briefly how these rules are justified. 2 marks
(a) Kirchhoffβs First law:
Ans. Junction rule: The algebraic sum of all the emf meeting at a point in an electrical circuit is always zero.Let the currents be πΌ1, πΌ2, πΌ3 πππ πΌ4
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Convention: Current towards the junction is always positive, while current away from the junction is negative
πΌ3 + (βπΌ1) + (βπΌ2) + (βπΌ) = 0
Kirchhoffβs Second law: Loop rule
In a closed loop, the algebraic sum of the emfβs is equal to the algebraic sum of the products of the resistance and current flowing through them.
For closed part BACB, πΈ1 β πΈ2 = πΌ1π 1 + πΌ2π 2 β πΌ3π 3
For closed part CADC, πΈ2 = πΌ3π 3 + πΌ4π 4 + πΌ5π 5
Demonstration:
Wheatstone Bridge: - The Wheatstone Bridge is an arrangement pf four resistance as shown in the following figure.
π 1, π 2, π 3 πππ π 4 are the four resistances.
Galvanometer(a) has a current πΌπflowing through it at
balanced condition πΌπ = 0
Applying junction rule at B,
β΄ πΌ2 = πΌ4
Applying junction rule at D,
β΄ πΌ1 = πΌ3
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Applying loop rule to closed loop ADBA,
βπ 1 + π + πΌ1π 2 = 0
β΄ πΌ1πΌ2
=π 2
π 1β¦(π)
Applying loop rule to closed loop CBDC,
π 1π 1 + π β π 1π 1 = π [πΌ3 = πΌ1, πΌ4 = πΌ2]
β΄ π 2
π 1=
π 4
π 3β¦(ππ)
From Eq. (i) and (ii)
π 2
π 1=
π 4
π 3
This is the required balanced condition of wheatstone bridge.
Q.9. Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions πΎπ and (πΎπ > πΎπ). On what factors does the (i) slope and (ii) intercept of the lines depend? 2 mark
Ans. The graph showing the variation of stopping potential (π0) with the frequency of incident radiation (π£0) for two different photosensitive materials having work functions W1 and W2(W1>W2) is shown in fig.
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(i) Slope of the line = βV
βπ£ =
β
π
[β΄ πβV = h βπ£] β΄ Slope of the line depends on the Planckβs constant h and the electronic charge π.
(ii) Intercept of graph A on the potential axis
=work function(W)
π= β
βπ£0
π
β΄ Intercept of the line depends upon Planckβs constant h, threshold frequency (π0) and the electronic charge (e).
Q.10. Using Rutherfordβs model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron? 2 mark
Ans. From Rutherfordβs model of the atom, the magnitude of this force is
πΉ =1
4 ππ0 .2π. (ππ)
π2
For hydrogen atom, Let, πΉπ βCentripetal force required to keep a revolving electron in orbit. Fe-Electrostatic force of attraction between the revolving electron and the nucleus. Then, for a dynamically stable orbit in hydrogen atom, where Z = 1
πΉπ = πΉπ
ππ£2
π=
(π)(π)
4ππ0π2 β¦ (π)
π =π
4ππ0ππ£2 β¦ (ππ)
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K.E. of electron in the orbit,
K =1
2ππ£2,
Form equation (π),
K =π2
8ππ0π
Potential energy of electron in
Orbit, U = (π)(π)
4ππ0π=
βπ2
4ππ0π
β΄ Total energy of electron in hydrogen atom
E = k + U = π2
8ππ0πβ
π2
4ππ0π; E = β
π2
8ππ0π
Here, negative sign indicates that the revolving electron is bound to the positive nucleus.
OR
Using Bohrβs postulates of the atomic model, derive the expression for radius of
nth electron orbit, thus obtaining the expression for Bohrβs radius.
Ans. Form de-Broglie hypothesis, wavelength associated with electron
π =β
ππ£
ππ£ =β
π
Substituting this value in ππ£π = πh
2π,
we get, β
π π = π
h
2π
2π π = ππ
π. π., circurmference (π = 2ππ) of nth permitted orbit for the electron can contains exactly π wavelength of de-Broglie wavelength associated with electron in that orbit.
SECTION - C
Q.11. A convex lens made up of glass of refractive index 1Β·5 is dipped, in turn, in (i) a medium of refractive index 1Β·65, (ii) a medium of refractive index 1Β·33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media? 3 marks
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Ans. Here,
ππ = 1.5π
Let ππππ be the focal length of the lens in air, Then, 1
ππππ( ππ
π β 1) (1
π 1β
1
π 2)
or (1
π 1β
1
π 2) =
1
πΉπππ( πππ β1)
=1
ππππ(1.5β1)
Or (1
π 1β
1
π 2) =
1
πΉπππ β¦. (i)
(i) When lens is dipped in medium A Here, ( ππ
π = 1.65)
Let πΉπ΄ be the focal length of the lens, when dipped in medium A. Then,
1
πΉπ΄( ππ
π β 1) (1
π 1β
1
π 2)
= ( ππ
π
πππ β 1)(
1
π 1β
1
π 2)
using the equation (i), we have
1
ππ΄= (
1.5
1.65β 1)Γ
1
ππππ= β
1
5.5ππππ
As the sign of ππ΄is opposite to that of ππππ the lens will behave as a diverging lens.
(ii) When lens is dipped in medium B:
Here, πππ = 1.33
Let πΉπ΅be the focal length of the lens, when dipped in medium B. Then,
1
πΉπ΅( ππ β 1π΅ ) (
1
π 1β
1
π 2) = (
πππ
πππ β 1)(
1
π 1β
1
π 2)
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Using the equation (i), we have
1
πΉπ΅= (
1.5
1.33β 1)Γ
2
ππππ=
0.34
1.33ππππ
Or ππ΅ = 3.91 ππππ
As the sign of ππ΅is same as that of ππππ the lens will behave as a converging lens.
Q.12. (a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm? 3 marks
(b) Which transition corresponds to emission of radiation of maximum wavelength?
Ans. (a) For element A
Ground state energy, πΈ1 = β2ππ
Exicted state energy, πΈ2 = 0 ππ
Energy of photon emitted, E = πΈ2 β πΈ1
= 0 β (β2) = 2ππ
β΄ Wavelength of photon emitted,
π =βπ
πΈ=
6.626Γ10β34Γ3Γ108
2Γ1.6Γ10β19=
19.878Γ108
3.2
6.211Γ10β7π = πππ. ππ§π¦
For element B
πΈπ = β4.5 π π, πΈ2 = 0π π
πΈ = 0 β (β4.5) = 4.5 π π
β΄ π =6.626Γ10β34Γ3Γ108
4.5Γ1.6Γ10β19
19.878Γ10β7
7.2= 2.760Γ10β7 = πππ π§π¦
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For element C
πΈπ = β4.5 π π, πΈ2 = β2π π
πΈ = β2 β (β4.5) = β2 + 4.5 = 2.5 ππ
β΄ π =6.626Γ10β34Γ3Γ108
2.5Γ1.6Γ10β19
19.878Γ10β7
4= 4.969Γ10β7π = πππ. π π§π¦
For element D
πΈπ = β10 π π, πΈ2 = β2π π
πΈ = β2 β (β10) = 8 ππ
β΄ π =6.626Γ10β34Γ3Γ108
8Γ1.6Γ10β19
=19.878Γ10β7
12.8= 1.552Γ10β7π = πππ. π π§π¦
β΄ Element B has a proton of wavelength πππ π§π¦
(b) Element A has radiation of maximum wavelength 621nm
Q.13. An air solenoid of length 0.3m, area of cross section is 1.2 x ππβπππand has 2500 turns. Around its central section, a coil of 350 turns is wound. The solenoid and the coil are electrically insulated from each other. Calculate the emf induced in the coil if the initial current of 3A in the solenoid is reversed in 0.25s. 3 marks
Ans. N1 = 2500
N2 = 350
A = 1.2 Γ10β3π2
l = 0.3m.
dl = 3-(-3) = 3+3 = 6A
dt = 0.25s
Since Mutual inductance,
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π = π0π1π2π΄
π=
4πΓ10β7Γ2500Γ350Γ1.2Γ10β3
0.3
=4Γ3.14Γ1.05Γ10β4
0.3= 43.96Γ10β4 = 4.39Γ10β3H
Induced πππ|E| = MdI
dt
=4.39Γ10β3Γ6
0.25= 105.36Γ10β3 = 0.10536 V.
Q.14. Draw a labelled ray diagram of a refracting telescope. Define its magnifying power write the expression for it. Write two important limitations of a refracting telescope over reflecting type
telescope. 3 marks
Ans. Refracting telescope:
Magnifying power- The magnifying power is in the ratio of the angle β subtended at the eye by the final image to the angle π½ which the object subtends at the lens or the eye.
π βπ½
ββ
β
ππ.π0β
=π0ππ
Limitations of refracting telescope over the reflecting type telescope β
(i) Refracting telescope suffers from chromatic aberration as it uses large sized lenses.
(ii) The requirements of big lenses tend to be very heavy and therefore difficult to make and support by their edges.
Q.15. In a Geiger-Marsden experiment, calculate the distance of closet approach to the nucleus of π = ππ, when an πΆ-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closet approach be affected when the kinetic energy of the πΆ-particle is doubled? 3 marks
Ans. Z=80, KE=8MeV.
Potential energy =πΎππ2
π0=
1
2 π1π£0
2
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π0 =πΎππ2
(π2π0
2
2)
=9 Γ 109 Γ 80 Γ (1.6Γ 1019)2
8 Γ 1.6 Γ 10β13
[β΅ 1πππ = 1.6Γ10β13π½]
=18 Γ 18 Γ 109 Γ 1.6 Γ 1010
8 Γ 106= 128.8 ππ
Since π0πΌ1
πΈπ
So, when kinetic energy is doubled the distance of closet to halved. If, the kinetic energy is doubled the distance of πΌ- particles is halved.
OR
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition form an energy level -0.85 eV to -3.4eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
Ans. πΈπ = β13.6
π2 ππ£. Here ground state energy for π = 1, πΈ1 = β13.6 ππ
Now electron transits form πΈπ = β0.85ππ π‘π πΈπ = β3.4ππ
β0.85 =β13.6
ππ2
ππ2 =
13.6
0.85= 16
Thus,
ππ = 4
Again, β3.4 = 13.6/ππ2
ππ2 =
13.6
3.4= 4
ππ = 2
Thus electron makes transition from n = 4 to n = 2. Hence, it is blamer series. Now
π = 1.0974 Γ 107
1
π= π (
1
22β
1
π2)
1
π= 1.0974Γ 107
(1
22β
1
42) =
1.09Γ107Γ12
4Γ16
1
2
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1
π= 0.2057 Γ 107
π = 4.861 Γ 107
π = 4861
intensity of light after passing through second polarizer π2 is given by πΌ = πΌ0 πππ 2π .
Q.16. An illuminated object and a screen are placed 90cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object. 3 marks
Ans.
m =π£
π’β 2 =
π£
π’β π£ = 2π’ β¦ . . (ππ)
Putting the value of v in (i), we get
u + 2u = 90 β π’ =90
30= 30
β΄ π£ = 2Γ30 = 60
Using lens formula, we get
1
π=
1
π£β
1
π’β
1
π=
1
60β
1
β30
β1
π=
1
60+
1
30β
1
π=
1 + 2
60=
3
60
β π =60
3β΄ π = 20 cm.
Q. 17. (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture πΓππβππ.The distance between the slit and the screen is 1.5.m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. 3marks
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Ans.(a). If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of two slits.
(b). Given that: Wavelength of the light beam
π1 = 590ππ = 5.9Γ10β7π
Wavelength of another light beam,
π2 = 596ππ = 5.96Γ10β7π
Distance of the slits from the screen = D = 1.5m
Distance of the two slits = a = 2Γ10β4π
For the first secondary maxima
sinπ =3π1
2π=
π₯1
π·
OR
π₯1 =3π1π·
2ππππ π₯1 =
3π2π·
2π
β΄ Spacing between the positions of first secondary maxima of two sodium lines
π₯1 β π₯2 =3π·
2π(π1 β π2) = 6.75 Γ10β5π.
Q.18. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity. 3 marks
Ans. (a)
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(b) Magnifying power: The linear magnification (π0) due to the objective is
π0 =π΄β²π΅β²
π΄π΅=
ββ²
ββ¦β¦(π)
Also tanπ½ =β
π0=
ββ²
πΏ
β΄ ββ²
β=
πΏ
π0β¦β¦ . . (ii)
From (i) and (ii) we have
π0 =πΏ
π0β¦β¦(πππ)
Where hβ is the size of the first image, the object size being h and π0 being the focal length of the objective and L be the distance between the second focal point of the objective and first focal point of the eye piece (focal length ππ) is called the tube length of compound microscope.
When the final image is formed at the near point, then the angular magnification (ππ) of the eye piece is
ππ = (1 +π
ππ)β¦β¦ . (ππ£)
β΄ Total magnification of compound microscope is
ππ = π0. ππ
π =1
π0+ (1 +
π·
ππ)
OR
(a) State Huygensβs principle. Using this principle draw a diagram to show how a
plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snellβs law of
refraction. (b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons: (i) Is the frequency of reflected and reflected light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave?
Ans.(a) Huygensβs Principle: It is based on the assumptions:
(i) Each point on the primary wave front acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does.
(ii) The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant
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Refraction on the basis of wave theory:
(i) Consider any point Q on the incident wave front. (ii) Suppose when disturbance from point P on incident wave front reaches point
pβ on the refracting surface XY. (iii) Since, PβAβ represents the refracted wave front, the time taken by light to
travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time taken by light to go from Q to Qβ will be
π‘ =ππΎ
π+
πΎπβ²
π£ β¦ . . (π)
In right angled βπ΄ππΎ,< ππ΄πΎ = π
β΄ ππΎ = π΄πΎ sin π ---(ii)
in right- =angled βπβ²πβ²πΎ,< πβ²πβ²πΎ = π πππ πΎπβ² = πΎπβ² sin π. ---(iii)
Substituting (ii) and (iii) in equation (i)
We get
π‘ =π΄πΎ sin π
π+
πΎπβ² sin π
π£
or,
π‘ =π΄πΎ sin π
π+
(π΄πβ² β π΄πΎ) sin π
π£(KPβ = APβ β AK)
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or,
π‘ =π΄πβ²
πsin π + π΄πΎ (
sin π
πβ
sin π
π£)β¦ . (ππ£)
The rays from different point on the incident wave front will take the same time to reach the corresponding points on the refracted wave from i.e., t given by equation
(iv) is independent of AK. It will happen so,
If
sin π
πβ
sin π
π£= 0
sin π
sin π=
π
π£= π =
sin π
sin π
This is the Snellβs law for refraction of light.
(b)(i). The frequency of refracted light remains same as the frequency of incident light frequency only depends on the source of light.
(ii) Since, the frequency remains same, hence there is no reduction in energy.
Q.19. State the law of radioactive decay. Plot a graph showing the number (N) of undecided nuclei as a function of time (t) for a given radioactive sample having half life π»Β½. Depict in the plot the number of undecided nuclei at (i) t = 3 π»Β½ and (ii) t = 5 π»Β½. 3 mark Ans. Radioactive decay Law: The number of atoms disintegrated per second at any instant is directly Proportional to the number of radioactive atoms actually present at that time. The following graph showing the number (N) of Undecided nuclei as a function of time (t) for a given radioactive sample having half-life. And undecided nuclei at (i) t = 2 π1/2 (ii) t = 4 π1/2 included also,
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Q.20. Three light rays red (R), green (G) and blue (B) are incident on a right angled prism 'abc' at face 'ab'. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Out of the three which color ray will emerge out of face 'ac' ? Justify your answer. Trace the path of these rays after passing through face 'ab'. 3 marks
Ans. The red light ray (R) will emerge out of face ac. The path of green (G) and blue (B) light rays will be as,
Q.21. (a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) write the basis features of photon picture of electromagnetic radiation on which Einsteinβs photoelectric equation is based. 3 marks
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Ans. (a) Wave nature of radiation cannot explain the following:
(i) The instantaneous ejection of photoelectrons. (ii) The existence of threshold frequency for a metal surface (iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of
light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave
nature of light.
(b) Photon picture of electromagnetic radiation on which Einsteinβs photoelectric equation is based on particle nature of light. Its basic features are:
(i) In interaction with matter, radiation behaves as if it is made up of particles called photons.
(ii) Each photon has energy E = hv and momentum π =βπ£
π and speed C, the speed of
light. (iii) All photons of light a particular frequency v, or wavelength π have the same energy
πΈ = βπ£ =βπ£
π and momentum π =
βπ£
π=
β
πwhatever the intensity of radiation may be.
(iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus photon energy is independent of intensity of radiation.
Photons are electrically neutral and are not deflected by electric and magnetic fields.
Q.22. You are given three lenses π³π, π³π and π³πeach of focal length 20 cm. An object is kept at 40 cm in front of π³π, as shown. The final real image is formed at the focus βIβ of π³π. Find the separations between π³π, π³π πππ π³π. 3 marks
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Ans.
1
π£1β
1
π’=
1
π1
1
π£1=
1
π1+
1
π’1
1
20+
1
β40=
1 β 2 β 7
β40
1
40
π£1 = 40ππ.
Here, image by πΏ3 is formed at focus. So the object should lie at infinity for πΏ3. Hence. πΏ2 will produce image at infinity. So we can conclude that object for πΏ2 should be at its focus.
But, we have seen above that image by πΏ1 is formed at 40 right of πΏ1which is at 20 cm left of πΏ2 focus of πΏ2.
So π1= distance between πΏ1 and πΏ2 = (40 + 20) cm = 60 cm
Again distance between πΏ2 and πΏ3 does not matter as the image by πΏ2 is formed at infinity so π2 can take any value.
SECTION - D
Q.23. With the help of a suitable ray diagram, derive the mirror formula for a concave mirror. 4 marks
Ans. Mirror Formula for Concave Mirror:
Let, AB be the length of an object placed beyond C in front of a concave mirror. The image AβBβ is real, inverted and between C and F.
Applying sign conventions, we have
Object distance PB = βπ’ image distance PBβ = βπ£ focal length PF = βπ and radius of curvature PC = β2π
In similar βs ABC and Aβ²Bβ²F
π΄π΅
π΄β²π΅β²=
π΅πΆ
π΅β²πΆ β¦ (π)
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and similar βs RSF and Aβ²Bβ²F
π π
π΄β²π΅β²=
ππΉ
π΅β²πΉ
β΅ π π = π΄π΅
π΄π΅
π΄β²π΅β²=
ππΉ
π΅β²πΉ β¦ (ππ)
From eq. (i) and (ii), we have
π΅πΆ
π΅β²πΆ=
ππΉ
π΅β²πΉ
Since, the aperture of the concave mirror is small so the point S and P coincides.
β΄π΅πΆ
π΅β²πΆ=
ππΉ
π΅β²πΉ
ππ΅ β ππΆ
ππΆ β ππ΅β²=
ππΉ
ππ΅β² β ππΉ
βπ’ + 2π
β2π + π£=
βπ
βπ£ + π
π’π£ β π’π β 2π£π + 2π2 = 2π2 β ππ£
β π’π£ = π’π + π£π
Dividing both side by π’π£π, we get
π’π£
π’π£π=
π’π
π’π£π+
π£π
π’π£π
β΄1
π=
1
π£+
1
π’
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SECTION β E
Q.24. (a) for a ray of light travelling from a denser medium of refractive index ππto a rarer medium of refractive index ππ, prove that
ππ
ππ= sin ππ ,
where ππis the critical angle of incidence for the media.
(b) Explain with the help of a diagram. how the above principle is used for transmission of video signals using optical fibers? 5 marks
Ans.(a) Relation between refractive index and critical angle: Let O be a point object in the denser medium of refractive index (π1). A ray incident along OA1 deviates away from normal and is refracted along A1 B1in the rarer medium of refractive index (π2). It increases with increase in the angle of incidence: For particular value of i = C, the critical angle, the incident ray OA2 is refracted at < r = 90Β° and goes along A2 B2.
Applying Snellβs Law at A2
n1 π ππ ic = n2 sin 900 βn1 π ππ ic = n2Γ1
β΄ π ππ ic =n2
n1 ππ
n2
n1= π ππ ic
(b) Optical Fiber: Optical fiber make use the phenomenon of total internal reflection. Optical fibers consist of many long high quality composite glass or quartz fibers. Each fiber consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding.
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Propagation of light through an optical fiber: When light is incident at one end of the fiber at a small angle, it suffers several total internal reflection at the glass boundary because the angle of incidence is greater than the critical angle. The intensity of the out coming beam is almost equal to that of the incident beam.
OR
(a) What is plane polarized light? Two polaroidβs are placed at πππ to each other
and the transmitted intensity is zero. What happens when one more polaroid is placed between these two bisecting the angle between them? how will the intensity of transmitted light vary on further rotting the polaroid?
(b) If a light beam shows no intensity variation when transmitted. Through a polaroid which is rotated, does it mean that the light is un β polarized? explain briefly.
Ans.(a) Plane polarized light: When polarized, light is passed through a tourmaline crystal cut with its face parallel to its crystallographic axis AB. Only those vibrations of light pass through the crystal, which are parallel to AB, all other vibrations are absorbed. The emerged light from the Crystal is said to be plane polarized light. If E is the amplitude of electric field component emanating-from 1st polaroid, then from 2nd polaroid at 45Β°.The amplitude of
electric field component is E1 = E cos 45Β° = E Γ1
β2Γ
E
β2
Again amplitude of electric field component coming from 3rd polaroid at 45Β° to 2nd polaroid would be
E2 = E1 cos 45Β° =
E
β2.1
β2=
E
2
= Half of E
As Intensity πΌE2
β΄ Intensity transmitted from three polaroidβs will be 1
4th of the intensity
transmitted from the first polaroid. (b) No, The light which is made up of electric field components Ex, Ey with 90Β° phase difference but equal amplitudes. The tip of electric vector executes uniform circular motion at the frequency of the light itself.
When such light is passed through a polaroid, which is rotated, the transmitted average intensity remains constant.
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Q.25. An a.c. source generating a voltage π = ππ π¬π’π§ππ is connected capacitor of capacitance C. find the expression for the current, i , through it. Plot a graph of v and i versus t to show that the current is π /π ahead of the voltage. A resistor of 200 π and a capacitor of ππ. π ππ Fare connected in series to a 220 V, 50 Hz a.c. source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? if yes, resolve the paradox. 5 marks
Ans. A.C source containing capacitor: Let a source of alternating πππ. π = ππ sinππ‘ be connected to a capacitor of capacitance C only
π = ππ sinππ‘ β¦ (π)
At every instant, the potential V is given by
π =π
πΆ β ππ
sinππ‘ =π
π β΄ π = πΆ ππ sinππ‘.
If I is instantaneous value of current in the circuit at instant t, then
π =ππ
ππ‘=
π
ππ‘ (πΆ ππ sinππ‘)
π = π ππ(cosππ‘). π =
1π1ππ
sin (ππ‘ +π
2)
The current will be maximum. When (sinππ‘ +π
2) = 1
β΄ ππ =ππ1
ππ
Γ 1 = ππ1
ππ
π = ππ sin (ππ‘ +π
2)
Therefore, alternating current I lead the alternating voltage by a phase angle of π
2.
Numerical: Here R = 200 Ξ©
πΆ = 15.0 ππΉ = 15 Γ 10β6πΉ, ππππ = 220 π,
π = 50 π»π§
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πΌπππ =?
ππ =1
ππ=
1
2πππ
=1
2 Γ227
Γ 50 Γ 15 Γ 10β6
=7 Γ 106
33000
= 212.12 β 212 Ξ©
β΄ π = βπ 2 + βππ2 = β(200)2 + (212)2
= β40000 + 44944
= β84944
= 291.45 Ξ©
β΄ πΌπππ =ππππ
π=
220
291.45
= 0.75 π΄
β΄ ππ = ππππ π = 0.75Γ200
= 125 π
ππ = πΌπππ . ππ
= 0.75 Γ 212 = 159 π
β΄ ππ + ππΆ = 150 + 159 = 309
β΄ ππ + ππΆ > π
This is because these voltages are not in same phase and they cannot be added like ordinary numbers.
β΄ π = βππ 2 + ππΆ
2
= β(150)2 + (159)2
= β47781
= 218.18 π.
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Q.26. State Faradayβs law of electromagnetic induction. Figure shows a rectangular conductor PORS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x=0 to x=b and is zero for x>b. Assume that only the arm PQ possesses resistance r, when the arm PQ is pulled outward from x=0 with constant speed v, absinthe expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 x 2b. 5 marks
Ans. Part I: Faradayβs law of induction: It states that the emf induced in a coil of N turns is directly related to the rate of change of flux through it.
β΄ π = βππβ π΅
ππ‘
Where β π΅ is the flux linked with one turn of the coil? If the circuit is closed, a
current I =π
π is set up in it.
Part II: refer to following fig (a). the arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and zero the situation when the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as joule heat. Sketch the variation of these quantities with distance.
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Let us first consider the forward motion from π₯ = 0 to x = 2b the flux β π linked with the circuit SPQR is
β π = π΅ππ₯ 0 β€ π₯ < π
= Blb b β€ π₯ < 2π
The induced emf is,
π = βπβ π
ππ‘= βπ΅ππ£ ; 0 β€ π₯ < π = 0 ; π β€< 2π
When the induced emf is non-zero, the current I is (in magnitude)
I =π΅ππ£
π
The force required to keep the arm PQ in constant motion is IlB. Its direction is to the left. In magnitude
F =π΅2π2π£
π= 0 ; 0 β€ π₯ < π = 0 ; π β€ π₯ < 2π
The joule heating loss is
ππ = I2π
=π΅2π2π£2
π 0 β€ π₯ < π
= 0 b β€ π₯ < 2
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OR
Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils?
How is the transformer used in large scale transmission and distribution of electrical energy over long distances?
Ans. Step up transformer: principle: It is a device which converts low voltage. A.C. into high voltage A. C. It is based upon the principle of mutual induction. When alternating current passed through a coil, an induced e. m. f. is set up in the neighboring coil.
Construction: A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as primary (p) is connected to A.C. supply. The other coil known as secondary (s) is connected to the load.
Working: When an alternating current is passed through the primary, the magnetic flux through the iron core changes which does two things. It produces e. m. f in the primary and an induced e. m. f is also set up in the secondary, if we assume that the resistance of primary is negligible, the back e. m. f will be equal to the voltage applied to the primary.
β΄ ππ = βππ
πβ
ππ‘
and ππ = βππ
πβ
ππ‘
Where ππand ππ are number of terms in the primary and secondary respectively. Vπ
and Vπ are their respective voltages.
ππ ππ
=Nπ
Nπ
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This ratio Nπ
Nπ called the turms ratio.
In a step β up transformer: Nπ > Nπ
So πs > Vπ
In a step β down transformer: Nπ < Nπ
So, Vπ > Vπ
Large scale transmission: The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-upβso that reduced and power loss I2R is cut down. It is then transmitted over long distances to an area sub- station near the consumers. There the voltage is stepped down. It is further stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.
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