cbse paper 2

8/10/2019 Cbse Paper 2

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JOSHI CLASSESA PREMIER INSTITUTE FOR QUALITY EDUCATION

Test (+2)

JC

Joshi Classes, SCO5, Top Floor, Sector 20/D, Chandigarh. M:- 08968081307, Ph:- 0172-4604567 Page 1

Duration: 3 Hr 02/02/14 Max. Marks: 70

One Marks Questions:

Q 1. A wire of resistance 4 R is bent to form a circle. What will be the equivalent resistance across diameter PQ?

Q 2. In the circuit shown here, what will happen to the brightness of the lamp L when a solid iron rod is inserted in

the air-cored coil C connected in the circuit?

Q 3. Out of red, blue and green, for which colour will the width of the central maxima in a single-slit diffraction

pattern by minimum?

Q 4. Half-life of a radioactive substance is 20 days. What percentage of a sample will decay in 10 days 25%,

less than 25% or more than 25%?

Q 5. In what medium should two charges be placed so that force between them may become 1/3rd of that in

vacuum?

Q 6. Write down following in decreasing order of their wavelengths: (a) rays, (b) Xrays, (c) microwaves.

Q 7. Show on a graph the variation of the de Broglie wavelength () associated with an electron with the square

root of accelerating potential (V).

Q 8. Name the type of biasing of a p-n junction diode so that the junction offers very high resistance.

Two Marks Questions:

Q 9. Two point charges 10 C and 40 C are placed 6 cm apart in vacuum. At what distance form 10 C charge

will the net electric field be zero?

OR

Two capacitors of capacitances 6 F and 12 F are connected in series with a battery. The voltage across the

6 F capacitor is 2 V. Compute the total battery voltage.

Q 10. An alternating voltage of frequency f is applied across a series LCR circuit. Let frbe the resonance frequencyfor the circuit. Will the current in the circuit lag, lead or remain in phase with the applied voltage when

(a) f > fr, (b) f < fr? Explain your answer in each case.

Q 11. Calculate the maximum energy that can be acquired by a particle of charge q and mass m in a cyclotron in

which there is a magnetic field B and whose dees radius is R.

Q 12. Why input characteristics for various values of VCEin common emitter configuration of n-p-n transistor gives

almost identical curves?

Q 13. Electromagnetic radiations with wavelength

(a) 1are used to kill germs in water purifiers,

(b) 2are used in TV communication systems,

(c) 3play an important role in maintaining the earths warmth.


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JOSHI CLASSES Test (+2)


Name the part of electromagnetic spectrum to which these radiations belong. Arrange these wavelengths in

decreasing order of their magnitude.

Q 14. What is dispersion of light? A ray of white light is incident on one face of a glass prism. Draw an appropriate

ray diagram to show dispersion and deviation of light by the prism. Label the ray diagram suitably.

Q 15. (a) Draw the block diagram of a communication system.

(b) What is meant by detection of a modulated carriers wave? Describe briefly the essential steps fordetection.

Q 16. The value of ground state energy of hydrogen atom is 13.6 eV.

(a) What does the negative sign signify?

(b) How much energy is required to take an electron in this atom from the ground state to the first excited

state?

Three Marks Questions:

Q 17. Figure shows a small magnetized needle P placed at a point O. The arrow shows the direction of its

magnetic moment. The other arrows show different positions of another identical magnetized needle Q.

(a) In which configuration is the system in (i) stable, and (ii) unstable equilibrium?

(b) Which configuration corresponds to the lowest potential energy among all the configurations shown?

Q 18. Draw the graphs showing the variation of photoelectric current with anode potential of a photocell for

(a) The same frequencies but different intensities I1> I2> I3of incident radiations,

(b) The same intensity but different frequencies v1> v2> v3of incident radiation. Explain why the saturation

current is independent of the anode potential.

Q 19. What is diffraction of light? In the case of diffraction at a single slit, explain the formation of central maxima

on a screen at a distance D from a slit of width d when the slit is illuminated with light of wavelength . Write

a formula for the width of the central maximum.

Q 20. Define the term resistivity of a conductor. Give its SI unit. Show the resistance R of a conductor is given by

2

ml

ne A where, symbols have their usual meanings.

Q 21. A wire frame of size 5 mm 2 mm is placed at a distance of 8 cm from a convex lens of focal length 10 cm

held close to the eye.

(a) How big will the wire frame appear to be?

(b) What is the angular magnification produced by the lens?

Q 22. Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions

where the nuclear force is

(i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.

Q 23. A square loop of wire of edge a carries a current i. Show that the value of B at the centre is given by

02 2 i

a


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OR

Two identical circular coils each of radius 10 cm are arranged concentrically with their plane perpendicular to

one another as shown in the figure. If current in each coil is 10 A, what is net magnetic field at the common

centre O?

Q 24. What is LOS Communication and what is its limitation? Name the waves used for it.

Q 25. AB is 2 m along uniform wire of 20 resistance. The other data are as shown in the circuit diagram given

below: Calculate (a) potential gradient along AB, and (b) length AO of the wire, when the galvanometer

shows no deflection. (i) 0.8 V/m (ii) 0.6 V; AO = 75 cm.

Four Marks Questions:

Q 26. Narayan was doing social work during vacation. He visited a village where there was no electricity. He

decided to help the villagers in purchasing solar panels. For this he awared them about the technology and

advantages. Villagers applied for solar panels and got it from government at nominal charges.

(a) What type of a person is Narayan?

(b)

What is the principle of solar cell? How does it work?

Five Marks Questions:

Q 27. You are given tow convex lenses of short aperture having focal lengths 4 cm and 8 cm respectively. Which

one of these will you use as an objective and which one as an eye piece for constructing a compound

microscope? Draw a ray diagram to show the formation of the image of a small object due to a compound

microscope. Derive an expression for its magnifying power.

OR

Derive lens makers formula by considering concave lens.

For a converging lens r1= r2= 24 cm and refractive index 1.6 (a) calculate its focal length in air and (b) if

the lens is split vertically into two identical parts, what is the focal length of each part?

Q 28. Derive an expression for the magnitude of electric field intensity at any point along the equatorial line of a

short electric dipole. Give the direction of electric field intensity at that point. For a short dipole what is the

ratio of electric field intensities at two equidistant points from the centre of dipole? One along the axial line

and another on the equatorial line.

Or

(a) On their inner faces, the plates are having equal and opposite charges (densities of opposite signs and

magnitude 17.0 1022c/m2).

(i)

In the outer region of 1stplate?

(ii) Between the plates in terms of ?


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(b) If the electric field between the plates is 2.25 104NC1and an oil drop having 12 excess electrons is

held stationary between the plates, estimate the radius of the drop.

(Density of the oil = 1.26 g/cm3)

Q 29. Figure given below shows how the reactance of a capacitor varies with frequency.

(a) Use the information on graph to calculate the value of capacity of the capacitor.

(b)

An inductor of inductance L has the same reactance as the capacitor at 100 Hz. Find the value of L.(c) Using the same axes, draw a graph of reactance against frequency for the inductor given in part (ii).

(d) If this capacitor and inductor were connected in series to a resistor of 10 , what would be the

impedance of the combination of 300 Hz?

OR

Draw the curves showing the variation of inductive reactance and capacitive reactance, with applied

frequency of an a.c. source.

A capacitor, resistor of 5 , and an inductor of 50 mH are in series with an a.c. source marked 100 V, 50 Hz.

It is found that voltage is in phase with the current. Calculate the capacitance of the capacitor and the

impedance of the circuit.


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Solutions

Paper - 2

Q 1. Across diameter; resistance of each wire is 2R.

1 1 1

R ' 2R 2R

R =2R

2= 2

Q 2. Brightness of the lamp will decrease. On introducing a solid iron rod in the air-cored coil, its self inductance

will increase. As a result, current it the circuit will decrease and therefore the brightness of the lamp will also

decrease.

Q 3. Width of central maxima in the diffraction pattern of a single slit will be minimum for blue colour.

Q 4. More than 25%.

Q 5. Fm =0

F;

3Now K = 0

m

F

F

K = 3. (Medium of dielectrics constant 3)

Q 6. Microwaves > X-rays > -rays.

Q 7. As =12.27

V

Q 8. Reverse biasing (i.e., n-type p-n junction is joined with higher potential and p-type is joined with lower

potential).

Q 9. Let q1= 10C = 10 106C, q2= 40 C = 40 10

6C and r = 6 cm.

The point where nit electric field due to q1and q2can be zero, will be on the right of q1. Let its distance from

q1be x.

For the net field at P to be zero.

1 2

2 2

0 0

q q1 1. .

4 x 4 r x

Taking q1and q2in C and r and x in cm

2 2 2 2

10 40 1 4, or

x x6 x 6 x

or

1 2

x 6 x

From hereEither 2x = 6 x, x = 2 cm,

or 2x = (6 x) x = 6 cm.


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As x has to be positive and less than 6 cm therefore electric field will be zero at a distance of 2 cm from q1on

the line segment joining q1and q2.

OR

Here total battery voltage is equal to the sum of potential across each capacitor.

V = V1+ V2

Given: C1= 6 F

C2= 12 F

V1= 2 V

Total charge flowing in the circuit is,

Q = C1V1= 6 1062

Q = 12 C

As C2is in series same amount of charge will also flow through it.

Now, V2=6

6

2

Q 12 10

C 12 10

V2= 1 volt.

Thus V = 2 + 1 = 3 volt

Total battery voltage is 3 volt.

Q 10. We know XLf and XC

1

f

As f increases XCdecreases and XLincreases. For a particular value of frequency called resonance frequency

(fr), XLbecomes equal to XC. In case we keep on increasing frequency after this XLbecomes greater than XC.

(a) Hence for f > frcircuit turns inductive in nature and current lags behind the voltage by certain phase

angle.

(b) For f < frcircuit is capacitive in nature as XC> XLand current leads the voltage by certain phase angle.

Q 11. Let v be the maximum speed acquired by the charged particle (charge q, mass m) in a cyclotron whose dees

radius is R.

Here qvB =2mv

,R

or v =qBR

m

Thus, the maximum kinetic energy acquired by the charged particle

=1

2mv2=

1

2(m)

2 2 2 2qBR q B R

m 2m

Q 12. Collector-emitter junction being kept at constant potential, collector current (IC) remains same. On changing

base-emitter voltage (VBE) only base current changes. When we fix value of collector-emitter voltage at a

higher value it changes collector current. Change in base current being very small the input characteristics is

almost same.

Q 13. (a) Ultraviolet rays

(b) Radiowaves

(c) Infrared radiations.


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of Radiowaves > of infrared rays > of ultraviolet rays.

Q 14. Dispersion of Light:

It is the phenomenon of splitting of white light into is constituent colours on passing through a glass prism.

Dispersion and deviation produced by a prism.

Q 15. (a) Block diagram of communication system

(b) Detection is the process of recovering the modulating signal from the modulated carrier wave.

Essential steps for detection

1. Detection of modulating signal

2. Rectification of the signal

3. Envelops detection (e.g. RC circuit)

Q 16. (a) The ve sign signifies that electrons are bounded with the nucleus and that of the force is attractive.

(b) In ground state

E1= 13.6 eV

In 1stexcited state

E2= 213.6

2

= 3.4 eV

Energy required = E2E1= 3.4 (13.6)

= 10.2 eVQ 17. Magnetic field at a point

On the axis of the magnetic needle p is

p0p 3

2MB

4 r

On the equatorial line of magnetic needle p is

p1 0p 3

2MB

4 r

(a) (i) A system is in stable equilibrium if its magnetic moment vector

QM and magnetic field vector of

magnetic needle are in the same direction.

Information

Source

Transmitter ChannelMessagesignal

Transmittedsignal

Re ceivedsignalReceiver

Message

signal

User of

Information

Noise


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(ii) System is in unstable equilibrium if its magnetic moment vector QM and magnetic field vector of

magnetic needle are in the opposite direction.

Thus,

Configuration 6thand 3rdcorrespond to stable and 4thand 5thcorrespond to unstable equilibrium.

As U = M.B, the lowest potential energy corresponds to = 0.

(b) Configuration 6thcorresponds to lowest potential energy, because = 0and BP< BP.

Q 18. (a) Variation in photoelectric current with anode potential of a photocell for same frequency but different

intensities I1> I2> I3of incident radiation.

(b) For same intensity but different frequencies v1> v2> v3of incident radiation.

Saturation current corresponds to the number of photoelectrons emitted by the cathode reaching the

anode at a constant intensity of incident radiation. Anode potential cannot change saturation current.

Q 19. Bending of light around the corner of an obstacle or an aperture is called diffraction.

Central maximum is formed between 1stsecondary minimum on both sides of the centre of the screen.

First minimum is obtained on the screen when

d sin = sin =d

Again tan =y

d. When is small,

y

D d

y =D

d


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Thus width of central maximum is 2y =2 D

d

Q 20. Resistivity is the resistance of the material of conductor of unit length and unit area of cross-section. Its SI

unit is m.

Consider a conductor of length l and area of cross-section A. Let a potential V be applied across it. An electric

field is developed across the conductor that derives the electrons in one direction and constitutes current i init

E =V

l .(i)

We know that

i = neAvd .(ii)

where n is number density of free electrons and vdis drift velocity of free electrons.

Force experienced by an electron is

F = eE

ma = eE

m dv

= eE

vd=eE

m

.(iii)

where is relaxation time.

Substitute the value of vdin equation (ii)

i = neA eE

m

i = neAeV

ml

2

V ml

i ne A

..(iv)

From Ohms law

V

i= R .(v)

On comparing equations (iv) and (v), we get

R =2

ml l

ne A A

Where =2

m

ne

is called resistivity of the conductor.

Q 21. Given: For a convex lens as a magnifier f = 10 cm, u = 8 cm

Size of the object (frame) = 5 mm 2 mm

Position of eye : close to the lens.

To find: (a) size of the frame as seen by the eye,

(b) angular magnification produced by the lens.

(a) As1 1 1

v u f


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1 1 1 1 1

v f u 10 8

,

or v = 40 cm.

So magnification, m =v 40

u 8

=5

Size of the image of frame = 25 mm 10 mm(b) Angular magnification (Magnifying Power)

Q 22.

Force is attractive for distances larger than 0.8 fm and repulsive for separation less then 0.8 fm.

Two characteristic features are

(i) Nuclear force is short-range force.

(ii) Nuclear force is much stronger than coulomb force and gravitational force.

Q 23. Magnetic field at O due to current element idl

is

dB

= 03

idl r

4 r

or dB = 02

idl sin

4 r

.(i)

From the figure sin (180) = sin =a / 2

r

and r2= l2+

2a

2

Therefore, equation (i) becomes

dB = 03/2

22

idla

4 a2 l

4

Magnitude of filed due to entire arm


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0

l 3/22 2

ia dlB

8 a / 4 l

As tan =l

a / 2

l = a/2 tan

dl =a

2sec2d

Bl= 2/4

0

3/22

/4 2

a / 2 sec dia

8 asec

4

On solving Bl=0i

2 a

Field due to entire loop will be four times the field due to one arm, i.e., B = 4Bl

Bl=

02 2 i

a

According to right-hand screw rule, the field is perpendicular to the page and directed into it.

OR

The magnetic field at common centre O due to each coil has same magnitude. The two fields are mutually

perpendicular

|B1| = |B2| =7

0

1

i 4 10 10T

2r 2 10

= 2105T

The net magnetic field B is

B = 2 21 2 1B B 2 B

= 52 2 10 T

B makes an angle of 45with the plane of other coil.

Q 24. LOS stands for line of sight communication. Space waves are used for it. In LOS communication waves travel

in a straight line. Signal more direct from transmitter to receiver. Due to curvature of earth LOS

communication is limited to a fixed distance.

Q 25. E = 4 V; R = 30 ; RAB= 20

Current through the wire i =4

50

Potential difference across wire =2 8

2025 5

V.

(a) Potential gradient =8

5 2= 0.8 V/metre

Current in the galvanometer circuit i' =3

3

(b) Potential difference across AO =3

30.6 = 0.6 V

Length AO =

0.6

0.8 100


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=3

4100 = 75 cm.

Q 26. (a) It seems Narayan is very social person. He wants to help the people with modern technology.

(b) It works on the principle of photovoltaic effect. When photons of appropriate frequency fall on the p-n

junction of solar cell they generate electron-hole pairs. These are separated out by electric field at the

junction. This way emf is developed across the solar cell.Q 27. Lens of focal length 4 cm is used as an objective and lens of focal length 8 cm is used as an eye piece.

Magnifying power of compound microscope is

Where, L = length of microscope tube

D = least distance of distinct vision.

m = e

1 e 1

A" B" / PB"tan A "B"

tan A B"/ P B" A B "

[ A1B = AB]

=A " B " A ' B '

.A 'B ' AB

= moce

But mo=u and me= 1 +

e

Df

m =e

D1

u f

As u fo, L, length of microscope.

m =o e

L D1

f f

Magnifying power can be increased by taking objective and eye piece of small focal lengths.

However, fe> foto increase resolution.

Hence, if focal length 4 cm is to be used as objective and that of focal length 8 cm is to be used as eye piece.

But mo=u

and me= 1 +e

D

f

=e

D1

u f

OR


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When refraction takes place from ratter to denser medium we have (through interface X1P1X1)

1 2 2 1

1 1

n n n n

u R

.(i)

For refraction from denser to rarer medium (through interface X2P2X2)

2 1 2 1

1 2 2

n n n n

R

.(ii)

Adding equation (1) and (2) we get

1 1

u

n1= (n2n1)1 2

1 1

R R

1

f= (n211)

1 2

1 1

R R

(a) Given: r1= r2= 24 cm

= 1.6

(i)1

f= (

1)

1 2

1 1

r r

Here, r1= 24 cm

r2= 24 cm

1

f= (1.6 1)

1 1 0.6 2 1

24 24 24 20

f = 20 cm

(b) We can consider it as a combination of two lenses in contact. If f is the focal length of each half.

1 1 1

f f ' f '

1 2

f f '

i.e., f = 2f

f = 40 cm

Q 28. Electric field at P due to charge at A (EA) = 2 2kq

a r

Electric field at P due to charge at B (EB) = 2 2kq

a r

Thus, EA= EB= 2 2kq

a r

Magnitude of resultant field at point P is


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E = EAcos + EBcos

= 2EAcos 2 2

acos

a r

E = 2

3/2

2 2

kp

a r

For r >> a

E =3

kp

r

The direction of electric field E is opposite to that of dipole moment p .

Electric field a axial line is double of the electric field on equatorial line for a short dipole. At two equidistant

points from the centre of dipole

i.e., Eaxial: Eequatorial: : 2 : 1.

OR

(a) (i) In the outer region of 1stplate electric fields are equal and opposite. So net field is zero.

(ii) Between the plates electric field are in the same directions. So net field is

E =0

=22

12

17.0 10

8.85 10

= 1.92 1010NC1

(b) As the drop is stationary

34 r3

g = Ene

r3

=

3Ene

4 g

r3=4 19

3

3 2.25 10 12 1.6 10

4 3.14 1.26 10 9.81

r = (0.94 1018)1/3

r = (0.94 1018)1/3m

Q 29. (a) XC=1

2 vC

C =C

1

2 vX

Form the graph we take frequency v = 100 Hz. The capacitive reactance corresponding to this frequency

is XC= 6

Thus C =1 1

2 100 6 1200

C = 2.65 104F

(b) According to question XL= XC

2vL = XC

L = 6

2 100 C

X 6

v 100 Hz

L = 9.54 103Hz

(c) XL= 2vL


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v XL

100

200

300

400

5.99 6

11.98 12

17.98 18

23.97 24

(d)

Z = 22

L CR X X

R = 10 , L = 9.54 103H

C = 2.65 104F, v = 300 Hz

XL= 2vL = 600 9.54 103

= 17.98 = 18

XC= 41 1

22 vC 600 2.65 1 0

Z = 100 256 356

Z = 18.9 19

OR

Numerical:

R = 5 , L = 50 mH, v = 50 Hz

As V and I are in phase

(resonance condition for LCR circuit)

XL= XC

C =2 2

1

4 v L=203 F

and Z = R = 5

Z = 2R = 5

cbse paper 2

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