1285066306 XII Phy TOPPERSamplePaperSolutions 30 - Copy

7/28/2019 1285066306 XII Phy TOPPERSamplePaperSolutions 30 - Copy

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TOPPER SAMPLE PAPER 3

Class XII- Physics

Solutions

Time: Three Hours Maximum Marks: 70

General Instructions

(a) All questions are compulsory.

(b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30carry five marks each.

(c) There is no overall choice.

(d) Use of calculations is not permitted.

(e) You may use the following physical constants wherever necessary.

19

8 1

34

7 1

23 1

23

27

1.6 10

3 10

6.6 10

4 10

1.38 10

6.023 10 /

1.6 10

o

B

A

n

e C

c ms

h JS

TmA

K JK

N mole

m kg

=

=

=

=

=

=

=

1.Is the force acting between two point electric charges q1 and q2 kept at some distance apart inair, attractive or repulsive when (i) q1q2 > 0 (ii) q1q2 < 0 ? (1 mark)

2. On a graph show the stopping potential for a given photosensitive surface varies with thefrequencies n1 and n2 of incident radiations where n1 >n2. Given that intensity is same for both lightradiations. (1 mark)

3. A message signal of frequency 10 kHz is used to modulate a carrier of frequency 1 MHz. Findthe side bands produced. (1 mark)

4. Is the diode in the following circuit forward biased or reverse biased ? Give reason for youranswer.


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(1 mark)

5. What is the frequency range of signals that are transmitted using optical fibers? (1 mark)

6. A spherical Gaussian surface encloses charges q1 and q2 with6 6

1 28.85 10 and q = - 8.85 10

= q C C

(i) Calculate the electric flux passing through the surface.(ii) How would the flux change if the spherical Gaussian surface is replaced with a cubicalGaussian surface and why? (1 mark)

7. For a given thermocouple, the emf generated across its ends is given by E = at + bt 2 where t inC is the temperature of the hot junction, the cold junction being at 0 C.If a = 10 mV /C and b = -0.02 mV/C2 , calculate the value of inversion temperature in C. (1

mark)

8. A 3.0 V battery is connected to an ammeter and a resistor of 3 ohms in series with it.What is the value of the current if the ammeter used is a galvanometer with a resistance of 60 ohm? (1 mark)

9. An electric bulb B and a parallel plate capacitor C are connected in series to the a.c. mains. Thebulb glows with some brightness. How will the glow of the bulb be affected on introducing adielectric slab between the plates of the capacitor? Give reasons in support of your answer. (2marks)

10. Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossedPolaroids. (2 marks)

11. Why is NAND gate called a universal gate. How can it to be used to realize a basic OR gate?(2 marks)

12. What is modulation? What are the different types of modulation? (2 marks)

13. What is a Wheatstone bridge? Derive the balance condition of a Wheatstone bridge. (2 marks)

14. Derive an expression for the equivalent emf and the equivalent internal resistance of a seriescombination of n cells in an electric circuit. (2 marks)

15. State Amperes circuital law. Write the expression for the magnetic field at the centre of acircular coil of radius R carrying a current I. Draw the magnetic field lines due to this coil. (2 marks)

16. Write the expression for the force acting on a charged particle of charge q moving with velocityv in the presence of a magnetic field B.Show that in the presence of this force(i) the kinetic energy of the particle does not change.(ii) its instantaneous power is zero. (2 marks)


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17. For a series LCR circuit, draw a graph showing the variation of current with angular frequencyof the source for a particular value of resistance. On the graph mark the angular frequencies forwhich the power is the maximum value. Also mark the bandwidth. (2 marks)

18. A 15.0 F capacitor is connected to a 220V, 50Hz source. Find the capacitive reactance andthe rms current in the circuit. If the frequency is doubled, what happens to the capacitive reactanceand the current? (2 marks)

19. Suppose that the electric field of an electromagnetic radiation wave in vacuum isE=(3.1cos[1.8rad/m]y+5.4 106 rad/s)t]

1): What is wavelength, ?

2): What is frequency,?3): What is magnitude of the magnetic field of the wave? (3 marks)

20. In Young's double slit experiment using monochromatic light of wavelength , the intensity at apoint on the screen where path difference is K units. What is the intensity of light at a point wherepath difference is /3? (3 marks)

21. An electron, -particle and a proton have the same de-Broglie wavelength. Which of theseparticles has (i) minimum kinetic energy, (ii) maximum kinetic energy, and why ? (3 marks)

22. State the law of radioactive decay. Establish a mathematical relation between half-life periodand disintegration constant of a radioactive nucleus. (3 marks)

23. In the fusion reaction 1H2 + 1H

2 ---------> 2He3 + 0n

1, the masses of deuteron, helium andneutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg deuterium undergoescomplete fusion, find the amount of total energy released. (3 marks)

24. We do not choose to transmit an audio signal by just directly converting it to an e.m. wave ofthe same frequency. Give three reasons for the same. (3 marks)

25. A parallel plate capacitor is charged to potential V by a source of emf. After removing thesource, the separation between the plates is doubled.How will the following change:i. electric fieldii. potential differenceiii. capacitance of the capacitor? Justify your answer. (3 marks)

26. For a transistor, the current amplification factor is 0.8 when the transistor is connected incommon emitter configuration. Calculate the change in the collector current when the base currentchanges by 6 mA. (3 marks)

27. When a transistor amplifier of current gain of 75 is given an input signal, V i= 2 sin (157t +/2),the output signal is found to be VO= 200 sin(157t +3/2 ).In which mode is the tarnsistor being used. Justify your result with proper explanation.Here the current gain is 75 i.e >1. (3 marks)


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28. a. Define the term dipole moment and its direction. State its SI unit. Deduce the expression forthe torque acting on it.b. In a particular situation, it has its dipole moment aligned with the electric field. Is the equilibriumstable or unstable? (5 marks)

29. a. State Lenzs law. Which conservation law can be used to explain this law?b. A wheel with 10 metallic spokes each 0.5m long is rotated with a speed of 120rev/min in a planenormal to the horizontal component of earths magnetic field at a place where the earths field is0.4x10-4G. What is the induced emf between the axle and the rim of the wheel? (5 marks)c. Two moving coil meters, M1 and M2 have the following particulars:R1 = 10 , N1 = 30,

A1 = 3.6 x 10-3 m2, B1 = 0.25 T

R2 = 14 , N2 = 42

A2 = 1.8 x 10-3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivityand (b) voltage sensitivity of M2 and M1.

30. a. Derive an expression for the magnifying power of an astronomical telescope. Draw a raydiagram showing image formation in it.b. An astronomical telescope consists of two thin lens set 36cm apart and has a magnifying power8. Calculate the focal length of the lens. (5 marks)c. A giant refracting telescope at an observatory has an objective lens of focal length 15 cm. If aneye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope? If thistelescope is used to view the moon, what is the diameter of the image of the moon formed by theobjective lens? The diameter of the moon is 3.48 106 m, and the radius of lunar orbit is 3.8 108m.


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TOPPER SAMPLE PAPER 3

Class XII- Physics

Solutions

Ans1. If q1q2 > 0 this means that q1 and q2 have the same sign. Either both charges are positive orboth are negative so the force between them will be repulsive. (1/2)q1q2 < 0 then force will be attractive (1/2)

Ans 2.

Ans 3. The side bands are at 1000+10 = 1010kHz (1/2)and 1000 -10 = 990kHz. (1/2)

Ans 4.The diode is reverse biased.(1/2)Because the p side is at a lower voltage than the n side. (1/2)

Ans 5. 1THz to 1000THz. (1)

Ans 6. i. The flux is zero as the net charge enclosed by the Gaussian surface will be zero.(1/2)ii. If the Gaussian surface is made cubical the flux is the same. It stays zero, this is because theflux depends only on the charge enclosed & is independent of the shape of the Gaussian surface.(1/2)


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Ans 7. The neutral temperature will be010 500

0.02n

aC

b = = = .(1/2)

The inversion temperature is twice the neutral temperature so it is 1000

0

C.(1/2)

Ans 8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I =3/63 =0.048A. (1)

Ans 9. On introducing a dielectric in the capacitor, its capacitance will increase. (1)

Then the total impedance of the circuit will decrease as

2

2 1Z RC

= +

. Hence the current in

the circuit increases and the brightness of the lamp increases.(1)

Ans10.Two crossed polaroids are placed perpendicular to each other.Let the intensity of incident light be I0 . The light transmitted by first Polaroid has intensity I0/2because if unpolarised light is incident on a polaroid the transmitted intensity is half the originalintensity. (1/2)

The light transmitted by second Polaroid has intensity ( )20 cos2

I where is the angle between

the axes of first and second Polaroid.The light transmitted by third Polaroid has intensity

( ) ( )2 2 0 2 2 20 0 0cos cos (90 ) cos (sin ) sin 22 2 8

I I I = = . (1)

This intensity will be maximum when 045 = (1/2)

Ans 11. NAND gate is called a universal gate because it can be used to obtain other basic gateslike AND, NOT and OR gates. (1)

NAND gates can be combined as shown below to realize a basic OR gate.

(1)

Ans 12. Some characteristic of the carrier signal is varied in accordance with the modulating ormessage signal. This is called modulation. (1)Amplitude modulation, frequency modulation and phase modulation of waves are different ways ofmodulation. (1)


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Ans 13.A Wheatstone bridge is an arrangement of four resistances that can be used to measure one ofthem in terms of the rest. (1/2 )

(1/2)I1 = I3 and I2 = I4.Next, we apply Kirchhoffs loop rule to closed loops ADBA and CBDC.The first loop ADBA givesI1 R1 + 0 - I2 R2 = 0 (Ig = 0)and the second loop CBDC gives, upon using I3 = I1, I4 = I2-I4 R4 + 0 + I3 R3 = 0 (1/2)Upon using I3 = I1, I4 = I2 & taking the ratio of the above two equations we obtain,

1 2

3 4

R R

R R= (1/2)

This last equation relating the four resistors is called the balance conditionfor the galvanometer togive zero or null deflection.

Ans 14.Consider first two cells in series, where one terminal of the two cells is joined together leaving theother terminal in either cell free.Let V(A), V(B), V(C) be the potentials at points A, B and C shown.

Then V(A) V(B) is the potential difference between the positive and negative terminals of thefirst cell.

We know

V(A) V(B)=1 Ir1 , where E1 is the emf of the Ist cell


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Similarly,V(B) V(C)=2 Ir2 , where E2 is the emf of the 2nd cell

Hence, the potential difference between the terminals A and C of the combination is

[ ] [ ]AC

1 2 1 2

V V(A) V(B) V(B) V(C)

E E I(r r )

= +

= + +(1)

If we wish to replace the combination by a single cell between A and C of emf E eqand internal

resistance req, we would haveVAC = EeqI req

Comparing the last two equations, we getand req = r1 + r2

The rule for series combination clearly can be extended to n number of cells:

(i) The equivalent emf of a series combination of n cells is just the sum of their individual emfs,and(ii) The equivalent internal resistance of a series combination of n cells is

just the sum of their internal resistances. (1)

Ans 15. Amperes law state that 0.B dl i=

where I refers to the current passing through

amperian loop S around the current element. (1)The magnitude of the magnetic field due to a circular coil of radius R carrying a current i at itscentre is

0

2

iB

R

= (1)

Magnetic field lines due to circular current carrying coil is

(1)


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Ans 19.

1. Wavelength: =2/k= 3.5m. (1)

2. Frequency, = /2 = 5.4x106/2 = 0.86MHz. (1)3. Magnetic field B0 = E0/c = 3.1/3x10

8 = 10nT. (1)

Ans 20.Intensity I= 4I0 cos

2/2When path difference is , phase difference is 2

I= 4Io cos2 = 4 Io = K (given) . (1)

When path difference is = /3, then the phase difference will be

'=2 /

=2 x / 3 = 2/3 (1)

Hence Intensity at a point where the path difference is /3, isI'=4I0 cos

2 2 /6 (since K = 4I0 )

= K cos2 /3= K x {1/2}2 = () K. (1)

Ans 21.The deBroglie wavelength

e e p p

h h

p 2mK

For the electron, proton and -particle, is same

m K m K m K cons tan t

= =

= = =

(1)

As mass of electron is minimum its kinetic energy will be maximum. (1)As mass of alpha-particle is maximum its kinetic energy is minimum. (1)

Ans 22.In any radioactive sample, the number of nuclei undergoing the decay per unit time is proportionalto the total number of nuclei in the sample. (1)

Radioactive decay law,

dNN

dt= .


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Integrating the above expression givest

0

0

1/2

1/2

N(t) N e

NPut N(t)

2and t=T

This gives

0.693T

=

=

=

(1)

(1)Ans 23.1amu = 931.5 MeV/c2

amum 004.0)009.1017.3()015.2(2 =+= (1)

Hence energy released per deuteron = (0.04 x 931.5)/2 = 3.726 /2 MeV (1)

The number of deuterons in 1kg =NA/2 =6.02x1023/2

Energy released =(3.02x1023)( 3.726x106/2)(1.6x10-19) =9.01013 J (1)

Ans 24.i) If the audio signal is directly transmitted, the size of antenna will be extra large which is not

practically feasible. This is because the size of the antenna required is proportional to / 4 . (1)

ii) Effective power radiated by antenna is proportional to square of frequency. For an audio

frequency wave the radiated power will be extremely small. (1)

iii) If different programmes, audio frequencies are directly transmitted then what you will hear atthe receiver will be a mixture of all these signals. (1)

Ans.25a. E remains same as it depends on the charge on the plates and the medium between the plates.b. Q remains same as charge does not change on the plates. (1)c. V =Ed, so as the distance is doubled V also doubles. (1)d. C=Q/V and V is doubled whereas Q remains same. So C is halved. (1)

Ans 26

Given

C b

=0.8

= /(1- ) (1)

= 0.8/(1-0.8) = 4 (1)

I = I = 6 4 = 24 mA (1)


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Ans 27.Here the current gain is 75 i.e >1. (1)Besides this there is a phase difference of between the signal at the input and the output. (1)Both these facts indicate that the amplifier is connected in common-emitter mode. (1)

Ans 28. Dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis fromq to q. (1)Its SI unit is C-m. (1)

Force of qE acts on charge q and force qE acts on charge q.

(1)

So the torque = qEsin = pE sin p E =

(1)

The equilibrium is stable as the torque acting on it will be zero. (1)

Ans 29.

a.Lenzs law states that the polarity of the induced emf is such that it tends to produce a currentwhich opposes the change in the magnetic flux that produces it.(1)Law of conservation of energy. (1)

b.2 4 2 5Induced emf = (1/2) BR (1/ 2) 4 0.4 10 (0.5) 6.28 10 = = V (1)

c.GivenR1 = 10 , N1 = 30, A1 = 3.6 x 10

-3 m2, B1 = 0.25 T

R2 = 14 , N2 = 42, A2 = 1.8 x 10-3 m2, B2 = 0.50 T

Current sensitivity,

Ratio of current sensitivities


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(1)

Ratio of voltage sensitivities

(1)

Ans 30.a. Magnifying power is the ratio of the angle subtended at the eye by the final image to the anglewhich the object subtends at the lens or the eye.

. o o

e e

f fhmf h f

= = = (1)

(1)

b. GivenMagnification m =fo/ fe & length of the tube, fo + fe = L8 = fo / fe & fo + fe =36

Hence 8fe + fe = 36Or fe = 4cm

Therefore, fo = 32cm (1)

c.

(a)

Angular magnification

15m 1500

0.01= =


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(1)

6

8

(b)

Let diameter of image be d. Then,

d 3.48 10

1500 3.8 10

d 13.7cm.

=

=

(1)

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1285066306 XII Phy TOPPERSamplePaperSolutions 30 - Copy

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