Physics CBSE solution 2012

155/1/1 P.T.O.

Series : SMA/1

Roll No.

Code No. 55/1/1Candidates must write the Code onthe title page of the answer-book.

Please check that this question paper contains 11 printed pages.

Code number given on the right hand side of the question paper should be written on the title page ofthe answer-book by the candidate.

Please check that this question paper contains 30 questions.

Please write down the Serial Number of the questions before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be distributed at10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will notwrite any answer on the answer script during this period.

PHYSICS (Theory)[Time allowed : 3 hours] [Maximum marks : 70]

General Instructuions:

(i) All questions are compulsory.

(ii) There are 30 questions in total. Questions numbered 1 to 8 are very short-answer questions andcarry 1 mark each.

(iii) Questions numbered 9 to 18 are short-answer questions and carry 2 marks each, Questionsnumbered 19 to 27 carry 3 marks each and Questions numbered 28 to 30 are long-answer questionsand carry 5 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in one question of twomarks, one question of three marks and all three questions of five marks each. You have toattempt only one of the choices in such questions.

(v) Use of calculators is not permitted. However, you may use log tables if necessary.

(vi) 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 0 = 4 × 10–7 T mA–1

0

14 = 9 × 109 Nm2C–2 m

e = 9.1 × 10–31 kg

Studymate Solutions to CBSE Board Examination 2011-2012

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1. When electrons drift in a metal from lower to higher potential, does it mean thatall the free electrons of the metal are moving in the same direction?

Ans. No. Eventhough, overall motion is from lower to higher potential, individually intheir motion randomness is present.

2. The horizontal component of the earth’s magnetic field at a place is B and angleof dip is 60°. What is the value of vertical component of earth’s magnetic field atequator?

Ans. At magnetic equator the angle of dip is zero. So, the vertical component of earth’smagnetic field is zero (It has been assumed that equator mentioned in questionpaper refers to magnetic equator).

3. Show on a graph, the variation of resistivity with temperature for a typicalsemiconductor.

Ans.

T

0

4. Why should electrostatic field be zero inside a conductor?

Ans. A conductor has free electrons. As long as electric field is not zero, the freecharge carriers would experience force and drift. In the static situation, the freecharges have so distributed themselves that the electric field is zero everywhereinside the conductor.

5. Name the physical quantity which remains same for microwaves of wavelength1 mm and UV radiations of 1600 Å in vacuum.

Ans. Velocity of wave in vacuum remains same for all kinds of E.M. Waves.

6. Under what condition does a biconvex lens of glass having a certain refractiveindex act as a plane glass sheet when immersed in a liquid?

Ans. If refractive index of glass equals the refractive index of liquid in which lens isimmersed, the lens behaves as plane glass sheet.

7. Predict a directions of induced currents in metal rings 1 and 2 lying in the sameplane where current I in the wire is increasing steadily.

1

2I

Ans. In ring 1, the direction of induced current is clockwise.

In ring 2, the direction of induced current is anti-clockwise.

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8. State de-Broglie hypothesis.Ans. De-Broglie hypothesis says that if radiation has a wave-particle nature then

moving particles of matter should also display wave-like properties under suitableconditions. He reasoned that nature was symmetrical and that the two basicphysical entities – matter and energy, must have symmetrical character.De-Broglie proposed that the wave length associated with a particle of momentump is given as

h h

p mv

where m is the mass of the particle and v its speed.

9. A ray of light, incident on an equilateral glass prism 3 moves parallel to

the base line of the prism inside it. Find the angle of incidence for this ray.

Ans. A = 60°. Since the light moves parallel to the base of prism, prism is underminimum angle of deviation position.

2r = A r = 2A

= 30°

Using = sinsin

i

r, we get

sin3

sin30i

sin i = 1

3 sin30 32

sin i = 32

i = 60°

10. Distinguish between ‘Analog and Digital signals’.OR

Mention the function of any two of the following used in communication system:

(a) Transducer

(b) Repeater

(c) Transmitter

(d) Bandpass Filter

Ans. (a) Analog - continuously varying signal represented generally by a sinusoidalfunction. Analog signals are susceptible to noise.

(b) Digital - Are discrete (non-continuous) in nature. Digital signals are relativelyimmune to noise.

OR(a) Transducer - converts one form of energy to another form.

(b) Repeater - picks up the signal from transmitter, amplifies it and re-transmitsit.

(c) Transmitter - sender of information in a communication system.

(d) Bandpass Filter - A band pass filter rejects low and high frequencies andallows a band of frequencies to pass through as per the requirement.

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11. A cell of emf E and internal resistance r is connected to two external resistancesR

1 and R

2 and a perfect ammeter. The current in the circuit is measured in four

different situations:

(i) without any external resistance in the circuit

(ii) with resistance R1 only

(iii) with R1 and R

2 in series combination

(iv) with R1 and R

2 in parallel combination

The currents measured in the four cases are 0.42 A, 0.15 A, 1.4 A and 4.2 A, butnot necessarily in that order. Identify the currents corresponding to the fourcases mentioned above.

Ans. (i) i = r

= 4.2 A

Because out of four situations resistance of circuit is minimum in this case.Hence, current is maximum.

(ii) i = 1r R = 1.05 A

Effective resistance of circuit is more than situation (i) and (iv) and lessthan (iii). So, current is 1.05 A.

(iii) i =

1 2r R R = 0.42 A

Effective resistance of circuit is maximum in this case. Hence, current is0.42 A.

(iv) i =

1 2

1 2

iR R

rR R

= 1.4 A

In parallel combination of R1 and R

2, the effective resistance is less than

either of resistors. So, effective resistance of circuit is more than situation(i) and less than (ii) and (iii). So, current is 1.4 A.

12. The susceptibility of a magnetic material is –2.6 × 10–5. Identify the type of magneticmaterial and state its two properties.

Ans. Diamagnetic materials have negative susceptibility. Hence, material isdiamagnetic.

Properties of Diamagnetic materials

(a) Shifts towards a weak field in a non-uniform field.

(b) When placed in a magnetic field, it is weakly magnetised in a directionopposite to that of the applied field.

13. Two identical circular wires P and Q each of radius R and carrying current ‘I’ arekept in perpendicular planes such that they have a common centre as shown inthe figure. Find the magnitude and direction of the net magnetic field at thecommon centre of the two coils.

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P

Q

I

I

Ans. Magnetic field due to horizontal loop P at its centre is

P

Q

I

I

B1 = 0

2i

r

in vertically upward direction.

Magnetic field due to vertical loop Q at centre.

B2 = 0

2i

r

in horizontal direction.

So, net magnetic field at centre

B = 2 2

0 02 21 2 2 2

i iB B

r r

B1

B2

B

= 45°

B =

0 022 2

i i

r rThe direction of magnetic field makes an angle 45°with horizontal.

14. When an ideal capacitor is charged by a dc battery, no current flows. However,when an ac source is used, the current flows continuously. How does one explainthis, based on the concept of displacement current?

Ans. When an ideal capacitor is charged by dc battery, charge flows till the capacitorgets fully charged. C

VWhen an ac source is connected then conduction current c

dqi

dt flows in the

connecting wire. Due to charging current, charge deposited on the plates of thecapacitor changes with time. Changing charge causes electric field between the

plates of capacitor to be varying, giving rise to displacement current 0 .

ed

di

dt

[As displacement current is proportional to the rate of flux variation].

Between the plates, electric fieldC

0 0

qE

A

Electric flux, 0

c A

qE A

A

So, 00

0

ed c

d d qA dqi i

dt dt A dt

Displacement current brings continuity in the flow of current between the platesof the capacitor.

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15. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential(V) with distance r due to a point charge Q.

Ans. Due to point charge,

electric field2

0

1

4

q

Er

and electric potential0

1

4

q

Vr

The field and potential can be plotted as shown below:

r

E

0

r

V

0

16. Define self-inductance of a coil. Show that magnetic energy required to build upthe current I in a coil of self inductance L is given by ½LI2.

Ans. Self induction is a property of a coil due to which the coil opposes any change inthe strength of current flowing through it by inducing an e.m.f. in itself.

As current i grows in an inductor a back emf of di

e Ldt

is developed.

In over coming the back emf, work is done.The work done = dW = edq for a charge dq

W e dq di

L idtdt

0

ILi di

21

2LI

This is stored as energy in the magnetic field of the inductor.

17. The current in the forward bias is known to be more (~mA) than the current inthe reverse bias (~µA). What is the reason, then, to operate the photodiode inreverse bias?

Ans. The current in the forward bias is due to majority carriers where as current inthe reverse bias is due to minority carriers. So current in forward bias is more(~mA) than current in reverse bias (~µA).

On illumination of photodiodes with light, the fractional change in the majoritycarriers would be much less than that in minority carriers. It implies fractionalchange due to light on minority carrier dominated reverse bias current is moreeasily measurable than fractional change in forward bias currnet. So photo-diodesare operated in reverse bias condition.

18. The metallic rod of ‘L’ length is rotated with angular frequency of ‘’ with one endhinged at the centre and the other end at the circumference of a circular metallicring of radius L, about an axis passing through the centre and perpendicular to

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the plane of the ring. A constant and uniform magnetic field B parallel to the axisis present everywhere. Deduce the expression for the emf between the centreand the metallic ring.

Ans.

L

As the rod is rotated, free electrons in the rod move towards the outer end due toLorentz force and get distributed over the ring. Thus, the resulting separation ofcharges produces an emf across the ends of the rod. At a certain value of emf,there is no more flow of electrons and a steady state is reached. The magnitudeof the em f generated across a length dr of the rod as it moves at right angles tothe magnetic field is given byd = Bvdr . Hence,

2

0 0 2

L L B Ld Bv dr B r dr

19. The figure shows a series LCR circuit with L = 5.0 H2, C = 80 µF, R = 40

connected to a variable frequency 240 V source. Calculate.

L

R

C

(i) The angular frequency of the source which drives the circuit at resonance.

(ii) The current at the resonating frequency.

(iii) The rms potential drop across the capacitor at resonance.

Ans. Given: L = 5.0 H; C = 80 F; R = 40 ; Vrms

= 240 V

(i) At resonance, angular frequency

w = 2f =

12

2 LC

w =

6

1 1

5 80 10LC

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w =

24

1 1/

2 104 10rad s

w = 50 rad/s.

(ii) At resonating frequency,

Impedance, Z = R ( XL = X

C)

Z = 40

Current, I = 24040

rmsVZ

I = 6 A

(iii) Reactance of capacitor, XC =

1wC

XC =

6 3

1 150 80 10 4 10

Potential drop across capacitor,V

C = I.X

C

VC =

3

16

4 10V

C = 1500 V

20. A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2 A. Astraight long wire carrying 5 A current is kept near the loop as shown. If the loopand the wire are coplanar, find

A

B C

D

4 cm

10 cm

1 cm

2AI = 5A

(i) the torque acting on the loop and

(ii) the magnitude and direction of the force on the loop due to the currentcarrying wire.

Ans. (i) Direction of magnetic dipole moment of loop due to current in loop isperpendicular inside the plane of paper.

Direction of magnetic field at loop due to current flow in straight wire isalso perpendicular inside the plane of paper.

Angle between M and

B = 0°

Torque on loop

M B

= MB sin

= MB sin 0°

= 0

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(ii) Force between two current carrying wires, F =

0 1 2.

2I I

lr

A

B C

D

4 cm

10 cm

1 cm

2AI = 5A

FAB FCD

Force on arm AB, FAB

=

0

2

5 2. (0.10)

2 1 10

FAB

=

0 100

2N [Attraction – towards the wire]

Force on arm CD, FCD

=

0

2

5 2(0.10)

2 5 10

FCD

=

0 20

2N [Repulsion – away from the wire]

Force on arm BC and AD, FBC

= FAD

= 0

Resultant force on loop,

F = FAB

– FCD

=

0 (100 20)

2 F = 2 × 10–7 × 80 = 16 × 10–6 N

[Towards the straight wire i.e. force of attraction]

21. (a) Using Bohr’s second postulate of quantization of orbital angular momentumshown that the circumference of the electron in the nth orbital state inhydrogen atom is n times the de Broglie wavelength associated with it.

(b) The electron in a hydrogen atom is initially in the third excited state. Whatis the maximum number of spectral lines which can be emitted when itfinally moves to the ground state?

Ans. (a) By Bohr’s postulate of quantization of orbital angular momentum

Angular momentum, mvrn =

2

nh

2n

nhr

mv

... (i)

Circumference of the electron in the nth orbital state in hydrogen atom

2rn = 2

2

nh

mv

=

h hn n

mv p ... (ii)

But de-broglie wavelength = h

p ... (iii)

Circumference = n × [from equations ii & iii]

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(b) 3rd excited state

2nd excited state

1st excited state

Ground state

From the above diagram, total number of spectral lines = 6

22. In the figure a long uniform potentiometer wire AB is having a constant potentialgradient along its length. The null points for the two primary cells of emfs

1 and

2 connected in the manner shown are obtained at a distance of 120 cm and 300

cm from the end A. Find (i) 1/

2 and (ii) position of null point for the cell

1.

How is the sensitivity of a potentiometer increased?

120 cm300 cm

1 2

1 2

A B

OR

Using Kirchoff’s rules determine the value of unknown resistance R in the circuitso that no current flows through 4 resistance. Also find the potential differencebetween A and D.

1

1

4

B

EF D

R

C3V9VA

6VI

Ans. Let potential gradient of wire = K volt/cm

Case 1: l1 = 120 cm

E1 – E

2 = Kl

1 = 120 K ...(i) [ cells are connected in opposite direction]

Case 2: l2 = 300 cm

E1 + E

2 = Kl

2 = 300 K ...(ii) [ cells are connected in same direction]

Equation (i) divided (ii)

1 2

1 2

120 2300 5

E E

E E

5E1 – 5E

2 = 2E

1 + 2E

2

3E1 = 7E

2

1

2

73

E

E

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Let null point for cell E1 is l

3.

E1 = Kl

3...(iii)

Equation (i) divided (iii)

1 2

1 3 3

120 120E E K

E Kl l

2

1 3

1201

E

E l

3

3 1201

7 l

3

4 1207 l

l3 = 210 cm

(i) 1

2

73

E

E

(ii) Position of null point for cell E1 = 210 cm

OR

1

1

4

B

EF D

R

C3V9VA

6VII

I I

II

Applying Kirchoff’s voltage law in closed loop AFEBA :

9 – I × 1 – I × 1 – 4 × 0 – 6 = 0

9 – 2I – 6 = 0

3 = 2I

Applying Kirchoff’s voltage law in closed loop BEDCB

6 + 4 × 0 – IR – 3 = 0

IR = 3

R = 3 3

23I

R = 2

Potential difference between A and D = Potential difference between A and E V

AD= I × 2= 1.5 × 2

VAD

= 3V

23. (i) What characteristic property of nuclear force explains the constancy of bindingenergy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A <170?

(ii) Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number A.

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Ans. (i) The constancy of the binding energy in the range 30 < A < 170 is aconsequence of the fact that the nuclear force is short-ranged. It will beunder the influence of only some of its neighbours, which come within therange of the nuclear force. If any other nucleon is at a distance more thanthe range of the nuclear force from the particular nucleon it will have noinfluence on the binding energy of the nucleon under consideration.

The property that a given nucleon influences only nucleons close to it isalso referred to as saturation property of the nuclear force.

(ii) Nuclear density = Mass of nucleus

Volume of nucleus

= 34

3

mA

R [where m = mass of one nucleon, A = mass number]

R = R0A1/3 = Radius of nucleus

= 1/3 3

0

34

( )3

mA m A

R A 304 R A

= 30

34

m

R

This expression is independent of mass number A and is constant.

24. Write any two factors which justify the need of modulating a signal.

Draw a diagram showing an amplitude modulated wave by superposing amodulating signal over a sinusoidal carrier wave.

Ans. Factors for modulating a signal(i) The energy (strength) of signal wave is low, so it cannot be transmitted

directly to large distance.

(ii) Height of signal antena (transmitter and receiver) 4

of signal wave is large Height of antenna will also be very large, which is practically not possible

Diagram for AM Modulation

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25. Write Einstein’s photoelectric equation. State clearly how this equation is obtainedusing the photon picture of electromagnetic radiation.

Write the three salient features observed in photoelectric effect which can beexplained using this equation.

Ans. KEmax

= h –

KEmax

energy of the most energetic photoelectron.

frequency of incident radiation.

work function.

h Planck’s constant.

When a photon falls on an electron, it transfers all of its energy (h) to theelectron. Electron being energized leaves the surface of the metal. But whileleaving the surface energy equal to is reduced in overcomming the bindingforce.

Hence, (KEmax

= h – )

0 =

h

The salient features are:

(i) KEmax

is independent of the intensity of incident radiation.

(ii) Below a threshold frequency, no photoelectric effect takes place.

(iii) KEmax

depends on frequency of radiation.

26. (a) Why are coherent sources necessary to produce a sustained interferencepattern?

(b) In Young’s double slit experiment using monochromatic light of wavelength, the intensity of light at a point on the screen where path difference is ,is K units. Find out the intensity of light at a point where path difference is/3.

Ans. (a) Coherent sources have constant phase difference between, i.e., phasedifference does not change with time. Hence, the intensity distribution onthe screen remains constant and sustained.

(b) I = 204 cos

2I

...(i)

I0 incident intensity.

I resultant intensity.

At a point where, path difference =

= 2

2

Putting in (i)

K = 4 I0 cos2

K = 4 I0

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I0 =

4

K

At a point where, path difference is 3

,

= 2

x

=

2 2

3 3

I2 = 4 I

0 cos2

2

= 24 cos

4 3

K

= 1

44 4 4

K K

27. Use Huygen’s principle to explain the formation of diffraction pattern due to asingle slit illuminated by a monochromatic source of light.When the width of the slit is made double the original width, how would thisaffect the size and intensity of the central diffraction band?

Ans. Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incidenton it. According to Huygen’s principle the diffraction pattern is the result ofsuperposition of a large number of waves, starting from different points ofilluminated slit.

P

C

N

90°O

M1

M2

A

B

Light from

source

At the central point C of the screen, the angle is zero. Hence the waves startingfrom all points of slit arrive in the same phase. This gives maximum intensity atthe central point C.

The observation point is now taken at P.

Minima: Now we divide the slit into two equal haves AO and OB, each of width

a

2. Now for every point, M1 in AO, there is a corresponding point M2 in OB, such

that M1M2 = a

2; Then path difference between waves arriving at P and starting

from M1 and M2 will be a

2 sin =

2

.

a sin =

In general,

a sin = n

Secondary Maxima : Similarly it can be shown that for maxima

1asin n

2

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The intensity pattern on the screen is shown in the given figure.

I0

a 20–2 –

a–3a

–4a a a

3a

4a

When the width of the slit is made double the original width, the size reduces byhalf according to the relation: size /d. Intensity increases four fold.

28. Explain the principle of a device that can built up high voltages of the order of afew million volts.

Draw a schematic diagram and explain the working of this device.

Is there any restriction on the upper limit of the high voltages set up in thismachine? Explain.

OR

(a) Define electric flux. Write its S.I. units.

(b) Using Gauss’s law, prove that the electric field at a point due to a uniformlycharged infinite plane sheet is independent of the distance from it.

(c) How is the field directed if

(i) the sheet is positively charged

(ii) negatively charged?

Ans. Van-de graff generator is based on the principle the charge given to a hollowconductor is transferred to the outer surface and uniformly distributed over it.

Potential inside conducting spherical shell of radius R carrying charge Q = constant

= 0

14

Q

RNow, as shown in figure, let us suppose that in some way we introduce a smallsphere of radius r, carrying some charge q, into the large one, and place it at thecentre. The potential due to this new charge clearly has the following values atthe radii indicated:

1655/1/1 P.T.O.

Potential due to small sphere of radius r carrying charge q

= 0

14

q

r at surface of small sphere

= 0

14

q

R at large shell of radius R.

Taking both charges q and Q into account we have for the total potential V andthe potential difference the values

V(R) = 0

14

Q q

R R

V(r) = 0

14

Q qR r

V(r) – V(R) = 0

1 14

q

r RWe see that, independent of the amount of charge Q that may have accumulatedon the larger sphere and even if it is positive, the inner sphere is always at ahigher potential: the difference V(r )–V(R) is positive.Smaller bulb is always at a higher potential than the bigger bulb. Thereforecharge continuously flows from the smaller bulb to the bigger bulb.Yes, the van-de-graft generator can only be charged upto a limit when the electricfield around it is less than breakdown field of the surrounding air.

OR(a) Electric flux is defined as the dot product of electric field intensity and area

vector. Its SI unit is Nm2/C.(b) Let be the uniform surface charge density of an infinite plane sheet.

Consider a Gaussian surface to be a rectangular parallelepiped of crosssectional area A.

z

y

x

x x

x

21

Surface charge density

E E

Gaussian surface for a uniformly charged infinite plane sheet

From figure, only the two faces 1 and 2 will contribute to flux. Electric fieldlines are parallel to the other faces and they do not contribute to the totalflux.The net flux through the Gaussian surface is 2EA. The charge enclosed bythe closed surface is A.By Gauss’s law,

0

A2EA

or

0E n

2

Since the expression does not contain ‘x’ electric field is independent of thedistance.

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(c) (i) The electric field is directed perpendicularly away from the sheet.

(ii) The electric field is directed perpendicularly towards the sheet.

29. Define magnifying power of telescope. Write is expression.

A small telescope has an objective lens of focal length 150 cm and an eye piece offocal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away,find the height of the final image when it is formed 25 cm away from the eyepiece.

OR

How is the working of a telescope different from that of a microscope?

The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5cm respectively. Find the position of the object relative to the objective in orderto obtain an angular magnification of 30 in normal adjustment.

Ans. Magnifying power of an astronomical telescope is defined as the ratio of theangle subtended at the eye by the final image of the angle subtended at the eye,by the object directly.

The expression for magnifying power: m =

0 1 e

e

f ff d

(a) In normal adjustment m = 0

e

f

f

(b) When final image is formed at infinity m =

0 1 e

e

f f

f df0 = 150 cm

fe = 5 cm

Using 1 1 1f v u for objective.

1 1 1150 3000v

1 1 1 20 1150 3000 3000v

v = 300019

m0 =

3000

119 3000

1

19

Using 1 1 1f v u for eye piece

1 1 15 25 u

1 1 125 5u

u = 1 5

25

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u = 256

cm

me =

v

u

me =

25

625

= 6

Total magnification = m0 × m

e = –0.315

hI = 0.315 × 100 = 31.578 cm

OR

Telescope Microscope1. Resolving power should be higher

for certain magnification.

2. Focal length of objective should be kept larger while eye piece focal length should be small for better magnification.

3. Objective should be of large aperture.

4. Distance between objective and eye piece is adjusted to focus the object at infinity.

1. Resolving power is not so large but the magnification should be higher.

2. Both objective and eye piece should have less focal length for better magnification.

3. Eye piece should be of large aperture.

4. Distance between objective and eye piece is fixed. For focusing an object distance of objective is changed.

m = .e

v D

u f

30 = 25v

u

5

5

6 = v

u v = –6u

Applying lens formula for objective

1 1 1v u f

1 1 1 100 46 1.25 125 5u u

1 6 46 5u

24u = –35

u = 35

cm24

u = –1.46 cm

30. Draw a simple circuit of a CE transistor amplifier. Explain the working. Show

that the voltage gain, AV, of the amplifier is given by A

V =

ac L

i

R

r, where

ac is the

current gain, RL is the load resistance and ri is the input resistance of the

transistor. What is the significance of the negative sign in the expression for thevoltage gain?

1955/1/1 P.T.O.

OR

(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode.Explain its working and show the output, input waveforms.

(b) Show the output waveforms (Y) for the following inputs A and B of

(i) OR gate (ii) NAND gate

A

B

t1 t2 t3 t4 t5 t6 t7 t8

Ans. Vcc = V

CE + I

cR

L

Likewise, the input loop gives

VBB

= VBE

+ IB R

B

When viis not zero, we get

IC

IB

RB

ViIE

RC V0

VBBVCC

C

E

C

B

VBE

+ vi = V

BE + I

B R

B + I

B (R

B + r

i)

The change in VBE

can be related to the input resistance riand the change in I

B.

Hence

vi= I

B (R

B + r

i )

vi= r I

B

The change in IB

causes a change in Ic. We define a parameter

ac, which is

similar to the dc

ac =

c c

B b

I i

I i

which is also known as the ac current gain Ai. Usually

ac is close to

dc in the

linear region of the output characteristics.

The change in Ic due to a change in I

B causes a change in V

CE and the voltage drop

across the resistor RL because V

CC is fixed.

These changes can be given by

VCC

= VCE

+ RLI

C= 0

or VCE

= –RLI

C

The change in VCE

is the output voltage v0. We get

v0 = V

CE= –

ac R

LI

B

The voltage gain of the amplifier is

Av =

0 CE

i B

v V

v r I

Av =

ac LR

rThe negative sign indicates that the output is phase reversed by an angle 180°.

OR(a) Full wave rectification: The circuit uses two diodes connected to the ends

of a centre tapped transformer. The voltage rectified by the two diodes is

2055/1/1 P.T.O.

half of the secondary voltage i.e., each diode conducts for half cycle of inputbut alternately so that net output across load comes as half sinusoids withpositive values only. For positive cycle diode D1 conducts (FB) but D2 is beingout of phase is reverse biased and does not conduct. Thus output across RL

is due to D1 only. In negative cycle of input D1 is R.B. but D2 is F.B. andconducts as with respect to centre tap point A is negative but B is positive.Hence output across RL is due to D2.

Input

D2

RL

D1A

B

X

Y

Centretaptransformer

Centretap

Output

Fig. Full wave rectification circuit

t

t

t

wav

efor

mat

Aw

avef

orm

atB

Outputwaveformacross RL

Due to Due to Due to Due toD1 D2 D1 D2

(b) (i) OR Gate (ii) NAND Gate

0 0 0

1 0 1

0 1 1

1 1 1

A B Y A B . ' .

0 0 0 1

1 0 0 1

0 1 0 1

1 1 1 0

A B A B Y A B

A

B

t1 t2 t3 t4 t5 t6 t7 t8

Y

Y

× · × · × · × · ×

2155/1/2 P.T.O.

1. Why must electrostatic field be normal to the surface at every point of a chargedconductor. [1]

Ans. Surface of a conductor is an equipotential surface, where the field goes normally.

6. Predict of the direction of induced current in a metal ring when the ring ismoved towards a straight conductor with constant speed v. The conductor iscarrying current I in the direction shown in the figure. [1]

v I

Ans. Clockwise using Fleming’s R.H. rule.

10. Derive the expression for the self inductance of a long solenoid of cross sectionalarea A and length l, having n turns per unit length. [2]

Ans. l

Magnetic field at a point inside the solenoid is B = 0NI

Where N is the total number of turns of the solenoid and l is its length. B isconstant throughout the length of the solenoid.

Magnetic flux through each turn = B × area of each turn.

1 = 0N

I × A where A is the area of each turn.

Total magnetic flux linked with the solenoid = = 0N

IA × N

But from the definition of self inductance (L), LI .

LI = 0N

IA × N L = 2

0N A

= µ0n2lA. Henry

14. The susceptibility of a magnetic material is 2.6 × 10–5. Identify the type of magneticmaterial and state its two properties. [2]

Ans. Paramagnetic, as susceptibility is +ve and not so large.

Paramagnetic material

(i) pull most of the field lines inside the material and leave some outside

(ii) Moves towards stronger field in a non-uniform field.


UNCOMMON QUESTION ONLY

Series : SMA/1 Code No. 55/1/2

2255/1/2 P.T.O.

16. Two identical circular loops, P and Q, each of radius r and carrying currentsI and 2I respectively are lying in parallel planes such that they have a commonaxis. The direction of current in both the loops is clockwise as seen from O whichis equidistant from the both loops. Find the magnitude of the net magnetic fieldat point O. [2]

2r

OP QI 2I

Ans. Field due to loop P = 2

2 2 3/2

µ

2( )o I r

r r towards P.

Field due to loop Q = 2

2 2 3/2

µ 2

2( )o I r

r r towards Q.

So net field at O = 2

2 3/22(2 )oµ I r

r towards Q

=

20

3/2 3

µ µ

2 2 4 2o I r I

r r towards Q.



L

R

C




Ans. (i) 26

1 1 150 rad/s

2 104 100 10LC

(ii) I = 240

460

rms rmsV VA

Z R

(iii) Vrms

across L = 4 LX

= 2

14 4.0

2 10

= 800 volt

21. (a) Why are coherent sources necessary to produce a sustained interferencepattern?

(b) In Young’s double slit experiment using monochromatic light of wavelength, the intensity of light at a point on the screen where path difference is ,is K units. Find out the intensity of light at a point where path difference is/3.

2355/1/2 P.T.O.

Ans. (a) Coherent sources have constant phase difference, i.e., phase differencedoes not change with time. Hence, the intensity distribution on the screenremains constant and sustained.

(b) I = 204 cos

2I

...(i)

I0 incident intensity.

I resultant intensity.

At a point where, path difference =

= 2

2

Putting in (i)

K = 4 I0 cos2

K = 4 I0

I0 =

4

K

At a point where, path difference is 2

3

,

= 2 2 4

3 3

Putting in (i)

I2 = 4 I

0 cos2 2

3

I2 = 4 I

0 ×

1

4 = I

0 =

4K

22. A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1A. Astraight long wire carrying 4 A current is kept near the loop as shown. If the loopand the wire are coplanar, find

2 cm

1A

1cm

1A

I = 4A

5 cm



Ans. (i) Torque = 0, as there will be forces on the plane of the rectangular coil andwire.

(ii) Forces on the loop = 1 2µ

2o I I

r * length (Newton)

2455/1/2 P.T.O.

= 7

2 2

4 1 4 12 10

1 10 3 10

× 5 × 10–2 towards the straight wire

= 5 4

2 10 43

× 5 × 10–2

= 80

3 × 10–7 newton towards the straight wire

27. Name the three different modes of propagation of electromagnetic waves. Explain,using a proper diagram the mode of propagation used in the frequency rangeabove 40 MHz. [3]

Ans. Three different modes of propagation of electromagnetic waves are

1. Ground (Surface) wave propagation

2. Sky wave propagation

3. Space wave propagation (LOS communication)

For Frequencies about 40 MHz, space wave propagation is being used as theionosphere will not reflect the signals and ground transmission is not possible.In space wave the transmitter and receiver must be on the line of sight (LOS)together.

Such wave propagation used for Television broadcast and Satellite communication.

In this propagation the uplink and downlink frequencies are kept different toavoid the mixingup of the signals.

× · × · × · × · ×

2555/1/3 P.T.O.

6. Why is electrostatic potential constant throughout the volume of the conductorand has the same value (as inside) on its surface ? [1]

Ans. E =0 in a conductor makes Qen

= 0 and charges stay on the surface. So thepotential is constant and will be the same as on the surface.

9. The relative magnetic permeability of a magnetic material is 800. Identify thenature of magnetic material and state its two properties. [2]

Ans. µ > > 1 for ferromagnetic

Ferromagnetic material

(i) Pulls all the field lines inside the material

(ii) Moves towards the strongest field in a non-uniform field.

12. Define mutual inductance between two long coaxial solenoids. Find out theexpression for the mutual inductance of inner solenoid of length l having theradius r1 and the number of turns n1 per unit length due to the second outersolenoid of same length and n2 number of turns per unit length.

Ans.

Solenoid S1

Solenoid S2

Let l be the length of each solenoid S1 and S2

N1 and N2 be the total number of turns of S1 and S2.

Magnetic field in S1 = B1 = 0 1 1N I

Magnetic flux linked with each turn of S2 = B1 × area of each turn = B1r12

Total magnetic flux linked with S2 = B1r12 N2

2 = 0 1 1N I

r12N2 =

20 1 2 1 1N N r I

But magnetic flux linked with S2 is due to I1

2 1I or 2 = M I1

Where M is the mutual inductance of S2 and S1

M I1 = 2

0 1 2 1 1N N r I

M = 20 1 2 1 n n r l

Code No. 55/1/3


UNCOMMON QUESTION ONLY

Series : SMA/1

2655/1/3 P.T.O.

21. Name the three different modes of propagation of electromagnetic waves. Ex-plain, using a proper diagram the mode of propagation used in the frequencyrange from a few MHz to 40 MHz.

Ans. Three different modes of propagation of electromagnetic waves are

1. Ground (Surface) wave propagation

2. Sky wave propagation

3. Space wave propagation (LOS communication)

The type of propagation used upto a frequency of 40 MHz is sky wave propagation.The ions in the ionosphere reflects the signal back to the surface of the earthmaking long distance communication possible. The maximum frequency f

c upto

which this is used depends on the ion density as fc=9 (N

max)1/2

23. A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1A. Astraight long wire carrying 2 A current is kept near the loop as shown. If the loopand the wire are coplanar, find

2.5 cm

1A

2cm

1A

I = 2A

4 cm



Ans. (i) Torque = 0, as there will be forces on the plane of the rectangular coil andwire.

(ii) Forces on the loop = 1 2µ

2o I I

r * length (Newton)

=

72 2

2 1 2 12 10

2 10 4.5 10× 4 × 10–2 towards the straight wire

2755/1/3 P.T.O.

=

5 22 10 1

5× 4 × 10–2

= 7 73 248 10 10

5 5= 4.8 × 10–7 newton towards the straight wire



L

R

C




Ans. (i) = 26

1 1 1

2 1010 40 10LC

= 50 rad/s

(ii) I = 240

460

rms rmsV VA

Z R

(iii) Vrms

across L = 4 LX

= 2

14 10

2 10

= 2000 volt

× · × · × · × · ×

Education

Physics CBSE solution 2012