ii p u c annual examination march-2018 scheme of … · 2018-07-13 · ii p u c annual examination...

II P U C ANNUAL EXAMINATION MARCH-2018

SCHEME OF EVALUATION

PHYSICS (33) (NEW SYLLABUS)

I PART-A

1 An equipotential surface is a surface with constant value of potential at all points on the surface

1

2 The average velocity with which the free electrons are drifted in a direction opposite to the applied field is called drift velocity

1

3 Cyclotron is used to accelerate the charged particles or ions to high energies 1

4 The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

1

5

√

1

6 Flux leakage/ resistance of the windings/ eddy currents/ hysteresis loss (any one)

1

7 P=P1+P2= 1.5-0.5= 1 D 1

8 The decay of proton to neutron is possible only inside the nucleus since proton has smaller mass than neutron.

1

9 The space charge region on either side of the p-n junction together is known as depletion region.

1

10 1 1

II PART-B

11 Area of the plates/ distance between the plates/ permittivity of the medium between the plates (any two)

2

12 Junction rule: At any junction the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. Loop rule: the algebraic sum of the changes in potential around any closed loop involving the resistors and cells in the loop is zero.

1 1

13 Declination: The angle between the true geographic north and the north shown by the compass needle is called declination. Dip: The angle between the direction of total intensity of magnetic field of earth and a horizontal line in the magnetic meridian.

1 1

14

= magnetic potential energy, = magnetic dipole moment, B= uniform magnetic field.

1 1

15 a) X-rays are used as diagnostic tool in medicine. b) To treat certain forms of cancer.

1 1

16 Myopia is a defect in human eye in which light from the distant objects arriving at eye lens get converged at a point in front of the retina OR unable to see the distant objects clearly Using concave lens

1 1

17

2

18 1. Strongest of all fundamental forces in nature. 2. Charge independent force 3. Short range force 4. Saturated force 5. Spin dependent force 6. Non-central force (any two)

2

III PART-C

19 1. Additivity of charges: charges can be added algebraically 2. Charge is conserved: the total charge of the isolated system is always

conserved. 3. Quantisation of charge: all free charges are integral multiples of a basic

unit of charge ‘e’.

1 1 1

20 Ampere’s circuital law: The integral of the product of magnetic field and length element is equal to times the total current passing through the surface.

∮

1

consider a point at a distance ‘r’ from a straight infinite current carrying wire.

From Ampere’s circuital law ∮

1 1

21. Hysteresis: The phenomenon of lagging behind of magnetic induction (B) with respect to the magnetizing field (H) is called hysteresis. Coercivity: it’s a phenomenon of completely demagnetizing the magnetic material by applying the magnetizing field in the opposite direction. Retentivity: the property of the magnetic material to retain magnetism even in the absence of the magnetizing field is known as retentivity.

1 1 1

22

AB = plane wave front

= direction of propagation of wave front

= surface of separation of medium 1 1nd 2

= speeds of light in medium 1 and 2 respectively

i = angle of incidence

r = angle of refraction

= time taken by the wave front to travel a distance BC

To determine the shape of the refracted wave front a sphere of radius is

drawn and = tangent plane if the refracted wave front.

1

Consider triangles ABC and AEC

------------ (1)

If C = speed of light in vacuum then,

,

(1)

1 1

23

1. An electron in an atom could revolve in certain stable orbits without the

emission of radiant energy.

2. Electron revolves around the nucleus only in those orbits for which the angular

momentum in some integral multiple of

where h = Planck’s constant.

.

3.When an electron make a transition from one non- radiating orbit to another of

lower energy a photon of energy equal to the energy difference between the two

states is emitted.

1 1 1

24

We have

when ⁄,

⁄

⁄

⁄

⁄

⁄

1 1 1

25

(any three)

n – type p – type

1 Obtained by doping pure

semiconductor by pentavalent

element like As, Sb, P.

Obtained by doping pure

semiconductor by trivalent

element like Al, B, In.

2 Majority charge carriers are

electrons and minority charge

carriers are holes

Majority charge carriers are holes

and minority charge carriers are

electrons.

3 Donor energy level lies close to

the conduction band.

Acceptor energy level lies close

to the valence band.

4 ne>>>nh nh>>> ne

3

26

3

IV PART-D

27 Electric potential is defined as the work done in bringing a unit positive charge

from infinity to that point against the direction of the field.

Consider a point ‘P’ at a distance ‘r’ from a positive charge +Q. To calculate the

potential at this point consider another point P| at a distance ‘r

|’ from +Q

Electrostatic force on a unit +ve charge at P| is given by

( ) Where is the unit vector along OP

|

Work done against this force from r| to r

|+Δ r

| is

1 1

{-ve sign because Δ r

|<0, Δw is +ve }

Total work done = ∫

( )

=

By definition work done in bringing unit positive charge from infinity to that

point is equal to potential

1 1 1

28

Consider two cells of emf andcorresponding internal resistances are

connected in parallel.

Since as much charge flows in as out, we have

Both the cells are at same potential since they are in parallel.

For first cell

For second cell

,

(

)

(

) ( ) ( )

( )

( ) ( )

1 1 1

( )

( )

( )

( )

( )

( )

1 1

29

Consider a current loop of radius R carrying a steady current I. let ‘P ‘ be a point

at a distance ‘x’ from the center of the current loop on its axis.

Let ‘dl’ be the current element as shown. The magnetic field due to it is given by

Biot- savart’s law.

| |

but

The displacement vector from to the axial point P is in the X-Y plane.

Hence

| |

| |

The component of magnetic field along X- direction is dBx= dB cosθ

( ) ⁄

( ) ⁄

( )

( ) ⁄

1 1 1

( ) ⁄ over the loop

Therefore the total magnetic field at P due to the loop

( ) ⁄

( ) ⁄

( ) ⁄

( ) ⁄

1 1

30

Consider a voltage source t connected to a series LCR with

inductance L, capacitance C and resistance R.

If q is the charge on the capacitor and ‘I’ the current at any time ‘t’ then from

Kirchhoff’s rule we have,

From the diagram ( ) where = phase difference between

voltage across the source and the current in the circuit.

are along same line but in opposite direction.

the magnitude

[ ]

1 1 1

( )

( )

[ ( )

]

√ ( ) √ ( )

√ ( )

1 1

31

OM = u = object distance

MI = v = image distance

MC = R = radius of curvature

Angle i = angle of incidence

Angle r = angle of refraction

ON = incident ray

NI = refracted ray

NC = normal & n1, n2 are the refractive indices

From the figure for small angles

In the triangle NOC, = exterior angle = sum of the interior opposite angles

Similarly

From Snell’s law

For small angles

Substituting the values of i& r we get

(

) (

)

1 1 1 1

Applying sign convention, OM = -u , MI = v MC = R

1

32 Rectifier is a device which converts alternating current (ac) into direct current (dc)

The circuit connections are as shown in the figure. In the positive half cycle of

input AC diode is forward biased and conducts. In the negative half cycle of

the input AC diode is forward biased and is reverse biased. gives output

across . The input and output waveforms are as shown. The output is a

pulsating DC.

1 1 1 2

33

( )

( )

( √ )

Resultant of EA& EC at D √

√ Total field at D due to all the charges ED + EB = 12.72 X 104 + 4.5 X 104 = 17.22 X 104 NC-1.

1 1 1 1 1

34

Density

length

Substituting the values given

( )

R= 5.71Ω

1 1 1 1 1

35 Emf induced in the coil = ( ) Area = Substitution E= (25 X 3.14 X 10 X 10-2 X 10 X 10-2 X 5 X 10-2 X 40) ( ) Induced emf = 1.57 ( ) V OR = 1.57 V

1 1 1 1

Current A OR {Note: since it is not mentioned either instantaneous or maximum or average value of induced emf in the question, if a student calculates by mentioning any of the above values marks can be awarded, also for the other values of θ=ωt}

1

36

Fringe width

Given the distance of the fifth dark fringe from the central bright fringe = 1.35 cm

Wavelength of light used λ=600 X 10-9 m Taking distance between the slits and the screen as 1m Calculation of fringe width = 2.142 X 10-3 m

1 1 1 1 1

37

or

( )

( )

Calculation of

1 1 1 1 1

NOTE: Any other alternate correct answers should be considered.