CBSE PHYSICS (Theory) – 2015 - Shivdas · PHYSICS (Theory) – 2015 (COMPTT. DELHI) 111 Shiv Das...

110 SHIV DAS SENIOR SECONDARY SERIES (XII)

Shiv Das i

λ

KE

Time allowed : 3 hours Maximum marks : 70

GENERAL INSTRUCTIONS

Same as in Delhi Board 2015.

SECTION A

Q.1. A particle of mass ‘m’ and charge ‘q’ moving with velocity ‘v’ enters the region of uniformmagnetic field at right angle to the direction of its motion. How does its kinetic energy getaffected? 1

Sol. Kinetic energy will NOT be affected.

*(When v→

is perpendicular to B→

, then magnetic field provides necessary centripetal force)

Q.2. Figure shows a current carrying solenoid movingtowards a conducting loop. Find the direction of thecurrent induced in the loop. 1

Sol. Clockwise on the side of observer.Q.3. When an object is placed between f and 2f of a concave mirror, would the image formed

be (i) real or virtual and (ii) diminished or magnified? 1Ans. (i) Real (ii) magnifiedQ.4. Draw a plot showing the variation of de Broglie wavelength of electron as a function of

its K.E. 1Sol.

Q.5. Why is the frequency of outgoing and incoming signals different in a mobile phone? 1Ans. To avoid overlapping of signals.

SECTION B

Q.6. Using the concept of drift velocity of charge carriers in a conductor, deduce therelationship between current density and resistivity of the conductor. 2

CBSE

PHYSICS (Theory) – 2015(COMPTT. DELHI)

PHYSICS (Theory) – 2015 (COMPTT. DELHI) 111

Shiv Das

Sol. The drift velocity is given by,

vd = emE τ …(i)

where e = charge of electron E = Intensity of electric fieldm = mass of electron τ = relaxation time

The current is given by,I = ne A vd …(ii)

Putting value of vd from equation (i) to equation (ii) we get,

I = ne A emE τ I =

ne

m

2A τE

Current density, J = IA

AA= ne

m

2τE J =

nem

2τ

⎛⎝⎜

⎞⎠⎟

E

J =

1ρ E ∴ ρ =

mne2τ

where ρ is the resistivity of conductor.Q.7. Distinguish between unpolarized and a linearly polarized light. Describe, with the help

of a diagram, how unpolarized light gets linearly polarized by scattering. 2Ans. Unpolarized light : A light wave, in which the electric vector oscillates in all possible

directions in a plane perpendicular to the direction of propagation is known as unpolarizedlight.Linearly polarized light : If the oscillations of the electric vectors are restricted to just onedirection, in a plane perpendicular to the direction of propagation, the corresponding lightis known as linearly polarized light.It is due to scattering of light by molecules of earth’satmosphere.Under the influence of the electric field of the incident(unpolarized) wave, the electrons in the molecules acquirecomponents of motion in both these directions. Charges,accelerating parallel to the double arrows, do not radiateenergy towards the observer since their acceleration has notransverse component.The radiation scattered by the molecules is therefore represented by dots, i.e., it is polarizedperpendicular to plane of figure.

Q.8. Why does white light disperse when passed through a glass prism?Using lens maker’s formula, show how the focal length of a given lens depends upon thecolour of light incident on it. 2

Ans. (i) The white light disperses when passed through a prism, because the refractive indexof the glass of the prism is different for different wavelengths (colours). Hence,different colours get bent along different directions.

(ii) Using lens maker’s formula,

1f = (n21 – 1)

1 1

1 2R R−⎛⎝⎜

⎞⎠⎟

…where ⎡

⎣⎢ n21 =

nn

2

1

As the refractive index of the medium (n2) (glass) with respect to air (n1) depends on thewavelength or colour of light, therefore focal length of the lens would change withcolour.

⎡

⎣⎢

To observer

Incident Sunlight(Unpolarized)

Scattered Light(Polarized)


Shiv Das

Q.9. Using the graph shown in the figure for stoppingpotential v/s the incident frequency of photons, calculatePlanck’s constant. 2

Sol. According to Einstein’s photoelectric equation,

V0 = he eν

φ− 0

In the given graph :Stopping potential, V0 = 1.23 VChange in frequency, Δν = (8 × 1014 – 5 × 1014) = 3 × 1014 Hz

∴ Slope of the line = he

∵

he

V= =

×0

141 23

3 10Δν. , ∵ e = 1.6 × 10–19 C

∴ h =

1 23 1 6 103 10

19

14. .× ×

×

− JS = 6.6 × 10–34 JS

Q.10. Complete the following nuclear reactions : 2

(a) 510

01

24B He++ ⎯⎯ →→⎯⎯ ++ ……n (b) 42

9412

4395Mo H Te++ ⎯⎯ →→⎯⎯ ++ ……

Sol. (a) 510

01

24

37B He Li+ ⎯ →⎯ +n (b) 42

9412

4395

01Mo H Te+ ⎯ →⎯ + n

OrIf both the number of protons and neutrons in a nuclear reaction is conserved, in whatway is mass converted into energy (or vice verse)? Explain giving one example.

Sol. Explanation for release of energy in a nuclear reaction : Since proton number and neutronnumber are conserved in a nuclear reaction, the total rest mass of neutrons and protons is thesame on either side of the nuclear reaction.But total binding energy of nuclei on the left side need not be the same as that on the righthand side. The difference in binding energy causes a release of energy in the reaction.Examples :

(i) 12

12

23

01H H He energy+ ⎯ →⎯ + +n

(ii) 92235

01

56144

3689

013U Ba Kr energy+ ⎯ →⎯ + + +n n

SECTION C

Q.11. An electric dipole of dipole moment p→→ is placed in a uniform electric field E

→→. Obtain the

expression for the torque ττ→→experienced by the dipole. Identify two pairs of perpendicularvectors in the expression. 3

Ans. (i) See Q. 28 (Or) (a), 2014 (I Outside Delhi). [Page 66(ii) Two pairs of perpendicular vectors are,

(a) τ→ is perpendicular to p→ (b) τ→ is perpendicular to E

→

Q.12. (a) Two spherical conductors of radii R1 and R2 (R2 > R1) are charged. If they areconnected by a conducting wire, find out the ratio of the surface charge densities onthem.

(b) A steady current flows in a metallic conductor of non-uniform cross-section. Which ofthese quantities is constant along the conductor : current, current density, electricfield, drift speed? 3

5 × 1014 H Z

2 4 6 8 10v(1014 Hz)

1.8

1.4

1.23

0.8

0.4

V0

(Vol

ts)


Shiv Das

Sol. (a) When two charged spherical conductors of Radii R1 and R2 respectively (R2 > R1) areconnected by a conducting wire, we know that the common potential (V) is given by,

V =

qc

qc

1

1

2

2=

∵ C for a spherical conductor, C = 4π∈0R,

We have,

q q1

0 1

2

0 14 4πε πεR R= ⇒

q q1

1

2

2R R= ⇒

qq

1

2

1

2=

RR

σσ

1

2=

qq

1

0 12

0 22

24

4

πε

πε

R

R× ∵ σ =

q4 0πε R2

=

qq

1

2

22

12×

R

R =

RR

R

R

RR

2

1

1

2

22

12× =

∴

σσσσ

1

2

2

1==

RR

(b) CurrentQ.13. In the two electric circuits shown in the figure, determine the reading of ideal ammeter (A)

and the ideal voltmeter (V). 3

Sol. (i) In circuit (a)Total emf = 15 V, Total Resistance = 2Ω

Current, i =

152

⎛⎝⎜

⎞⎠⎟

A = 7.5 A

Potential difference between the terminals of 6 V batteryV = E – iR = [6 – (7.5 × 1) = – 1.5 V

(ii) In circuit (b)Effective emf = (9 – 6)V = 3V, Total Resitance = 2Ω

Current, i =

32

⎛⎝⎜

⎞⎠⎟

A = 1.5 A

Potential Difference across 6 V cell,V = E – (– I)R = V + IR ( ∵Current is in opposite direction to 6 V cell) = 6 + 1.5 × 1 = 7.5 V

Or

In the circuit shown in the figure, find the current througheach resistor.

Sol. Total emf of the circuit = 8 V – 4 V = 4 VSince two resistors 3Ω and 6Ω are connected in parallel,their combined resistance is

1

R′ =

1 1

1 2R R+ = 13

16

36

12+ = = R′ = 2Ω

–+– +V

1Ω6V

A

1Ω9V

V

1Ω6V

A

1Ω9V

(a) (b)

4V 8V 1.0Ω

0.5Ω

3.0Ω

4.5Ω

6.0Ω

i i


Shiv Das

Since all resistors are in series, R = 0.5 + 4.5 + 2 + 1 = 8Ω

Current through circuit, I = VR = 4

8 = 0.5 A

0.5ΩΩΩΩΩ A current will flow in 3 resistors, i.e., 0.5ΩΩΩΩΩ , 4.5ΩΩΩΩΩ and 1ΩΩΩΩΩ.Current through resistors 3Ω and 6Ω

V = I1 × 3 = I2 × 6 = (I – I1) × 6 I1 = 13 A

[I2 = I – I1]

I2 = (I – I1) =

12

13− =⎛

⎝⎜⎞⎠⎟

16 A

Q.14. Define the following using suitable diagrams : (i) magnetic declination and (ii) angle ofdip. In what direction will a compass needle point when kept at the (i) poles and(ii) equator? 3

Ans. (i) Magnetic declination : Angle betweenmagnetic meridian and geographical meridian

(ii) Angle of dip : It is the angle which themagnetic needle makes with the horizontal inthe magnetic meridian .

(iii) (a) Direction of compass needle is vertical to the earth’s surface at poles.(b) Parallel to the earth’s surface at equator.

Q.15. Derive the expression for the magnetic energy stored in a solenoid in terms of magneticfield B, area A and length l of the solenoid carrying a steady current I. How does thismagnetic energy per unit volume compare with the electrostatic energy density stored ina parallel plate capacitor? 3

Sol. Magnetic energy in a solenoid,

Rate of work done, ddtW = |ε| I = (LI)

dIdt

∵ dW = LIdI

Total amount of work done,

d dW LI I= ∫∫ , W = 12 LI2

For the solenoid, we knowInductance, L = μ0n2Al; also, B = μ0nI

∴ W = UB = 12 LI2 =

12 ( μ0n2Al)

Bμ0

2

n⎛⎝⎜

⎞⎠⎟

=

B A2

02l

μ =

B V2

02μμ

∵ Magnetic energy per unit volume (EM) =

B2

02μ =

12 0μμ

B2 …(i)

N

S

Dec

linat

ion

NmP

ZE

BEI

HE

D

True North

⎯ →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

→⎯⎯⎯⎯⎯⎯⎯⎯


Shiv Das

The electrostatic energy stored per unit volume for a parallel plate capacitor,

ES = 12 ∈0E2 …(ii)

These two expressions are similiar in nature.Q.16. A circuit containing an 80 mH inductor and a 250 μμμμμF capacitor in series connected to a 240

V, 100 rad/s supply. The resistance of the circuit is negligible.(i) Obtain rms value of current.

(ii) What is the total average power consumed by the circuit? 3Sol. (i) XL = ωL = 100 × 80 × 10–3 = 8 Ω

XC =

1ωC =

1100 250 10 6× × − Ω = 40 Ω

Total impedence, (Z) = XC – XL = 32 Ω

Irms = V

Rrms = 240

32 = 7.5 A

(ii) Average power consumed = 0 (Zero) (As there is no ohmic resistance in the current)Q.17. Answer the following questions:

(i) Why is the thin ozone layer on top of the stratosphere crucial for human survival?Identify to which a part of electromagnetic spectrum does this radiation belong andwrite one important application of the radiation.

(ii) Why are infrared waves referred to as heat waves? How are they produced? What roledo they play in maintaining the earth’s warmth through the greenhouse effect? 3

Ans. (i) The thin ozone layer on the top of stratosphere is crucial for human survival, becauseit absorbs ultraviolet radiations from the Sun and thus prevents them from reaching theearth’s surface causing damage to life.Identification : ultraviolet radiations.Its correct application is Sanitization.

(ii) (a) Water molecules present in most materials readily absorb infra red waves. Hence,their thermal motion increases. Therefore, they heat their surroundings and are hencereferred to as heat waves.

(b) They are produced by hot bodies and molecules.(c) Incoming visible light is absorbed by earth’s surface and radiated as infra red

radiations. These radiations are trapped by green house gases.Q.18. Define the term ‘critical angle’ for a pair of media.

A point source of monochromatic light ‘S’ is kept at the centre of the bottom of a cylinderof radius 15.0 cm. The cylinder contains water (refractive index 4/3) to a height of 7.0 cm.Draw the ray diagram and calculate the area of water surface through which the lightemerges in air. 3

Ans. (a) Critical Angle : For an incident ray, travelling from an optically denser medium tooptically rarer medium, the angle of incidence, for which the angle of refraction is 90°,is called the critical angle.

μ =

1sin ic

or ic = sin–1

1μ

⎛⎝⎜

⎞⎠⎟

(b)

C

A

B

3 (ic)4

S15 cm

x

←→

⎯⎯7 cmic

L = 80 mH C = 250 μF

240 V, 100 Rads

7


Shiv Das

μ =

1sin ic

= 43

∴ sin ic = 34 = AB

AC Q BC = 4 3 16 9 72 2− = − =

cos ic = BCAC = 7

4 tan ic =

ABBC = 3

7From figure,

tan ic =

x x x7

37 7

3 7⇒ = ⇒ = cm

Area = πx2 = 63π cm2

∴ The area of water surface through which the light emerges in air is 63πππππ cm2.Q.19. Which two of the following lenses L1, L2 and L3 will you select as objective and eyepiece

for constructing best possible (i) telescope (ii) microscope? Give reason to support youranswer.

Lens Power (P) Aperture (A)L1 6D 1 cmL2 3D 8 cmL3 10 D 1 cm 3

Ans (i) Telescope : L2 : objective, L3 = eyepieceReason : Light gathering power and magnifying power will be larger.

(ii) Microscope : L3 : objective, L1 = eyepieceReason : Angular magnification is more for short focal length of objective and eyepiece.

Q.20. Explain by drawing a suitable diagram that the interference pattern in a double slit isactually a superposition of single slit diffraction from each slit.Write two basic features which distinguish the interference pattern from those seen in acoherently illuminated single slit. 3

Ans. See Q. 20, 2017 Comptt. (II Outside Delhi). [Page 269Q.21. Distinguish between n-type and p-type semi-conductors on the basis of energy band

diagrams. Compare their conductivities at absolute zero temperature and at roomtemperature. 3

Sol. Distinction between n-type and p-type semiconductors on the basis of energy leveldiagram : See Q. 20, 2015 Comptt. (Outside Delhi). [Page 131

Q.22. (a) Given a block diagram of a generalized communication system.

Identify the boxes ‘X’ and ‘Y’ and write their functions.(b) Distinguish between “Point to Point” and “Broadcast” modes of communication.

(Not in Syllabus) 3Ans. (a) X : Transmitter Y : Channel

Their function:Transmitter : To convert the message signal into suitable form for transmission throughchannel.Channel : It sends the signal to the receiver.

InformationSource ⎯ →⎯⎯⎯

Message

SignalYX ⎯ →⎯⎯⎯ ⎯ →⎯⎯⎯ Receiver ⎯ →⎯⎯⎯ User

Message

Signal

Transmitted

Signal


Shiv DasO

IN

90°

M

I

(b) In point to point mode, communication takes place between a single transmitter andreceiver while in broadcast mode, large number of receivers are connected to a singletransmitter.

SECTION D

Q.23. Ameen had been getting huge electricity bill for the past few months. He was upset aboutthis. One day his friend Rohit, an electrical engineer by profession, visited his house.When he pointed out his anxiety about this to Rohit, his friend found that Ameen wasusing traditional incandescent lamps and using old fashioned air conditioner. In additionthere was no proper earthing in the house. Rohit advised him to use CFL bulbs of 28 Winstead of 1000 W – 220 V and also advised him to get proper earthing in the house. Hemade some useful suggestion and asked him to spread this message to his friends also.*(i) What qualities/values, in your opinion did Rohit possess?(ii) Why CFLs and LEDs are better than traditional incandescent lamps?

(iii) In what way earthing reduces electricity bill? 4Ans. *(i) Value Based Questions will not be asked in the Board Examination.

(ii) (a) Low operational voltage and less power.(b) Fast action and no warm up time required.

(iii) In the absence of proper earthing, the consumer can get bill for charges for the electricalenergy NOT consumed by the devices in his premises.

SECTION E

Q.24. (a) Use Biot-Savart law to derive the expression for the magnetic field due to a circularcoil of radius R having N turns at a point on the axis at a distance ‘x’ from its centre.Draw the magnetic field lines due to this coil.

(b) A current ‘I’ enters a uniform circular loop ofradius ‘R’ at point M and flows out at N as shownin the figure.Obtain the net magnetic field at the centre of theloop. 5

Sol. (a) Biot-Savart Law. See Q. 25 (a) (Or), 2015 (Outside Delhi). [Page 106Magnetic field lines due to circular coil.


Shiv Das

(b) Let current I be divided at point M into two partsI1 and I2 ; in bigger and smaller parts of the looprespectively. Magnetic field of current I1(clockwise) at point O

B1

→ =

μ0 12

14

IR × ⊗

Magnetic field of current I2 (anticlockwise) at point O

B2

→ =

μ0 22

34

IR × .

Net magnetic field, B→

= B1

→ + B2

→

| B→

| = μ μ0 1 0 28

38

IR

IR− …(i)

But I1 = 3I2 (As resistance of bigger part is three times that of the smaller part of the loop.)

Substituting I1 = 3I2 in equation (i), we get | B→

| = 0∴ Magnetic field at the centre of loop is zero.

Or(a) Show how Biot-Savart law can be alternatively expressed in the form of Ampere’s

circuital law. Use this law to obtain the expression for the magnetic field inside asolenoid of length ‘l’, cross-sectional area ‘A’ having ‘N’ closely wound turns andcarrying a steady current ‘I’.Draw the magnetic field lines of a finite solenoid carrying current I.

(b) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended bytwo vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

Find the magnitude and direction of the magnetic field which should be set up inorder that the tension in the wire is zero.

Ans. (a) (i) Biot-Savart law and Ampere’s Circuital law. Biot-Savart law can be expressed asAmpere’s Circuital law by considering the surface to be made up of a largenumber of loops. The sum of the tangential components of the magnetic fieldmultiplied by the length of all such elements leads to integral. Ampere’s circuitallaw states that this integral is equal to μ0 times the total current passing throughthat surface, i.e.,

B→→

∫∫→→

. dl = μμμμμ0 I

(ii) Expression for magnetic field inside asolenoid. Let ‘n’ be the number of turnsper unit length. Then total number ofturns in the length ‘h’ is nh.Hence, total enclosed current = nhIUsing Ampere’s circuital law,

B dl→ →

∫ . = μ0nhI

Bh = μ0nhIB = μ0nI

B P

ω

b

chQ

d

a

I2

O90°

NI

I1

MI


Shiv Das

(iii) Magnetic field lines of a finite solenoid

(b) As per the figure given, magnetic field must bevertically inwards, to make tension zero. Therefore,force on current carrying conductor and the weightof conductor are equal and opposite; and balanceeach other.∴ IlB = mg

or B =

mglI

= × ××

−60 10 9 85 0 0 45

3 .. .

T = 0.26 T

Direction : Perpendicular to the direction of both current and force and inwards.Q.25. (a) Draw a schematic arrangement of Geiger-Marsden experiment showing the scattering

of ααααα-particles by a thin foil of gold. Why is it that most of the ααααα-particles go rightthrough the foil and only a small fraction gets scattered at large angles?Draw the trajectory of the ααααα-particle in the coulomb field of a nucleus. What is thesignificance of impact parameter and what information can be obtained regarding thesize of the nucleus?

(b) Estimate the distance of closest approach to the nucleus (Z = 80) if a 7.7 MeVααααα-particle before it comes momentarily to rest and reverses its direction. 5

Ans. (a) (i) Geiger-Marsden experiment

(ii) For most of the α-particles, impact parameter is large, hence they suffer very smallrepulsion due to nucleus and go right through the foil.

P•

• Q

l = 0.45 m

T TIlB

I(5A)

mg(60 × 10–3 × 9.8)

Thin gold foil

Small anglescattering of most

Lead bricks

•Source ofα-particles

Beam ofα-particles

ZnS screen

θ

Large anglescattering of someBackward scattering

of a very smallfraction (1 in 8000 or so)

Detector(Microscope)


Shiv Das

(iii) Trajectory of ααααα-particles : See Q.26 (Or) (a), 2015 Comptt. (Outside Delhi). [Page 135(b) K.E. of the α-particle = Potential energy possessed by beam at distance of closest

approach.

12

14

2

0 0mv e e

r=

∈π. ( )( )2 Z

7.7 × (1.6 × 10–13) =

( ) ( . )9 10 2 1 6 10 809 19 2

0

× × × × ×−

r

r0 =

9 10 2 2 56 10 807 7 1 6 10

9 38

13× × × × ×

× ×

−

−.

. . m

= 299 × 10–16 m = 29.9 × 10–15 m ≈ 30 × 10–15 mOr

(a) Write two important limitations of Rutherford model which could not explain theobserved features of atomic spectra. How were these explained in Bohr’s model ofhydrogen atom?Use the Rydberg formula to calculate the wavelength of the Hααααα line.(Take R = 1.1 ××××× 107 m–1).

(b) Using Bohr’s postulates, obtain the expression for the radius of the nth orbit in hydrogenatom.

Ans. (a) Limitations of Rutherford Model:(i) Electrons moving in a circular orbit around the nucleus would get accelerated,

therefore it would spiral into the nucleus, as it looses its energy.(ii) It must emit a continuous spectrum.

Explanation according to Bohr’s model of hydrogen atom(i) Electron in an atom can revolve in certain stable orbits without the emission of

radiant energy.(ii) Energy is released/absorbed only, when an electron jumps from one stable orbit to

another stable orbit. This results in a discrete spectrum.Wavelength of Hααααα lineHα line is formed when an electron jumps from nf = 3 to ni = 2 orbit. It is the Balmer series

1 12

132 2λ = −⎛

⎝⎜⎞⎠⎟

R

1 14

191 107

λ = × −⎛⎝⎜

⎞⎠⎟

1. or λ = 656.3 nm

(b) Radius of nth orbit

We have,

mvr

ern n

2

0

2

21

4= πε . ⇒ rn =

evn

2

024πε

…(i)

From Bohr’s Postulates: mvnrn = nh2π , vn =

nhmrn2π

Substituting for vn, in equation (i), we get

rn =

εεππ0

2 2

2n hme

Q.26. (a) Figure shows the input waveform which is converted by adevice ‘X’ into an output waveform. Name the device andexplain its working using the proper circuit. Derive theexpression for its voltage gain and power gain.

(Not in Syllabus)

DeviceX


Shiv Das

(b) Draw the transfer characteristic of a base biased transistor in CE configuration.Explain clearly which region of the curve is used in an amplifier.

(Not in Syllabus) 5Ans. (a) (i) Name of device is common emitter amplifier.

(ii) Circuit diagram. See Q. 20 2015 (Outside Delhi). [Page 99(iii) Expression for voltage gain and power gain

Power gain, Ap = Current gain × Voltage gain

= (βac) (Av) = βac

βac rRL⎛

⎝⎜⎞⎠⎟

= βac r

2 RL

(b) Transfer characteristic of base-biassed transistor in CE configuration:Transistor as a switch. The circuit diagram of transistor as a switch is shown inFigure 1.Transfer characteristics. The graph between V0 and Vi is called the transfercharacteristics of the base-biased transistor, shown in Figure 2.

When the transistor is used in the cut off or saturation state, it acts as a switch.As long as Vi is low and unable to forward bias the transistor, then V0 is high. If Vi ishigh enough to drive the transistor into saturation, then V0 is low. When the transistoris not conducting, it is said to be switched off and when it is driven into saturation, itis said to be switched on. This shows that a low input switches the transistor off anda high input switches it on.Transistor acts as an amplifier in the active region.

Or(a) Explain briefly, with the help of circuit diagram, the working of a full wave rectifier.

Draw its input and output waveforms.(b) Identify the logic gate equivalent to the circuit

shown in the figure.Draw the truth table for all possible values ofinputs A and B. (Not in Syllabus) 5

Ans. (a) p–n junction diode as full waverectifier. A full wave rectifier consists oftwo diodes and special type oftransformer known as centre taptransformer as shown in the circuit.The secondary of transformer gives thedesired a.c. voltage across A and B.During the positive half cycle of a.c.input, the diode D1 is in forward bias

VBB

RB

IB

C

ERC

VCC

ICV0

+

–

IE

Transfer characteristicFigure 2

← →⎯⎯⎯← →⎯⎯⎯V0

AV

Vi

← →⎯⎯⎯

Saturationregion

Cutoff

region Activeregion

Base-biased transistor in CE configurationFigure 1

B

Vi

Diode D1A

Secondary coil

RLX Y

BDiode D2

Centre taptransformer

Primarycoil

Figure 1

A

B

Y′


Shiv Das

Input Output

A B Y0 0 00 1 01 0 01 1 1

and conducts current while D2 is inreverse biased and does not conductcurrent. So we get an output voltageacross the load resistor RL.During the negative half cycle of a.c.input, the diode D1 is in reverse biasedand does not conduct current whilediode D2 is in forward biased andconducts current. So we get an outputvoltage across the load resistor RL.

(b) Identification: AND GateTruth table:

Please Note : The questions asked in Set II and Set III are identical to Set I. Simply the serialnumbers of questions were changed.

Due toD2

Output waveform

Inpu

t vo

ltage

Vol

tage

acr

oss

RL

Input waveformt

t

Due toD1

Due toD2

Due toD1

Figure 2

PHYSICS (Theory) – 2015 (COMPTT. OUTSIDE DELHI) 123

Shiv Das

XC

BA

YA B Y

0 0 10 1 11 0 11 1 0

ω




SECTION A

Q.1. Why are electric field lines perpendicular at a point on an equipotential surface of aconductor? 1

Ans. If the electric field lines were not normal to the equipotential surface, it would have a non-zero component along the surface. To move a unit test charge against the direction of thecomponent of the field, work would have to be done which means this surface cannot beequipotential surface.Hence, electric field lines are perpendicular at a point on an equipotential surface of aconductor.

Q.2. A variable frequency AC source is connected to a capacitor. Will the displacement currentchange if the frequency of the AC source is decreased? 1

Sol. On decreasing the frequency of AC source, reactance, XC =

1ωC will increase, which will lead

to decrease in conduction current. In this caseID = IC

Hence, displacement current will decrease.Q.3. Draw the logic symbol of NAND gate and give its Truth Table. (Not in Syllabus) 1

Ans. Symbol of NAND GateTruth table

Q.4. Plot a graph showing variation of capacitive reactance with the change in the frequencyof the AC source. 1

Ans. Graph showing a variation of capacitive reactance with thechange in frequency of AC source.

CBSE

PHYSICS (Theory) – 2015(COMPTT. OUTSIDE DELHI)


Shiv Das

A

B C

1Ω 5Ω D 4Ω

2Ω

2Ω

4V

4V

D

B

2Ω = R

4Ω = S

5Ω2Ω = Q

A1Ω

= P

I1I2

I1

C

I2

2A

1A

20 cm

20 cm

I

30 cm

10 cm

Q.5. Distinguish between amplitude modulation and frequency modulation.(Not in Syllabus) 1

Ans. The amplitude modulation provides a larger coverage area, while frequency modulationprovides a better quality transmission.

SECTION B

Q.6. Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field

E→→

== ××3 103 i N/C through a square of side 10 cm, when it is held normal to E→→

? 2Sol. See Q. 24 (a), 2015 (I Outside Delhi). [Page 103

Given : E = 3 103× i N/C

A = 10 × 10 cm2 = 10

10010

100× m2

φ = E A→ × → = EA cos θ ∵ θ = 0 and cos θ = 1

= EA

= (3 × 103) ×

10100

10100×⎛

⎝⎜⎞⎠⎟

= 30 Nm2 C–1

Q.7. Calculate the current drawn from the battery by thenetwork of resistors shown in the figure. 2

Ans. Given : Circuit diagram can be rearranged as shownbelow :It forms a wheatstone’s bridge

PQ

RS= ⇒

12

24

12= =

It is the condition of null point when no current flowsthrough BD arm, i.e. 5 Ω.Resistances P = (1 Ω) and R = (2 Ω) are in series;

R1 = 1 + 2 = 3 ΩSimilarly, Resistances Q = (2 Ω) and S = (4 Ω) are in series,

R2 = 2 + 4 = 6 ΩNow, R1 and R2 are in parallel,

1 1 1 13

16

121 2R R R= + = + = ⇒ R = 2 Ω

I = VR = 4

2 = 2A∴ Current in the circuit is 2A.

Q.8. A square loop of side 20 cm carrying current of 1A iskept near an infinite long straight wire carrying acurrent of 2A in the same plane as shown in the figure.Calculate the magnitude and direction of the net forceexerted on the loop due to the current carryingconductor. 2

Sol. Given : l = 20 cm = (20 × 10–2) mI1 = 1 A, r1 = 10 cm = 10 × 10–2 mI2 = 2 A, r2 = 30 cm = 30 × 10–2 m

F = μ0I1 I2l

1 1

1 2r r−⎛⎝⎜

⎞⎠⎟


Shiv Dasn = 4

n = 3

n = 2

n = 1

= (2 × 10–7) × (1) × (2) × (20 × 10–2)

110 10

130 102 2× ×

⎡⎣⎢

⎤⎦⎥− −−

= 5.3 × 10–7 NThe direction of force is towards the infinitely long straight wire.

OrA square shaped plane coil of area 100 cm2 of 200 turns carries a steady current of 5A. Itis placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil.Calculate the torque on the coil when its plane makes an angle of 60° with the directionof the field. In which orientation will the coil be in stable equilibrium?

Sol. Given : A = 100 cm2 = 100 × (10–4) m2 = 10–2 mN= 200 turns, I = 5A, B = 0.2 Tθ = 90° – 60° = 30°τ = NIAB sin θ

= (200) × (5) × (10–2) × (0.2) ×

12

⎛⎝⎜

⎞⎠⎟

sin 3012° =⎡

⎣⎢⎤⎦⎥

= 1 NmThe coil will be in stable equilibrium when it is parallel to the magnetic field.

Q.9. Name the types of e.m. radiations which (i) are used in destroying cancer cells, (ii) causetanning of the skin and (iii) maintain the earth’s warmth.Write briefly a method of producing any one of these waves. 2

Ans. (i) γ-rays (ii) Ultraviolet rays (iii) Infrared raysMode of production

(i) γγγγγ-rays are produced by radioactive decay of nucleus.(ii) Ultraviolet rays are produced when inner shell electrons in atoms move from one

energy level to another energy level.(iii) Infrared rays are produced due to vibration of atoms and molecules.

Q.10. The figure shows energy level diagram of hydogenatom.(a) Find out the transition which results in the

emission of a photon of wavelength 496 nm.(b) Which transition corresponds to the emission of

radiation of maximum wavelength? Justify youranswer. 2

Sol. (a) Transition emitting wavelength λ = 496 nmThe given wavelength lies in visible region (Balmer series) when,

n1 = 2, n2 = 4

1 1 1

12

22λ = −

⎛

⎝⎜⎞

⎠⎟R

n n[R = 1.097 × 107 m–1]

1 12

14

3161 097 10 1 097 107

2 27

λ = × − = × ×⎛⎝⎜

⎞⎠⎟

. .

⇒ λ = 163

101 097

7×

−

. = 486 nm ≈ 496 nm

Hence the transition from, n2 = 4 to n1 = 2(b) Transition corresponding to emission of maximum wavelength,

E = hv = hcλ or E ∝

1λ


Shiv Das

0 5 10

5

I(A)

t(s)

which means that the maximum wavelength emission will be there when the energylevel difference is minimum. From the given energy level diagram, it corresponds to :

n2 = 4 to n1 = 3.

SECTION C

Q.11. (a) Deduce the relation between current I flowing through a conductor and drift velocity

Vd→→

of the electrons.

(b) Figure shows a plot of current ‘I’ flowing through thecross-section of a wire versus the time ‘t’. Use the plotto find the charge flowing in 10s through the wire. 3

Ans. (a) Relation between electric current and drift velocity : See Q. 26 (i), 2017 (I Delhi).[Page 207

(b) Charge = Current × time

(i) For 0 to 5 second, the current = 0 5

2+ = 2.5 A

Hence, q1 = (2.5) × (5) = 12.5 C(ii) For 5 sec to 10 sec, the current = 5A

Hence, q2 = 5 × (10 – 5) = 5 × 5 = 25 Cq = q1 + q2 = 12.5 + 25 = 37.5 C

Q.12. Draw a circuit diagram of a potentiometer. State its working principle. Derive thenecessary formula to describe how it is used to compare the emfs of the two cells. 3

Ans. Principle : The basic principle of a potentiometer is that “when a constant current flows througha wire of uniform cross-sectional area and composition, the potential drop across any length of thewire is directly proportional to that length.”

v ∝ l.Comparison of emf’s of two cells: First of all the ends of potentiometer are connected toa battery B1, key K and rheostat Rh such that the positive terminal of battery B1 is connectedto end A of the wire. This completes the primary circuit.

10


Shiv Das

Now the positive terminals of the cells C1 and C2 whose emfs are to be compared areconnected to A and the negative terminals to the jockey J through a two-way key and agalvanometer (fig). This is the secondary circuit.Method:

(i) By closing key K, a potential difference is established and rheostat is so adjusted thatwhen jockey J is made to touch at ends A and B of wire, the deflection in galvano-meteris on both sides. Suppose in this position the potential gradient is k.

(ii) Now plug is inserted between the terminals 1 and 3 so that cell C1 is included in thesecondary circuit and jockey J is slided on the wire at P1 (say) to obtain the null point.The distance of P1 from A is measured. Suppose this length is l1 i.e. AP1 = l1∴ The emf of cell C1, ε1 = kl1 …(i)

(iii) Now plug is taken off between the terminals 1 and 3 and inserted in between the terminals2 and 3 to bring cell C2 in the circuit. Jockey is slided on wire and null deflection positionP2 is noted. Suppose distance of P2 from A is l2 i.e. AP2 = l2∴ The emf of cell C2, ε2 = kl2 …(ii)

Dividing (i) by (ii), we get

εε

1

2

1

2= l

l…(iii)

Thus emf’s of cells may be compared. Out of these cells if one is standard cell, then theemf of other cell may be calculated.

OrWith the help of the circuit diagram, explain the working principle of meter bridge. Howis it used to determine the unknown resistance of a given wire? Write the necessaryprecautions to minimize the error in the result.

Ans. See Q. 26 (a), 2017 Comptt. (I Delhi). [Page 247Precautions:

(i) In this experiment the resistance of the copper strips and connecting screws havenot been taken into account. These resistances are called end-resistances. Thereforevery small resistances cannot be found accurately by metre bridge. The resistanceS should not be very small.

(ii) The current should not flow in the metre bridge wire for a long time, otherwise thewire will become hot and its resistance will be changed.

Q.13. (a) Why is the magnetic field radial in a moving coil galvanometer? Explain how it isachieved.

(b) A galvanometer of resistance ‘G’ can be converted into a voltmeter of range (0 – V)volts by connecting a resistance ‘R’ in series with it. How much resistance will berequired to change its range from 0 to V/2? 3

Sol. (a) The magnetic field in a moving coil galvanometer is made ‘radial’ to keep the magneticfield ‘normal’ to the area vector of the coil. It is done by taking the cylindrical soft ironcore. The torque acting on the coil is maximum (sin θ = 1, when, θ = 90°)

(b) Given : resistance of galvanomter = G ΩRange of voltmeter (RL) = (0 – V) voltsResistance to be connected in parallel = R

R′ = ?, where range is

0 2−⎛⎝⎜

⎞⎠⎟

V Volts

In first case, ig =

VR G+ …(i)


Shiv Das

In second case, ig =

V / 2R G′ + …(ii)

[ig is the maximum current which can flow through galvanometer]From equation (i) and (ii), on solving we get

R′′′′′ =

R G−−⎛⎛⎝⎝⎜⎜

⎞⎞⎠⎠⎟⎟2

Q.14. A source of ac voltage V = V0 sin ωωωωωt is connected to a series combination of a resistor ‘R’and a capacitor ‘C’. Draw the phasor diagram and use it to obtain the expression for(i) impedance of the circuit and (ii) phase angle. 3

Ans. Phasor diagram and circuit diagram for the given circuit are,

Expression for Impedance and phase angle : A resistor and a capacitor are connected inseries to a source of alternating current, V = V0 sin ωtLet ‘I’ be the instantaneous value of current in this circuit.

(i) Voltage across ‘R’ = VR = RI (in phase)(ii) Voltage across ‘C’ = VC = XCI (lags by 90°)

V = OA OD2 2+ =

V VR L2 2+ = I

R

C)2

21+⎛

⎝⎜⎞⎠⎟(ω

VI =

R

C)2

21++

(ωω = Z ( ∵ XC =

1

ωC )

Which is the effective resistance of L – C circuit and is called its ‘impedance’.Phase angle :

tan φ = XR

C

φ = tan–1

XR

C⎛⎝⎜

⎞⎠⎟

= tan–1

1ωCR

⎛⎝⎜

⎞⎠⎟ ∴ φφφφφ = tan–1

1

R CωωQ.15. A closely wound solenoid of 2000 turns and cross sectional area 1.6 ××××× 10–4 m2 carrying a

current of 4.0 A is suspended through its centre allowing it to turn in a horizontal plane.Find (i) the magnetic moment associated with the solenoid, (ii) magnitude and directionof the torque on the solenoid if a horizontal magnetic field of 7.5 ××××× 10–2 T is set up at anangle of 30° with the axis of the solenoid. 3

Sol. Given : n = 2000 turns, A = 1.6 × 10–4 m2

I = 4.0 A, B = 7.5 × 10–2 T

θ = 30°,

sin 30 12° =⎛

⎝⎜⎞⎠⎟

(i) Magnetic moment (M)M = NIA = (2000) × (4) × (1.6 × 10–4) = 1.28 Am2

CR

I

V = V0 sinωt

VL

O Iφ

V

PD

← →⎯⎯⎯⎯VR A

V


Shiv Das

(ii) Magnitude of torque

τ→ = M B→ × → = MB sin θ

= (1.28) × (7.5 × 10–2) × 12 = 48 ××××× 10–3 Nm

(iii) Direction of torque is perpendicular to both the planes of the solenoid and themagnetic field.

Q.16. (a) The ratio of the widths of two slits in Young’s double slit experiment is 4 : 1. Evaluatethe ratio of intensities at maxima and minima in the interference pattern.

(b) Does the appearance of bright and dark fringes in the interference pattern violate, inany way, conservation of energy? Explain. 3

Sol. (a) Given :

ww

1

2

41=

ww

aa

1

2

12

22

41= = ∴

aa

1

2

21= = 2

We know,

II

I II I

max

min

( )( )= =

+−

+−

⎛⎝⎜

⎞⎠⎟

1 2

1 2

1 2

1 2

2a aa a

Dividing both numerator and denominator by a2, we get

aaaa

1

2

1

2

2

21

1

2 12 1

91

+

−

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

+−

⎛⎝⎜

⎞⎠⎟

= = ∴

II

max

min== 9

1

(b) The appearance of bright and dark fringes in the interference pattern does not violatethe principle of conservation of energy, because the light energy is distributed. If itreduces in one region, producing a dark fringe, it increases in another region, producinga bright fringe. There is no gain or loss of energy.

Q.17. (a) Write the factors by which the resolving power of a telescope can be increased.(b) Estimate the angular separation between first order maximum and third order

minimum of the diffraction pattern due to a single slit of width 1 mm, when light ofwavelength 600 nm is incident normal on it. 3

Ans. (a) Factors for increasing the resolving power of telescope : The resolving power of atelescope is given by,

R.P =

d1 22. λ where, d = diameter of aperture of objective lens

λ = Wavelength of light usedFrom the above expression, we conclude :

(i) R.P ∝ d, so by increasing aperture of objective lens, R.P is increased.

(ii) R.P ∝ 1λ , so by decreasing the wavelength of light used, R.P is increased.

(iii) Resolving power is independent of focal length of objective lens.(b) Given : a = 1 mm = 1 × 10–3 m, λ = 600 nm = 600 × 10–9 m

(i) Condition for minima is,a sin θ1 = nλ or a sin θ1 = 3λ [ ∵ n = 3]

θ1 = 3λa …(i) [ ∵ for small angles, sin θ ≈ θ]

⎡

⎣⎢⎢


Shiv Das

(ii) Condition for maxima is,

a sin θ2 = (2n + 1) λ2 [ ∵ n = 1]

a sin θ2 = 32λ

θ2 = 32λa …(ii) [ ∵ for small angles, sin θ ≈ θ]

From the equations (i) and (ii), we have

Δθ = (θ1 – θ2) = 3 3

232

λ λ λa a a− =

Putting the value of λ and a, we have

Δθ =

3 600 102 10

9

3× ×

×

−

−( ) = 9 × 10–4 radian

Angular separation = 9 ××××× 10–4 radianQ.18. (a) Good quality sun-glasses made of polaroids are preferred over ordinary coloured

glasses. Justify your answer.(b) Two polaroids P1 and P2 are placed in crossed positions. A third polaroid P3 is kept

between P1 and P2 such that pass axis of P3 is parallel to that of P1. How would theintensity of light (I2) transmitted through P2 vary as P3 is rotated? Draw a plot ofintensity ‘I2’ vs the angle ‘θθθθθ’ between pass axes of P1 and P3. 3

Ans. (a) Such good quality goggles made of polaroids are preferred over ordinary colouredglasses because they protect the eyes from the glare.

(b) See Q. 9 (Or), 2015 (I Delhi). [Page 80

I′′ = I04 . sin2 2θ, I2 =

I04 . sin2 2θ [ ∵ I′′ = I2]

From here, we have corresponding values of I2 with θ :θ → 0 45° 90° 135° 180°

I2 → 0 I04 0

I04 0

Hence graph will be

P1

P3

P2

90 –

θ

θ

I2 = I0I

IncidentIntensity

45° 90° 135° 180°

⎯ →⎯ θ

I2 ⎯→

⎯

I0/4


Shiv DasEC

EgEA

≈0.01 – 0.05 eVEV

(b) T > OK(a) T > OK

EC

Eg

EV

≈0.01 eVED

⎯→

⎯⎯

⎯⎯

⎯⎯⎯

Ele

ctro

n en

ergy

One thermally generated electron-holepair +9 electrons from donor atoms

Q.19. Complete the following nuclear reactions : 3

(a) (i) 84208

82204Po Pb⎯⎯ →→⎯⎯ ++ …… (ii) 15

321632P S⎯⎯ →→⎯⎯ ++ ……

(b) Write the basic process involved in nuclei responsible for (i) βββββ– and (ii) βββββ+ decay.(c) Why is it found experimentally difficult to detect neutrinos?

Sol. (a) (i) 84208

82204

24Po Pb He⎯ →⎯ + (α-decay process)

(ii) 1532

1632

10P S⎯ →⎯ + − e (β-decay process)

(b) β– decay takes place according to

01

11

10n p e⎯ →⎯ + −

“one neutron is converted into proton”while β+ decay occurs according to

01

01

10p n e⎯ →⎯ + +

“one proton is converted into neutron”(c) See Q. 6, 2014 (I Outside Delhi). [Page 55

Q.20. Draw the energy band diagram of (i) n-type and (ii) p-type semiconductor at temperature,T > 0K.In the case n-type Si semiconductor, the donor level is slightly below the bottom ofconduction band whereas in p-type semiconductor, the aceceptor energy level is slightabove the top of the valence band. Explain, what role do these energy levels play inconduction and valence bands. 3

Ans. Distinction between n-type and p-type semiconductors on the basis of energy leveldiagram :

Role of energy levels in conduction and valence bonds : In the energy bond diagram of n-type Si semiconductor, the donor energy level ED is slightly below the bottom EC of theconduction band and electrons from this level moves into conduction band with very smallsupply of energy. At room temperature, most of the donor atoms get ionised, but very few(~ 10–12) atoms of Si atom get ionised. So the conduction band will have most electronscoming from donor impurities, as shown in the figure.For p-type semiconductor, the acceptance energy level EA is slightly above the top EV of thevalence bond. With very small supply of energy, an electron from the valence bond can jumpto the level EA and ionise the acceptor negatively. At room temperature, most of the acceptoratoms get ionised leaving holes in the valence bond.

Q.21. Draw a plot of transfer characteristic (V0 vs Vi) and show which portion of thecharacteristic is used in amplification and why?Draw the circuit diagram of base bias transistor amplifier in CE configuration and brieflyexplain its working. (Not in Syllabus) 3


Shiv Das

Sol. (i) Transistor as a switch. The circuit diagram of transistor as a switch is shown inFigure 1.Transfer characteristics. The graph between V0 and Vi is called the transfer characteris-tics of the base-biased transistor, shown in Figure 2.When the transistor is used in the cut off or saturation state, it acts as a switch.As long as Vi is low and unable to forward bias the transistor, then V0 is high. If Vi is highenough to drive the transistor into saturation, then V0 is low. When the transistor is notconducting, it is said to be switched off and when it is driven into saturation, it is saidto be switched on. This shows that a low input switches the transistor off and a high inputswitches it on.

(ii) See Q. 20, 2015 (I Outside Delhi). [Page 99Q.22. Explain the following terms in relation to the use of internet :

(i) Internet surfing (ii) Social networking (iii) E-mail 3Ans. Not in syllabus

SECTION D

Q.23. Immediately after school hour, as Bimla with her friends came out, they noticed that therewas a sudden thunderstorm accompanied by the lightening. They could not find anysuitable place for shelter. Dr. Kapoor who was passing thereby in his car noticed thesechildren and offered them to come in his car. He even took care to drop them to thelocality where they were staying. Bimla’s parents, who were waiting, saw this andexpressed their gratitude to Dr. Kapoor.*1. What values did Dr. Kapoor and Bimla’s parents display?2. Why is it considered safe to be inside a car especially during lightening and

thunderstorm?3. Define the term ‘dielectric strength’. What does this term signify? 4

Ans. *1. Value Based Questions will not be asked in the Board Examination.2. During thunder storm and lightening, a strong electric field exists in the atmosphere. A

car can be compared with a conductor with cavity. The electric field inside the cavity ofa conductor is zero, hence inside the car, they are electrostatically shielded.

3. Dielectric strength : It is the intrinsic property of a material and is the ability of thematerial to act as an insulator. The delectric strength is the maximum electrical potentialgradient that a material can withstand without breakdown (rupture).Its SI unit is Vm–1 and is expressed generally in megavolt per meter.Its values for some known materials in MV m–1 is :

Air : 3.0, Alumina : 13.4, Water : 65, Mica : 118It signifies that the materials start conducting and cease to be insulators beyond theirdielectric strength.

VBB

RB

IB

C

ERC

VCC

ICV0

+

–

IE

Transfer characteristicFigure 2

← →⎯⎯⎯← →⎯⎯⎯V0

AV

Vi

← →⎯⎯⎯

Saturationregion

Activeregion

Base-biased transistor in CE configurationFigure 1

B

Vi

Cut offregion


Shiv Das

SECTION E

Q.24. (a) State Lenz’s law. Use it to predict the polarity ofthe capacitor in the situation given :

(b) A jet plane is travelling towards west at a speed of 1800 km/h.(i) Estimate voltage difference developed between the ends of the wing having a span

of 25 m if the earth’s magnetic field at the location has a magnitude of 5 ××××× 10–4 T anddip angle is 30°.

(ii) How will the voltage developed be affected if the jet changes its direction fromwest to north? 5

Ans. (a) Lenz’s law : It states that “the direction of induced emf is such that it tends to produce acurrent which opposes the change in magnetic flux producing it.”Polarity of capacitor : Current induced in the coil will oppose the approach of themagnet, therefore, left face of the coil will act as N-pole and right face S-pole. For thisthe current in coil will be anticlockwise, as seen from left; and hence the plate ‘A’ of thecapacitor will be positive and plate ‘B’ will be negative.

(b) (i) Given: v = 1800 km h–1 = 1800 × 103 × (60 × 60)–1 = 500 ms–1

l = 25 m, B = 5 × 10–4 T, θ = 30°e = Bv lv = [(5 × 104) (sin 30°)] × 25 × 500

= 3.125 V [sin 30° = 12

(ii) When the jet changes its direction from west to north, the voltage developed will notbe affected.

OrDefine mutual inductance of a pair of coilsand write on which factors does it depend.A square loop of side 20 cm is initially kept 30cm away from a region of uniform magneticfield of 0.1 T as shown in the figure. It is thenmoved towards the right with a velocity of 10cm s–1 till it goes out of the field.Plot a graph showing the variation of

(i) magnetic flux (φ) through the loop with time (t)(ii) induced emf (ε) in the loop with time t.

(iii) induced current in the loop if it has resistance of 0.1 ΩΩΩΩΩ. 5Ans. Mutual inductance: See Q. 24 (Or) (a), 2015 (I Delhi). [Page 88

Factors :(i) Separation between the coils

(ii) Relative orientation of coils(iii) Number of turns in the coils

Given : Side of square loop (l) = 20 cm = 20 × 10–2 mB = 0. 1 T, R = 0.1 Ω, v = 10 cm s–1 = 10 × 10–2 ms–1

AB

S N

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×80 cm

10 cm

30 cm

20 cm

B→

⊥ down


Shiv Das

d1 = 30 cm, d2 = 80 cm

t1 = 3010 = 3 sec, t2 =

8010 = 8 sec, t3 =

2010 = 2 sec

φ = BA = (0.1) × (20 × 10–2)2 = 0.1 × 20 × 20 × 10–4 = 10–3 Weber [ ∵ Area = (side)2

e = Blv = (0.1) × (20 × 10–2)(10 × 10–2) = 2 × 10–3 V

i = eR = × −2 10

0 1

3

. = 20 × 10–3 A

(i) (ii) (iii)

Q.25. (a) Define a wavefront. How is it different from a ray?(b) Depict the shape of a wavefront in each of the following cases.

(i) Light diverging from point source.(ii) Light emerging out of a convex lens when a point source is placed at its focus.

(iii) Using Huygen’s construction of secondary wavelets, draw a diagram showing thepassage of a plane wavefront from a denser into a rarer medium. 5

Ans. (a) Wavefront is the locus of all points, which oscillate in phase or it is the surface ofconstant phase.A ray is defined as the path of energy propagation in the limit of wavelength tending to zero.

(b) Shapes of wave-front

(i) from a point source (ii) from a convex lens (iii) from denser towhen source kept rarer mediumat its focus

Or(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain

expression for total magnification when the image is formed at infinity.(b) How does the resolving power of a compound microscope get affected, when

(i) focal length of the objective is decreased. (ii) the wavelength of light is increased?Give reasons to justify your answer. 5

Ans. (a) See Q. 17 (a), 2017 (III Outside Delhi). [Page 233(b) See Q. 17 (c), 2017 (III Outside Delhi). [Page 233

B C

A

φ

D E

30 5 11 0 1 2 3 4 5 6 7 8 9 101112 1314

B C

A D E

t 0 1 2 3 4 5 6 7 8 9 10111213 14

i

B C

A D

t

13t

e

RaysS Parallelrays

N

A

N′

B

P

X

C1

C2

A′Y

D

ii

rr


Shiv Das

Q.26. (a) Write three observed features of photoelectric effect which cannot be explained by wavetheory of light.Explain how Einstein’s photoelectric equation is used to describe these featuressatisfactorily.

(b) Figure shows a plot of stopping potential (V0) withfrequency (v) of incident radiation for two photosensitivematerials M1 and M2. Explain

(i) why the slope of both the lines is same?(ii) for which material emitted electrons have greater

kinetic energy for the same frequency of incidentradiation?. 5

Ans. (a) (i) The maximum kinetic energy of the emitted electron should be directlyproportional to the intensity of incident radiations but it is not observedexperimentally. Also maximum kinetic energy of the emitted electrons should notdepend upon incident frequency according to wave theory, but it is not so.

(ii) According to wave theory, threshold frequency should not exist. Light of allfrequencies should emit electrons provided intensity of light is sufficient forelectrons to eject.

(iii) According to wave theory, photoelectric effect should not be instantaneous.Energy of wave cannot be transferred to a particular electron but will bedistributed to all the electrons present in the illuminated portion. Hence, therehas to be a time lag between incidence of radiation and emission of electrons.

Basic features of photon picture of electromagnetic radiation :(i) Radiation behaves as if it is made of particles like photons. Each photons has energy

E = hν and momentum p = h/λ.(ii) Intensity of radiation can be understood in terms of number of photons falling per

second on the surface. Photon energy depends only on frequency and isindependent of intensity.

(iii) Photoelectric effect can be understood as the result of the one to one collisionbetween an electron and a photon.

(iv) When a photon of frequency (ν) is incident on a metal surface, a part of its energyis used in overcoming the work function and other part is used in imparting kineticenergy, so KE = h(ν – ν0).

(b) (i) The slope (V0/v) of both the lines is the same and represents the universal constantknown as ‘Planck’s constant’ (h) = 6.62 × 10–34 JS.

(ii) For the same frequency of incident radiations, M1 will have greater kinetic energy,because the value of V0 is greater for M1 material. It can be easily seen by drawinga vertical line (frequency being the same) and intersecting M1 and M2 at differentpoints (V0 for M1 is higher).

Or(a) In Rutherford scattering experiment, draw the trajectory traced by ααααα-particles in the

coulomb field of target nucleus and explain how this led to estimate the size of thenucleus.

(b) Describe briefly how wave nature of moving electrons was established experimentally.(c) Estimate the ratio of de-Broglie wavelength associated with deuterons and ααααα-particles

when they are accelerated from rest through the same accelerating potential V.

V0

M1 M2

v


Shiv DasH.T.

F A Electron BeamNickeltarget

←⎯⎯

←⎯VacuumChamber

Diffractedelectronbeam

←⎯

⎯

←⎯

←⎯

To galvanometer

Movablecollector

← →⎯⎯⎯Electrongun

θ

L.T.

– +

θb↓

↑ Target nucleus

Ans. (a) Trajectory of ααααα-particles

It gives an estimate of the size of nucleus, that it relatively very very small as comparedto the size of atom.

(b) Davisson and Germer experiment for wave nature of moving electrons : The experimentalarrangement used by Davisson and Germer is schematically shown in the fig. It consistsof an electron gun which comprises of a tungsten filament F, coated with barium oxideand heated by a low voltage power supply (L.T. or battery). Electrons emitted by the filamentare accelerated to a desired velocity by applying suitable potential/voltage from a highvoltage power supply (H.T. or battery). They are made to pass through a cylinder withfine holes along its axis, producing afine collimated beam. The beam ismade to fall on the surface of a nickelcrystal. The electrons are scattered inall directions by the atoms of thecrystal. The intensity of the electronbeam, scattered in a given direction, ismeasured by the electron detector(collector). The detector can be movedon a circular scale and is connected toa sensitive galvanometer, whichrecords the current. The deflection ofthe galvanometer is proportional tothe intensity of the electron beamentering the collector. The apparatus isenclosed in an evacuated chamber.By moving the detector on the circular scale at different positions, the intensity of thescattered electron beam is measured for different values of angle of scattering θ whichis the angle between the incident and the scattered electron beams. The variation of theintensity (I) of the scattered electrons with the angle of scattering θ is obtained fordifferent accelerating voltages.The experiment was performed by varying the accelarating voltage from 44 V to 68 V.It was noticed that a strong peak appeared in the intensity (I) of the scattered electronfor an accelarating voltage of 54V at a scattering angle θ = 50°.The appearance of the peak in a particular direction is due to the constructiveinterference of electrons scattered from different layers of the regularly spaced atoms ofthe crystals. From the electron diffraction measurements, the wavelength of matterwaves was found to be 0.165 nm.


Shiv Das

The de Broglie wavelength λ associated with electrons, is given by

λ =

hp = 1 227.

V nm …(i)

For V = 54 volts,

λ =

1 227.54

nm = 0.167 nm

Thus, there is an excellent agreement between the theoretical value and the experimentallyobtained value of de Broglie wavelength. Davisson-Germer experiment thus strikinglyconfirms the wave nature of electrons and the de Broglie relation.

(c) From de Broglie relation, we have, λ =

hmv

λd =

hm vd d

…(i)

λα =

hm vα α

…(ii)

Because both the particles are accelerated through the same potential V, kinetic energy ofthe both will be the same.

Ed = Eα

12 mdvd

2 = 12 mαvα

2

v

vmm

d

d

2

2α

α= or

vv

mm

d

dα

α= …(iii)

Dividing (i) by (ii) we get

λλα

α α α

α

αd

d d d

d

d

m vm v

mm

mm

mm= = = =

2 21

mm

d

d= Qmα = 2md

λλλλλd : λλλλλααααα : : 2 : 1


PHYSICS (Theory) – 2016 COMPTT. (DELHI) 167

Shiv Das




SECTION A

Q.1. A coil, of area A, carrying a steady current I, has a magnetic moment, m→

, associated with

it. Write the relation between m→

, I and A in vector form. 1

Ans. Relation for magnetic moment = m→

= I A→

Q.2. Name any two applications where eddy currents are used to advantage. 1Ans. Applications of Eddy currents :

(i) Electromagnetic Damping (ii) Magnetic Breaking(iii) Induction Furnace (iv) Electric Power metres (any two)

Q.3. Identify the logic gate whose output equals 1 when both of its inputs are 0 each.(Not in Syllabus) 1

Sol. NAND gate or NOR gate.Q.4. Why is a potentiometer preferred over a voltmeter for determining the emf of a cell? 1

Ans. Potentiometer does not draw any (net) current from the cell; while Voltmeter draws somecurrent from cell, when connected across it, hence it measures terminal voltage. It is why apotentiometer is preferred over a voltmeter to measure emf.

Q.5. Write the full forms of the terms :(i) LAN (ii) WWW (Not in Syllabus)

Sol. (i) Local Area Networking (ii) World Wide Web

SECTION B

Q.6. The work function (W), of a metal X, equals 3 ××××× 10–19 J. Calculate the number (N) ofphotons, of light of wavelength 26.52 nm, whose total energy equals W. 2

Ans. Given : W = 3 × 10–19 J, λ = 26.52 nm = 26.52 × 10–9 m n = ?Let n be the number of photons, then

Work function, W =

nhcλ

n =

Wcλ

h =

( ) ( . )( . ) ( )

3 10 26 52 106 6 10 3 10

19 19

34 8× × ×

× × ×

− −

− = 4 ××××× 10–2.

Q.7. Distinguish between ‘Sky wave’ and ‘Space wave’ modes of propagation in acommunication system. (Not in Syllabus) 2

CBSE

PHYSICS (Theory) – 2016(COMPTT. DELHI)


Shiv Das

Can take place (even) beyond 40MHz frequency.

Space waves travel in a straightline, either direct from transmittingantenna to receiving antenna orthrough satellite.

B

A

45°

45°

90° 45°

i

B′ A′

Sol. Distinction between ‘Sky Wave’ and ‘Space Wave’

S.No. Sky Wave Space Wave

1. Range of frequencies Restricted upto a few MHzfrequency (30 to 40 MHz).

2. Mode of propagation Waves are reflected back fromionosphere.

Q.8. The following data was obtained for a given transistor :

VCE → 10.0V 10.0 VVBE → 0.82 V 0.72VIB → 80 μμμμμA 30 μμμμμA

For this data, calculate the input resistance of the given transistor. 2Sol. Given : (VCE)1 = 10.0 V (VCE)2 = 10.0 V [∴∴∴∴∴ (VCE = constant)]

(VBE)1 = 0.82 V (VBE)2 = 0.72 V(IB)1 = 80 μA (IB)2 = 30 μA

Calculation of input resistance of the given transistor

Input Resistance r1 =

ΔΔVIBE

B

⎛⎝⎜

⎞⎠⎟

=

( . . )( )

0 82 0 7280 10 30 106 6

−× − ×− −

=

( . . )( )

0 82 0 7280 30 10 6

−− × −

V =

0 1050 10 6

.× −

=

100 1050

3× = 2000ΩΩΩΩΩ

Q.9. Draw a ray diagram to show how a right angled isosceles prism may be used to “bendthe path of light rays by 90°”.Write the necessary condition in terms of the refractive index of the material of thisprism for the ray to bend to 90°. 2

Ans. (i) Ray diagram :

(ii) For total internal reflection : 45° > ic∴ sin 45 > sin ic

1

2 > sin ic >

1μ

⇒ necessary condition is μμμμμ > 2


Shiv Das

a d

imageobject

R

Cimageobject

a d

OrThe image of an object, formed by acombination of a convex lens (of focallength f) and a convex mirror (of radius ofcurvature R), set up, as shown is observedto coincide with the object.Redraw this diagram to mark on it theposition of the centre of curvature of themirror.Obtain the expression for R in terms of the distances, marked as a and d, and the focallength, f, of the convex lens.

Sol. (i) Redrawn Ray diagram :

(ii) Expression for radius of curvature (R) of convex mirror :For convex lens u = – a, v = R + d

1 1 1f v u

= −

1 1 1 1 1f d a d a

= − = ++ − +( ) ( ) ( )R R

On solving, we get

⇒ R =

afa f−

⎛⎝⎜

⎞⎠⎟

– d

Q.10. The electron, in a hydrogen atom, is in its second excited state.Calculate the wavelength of the lines in the Lyman series, that can be emitted through thepermissible transitions of this electron.(Given the value of Rydberg constant, R = 1.1 ××××× 107 m–1) 2

Sol. Given : For second excited state, n = 3, R = 1.1 ××××× 107 m–1

Hence two possible transitions of the Lyman series : 3 → 1 and 2 → 1.(i) Wavelength for transition 3 → 1, nf = 1, ni = 3

1 1 12 2λ

= −⎛

⎝⎜⎜

⎞

⎠⎟⎟R

n nf i

1 11

19

1 1 107λ

= × −⎛⎝⎜

⎞⎠⎟

( . ) = 1.1 × 107

89

⎛⎝⎜

⎞⎠⎟

⇒ λ =

98 1 1 107× ×( . )

= 1.023 × 10–7 = 102.3 nm

(ii) For transition 2 → 1, nf = 1, ni = 2

1 14

1 1 10 17λ

= × −⎛⎝⎜

⎞⎠⎟

( . ) = (1.1 × 107)

34

⎛⎝⎜

⎞⎠⎟

⇒ λ = 121 nm


Shiv Das

ab

qQ

A

K1

K2

X

R

S

ε

+ –

SECTION C

Q.11. Two thin concentric and coplanar spherical shells, of radii a and b (b > a) carry charges,q and Q, respectively. Find the magnitude of the electric field, at a point distant x, fromtheir common centre for

(i) 0 < x < a (ii) a ≤ x < b (iii) b ≤ x < ∞ 3Sol. Magnitude of Electric field : Two thin concentric and

coplanar spherical shells of radii ‘a’ and ‘b’ (b > a) carrycharges ‘q’ and ‘Q’ respectively.

(i) For 0 < x < aPoint lies inside both the spherical shells. Hence, E(x) = 0

(ii) For a ≤ x < bPoint is outside the spherical shell of radius ‘a’ but inside the spherical shell of radius ‘b’.

∴ E(x) =

14 0

2πε. q

x(iii) For b ≤ x < ∞

Point is outside of both the spherical shells. Total effective charge at the centre equals(Q + q).

∴ E(x) =

14 0

2πε. ( )q

x+ Q

Q.12. The reading of the (ideal) ammeter, in the circuitshown here, equals :

(i) I when key K1 is closed but key K2 is open.

(ii)

I2

when both keys K1 and K2 are closed.

Find the expression for the resistance of X interms of the resistances of R and S. 3

Ans. Finding the expression for the resistance X(i) Current I when K2 is open and K1 is closed

I =

ER X+

…(i)

(ii) Current I′ when both the keys K1 and K2 are closed

I′ =

E

RSX

S X

E(S X)R(S X) SX

++

⎛⎝⎜

⎞⎠⎟

⎧⎨⎩

⎫⎬⎭

++ +

⎧⎨⎩

⎫⎬⎭

= …(ii)

∴ S and X are in parallel.Current flowing through X,

I I SS X

E(S X)R(S X) SX

SS X2

= =′+

++ +

⎧⎨⎩

⎫⎬⎭ +( )

…(iii)

Putting the value of I, we get on equating

ER X

ESR(S X) SX2( )+ + +

=

⇒ 2(R + X)S = R(S + X) + SX2RS + 2XS = RS + RX + SXRS = RX – SX

X =

RSR S−


Shiv Das

Q.13. Three long straight parallel wires are kept as shown in thefigure. The wire (3) carries a current I

(i) The direction of flow of current I in wire (3), is such thatthe net force, on wire (1), due to the other two wires, iszero.

(ii) By reversing the direction of I, the net force, on wire (2),due to the other two wires, becomes zero. What will bethe directions of current I, in the two cases? Also obtainthe relation between the magnitudes of currents I1, I2and I. 3

Sol.

(i) Net force experienced by wire (1) can be zero only, when the current in wire (3) flows

along – J i.e. downwards, it means that the forces acting on wire (1) due to wire (3) andwire (2) are equal and opposite.

μπ

μπ

0 0 2I I I I1 12 2 2( ) ( )a a

= ∴ I = 2I2

(ii) When direction of current in wire (3) is reversed then current should be along + J i.e.upwards.For this case net force on wire (2) becomes zero, which means that the forces due to wire(1) and wire (3) are equal and opposite.

μπ

μπ

0 2 0I I I I1 22 2a a

=( )

∴ I = I1 ⇒ I = I1 = 2I2Q.14. Derive the expression for the average power dissipated in a series LCR circuit for an ac

source of a voltage, ννννν = νννννm sinωωωωωt, carrying a current, i = imsin(ωωωωωt + φ).Hence define the term “Wattless current”. State under what condition it can be realized ina circuit. 3

Sol. See Q. 23, 2014 (I Outside Delhi). [Page 63Or

Obtain the expression for the magnetic energy stored in an ideal inductor of selfinductance L when a current I passes through it.Hence obtain the expression for the energy density of magnetic field B produced in theinductor.

Sol. Expression for Magnetic Energy density in an ideal inductor :Instantaneous induced emf in an inductor when current changes through it

e = – L

ddt

I

Hence instantaneous applied voltage

e = V = L

ddt

I

Y

O XZI

I1 I2

(1) (2) (3)a a

I

Y

X

Z(1) (2) (3)a a

I1 I2


Shiv Das

Work done dW = V.dq = VIdt

dW =

L Iddt

⎛⎝⎜

⎞⎠⎟

Idt

∴ dW = LIdI

⇒

d dW LI II

= ∫∫0

W =

12

LI2

Energy Density (u) =

Total Energy StoredVolume

u =

12

12

2⎛⎝⎜

⎞⎠⎟ =

LI

A

LI I

Al l

( )…(i)

Flux = NBA = (LI) …(ii)

and B =

μμ

0

0

NI BN

Il

l⇒ = …(iii)

Putting the values of (LI) and (I) from equations (ii) and (iii) in equation (i), we have

u =

12 0

( ).NBAB

NA

l

lμ =

B2

02μQ.15. The graphs, drawn here, are for the phenomenon of photoelectric effect.

(i) Identify which of the two characteristics (intensity/frequency) of incident light, isbeing kept constant in each case.

(ii) Name the quantity, corresponding to the, mark, in each case.(iii) Justify the existence of a ‘threshold frequency’ for a given photosensitive surface. 3

Sol. (i) (a) In graph 1 : intensity is being kept constant.(b) In graph 2 : frequency is being kept constant.

(ii) (a) In graph 1 : Saturation current(b) In graph 2 : Stopping potential.

(iii) For a given photo-sensitive surface electrons need a minimum energy to be emitted,this is called work function of the surface W.∴ Photons energy hν should be greater/equal to the work function.

∴ hν ≥ W or ν ≥

Wh

∴ Minimum frequency for photo emission

ν0 =

Wh

which is justified to be called as threshold frequency.

↑PhotoelectricCurrent

–V03 –V02 –V01 0 Collector Plate Potential →

Graph 1

–V00

Graph 2

↑Photocurrent

Collector Plate Potential →


Shiv Das

Q.16. Obtain the relation N = N0e–λλλλλt for a sample of radioactive material having decay constantλλλλλ, where N is the number of nuclei present at instant t. Hence obtain the relation between

decay constant λλλλλ and half life

T12

of the sample. 3

Sol. (i) Derivation of radioactive decay law : See Q. 27(a), 2014 (I Delhi). [Page 44

(ii) Relation between λλλλλ and

T12

After one half life, Number of nuclei becomes

N02

⇒

N02

= NT

0

12e

− λ

⇒ 2 = eλT1

2

⇒ loge 2 =

λT12

⇒

λT12

= 0.6931

⇒

T12

0 6931= .λ

Q.17. Give reasons for the following :(i) High reverse voltage do not appear across a LED.

(ii) Sunlight is not always required for the working of a solar cell.(iii) The electric field, of the junction of a Zener diode, is very high even for a small

reverse bias voltage of about 5V. 3Ans. (i) It is because reverse breakdown voltage of LED is very low, i.e., nearly 5V.

(ii) Solar cell can work with any light whose photon energy is more than the band gapenergy.

(iii) The heavy doping of p and n sides of pn junction makes the depletion region very thin,hence for a small reverse bias voltage, electric field is very high.

Q.18. What does the term ‘Modulation’, used in communication system, mean?

Identify the two types of modulation shown here. Give two advantages of any one ofthese over the other. (Not in Syllabus) 3

Sol. (i) Modulation is a process in which one of the characteristics (amplitude, frequency,phase) of a high frequency carrier wave is made to change in accordance with a givenlow frequency message signal.

1

0

–1

Carrier wave→

1

0

–1

Modulating →signal

1

0

–1

(Modulatedwave 1)

2

0

–2

(Modulatedwave 2)

0 0.5 1 1.5 2 2.5 3

0 0.5 1 1.5 2 2.5 3

0 0.5 1 1.5 2 2.5 3

0 0.5 1 1.5 2 2.5 3


Shiv Das

(ii) (a) Modulated wave 1 : Frequency Modulation (FM)(b) Modulated wave 2 : Amplitude Modulation (AM)

Two advantages of FM over AM :(i) Lower noise, better power efficiency.

(ii) Higher operating range.(iii) Higher fidelity reception.Two advantages of AM over FM :

(i) Simple circuits are required.(ii) Lower frequency space for transmission.

Q.19. The figure, drawn here, shows the geometry of pathdifferences for diffraction by a single slit of width a.Give appropriate ‘reasoning’ to explain why theintensity of light is

(i) maximum of the central point C on the screen.(ii) (nearly) zero for point P on the screen when

θθθθθ ≈≈≈≈≈

λa

.

Hence write an expression for the total linear width ofthe central maxima on a screen kept at a distance Dfrom the plane of the slit. 3

Sol. (i) At central point C, angle θ is zero, all path differences are zero. Hence, all the parts ofthe slit contribute in same phase. This gives maximum intensity at point C.

(ii) From the diagram :NP – LP = NQ

= a sin θ = aθ( Q sin θθθθθ ≈≈≈≈≈ θθθθθ for small values of θθθθθ)

When θ =

λa

Path difference NP – LP = aθ = λHence, MP – LP ≈ NP – MP ≈

λ2

It implies that the contribution from correspondingpoints in two halves of the slit have a phase differenceof π. Therefore, contributions from two halves canceleach other in pairs, resulting in a zero net intensity atpoint P on the screen.

Half angular width of central maxima =

λa

⇒ Half Linear width =

λDa

⇒ Linear width of central maximum on the screen =

2λDa

Q.20. A circular coil, having 100 turns of wire, of radius (nearly) 20 cm each, lies in the XY planewith its centre at the origin of co-ordinates. Find the magnetic field, at the point

(0, 0, 20 3 cm), when this coil carries a current of

2π

⎛⎝⎜

⎞⎠⎟

A. 3

Sol. Given : N = 100 turns, R = 20 cm = 20 × 10–2 m = 0.2 m

I =

2π

⎛⎝⎜

⎞⎠⎟

A, z = 20 3 cm = 20 3 × 10–2 m = 0.2 3 m, B = ?

FromsourceS

θθθ

a

To P

To C

D

To P

To Cθ

θ

θ • M2

LM1MM2

N

QFrom S

•••


Shiv Das

The plane of coil is XY plane and field point is on the Z-axis.

∴ Magnetic field on the axial point B =

μ02

2 2322

IR N

R( )+ z

=

( ) ( . )

[( . ) ( . ) ]

4 102

0 2 100

2 0 2 0 2 3

7 2

2 232

ππ

× × ⎛⎝⎜

⎞⎠⎟

× ×

+

−

T

=

8 0 04 10 1002 0 04 8 0 2

7× × ×× × ×

−.. .

T = 25 × 10–6 = 25 μμμμμT

Q.21. The temperature coefficient of resistivity, for two materials A and B, are 0.0031/°C and0.0068/°C, respectively.Two resistors, R1 and R2, made from materials A and B, respectively, have resistances of200Ω and 100Ω at 0°C. Show on a diagram, the ‘colour code’, of a carbon resistor, that wouldhave a resistance equal to the series combination of R1 and R2, at a temperature of 100°C.(Neglect the ring corresponding to the tolerance of the carbon resistor) 3

Ans Given : α1 = 0.0031/°C, α2 = 0.0068/°CR1 = 200 Ω, R2 = 100 ΩRt = R0(1 + αΔt)

For Resistance R1R1′ = R1(1 + αΔt)

= 200(1 + 0.0031 × 100)= 262 ΩΩΩΩΩ …(i)

For Resistance R2R2′ = 100(1 + 0.0068 × 100)

= 168 ΩΩΩΩΩ …(ii)Hence, Total Resistance in series combination of R1 and R2 at 100° C :

R′ = R1′ + R2′= 262 Ω + 168 Ω= 430 ΩΩΩΩΩ = 43 ××××× 101 ΩΩΩΩΩ

Q.22. Two polaroids, P1 and P2, are ‘set-up’ so that their ‘pass-axis’ are ‘crossed’ with respect toeach other. A third polaroid, P3, is now introduced between these two so that its ‘pass-axis’makes an angle θθθθθ with the ‘pass-axis’ of P1.A beam of unpolarised light, of intensity I, is incident on P1. If the intensity of light, that

gets transmitted through this combination of three polaroids, is I′′′′′, find the ratio

II′⎛

⎝⎜⎞⎠⎟

when

θθθθθ equals : (i) 30°, (ii) 45° 3Ans. A beam of unpolarised light of intensity I is incident on P1

I1(through P1) =

I2

and I3(through P3) = I1 cos2 θ

I′(through P2) = I3 cos2 (90 – θ) = (I1 cos2 θcos2(90 – θ) =

I2

cos2 θ sin2 θ

=

I2

44 8

2 2cos sinθ θ⎛

⎝⎜⎞

⎠⎟= I (2 sin θ cos θ)2 =

I8

(sin 2θ)2

I′ =

I8

(sin 2θ)2 or

′ =II

I8

(sin 2θθθθθ)2

(i) When θθθθθ = 30°

Then I′ =

I8

(sin 60°)2 ⇒

II′ = 3

32 Q sin 60° =

32

(ii) When θθθθθ = 45°, then I′ =

I8

(sin 90°)2 ⇒

II′ = 1

8 Q sin 90° = 1

⎯→

⎯

orange (3)

⎯→

⎯

yellow (4)

⎯→

⎯

brown (101)


Shiv Das

P

– q0

r

2a

+ q

OQ r

r′

P

P′

Δr′ +1C

∞

SECTION D

Q.23. Rakesh and Rajesh are 8th class students in a school. They are fond of watching cricket match,particularly when it is played between Australia and India. They observed that most of theplayers, when they are in the field, apply a cream on their face. They did not know its reason.One day they asked this question to their teacher. The teacher thought it to be a good questionand explained the reason for applying this cream to the whole class.Based on this paragraph, answer the following questions :(a) In your opinion, what explanation did the teacher offer to the students in the class?(b) Why is small ozone layer on top of the stratosphere considered crucial for human

survival?*(c) Write any two values displayed by Rakesh and Rajesh and their class teacher. 4

Ans. (a) The cream used by the players helps to protect the skin of the face from harmful raysof the sun.

(b) Ozone layer is crucial because it absorbs harmful ultraviolet radiations.*(c) Value Based Questions will not be asked in the Board Examination.

SECTION E

Q.24. (a) Obtain the expression for the potential due to a point charge.(b) Use the above expression to show that the potential, due to an electric dipole (length

2a), varies as the ‘inverse square’ of the distance r of the ‘field point’ from the centreof the dipole for r > a. 5

Ans. (a) Consider a point charge ‘Q’ kept at point O.Let P be the field point at distance r.At some point p′, the electrostatic force on theunit positive charge is

=

Q ×′1

4 02πε r

Work done against this force r′ to r′ + Δr′ is

ΔW = –

Q4 0

2πε r′Δr′

Total Work done ‘W’ by the External Force from ∞ to r

W = –

Q4 0

2πε r

r

′∞

−

∫ Δr′ = –

Q4 0πε r

r⎡

⎣⎢

⎤

⎦⎥∞

−

W =

Q4 0πε r

Hence potential at this point

V = W =

Q4 0πε r

(b) Potential at point P due to charge (– q)

V1 =

−+

14 0πε

qr a( )

Potential due to charge + q

V2 =

14 0πε

qr a( )−


Shiv Das

t →

C1 C2 C3 C4 Cn

V

t →T/2

φ→

0T/4

3T/4

T

e→

0 T/4

T/2

3T/4

T

Hence total potential at point P

V = V1 + V2 =

qr a r a4

1 1

0πε−

+ −⎡⎣⎢

⎤⎦⎥

+( ) ( )

=

q ar a×

−2

4 02 2 2πε ( )

V =

14 0

2 2πε.( )

p

r a− where p

→ = q × 2a = dipole moment

For r >> a

⇒ V =

14 0

2πε. p

r⇒ V ∝∝∝∝∝

1

r2

Or(a) Define the SI unit of capacitance.(b) Obtain the expression for the capacitance of a parallel plate capacitor.(c) Derive the expression for the affective capacitance of a series combination of n

capacitors.Ans. (a) When a charge of one coulomb produces a potential difference of one volt between the

plates of capacitor, the capacitance is one farad.

Q = CV ⇒ C =

QV

When Q = 1 coulomb, V = 1 volt …[1 farad =

1coulomb1 volt

(b) See Q. 24 (Or) (a), 2014 (I Delhi). [Page 42(c) In series combination, charge on each capacitor is same.

Let it be Q.

V1 =

QC1

V2 =

QC2

.

.

.Vn =

QCn

Total potentialV = V1 + V2 + V3 + … + Vn

V =

QC

QC

QC

QC1 2 3

+ + + … +n

VQ

1C

1C

1C

1C

= + + + … +1 2 3 n

⇒

1C

1C

1C

1C

1C

= + + + … +1 2 3 n

Q.25. Discuss how Faraday’s law of e.m. induction is applied in an ac-generator for convertingmechanical energy into electrical energy.Obtain an expression for the instantaneous value of the induced emf in an ac generator.Draw graphs to show the ‘phase relationship’ between the instantaneous (i) magnetic flux(φ) linked with the coil and (ii) induced emf (ε) in the coil. 5

Ans. Faradays law of e.m. induction and expression for instantaneous value of induced emf.See Q. 25 (Or) (a), 2017 (I Outside Delhi). [Page 226Graph : Graph for magnetic flux

φ = NBAcosωωωωωt Graph for induced e.m.f.e = NBAωωωωωsinωωωωωt


Shiv Das

OrDraw an arrangement for winding of primary and secondary coils in a transformer withtwo coils on a separate limb of the core.State the underlying principle of a transformer. Deduce the expression for the ratio ofsecondary voltage to the primary voltage in terms of the ratio of the number of turns ofprimary and secondary winding. For an ideal transformer, obtain the ratio of primary andsecondary currents in terms of the ratio of the voltages in the secondary and primaryvoltages.Write any two reasons for the energy losses which occur in actual transformers.

Ans. See Q. 24 (Or), 2016 (I Delhi). [Page 148Q.26. (a) A point object, O is on the principal axis of a spherical surface having a radius of

curvature, R. Draw a diagram to obtain the relation between the object and imagedistances, the refractive indices of the media and the radius of curvature of thespherical surface.

(b) Write the Lens Maker’s formula and use it to obtain the range of values of μμμμμ (therefractive index of the material of the lens) for which the focal length of anequiconvex lens, kept in air, would have a greater magnitude than that of the radiusof curvature of its two surfaces. 5

Ans. (a) See Q. 25 (Or) (a), 2015 (I Delhi). [Page 89

(b) Lens Maker’s formula,

1 1 121

1 21

fn= − −

⎛⎝⎜

⎞⎠⎟

( )R R

For equiconvex lens :R1 = + ve = R R2 = – ve = – R

⇒

1 21f

= − ⎛⎝⎜

⎞⎠⎟

( )μR

For f to be greater than R2(μ – 1) < 1

⇒ 2μ – 2 < 1 or 2μ < 3μ < 1.5

Hence required range is 1.0 < μμμμμ < 1.5Or

(a) Draw a diagram showing the ‘Young’s arrangement’ for producing ‘a sustainedinterference pattern’. Hence obtain the expression for the width of the interferencefringes obtained in this pattern.

(b) If the principal source point S were to be moved a little upwards, towards the slit S1from its usual symmetrical position, with respect to the two slits S1 and S2, discusshow the interference pattern, obtained on the screen, would get affected.

Ans. (a) Consider two coherent sources S1 and S2 separatedby a distance d. Let D be the distance between thescreen and the plane of slits S1 and S2.Light waves emitted from S1 and S2 reach point Oon the screen after travelling equal distances. Sopath difference and hence phase differencebetween these waves is zero. Therefore, they meetat O in phase and hence constructive interferencetakes place at O. Thus O is the position of thecentral bright fringe.

P

y

O

B

Screen

d/2

d/2

S1

d

S2

x + Δx

AΔx

x

D


Shiv Das

Expression for the fringe width.Let d = distance between slits S1 and S2

D = distance of screen from two slitsx = distance between the central maxima O and observation point P

Light waves spread out from S and fall on both S1and S2. The spherical waves emanating from S1 andS2 will produce interference fringes on the screenMN.In rt ΔS1AP, we have

(S1P)2 = (S1A)2 + (AP)2

S1P =

D22

2+ −⎛

⎝⎜⎞⎠⎟

x d =

DD

2

2

21 2+

−⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

x d

⇒ S1P = D

1 2

2

2

12

+−⎛

⎝⎜⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

x d

D

By Binomial theorem and neglecting higher terms, we have

S1P = D

1 12

2

2

2+

−⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

x d

D = D +

x d−⎛⎝⎜

⎞⎠⎟2

2

2

D

Similarly, S2P = D +

x d+⎛⎝⎜

⎞⎠⎟2

2

2

DHence, path difference = S2P – S1P

= D +

x d+⎛⎝⎜

⎞⎠⎟2

2

2

D – D –

x d−⎛⎝⎜

⎞⎠⎟2

2

2

D

=

12 4 4

22

22

Dx xd x xdd d+ + − − +⎡

⎣⎢

⎤

⎦⎥ =

12D

. 2xd =

xdD

Let the waves emitted by S1 and S2 meet at point P and the screen at a distance y fromthe central bright fringe.The path difference between these waves at P is given by

Δx = S2P – S1P …(i)From right angled triangle S2BP,

S2P = [S2B2 + PB2]1/2 ⇒ S2P =

D DD

22 1 2

22

2

1 2

21+ +⎛

⎝⎜⎞⎠⎟

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

= ++⎡

⎣⎢⎢

⎤

⎦⎥⎥

yd y d

/ /( )

⇒ S2P = D

1 22

2++⎡

⎣⎢⎢

⎤

⎦⎥⎥

( )y d

2D∴ S2P = D +

( )y d+ 22

2D…(ii)

Similarly from right angled ΔS1AP, S1P = D +

( )y d− 22

2D…(iii)

Substituting these values in (i), we get

S1

S2

SAO

P

d

x

Sou

rce

ofm

onoc

hrom

atic

lig

ht

D

M

NB


Shiv Das

Δx = D +

( ) ( )y yd d+− −

−22

22

2 2DD

D

=

( ) ( )y yd y yd yd ydd d24

24

2 2

222

+ + − + −= =

D D D…(iv)

For constructive interference/maxima:If path difference is an integral multiple of λ, then bright fringe will be formed at P

i.e.,

ydD

= mλ or y =

mdλD …(v) …where [m = 1, 2, 3 …

which is the position of mth bright fringe from the central bright fringe.Fringe width (βββββ) : The distance between any two successive bright fringes (or successivefringes) is called fringe width.

β = y2 – y1 =

2λ λD Dd d

− =

λDd

Destructive interference/Minima : If path difference is odd multiple of λ/2, then darkfringe is formed at P

i.e.,

ydm

D= +⎛

⎝⎜⎞⎠⎟

12

∴ y =

m

d

+⎛⎝⎜

⎞⎠⎟

12

λD…where [m = 1, 2, 3 …

which is the position of mth dark fringe from the central bright fringe.

βββββ (fringe width) = y1 – y0 =

32 2

22

λ λ λD D Dd d d

− = =λλDd

(b) On shifting principal source point ‘S’ little upwards i.e. towards S1, the position of thecentral maximum on the screen will shift downwards on the screen, i.e. below itsprevious position. Hence, whole interference pattern will get shifted little downwardsbut fringe width will remain the same as that of the initial arrangement.


PHYSICS (Theory) – 2016 COMPTT. (OUTSIDE DELHI) 181

Shiv DasE

V

r




SECTION A

Q.1. Show on a plot the nature of variation of the (i) Electric field (E) and (ii) potential (V), ofa (small) electric dipole with the distance (r) of the field point from the centre of thedipole. 1

Ans.

Q.2. For an ideal inductor, connected across a sinusoidal ac voltage source, state which one ofthe following quantity is zero :

(i) Instantaneous power(ii) Average power over full cycle of the ac voltage source 1

Ans. Average power over full cycle of the ac voltage source is zero, when connected with an idealinductor.

Q.3. A beam of unpolarised light is incident, on the boundary between two transparent media,at an angle of incidence = iB, the Brewester’s angle. At what angle does the reflected lightget polarised? 1

Sol. At an angle of incidence = iB, the reflected light gets polarised.Q.4. Name and define, the SI unit for the ‘activity’, of a given sample of radioactive nuclei.

1Ans. (i) becquerel is the unit of ‘activity’ of a nuclear sample.

(ii) One becquerel activity corresponds to ‘one decay/disintegration per second’.Q.5. Name the two basic modes of communication system. (Not in Syllabus) 1

Ans. (i) Point to Point Communication and (ii) Broadcast are the two basic modes ofcommunication.

SECTION B

Q.6. An ααααα-particle moving with initial kinetic energy K towards a nucleus of atomic numberz approaches a distance ‘d’ at which it reverses its direction. Obtain the expression for thedistance of closest approach ‘d’ in terms of the kinetic energy of ααααα-particle K. 2

CBSE

PHYSICS (Theory) – 2016(COMPTT. OUTSIDE DELHI)


Shiv Das

Ans. At the distance d, the K.E. (K) gets converted into P.E. (P) of the system.

∴ P.E. at distance (d) =

14

2

0πεe e

d× Z

∴

14

2

0

2

πεZed

= K

∴ d =

14

2

0

2

πεZK

e

OrFind the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmerand Paschen series of the hydrogen spectrum.

Ans. As per Bohr’s Model, the wavelengths are given by

1 1 1

12

22λ

= −⎛⎝⎜

⎞⎠⎟

Rn n

[[[[[ ∵ R = Rydberg’s constant

For most energetic spectral lines, electron should jump from infinity.(i) For the most energetic line of Balmer Series :

1 12

142 2λB

R R= − →∞

⎛⎝⎜

⎞⎠⎟ …(i) ∵ n1 = 2

(ii) Similarly for the Paschen Series :

1 13

192 2λP

R R= − →∞

⎛⎝⎜

⎞⎠⎟ …(ii) ∵ n2 = 3

∴

λλ

P

B= 9

4Q.7. For a plane electromagnetic wave, propagating along the Z-axis, write the two (possible)

pairs of expression for its oscillating electric and magnetic fields. How are the peak valuesof these (oscillating) fields related to each other? 2

Ans. For the e.m. wave, propagating along the z-axis, we haveE = E0 sin ( kz ∓ ωt) and B = B0 sin ( kz ∓ ωt)

The two possible forms for electric and magnetic fields are :Ex = E0 sin ( kz – ωt) By = B0 sin ( kz – ωt) and

and Ey = E0 sin ( kz + ωt) Bx = B0 sin ( kz + ωt)The peak values of these two fields are related by

EB

0

0 = C

Q.8. An e.m. wave, Y1, has a wavelength of 1 cm while another e.m. wave, Y2, has a frequencyof 1015 Hz. Name these two types of waves and write one useful application for each. 2

Sol. (i) Y1→→→→→ MicrowavesApplications : Microwaves are used in Microwave ovens, Aircraft Navigators etc.

(ii) Y2→→→→→ Ultraviolet wavesApplications : Ultraviolet rays are used in sterilizing surgical instruments, foodpreservation etc.

Q.9. The Kinetic Energy (K.E.), of a beam of electrons, accelerated through a potential V,equals the energy of a photon of wavelength 5460 nm. Find the de Broglie wavelengthassociated with this beam of electrons. 2


Shiv Das

Ans. Given : λ = 5460 nm = 5460 × 10–9 m λB = ?

Energy of the photon (K) =

hcλ

…(i)

de-Broglie wavelength, (λB) =

hp

hmk

=2

…(ii)

∴ λB =

h

mhc

2 .λ

=

hmcλ

2

=

( . ) ( )( . ) ( )

6 63 10 5460 102 9 1 10 3 10

34 9

31 8

12× × ×

× × × ×⎡

⎣⎢

⎤

⎦⎥

− −

− = 25.75 ××××× 10–10 m

Q.10. If both the number of protons and the number of neutrons are conserved in each nuclearreaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction?Explain. 2

Ans. The number of protons and neutrons in a nuclear reaction are conserved but the total massis not conserved.The total mass of the free protons and neutrons is more than their total mass within thenucleus.The lost mass (= Δm) known as ‘mass defect’, gets converted into energy as per the relation

E = (Δm)c2 (c is the velocity of light)

SECTION C

Q.11. A charge +Q, is uniformly distributed within a sphere of radius R. Find the electric field,due to this charge distribution, at a point distant r from the centre of the sphere where :

(i) 0 < r < R and (ii) r > R 3Sol. We have

E =

14 0

2πεqr

, for a point charge

Now (volume) charge density

ρ =

Q

R43

3π⎛⎝⎜

⎞⎠⎟

(i) ∴ Charge contained within a sphere of radius r, (0 < r < R)

Q′ = ρ.

43π

.r3 =

Q

R43

3π⎛⎝⎜

⎞⎠⎟

×

43π

.r3 = Q

r3

3R

∴ Electric Field

E =

14

14

140

20

2

3

30

3πε πε πεQ

RQR

Q′ ⎛

⎝⎜⎞

⎠⎟⎛⎝⎜

⎞⎠⎟

= × =r r

r r( ) . . =

14 0

3πεQ

R

r

(ii) For r > RElectric field due to a point charge (assumed to be kept at centre) at distance ‘r’ :

E =

14 0

2πεQ

r


Shiv Das

Q.12. A student connects a cell, of emf E2 and internal resistance r2 with a cell of emf E1 andinternal resistance r1, such that their combination has a net internal resistance less than r1.This combination is then connected across a resistance R.Draw a diagram of the ‘set-up’ and obtain an expression for the current flowing throughthe resistance R. 3

Ans. Since the net internal resistance of two cells is less than r1,it means that the two cells must have been connected in parallelThe diagram of the setup is as shown

Equivalent internal resistance r =

r rr r

1 2

1 2( )+…(i)

∴ = +

⎛⎝⎜

⎞⎠⎟

1 1 1

1 2r r r

Equivalent emf e, is given by

e =

er

er

1

1

2

2+

⎛⎝⎜

⎞⎠⎟

r …(ii)

The current flowing through R, is

I =

er( )R +

…(iii)

Where e and r are given by the equations (ii) and (i) respectively.

Q.13. Write the expression for the magnetic force F→

acting on a charged particle q moving with

velocity v→

in the presence of the magnetic field B→

in a vector form. Show that no workis done and no change in the magnitude of the velocity of the particle is produced by thisforce. Hence define the unit of magnetic field. 3

Sol. (i) The required expression is F B→

= ×→→

q v( )(ii) The magnetic force, at all instants, is, therefore, perpendicular to the instantaneous

direction of v→

, which is also the instantaneous direction of displacement ( )ds→

.

Since, F→

is perpendicular to ( )ds→

, at all instants, work done ( . )=→ →F ds is zero.

There being no work done, there can be no change in the magnitude of v→

.

From F B→

= ×→→

q v( ) , we get

| F→

| = F = qvB sin θ

(iii) ∴ F = B if q = 1, v = 1 and θ =

π2

Hence, the magnetic field B→

, at a point equals one tesla if a charge of one coulomb,

moving with a velocity of 1 m/s, along a direction perpendicular to the direction of B→

,experiences a force of one newton.

ε2

r2

r1

R

ε1

+ –

+ –


Shiv Das

Q.14. A long straight wire, of circular cross section (radius = a) carries a current I which isuniformly distributed across the cross section of the wire.Use Ampere’s circuital law to calculate the magnetic field B(r), due to this wire, at a pointdistance r < a and r > a from its axis. Draw a graph showing the dependence of B(r)on r. 3

Sol. As per Ampere’s Circuital law :

B I→

=∫→

.d l μ0

(i) For r < aIe = current enclosed by Amperian circuital loop of radius r

= I.

ππ

ra

ra

2

2

2

2= I

∴ B

dl er

a∫ = =μ μ0

02

2.II

or B =

μπ

μπ

02

20

21

2 2I Ir

a r ar.⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= or B ∝∝∝∝∝ r

∵ dl rr

=⎡

⎣⎢⎢ ∫ 2

0

π

(ii) For r > a

F I→

=→

∫ .d l μ0

∴ B. 2πr = μ0I

or B =

μπ0

2Ir

or B ∝∝∝∝∝

1r

The graph of, B(r) vs r, is as shown in the adjoing figure.Or

Derive the expression for the torque τττττ acting on a rectangular current loop of area A placed

in a uniform magnetic field B. Show that τ→ →

= ×→

m B where m→

is the magnetic moment of

the current loop given by m→

=→

I A .Sol. Let I = current through the coil

a, b = sides of the rectangular loopA = ab = area of the loopn = Number of turns in the loopB = Magnetic fieldθ = angle between magnetic field

B→

and area vector A→

Force exerted on the arm DA inwardF1 = I b B …[ ∵ F= ILB]

Force exerted on the arm BC outwardF2 = I b B ∴ F2 = F1

Thus net force on the loop is zero∴ Two equal and opposite forces form a couple which exerts a torque

B

B ∝ r

a

B ∝

1/r

r

D C

SF1

F2

θm

A B

N

a

b

←→

⎯⎯⎯⎯

⎯⎯⎯b


Shiv Das

∴ Magnitude of the torque on the loop is,

τ = F1

a2

sin θ + F2

a2

sin θ

= (F1 + F2)

a2

sin θ = (IbB + IbB)

a2

sin θ

⇒⇒⇒⇒⇒ τ = 2IbB

a2

sin θ = IabB sin = IAB sin θ

∴ Magnetic moment of the current loop is,M = I A

∴ τ = MB sin θ ⇒⇒⇒⇒⇒ τ→ → →

= ×m B

Q.15. (a) Define self-inductance of a coil and hence write the definition of ‘Henry’.(b) Write any two factors each on which the following depends :

(i) Self inductance of a coil.(ii) mutual inductance of a pair of coils. 3

Ans. (a) The self inductance (L) of a coil equal the magnetic flux linked with it, when a unitcurrent flows through it.One henry is the self inductance of a coil for which the magnetic flux, linked with it,due to a current of 1A, flowing in it, equals one weber.

(b) (i) Self inductance of a coil depends on• Its geometry (area and length of a coil.• Number of turns• Medium within the coil

(ii) Mutual inductance of a given pair of coils depends on• Their geometries• Their distance of separation• Number of turns in each coil.• Nature of medium in the intervening space.

Q.16. The current, in the LCR circuit shown in the figure isobserved to lead the voltage in phase. Withoutmaking any other change in the circuit, a capacitor, ofcapacitance C0, is (appropriately) joined to thecapacitor C. This results in making the current, in the‘modified’ circuit, flow in phase with the appliedvoltage.Draw a diagram of the ‘modified’ circuit and obtainan expression for C0 in terms of ωωωωω, L and C. 3

Sol. Since the current leads the voltage in phase, hence, XC > XLFor resonance, we must haveNew value of X′C = XL

We, therefore, need to decarease the value of XC =

1ωC

⎛⎝⎜

⎞⎠⎟

.

This requires an increase in the value of C. Hence, capacitor C0 should be connected inparallel across C.

F1

F2

m

βθθ

sin θa

2

L

C R

V = V0 sin ωt


Shiv Das

The diagram of the modified circuit is shown.For resonance, we then have

1

0ω( )C C+ = ωL ⇒ ω(C + C0) × ωL = 1

⇒ ω2L(C + C0) = 1

∴ C0 =

12ω L

C−⎡⎣⎢

⎤⎦⎥

Q.17. Point out two distinct features observed experimentally in photoelectric effect whichcannot be explained on the basis of wave theory of light. State how the ‘photon picture’of light provides an explanation of these features. 3

Sol. See Q. 14, 2017 (I Outside Delhi). [Page 220Q.18. It is required to design a (two-input) logic gate, using an appropriate number, of :

(a) NAND gates that gives a ‘low’ output only when both the inputs are ‘low’.(b) NOR gates that gives a ‘high’ output only when both the inputs are ‘high’.

Draw the logic circuits for these two cases and write the truth table, corresponding to eachof the two designs. (Not in Syllabus) 3

Sol. (a) The ‘NAND’ gate that gives a ‘low’ output only when both its inputs are low, is an ‘OR’ gateThe required design and the truth table are as follow :

(b) The ‘NOR’ gate that gives a high output only when both the inputs are high, is an ‘AND’gate.The required design and the ‘truth table’ are as follow :

Q.19. Give (brief) reasons for the following :(a) We use the ‘sky wave’ mode of propagation, of electromagnetic waves, only for

frequencies up to 30 to 40 MHz.(b) The LOS communication, via space waves, has a (fairly) limited range.(c) A mobile phone user gets an ‘uninterrupted link to talk’ while walking.

(Not in Syllabus) 3

C R

V = V0 sin ωt

C0

L

Truth Table

A B Y

0 0 00 1 11 0 11 1 1

A

BY

A

B

A

B

Y

Truth Table

A B Y

0 0 00 1 01 0 01 1 1


Shiv Das

Ans. (a) The ionosphere can act as a ‘reflector’ only for e.m. waves of frequencies upto 30 to 40MHz. Higher frequency e.m. waves penetrate into the atmosphere and escape.

(b) The range is (fairly) limited because the e.m. waves lose energy (fairly rapidly) whenthey glide over the surface of the earth.

(c) This is because of the presence of a network of base stations’/cells’ which keep onpassing the signals from one base station/cell to the other.

Q.20. A parallel plate capacitor, of capacitance 20μμμμμF, is conneted to a 100 V supply. Aftersometime the battery is disconnected, and the space, between the plates of the capacitoris filled with a dielectric, of dielectric constant 5. Calculate the energy strored in thecapacitor (i) before (ii) after the dielectric has been put in between its plates. 3

Sol. Given : C = 20 μF = 20 × 10–6 F, V = 100 V, K = 5E1 = ?, E2 = ?

Charge stored, Q = CV = (20 × 10–6) × (100) = 2000 μCNew value of capacitance (C′) = 5 × 20 μF = 100 μF

Energy stored in a capacitor, (E) =

12

QC

2

(i) ∴ Energy stored before dielectric is introduced

E1 =

12

2000 10 2000 1020 10

6 6

6× × × ××

− −

−( ) ( )

( ) = 0.1 J

(ii) Energy stored after the dielectric is introduced(∴ there is no change in the value of Q )

E2 =

12

2000 10 2000 10100 10

6 6

6× × × ××

− −

−( ) ( )

( ) = 0.02 J

Q.21. A convex lens, of focal length 25 cm, and a convex mirror, of radius of curvature 20 cm,are placed co-axially 40 cm apart from each other. An incident beam, parallel to theprincipal axis, is incident on the convex lens. Find the position and nature of the imageformed by this combination. 3

Sol. The given ‘setup’ is as shownThe object, being at infinity, the image formed by theconvex lens, is at its focus, i.e. 25 cm from the lens. Thisimage becomes the ‘object’ for convex mirror.Now, for the convex mirror

Object distance = (40 – 25) cm = 15 cmRadius of curvature, R = 20 cm

u = –15 cm, R = +20 cmUsing the mirror formula :

1 1 1 2v u f

+ = =R

,

We get

1 220

115v

= ⎛⎝⎜

⎞⎠⎟

− − =

110

115

16

+ = ∴ v = + 6 cm

CFMF2

f = 25 cm R = 20 cm

← →⎯⎯⎯⎯⎯⎯⎯⎯40 cm


Shiv Das

The final image is, therefore, a vitrual image that appears to be formed (behind the convexmirror) at a distance of 6 cm from it.

Q.22. A 200 mH (pure) inductor, and a 5μμμμμF (pure) capacitor, are connected, one by one, across asinusoidal ac voltage source V = [70.7 sin (1000 t)] voltage. Obtain the expressions for thecurrent in each case. 3

Sol. Given : For the applied voltage V = 70.7 sin (1000 t), we have V0 = 70.7 volts, ω = 1000 s–1

(i) For the inductor, the peak value of current (i0) is :

i0 =

VL0

370 7

1000 200 10ω=

× × −.

( ) ( ) A

= 35.35 × 10–2 A = 0.3535 A

∴ Expression for current is i = (0.3535) sin

10002

t −⎛⎝⎜

⎞⎠⎟

π Q

i i t= −⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥0 2

sin ω π

(ii) For the capacitor, the peak value of current (i0) is

i0 =

V01

ωc⎛⎝⎜

⎞⎠⎟

= V0.ωc

= (70.7) × (1000) × (5 × 10–6 A)= 353.5 × 10–3 A = 0.3535 A

∴ Expression for current is i = (0.3535) sin

10002

t +⎛⎝⎜

⎞⎠⎟

π Q

i i t= +⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥0 2

sin ω π

SECTION D

Q.23. Ravi is a student of mechanical engineering studying in one of the engineering colleges.The other day he saw an old man who suddenly collapsed as he walked out of the housein his neighbourhood. Ravi rushed towards him, immediately made a call to the nearbyhospital, asked for the ambulance and took him to the emergency ward of the hospital. Ongetting the medical aid, the old man soon recovered. He did not forget to thank Ravi forthe timely help he rendered. He was wondering that in his times to get the telephoneconnection, one had to wait for years whereas these days it takes no time to get theconnection. Ravi told him it was all because of the technological progress/developmentdue to which the simple phenomenon in Physics could be easily used.Answer the following questions based on the above :(a) To which phenomenon in physics was Ravi referring to, which made the land line

links so easily accessible?(b) What are the essential conditions required to observe this phenomenon?

*(c) Write two values displayed by Ravi and the old man in this episode. 4Ans. (a) Ravi was referring to the phenomenon of total Internal Reflection.

Alternatively:Development of techniques for rapid transmission of e.m. waves/setting up of linkingnetwork.(or development of mobile telephony)


Shiv Das

i

δ

δm

• FIncidentplane wavefront

Spherical wavefrontof radius f

(b) Essential conditions1. Angle of incidence > Critical Angle2. Light should travel from denser to rarer medium.

Alternatively:(i) Setting up appropriate transmission cable network/antennas

(ii) Availability of appropriate transducers.(or setup of mobile cell network)

*(c) Value Based Questions will not be asked in the Board Examination.

SECTION E

Q.24. (a) The relation, between the angle of incidence (i)and the corresponding, angle of deviation (δδδδδ), fora certain optical device, is represented by thegraph shown in the figure. Identify this device.Draw a ray diagram for this device and use it forobtaining an expression for the refractive indexof the material of this device in terms of an anglecharacteristic of the device and the angle, markedan δδδδδm, in the graph.

(b) Based on Huygen’s construction, draw the shape of a plane wavefront as it getsrefracted on passing through a convex lens. 5

Ans. (a) (i) The device identified is a ‘PRISM’.(ii) For ray diagram and derivation : See Q. 26 (Or), 2015 (I Outside Delhi).

[Page 108

(b) The required shape of the wavefront is asshown

OrWhen a plane wave front, of light, of wavelength λλλλλ,is incident on a narrow slit, an intensitydistribution pattern, of the form shown is observedon a screen, suitably kept behind the slit. Name thephenomenon observed.(a) Obtain the conditions for the formation of central maximum and secondary maxima

and the minima.(b) Why is there significant fall in intensity of the secondary maxima compared to the

central maximum, where as in double slit experiment all the bright fringes are of thesame intensity?


Shiv Das

(c) When the width of the slit is made double the original width, how is the size of thecentral band affected?

Ans. The phenomenon observed is the phenomenon of ‘diffraction’.

(a) At the central maximum : The contributions due to the secondary wavelets, from allparts of the wave front (at the slit), arrive in phase at the central maxima, At the centralmaxima θ = 0.At the secondary maxima : It is only the contributions from (nearly) 1/3 (or 1/5, or1/7, …) of the incident wavefront that do not get cancelled at the locations of thesecondary maxima. These occur at points for which

θ ≅

na

+⎛⎝⎜

⎞⎠⎟

12

λ (n = 0, 1, 2, 3 …)

At the secondary minima : The contributions, from ‘corresponding pairs’, of the sub-parts of the incident wavefront, cancel each other and the net contribution, at thelocation of the minima, is zero. The minima occur at points for which

θ = n

λa

(n = 1, 2, 3, …)

(b) There is a significant fall in intensity at the secondary maxima because the intensitythere, is only due to the contribution of (nearly) (1/3 or 1/5 or 1/7, …) of the incidentwavefronts.

(c) The size of the central maximum would get halved when width of the slit is doubled.

Q.25. (a) Obtain the condition under which thecurrent flowing, in the ‘current detectingdevice’, used in the circuit shown in thefigure, becomes zero.

(b) Describe briefly the device, based on theabove condition. Draw a circuit diagramfor this device and discuss, in brief, how itis used for finding an unknownresistance. 5

Ans. (a) The given circuit can be redrawn as :It is, therefore, a wheatstone bridgeUsing Kirchoff’s laws, we get (when ig = 0)

i1 = i3 and i2 = i4For the loop ABDA, we have

– i1P + i2X = 0 or i1P = i2XFor the loop BCDB, we have

– i3Q + i4R = 0 or i3Q = i4RDividing we get,

ii

ii

1

3

2

4

PQ

XR

=

or

PQ

XR

= ( ∵i1 = i3 and i2 = i4)

P Q

X R

+ –

(Resistance= R0)

Currentdetecting

device

+ –

A

B

C

R

D

X

P

ii4

i2

i1

igi3

Q


Shiv Das

(b) A simple device, based on the above condition is ‘Meter Bridge’.Metre Bridge is special case of Wheatstone Bridge.It is a device based on Wheatstone bridge to determine the unknown resistance of awire.Principle : Meter bridge is based on the principle of wheatstone bridge, i.e. when bridgeis balanced

ll( )100 − = R

S or S = R( )100 − l

l

Circuit : To find the unknownresistance S, the circuit is completed asshown in figure. The unknownresistance wire of resistance S isconnected across the gap betweenpoints C and D and a resistance box(R) is connected across the gapbetween the points A and D. A cell, arheostat and a key (K) is connectedbetween the points A and C by meansof connecting screws. In the experiment when the sliding jockey touches the wire AC atany point, then the wire is divided into two parts. These two parts AB and BC act asthe resistances P and Q of the Wheatstone bridge. In this way the resistances of arms AB,BC, AD and DC form the resistances P, Q, R and S of Wheatstone bridge. Thus the circuitof meter bridge is the same as that of Wheatstone bridge.Method : To determine the unknown resistance first of all key K is closed and aresistance R is taken in the resistance box in such a way that on pressing jockey B at endpoints A and C, the deflection in galvanometer is on both the sides. Now jockey is slidedon wire at such a position that on pressing the jockey on the wire at that point, there isno deflection in the galvanometer G. In this position the points B and D are at the samepotential, therefore the bridge is balanced. The point B is called the null point. Thelength of both parts AB and BC of the wire are read on the scale. The condition ofbalance of Wheatstone bridge is,

PQ

RS=

∴ Unknown resistance, S =

QP

⎛⎝⎜

⎞⎠⎟

R …(i)

If r is the resistance per cm length of wire AC and l cm is the length of wire AB, thenlength of wire BC will be (100 – l) cm∴ P = resistance of wire AB = lr

Q = resistance of wire BC = (100 – l)rSubstituting these values in equation (i), we get

S = ( )100 − l r

lr × R ⇒ S = 100 − l

l R …(ii)

As the resistance (R) of wire (AB) is known, the resistance S may be calculated.A number of observations are taken for different resistances taken in resistance box andS is calculated each time and the mean value of S is found.

Resistance box Resistance wire

(R) (S)

D

GA (P) B (Q)

l cm (100 – l) cm

C

(Rh)

RheostatK

+ –Cell


Shiv Das

Or(a) Why do the ‘free electrons’, in a metal wire, ‘flowing by themselves’, not cause any

current flow in the wire?Define ‘drift velocity’ and obtain an expression for the current flowing in a wire, interms of the ‘drift velocity’ of the free electrons.

(b) Use the above expression to show that the ‘resistivity’, of the material of a wire, isinversely proportional to the ‘relaxation time’ for the ‘free electrons’ in the mwtal.

Ans. (a) (i) The free electrons, in a metal, (flowing by themselves), have a random distributionof their velocities. Hence the net charge crossing any cross section in a unit time iszero.

(ii) The drift velocity equals the average (time dependent) velocity acquired by freeelectrons, under the action of an applied (external) electric field.

(iii) Expression for current in terms of drift velocity :

(b) Relation between electric current anddrift velocity : Consider a uniformmetallic wire XY of length l and cross-sectional area A. A potential differenceV is applied across the ends X and Y ofthe wire. This causes an electric field ateach point of the wire of strength :

E = Vl …(i)

Due to this electric field, the electronsgain a drift velocity Vd opposite todirection of electric field. If q be the charge passing through the cross-section of wire in

t seconds, then, current in wire, I = qt

The distance transversed by each electron in time (t) = drift velocity × time = Vd tIf we consider two planes P and Q at a distance Vd t in a conductor, then the totalcharge flowing in time t will be equal to the total charge on the electrons present withinthe cylinder PQ.The volume of this cylinder = cross sectional area × length = A Vd tIf n is the number of free electrons in the wire per unit volume, then the number of freeelectrons in the cylinder = n(AVd t)

If charge on each electron is – e (e = 1.6 × 10–19 C), then the total charge flowing througha cross-section of the wire,

q = (nA Vd t) (– e) = – neA Vd t …(iii)

∴ Current flowing in the wire,

I =

qt

ne tt

d= − AV

i.e. Current I = – neA Vd …(iv)

Current(I)

vdt

E⎯ →⎯

V

+ –

II

X P Q Y


Shiv Das

This is the relation between electric current and drift velocity. Negative sign shows thatthe direction of current is opposite to the drift velocity.

(b) We know from aboveI = – neA vd …(i)

Also we know, vd =

emE→⎛

⎝⎜⎜

⎞

⎠⎟⎟

τ …(ii)

Putting the value of vd in equation (i) from equation (ii) we have

I = – neA

emE τ

QE

V=⎡⎣⎢ l

or

EI A

= − mne2 τ

Q ρ =⎡⎣⎢

RAl

ρ = − m

ne2τ

ρ ∝

1τ

Q.26. Draw the ‘Energy bands’, diagrams for a (i) pure semiconductor (ii) insulator.How does the energy band, for a pure semiconductor, get affected when thissemiconductor is doped with (a) an acceptor impurity (b) donor impurity? Hence discusswhy the ‘holes’, and the ‘electrons’ respectively, become the ‘majority charge carriers’ inthese two cases?Write the two processes involved in the formation of p-n junction. 5

Ans. ‘Energy Band’ diagrams : See Q. 24, 2014 (I Outside Delhi). [Page 63(a) When the semiconductor is doped with an acceptor impurity, thereby results in an

additional energy level a little above the top of the valence band.(b) The donor impurity results in an additional energy level a little below the bottom of the

conduction band.In the first case, electrons from the valence band, easily jump over to the acceptor lelvel,leaving ‘holes’ behind. Hence, ‘holes’ becomes the majority charge carriers.In the second case, electrons from the donor level, easily ‘jump over’ to the conductionband. Hence, electrons become the majority charge carriers.

The two processes involved in the formation of the p-n junction are :(i) Diffusion (ii) Drift

Or(a) Draw the diagram of the ‘circuit arrangement used for studying the ‘input’ and

‘output’ characteristics of an n-p-n transistor in its CE configuration’. Briefly explainhow these two types of characteristics are obtained and draw these characteristics.

(b) ‘Define’ the terms, (i) Input resistance (ii) Output resistance (iii) Currentamplification factor, for a given transistor. (Not in Syllabus)

Ans. (a) See Q. 19 (ii), 2016 (I Delhi). [Page 144


Shiv Das

(b) (i) Input Resistance : ri =

ΔΔVIBE

B VCE

⎛⎝⎜

⎞⎠⎟

(ii) Output Resistance : r0 =

ΔΔVICE

C VB

⎛⎝⎜

⎞⎠⎟

(iii) Current amplification factor : β =

ΔΔ

IIC

B VCE

⎛⎝⎜

⎞⎠⎟


236 shiv das senior secondary series (Xii)

Shiv Das

set i

section A

Q.1. Draw a plot showing variation of electric field with distance from the centre of a solid conducting sphere of radius R, having a charge of +Q on its surface. 1

Ans. Plot between E and r

Q.2. State one factor which determines the intensity of light in the photon picture of light. 1 Ans. The factor determining the intensity of light is number of electrons emitted per second. Q.3. An iron-cored solenoid has self-inductance 2.8 H. When the core is removed, the self

inductance becomes 2 mH. What is the relative permeability of the core used? 1 Ans. Given : L = 2.8 H, L0 = 2 mH = 2 × 10–3 H

Relative permeability, μr = L

L0 32 8

2 10=

× −.

= 1400

Q.4. An object is kept in front of a concave lens. What is the nature of the image formed? 1 Ans. When an object is kept in front of a concave lens, the nature of image formed is virtual,

erect and diminished. Q.5. When light travels from a rarer medium to denser medium, the speed of light decreases.

Does the reduction in speed imply a reduction in the energy? 1 Ans. The reduction in speed, due to light travelling from a rarer to denser medium does not

imply reduction in the energy.

section B

Q.6. How is electromagnetic wave produced? Draw a sketch of a plane e.m. wave propagating along X-axis depicting the directions of the oscillating electric and magnetic fields. 2


General InstructIons Same as in Set-1, Delhi Board 2016.

cbse

physics (Theory) — 2017Comptt. (DELHI)

R

Er > R

r →

physics (theory) – 2017 comptt. (delhi) 237

Shiv Das

Ans. Electromagnetic waves are produced due to oscillating/accelerating charged particles. Sketch of e.m. wave :

y or z

z or yB

E

x

Q.7. A charge q of mass m is moving with a velocity of v, at right angles to a uniform magnetic field B. Deduce the expression for the radius of the circular path it describes. 2

Ans. Force experienced by charged particle in magnetic field

F B→ → →

= ×q v( )

As v and B are perpendicular, F = qvB This force is perpendicular to the direction of velocity and hence acts as

centripetal force mv

r

2

mvr

2= qvB

On solving,

r = mvqB

Q.8. Calculate the shortest wavelength of light emitted in the Paschen series of hydrogen spectrum. Which part of the electromagnetic spectrum, does it belong to ? (Given : Rydberg constant, R = 1.1 × 107 m–1) 2

Ans. Given : For shortest wavelength, ni = ∞, nf = 3 R = 1.1 × 107 m–1

λ–1 = R 1 12 2n

if n−

= 1.1 × 107

13

12 −

∞

∴ λ–1 = 1.1 × 107 19

∴ λ = 8.18 × 10–7 m = 818 nm

It belongs to the Infrared part of the spectrum. Q.9. A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive

index μ upto a height H. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut-off the light from the bulb. 2

Ans. It is only the light coming out from a cone of semi-vertical angle ic (ic = sin–1 1µ

= critical

angle) that needs to be stopped by the opaque disc.


Shiv Das

Now, sin ic = 1µ

∴ cos ic = 1 12−

µ …[ cos θ = 1 2− sin θ

Also, tan ic = rH

⇒ r = H tan ic = Hsincos

iic

c = H.

1

112

µ

µ−

r = H

µ2 1−

∴ Diameter of the opaque disc = 2r = 2H

µ2 1−

Or A ray of light is incident on a glass prism of refractive index μ and refractive angle A. If

it just suffers total internal reflection at the other face, obtain an expression relating the angle of incidence, angle of prism and critical angle. 2

Ans. μ = sinsin

ir

and 1µ= =sin

sinsini

ec

ci

∠ = ∠∠ = ∠∠ = ° ° =

i i

r e

e

c

90 90 1; sin ∠A + ∠ P = 180° and ∠r + ∠ic = 180° – ∠P = ∠A ⇒ ∠r = ∠A – ∠ic

⇒ μ = sin

sin( )i

icA −

∴ 1sin

sinsin( )i

iic c

=−A

Q.10. Depict the behaviour of magnetic field lines near (i) diamagnetic and (ii) paramagnetic substances. Justify, giving reasons. 2

Ans. Behaviour of magnetic lines of force near (i) diamagnetic substances (ii) paramagnetic substances

Justification : The field lines are repelled or expelled and the field inside the material is reduced near diamagnetic substances.

ic

A

P

e = 90°

μ

r∠i

R = 90°r

H

μ

H

ic

icic


Shiv Das

In the presence of magnetic field, the individual atomic dipoles can get aligned in the direction of the applied magnetic field. Therefore, field lines get concentrated inside the material and the field inside is enhanced near paramagnetic substances.

section c

Q.11. Draw a graph showing the variation of de Broglie wavelength of a particle of charge q and mass m with the accelerating potential. Proton and deuteron have the same de Broglie wavelengths. Explain which has more kinetic energy. 3

Ans. We have, λ = hmq

hm2 2V K

= (K = qV = K.E.)

Mass of deuteron is more than that of proton,

i.e., md > mp

∴ For same λ, we must have Kp > Kd i.e., the proton has more kinetic energy. Q.12. Explain the term, ‘amplitude modulation’ of a signal. For an amplitude modulated

wave, the maximum amplitude is 10 V and the minimum amplitude is 2 V. Calculate the modulation index. (Not in Syllabus) 3

Ans. Amplitude modulation is the process of superposition of a message signal over a carrier wave in which amplitude of the carrier wave is varied in accordance with the message/information signal.

Given : (A) maximum = 10 V, (A) minimum = 2V, μ = ? ac + am = 10 ac – am = 2 ∴ 2ac = 12 ⇒ ac = 6V ∴ am = 4V

μ = aam

c= =4

623

∴ Modulation index (μ) = 23

Q.13. State the Lorentz’s force and express it in vector form. Which pair of vectors are always perpendicular to each other? Derive the expression for the force acting on a current carrying conductor of length L in a uniform magnetic field ‘B’. 3

Ans. Lorentz’s magnetic force is force experienced by a charged particle of charge ‘q’ moving in

magnetic field B→

with velocity v→

.

F Bm q v→

= ×→ →

( )

Two pairs F andm v→ →

and F and Bm

→ →are always perpendicular to each other.

Let us consider a conductor of uniform cross-sectional area A and length ‘L’ having number density of electrons as ‘n’.

Total force on charge carriers in the conductor, F = ( AL) B

→ →→×[ ]n q vd

λ

V


Shiv Das

But since nq vdA L = I L→→ →

∴ F I ( L B→ → →

= × )

Q.14. An optical instrument uses eye-lens of power 16 D and objective lens of power 50 D and has a tube length of 16.25 cm. Name the optical instrument and calculate its magnifying power if it forms the final image at infinity. 3

Ans. The optical instrument is Compound Microscope. Given : Pe = 16 D, L = 16.25 cm, Po = 50 D

We know, f = 1P

fe = 10016

= 6.67 cm, fo = 10050

= 2 cm

D = 25 cm The magnifying power of compound microscope is

m = L Df fo e× = 16 25

2 025

6 67.. .

× = 30.45

Q.15. Explain the two processes involved in the formulation of a p-n junction diode. Hence define the term ‘barrier potential’. 3

Ans. See Q. 12, 2017 (I Delhi). [Page 198Q.16. (a) Write two properties by which electric potential is related to the electric field. (b) Two point charges q1 and q2, separated by a distance of r12 are kept in an external

electric field. Derive an expression for the potential energy of the system of two charges in the field. 3

Ans. (a) Two properties which relate electric potential to electric field— (i) Electric field is in the direction in which potential decreases at the maximum

rate. (ii) Magnitude of electric field is given by change in the magnitude of potential per

unit displacement normal to a charged conducting surface, or E = ddrV

(b) Work done in bringing the charge q1 to a point against external electric field,

W1 = q1V ( )r1→

Work done in bringing the charge q2 against the external electric field and the electric field produced due to charge q1

W2 = q2V ( ) .r q qr2

0

1 2

12

14

→+

πε

Therefore, Total work done = Electrostatic potential energy

U = q qq qr

r r1 1 2 20

1 2

12

14

V V( ) ( ) .→ →+ +πε


Shiv Das

Or State Gauss’s law in electrostatics. Derive an expression for the electric field due to an

infinitely long straight uniformly charged wire. 3

Ans. Gauss’s Theorem : The surface integral of electric field over a closed surface is equal to 1

0ε

times the charge enclosed by the surface.

Alternatively, E→ →

∫ =

. ds qε0

Expression for electric field

Flux through the Gaussian surface

= Flux through the curved cylindrical part of the surface

= E × 2πrl

Charge enclosed by the surface = λl

⇒ E × 2πrl = λε

l

0 or E =

λπε2 0r

Q.17. State Lenz’s law. Explain, by giving examples that Lenz’s law is a consequence of conservation of energy. 3

Ans. Lenz’s law states that “The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.”

As shown in the diagram (a) given, when the north pole of a bar magnet is pushed towards the close coil, the magnetic flux through the coil increases and the current is induced in the coil in such a direction that it opposes the increase in flux. This is possible when the induced current in the coil is in the anticlockwise direction.

Similarly as shown in the diagram (b), just the opposite happens when the north pole is moved away from the coil.

In either case, it is the work done against the force of magnetic repulsion or attraction that gets ‘converted’ into the induced emf.

Q.18. A capacitor of unknown capacitance is connected across a battery of V volt. A charge of 360 μC is stored in it. When the potential across the capacitor is reduced by 120 V, the charge stored in the capacitor becomes 120 μC. Calculate V and the unknown capacitance. What would have been the charge on the capacitor if the voltage were increased by 120 V? 3

Ans. Given : q1 = 360 μC = 360 × 10–6 C, q2 = 120 μC = 120 × 10–6 C

C = q

1

1V. Also C =

q2

2V

and C = q

3

3V

[ Capacitor is the same

q q1

1

2

2V V

=

P

E

++++++++++++++++++++

r

rl

N

N

(a)

(b)


Shiv Das

⇒ ( ) ( )

( )360 10 120 10

120

6 6× ×−

− −=

V V

On solving, V = 180 V

C = 360 10180

6× − = 2μF is the unknown capacitance.

Now the voltage has been increased by 120 V, then V = 180 + 120 = 300 V

C = q

3300

= 2μF

q3 = 300 × μC

q3 = 600 μC would be the charge on the capacitor if voltage were increased by 120 V. Q.19. A plane wavefront propagating from a denser into a rarer medium is incident at an angle

of incidence i on a refracting surface. Draw a diagram showing incident wavefront and refracted wavefront. Hence verify Snell’s laws of refraction. 3

Ans. See Q. 18, 2015 (I Outside Delhi). [Pages 98-99 Q.20. Distinguish between sky wave and space modes of communication. What is the main

limitation of space wave mode? Write the expression for the optimum separation between the transmitting and receiving antenna for effective reception of signals in this mode of communication. (Not in Syllabus) 3

Ans. (i) In sky wave mode of communication, waves reach from transmitting antenna to receiving antenna through reflections from ionosphere; while in space wave mode of communication, waves travel either directly from the transmitter to receiver or through satellites.

(ii) Direct waves get blocked at some point due to the curvature of earth. (iii) Optimum distance between transmitting and receiving antenna = 2 2h hT RR R+Q.21. Using the wave forms of the input A and B , draw the output waveform of the given logic

circuit. Identify the logic gate obtained. Write also the truth table. (Not in Syllabus) 3

AOutputB

Input A

O

Input B

OLogic Circuit

Ans. (i) Output wave form (ii) NAND GATE

0

1

(iii) Truth Table

Inputs Output

A B 1 1 0 0 0 1 1 1 1 0 0 1


Shiv Das

Q.22. Derive the expression for the current density of a conductor in terms of the conductivity and applied electric field. Explain, with reason how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant. 3

Ans. (i) Derivation of expression for current density— Using Ohm’s law,

V = IR = IA A

Iρ ρl l=

…(i)

Potential difference (V), across the ends of a conductor of length ‘l’ where field ‘E’ is applied, is given by

V = El …(ii) From equations (i) and (ii),

∴ El = IAρl

But current density J = IA

El = Jρl = Jlσ

ρσ

=

1

⇒ J = σE

(ii) Mobility, μ = vE

v

l

d d= V

So, as potential is doubled, drift velocity also gets doubled, therefore, there will be no change in mobility.

section D

Q.23. Ram was a daily wage worker in a factory. He was suffering from cancer. On hearing this, most of his co-workeers, started avoiding him under the impression that it was a contagious disease. When Prof. Srivastava came to know about this case, he took him to a leading radiologist, who examined him and told that it was at the beginning stage. He advised that it could be easily cured and also certified that it was not a communicable disease. After this, Ram was given proper treatment by the doctor and got cured completely.

*(a) What moral values did Prof. Srivastava display? (b) How is mean life of a radioactive element related to its half life? (c) A radioactive sample has activity of 10,000 disintegrations per second after 20 hours.

After next 10 hours its activity reduces to 5,000 dps. Find out its half-life and initial activity. 4

Ans. *(a) Value Based Questions will not be asked in the Board Examination. (b) Mean life = (half life/0.693) or (1.44 times half life)


Shiv Das

(c) Half life = 10 hours [because activity reduces to half in 10 hours

R = R RR0

012

2

⇒ =

nn( )

R10,000

0 = (2)2 [R = 10,000, n = 2

⇒ R0 = 40000 dps section e

Q.24. In the given circuit, calculate (a) the capacitance of the capacitor, if the power factor of the circuit is unity, (b) the Q-factor of this circuit. What is the significance of the Q-factor in a.c. circuit? Given the angular frequency of the a.c. source to be 100/s. Calculate the average power dissipated in the circuit. 5

Ans. Given : R = 10 Ω, L = 100 mH = 100 × 10–3 H, V = 50 V, ω = 100 s–1

(a) As power factor is unity, ∴ XL = XC

⇒ ω = 1LC

100 = 1

200 10 3( )× ×− C

Squaring both sides, we have

104 × (2 × 102) × 10–3 × C = 1

C = 12 103×

F = 0.5 × 10–3 F = 0.5 mF

(b) Quality factor, Q = 1R

LC

= 110

200 100 5 10

3

3( )( . )

××

−

− = 1

10× 20 = 2

Significance: It measures the sharpness of resonance. Average Power dissipated— P = Vrms Irms cos φ

= 50 × 5010

× 1 [cos φ = 1 for power factor to be unity]

= 250 watts Or

(a) Prove that the current flowing through an ideal inductor connected across a.c.

source, lags the voltage in phase by π2

.

(b) An inductor of self inductance 100 mH, and a bulb are connected in series with a.c. source of rms voltage 10 V, 50 Hz. It is found that effective voltage of the circuit

C

V = 50 V

R

10 Ω

L

200 mH


Shiv Das

leads the current in phase by π4

. Calculate the inductance of the inductor used and

average power dissipated in the circuit, if a current of 1 A flows in the circuit. 5

Ans. (a) induced emf, e = –L ddtI

Hence net voltage in the circuit = V L I−

ddt

According to Kirchoff’s Rule,

V – L ddtI = 0

Vm sin ωt = L ddtI

d t dtmI

IVL

0 0∫ ∫=

ωωsin

I = − VLm

ω cos ωt = V

Lm t

ωπω

−sin

2 V

Lm

ω = Im

∴ I = Im sin ω πt −

2

Hence, current lags by π2

.

(b) Given : L = 100 mH = 100 × 10–3 H, Vrms = 10 V Frequency = 50 Hz, φ = π

4, I = 1A

Average power dissipation P = Vrms Irms cos φ

= 10 × 1 × cos π4

=10

2W = 5 2 Watts cos π

412

=

Q.25. Explain with diagram, how plane polarized light can be produced by scattering of sunlight. An incident beam of light of intensity Io is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced mid-way between A and B and is so oriented that its axis bisects the angle between the axes of A and B. Calculate the intensity of light transmitted by A, B and C. 5

Ans. (i) It is due to scattering of light by molecules of earth’s atmosphere.

Under the influence of the electric field of the incident (unpolarized) wave, the electrons in the molecules acquire components of motion in both these directions. Charges, accelerating parallel to the double arrows, do not radiate energy towards the observer since their acceleration has no transverse component.

The radiation scattered by the molecules is therefore represented by dots, i.e., it is polarized perpendicular to plane of figure.

E L

To observer

Incident Sunlight (Unpolarized)

Scattered Light (Polarized)


Shiv Das

(ii) •IntensityoflighttransmittedthroughA= I02

•TransmittedthroughPolaroid‘C’:

I’ = I02

cos2 45° = I0

2

212

×

= I0

4

•TransmittedthroughPolaroid‘B’:

I” = I04

cos2 45° = I02

412

×

= I08

Or (a) In Young’s double slit experiment a monochromatic source of light S is kept

equidistant from the slits S1 and S2. Explain the formation of dark and bright fringes on the screen.

(b) A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in Young’s double-slit experiment.

(i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm?

(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Given : the separation between the slits is 4 mm and the distance between the screen and plane of the slits is 1.2 m. 5

Ans. (a) See Q. 29 (a), 2014 (I Outside Delhi). [Page 67 (b) (i) Given : n = 3, λ = 650 nm = 650 × 10–9 m, D = 1.2 m, d = 4 mm = 4 × 10–3 m We know χn = n

dλD

= 3 650 10 1 2

4 10

9

3× × ×

×

−

−( ) .

( )

= 585 × 10–6 m = 0.585 mm (ii) Given : λ1 = 650 nm, λ2 = 520 nm, D = 1.2 m, d = 4 mm = 4 × 10–3 m Since both fringes coincide, we have

nd

nd

1 1 2 2λ λD D=

⇒ n1λ1 = n2λ2

nn

1

2

2

1

520650

45

= = =λλ

Therefore, 4th bright fringe of λ = 650 mm will coincide with 5th bright fringe of 520 mm. Least distance from central maximum where bright fringes of both wavelengths

coincide

χ = 4 650 10 1 2

4 10

9

3× × ×

×

−

−( ) .

( ) m = 780 × 10–6 m = 0.78 nm

I0/2I0

A

B

45°45°

C


Shiv Das

Q.26. (a) Draw a circuit diagram of a meter bridge used to determine the unknown resistance R of a given wire. Hence derive the expression for R in terms of the known resistance S.

(b) What does the term ‘end error’ in a metre bridge circuit mean and how is it corrected? How will the balancing point be affected, if the positions of the battery and galvanometer are interchanged in a metre bridge experiment? Give reason for your answer. 5

Ans. (a) Metre Bridge is special case of Wheatstone Bridge. It is a device based on Wheatstone bridge to determine the unknown resistance of a

wire. Principle : Meter bridge is based on the principle of wheatstone bridge, i.e. when

bridge is balanced

ll( )100 −= R

S or S = R( )100 − l

l

Circuit : To find the unknown resistance S, the circuit is completed as shown in figure. The unknown resistance wire of resistance S is connected across the gap between points C and D and a resistance box (R) is connected across the gap between the points A and D. A cell, a rheostat and a key (K) is connected between the points A and C by means of connecting screws. In the experiment when the sliding jockey touches the wire AC at any point, then the wire is divided into two parts. These two parts AB and BC act as the resistances P and Q of the Wheatstone bridge. In this way the resistances of arms AB, BC, AD and DC form the resistances P, Q, R and S of Wheatstone bridge. Thus the circuit of meter bridge is the same as that of Wheatstone bridge.

Method : To determine the unknown resistance first of all key K is closed and a resistance R is taken in the resistance box in such a way that on pressing jockey B at end points A and C, the deflection in galvanometer is on both the sides. Now jockey is slided on wire at such a position that on pressing the jockey on the wire at that point, there is no deflection in the galvanometer G. In this position the points B and D are at the same potential, therefore the bridge is balanced. The point B is called the null point. The length of both parts AB and BC of the wire are read on the scale. The condition of balance of Wheatstone bridge is,

PQ

RS

=

∴ Unknown resistance, S = QP

R …(i)

If r is the resistance per cm length of wire AC and l cm is the length of wire AB, then length of wire BC will be (100 – l) cm

∴ P = resistance of wire AB = lr

Resistance box Resistance wire

(R) (S)

D

GA (P) B (Q)

l cm (100 – l) cm

C

(Rh)

RheostatK

+ –Cell


Shiv Das

Q = resistance of wire BC = (100 – l)r

Substituting these values in equation (i), we get

S = ( )100 − l rlr

× R ⇒ S = R …(ii)

As the resistance (R) of wire (AB) is known, the resistance S may be calculated. A number of observations are taken for different resistances taken in resistance box and

S is calculated each time and the mean value of S is found. (b) The error which arises on account of resistance of copper strips and the connecting

wire at both ends of the meter bridge is called end error. It is minimized by adjusting the balance point near the middle point of the bridge.

•Thebalancingpointwillnot be affected, as the bridge remains balanced. Or

(a) State the working principle of a potentiometer with help of a circuit diagram, explain how the internal resistance of a cell is determined.

(b) How are the following affected in the potentiometer circuit when (i) the internal resistance of the driver cell increases and (ii) the series resistor connected to the driver cell is reduced? Justify your answer. 5

Ans. (a) Potentiometer principle : When a constant current flows through a wire of uniform cross sectional area, the potential difference, across any length, is directly proportional to the length.

V ∝ L

ε = φ l1 …(i) V = φl2 … (ii)

εV= l

l1

2 … (iii)

Since ε = I(r + R) and V = IR

Therefore, εV

RR

= +r …(iv)

From (iii) and (iv),

r = R ll1

21−

(b) As per CBSE guidelines this question is incomplete hence NOT attempted.

set ii

Note: Except for the following questions, all the remaining questions have been asked in Set-I. Q.5. State one reason to explain why wave theory of light does not support photoelectric

effect. 1 Ans. One reason why wave theory of light does not support photoelectric effect is that the

kinetic energy of photo electrons does not depend on the intensity of incident light. Q.9. A proton and an alpha particle having the same kinetic energy are, in turn, passed

through a region of uniform magnetic field, acting normal to the plane of the paper and

B

R

K

K

GN2

N1

A

C

RB


Shiv Das

travel in circular paths. Deduce the ratio of the radii of the circular paths described by them. 2

Ans. Given : Ep = Eα = E

We know, mvr

2= qvB

or r = 2mq

EB

E = 12

mv2

rp = 2m

qp

p

E

B and rα =

2mq

α

α

EB

m m

q qp

p

α

α

=

=

4

2

r

rp

α =

2

2

m

q Bqm

m

mqq

p

p

p

p

E BE

× = ×α

α αα =

m

m

q

qp

p

p

p4

2 12

2 1× = × =

rp : rα :: 1 : 1

Q.10. Calculate the shortest wavelength of photons emitted in the Bracket series of hydrogen spectrum. Which part of the e.m. spectrum, does it belong? [Given Rydberg constant, R = 1.1 × 107 m–1] 2

Ans. In bracket series, nf = 4 For the shortest wavelength, ni = ∞

1λ

= R 1

41

2 2−∞

( )

1λ

= R16

∴ λ = 16 161 1 107R

=×.

= 14.54 × 10–7 m = 1454 nm

The wavelength corresponds to “infrared” part of spectrum. Q.15. (a) Draw the graph showing the variation of de Broglie wavelength λ of a particle of

charge q and mass m with the accelerating potential. (b) An electron and proton have the same de Broglie wavelengths. Explain, which of

the two has more kinetic energy. 3

Ans. (a) For Graph : See Q. 11 of Set I.

(b) de Broglie wavelength, λ = hp

hmv

hm

= =2 E

E

or E

=

=

12

2

2

mv

mv m

Given : λe = λp , qp = qe

h

me e2 E = h

mp p2 E

meEe = mpEp

EE

e

p

p

e

m

m= or E ∝ 1

m


Shiv Das

Since the mass of electron is less than that of proton, hence electron will have more kinetic energy.

Q.16. Explain the meaning of terms : Attenuation and Demodulation. For an amplitude modulated wave, the maximum amplitude is 12 V and the minimum amplitude is 2 V. Calculate the modulation index. (Not in Syllabus) 3

Ans. Attenuation : It is the loss of strength of signal, while propagating through a medium. Demodulation : It is the process of recovery of audio signal from the modulated wave. Given : (A)max = 12 V, (A)min = 2V ac + am = 12 ac – am = 2

ac = 7 V, am = 5 V ∴ μ = aam

c= 5

7

Q.17. Define the term magnetic moment of a current loop. Derive an expression for the magnetic field at any point along the axis of a solenoid of length 2l, and radius a, and number of terms per unit length n. 3

Ans. (i) Definition of magnetic moment : Magnetic moment of a current loop is equal to the product of current flowing in the loop and its area; and its direction is along area vector as per the right handed screw rule.

(ii) Magnetic field for solenoid : Using Ampere’s circuital law

B→ →

∫ =

. dl nhIµ0

m→ →

= I A

Bh = μ0nhI ⇒ B = μ0nI

set iii

Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set II.

Q.4. If the distance between the source of light and the cathode of a photo cell is doubled, how does it affect the stopping potential applied to the photo cell? 1

Ans. Stopping potential remains unchanged, if the distance between the light source and cathode is doubled.

Q.10. Calculate the longest wavelength of the photons emitted in the Balmer series of hydrogen spectrum. Which part of the e.m. spectrum, does it belong to? [Given Rydberg constant, R = 1.1 × 107 m–1]. 2

Ans. Balmer series is produced when an electron jumps from higher orbits to second stationary orbit (nf = 2).

Thus for this series,

1 1

21

2 2λ=

−R

ni [ni = 3, 4, 5, 6]

For longest wavelength, ni = 3;

B

Q

P

a b

cw

dh


Shiv Das

1 12

13

5362 2λmax

=

− =R R

λmax = 365

365 1 1 107R

=× ×( . )

= 6.563 × 10–7 = 6563Å

It belongs to ‘Visible’ part of spectrum. Q.11. A plane wavefront is incident at an angle of incidence i on a reflecting surface. Draw

a diagram showing incident wavefront, reflected wavefront and verify the laws of reflection. 3

Ans. Diagram of a plane wave front for Reflection :

AM

EB

CN

Incidentwavefront

Reflectedwavefront

r

i

i

Since time taken by waves from point B to C and from A to E is same ∴ BC = AE = vτ In ΔABC and ΔAEC, AC = AC (common) ∠ABC = ∠AEC (90° each) AE = BC ∴ ΔABC ≅ ΔAEC Hence ∠BAC = ∠ECA ∠i = ∠r i.e., Angle of incidence = Angle of reflection Q.15. Draw a graph showing the variation of de Broglie wavelength λ of a particle of charge

q and mass m, with the accelerating potential V. An α-particle and a proton have the same de Broglie wavelength equal to 1Å. Explain with calculations, which of the two has more kinetic energy. 3

Ans. For graph : See Q. 11 of Set I.

de Broglie hypothesis—

λ = hp

hmv

hm

= =2 E E or E= =

1

22 2mv mv m

λα = λp = 1 Å

λα = h

m2 α αE and λp = h

mp p2 E

λα = λp


Shiv Das

h

m2 α αE=

hmp p2 E ⇒

E

Ep

α=

mm

m

mp

p

p

α =4

= 2 ⇒ Ep = 2Eα

∴ Proton has double energy than that of α-particle. Q.16. Write the functions of (i) Repeater and (ii) Receiver. For an amplitude modulated wave,

the maximum amplitude is 15 V and the minimum amplitude is 3 V. Calculate the modulation index. (Not in Syllabus) 3

Ans. Repeater is a combination of a receiver and a transmitter. It is used to increase the range of communication of signals. A repeater picks up the signal from the transmitter, amplifies it and retransmits it to the receiver.

Receiver extracts the desired message signals from the received signals at the channel output. It is the combination of receiving antenna, amplifier, intermediate frequency converter, demodulator and amplifier of audio signal.

Given : (A)max = 15 V, (A)min = 3V ac + am = 15 ac – am = 3

On solving we get, ac = 9, am = 6 μ = aam

c= =6

923

physics (theory) – 2017 comptt. (outside delhi) 253

Shiv Das

set i

section A

Q.1. A point charge +Q is placed in the vicinity of a conducting surface. Draw the electric field lines between the surface and the charge. 1

Ans. Electric field lines between surface and charge

+Q

Q.2. Define modulation index. Why is it generally kept less than one? (Not in Syllabus) 1 Ans. Modulation index is defined as the ratio of amplitude of modulating signal (Am) to

amplitude of carrier wave (Ac)

μ = AA

m

c

It is kept less than one to avoid distortion.

Q.3. In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. 1

Ans. The polarity of plate is positive; while that of plate B is negative.

Q.4. The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1 : 2. Compare the resolving powers of the two telescopes. 1

Ans. Ratioofresolvingpower=1:1 Resolving power is same because it does not depend on focal length of the objective.

Q.5. Define the conductivity of a conductor. Write its SI unit. 1


General InstructIons Same as in Set-1, Delhi Board 2016.

cbse

physics (Theory) — 2017Comptt. (outsIDE DELHI)

BS N S N

A


Shiv Das

Ans. Conductivity is defined as the reciprocal of resistivity, i.e., σ = 1ρ

Its SI unit is S(siemen).

section B

Q.6. (i) Define refractive index of a medium. (ii) In the following ray diagram, calculate the

speed of light in the liquid of unknown refractive index. 2

Ans. (i) The refractive index of a medium is defined as the ratio of speed of light in vacuum (c) to the speed of light in the medium (v)

μ = cv

(ii) Given : Width of liquid surface (w) = 30 cm Depth of liquid (d) = 40 cm Since it is a critical case of total internal reflection, when refracted ray gazes along the

liquid surface.

∴ μ = 1

sin ic

From the given ray diagram, the length of ray inside liquid is 50 cm

30 402 2+ ( ) = 50

∴ sin ic = perpendicular

hypotenuse= 30

50

Also from the definition

μ = cv ic= 1

sin

or v = (c) sin ic = (3 × 108) 3050

= 1.8 × 108 ms–1

Q.7. Electrons are emitted from the cathode of a photocell of negligible work function, when photons of wavelength are incident on it. Derive the expression for the de Broglie wavelength of the electrons emitted in terms of the wavelength of the incident light. 2

Ans. Expression for wavelength : When work function is negligible, we have, from Einstein’s equation

12

mv2 = hcλ

∴ v = 2hcmλ

…(i)

From de Broglie hypothesis, we know

λdB = hmv

Liquid

30 cm

40 c

m


Shiv Das

∴ λdB = hm

mhc

λ2

(Putting the value of v from equation (i))

∴ λdB = hmcλ

2

Or Derive the Bohr’s quantisation condition for angular momentum of the orbitting of

electron in hydrogen atom, using de Broglie’s hypothesis. Ans. By de Broglie’s hypothesis

We have λ = hp

hmvn

=

From Bohr’s postulate, 2πrn = nλ (n = 1, 2, 3)

∴ 2πrn = nh

mvn

mvnrn = nh2π

∴ Angular momentum = (mvnrn) = nh2π

Q.8. (a) Write two characteristic features of nuclear force. (b) Draw a plot of potential energy of a pair of nucleons as a function of their

separation. 2 Ans. See Q. 21, 2015 (I Outside Delhi). [Page 100 Q.9. State the two points to distinguish between sky wave and space wave modes of

propagation. (Not in Syllabus) 2 Ans. Distinction between sky wave and space wave propagation : (i) Sky wave propagation uses reflection from ionosphere; whereas space wave

propagation uses line of sight of propagation. (ii) Sky wave propagation is for waves of frequency between 3 to 30 MHz; whereas space

wave propagation is preferred for waves of frequency more than 40 MHz.. Q.10. The figure shows a plot of terminal voltage ‘V’ versus

the current ‘i’ of a given cell. Calculate from the graph (a) emf of the cell and (b) internal resistance of the cell . 2

Ans. Fromthegraph: (a) E = V = 6V for when I = 0 (no current is drawn) (b) E = V + Ir [ From the graph for I = 1 A, V = 4V] 6 = 4 + 1(r) or r = 2 Ω

i

V

6 V

4 V

0 1.0 A 2.0 A


Shiv Das

section c

Q.11. A parallel plate capacitor of capacitance C is charged to a potential V by a battery. Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates of the capacitor. Explain giving reasons, how will the following be affected:

(i) capacitance of the capacitor (ii) charge on the capacitor, and (iii) energy density of the capacitor. 3

Ans. Given : d′ = 3d, K = 10, C′ = ?, Q′ = ?, Ud′ = ? (i) For parallel plate capacitor

C = ∈0 Ad

Let the new capacity be C′

C′ = K A A∈′

× ∈=0 0103d d( ) [K = 10, d′ = 3d]

⇒ 103

0

∈

Ad

= 103

C

⇒ C′ = 103

C

(ii) Since V remains the same as the battery is not disconnected, Q′ = C′V

Q′ = 103

103

103

C V CV Q

= =

⇒ Q′ = 103

Q

(iii) We know energy density (Ud) = 12

∈0 E2 …(i)

Ud′ = 12

K ∈0E′2

= 12 0

2K V∈

′

d

E V=

d

= 102 3

02

2

∈

V( )d

= 102

190

2

∈ × V

d

= 109

12 0

2

∈ E

= 109

Ud ( as per equation (i))

∴ Ud′ = 109

Ud

Q.12. (a) Draw a graph showing the variation of binding energy per nucleon (BE/A) vs mass number A for the nuclei in 20 ≤ A ≤ 170.


Shiv Das

(b) A nucleus of mass number 240 and having binding energy/nucleon 7.6 MeV splits into two fragments Y, Z of mass numbers 110 and 130 respectively. If the binding energy/nucleon of Y, Z is equal to 8.5 MeV each, calculate the energy released in the nuclear reaction. 3

Ans. (a) Binary Energy Curve:

20 40 60 80 100 120 140 160 180 200 220 240

9

8

7

6

5

4

3

2

1

7.6 mMeV

8.8 MeV

Bin

ary

ener

gy p

er n

ucle

on (M

eV)

Mass Number (A)

816O 26

56Fe 92235U

24He

1 1H

Conclusions : (i) Theforceisattractiveandsufficientlystrongtoproduceabindingenergyofafew

MeVpernucleon. (ii) Formassnumber2 to20, therearesharplydefinedpeakscorresponding to 2He4,

6C12, 8O16 etc. Thepeak indicates that their nuclei are relativelymass stable thantheothernuclei in theirneighbourhood.

(b) Given : MX = 240, BX = 7.6 MeV, MY = 110, MZ = 130 BY = BZ = 8.5 MeV, E = ? Energy released E = (MY + MZ)BZ – (MX × BX) = (110 + 130)8.5 – (240 × 7.6) = (240 × 8.5) – (240 × 7.6) = 240(8.5 – 7.6) = 240 × 0.9 = 216 MeVQ.13. (a) In a Young’s double slit experiment, the two slits are illuminated by two different

lamps having same wavelength of light. Explain with reason, whether interference pattern will be observed on the screen or not.

(b) Light waves from two coherent sources arrive at two points on a screen with path differences of 0 and λ/2 . Find the ratio of intensities at the points. 3

Ans. (a) Interference pattern in Young’s double slit experiment will not be observed, because two independent lamps do not constitute ‘coherent sources’.

(b) (i) When path difference is zero, corresponding phase difference will also be zero. We know, I = 4I0

2 cos2 ϕ I1 = 4I2 cos2 0 = 4I2 × 1 = 4I2

(ii) When path difference is λ2

, the corresponding phase difference will be π.

I2 = 4I2 cos2 π = 4I2 × 1 = 4I2

∴ II

II

1

2

2

244

= =1:1


Shiv Das

Q.14. Using Bohr’s postulates, derive the expression for the total energy of the electron revolving in nth orbit of hydrogen atom. Find the wavelength of Hα line, given the value of Ryderg constant, R = 1.1 × 107 m–1. 3

Ans. •For total energy of the electron orbiting in nth orbit : Total energy of electron in Bohr’s stationary orbit K.E. which is due to velocity and P.E.

due to position of electron. From the first postulate of Bohr’s atom modal

mvr

er

2 2

2= KZ ∴ 12

12

22

mv er

= KZ

i.e., K.E. of electron = 12 2

22

mv er

= KZ …(i)

Potential due to the nucleus = KZer

∴ P.E. of electron = Potential × Charge = KZ Ze er

K er

( )− −=2 …(ii)

P.E. of electron in the orbit, E = K.E. + P.E.

= KZ Z KZ KZ KZe

rK e

re

re

rer

2 2 2 2 2

212 2

+ = − = −−

Putting the value of r = n hm e

2 2

2 24π KZ, we get (By Bohr’s formula)

E = − ×KZ K Ze m en h

2 2 2 2

2 224π = − 2 2 2 2 4

2 2π m e

n hK Z

•Given : R = 1.1 × 107 m–1

λ for Hα = ?

For Hα line, ni = 3, nf = 2

We know 1λ

= R 1 12 2n nf i−

1λ

= 1.1 × 107 1

2

1

32 2−

= 1.1 × 107 14

19

−

= 1.1 × 107 × 536

λ = 365 5.

× 10–7 = 6.55 × 10–7 m = 655 × 10–9 m = 655 nm

Q.15. Name the e.m. waves in the wavelength range 10 nm to 10–3 nm. How are these waves generated? Write their two uses. 3

Ans. •e.m. waves in the wavelength range 10 nm to 10–3 nm are X-rays. •X-raysaregeneratedbybombardingametaltargetwithhighenergyelectrons. •Uses : (i) Diagnosis of bone fractures. (ii) Treatment of some forms of cancer.


Shiv Das

ix

x

XO

P

Y O′

Q.16. (a) Draw the pattern of magnetic field lines for a circular coil carrying current.

(b) Two identical circular loops X and Y of radius R and carrying the same current are kept in perpendicular planes such that they have a common centre at P as shown in the figure. Find the magnitude and direction of the net magnetic field at the point P due to the loops. 3

Ans. (a) Patternofmagneticfieldlinesforacircularcoilcarryingcurrent: See Q. 22 (ii), 2016 (I Outside Delhi). [Page 161 (b) The magnetic field due to a circular coil at a point carrying current is given by

B = µ0

2

2 2 3 22ixR

R( ) /+

Since these two circular coils are identical and carrying the same current,

BX = BY = µ0

2

2 2 3 22ixR

R( ) /+= B

Resultant magnetic field (BR)

= B BX Y2 2+ = B B2 2+ = B ( )1 1 2+ = B

= 2

2

2

2 2 3 2µai

xR

R( ) /+ and making 45° with BX or BY.

Q.17. State the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode, has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of wavelength of 400 nm incident on it. 3

Ans. See Q. 19, 2015 (I Delhi). [Page 85 Detection is possible if Ep > Eg

Ep = hcλ

J

= hceλ

eV

= ( . ) ( )

( . ) ( )6 63 10 3 10

1 6 10 400 10

34 8

19 9

× × ×× × ×

−

− − = 3.1 eV > (Eg = 1.2 eV)

∴ It can detect this light.Q.18. Draw the circuit diagram of a common emitter transistor amplifier. Write the expression for

its voltage gain. Explain, how the input and output signals differ in phase by 180°. 3 (Not in Syllabus)

Ans. See Q. 20, 2015 (I Outside Delhi). [Page 99 From the circuit diagram, we find VCC = VCE + IC RL ∴ ΔVCC = ΔVCE + RL ΔIC = 0 ∴ ΔVCE = –RL ΔIC Hence, change in output is negative when the input signal is positive. This shows that input and output signals differ in phase by 180°.


Shiv Das

Or Draw the circuit diagram of a full wave rectifier. Explain its working principle. Draw the

input and output waveforms. 3 Ans. Working of a full wave rectifier : 1. A full wave rectifier uses two diodes

and gives the rectified output voltage corresponding to both the positive and negative half-cycle of alternating current.

2. The p-side of the two diodes are connected to the ends of the secondary of the transformer and, the n-sides of the diodes are connected together.

3. Output is taken from between the common-point of the two diodes and secondary of the transformer. Hence, the secondary of the transformer is provided with center tapping and is also called the centre-tap transformer.

4. Let, the input voltage to A with respect to the centre be positive and, at the same instant, voltage at B being out-of-phase will be negative. Therefore, diode D1 is forward biased and starts conducting whereas, D2 being reverse biased does not conduct.

5. Thus, we get an output current and an output voltage across the load resistance RL in the first positive half-cycle.

6. During the course of the negative half-cycle, that is, when voltage at A becomes negative and voltage at B becomes positive, we will have D1 as reverse biased and D2 forward biased.

7. In the negative part of the cycle, only diode D2 will conduct giving an output current and output voltage across RL.

8. For both positive and negative half cycle we will get the output voltage. This rectified output voltage has the shape of half sinusoids.

Q.19. Briefly explain the three factors which justify the need of modulating low frequency signal into high frequencies. (Not in Syllabus) 3

Ans. See Q. 7, 2015 (I Outside Delhi). [Page 94

Q.20. Define the term current sensitivity of a galvanometer. In the circuits shown in the figures, the galvanometer shows no deflection in each case. Find the ratio of R1 and R2. 3

G

G3 V

3 V4 Ω 6 Ω

8 Ω

12 Ω

6 Ω 4 Ω

R1

R2

→

Centre-TapTransformer Diode I(D)

CentreTap

A

(a)

(i)

(ii)

(b)

(c)t

t

t

B

Diode 2(D2)RL Output

X

Due toD1

Due toD2

Due toD1

Due toD2

Out

put

wav

efor

m(a

cros

s R

L)

Wav

efor

m a

t B

Wav

efor

m a

t A


Shiv Das

Ans. •Currentsensitivity of a galvanometer is “deflection per unit current”.

IS = φI

NABK

=

•Asperwheatstonebridge, PQ

RS

=

For circuit (1), we have

46

1= R4

⇒ R1 = 83

Ω

Similarly for circuit (2)

6 12

82R= ⇒ R2 = 4 Ω

∴ RR

1

2

834

23

= =

Q.21. The current through two inductors of self-inductance 12 mH and 30 mH is increasing with time at the same rate. Draw graphs showing the variation of the

(a) emf induced with the rate of change of current in each inductor (b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the

same. 3 Ans. (a) Graph for induced emf :

dI/dt

i

(L1 = 12 mH)

(L2 = 30 mH)

(b) Graph for energy stored

u

L2 = 30 mH L1 = 12 mH

I

Comparison of energy stored :

uu

1

2 =

1212

1 12

2 22

L

L

i

i

But ε1i1 = ε2i2 ( power dissipated is the same, as given in the question)


Shiv Das

∴ ii1

2

2

1

2

1= =ε

εLL

dIdt

dIdt

is same and also Lε = −

∴ uu

1

2

1

2

2

1

22

1

3012

= = =

LL

LL

LL

= 2.5

Q.22. (a) Explain how the intensity of diffraction pattern changes as the order (n) of the diffraction band varies.

(b) Two wavelengths of sodium light 590 nm and 596 nm are used in turn to study the diffraction at a single slit of size 4 mm. The distance between the slit and screen is 2 m. Calculate the separation between the positions of the first maximum of the diffraction pattern obtained in the two cases. 3

Ans. (a) Intensity of diffraction pattern drops rapidly with order n, because every higher order

maxima gets intensity only from 1

2 1n +

part of the slit. The central maxima gets

intensity from the whole slit (n = 0). 1st secondary maxima gets its intensity only from 1/3 of slit. 2nd secondary maxima gets its intensity only from 1/5 of slit and so on. (b) Given : λ1 = 590 nm, λ2 = 596 nm a = 4 mm = 4 × 10–3 m, D = 2 m

Position of 1stmaximaonthescreen:

x1 = 32

1λaD:λ1 = 590 nm …

x na

= +

2 12

λD

x2 = 32

2λaD:λ2 = 596 nm

Separation Δx = (x2 – x1)

= 32

Da

(λ2 – λ1)

= 32

24 10 3×

− × (596 × 10–9) – (590 × 10–9)

= 32

24 10 3

96 10×

−

−×( )

= 4.5 × 10–6 m

section D

Q.23. Mr Kamath, the chief mechanical engineer, in Northern railways went to Tokyo to attend a seminar on fast moving trains. His friend Mr Hiorki explained how Japanese people are concentrating on energy conservation and saving fossil fuels using maglev trains. Mr Kamath travelled from Tokyo to Osaka in maglev train and found that the sound is less, travel is smooth and understood the Japanese technology in mass transporting systems. Maglev trains work on the principle of Meissner’s effect.

*(a) Mention two values which Mr Kamath found in Mr Hiorki.


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(b) Which values in Mr Kamath do you appreciate? (c) What is Meissner’s effect? Write the value of magnetic permeability for perfect

diamagnetism. 4 Ans. *(a) Value Based Questions will not be asked in the Board Examination. (b) Values of Mr Kamath worth appreciating are : Eager to learn, a good observer, open

minded, appreciating good ideas. (c) Phenomenon of perfect diamagnetism in super conductors is known as ‘Meissner’s

effect’. —Magnetic permeability (μ) for perfect diamagnetism = 0 (Zero)

section e

Q.24. (a) State Gauss’ law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density λ.

(b) A wire AB of length L has linear charge density λ = kx, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface. 3+2

Ans. (a) See Q. 16 (Or), 2017 Comptt. (I Delhi) [Page 241 (b) Given : Length of wire = L, Charge density (λ) = kx, ϕ = ? We know dq = λ dx = kx dx

Q = dq kx dx kq

0 0

21

2∫ ∫= =L

L

∴ ϕ = Q∈0

= kL2

02 ∈

Or (a) Derive the expression for the electric potential at any point P, at distance r from the

centre of an electric dipole, making angle α, with its axes. (b) Two point charges 4 μC and +1 μC are separated by a distance of 2 m in air. Find

the point on the line-joining charges at which the net electric field of the system is zero. 3+2

Ans. (a) Electric potential : See Q. 22 (i), 2017 (II Delhi). [Page 211 (b) Given : q1 = 4 μC, q2 = 1 μC, r = 2 m Let us say that at a point P (distance x) the net field is zero.

E1 = 1

44

0 2π ∈ x …(i) E =

∈

14 0 2π

q

q

E2 = 1

41

20 2π ∈ −( )x …(ii)

E1 = E2

4 1

22 2x x=

−( )


Shiv Das

2 2

x

=

12

2

−

x

Taking square root on both sides,

2x

= 12 − x

x = 43

m

At this point, the net electric field of the system is zero. Q.25. (a) Prove that an ideal capacitor in an ac circuit does not dissipate power. (b) An inductor of 200 mH, capacitor of 400 μf and a resistor of 10 Ω are connected in

series to ac source of 50 V of variable frequency. Calculate the (i) angular frequency at which maximum power dissipation occurs in the circuit

and the corresponding value of the effective current, and (ii) value of Q-factor in the circuit. 2+3 Ans. (a) Power dissipated in AC circuit is given by

(P) = Vrms Irms cos ϕ …where [cos ϕ = RZ

For an ideal capacitor, R = 0, cos ϕ = RZ

= 0

P = Vrms Irms × (0) = 0(Zero)

i.e., power dissipated in an ideal capacitor is zero. (b) Given : L = 200 mH, C = 400 μF, R = 10 Ω, V = 50 V

(i) angular frequency (W0) = 1LC

= 1

200 10 400 10

1

8 10

10803 6 5

3

( ) ( )× × × ×− − −= =

= 10008 9.

= 112 s–1

Corresponding value of current (I) = VR

= 5010

= 5A

(ii) Q factor (Q) = 1 1

10200 10400 10

3

6RLC

= ××

−

−( )( ) = 5

Or (a) A metallic rod of length l is moved perpendicular to its length with velocity v in a

magnetic field B→

acting perpendicular to the plane in which rod moves. Derive the expression for the induced emf.

(b) A wheel with 15 metallic spokes each 60 cm long, is rotated at 360 rev/min in a place normal to the horizontal component of earth’s magnetic field. The angle of dip at that place is 60°. If the emf induced between rim of the wheel and the axle is


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400 mV, calculate the horizontal component of earth’s magnetic field at the place. How will the induced emf change, if the number of spokes is increased? 2+3 Ans. (a) The magnitude of the emf, generated across a

length dr of the rod, as it moves at right angles to the magnetic field, is given by

dε = Bνdr

∴ ε εων ω= =∫ ∫ ∫= =d dr rdrB B

R RB R

0 0

2

2

(b) Given : Number of spokes (n) = 15, l = 60 cm = 60 × 10–2 m

rpm = 360, ϕ = 60°, e = 400 mV, BH = ?

We know, e = 12

BHl2w

BH = 2 2 400 10

60 10 360260

2

3

2 2

e

l w= × ×

× × ×

−

−

( )

( )π

= 527π

= 0.06 T

•There would be no effect on induced emf, if number of spokes is increased. Q.26. (a) Explain with reason, how the power of a diverging lens changes when (i) it is kept

in a medium of refractive index greater than that of the lens. (ii) incident red light is replaced by violet light.

(b) Three lenses L1, L2, L3 each of focal length 30 cm are placed co-axially as shown in the figure. An object

is held at 60 cm from the optic centre of Lens L1. The final real image is formed at the focus of L3. Calculate the separation between (i) (L1 and L2) and (ii) (L2 and L3). 3+2

Ans. (a) The power of a lens is the reciprocal of its focal length.

P = 1f (when ‘f ‘ is taken in meters)

P = n n

n2 1

1 1 2

1 1−

−R R

= n n

n2 1

1

2−

−

R …

1 1 1

1 2R R R= =

= negative … [ R1 is negative for diverging lens

(i) If n1 > n2 (given)

n n

n2 1

1

− becomes negative

L1 L3L2 IO

30 cm60 cm

O

Metallic ringR

Pθ = ωt

O

ω

where [R is the radius of the circle]


Shiv Das

∴ P = 1f becomes positive

or lens becomes converging

(ii) (n2)violet > (n2)red which means that the power of lens increases on changing to violet light from

red light. (b) Given : f1 = f2 = f3 = 30 cm u1 = 60 cm, I3 is formed at the focus of L3. L1 L2 = ?; L2L3 = ? Since final image (I3) is formed at the focus of L3, the rays emerging from L2 and

incident on L3 have to be parallel to principal axis. Since the object is placed at a distance of 60 cm from L1, i.e., at 2F; the image will be

formed at 2F on the other side of L1 (60 cm). This image I1 will be at the focus of L2, because rays emerging out from L2 are parallel to principal axis.

(i) ∴L1L2 = 2f1 + f2 = (2 × 30) + 30 = 90 cm (ii) L2L3 can be any distance.

Or (a) Deduce the expression, by drawing a suitable ray diagram, for the refractive index

of triangular glass prism in terms of the angle of minimum deviation (D) and the angle of prism (A).

Draw a plot showing the variation of the angle of deviation with the angle of incidence.

(b) Calculate the value of the angle of incidence when a ray of light incident on one face of an equilateral glass prism produces the emergent ray, which just grazes along the adjacent face. Refractive index of the prism is 2 . 3+2

Ans. (a) See Q. 26 (Or) (a), 2015 (I Outside Delhi). [Page 108 (b) Given μ = 2 , ∠i = ? Since the emergent ray just grazes along the adjacent face of an equilateral glass

prism,

sin c = 1 1

2µ= = sin 45° ⇒ c = 45°

Since r + c = 60° ∴ r = 15°

μ = sinsin

sinsin

ir

i=°15

= 2

⇒ sin i = 2 sin 15° ∴ ∠i = sin–1[ 2 sin 15°]

set ii

Note: Except for the following questions, all the remaining questions have been asked in Set-I. Q.1. Why is sky wave propagation of signals restricted to a frequency of 30 MHz?

(Not in Syllabus) 1


Shiv Das

Ans. Sky wave propagation of signals is restricted to a frequency of 30 MHz, because waves of frequency greater than 30 MHz get penetrated through the ionosphere and thus they do not get reflected by it.

Q.9. State two properties of nuclear forces. Write the relation between half life and decay constant of a radioactive nucleus. 2

Ans. •Properties of nuclear forces : See Q. 21 (a), 2015 (I Outside Delhi). [Page 100

• T1/2 = ln2

λ = 0 693.

λ

Or Calculate the kinetic energy of an electron having de Broglie wavelength of 1Å. 2 Ans. Given : λ = 1Å = 1 × 10–10 m, K = ? From de Broglie equation, we have

λ = hp

…(i) [p = mv]

K = 12

mv2 or mv2 = 2K or m2v2 = 2mK

⇒ p2 = 2 mK or p = 2mK

⇒ λ = hm2 K

or K = hm

2

21

2λ

K = ( . )

( ) ( . )6 63 10

1 10 2 9 1 10

34 2

10 2 31×

× × × ×

−

− −

= 4 8 102

17. × − J

= 2.4 × 10–17 J = 1.5 × 102 eV [ 1 eV = 1.6× 10–19 J] = 150 eVQ.10. Two radioactive samples, X, Y have the same number of atoms at t = 0. Their half lives

are 3h and 4h respectively. Compare the rates of disintegration of the two nuclei after 12 hours. 2

Ans. Given : NX = NY at t = 0

(TX)1/2 = 3h, (TY)1/2 = 4h, t = 12h, RR

X

Y = ?

The decay equation is ddtN = – λN = N0 e–λt

Given time = 12 h = 4(TX)1/2 = 3(TY)1/2 …(i)

∴ NN

X

0

412

116

= =

or NX = N0

16


Shiv Das

and NN

Y

0

312

18

= =

or NY = N0

8

RX = ddtN

TN

X X

= 0 693

161 2

0.( )/

.

RY = ddtN

TN

Y Y

= 0 693

81 2

0.( )/

.

∴ RR

TT

X

Y

Y

X= == ×1

212

43

23

1 2

1 2

( )( )

/

/ …[From (i)

The ratio of disintegration is 2 : 3 Q.11. Why are coherent sources necessary to produce interference in Young’s double slit

experiment? Light waves from two coherent sources have intensities in the ratio of 4 : 9. Find the ratio of intensities of maxima and minima in the interference pattern. 3

Ans. If sources are not coherent, the superposition pattern (intensity pattern) is not stable. It keeps on changing with time and hence it is necessary to have coherent sources to observe interference in Young’s double slit experiment.

Given : I1 = 4x, I2 = 9x

IImax

min = ?

We know the resultant intensity is

I = I1 + I2 + 2 I I1 2 cos φ

Imax = I1 + I2 + 2 I I1 2 [ ϕ = 0, cos 0 = 1]

Imin = I1 + I2 + 2 I I1 2 × (–1) [ ϕ = 180°, cos 180 = –1]

= I1 + I2 – 2 I I1 2

IImax

min =

I I I I

I I I I1 2 1 2

1 2 1 2

2

2

+ ++ −

= 4 9 2 4 94 9 2 4 9

x x x xx x x x+ + ×+ − ×

= 4 9 124 9 12

x x xx x x+ ++ −

= 251

251

xx

=

Hence the ratio is 25 : 1 Q.13. The photon emitted during the de-excitation from the 1st excited level to the ground

state of hydrogen atom is used to irradiate a photo cathode of a photocell, in which stopping potential of 5 V is used. Calculate the work function of the cathode used. 3

Ans. Given : n1 = 2, n2 = 1, V0 = 5V, eV0 = 5eV, (ϕ0) = ? Energy of photon (E) = (13.6) – (3.4) eV = 10.2 eV According to photoequation, E = eV0 + ϕ0

ϕ0 = E – (eV0) = (10.2) – (5) ∴ ϕ0 = 5.2 eV


Shiv Das

Q.16. Name the type of e.m. waves having a wavelength range of 0.1 m to 1 mm. How are these waves generated? Write their two uses. 3

Ans. •E.m. waves having a wavelength range 0.1 m to 1 mm are MICROWAVES. •Microwaves are generated by special vacuum tubes such as klystron,magnetron and

gunn diodes. •Microwaves are used in : (a) Radar system in aircraft navigation (b) Ovens for heating

and cooking. Q.19. The current through two inductors of self-inductance 15 mH and 25 mH is increasing

with time at the same rate. Draw graphs showing the variation of the (a) emf induced with the rate of change of current (b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the

same. 3 Ans. Given : L1 = 15 mH and L2 = 25 mH (a) dI/dt

L1 = 15 mH

L2 = 25 mH

i

(b)

u

L2 = 25 mH L1 = 15 mH

I

uu

1

2 =

1212

1 12

2 22

L

L

i

i

But ε1i1 = ε2i2 ( power dissipated is same)

∴ ii1

2

2

1

2

1= =ε

εLL

dIdt

dIdt

is same and Lε = −

∴ uu

1

2

1

2

2

1

22

1

2515

= = =

LL

LL

LL

= 1.66

Q.20. State the two features to distinguish between interference and diffraction phenomena. Two wavelengths of light 600 nm and 610 nm are used in turn, to study the diffraction at a single slit of size 2 mm. The distance between the slits and screen is 2 m. Calculate the separation between the positions of the second order maximum of the diffraction pattern obtained in the two cases. 3


Shiv Das

Ans. Interference pattern and Diffraction pattern : The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :

(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of λa

.

At the same angle of λa

, we get a maxima for two narrow slits separated by a distance ‘a’.

Given : λ1 = 600 nm = 600 × 10–9 m, λ2 = 610 nm = 610 × 10–9 m, a = 2 mm = 2 × 10–3 m, D = 2 m Position of second order maximum is

x1 = 52

1λ Da

for λ1 = 600 nm

2 12

2n n+

=, given ( )

x2 = 52

2λ Da

for λ2 = 610 nm [ 2nd order]

Δx = x2 – x1 = 52

610 10 22 10

52

600 10 22 10

9

3

9

3× − ×× ××

× ××

−

−

−

−( )

( )( )

( )

= 52

2 102 10

5 10 102 10

9

3

9

3610 600× −××

= × ××

−

−

−

−( ) ( )[ ]

Δx = 2.5 × 10–5 m

set iii

Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set II.

Q.1. Why must both the objective and the eye piece of a compound microscope have short focal lengths? 1

Ans. For getting higher magnification in compound microscope, both objective and eyepiece must have short focal length, because

m ∝ 1

f fo e

Q.3. State two reasons why high frequency carrier waves are needed in transmitting a

message signal. (Not in Syllabus) 1

O– 2 – 1 1 2

I0

Intensity

(Units of λ/a)


Shiv Das

Ans. For the following reasons, high frequency carrier waves are needed in transmitting a messagesignal:

(i) Length of transmitting antenna is short. (ii) Power radiated is more. (iii) Mixing of signals can be avoided. Q.6. State two properties of photons. For a monochromatic radiation incident on a

photosensitive surface, why do all photoelectrons not come out with the same energy? Give reason for your answer. 2

Ans. •Two properties of photons : (a) photon is electrically neutral. (b) photon has an energy equal to hv •Foramonochromatic radiation incidentonaphotosensitivesurface,allphotoelectrons

do not come out with the same energy, because in addition to the work done to free electrons from the surface, different (emitted) photoelectrons need different amount of work to be done on them to reach the surface.

Or An electron, an α-particle and a proton have the same kinetic energy. Which of these

particles has the shortest de Broglie wavelength?

Ans. For a particle, de Broglie wavelength, λ = hp

Kinetic energy, K = pm

2

2

Then, λ = hm2 K

For the same kinetic energy K, de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.

A proton (11H) is 1836 times massive than an electron and α-particle (2

4 He) four times that of a proton. Hence, α-particle has the shortest de Broglie wavelength.

Q.7. Distinguish between nuclear fission and fusion. Explain how the energy is released in both the processes. 2

Ans. In nuclear fission a heavy nucleus breaks up into smaller nuclei accompanied by release of energy; whereas in nuclear fusion two light nuclei combine to form a heavier nucleus accompanied by release of energy.

Inboththecases,somemass(=massdefect)getsconvertedintoenergyaspertherelation: E = Δm × c2 Q.8. A cell of emf 4 V and internal resistance 1 Ω is connected to a d.c. source of 10 V through

a resistor of 5 Ω. Calculate the terminal voltage across the cell during charging. 2 Ans. E1 – E2 = I(R + r) 10 – 4 = I(5 + 1) 6 = I × 6 I = 1A Terminal voltage across the cell = (E2 + Ir) = [(4) + (1 × 1)] = (4 + 1) = 5V

+

+1 Ω 5 ΩE2 = 4 V

E1 = 10 VI

Rr

–


Shiv Das

Q.9. Distinguish between point-to-point and broadcast modes of communication. Give one example for each. (Not in Syllabus) 2

Ans. Point to point communication takes place between a single transmitter and a receiver; While in broadcast mode, a large number of receivers can receive signal from a single

transmitter. Example of point to point mode : telephony Example of broadcast mode : Radio/TV Q.11. Draw the V-I characteristic of an LED. State two advantages of LED lamps over

conventional incandescent lamps. Write the factor which controls (a) wavelength of light emitted, (b) intensity of light emitted by an LED. 3

Ans. LED V-I characteristic Light Emitting Diode (LED) : A light emitting

diode is simply a forward biased p-n junction which emits spontaneous light radiation. When forward bias is applied, the electron and holes at the junction recombine and energy released is emitted in the form of light. V-I characteristics of LED are similar to that of Si junction diode but the threshold voltages are much higher and slightly different for each colour. No conduction or light emission occurs for reverse bias which, if it exceeds 5V, may damage the LED.

Advantages of LED over conventional lamps : (i) Low operational voltage. (ii) Less power consumption. (iii) Long life. (iv) Ruggedness Controlling factors : (a) Energy band gap controls the wavelength of

light emitted. (b) Forward current controls the intensity of

emitted light. Q.12. A toroidal solenoid of mean radius 20 cm has 4000 turns of wire wound on a ferromagnetic

core of relative permeability 800. Calculate the magnetic field in the core for a current of 3A passing through the coil. How does the field change, when this core is replaced by a core of Bismuth? 3

Ans. •Given : r = 20 cm = 20 × 10–2 m, N = 4000 turns, μr = 800, I = 3A

B = μrμ0nI [n = number of turns per unit length

= (800) × (4π × 10–7) × 4000

2 20 10 2( )π × ×

− × 3

= 9.6 T •Sincebismuthisdiamagnetic, itsμr < 1, therefore the magnetic field in the core will be

very much reduced. Q.13. Name the type of e.m. waves having a wavelength range 10–7 m to 10–9 m. How are these

waves generated? Write their two uses. 3

+R

LED

Light

p n

LED symbol

Y

50

40

30

20

10

O 1 2 3 4 5X

Forward voltage V(V)

Forw

ard

cur

rent

I (

mA

) Infr

a-R

ed Red

Am

ber

Gre

enB

lue

Whi

te


Shiv Das

Ans. •e.m. waves having a wavelength range 10–7 m to 10–9 m are ultra violet rays. •Sun is an important source ofUV rays. Some special lamps and very hot bodies also

produce UV rays. •Uses : (i) UV rays are used in lasik eye surgery. (ii) UV lamps are being used to kill germs in water purifiers. Q.14. An electron microscope uses electrons accelerated by a potential difference 50 kV.

Calculate the de Broglie wavelength of the electrons. Compare the resolving power of an electron microscope with that of an optical microscope, which uses visible light of wavelength 550 nm. Assume the numerical aperture of the objective lens of both microscopes are the same. 3

Ans. Given : V = 50 kV, (a)EM = (a)OM,

λ for optical microscope (λ)OM = 550 nm = 550 × 10–9 m

λDB = ?, ( )( )RR

EM

OM = ?

We know λDB =

1 227 1 227

50 103

. .V

=×

= 0.005 nm

Resolving power of telescope is given by

RP = 21 22n sin.

βλ

( )( )RPRP

EM

OM =

21 22

21 22

n nsin. ( )

sin. ( )

βλ

βλEM OM

= ( )( )

( )( . )

λλ

OM

EM= ×

×

−

−550 10

0 005 10

9

9 = 1,10,000

Q.21. Write the functions of (i) Transducer, (ii) Transmitter, (iii) Repeater used in communication systems. (Not in Syllabus) 3

Ans. Functions of : (i) Transducer : It converts one form of energy into another. (ii) Transmitter : It processes the incoming message so as to make it suitable for

transmission through a channel. (iii) Repeater : It picks up signal from the transmitter, amplifies and retransmits it to the

receiver sometimes with a change in carrier frequency.

PHYSICS (Theory) – 2018 (COMPTT.) 291

Shiv Das

CBSE

PHYSICS (Theory) – 2018 (COMPTT.)

(Common for Delhi & Outside Delhi)


GENERAL INSTRUCTIONS :

(i) All questions are compulsory. There are 26 questions in all.(ii) This question paper has five sections: Section A, Section B, Section C, Section D and

Section E.(iii) Section A contains five questions of one mark each, Section B contains five questions of two

marks each, Section C contains twelve questions of three marks each, Section D contains onevalue based question of four marks (*) and Section E contains three questions of five 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 the three questions of five marks weightage. You haveto attempt only one of the choices in such questions.

(v) 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, μ0 = 4π × 10–7 TmA–1,

ε0 = 8.854 × 10–12 C2N–1m–2,

14 0πε

= 9 × 109 Nm2C–2,

Mass of Electron (me) = 9.1× 10–31 kg, Mass of neutron = 1.675 × 10–27 kg

Mass of Proton = 1.673× 10–27 kg, Avogadro's number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10–23 JK–1

SECTION A

Q.1. An electron is accelerated through a potential difference V. Write the expression for itsfinal speed, if it was initially at rest. 1

Ans. Expression for final speed

v =

2emV …where

e

m

v

==

==

⎡

⎣

⎢⎢⎢⎢⎢

Charge of electronMass of electronVelocity of electron

V Potential difference

Q.2. Two protons of equal kinetic energies enter a region of uniform magnetic field. The firstproton enters normal to the field direction while the second enters at 30° to the fielddirection. Name the trajectories followed by them. 1

Ans. (i) At normal, it will follow circular path.(ii) At an angle of 30°, it will follow helical path.


Shiv Das

Q.3. Draw a graph showing the intensity distribution of fringes due to diffraction at single slit.1

Ans. Intensity Distribution for Single slit diffraction :

Q.4. Write the range of frequencies of electromagnetic waves which propagate through skywave mode. 1

Ans. From few MHz to 30-40 MHz is the range of frequencies of electromagnetic wavespropagating through sky wave mode.

Q.5. Define the power of a lens. Write its S.I. unit. 1Ans. The power of a lens equals to the reciprocal of its focal length (in meter)

i.e., p =

1f ( )meter

The S.I. Unit of power of a lens is Dioptre (D).

SECTION B

Q.6. A rectangular frame of wire is placed in a uniformmagnetic field directed outwards, normal to the paper. ABis connected to a spring which is stretched to A′′′′′B′′′′′ andthen released at time t = 0. Explain qualitatively howinduced e.m.f. in the coil would vary with time.(Neglect damping of oscillations of spring) 2

Ans. The wire AB would oscillate in a simple harmonic wayunder the restoring force developed in the spring due todisplacement∴ x = – a cos ωt (as x = – a at t = 0)Therefore, Instantaneous magnetic Flux

φ(t) = Blx [l = AB

Instantaneous induced emf e(t) =

ddt

ddt

ddt

lxφ = =( )B (– Bl a cos ωt)

e(t) = –

ddtφ = (a Bl ω) sin ωt = e0 sin ωt [where e0 = a Blω]

The induced emf, therefore varies with time sinusoidally.Q.7. (a) Define the term magnetic susceptibility and write its relation in terms of relative

magnetic permeability.(b) Two magnetic materials A and B have relative magnetic permeabilities of 0.96 and

500. Identify the magnetic materials A and B. 2Ans. (a) The magnetic susceptibility of a magnetic substance is defined as the ratio of the

intensity of magnetisation (I) to the strength of the magnetic field (H).

–3λ λ–2λ 2λ–λ λ

Intensity

Distance from center

A′A

DB′

B

C

←←⎯⎯


Shiv Das

Magnetic susceptibility measures the response of magnetic material to an externalmagnetic field.It is denoted by χm.

χm =

IH

Relation in term of relative magnetic permeabilityχχχχχm = (μμμμμr – 1) (where μr = relative magnetic permeability

(b) μr = 0.96 is diamagnetic material because relative magnetic permeability of diamagneticsubstances is always less than unity, i.e., μr < 1.μr = 500 is ferromagnetic material because relative magnetic permeability of ferromag-netic material is very large and positive.

Q.8. (a) Give one use of electromagnetic radiations obtained in nuclear disintegrations.(b) Given one example each to illustrate the situation where there is (i) displacement

current but no conduction current and (ii) only conduction current but nodisplacement current. 2

Ans. (a) Electromagnetic radiations are used to destroy cancer cells.(b) (i) The region, between the plates of a capacitor, connected to time varying voltage

source, has a displacement current but no conduction current.(ii) The wires, connected to the plates of a capacitor, joined to a time varying or steady

voltage source, carry a conduction current but no displacement current.Q.9. Find the frequency of light which ejects electrons from a metal surface, fully stopped by

a retarding potential of 3.3 V. If photo electric emission begins in this metal at a frequencyof 8 ××××× 1014 Hz, calculate the work function (in eV) for this metal. 2

Ans. Given : Vs = 3.3 V, ν0 = 8 × 1014 Hz, h = (6.63 × 10–34), e = 1.6 × 10–19, W = ?, ν = ?We have

W = hν0= (6.63 × 10–34) × (8 × 1014) J [Given h = 6.63 × 10–34)

=

( . )( . )

6 63 10 81 6 10

20

19× ×

×

−

− = 3.315 eV

We havehν = W + Vs

= (3.315 + 3.3) eV

∴ ν =

( . ) ( . )( . )

6 615 1 6 106 63 10

19

34× ×

×

−

− Hz

= 1.596 ××××× 1015 HzOr

Monochromatic light of frequency 6.0 ××××× 1014 Hz is produced by a laser. The power emittedis 2.0 ××××× 10–3 W.Calculate the (i) energy of a photon in the light beam and (ii) number ofphotons emitted on an average by the source. 2

Ans. Given : ν = 6.0 × 1014 Hz, P = 2.0 × 10–3 W, E = ? N = ?Energy of photon = hν

= (6.63 × 10–34) × (6.0 × 1014) J (h = 6.63 × 10–34)

= 3.978 × 10–19 J =

3 978 101 6 10

19

19.

( . )×

×

−

−

≅ 2.49 eV


Shiv Das

Number of photons emitted per second (N) =

PowerEnergy of photon

=

(2.0 10 J s(3.978 10 J

-3

-19×

×) /

) =

2 0 103 978

16..×⎛

⎝⎜⎞

⎠⎟

=

2 0 103 978.

.×⎛

⎝⎜⎞⎠⎟

× 1015

= 5.03 ××××× 1015 photons/second

Q.10. Calculate the ratio of the frequencies of the radiation emitted due to transition of theelectron in a hydrogen atom from its (i) second permitted energy level to the first level and(ii) highest permitted energy level to the second permitted level. 2

Ans. We know ΔE = hν = (Ef – Ei) =

E E02

02n nf i

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

(i) hν1 = E0

11

12

342 2 0−⎛

⎝⎜⎞⎠⎟

= ×E (nf = 1, ni = 2)

(ii) hν2 = E0

12

1 142 2 0−

∞⎛⎝⎜

⎞⎠⎟

= ×E (nf = 2, ni = ∞)

hhνν

1

2

0

0

3414

=×

×

E

E

νν

1

2

31

= ∴ ννννν1 : ννννν2 :: 3 : 1

SECTION C

Q.11. Define electric flux and write its SI unit.The electric field components in the figureshown are : Ex = αααααx, Ey = 0, Ez = 0 where

ααααα =

100NCm

. Calculate the charge within the cube,

assuming a = 0.1 m. 3

Ans. Electric Flux is the dot product of electric field and area vector.

φ =

E→→ →→

∫∫ . ds

SI Unit: Nm2/C or volt-meter

For a given case α = 100 NC–1 m–1, a = 0.1 m [φ = E→ →

. ds ]φ = φ1 + φ2 = [Ex(at x = 2a) – Ex(at x = a)]a2

= [α(2a) – α(a)]a2

= αa3

φ = (100) × (0.1)3 = 0.1 = 10–1 Nm2/C

x

aa

O

y

z

Ex→→


Shiv Das

But φ =

qε0

∴ q = ε0φ = (8.854 × 10–12) × 10–1 = 8.854 × 10–13 = 0.8854 pCOr

An electron falls through a distance of 1.5 cmin a uniform electric field of magnitude2.0 ××××× 104 N/C (Fig. a)Calculate the time it takes to fall through thisdistance starting from rest.If the direction of the field is reversed (Fig. b)keeping its magnitude unchanged, calculatethe time taken by a proton to fall through thisdistance starting from rest. 3

Ans. Given : s = 1.5 cm = 1.5 × 10–2 m, E = 2.0 × 104 NC–1, u = 0me = 9.1 × 10–31 kg, mp = 1.67 × 10–27 kg, q = 1.6 × 10–19 C

We have F = qE or ma = qE

Acceleration (a) =

qmE

Also from equation of motion, we know

s =

12

at2

∴ t =

2 2 2sa

sq m

s mq

= = ×E E/

…(i)

(i) For the electron, using above equation (i), we have

∴ t =

2 1 5 10 9 1 101 6 10 2 0 10

2 31

19 4× × × ×

× × ×

− −

−( . ) ( . )

( . ) ( . ) = 2.92 × 10–9 s = 2.92 ns

(ii) For proton

∴ t =

2 1 5 10 1 67 101 6 10 2 0 10

2 27

19 4× × × ×

× × ×

− −

−( . ) ( . )

( . ) ( . ) = 0.125 × 10–6 s = 0.125 μμμμμs

Q.12. Two cells of emf εεεεε1 and εεεεε2 and internal resistances r1 and r2 respectively are connected inparallel. Obtain expressions for the equivalent

(i) resistance and (ii) emf of the combination. 3Ans. When two cells of emf ε1 and ε2 and internal resistances r1 and r2 are connected in parallel,

the following will be circuit diagram :The equivalent internal resistance (r) is :

1 1 1

1 2r r r= +

∴ r =

r rr r

1 2

1 2++…(i)

We know V = ε – Ir

(a)

– e

(b)

p

A I B1

I1 I1

I2I2

B2

E1

I Cr1E2

r2


Shiv Das

or I =

ε −⎛⎝⎜

⎞⎠⎟

Vr

∴ I = I1 + I2 =

ε ε1

1

2

2

−⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

+V Vr r

or V =

ε ε1 2 2 2

1 2

1 2

1 2

r rr r

r rr r

++

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

− I

also V = (ε – Ir) ⇒

ε ε εr r r

= +⎛⎝⎜

⎞⎠⎟

1

1

2

2

εεεεε =

εε εε1

1

2

2r r++

⎡⎡

⎣⎣⎢⎢

⎤⎤

⎦⎦⎥⎥r …(ii)

Q.13. (a) Define SI unit of current in terms of the force between two parallel current carryingconductors.

(b) Two long straight parallel conductors carrying steady current Ia and Ib along the samedirection are separated by a distance d. How does one explain the force of attractionbetween them? If a third conductor carrying a current Ic in the opposite direction isplaced just in the middle of these conductors, find the resultant force acting on thethird conductor. 3

Ans. (a) S.I. unit of current: AmpereAmpere : Ampere is that current which if maintained intwo infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes aforce of 2 × 10–7 N on each metre of the other wire.Then current flowing is 1A

F =

μπ

0 1 12 1

× ××

∴ F =

μπ0

2 = 2 ××××× 10–7 N

(b) The wire (b) experiences a force due to themagnetic field caused by the current flowing inwire (a).The magnetic field at any point on the wire (b)due to the current in wire (a) is perpendicularto the plane of two wires and pointing inwardsand hence force on it will be towards wire (a).Similarly force on wire (a) will be towards wire(b). Hence two wires carrying currents in samedirection attract each other.Force on wire (3) due to wire (1)

=

μ

π

0

22

I Ia cd⎛

⎝⎜⎞⎠⎟

towards right

Force on wire (3) due to wire 2

=

μ

π

0

22

I Ib cd⎛

⎝⎜⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

towards left

Resultant force on wire 3 due to wires 1 and 2 =

μμππ0Icd

[Ia – Ib] towards right.

d = 1mVaccum

I1 = 1A I2 = 1A

μd = 1

da

b

I1

I2

L

B

F

1 2 3

d/2d/2

Ia Ib Ic


Shiv Das

Q.14. (a) When an unpolarized light of intensity I0 is passed through a polaroid, what is theintensity of the linearly polarized light? Does it depend on the orientation of thepolaroid? Explain your answer.

(b) A plane polarized beam of light is passed through a polaroid. Show graphically thevariation of the intensity of the transmitted light with angle of rotation of thepolaroid in complete one rotation. 3

Ans. (a) The intensity of the linearly polarized light would be

I02

.

No; it does not depend on the orientation, because the polaroid will let the componentof the unpolarized light, parallel to its pass axis, to pass through it irrespective of itsorientation.

(b) See Q. 12 (i & ii), 2016 (I Outside Delhi). [Page 157Q.15. State Einstein’s photoelectric equation explaining the

symbols used.Light of frequency ννννν incident is on a photosensitivesurface. A graph of the square of the maximum speed of

the electrons ( )maxv2 vs. ννννν is obtained as shown in the

figure. Using Einstein’s photoelectric equation, obtainexpressions for (i) Planck’s constant (ii) work function ofthe given photosensitive material in terms of parametersl, n and mass of the electron m. 3

Ans. Einstein‘s photoelectric equation is

E = hν = W +

12

2mvmax

E = hν = hν0 +

12

2mvmax [W = hv0]

(i) From Einstein’s photoelectric equation, we have

hν = W +

12

2mvmax

∴ vmax2 =

2m

(hν – W) =

2 2hm m

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

+ν W

(ii) Scope of the given graph =

ln

Intercept on y-axis = – l

∴

ln

hm

= 2 or h =

mln2

and – l =

− 2Wm

or W =

ml2

Q.16. (a) Draw a plot showing the variation of potential energy of a pair of nucleons as afunction of their separation. Mark the regions where the nuclear force is(i) attractive and (ii) replusive

n

νl

O

Where: ν = frequency of incident lightν0 = threshold frequencyW = work function

12

2mvmax = maximum kinetic energy

h = Planck’s constant

⎡

⎣

⎢⎢⎢⎢⎢⎢

( )maxv2


Shiv Das

(b) In the nuclear reaction

n U Xe Sr n++ ⎯⎯ →→⎯⎯ ++ ++92

23554

94 2ab

determine the values of a and b. 3Ans. (a) A plot of the potential energy between two nucleons as a function of distance is shown

in the diagram.

Important conclusions from the graph :(i) The nuclear force is much stronger than the Coulomb force acting between charges

or the gravitational forces between masses. The nuclear binding force has todominate over the Coulomb repulsive force between protons inside the nucleus. Thishappens only because the nuclear force is much stronger than the coulomb force. Thegravitational force is much weaker than even Coulomb force.

(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is morethan a few femtometers. This leads to saturation of forces in a medium or a large-sized nucleus, which is the reason for the constancy of the binding energy pernucleon.

(iii) The nuclear force between neutron-neutron, proton-neutron and proton-proton isapproximately the same. The nuclear force does not depend on the electric charge.

For r > r0, the force is attractive, while for r < r0, the force is repulsive(b) In the given equation

n U Xe Sr n+ ⎯ →⎯ + +92235

5494 2ab

(i) Using mass number of the given nuclei, we have1 + 235 = a + 94 + (2 × 1)

or a = (235 + 1) – (94 + 2) = 236 – 96 = 140(ii) Using atomic number of the given nuclei, we have

0 + 92 = 54 + b + (2 × 0)or b = 92 – 54 = 38

Q.17. (a) Write the truth table for thecombination of the gates shown in thefigure.

(b) Explain briefly how a photo diodeoperates. (Not in Syllabus) 3

100

Pot

entia

l en

ergy

(M

eV)

0

–100

1 2 3r0r (fm)

A

BC


Shiv DasC F B

A M

P

A′

B′

Ans. (a) The inputs of the third gate are A and B . Hence the truth table is as given below :

A B A B C0 0 1 1 0

0 1 1 0 0

1 0 0 1 0

1 1 0 0 1

(b) A photodiode is a special purpose p – n junction diode fabricated with a transparentwindow to allow light to fall on the diode.Incident light, with photon energy greater than the energy gap of the semiconductor,generates electron–hole pairs. The magnitude of the photo current depends on theintensity of incident light.The photodiode is usually operated under reverse bias condition, because this makes iteasier to detect changes in light intensity and makes the photodiode work as a detectorof optical signals.

Q.18. Draw a labelled circuit diagram of n-p-n germanium transistor in common emitterconfiguration. Explain briefly, how this transistor is used as a voltage amplifier. 3

(Not in Syllabus)Ans. See Q. 14, 2017 (III Outside Delhi). [Page 232Q.19. (a) With the help of a ray diagram, show how a concave mirror is used to obtain an erect

and magnified image of an object.(b) Using the above ray diagram, obtain the mirror formula and the expression for linear

magnification. 3Ans. (a) Ray diagram

(b) From similar triangles A′B′F and MPF, we have

B APM

B FFP

′ ′ ′= or

B ABA

B FFP

′ ′ ′= …(i) (since PM = BA)

From similar triangles A′B′P and ABP, we have

B ABA

B PBP

′ ′ ′= …(ii)

Hence, from equations (i) and (ii), we have

B FFP

B PBP

′ ′= …(iii)

Now B′F = B′P + PF = (+ v) + (– f) = (v – f), BP = – uHence from equation (iii), we have

∴

v ff

vu

−−

+−

= or

− −+ =vf

vu

1 ∴

1 1 1v u f

++ ==


Shiv Das

This is the mirror formula.

We know Linear magnification (m) =

B ABA′ ′

From similar triangles A′B′P and ABP, we get

B ABA

B PBP

′ ′ ′=

∴ Linear magnification (m) =

B PBP′ +

−= v

u =

−− v

u

Q.20. (a) State Biot-Savart law and express it in the vector form.(b) Using Biot-Savart law, obtain the expression for the magnetic field due to a circular

coil of radius r, carrying a current I at a point on its axis distant x from the centre ofthe coil. 3

Ans. See Q. 25 (Or), 2015 (I Outside Delhi). [Page 106

Q.21. Using Kirchhoff’s rules, calculate the potentialdifference between B and D in the circuit diagramas shown in the figure. 3

Ans. The given circuit diagram may be simplified asshown below :Using Kirchhoff’s voltage rule, we have :For loop DABD

I1 × 1 + (1) + (– 2) + 2I1 + 2(I1 + I2) = 0or 5I1 + 2I2 = 1 …(i)For loop DCBD

I2 × 3 + (3) + (– 1) + I2 + 2(I1 + I2) = 0or 2I2 + 6I2 = 2 …(ii)Multiplying equation (i), by 3 we have

15I1 + 6I2 = 3 …(iii)Solving (ii) and (iii), we get

I1 =

513

A I2 =

− 613

∴ Current through DB(I) = I1 + I2 =

513

613

113

+ = −−⎛⎝⎜

⎞⎠⎟

A (direction is reverse)

∴ Potential difference between B and D = 0.154 V = (IR) =

113

2×⎛⎝⎜

⎞⎠⎟

= 0.154 V

Q.22. (i) Write two points to distinguish between interference and diffraction fringes.(ii) In a Young’s double slit experiment, fringes are obtained on a screen placed a certain

distance away from the slits. If the screen is moved by 5 cm towards the slits, thefringe width changes by 30 μμμμμm. Given that the slits are 1 mm apart, calculate thewavelength of the light used. 3

A B

D

2V, 2Ω

2Ω1V,1Ω

C

1V,1Ω

3V, 3Ω

2V, 2Ω

AI1

1V, 1Ω

DI1 + I2

I2 3V, 3Ω C 1V, 1Ω

B2Ω


Shiv Das

Ans. (i) Interference Diffraction

Fringes are equally spaced.

Intensity is same for all maxima.

Superposition of two waves originatingfrom two narrow slits.

Maxima along an angle λ/a for twonarrow separated by a distance a.

(ii) Given : Δβ = 30 μm = 30 × 10–6 m, ΔD = 5 cm = 5 × 10–2 md = 1 mm = (1 × 10–3) m, λ = ?

We have, Fringe width β =

λDd

In the first case, β1 =

λD1d

or β1d = λD1 …(i)

In the second case, β2d = λD2

β1 – (30 × 10–6) =

λ( . )D1 0 05−d

or β1 – (30 × 10–6)d = λ(D1 – 0.05) …(ii)

Subtracting (ii) from (i) we get(30 × 10–6) × d = λ × 0.05

∴ λ =

( ) ( )30 10 1 105 10

6 3

2× × ×

×

− −

− = 6 × 10–7 m = (6 × 102) × 10–9 m

= 600 × 10–9 m = 600 nm

SECTION D

Q.23. Mrs. Rajlakshmi had a sudden fall and was thereafter unable to stand straight. She wasin great pain. Her daughter Rita took her to the doctor. The doctor took a photograph ofMrs. Rajlakshmi’s bones and found that she had suffered a fracture. He advised her to restand take the required treatment.(a) Name the electromagnetic radiation used to take the photograph of the bones.(b) How is this radiation produced?(c) Mention the range of the wavelength of this electromagnetic radiation.

*(d) Write two values displayed by Rita. 4Ans. (a) X-rays are used to take the photographs of the bones.

(b) X-rays are produced by bombarding a metal target with high energy electrons.(c) Wavelength range of X-rays is from about (10 nm to 10–4 nm)

*(d) Value Based Questions will not be asked in the Board Examination.

Fringes are not equally spaced.

Intensity falls as we go to successivemaxima away from the centre.

Superposition of a continuous family ofwaves originating from each point on asingle slit.

Minima at an angle of λ/a for a single slitof width a.


Shiv Das

SECTION E

Q.24. Two point charges q and – q are located at points (0, 0, – a) and (0, 0, a) respectively.(a) Find the electrostatic potential at (0, 0, z) and (x, y, 0)(b) How much work is done in moving a small test charge from the point (5, 0, 0) to

(– 7, 0, 0) along the X-axis ?(c) How would your answer change if the path of the test charge between the same

points is not along the x-axis but along any other random path?(d) If the above point charges are now placed in the same positions in a uniform external

electric field E→→

, what would be the potential energy of the charge system in itsorientation of unstable equilibrium? Justify your answer in each case. 5

Ans. (a) We have, for a point charge,

V =

14 0πε

. qr

…(i)

(i) At point (0, 0, z) :Potential due to the charge (+ q), located at point (0, 0, – a)

V+ =

14 0πε

.( )

qz a−

…(ii)

Potential due to the charge (– q), located at point (0, 0, a)

V– =

14 0πε

. ( )( )

−+q

z a…(iii)

Total potential at (0, 0, z) =

qz a z a4

1 1

0πε ( ) ( )− +⎡⎣⎢

⎤⎦⎥

−

=

2

4 02 2

qa

z aππεε ( )−−

(ii) At point (x, y, 0)Potential due to the charge (+ q), located at point (0, 0, – a)

V+ =

14 0 2 2 2πε

. q

x y a+ +

Potential due to the charge (– q), located at point (0, 0, a)

V– =

14 0 2 2 2πε

. −

+ +

q

x y a

Total potential at (x, y, 0)

V =

q

x y a x y a41 1

0 2 2 2 2 2 2πε + + + +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

− = 0

It is evident that the point (x, y, 0) is equidistant from charges + q and – q, the totalpotential will be zero.


Shiv Das

(b) Work done = q[V1 – V2] Q V1 = 0 and V2 = 0∴ work done = 0Where V1 and V2 are the total potential due to dipole at point (5, 0, 0) and (– 7, 0, 0).

(c) There would be no change in the work done, this is because the electrostatic field is aconservative field. The work done, in moving a test charge between two given points isindependent of the path taken.

(d) The two given charges make an electric dipole of dipole moment p q a→

=→

. 2 . The

potential energy in position of unstable equillibrium (where p→

and E→

are antiparallelto each other)

= + pE = (q.2a)E = 2aqEOr

A capacitor of capacitance C1 is charged to a potential V1 while another capacitor ofcapacitance C2 is charged to a potential difference V2. The capacitors are now disconnectedfrom their respective charging batteries and connected in parallel to each other.(a) Find the total energy stored in the two capacitors before they are connected.(b) Find the total energy stored in the parallel combination of the two capacitors.(c) Explain the reason for the difference of energy in parallel combination in comparision

to the total energy before they are connected.Ans. (a) We know that

Energy stored in a capacitor =

12

CV2

∴ Energy stored in the given two charged capacitors E1 =

12

C1V12 and E2 =

12

C2V22

∴ Total energy stored =

12

C1V12 +

12

C2V22

(b) Let V be the potential difference across the parallel combination.Equivalent capacitance = (C1 + C2)

Since charge is a conserved quantity, we have

C QV

=⎡⎣⎢

⎤⎦⎥

(C1 + C2)V = C1V1 + C2V2

⇒ V =

C V C VC C

1 1 2 2

1 2

++

⎡

⎣⎢

⎤

⎦⎥( )

∴ Total energy stored in the parallel combination

=

12

(C1 + C2)V2 =

12

(C1 + C2)

( )( )

C V C VC C

1 1 2 2

1 2

2++

⎡

⎣⎢

⎤

⎦⎥

=

12

.

( )( )

C V C VC C

1 1 2 22

1 2

++++

(c) Total energy of the parallel combination is different (less) from the total energy beforethe capacitors are connected because some energy gets used up due to the movement ofcharges during sharing of charge.


Shiv Das

CapacitiveReactance(Ohm)

Frequency (Hz)

InductiveReactance(Ohm)

Frequency (Hz)

⎯→

⎯

I or VI

V

ωt

ωt

θ

Q.25. (a) Draw graphs showing the variations of inductive reactance and capacitive reactancewith frequency of the applied ac source.

(b) Draw the phasor diagram for a series RC circuit connected to an ac source.(c) An alternating voltage of 220 V is applied across a device X, a current of 0.25 A flows,

which lag behind the applied voltage in phase by

ππ2

radian. If the same voltage is

applied across another device Y, the same current flows but now it is in phase withthe applied voltage.

(i) Name the devices X and Y(ii) Calculate the current flowing in the circuit when the same voltage is applied

across the series combination of X and Y. 5Ans. (a) The two graphs are as shown

(b) The current leads the voltage by an angle θ

where 0 < θ <

π2

.

The required phasor diagram is shown.

Here tan θ =

XR

CR CR

C = =1

1ωω

(c) In device X, current lags behind the voltage by

π2

∴ X is an inductor.In device Y, current is in phase with the applied voltage∴ Y is a resistor

We are given that V = (XL)I or XL =

VI

or XL =

2200 25.

Ω = 880 Ω

Also 0.25 =

220XR

V = RI or R =

VI

∴ R =

2200 25.

Ω = 880 Ω

For the series combination of X and Y,


Shiv Das

Equivalent impedance (Z) = X RL2 2+ = (880 2 ) Ω = ( ) ( )880 8802 2+ = 880 2

Q Current (I) =

VZ

= 220880 2

= 0.177 A

Or(a) State the principle of working of a transformer.(b) Define efficiency of a transformer.(c) State any two factors that reduce the efficiency of a transformer.(d) Calculate the current drawn by the primary of a 90% efficient transformer which steps

down 220 V to 22 V, if the output resistance is 440 ΩΩΩΩΩ.Ans. (a) Principle of working of a transformer. See Q. 24 (Or)(i), 2016 (I Delhi). [Page 148

(b) The efficiency of a transformer equals the ratio of the output power to the inputpower.

Efficiency =

Output powerInput power

V IV I

= s s

p p

(c) Factors reduce the efficiency of a transformer(i) Eddy current losses

(ii) joule heat losses(iii) hysteresis losses(iv) magnetic flux leakage losses

(d) Given : efficiency = 90% = 0.9, Vs = 22 V, Vp = 220 V, Rs = 440 ΩWe have

V IV I

s s

p p = 90% = 0.9

∴

22220

. II

s

p = 0.9 or

II

s

p= 0 9

0 1..

= 9

∴ Ip =

I As9

224409

1180

= =

⎛⎝⎜

⎞⎠⎟ = 0.0056 A

I

VRs

s

s= =

⎡

⎣⎢

⎤

⎦⎥

22440

Q.26. (a) Explain with the help of suitable diagram, the two processes which occur during theformations of a p-n junction diode. Hence define the terms (i) depletion region and(ii) potential barrier.

(b) Draw a circuit diagram of a p-n juncion diode under forward bias and explain itsworking. 5

Ans. (a) See Q. 12, 2017 (I Delhi). [Page 198

(b) The circuit diagram is as shown:Voltmeter (V)

p nMiliameter

(mA)Switch

+ –


Shiv Das

Working:In forward bias condition, the direction of the applied voltage is opposite to the barrierpotential. This reduces the width of the depletion as well as the height of the barrier. Acurrent can, therefore, flow through the circuit. This current increases (non-linearly) withincrease in the applied voltage.

Or(a) Describe briefly three factors which justify the need for modulation of audio

frequency signals over long distances in communication.

(b) Draw the wavefroms of (i) carrier wave, (ii) a modulating signal and (iii) amplitudemodulated wave. (Not in Syllabus)

Ans. For (a & b) : See Q. 22, 2018 (I Delhi). [Page 284

Please Note : The questions asked in Set II and Set III are identical to Set I. Only the serialnumbers of questions were changed.

CBSE PHYSICS (Theory) – 2015 - Shivdas · PHYSICS (Theory) – 2015 (COMPTT. DELHI) 111 Shiv Das...

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