96inter 2nd Year Physics Paper E

8/12/2019 96inter 2nd Year Physics Paper E

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Physics II year Model question paper 2013 2014

Section A

I Very short answer questions 10 2 = 20 M

Answer all questions, each carries two marks.

(1) What are beats?(2)Can there be electric intensity at a point with zero electric potential? Give example.

(3) define resistivity (or) Specific resistance.

(4)Magnetic lines form continuous closed loops. Why?

(5)State Lenzs law.

(6)What is transformer ratio?

(7)Give two uses of infrared rays.

(8) What is work function?

(9) Draw the circuit symbol for p n p and n - p n transistors.

(10) Define modulation. Why is it necessary?

Section B

II Short Answer questions

(i)Answer any six questions. 64 = 24 M

(ii) Each question carries four marks.

(11)What is Doppler Effect? Obtain an expression for the apparent frequency of sound guard when the source

is in motion with respect to an observer at rest.

(12)Explain polarization by refraction.

(13)State Gausss law in electrostatics and explain its importance.

(14)Derive an expression for the equivalent capacity when capacitors are connected in series.

(15)State Kirchhoffs law for an electrical network. Using these laws deduce the condition for balance in wheat

stone bridge.

(16)Derive an expression for the magnetic induction at a point on the equatorial line of a bar magnet.

(17)Discuss Bohrs theory of the spectrum of hydrogen atom.

(18)What is rectification? Explain the working of a Full wave rectifier.

Section C

IIILong answer4 questions 28 = 16 M

(i) Answer any two questions (ii) Each question carries eight marks

(19) Define Snells law. Using a neat labeled diagram derive an expression for the refractive index of the

material of an equilateral prism.

(b)A ray of light, after passing through a medium, meets the surface separating the medium from air at an

angle of 45and is just not refracted. What is the refractive index of the medium?

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E

C

B

E C

B

L

L + Vs To VsTo

VsO

S1

S2

P n p transistor n- p n transistor

(10) The process of combining audio frequency signal with high frequency signal is called modulation.

Necessity: -

1) Low frequencies cannot be transmitted to long distances. Hence the modulation is necessary.

2) To reduce the size of antenna

3) To avoid mixing up of signals from different transmitters.

Section B

(11)Due to relative motion b/w the source and observer, the frequency received by the observer is different

from that of the source. This phenomenon is known as Doppler Effect.

Ex: - The whistle of the approaching train towards an observer at rest appears to be higher than the

original frequency.

Source in motion Observed at rest: -

Let us consider an observer at O who is stationary in a frame in which the medium is also at rest. Let the

source of sound is initially S1 and moving away form the observer with a velocity Vs

Let the angular frequency of sound wave be W. Time period be T0 and speed of sound wave be V.





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Reflected

Transmitted

Plane polarized light

Unpolarized

32.5

When t = 0, the source is at S1, and the distance of L from the observer and emit a crest. This reaches the

observer at time t1= L/V when t = T0the source has move a distance VsTo and is at a points S2

= +

When t = nTothe source emits its (n + 1)th crest and this reaches the observer at time

= + Hence, in a time interval +

the observer receives n crests and the observer record the

period of the wave as T = + /n

The inverse of T gives the frequency received by the observer at rest.

=

1+

=

= , =

By neglecting higher powers the above equation may be approximated.

This gives the frequency received by the stationary observer, when the source is moving away

from the observers with a velocity. If the source is approaching the observer, VSis to be replaced by - VS, thenthe frequency received by the observers is =

(12) Pile of plates: -A number of glass plates are arranged parallel to each other. This is known as pile of

plates.

These piles of plates are fixed in a suitable tube with an inclination of 32.5 to the axis of the table.When unpolarised light is allowed to incident on the first plate at the angle of polarization (57.5 ) part of the

light reflects and remaining part refracts.

After many refractions, the final refracted rays are free from vibrations parallel to the plane of incidence.

Hence plane polarized light is said to be produced due to refraction.





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s

+-

C1 C2 C3

+q-q-q+q

+q

(13)Gauss law: -Total electric flux through a closed surfaced is equal to

times the total charge enclosed by

the surface.

. =

Where q is the total charge enclosed by the surface is the permittivity of free space.Importance: -1) Gauss theorem is vaid for stationary charges as well as for rapidly moving charge.

2) Gauss theorem holds good for any closed surface of any shape.

3) Gauss law gives the relation between the electric field and the charge.

(14)Condensers in series: -(or) Capacitors in series: -

If number of capacitors is connected in such a way so that the charge on the plates of every one of them is

same, then the condensers are said to be connected in series.

Let three condensers of capacities,, are connected in series as shown fig. Thecharge on the plates of the capacitors is same but the P.D across the capacitors of capacities ,,are,,respectively. And V be the P.D. across the combination of the capacitors.

= + + (1)But = , =

, = V =

,

=

+

+

=

+

+

(2)

The reciprocal of the equivalent capacity is equal to the sum of reciprocal values of the capacities of thecombination.

(15)Kirchhoffs first law (or) current law: -

The sum of current flowing into Junction is equal to sum of currents flowing out of same junction.

V





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i1i2

i3

i4i5

i1+i2+ i4+ = i3+ i5

i1+i2 - i3+i4- i5= 0

i = 0

i.e. algebraic sum of current meeting at a junction is equal to zero.

Kirchhoffs second law (or) voltage law: -

In any closed circuit, the algebraic sum of all potential difference is zero.

V = 0 iR = VWheatstone bridge: -

Wheat stone bridge is an electric circuit used to compare resistances or to find the value of unknown

resistance.

It consists of four resistances P, Q, R, S that are connected to form four sides of a quadrilateral.

These four sides are referred as arms of the bridge. Four junctions are formed at A, B, C and D.

A batter y of emf is connected between two junctions A and B.

A galvanometer resistance G ohm is connected between C and D.

Applying Kirchhoffs first law

At junction i1= ig+i3 (1)At junction i2+ig= i4 (2)

Applying Kirchhoffs second law for the loop ACDA

i1p- igG+ i2R=0 (3)





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R

N

d

Q

E

d2+l2

S0l l

Applying Kirchhoffs second law for the loop CBDC

i3Q+ i4S+ igG=0 (4)The values of P, Q, R, and S are suitably adjusted so that no current should pass through galvanometer. Then

wheat stone bridge is said to be balanced. Substituting ig= 0 in equation (1), (2), (3), and (4) we get

i1 = i3 (5) and i2 = i4 (6)

i1p= i2R (7) and i3Q= i4S (8)

Using equations (5) and (6) and dividing eq (7) by (8) we get=

this is the principle of Wheat Stone bridge.

(16)Let us consider a bar magnet of pole strength m and length 2l. O be the mid point on the bar magnet

Line drawn at perpendicular to NS is the equatorial line of the bar magnet. Let E be the point on the

equatorial line and it is at a distance ofd from O.Now NE = SE = + and the magnetic induction at E due to morth pole is given by BN and is

the direction of NP

=

()

=

()

Similarly, the magnetic induction at E due to the South Pole is B and is in the direction of ES

BS=04

m

(SE)2=

04

m

d2+l2

BNand BSare adjacent sides of a parallelogram EPRQ. The ER the diagonal of the parallelogram gives the

resultant field induction BE from similar triangles EPR and NES. We have =

(Or) =

() =

= 04

m

d2+l2

= 04 ()/ ( = 2)The direction of the field induction BE on the equatorial line is always opposite to the direction of the

magnetic moment.

(17)Bohrs theorem of spectrum of hydrogen atom: -

Hydrogen atom has one proton and one electron in the ground state. By the absorption of energy, electron

jumps from the ground state to the higher energy level ( =)with the absorption of 13.58 ev. An atom

P





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becomes unstable. In order to get the stability the electron should comes back to lower energy level with

mission of energy

While de-excitation of electron, the electron directly comes to the first energy level (or) it may

while de-excitation of electron, the electron directly comes to the first energy levels.

Since many atoms are involved, hence it produces large number of spectral lines in the hydrogen

spectrum. Five series of spectral lines are formed. They are: -

The wave number and wave length of these spectral lines can be calculated by using Rydbergs equation.

=

R Rydbergs constant = 109677

n = 7

n = 6

n = 5 P fond series

n = 4 Brackelt series

n = 3 Pascher seriesn = 2 Balmer series

n = 1 lyman series (U.V region)

(18)Rectification: -The process of converting an alternating current into a direct current is called rectification.

Full wave rectifier: -A rectifier which rectifier both half cycles of the ac input is called full wave rectifier. A full

wave rectifier can be constructed with two diodes D1 and D2, centre tap transformer load resistance R2and

a.c input as in centre tap transformer, the secondary coil is wound into two equal parts.

Name of the series n1 n2 Spectral region

1) Lyman series 1 2, 3, 4, 5 ----- UV region

2) Blamer series 2 3, 4, 5, 6 ------- Visible region

3) Paschem series 3 4, 5, 6, 7-------- Near IR

4) Brackett series 4 5, 6, 7 ---------- Middle IR

5) P fund series 5 6, 7, 8, 9--------- Far IR





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During positive half cycled, end A becomes positive and end B becomes ve. This makes diode D 1 forward

bias and diode D2 reverse bias. Than D1 conducts and D2does not. So o/p voltage is obtained through load

resistance (R2) due to diode D1

During negative half cycle, end A becomes negative and B becomes positive. The diode D1reverses bias and

D2 forward bias. So D2conduct and D1does not. So output obtained across R2 is due to D2

The full wave rectifier 81.2% of a.c is converted into D.C

The efficiency of the full wave rectifier

= . where = diode resistance

= load resistance

Section C

(19) (a) Snells law: -The ratio of the size of the angle of incidence to the size of angle of refraction is constantThis is constant is known as refractive index of the medium.

= Where i is the angle of incidence, r is the angle of refraction.

Expression for the refractive index: -

Let us consider an equilateral triangular prism ABC.





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30 40 50 60700

20

40

60

i = e

A

BC

SP

Q R

M

i er1 r2

N

A ray PQ is incident on the first surface AB and makes and angle of incidence i with the normal drawn to

AB. It refracts into the prism and makes an angle r called angle of refraction, with the normal drawn to AB

The light ray travels along QR in the prism and makes angle of incidence r2on second surface AC of the prism.

Finally after refracting from prism the light ray with the normal to AC and is called

angle of emergence. The angle b/w the emerged ray RS and the direction of the incident ray PQ is called the

angle of deviation.

The quadrilateral AQNR two of the angle are right angles. Therefore, the sum of the

other angles of the quadrilatera180

A + QNR = 180 (1)

From the triangle QNR, r1+ r2+ QNR = 180 (2)

Comparing eq (1) and (2) we get r1+ r2= A (3)

The total deviation is the sum of deviations at the two faces. = (i r1) + (e r2)

i.e. = i+ e - A (4)

A plot between the angle of deviation and angle of incidence

From the graph, it can be under stood that for any given value, of except for i = e, corresponds to two values i

and hence of e. remains the same if i and e are inter changed.At minimum deviation Dmthe refracted ray in side the prism becomes parallel to its base.

Then, = Dm i = e and r1= r2equation (3) gives, 2r = A (or) r = A/2

equation (4) gives Dm= 2i A (or) i = (A + Dm)/2





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0

A

B

C

D

P

Y

i

Y P

0

Q

R

B

= Bil

= Bil

= Bil = Bil cos = Bil = Bil cos

the refractive index of the prism = =()/

/

(b)Angle made by the ray with the surface = 45

Angle of incidence (or) critical angle = 90 45 = 45

Angle of refraction r = 90

It is just not refracted.

Refractive index = = = / = 2 = 1.414

(20)

Torgue on current loop

Consider a rectangular current loop ABCD having length AB = CD = l and breadth AD = BC = b and carrying

current be suspended a magnetic field of induction (or) flux density B with normal on to its field direction.

There are four forces and are acting on the four sides AB, BC, CD, and AD of the loopmagnitude of these forces are

By fleming left hand rule, force and acts on sides BC, and AD exactly in opposite direction along sameline. Hence they cancel.

By Flemings left hand rule the two equal forces

and acts on vertical sides of the loop in opposite

directions at different points on the same loop, if the coil is made to rotate about vertical axis YY

The moment of couple (or) torque,

= force perpendicular distance between the two forcesHere, F = = = Bil sin

= PR= Bilsin sin = = sin = sin

= BilsinIf area of the loop is A = lb, then = Bilsin





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i Hence the deflection of the coil is directly proportional to the steady current passing through it.

(21) Principle A nuclear reactor is a device in which nuclear fission can be carried out through a self sustained

and a controlled chain reaction. It is also called an atomic pile.

Neutron multiplication factor (k): -The ratio of the number of neutrons generated in the present stage to the

number of neutrons generated in the previous stage is called as neutron multiplication factor.

Coolant

Refractor

Steam to turbine

Control rods head exchanger

Water from condenser

The essential component of a nuclear reactor is

(1) Fuel (2) Modular (3) control rods (4) Radiation shielding (5) Coolant

(1)Fuel: -The fissionable material used in the reactor is called nuclear fuel. The nuclear fuel is taken in the

form of rods and sealed in aluminum cylinders.

Ex: Enriched branium, U235

enriched plutonium (or) U233

etc.

(2)Moderator: -Moderator is a substance which slows down the fast moving neutrons produced during the

nuclear fission process. The average energy of neutrons released in the fission process is 2 Mev. They are slow

down to thermal neutrons (0.025ev) by moderator.

Ex: - Heavy water, graphite, Hydrocarbon plastics etc.

(3)Control rods: -The fission rate in the reactor is controlled by using the neutron absorbing materials.

Ex: - Cadmium and Boron rods.(4) Shielding: -During fission reaction ()and gama() radiation are emitted along with neutrons. Suitanleshielding such as steel lead and concrete walls are provide.

(5)Coolant: -The heat generated in fuel elements is removed by using a suitable coolant to flow around them

The coolent used are water at high pressures, molten Sodium etc.

Working: -Uranium fuel rods are placed in the aluminum cylinders which are separated by some distance. The

graphite moderator is placed in between the fuel cylinders.

The control rod is kept in the holes of the graphite block. When a few U 235 nuclei undergo fission, fast

neutrons are liberated.

Core





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These matrons pass through graphite moderator and loose their energies to become thermal neutrons. These

thermal neutrons participate in further fission process.

By proper adjustment of the control rods in the core the fission events are suitably controlled.

The heat generated is used for heating the coolant head is transferred to water and steam is produced.

This steam is used to turn the steam turbine. This steam turbine is used to drive a generator to produce

electricity or to convert into mechanical energy.



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96inter 2nd Year Physics Paper E

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