6 cbse solved paper 2019 solution8 cbse solved paper 2019 7. by using lens maker formulae for given...

6 CBSE Solved Paper 2019

SolutionSection A

1. Equal charge of opposite nature induces in the surface of conductor nearer to the source charge

+ + + + + + + + + +q– – – – – – – – – –q

Q

E

(1)

Electric lines of forces should fall / normally 90° away on / from the conducting plate.2. If the potential difference applied across the conductor is doubled, keeping the length and temperature of the

conductor, mobility of electron remains unchanged because mobility () is independent of applied potentialdifference. (1)

3. Threshold Frequency : The minimum frequency of incident light below which the photoelectric phenomenon doesnot occur. (1)

ORIntensity : Intensity is defined as the radiation energy incident per unit area per unit time. The SI unit of intensity isW/m2. (1)

4. By Brewster's law, = tan ip (½)

1

tan 303

c 1V 3

c 1V 3

8

8

V C 3 3 10 3

3 3 10 m/s(½)

5. In sky wave mode of propagation, the ionosphere layer of atmosphere acts as a reflector for range frequencies below30 MHz. Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.

(1)OR

The range of coverage in ground wave propagation depends on the following factors :–(i) Transmitted power (½)(ii) Frequency of signal (½)

Section B6. Let (P1, V) is rating of bulb B1 and (P2, V) is for bulb B2, then

P = 2V

RFor bulb B1,

R1 = 2

1

VP

...(i)

CBSE Solved Paper 2019 7and for bulb B2,

R2 = 2

2

VP ...(ii)

Now, in series combination,

Req = R1 + R2 = 2 2

1 2

V VP P

Pseries = 2 2

2 2eq

1 2

V VR V V

P P

V

(1)

= 2

1 2

2 1 2

1 2

V P PP P1 1

VP P

Again, in parallel combination,

eq

1R

= 1 2

1 1R R

= 1 2 1 22 2 2 2 2

1 2

1 1 P P P PV V V V VP P

V

(1)

Pparallel = 2

2 1 22

eq

V P PV

R V

= P1 + P2


7. By using lens maker formulae for given equi – concave lens,

2

1

n1 1 11

f n R R

2

1

n1 21

f n R ... (i)

where n2 and n1 are the refractive index of the lens and medium, respectively.Given, n2 = 1.5 and n1 = 1.4 Power of lens = – 5 D

100

5f

100f 20cm

5(1)

On putting all the values in equation (i),

1 1.5 2

120 1.4 R

1 0.1 2

20 1.4 R

1 1 2

20 14 R

R 120 7

20

R 2.85cm7 (1)

Hence the radius of curvature of an equi – concave lens is 2.85 cm.OR

When the prism is kept in another medium , we have to take the refractive index of eh prism with respect to the givenmedium.

prism

prismmedium

medium

prism

medium

A Dmsin

2A

sin2

60 Dmsin

1.6 2604 2 sin

25

CBSE Solved Paper 2019 9

60 Dmsin

8 214 22

2 60 Dm

2sin2 2

1 60 Dm

sin2 2

1 1 60 Dm

sin2 2

60 Dm45

2

Dm = 90° – 60° = 30°Therefore, the angle of minimum deviation of the prism is 30°

8. When a charged particle enters in the magnetic field at right angle, then the particle follows a circular path.

x x x x xx x x x x

x x x x xx x x x x

pProton

–particle

–particle(1)

Radius of the circular path mv

rqB

For same speed v, magnitude of charge and magnetic field.

p pr m

r mr m

m = 4 mp

p p

p

r m 1r 4m 4

p

r 4Hence,

r 1 (1)

9. According to Bohr's quantization condition, the angular momentum of orbiting electron is integral multiple of

h2

.

where 'h' is the plank's constant. Thus the angular momentum (L) of the orbiting electron is quantised.

h

L n2

where n number of orbitIn Brackett series for minimum wavelength.


n1 = 4, n2 = 2

min

1 1 1R

4

min

1 R16

min16R

(1)

where, R = Rydberg constant= 1.09 × 107 m–1

1

Å912

min = 16 × 912 = 14592 Å

(1)Hence, this wavelength is associated with infrared region of electromagnetic spectrum.

ORIn the first excited state of hydrogen atom n = 2

n

nn

2 rT

V (1)

210

6

n2 0.53 10

zZ

2.19 10n

3

162

n1.5 10

Z

nd

316

2

2T 1.5 10

1

= 1.5 × 10–16 ×8 = 1.2 × 10–15 seconds (1)10. Above 40 MHz frequencies, communication is essentially limited to the path of line of sight . These direct waves get

blocked at some point by the curvature of earth.

Ranged hRhT

R

R(1)

CBSE Solved Paper 2019 11where R Radius of earthand h Height of antenna

Therefore, range T R2Rh 2Rh (1)

11. The wave theory of electromagnetic radiation is not able to explain photoelectric effect due to the followingreasons :(i) According to wave theory, electromagnetic wave of any frequency should be able to cause photoelectric effect overa sufficient time However, it is observed that there is a threshold frequency below which photoelectric effect is notobserved. (1)(ii) It is observed that even with low intensity radiation of frequency above threshold frequency, emission ofphotoelectrons is almost immediate. This is contrary to prediction of wave theory.

According to the photon picture if frequency of photon (f) is greater than work function

h, the photoelectric

emission will happen almost instantaneously even if intensity is low.(1)

12. To plot the graph between two quantities, first of all we have to find the relation between the two. So de - Broglie wavelength of accelerated charged particle of mass m is given by

O

1V

(1)

h

2mqV

h

V constan t2mq

1V

slope of the line

h

2mq

2 2

22

h hslope charge(q)

2mq 2m slope (1)

SECTION C

13. (a) Equipotential surfaces due to an electric dipole :

+ –

Equipotentialsurface

q q (1)


(b) Let distance of point p where field has to be calculate be y from axial line,

E+q

y

p

+q–q

E –q

2a

y

X

–q 2 2

o

q1E

4 y a

q 2 2

o

q1and E

4 y a (1)

q qE E EDue to symmetry electric field in y direction will cancel out.

q 3/22 2

0

ˆ2qa iˆE 2 E cos ( i)4 a y

for y >>a

3 3

0 0

2qa pˆE i4 y 4 y

(1)

14. A

D

E(I + I )1 2 40V

10

20

40

80V

C

B

F

I2

I1

(1)

On applying kirchhoff's voltage Law in FEDC,–10 (I1 + I2) + 40 – 40 I1 = 0 – 10 I1 – 10I2 + 40 – 40I1 = 0 – 50 I1 – 10I2 = – 40 5I1 + I2 = 4 .... (i) (½)

Again in CDAB,40I1 + 80 – 20 I2 = 0 40I1 – 20I2 = – 80 I2 – 2I1 = 4 ... (ii) (½)

CBSE Solved Paper 2019 13On solving eqn (i) and (ii), we getI1 = 0 and I2 = 4 ATherefore, current through 20 resistor is 4 A and current through 40 resistor is zero. (1)

OREnd Error : Shifting of zero of scale at different point as well as the stray resistance give rise to the end error in meterbridge wire.It can be overcome by interchanging resistance in resistance box or can be overcome by using end correction inmeasurement. (1) From the given meter bridge,

1

1

lRS 100 l ...(i)

1

1

1.5l2RS 100 1.5l ... (ii)

From eqn (i) and (ii),

1 1

1 1

1.5l l2 100 1.5l 100 l

150 – 1.5 l1 = 200 – 3 l1 3 l1 – 1.5 l1 = 200 – 150 (1)1.5 l1 = 50

From eqn (ii)

2R 50 50S 100 50 50

2R

1S

S = 2 R = 2 ×5=10 15. (a) (i) Microwaves are used for 'Radar' system Frequency range of microwaves are from 1 GHz to 2 GHz.

(ii) Ultraviolet rays (UV rays) are used in eye surgery Frequency range of UV rays are from 1015 Hz to 1017Hz.(1)

(b) For a travelling electromagnetic wave equations can be written asE = E0 sin (kx – t )

and B = B0 sin (kx – t )

We get,

O

O 0 0

E 1C andC

B

Now, energy density of electric field.

E

20

1p E

2

2 20 0

1E sin kx wt

2

Average of sin2 (kx – t ) will come out to be 12 over a long time.


E

20 0

1p E

4 ... (i)

Now, energy density of magnetic field.

B

2 2 20

0 0

1 1p B B sin kx t

2 2 (1)

2 20 0

0 0

B B1.

2 2 4 ... (ii)

00

EB

C

B

2 20 0 0 0

200

E Ep

44 C

20 0

1E

4 ... (iii)

Therefore, from equation (i) and (ii) average electric field energy density and average magnetic field energy densityare equal. (1)

16. In a region where wave is propagating, geometrical structure formed by joining all the points which are vibrating insame (absolute) phase are said to be wavefront.Lasws of reflection on the basis of Huygens’ wave theory:As shown in the figure, consider a plane wavefront AB incident on the reflecting surface XY, both the wavefront andthe reflecting surface being perpendicular to the plane of paper.

ii

x

S

Y

N

A

B C

r

D

N1

Incident Wavefront Reflected Wavefront

r

(1)

In accordance with Huygens’ principle, from each point on plane BC, secondary wavelets start growing with the speedC. During the time the disturbance from A reaches the point C the secondary wavelets from B must have spread overa hemisphere of radius BD = AC = ct, where t is the time taken by the disturbance to travel from A to C.Let angles of incidence and reflection be i and r, respectively. In ABC and DCB, we have

BAC CDB 90 BC = BC and AC = BD = ct ABC DCB

Hence, ABC DCB i r (1)

ORRefractive Index : The factor by which speed of light gets reduced with respect to the "speed of light in vacuum" issaid to be the refractive index for the medium.


C CorV

V

Here, V is the speed of light in medium, c is the speed of light in vacuum and is the defined refractive index for themedium.

A1

A2

B2

B1

ri

V1T

V2T

I

II

(1)

In the above figure, let B1B2 is the incident wavefront and A1A2 is the refracted wavefront in other medium with V2as the wave speed. When the point B1 traverse a distance B1A1 = V1 T in medium I, the wavefront originated from pointB2 must have travelled a distance of V2T in medium II.Now,

1 1 2 2

1 2 1 2

B A A Bsin i and sin r

A B A B

1 1 1 1

2 2 2 2

B A V T Vsin isin r A B V T V

2

1

sin iHence,

sin r (1)

17. (a) Mutual Inductance is a phenomenon where the magnetic field generated by a coil of wire induces voltage in anadjacent coil of wire. A transformer is a device constructed of two or more coils in close proximity to each other withthe express purpose of creating a condition of mutual inductance between the coils. (1)(b) According to the right land screw rate, the magnetic field will be into the plane across the loop. The square loopABCD and infinitely long straight wire work as two parallel current carrying conducting wires.

a

F1F2

a

F4

F3

a

a

A

D

I1

xI2

B

C

(1)

16 CBSE Solved Paper 2019Now we know that the magnetic force between the two parallel current carrying conducting wire

0 1 2I I LF

2 r

, where L is length of wire, r is the distance between the wire.

Force on length AD,

0 1 21

I I aF .....(i)[Toward right]

2 xForce on length BC,

0 1 2

3I I a

F .....(ii)[Toward left]2 x a

As we seen above the diagram, the force on AB and CD will be equal in magnitude but opposite direction, so they willcancelled to each other. (1)

18. (a) Let I be the current through the current carrying loop. Area of the loop = A = ab and number of turns in the loop= N, a, b sides of rectangular loop, then

Q R

BI

SP

N SI F1

F2

Iup

Idown

M

b

F2

b sin

(1)

According to fleming's left – hand rule,F1 = IaB and F2 = Ia BThese two equal and opposite forces form a couple which exerts a torque given as = Force × Perpendicular distance.= IaB × b sin = IBA sin [ A = ab]as the coil has N turns,

NIABsin (1)(b) Significance of a radial magnetic field : In radial magnetic field the plane of the coil is always parallel to the planof the magnetic field and area vector of coil is perpendicular to magnetic field. It is always exerts maximum torque onthe coil. (1)

19.

Eyepiece

Tube Length

ObjectiveLens

ImageD

(1)

Now here fo = focal length of objective lens = 15 mand fe = focal length of eyepiece lens = 1 cm

CBSE Solved Paper 2019 17Image formed by objective lens is in its focal plane

MoonImage

ObjectiveLens

fo (1)

imagemoon

moon focal plane

ddr r

6image

8

d3.48 10153.8 10

2image

3.48 15d 10 m

3.8

= 13.74 ×10–2m (1) = 13.74 cm

20. (a) Gauss's Law of Magnetism: It states that net magnetic flux through a closed surface or gaussian surface is zeroMathematically it can be represented as –

B.ds 0 (1)

Gauss's law for magnetism is one of the four maxwell's equations that underline classical electrodynamics. It tells usthat magnetic monopoles do not exist.(b) Magnetic field lines are imaginary lines which give pictorial representation for the magnetic field inside and aroundthe magnet. (1)Properties of magnetic field lines due to a bar magnet :(i) These lines form continuous closed loops.(ii) The tangent to the field line gives direction of the field at that point(iii)Larger the density of the lines, stronger will be the magnetic field.(iv)These lines do not intersect one another. (1)

OR

Property Paramagnetic Materials

Diamagnetic Materials Ferromagnetic Materials

(i)Susceptibility (Xm)

Small and Positive O < Xm < = small number

small and negative -1 < Xm < 0

Very large Xm > 1000

(ii) Relative Permeability

1 < µr < (1 + ) = small number

Positive and less than one 0 < r < 1

Large value r > 1000

a(iii) Effect of temperature m

1X

T

Ex: Al, Na, Ca

Independent with temperature Ex: Bi, Cu, Si,

mc

1X

T T

(T > TC) Ex – Iron, Ni, Co, Gd (1 + 1 +1)

18 CBSE Solved Paper 201921. Decay Constant : The radioactive decay constant may be defined as the reciprocal of the time during which the

number of atoms in a radioactive substance reduces to 86.8 % of their initial number.

o10

N2.303log

t N (½)

Let No be the initial number of nuclei, be the decay constant and t 12 be the half – life, then

The instantaneous activity of radioactive material is given by, (½)A = A0 e –t

where, A0 is activity at t = 0Given, activity after 20 hours is 10, 000 disintegrations per second and activity after 30 hours is 5, 000 disintegrationsper second.

10,000 = A0 e– (20 × 3600) ...(i)and 5000 = A0 e – (30 × 3600) ... (ii)On dividing eqn (i) by (ii), we get,

20 3600o

30 3600o

A e10,0005000 A e

2= e (10 × 3600)

2= e (36000)

log 2 = log e (36000)

log 2 = 36000 × log e 36000 × = log 2

5log 21.92 10

36000

log 2Half life 36000S 10hours (1)

we know that,dN

Ndt

10,000 = 1.92 × 10–5 × N1

81 5

10,000N 5.208 10

1.92 10

Half life = 10 hours Initial numbers of nuclei N0 = 2 N1 = 2×5.208 ×108 (1) = 10.416 ×108

22. (a) Given, Band gap of diode D1 = 2.5 eVBand gap of diode D2 = 2 eVand Band gap of diode D3 = 3 eVWavelength of light incidented = 600 nm Energy associated with a photon of incident light,

hc 1240E 2.1eV

600

(½)

To detect the light, the band gap should be less than E. Band gap < EThere fore, only diode D2 will be able to detect the light. (1)(b) The photodiodes are operated in reverse biased. In this case, the depletion region is large, electric field strengthis very high and current through the circuit is very small. All these things, facilitate the easy detection of light.

CBSE Solved Paper 2019 19When light photons get incidental on depletion region, they generate new pairs of holes and electrons. Due to strongelectric field, they separate out and constitute rise in electric current in the circuit, which indicates detection.

(1)

p n

(½)

23. (a) In n–p–n transistor, two segments of n–type semiconductor (emitter and collector) are separated by a segmentof p – type semiconductor (base).The function of three segments of a n–p–n transistor are : –(i) Emitter : It emmits the majority charge carriers.(ii) Collector : It collects the majority charge carriers (1)(iii)Base : It provides the interaction between the collector and emitter in n–p–n transistor.(b) Common–emitter Transistor : In n–p–n transistor, base–emitter circuit is forward biased with VBE and emitter– collector circuit is reverse biased with battery Vcc.

IC

VCC

VCE

IEVBB

C mA

+ –

IB

VBE

IB +–+

–

+–

µA

(1)

Collector or Output characteristics : A graph showing the variation of collector current (IC) with collector emittervoltage VCE at constant base current Ib is called the output characteristic of transistor.A study of these curve reveals the following:(i) Initially the collector current IC increases rapidly with increase in VCE and becomes constant afterwards.(ii) Saturation collector current depends entirely on base current Ib.

IC (mA)

Ib= 30 µA

Ib= 20 µAIb= 10 µA

VCE (Volts)

Output resistance (rO) CE

C Ib

VI

(1)

20 CBSE Solved Paper 2019OR

Full wave rectifier : A full wave rectifier consists of two diodes connected in parallel across the ends of secondarywinding of a centre tapped step down transformer. The load resistance RL is connected across secondary winding andthe diodes between A and B as shown in the circuit.

A

B

x

y

R (Output)L

Centre-TopTransportion

D1

D2

Centre Tap

(1)

During positive half cycle of input a.c, end A of the secondary winding becomes positive and end B negative. Thusdiode D1 becomes forward biased, where as diode D2 reverse biased. So, diode D1 allows the current to flow throughit, while diode D2 does not, and current in the circuit flows from D1 and through load RL form x to y. (1)Again during negative half cycle of input a.c, end A of the secondary winding becomes negative and end B positive,thus diode D1 becomes reverse biased. whereas diode D2 forward biased.

Vin

Vout

t

t

O

Input-Output Waveforms

Because , in both the half cycle of input a.c. electric current through load RL flows in the same direction, so. d.c isobtained across RL . Although direction of electric current through RL remains same, but its magnitude changes withtime, so it is called pulsating d.c. (1)

24. (a) Let the amplitude modulated signal isCm (t) = (AC + Am sin mt) sinct A = AC + Am ... (i)and B = AC – Am ... (ii)From eq. (i) and (ii)

c m c mA A A AA B

2 2(1)

c

c2AA B

A2 2

and

c m c mA A A AA – B2 2

mm

2AA

2

CBSE Solved Paper 2019 21Therefore, the modulation index is given as

m

c

A A B2A

A B2

(1)

A BA B

(b) We know that,

Modulation index, m

c

AA

Given. Am = 10V and Ac = 15 V

10 215 3

Therefore, modulation index () is generally kept less than one to avoid distortion. (1)SECTION –D

25. (a)

AC source ofvariable frequency

I

VR VL VC (1)

(i) In a series LCR circuit current will be same across all components.(ii)Voltage drop across resistance, VR = IR

Voltage droop across inductor, VL = IXLand Voltage drop across capacitor VC = IXC(iii) Voltage across inductor leads current by /2and voltage across capacitor lags current by /2Therefore, phasor diagram is as follows.

VL

VR

VC

(V – V )L CVL

VC

AX

2 22 2 2R L C L CV V V V IZ I R IX IX


22L CZ R X X

22 1

R Lc

Variation of impedance (z) with frequency of ac source

Impedance(z)

frequency(f)R

f o

where fo is a resonance frequency, at this frequency impedance will be minimum and equal to the resistance in thecircuit. (1)(b) The phase difference between voltages across inductor and the capacitor at resonance in the LCR circuit is .

(1)(c) when an inductor is connected to a 200 V dc source,

L R

I

200 V

dc

Impedance of real inductor 2 2LZ R X (1)

For dc source XL = O Z = R

NowV V

IZ R

2001

R

R = 200 when an inductor is connected to a 200 V ac source,

L R

0.5

200 V

CBSE Solved Paper 2019 23V

IZ

200 2000.5 Z 400

Z 0.5 (1)

Therefore, impedance for ac source is more than dc source. Current in ac source is less.Now,

2 2 2L LZ R X 160000 40000 X

2LX 120000

LX 200 3

L 200 3

2 fL 200 3 (1)

200 3 2 3L Henry.

2 50OR

(a) Working Principle of Transformer : A transformer is based on the principle of mutual induction, i.e. wheneverthe amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

ACSource P S Load

SecondaryWinding

PrimaryWinding Laminated

core

(1)

Four sources of energy losses in transformer :(i) Eddy current Loss : Energy loss in the form of heat due to eddy current. It can be minimised by taking laminatedcore consisting of insulated rectangular sheets.(ii) Flux Leakage : Total fluxes linked with the primary do not completely pass through the secondary which denotesthe loss in the flux.(iii) Copper loss : Due to heating energy loss takes place in copper wires of primary and secondary coils.(iv) Hysteresis Loss : The energy loss takes place in magnetising and demagnetising the iron core over every cycle.

(2)(b) Total electric power requiredp = 1200 kW = 1200 ×103 wattand supply voltage V = 220 voltVoltage at which electric plant is generating power V = 440 voltDistance between the four and power generation station (d) = 20 kmResistance of two wire lines carrying power = 0.5 /km.Total resistance of wires (R) = (20 + 20)×0.5

= 40 × 0.5 = 20


Input voltage of a step – down transformer, V1 = 4000 voltand output voltage V2 = 220 volt rms current in the wire is give as

3

1

p 1200 10I 300A

V 4000

Power loss = I2R = (300)2×20 = 1800 Kw (2)Hence, Line power loss in the form of neat = 1800 Kw

26. (a) Difference between Interference and diffractionInterference Diffraction

(i) Interference is caused by (i) Diffraction is caused by superpositionSuper position of finite number of infinite number of wavesof waves

(ii) All fringes are of equal width (ii) All fringes are not of equal width (1)

Expression for the intensity :Let displacement produced by S1 and S2 arey 1 = acos t and y2 = a cos (t + ) (1)then amplitude of resultant displacement isy = y1 + y2 = a cos t + a cos (t + )= a [ cos t + cos (t + )]

2acos cos t2 2

A 2acos2

Therefore, intensity at any point is

2 2

o oI 4I cos I A2 (1)

(b) Given, wavelength () = 6.20 × 10–7m = 620 nmwidth of slit (a) = 3 mm = 3× 10–3mDistance of screen from slit (D) = 1.5 m

Fringe width ()

Da

(1)

Now, separation between 1st minimum and 3rd order

7 D D 5 Dmax ima is y

2 a a 2 a

7

43

5 6.20 10 1.57.75 10 m

2 3 10

(1)

= 0.775 mmOR

(a) when light travel from denser medium to rarer medium it moves away from the normal. If we keep on increasingthe angle of incidence then at a particular angle of incidence, angle of refraction become 90°. This angle of incidenceis known as critical angle. Beyond this angle of incidence, light ray does not move to other medium it gets reflectedto the same medium. This phenomenon is known as total internal reflection (TIR). According to snell’s law (1)

1 sin i = 2 sin r 1 sin C = 2 sin 90°


C1

sin

N

R

D

1

2

c

(1)

where 1 Refractive index of denser medium2 Refractive index of rarer medium.

(b) +10 cm –10 cm +30 cm

O

10 cm5 cm30 cm

(1)

From lens formula,

1 1 1 1 1 1 1 1 1 2v u f v 30 10 v 10 30 30

v = 15 cmNow, for middle lens. it will act as an object = + 10 cm and f = – 10 cmAgain, from lens formula.

1 1 1v u f

1 1 1v 10 10

10 v

v

Therefore, rays will become parallel to principal axis and for third lens object is at infinite1 1 1v u f

1 10 v 30cm

v 30

Hence, final image is 30 cm right of the third lens. (2)27. (a) When plates A and B of a parallel plate capacitor are connected to the terminals of a battery, the battery sets up

electric field in the connecting wires. The electric field drives charge Q from plate B, connected to the negativeterminal of the battery to plate A connected to the positive terminal. (1)

26 CBSE Solved Paper 2019The charge transfer takes place till the potential difference across the capacitor plates equals the EMF of the battery.

+ –

+Q –QA B

++++

––––

–

+(1)

Potential difference across the capacitor platesQ

VC

Small amount of work done in giving an additional charge dq to the capacitor is

qdW dq

c

Total work done

QQ 2

0 0

q q1W dq

c C 2 2Q

2C

Hence energy stored in a capacitor2Q

U2C

(1)

(b) Let charge q be transferred from one capacitor to the other, then equating potential difference across thetwo capacitors,

CV q qC C

CV–q = q CV = 2 q

CVq

2

+CV –CV

SC

+ –++++

––––

–

+

(1)

Now, energy stored in the combination,

22 2

fCVq CV

U 22C 4C 4

Initial energy , 2i

1U CV

2

2

f2

i

CVU 14U 2CV

2


q

(CV – q)+ –

– q

(CV – q)

(1)

OR(a) Let P be an equatorial point for a dipole consisting of charges –q and + q with a separation 2a, then

E

= electric field due to the charge + q

P

r

+q–q2a

a

E

E

netE

(1)

2 20

q1E

4 a r

... (i)

Again,

E

= electric field due to the charge – q

2 20

q14 a r

... (ii)

By superposition principle,

netE E E

net 2 2 2 20

q2 aE 2E cos

4 a r a r (1)

3/22 20

2aq14 a r

In vector form,

net 3/22 20

pE

4 a r

28 CBSE Solved Paper 2019For short dipole, r >> a, then

net 3

0

pE

4 r (1)

(b) To overcome the electrostatic repulsion between the point charges q each, Q must be at the centre of the linesegment joining the two. The sign of Q must be opposite to that of q.

2

2 2kq kqQ

02 1

2kq kqQ4 1

(1)

q1 m

Q

q1 m

qQ

4

qQ

4

Hence, Q must be at a distance of 1 m from each charge q. (1)

6 cbse solved paper 2019 solution8 cbse solved paper 2019 7. by using lens maker formulae for given...

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