CBSE 12th Physics 2017 Guess Paper By 4ono · CBSE 12th Physics 2017 Guess Paper By 4ono.com TIME - 3HR. | QUESTIONS - 26 THE MARKS ARE MENTIONED ON EACH QUESTION _____ SECTION –

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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 – 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 (𝝓) 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 𝜙 = LI ⇒ L = 𝜙

1 = slope of 𝜙 − I graph ∴ 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 (λ) 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 Ω 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 𝐼 =ε

𝑅.

𝐼 =190

38= 5𝐴

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

∴ 𝜌𝑇 = 𝜌0[1 + 𝛼(𝑇 − 𝑇0)]

(ii) The resistivity of a conductor is the reciprocal of electrical conductivity

∴ 𝜌 =1

𝜎




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 − 𝐼.

(ii) 𝐴𝐽 = 𝑙.

From the principle of Wheat Stones Bridge,

𝑅

𝑋=

𝑙

100 − 𝑙

𝑋 = 𝑅 (100 − 𝑙

𝑙)

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) Kirchhoff’s 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 𝐼1, 𝐼2, 𝐼3 𝑎𝑛𝑑 𝐼4




Convention: Current towards the junction is always positive, while current away from the junction is negative

𝐼3 + (−𝐼1) + (−𝐼2) + (−𝐼) = 0

Kirchhoff’s Second law: Loop rule

In a closed loop, the algebraic sum of the emf’s is equal to the algebraic sum of the products of the resistance and current flowing through them.

For closed part BACB, 𝐸1 − 𝐸2 = 𝐼1𝑅1 + 𝐼2𝑅2 − 𝐼3𝑅3

For closed part CADC, 𝐸2 = 𝐼3𝑅3 + 𝐼4𝑅4 + 𝐼5𝑅5

Demonstration:

Wheatstone Bridge: - The Wheatstone Bridge is an arrangement pf four resistance as shown in the following figure.

𝑅1, 𝑅2, 𝑅3 𝑎𝑛𝑑 𝑅4 are the four resistances.

Galvanometer(a) has a current 𝐼𝑔flowing through it at

balanced condition 𝐼𝑔 = 0

Applying junction rule at B,

∴ 𝐼2 = 𝐼4

Applying junction rule at D,

∴ 𝐼1 = 𝐼3




Applying loop rule to closed loop ADBA,

−𝑅1 + 𝑂 + 𝐼1𝑅2 = 0

∴ 𝐼1𝐼2

=𝑅2

𝑅1…(𝑖)

Applying loop rule to closed loop CBDC,

𝑅1𝑅1 + 𝑂 − 𝑅1𝑅1 = 𝑂 [𝐼3 = 𝐼1, 𝐼4 = 𝐼2]

∴ 𝑅2

𝑅1=

𝑅4

𝑅3…(𝑖𝑖)

From Eq. (i) and (ii)

𝑅2

𝑅1=

𝑅4

𝑅3

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 𝑾𝟏 and (𝑾𝟏 > 𝑾𝟐). 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 (𝑉0) with the frequency of incident radiation (𝑣0) for two different photosensitive materials having work functions W1 and W2(W1>W2) is shown in fig.




(i) Slope of the line = ∆V

∆𝑣 =

ℎ

𝑒

[∴ 𝑒∆V = h ∆𝑣] ∴ Slope of the line depends on the Planck’s constant h and the electronic charge 𝑒.

(ii) Intercept of graph A on the potential axis

=work function(W)

𝑒= −

ℎ𝑣0

𝑒

∴ Intercept of the line depends upon Planck’s constant h, threshold frequency (𝑉0) and the electronic charge (e).

Q.10. Using Rutherford’s 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 Rutherford’s model of the atom, the magnitude of this force is

𝐹 =1

4 𝜋𝜖0 .2𝑒. (𝑍𝑒)

𝑟2

For hydrogen atom, Let, 𝐹𝑐 −Centripetal 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

𝐹𝑐 = 𝐹𝑒

𝑚𝑣2

𝑟=

(𝑒)(𝑒)

4𝜋𝜀0𝑟2 … (𝑖)

𝑟 =𝑒

4𝜋𝜀0𝑚𝑣2 … (𝑖𝑖)




K.E. of electron in the orbit,

K =1

2𝑚𝑣2,

Form equation (𝑖),

K =𝑒2

8𝜋𝜀0𝑟

Potential energy of electron in

Orbit, U = (𝒆)(𝒆)

4𝜋𝜀0𝑟=

−𝑒2

4𝜋𝜀0𝑟

∴ Total energy of electron in hydrogen atom

E = k + U = 𝑒2

8𝜋𝜀0𝑟−

𝑒2

4𝜋𝜀0𝑟; E = −

𝑒2

8𝜋𝜀0𝑟

Here, negative sign indicates that the revolving electron is bound to the positive nucleus.

OR

Using Bohr’s postulates of the atomic model, derive the expression for radius of

nth electron orbit, thus obtaining the expression for Bohr’s radius.

Ans. Form de-Broglie hypothesis, wavelength associated with electron

𝜆 =ℎ

𝑚𝑣

𝑚𝑣 =ℎ

𝜆

Substituting this value in 𝑚𝑣𝑟 = 𝑛h

2𝜋,

we get, ℎ

𝜆 𝑟 = 𝑛

h

2𝜋

2𝜋 𝑟 = 𝑛𝜆

𝑖. 𝑒., circurmference (𝑆 = 2𝜋𝑟) of nth permitted orbit for the electron can contains exactly 𝑛 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 1·5 is dipped, in turn, in (i) a medium of refractive index 1·65, (ii) a medium of refractive index 1·33.

(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




Ans. Here,

𝜇𝑔 = 1.5𝑎

Let 𝑓𝑎𝑖𝑟 be the focal length of the lens in air, Then, 1

𝑓𝑎𝑖𝑟( 𝜇𝑔

𝑎 − 1) (1

𝑅1−

1

𝑅2)

or (1

𝑅1−

1

𝑅2) =

1

𝐹𝑎𝑖𝑟( 𝜇𝑔𝑎 −1)

=1

𝑓𝑎𝑖𝑟(1.5−1)

Or (1

𝑅1−

1

𝑅2) =

1

𝐹𝑎𝑖𝑟 …. (i)

(i) When lens is dipped in medium A Here, ( 𝜇𝑎

𝑎 = 1.65)

Let 𝐹𝐴 be the focal length of the lens, when dipped in medium A. Then,

1

𝐹𝐴( 𝜇𝑔

𝑎 − 1) (1

𝑅1−

1

𝑅2)

= ( 𝜇𝑔

𝑎

𝜇𝑎𝑎 − 1)(

1

𝑅1−

1

𝑅2)

using the equation (i), we have

1

𝑓𝐴= (

1.5

1.65− 1)×

1

𝑓𝑎𝑖𝑟= −

1

5.5𝑓𝑎𝑖𝑟

As the sign of 𝑓𝐴is opposite to that of 𝑓𝑎𝑖𝑟 the lens will behave as a diverging lens.

(ii) When lens is dipped in medium B:

Here, 𝜇𝑔𝑎 = 1.33

Let 𝐹𝐵be the focal length of the lens, when dipped in medium B. Then,

1

𝐹𝐵( 𝜇𝑔 − 1𝐵 ) (

1

𝑅1−

1

𝑅2) = (

𝜇𝑔𝑎

𝜇𝑏𝑎 − 1)(

1

𝑅1−

1

𝑅2)




Using the equation (i), we have

1

𝐹𝐵= (

1.5

1.33− 1)×

2

𝑓𝑎𝑖𝑟=

0.34

1.33𝑓𝑎𝑖𝑟

Or 𝑓𝐵 = 3.91 𝑓𝑎𝑖𝑟

As the sign of 𝑓𝐵is same as that of 𝑓𝑎𝑖𝑟 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, 𝐸1 = −2𝑒𝑉

Exicted state energy, 𝐸2 = 0 𝑒𝑉

Energy of photon emitted, E = 𝐸2 − 𝐸1

= 0 − (−2) = 2𝑒𝑉

∴ Wavelength of photon emitted,

𝜆 =ℎ𝑐

𝐸=

6.626×10−34×3×108

2×1.6×10−19=

19.878×108

3.2

6.211×10−7𝑚 = 𝟔𝟐𝟏. 𝟏𝐧𝐦

For element B

𝐸𝟏 = −4.5 𝑒 𝑉, 𝐸2 = 0𝑒 𝑉

𝐸 = 0 − (−4.5) = 4.5 𝑒 𝑉

∴ 𝜆 =6.626×10−34×3×108

4.5×1.6×10−19

19.878×10−7

7.2= 2.760×10−7 = 𝟐𝟕𝟔 𝐧𝐦




For element C

𝐸𝟏 = −4.5 𝑒 𝑉, 𝐸2 = −2𝑒 𝑉

𝐸 = −2 − (−4.5) = −2 + 4.5 = 2.5 𝑒𝑉

∴ 𝜆 =6.626×10−34×3×108

2.5×1.6×10−19

19.878×10−7

4= 4.969×10−7𝑚 = 𝟒𝟗𝟔. 𝟗 𝐧𝐦

For element D

𝐸𝟏 = −10 𝑒 𝑉, 𝐸2 = −2𝑒 𝑉

𝐸 = −2 − (−10) = 8 𝑒𝑉

∴ 𝜆 =6.626×10−34×3×108

8×1.6×10−19

=19.878×10−7

12.8= 1.552×10−7𝑚 = 𝟏𝟓𝟓. 𝟐 𝐧𝐦

∴ Element B has a proton of wavelength 𝟐𝟕𝟓 𝐧𝐦

(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 𝟏𝟎−𝟑𝒎𝟐and 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 ×10−3𝑚2

l = 0.3m.

dl = 3-(-3) = 3+3 = 6A

dt = 0.25s

Since Mutual inductance,




𝑀 = 𝜇0𝑁1𝑁2𝐴

𝑙=

4𝜋×10−7×2500×350×1.2×10−3

0.3

=4×3.14×1.05×10−4

0.3= 43.96×10−4 = 4.39×10−3H

Induced 𝑒𝑚𝑓|E| = MdI

dt

=4.39×10−3×6

0.25= 105.36×10−3 = 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 ∝ subtended at the eye by the final image to the angle 𝛽 which the object subtends at the lens or the eye.

𝑚 ≈𝛽

∝≈

ℎ

𝑓𝑒.𝑓0ℎ

=𝑓0𝑓𝑒

Limitations of refracting telescope over the reflecting type telescope –

(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 𝒁 = 𝟖𝟎, when an 𝜶-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 𝜶-particle is doubled? 3 marks

Ans. Z=80, KE=8MeV.

Potential energy =𝐾𝑍𝑒2

𝑑0=

1

2 𝑚1𝑣0

2




𝑑0 =𝐾𝑍𝑒2

(𝑀2𝑉0

2

2)

=9 × 109 × 80 × (1.6× 1019)2

8 × 1.6 × 10−13

[∵ 1𝑀𝑒𝑉 = 1.6×10−13𝐽]

=18 × 18 × 109 × 1.6 × 1010

8 × 106= 128.8 𝑓𝑚

Since 𝑑0𝛼1

𝐸𝑘

So, when kinetic energy is doubled the distance of closet to halved. If, the kinetic energy is doubled the distance of 𝛼- 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. 𝐸𝑛 = −13.6

𝑛2 𝑒𝑣. Here ground state energy for 𝑁 = 1, 𝐸1 = −13.6 𝑒𝑉

Now electron transits form 𝐸𝑞 = −0.85𝑒𝑉 𝑡𝑜 𝐸𝑞 = −3.4𝑒𝑉

−0.85 =−13.6

𝑛𝑝2

𝑛𝑝2 =

13.6

0.85= 16

Thus,

𝑛𝑝 = 4

Again, −3.4 = 13.6/𝑛𝑞2

𝑛𝑞2 =

13.6

3.4= 4

𝑛𝑞 = 2

Thus electron makes transition from n = 4 to n = 2. Hence, it is blamer series. Now

𝑅 = 1.0974 × 107

1

𝜆= 𝑅 (

1

22−

1

𝑛2)

1

𝜆= 1.0974× 107

(1

22−

1

42) =

1.09×107×12

4×16

1

2




1

𝜆= 0.2057 × 107

𝜆 = 4.861 × 107

𝜆 = 4861

intensity of light after passing through second polarizer 𝑃2 is given by 𝐼 = 𝐼0 𝑐𝑜𝑠2𝜃 .

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 =𝑣

𝑢⇒ 2 =

𝑣

𝑢⇒ 𝑣 = 2𝑢 … . . (𝑖𝑖)

Putting the value of v in (i), we get

u + 2u = 90 ⇒ 𝑢 =90

30= 30

∴ 𝑣 = 2×30 = 60

Using lens formula, we get

1

𝑓=

1

𝑣−

1

𝑢⇒

1

𝑓=

1

60−

1

−30

⇒1

𝑓=

1

60+

1

30⇒

1

𝑓=

1 + 2

60=

3

60

⇒ 𝑓 =60

3∴ 𝑓 = 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 𝟐×𝟏𝟎−𝟒𝒎.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




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

𝜆1 = 590𝑛𝑚 = 5.9×10−7𝑚

Wavelength of another light beam,

𝜆2 = 596𝑛𝑚 = 5.96×10−7𝑚

Distance of the slits from the screen = D = 1.5m

Distance of the two slits = a = 2×10−4𝑚

For the first secondary maxima

sin𝜃 =3𝜆1

2𝑎=

𝑥1

𝐷

OR

𝑥1 =3𝜆1𝐷

2𝑎𝑎𝑛𝑑 𝑥1 =

3𝜆2𝐷

2𝑎

∴ Spacing between the positions of first secondary maxima of two sodium lines

𝑥1 − 𝑥2 =3𝐷

2𝑎(𝜆1 − 𝜆2) = 6.75 ×10−5𝑚.

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)




(b) Magnifying power: The linear magnification (𝑚0) due to the objective is

𝑚0 =𝐴′𝐵′

𝐴𝐵=

ℎ′

ℎ……(𝑖)

Also tan𝛽 =ℎ

𝑓0=

ℎ′

𝐿

∴ ℎ′

ℎ=

𝐿

𝑓0…… . . (ii)

From (i) and (ii) we have

𝑚0 =𝐿

𝑓0……(𝑖𝑖𝑖)

Where h’ is the size of the first image, the object size being h and 𝑓0 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 𝑓𝑒) is called the tube length of compound microscope.

When the final image is formed at the near point, then the angular magnification (𝑚𝑒) of the eye piece is

𝑚𝑒 = (1 +𝑑

𝑓𝑒)…… . (𝑖𝑣)

∴ Total magnification of compound microscope is

𝑚𝑒 = 𝑚0. 𝑚𝑒

𝑚 =1

𝑓0+ (1 +

𝐷

𝑓𝑒)

OR

(a) State Huygens’s 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 Snell’s 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) Huygens’s 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




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

p’ on the refracting surface XY. (iii) Since, P’A’ 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 Q’ will be

𝑡 =𝑄𝐾

𝑐+

𝐾𝑄′

𝑣 … . . (𝑖)

In right angled ∆𝐴𝑄𝐾,< 𝑄𝐴𝐾 = 𝑖

∴ 𝑄𝐾 = 𝐴𝐾 sin 𝑖 ---(ii)

in right- =angled ∆𝑃′𝑄′𝐾,< 𝑄′𝑃′𝐾 = 𝑟 𝑎𝑛𝑑 𝐾𝑄′ = 𝐾𝑃′ sin 𝑟. ---(iii)

Substituting (ii) and (iii) in equation (i)

We get

𝑡 =𝐴𝐾 sin 𝑖

𝑐+

𝐾𝑃′ sin 𝑟

𝑣

or,

𝑡 =𝐴𝐾 sin 𝑖

𝑐+

(𝐴𝑃′ − 𝐴𝐾) sin 𝑟

𝑣(KP’ = AP’ − AK)




or,

𝑡 =𝐴𝑃′

𝑐sin 𝑟 + 𝐴𝐾 (

sin 𝑖

𝑐−

sin 𝑟

𝑣)… . (𝑖𝑣)

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 𝑖

𝑐−

sin 𝑟

𝑣= 0

sin 𝑖

sin 𝑟=

𝑐

𝑣= 𝜇 =

sin 𝑖

sin 𝑟

This is the Snell’s 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 𝑻½. Depict in the plot the number of undecided nuclei at (i) t = 3 𝑻½ and (ii) t = 5 𝑻½. 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 𝑇1/2 (ii) t = 4 𝑇1/2 included also,




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 Einstein’s photoelectric equation is based. 3 marks




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 Einstein’s 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 𝑝 =ℎ𝑣

𝑐 and speed C, the speed of

light. (iii) All photons of light a particular frequency v, or wavelength 𝜆 have the same energy

𝐸 = ℎ𝑣 =ℎ𝑣

𝑐 and momentum 𝑝 =

ℎ𝑣

𝑐=

ℎ

𝜆whatever 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 𝑳𝟏, 𝑳𝟐 and 𝑳𝟑each of focal length 20 cm. An object is kept at 40 cm in front of 𝑳𝟏, as shown. The final real image is formed at the focus ‘I’ of 𝑳𝟑. Find the separations between 𝑳𝟏, 𝑳𝟐 𝒂𝒏𝒅 𝑳𝟑. 3 marks




Ans.

1

𝑣1−

1

𝑢=

1

𝑓1

1

𝑣1=

1

𝑓1+

1

𝑢1

1

20+

1

−40=

1 − 2 − 7

−40

1

40

𝑣1 = 40𝑐𝑚.

Here, image by 𝐿3 is formed at focus. So the object should lie at infinity for 𝐿3. Hence. 𝐿2 will produce image at infinity. So we can conclude that object for 𝐿2 should be at its focus.

But, we have seen above that image by 𝐿1 is formed at 40 right of 𝐿1which is at 20 cm left of 𝐿2 focus of 𝐿2.

So 𝑋1= distance between 𝐿1 and 𝐿2 = (40 + 20) cm = 60 cm

Again distance between 𝐿2 and 𝐿3 does not matter as the image by 𝐿2 is formed at infinity so 𝑋2 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 A’B’ is real, inverted and between C and F.

Applying sign conventions, we have

Object distance PB = −𝑢 image distance PB’ = −𝑣 focal length PF = −𝑓 and radius of curvature PC = −2𝑓

In similar ∆s ABC and A′B′F

𝐴𝐵

𝐴′𝐵′=

𝐵𝐶

𝐵′𝐶 … (𝑖)




and similar ∆s RSF and A′B′F

𝑅𝑆

𝐴′𝐵′=

𝑆𝐹

𝐵′𝐹

∵ 𝑅𝑆 = 𝐴𝐵

𝐴𝐵

𝐴′𝐵′=

𝑆𝐹

𝐵′𝐹 … (𝑖𝑖)

From eq. (i) and (ii), we have

𝐵𝐶

𝐵′𝐶=

𝑆𝐹

𝐵′𝐹

Since, the aperture of the concave mirror is small so the point S and P coincides.

∴𝐵𝐶

𝐵′𝐶=

𝑃𝐹

𝐵′𝐹

𝑃𝐵 − 𝑃𝐶

𝑃𝐶 − 𝑃𝐵′=

𝑃𝐹

𝑃𝐵′ − 𝑃𝐹

−𝑢 + 2𝑓

−2𝑓 + 𝑣=

−𝑓

−𝑣 + 𝑓

𝑢𝑣 − 𝑢𝑓 − 2𝑣𝑓 + 2𝑓2 = 2𝑓2 − 𝑓𝑣

⇒ 𝑢𝑣 = 𝑢𝑓 + 𝑣𝑓

Dividing both side by 𝑢𝑣𝑓, we get

𝑢𝑣

𝑢𝑣𝑓=

𝑢𝑓

𝑢𝑣𝑓+

𝑣𝑓

𝑢𝑣𝑓

∴1

𝑓=

1

𝑣+

1

𝑢




SECTION – E

Q.24. (a) for a ray of light travelling from a denser medium of refractive index 𝒏𝟏to a rarer medium of refractive index 𝒏𝟐, prove that

𝒏𝟐

𝒏𝟏= sin 𝒊𝒄 ,

where 𝒊𝒄is 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 (𝑛1). A ray incident along OA1 deviates away from normal and is refracted along A1 B1in the rarer medium of refractive index (𝑛2). 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 = 90° and goes along A2 B2.

Applying Snell’s Law at A2

n1 𝑠𝑖𝑛 ic = n2 sin 900 ⇒n1 𝑠𝑖𝑛 ic = n2×1

∴ 𝑠𝑖𝑛 ic =n2

n1 𝑜𝑟

n2

n1= 𝑠𝑖𝑛 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.




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 polaroid’s are placed at 𝟗𝟎𝟎 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 – 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 45°.The amplitude of

electric field component is E1 = E cos 45° = E ×1

√2×

E

√2

Again amplitude of electric field component coming from 3rd polaroid at 45° to 2nd polaroid would be

E2 = E1 cos 45° =

E

√2.1

√2=

E

2

= Half of E

As Intensity 𝛼E2

∴ Intensity transmitted from three polaroid’s 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 90° 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.




Q.25. An a.c. source generating a voltage 𝒗 = 𝒗𝒎 𝐬𝐢𝐧𝝎𝒕 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/𝟐 ahead of the voltage. A resistor of 200 𝛀 and a capacitor of 𝟏𝟓. 𝟎 𝝁𝑭 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 𝑒𝑚𝑓. 𝑉 = 𝑉𝑚 sin𝜔𝑡 be connected to a capacitor of capacitance C only

𝑉 = 𝑉𝑚 sin𝜔𝑡 … (𝑖)

At every instant, the potential V is given by

𝑉 =𝑞

𝐶 ⇒ 𝑉𝑚

sin𝜔𝑡 =𝑞

𝑐 ∴ 𝑞 = 𝐶 𝑉𝑚 sin𝜔𝑡.

If I is instantaneous value of current in the circuit at instant t, then

𝑖 =𝑑𝑞

𝑑𝑡=

𝑑

𝑑𝑡 (𝐶 𝑉𝑚 sin𝜔𝑡)

𝑖 = 𝑐 𝑉𝑚(cos𝜔𝑡). 𝜔 =

1𝑚1𝜔𝑐

sin (𝜔𝑡 +𝜋

2)

The current will be maximum. When (sin𝜔𝑡 +𝜋

2) = 1

∴ 𝑖𝑚 =𝑉𝑚1

𝜔𝑐

× 1 = 𝑉𝑚1

𝜔𝑐

𝑖 = 𝑖𝑚 sin (𝜔𝑡 +𝜋

2)

Therefore, alternating current I lead the alternating voltage by a phase angle of 𝜋

2.

Numerical: Here R = 200 Ω

𝐶 = 15.0 𝜇𝐹 = 15 × 10−6𝐹, 𝑉𝑟𝑚𝑠 = 220 𝑉,

𝜐 = 50 𝐻𝑧




𝐼𝑟𝑚𝑠 =?

𝑋𝑐 =1

𝜔𝑐=

1

2𝜋𝜐𝑐

=1

2 ×227

× 50 × 15 × 10−6

=7 × 106

33000

= 212.12 ≈ 212 Ω

∴ 𝑍 = √𝑅2 + √𝑋𝑐2 = √(200)2 + (212)2

= √40000 + 44944

= √84944

= 291.45 Ω

∴ 𝐼𝑟𝑚𝑠 =𝑉𝑟𝑚𝑠

𝑍=

220

291.45

= 0.75 𝐴

∴ 𝑉𝑅 = 𝑉𝑟𝑚𝑠 𝑅 = 0.75×200

= 125 𝑉

𝑉𝑐 = 𝐼𝑟𝑚𝑠. 𝑋𝑐

= 0.75 × 212 = 159 𝑉

∴ 𝑉𝑅 + 𝑉𝐶 = 150 + 159 = 309

∴ 𝑉𝑅 + 𝑉𝐶 > 𝑉

This is because these voltages are not in same phase and they cannot be added like ordinary numbers.

∴ 𝑉 = √𝑉𝑅2 + 𝑉𝐶

2

= √(150)2 + (159)2

= √47781

= 218.18 𝑉.




Q.26. State Faraday’s 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: Faraday’s 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.

∴ 𝜀 = −𝑁𝑑∅𝐵

𝑑𝑡

Where ∅𝐵 is the flux linked with one turn of the coil? If the circuit is closed, a

current I =𝜀

𝑅 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.



http://api.gmath.guru/cgi-bin/gmath?/le

http://api.gmath.guru/cgi-bin/gmath?/le


Let us first consider the forward motion from 𝑥 = 0 to x = 2b the flux ∅𝑏 linked with the circuit SPQR is

∅𝑏 = 𝐵𝑙𝑥 0 ≤ 𝑥 < 𝑏

= Blb b ≤ 𝑥 < 2𝑏

The induced emf is,

𝜀 = −𝑑∅𝑏

𝑑𝑡= −𝐵𝑙𝑣 ; 0 ≤ 𝑥 < 𝑏 = 0 ; 𝑏 ≤< 2𝑏

When the induced emf is non-zero, the current I is (in magnitude)

I =𝐵𝑙𝑣

𝑟

The force required to keep the arm PQ in constant motion is IlB. Its direction is to the left. In magnitude

F =𝐵2𝑙2𝑣

𝑟= 0 ; 0 ≤ 𝑥 < 𝑏 = 0 ; 𝑏 ≤ 𝑥 < 2𝑏

The joule heating loss is

𝑝𝑗 = I2𝑟

=𝐵2𝑙2𝑣2

𝑟 0 ≤ 𝑥 < 𝑏

= 0 b ≤ 𝑥 < 2




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.

∴ 𝑉𝑝 = −𝑁𝑝

𝑑∅

𝑑𝑡

and 𝑉𝑠 = −𝑁𝑠

𝑑∅

𝑑𝑡

Where 𝑁𝑝and 𝑁𝑠 are number of terms in the primary and secondary respectively. V𝑝

and V𝑠 are their respective voltages.

𝑉𝑠𝑉𝑝

=N𝑠

N𝑝




This ratio N𝑠

N𝑝 called the turms ratio.

In a step − up transformer: N𝑠 > N𝑝

So 𝑉s > V𝑝

In a step − down transformer: N𝑠 < N𝑝

So, V𝑠 > V𝑝

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-up—so 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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CBSE 12th Physics 2017 Guess Paper By 4ono · CBSE 12th Physics 2017 Guess Paper By 4ono.com TIME - 3HR. | QUESTIONS - 26 THE MARKS ARE MENTIONED ON EACH QUESTION _____ SECTION –