physics 2008 set 1.1 sol

Physics 2008 Set 1 Close

Subjective Test

(i) All questions are compulsory.

(ii) There are 30 questions in total.

Questions 1 to 8 carry one mark each,

Questions 9 to 18 carry two marks each,

Question 19 to 27 carry three marks each and

Question 28 to 30 carry five marks each.

(iii) There is no overall choice. However, an internal choice has been provided.

(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

(v) Use of calculators is not permitted.

Question 1 ( 1.0 marks)

What is the direction of the force acting on a charged particle q, moving with a velocity in a uniform

magnetic field ?

Solution:

The force acting on a charged particle q moving in a uniform magnetic field with velocity is given

by the relation:

This relation involves the cross product of and . Hence, magnetic force is always normal to both and

.


Name the part of the electromagnetic spectrum of wavelength 10−2 m and mention its one application.

Solution:

The part of the electromagnetic spectrum which ranges from 0.1 m to 10−3 m is known as microwave.

Microwaves are used in radar systems for aircraft navigation.


An electron and alpha particle have the same de−Broglie wavelength associated with them. How are their

kinetic energies related to each other?

Solution:

Kinetic energy of a particle of mass m and velocity v is given as:

2/3/2011 Subjective Test Paper - Physics - Meritn…

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de-Broglie wavelength associated with a particle of momentum p is given as:

It is given that an electron and an alpha particle have the same de-Broglie wavelength. Hence, we can

write:

Hence, the kinetic energy of the electron is greater than that of the alpha particle.


A glass lens of refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the

liquid in order to make the lens disappear?

Solution:

The lens will not be visible if no refraction occurs at the liquid−glass interface. This means that the incident

ray should go through the glass without any deviation. For this condition to be fulfilled, the refractive index

of the liquid must be equal to 1.5.


A 500 µC charge is at the centre of a square of side 10 cm. Find the work done in moving a charge of 10

µC between two diagonally opposite points on the square.

Solution:

The 500 µC charge is placed at the centre of a square. This charge is, therefore, at the same distance from

all the corners of the square. The opposite corners, say A and C, will have the same potential i.e.,

.

Work done in moving a charge q between points A and C is given as:

W = q(VC − VA) = q × 0 = 0

Hence, no work is done in moving the charge between two diagonally opposite points on the square.


State the reason, why heavy water is generally used as a moderator in a nuclear reactor.

Solution:

In nuclear reactors, heavy water is generallyused as a moderator because unlike normal water, which



In nuclear reactors, heavy water is generallyused as a moderator because unlike normal water, which

absorbs neutron, it slows down neutron without absorbing it.


How does the fringe width of interference fringes change, upon the whole apparatus of Young’s

experiment is kept in a liquid of refractive index 1.3?

Solution:

The fringe width will decrease.

When light enters a denser medium, its wavelength decreases by a factor 1.3 and hence the fringe width

also decreases by a factor 1.3.


The plot of the variation of potential difference across a combination of three identical cells in series versus

current is as shown below. What is the emf of each cell?

Solution:

It can be inferred from the given graph that for zero current, equivalent emf is 6 V. Since three cells are

connected in series, emf of each cell will be


Derive the expression for the electric potential at any point along the axial line of an electric dipole?

Solution:

The following figure shows an electric dipole of length 2a.

Let P be an axial point at distance r from the centre of the dipole. Electric potential at point P is given as:

Question 10




Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the

other having negative susceptibility. What does negative susceptibility signify?

Solution:

Magnetic susceptibility of a material is defined as the ratio of the intensity of magnetisation (I) induced in

the material to the magnetisation force (H) applied on it.

Magnetic susceptibility is represented as:

Substances such as copper, lead, etc. have negative susceptibility. These substances are called

diamagnetic substances.

Substances such as aluminium, calcium, etc. have positive susceptibility. These substances are called

paramagnetic substances.

Substances with negative susceptibility signify that they are repelled by magnets.


The oscillating magnetic field in a plane electromagnetic wave is given by

(i) Calculate the wavelength of the electromagnetic wave.

(ii) Write down the expression for the oscillating electric field.

Solution:

The general equation of a magnetic field varying sinusoidally with x and t is given as:

The given equation is:

On comparing equations (1) and (2), we obtain

(i) Wavelength of the given electromagnetic wave is 0.0067 m.

(ii) Electric field amplitude is given as:

The general equation for an oscillating electric field varying sinusoidally with x and t will be along the z-

component. It is given as:




Prove that an ideal capacitor, in an a.c. circuit does not dissipate power.

OR

Derive an expression of the impedance of an ac circuit consisting of an inductor and a resistor.

Solution:

In a capacitative circuit, the current leads the emf by a phase angle of .

For emf, E = E0 sin ωt, the current flowing in a capacitative circuit will be:

Work done in one complete cycle of ac is written as:

∴Average power dissipated, P

Therefore, the average power dissipated by an ideal capacitor over a complete cycle of ac is zero.

OR

Let an inductor L and resistor R be connected in series to a source of alternating emf as shown in the



Let an inductor L and resistor R be connected in series to a source of alternating emf as shown in the

following figure.

The maximum voltage across resistor R is given as:

In a resistor circuit, voltage is in phase with current . It is represented by along OX.

The maximum voltage across inductor L is given as:

In an inductive circuit, voltage across the inductor leads the current by 90°. It is represented by

along OY.

The vector sum of potentials and is the voltage phasor . It is represented by that is

making an angle Φ with the current phasor I0.

The impendence of the circuit is given by:

This is the expression for the impedance of an ac circuit consisting of an inductor and a resistor.


A nucleus undergoes β− decay and becomes . Calculate the maximum kinetic energy of

electrons emitted assuming that the daughter nucleus and anti − neutrino carry negligible kinetic energy.

Solution:

β− decay of is given as:

Where,

Q = Kinetic energy of the daughter nucleus



Ignoring the rest mass of the anti-neutrino and the mass of the electron , the mass defect involved

in the nuclear reaction is given as:

Hence, when the energy carried by is zero, the maximum kinetic energy of β− particle is 4.374 MeV.


Distinguish between an intrinsic semiconductor and p-type semiconductor. Give reason, why a p-type

semiconductor crystal is electrically neutral, although nh >> ne?

Solution:

Intrinsic semi-conductor is a pure semi-conducting material. No impurity atoms are present in it or added

to it. A p-type semi-conductor is an extrinsic semi-conductor obtained by doping the impurity of trivalent

atoms such as Al, B, In, etc. to an intrinsic semi-conductor.

The charge of additional charge carriers present in a p-type semi-conductor are just equal and opposite to

that of the ionised cores in the lattice. These additional charge carriers are balanced out by the ionised

cores. Hence, a p-type semi-conductor is electrically neutral.


Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting

telescope.

Solution:

The following figure shows the ray diagram of the image formation by a reflecting telescope.

Advantages of a reflecting telescope over a refracting telescope:

1. Image formed by a reflecting telescope is brighter than that formed by a refracting telescope.

2. As the objective in a reflecting telescope is a mirror, the image formed by this telescope does not

undergo any chromatic aberration.


A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation

when angle of incidence is of the angle of prism. Calculate the speed of light in the prism.

Solution:



It is given that the angle of incidence (i) is of the angle of prism (A) i.e.,

Angle of prism of an equilateral triangular glass prism, A = 60°

Hence, angle of incidence,

Using the prism formula, we can obtain the relation for the refractive index of the material of the prism as:

Hence, the speed of light in the prism is 2.12 × 108 m/s.


The given inputs A, B are fed to a 2-input NAND gate. Draw the output wave form of the gate.

Solution:

For NAND gates, the Boolean expression is written as:

For the given wave form, we can list the values of inputs A and B and output Y as shown in the following

table.



Time interval Input A Input B Output Y =

t < t1 A = 1 B = 1 0 + 0 = 0

t1 <t< t2 A = 0 B = 0 1 + 1 = 1

t2 < t< t3 A = 0 B = 1 1 + 0 = 1

t3 < t< t4 A = 1 B = 0 0 + 1 = 1

t4 < t< t5 A = 1 B = 1 0 + 0 = 0

t5 < t< t6 A = 0 B = 0 1 + 1 = 1

t6 < t< t7 A = 0 B = 1 1+ 0 = 1

Thus, the waveform for output Y can be shown as:


A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is

49 m. What is the maximum distance between them, for satisfactory communication in the LOS mode?

(Radius of earth = 6400 km)

Solution:

It is given that:

Height of the transmitting antenna, hT = 36 m

Height of the receiving antenna, hR = 49 m

Radius of earth, R = 6400 km = 6400000 m

The maximum line-of-sight (LOS) distance dm between the two antennas for a satisfactory communication

is given by the relation:

Hence, the maximum distance between the two antennas should be 46.51 km.


How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a

plane wave refracting at a plane surface separating two media. Hence verify Snell’s law of refraction.

Solution:



The locus of all the particles of a medium, which are vibrating in the same phase, is called a wavefront.

Snell’s law of refraction:

Let us consider the following figure.

Let PP’ represent the surface separating media 1 and 2.

Let v1 and v2 represent the speeds of light in media 1 and 2 respectively. A plane wavefront AB is

propagating in the direction A’A, incident on the interface at an angle i. Let t be the time taken by the

wavefront to travel the distance BC.

Distance = Speed × Time

∴BC = v1 t

In order to determine the shape of the refracted wavefront, we draw a sphere of radius v2 t from the point

A in the second medium.

Let CE represent a tangent plane drawn from point C.

Then, AE = v2 t

Hence, CE would represent the refracted wavefront.

In ∆ABC and ∆AEC, we have:

If c is the speed of light in vacuum, then:



Where, n1 and n2 are the refractive indices of media 1 and 2 respectively

This is Snell’s law of refraction.


A metallic rod of length l is rotated at a constant angular speed ω, normal to a uniform magnetic field B.

Derive an expression for the current induced in the rod, if the resistance of the rod is R.

Solution:

The given situation can be shown as:

Let θ be the angle traced by the free end of the rod in time t. The area swept-out by the rod in time t is

given as:

Since the angle between the area vector and the magnetic field vector is zero, the magnetic flux linked to

this area is given as:

According to Faraday’s laws of electromagnetic induction, induced emf (e) is given as:

Hence, the current induced in the rod is given as:


The figure below shows the V-I characteristic of a semiconductor diode.



(i) Identify the semiconductor diode used.

(ii) Draw the circuit diagram to obtain the given characteristic of this device.

(iii) Briefly explain how this diode can be used as a voltage regulator.

Solution:

(i)The V-I characteristic shown has a breaking voltage (Vbr) at −90 V during reverse biasing. During

forward biasing, the semi-conductor shows a steep rise at 0.8 V. Hence, the semi-conductor diode is a

Zener diode.

(ii)In order to show the given V-I characteristic, a circuit diagram can be drawn as follows:

(iii) Zener diode as a voltage regulator:

When the input dc voltage increases beyond a certain limit, the voltage across Zener diode becomes

constant, equal to the Zener breakdown voltage (Vbr). The current through the Zener diode circuit sharply

increases as the dynamic resistance of the Zener diode becomes almost zero after the Zener breakdown

voltage. As a result, there is an increase in the voltage drop across resistor RS. Since RL is connected in

parallel to the Zener diode, the voltage across load resistance RL remains the same as that of the Zener

breakdown voltage. The output voltage remains constant because the current through the Zener increases

in order to reduce the output voltage. Thus, a regulated voltage is obtained across the load resistance.


An inductor 200 mH, a capacitor 500 µF and a resistor 10 Ω are connected in series with a 100 V

variable frequency a.c. source. Calculate the

(i) frequency at which the power factor of the circuit is unity.

(ii) current amplitude at this frequency.

(iii) Q-factor.

Solution:

It is given that:

Inductance of the inductor, L = 200 mH = 200 × 10−3 = 0.2 H

Capacitance of the capacitor, C = 500 µF = 500 × 10−6 = 5 × 10−4 F

Resistance of the resistor, R = 10 Ω

Voltage of the source, V = 100 V

Power factor of the LCR circuit, cosΦ = 1

(i) Power factor of the LCR circuit is given as:



(i) Power factor of the LCR circuit is given as:

Hence, the power factor of the series LCR circuit will be unity at frequency 16 Hz.

(ii) Current amplitude at resonance frequency is calculated as:

This is the rms value of current.

Current amplitude is the peak current which can be calculated by multiplying the rms value by .

Hence, the amplitude of current is:

A

(iii)


Prove that the current density of a metallic conductor is directly proportional to the drift speed of

electrons.

OR

A number of identical cells, n, each of emf E, internal resistance r connected in series are charged by a



A number of identical cells, n, each of emf E, internal resistance r connected in series are charged by a

d.c. source of emf E′, using a resistor R.

(i) Draw the circuit arrangement.

(ii) Deduce the expressions for (a) the changing current and (b) the potential difference across the

combination of the cells.

Solution:

Consider a conductor of length l and area of cross-section A, having n electrons per unit length, as shown

in the following figure.

Volume of the conductor = Al

Total number of electrons in the conductor = Volume × Electron density = Aln

Since e is the charge of an electron, the total charge contained in the conductor:

Let a potential difference V be applied across the conductor. The resulting electric field in the conductor is

given by:

Hence, free electrons begin to drift in a direction opposite to that of the electric field E. The time taken by

the free electrons to cross-over the conductor is given as:

Current flowing through the conductor is given by:

OR

(i) The following figure shows a simple circuit diagram that consists of n identical cells connected in series,

each having an emf E and internal resistance r. The cells are charged by a dc source of emf E’ using a

load of resistance R.



Equivalent emf of n cells connected in series is given as:

E′ = E + E + … up to n terms = nE

Equivalent internal resistance of n cells connected in series is given as:

r’ = r + r + … up to n terms = nr

Total resistance of the resistance R is given by:

R’ = r’ + R = nr + R

(ii) (a) Current flowing through resistor R is given by:

(b) Potential difference across the combination of cells is given as:


A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V as shown in the figure. When a

cell of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this

cell and using a cell of unknown emf, the balance point shifts to 80 cm.

(i) Calculate unknown emf of the cell.

(ii) Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1 V.

(iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer.

Solution:

(i) It is given that:

E1 = 1.5 V

l1 = 60 cm = 0.60 m

l2 = 80 cm = 0.80 m

E2 = emf of the unknown cell

For a potentiometer, we have the following formula for the ratio of emfs:



Hence, the emf of the unknown cell is 2 V.

(ii) The circuit will not work because there will be little drop of potential across the potentiometer wire in

comparison to the emf of the unknown cell. Hence, the balance point will not be obtained on the

potentiometer wire. Thus, for the functioning of the potentiometer, the emf of the driver cell should be

greater than the emf of the unknown cell.

(iii) No, the balance point will not be affected by any change in R. This is because no current passes

through the resistor R at the time when balance is achieved. R has, therefore, no role in changing the

balance point.


An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work

function. If the photo-electrons emitted from this surface have the de-Broglie wavelength λ1, prove that

.

Solution:

Einstein’s photoelectric effect for a photosensitive metal surface of work function Φis given as:

de-Broglie wavelength of the emitted photoelectrons = λ1



It is given as:


The energy level diagram of an element is given below. Identify, by doing necessary calculations, which

transition corresponds to the emission of a spectral line of wavelength 102.7 nm.

Solution:

For element A:

Energy State (E1) = −1.5 eV

Energy State (E2) = −0.85 eV

Energy of the emitted photon is given as:

E = E2 − E1= −0.85 − (−1.5) = − 0.85 + 1.5 = 0.65

Hence, the wavelength of the emitted photon is given by:

For element B:

Ground state energy, E1 = − 3.4 eV

Excited state energy, E2 = − 0.85 eV


E = E2 − E1= − 0.85 − (− 3.4) = − 0.85 + 3.4 = 2.55 eV



For element C:

Ground state energy, E1 = − 3.4 eV

Excited state energy, E2 = − 1.5 eV


E= E2 − E1 = − 1.5 − (− 3.4) = − 1.5 + 3.4 = 1.9 eV

For element D:

Ground state energy, E1 = −13.6 eV

Excited state energy, E2 = −1.5 eV


E = E2 − E1= − 1.5 − (− 13.6) = 12.1 eV

Hence, element D corresponds to a spectral line of wavelength 102.7 nm.


Draw a plot of the variation of amplitude versus ω for an amplitude modulated wave. Define modulation

index. State its importance for effective amplitude modulation.

Solution:

A plot of amplitude (A) versusω for an amplitude modulated wave is shown in the following figure.

Where,

Ac = Amplitude of the carrier voltage

µ = Index of modulation

ωc = Carrier frequency

ωm = Modulated frequency

Modulation index is the ratio of the amplitude of modulating voltage (Am) to that of the carrier voltage (Ac)

i.e.,



The quality of the transmitted signal is determined by the index of modulation (µ).

The variation in the carrier amplitude Ac is small for a small modulation index and vice-versa.

Hence, for effective modulation, the value of µshould be checked and it should be ensured that the

modulation index is never greater than unity.


(a) Using Biot-Savart’s law, derive an expression for the magnetic field at the centre of a circular coil of

radius R, number of turns (, carrying current i.

(b) Two small identical circular coils marked 1, 2 carry equal currents and are placed with their geometric

axes perpendicular to each other as shown in the figure. Derive an expression for the resultant magnetic

field at O.

OR

Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the

function of the electric and magnetic fields applied on charged particle.

Deduce an expression for the period of revolution and show that it does not depend on the speed of the

charged particle.

Solution:

(a)According to Biot-Savart’s law, the magnetic field due to a current element at a point with position

vector is given by the relation:

Where,

µ0 is the permeability of free space

Consider a circular loop of wire having radius r and carrying a current I. Take a current element on the

loop, as shown in the following figure.

The direction of element dl is along the tangent at a point on the loop.

∴dl⊥r

According to Biot-Savart’s law, the magnetic field at the centre O due to this current element is given as:



The magnetic field due to all such current elements will point onto the plane of paper at centre O. Hence,

the total magnetic field at centre O is given by the integration:

For a coil having ( turns, the magnetic field at centre,

(b) Magnetic field at O due to loop 1 is given as:

, acting towards left

Magnetic field at O due to loop 2 is given as:

, acting vertically upwards

Hence, the resultant field at point O will be:

The resultant field acts at an angle of 45° with the axis of loop 1.

OR

Cyclotron:It is a device using which positively-charged particles such as protons, deuterons, etc. are

accelerated to high speeds.

Principle of a cyclotron:

A positively-charged particle can be accelerated by making it cross the same electric field repeatedly with

the help of a strong magnetic field.

Construction of a cyclotron:

The construction of a simple cyclotron is shown in the following figure.

It consists of two semi-cylindrical boxes D1 and D2, called ‘dees’, enclosed in an evacuated chamber. The

chamber is kept between the poles of a powerful magnet so that a uniform magnetic field acts

perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the dees by

using a high-frequency oscillator. The electric field is zero inside the dees.



Working and theory of a cyclotron:

At a certain instant, let D1 be positive and hence D2 is negative. A proton from an ion source initially

present at the centre of the dees is accelerated towards D2. It describes a semi-circular path with a

constant speed and is acted upon only by the magnetic field when it is inside either of the Ds.

The function of electric field is to accelerate the charged particle.

The function of magnetic field is to turn the accelerated particle in a circular motion such that it interacts

with the electric field between the gaps again and again.

The radius of the circular path is given by the relation:

Period of revolution of proton,

Frequency of revolution of proton,

It is called cyclotron frequency.

It can be inferred that cyclotron frequency is independent of both velocity v and radius r of the proton. It

implies that even if the proton is accelerated in velocity and it increases its radius, the frequency with which

it completes its path does not change. Hence, we are able to use a constant frequency oscillator to reverse

the electric field at the time when the proton tries to cross the gap. If the frequency of applied ac is equal

to cyclotron frequency, then every time the proton reaches the gap between the dees, the direction of

electric field is reversed and the proton receives a push, finally gaining very high kinetic energy. The proton

follows a spiral path between the dees and ejects from an appropriate opening.


(a) For a ray of light travelling from a denser medium of refractive index n1 to a rarer medium of refractive

index n2, prove that where ic is the critical angle of incidence for the media.

(b) Explain with the help of a diagram, how the above principle is used for transmission of video signals

using optical fibres.

OR

(a) What is plane polarized light? Two Polaroid are placed at 90° to each other and the transmitted

intensity is zero. What happens when one more Polaroid is placed between these two, bisecting the angle

between them? How will the intensity of transmitted light vary on further rotating the third Polaroid?

(b) If a light beam shows no intensity variation when transmitted through a Polaroid which is rotated, does

it mean that the light is un-polarized? Explain briefly.

Solution:

(a) Relation between refractive index and critical angle:

Let O be a point object in the denser medium of refractive index (n1). A ray incident along OA1 deviates

from the normal and is refracted along A1B1 in the rarer medium of refractive index (n2), as shown in the

following figure.



For a particular value of i = ic (the critical angle), the incident ray OA2 is refracted at angle r = 90° and

goes along A2B2.

Applying Snell’s law at A2, we can write:

This expression shows the relation between the refractive indices and the critical angle.

(b) Optical Fibre:

The working of optical fibres involves the phenomenon of total internal reflection. Optical fibres consist of

many long, high quality composite glass or quartz fibres. Each fibre consists of a core and cladding. The

refractive index of the material of the core is higher than that of the cladding, as shown in the following

figure.

Propagation of light through an optical fibre:

When light is incident at one end of the fibre at a small angle, it suffers several total internal reflectionsat the

glass boundary because the angle of incidence is greater than the critical angle of the core-cladding

combination. The beauty of an optical fibre is that the intensity of the emergent beam is almost equal to that

of the incident beam (almost negligible loss of signal).

OR

(a) Plane polarised light:

When an unpolarised light is passed through a tourmaline crystal cut with its face parallel to its

crystallographic axis AB, only those vibrations of light pass through the crystal which are parallel to AB.

All other vibrations are absorbed. The light emerging from the crystal is said to be plane polarised light.

Let E be the amplitude of electric field component emanating from the 1st Polaroid. The emergent light will

be at 45° from the 2nd Polaroid.

The amplitude of the electric field component is given as:

Again, the amplitude of the electric field component coming from the 3rd Polaroid at 45° to the 2nd

Polaroid would be:



Intensity ∝ (Electric Field)2

Therefore, the intensity transmitted from the three Polaroids will be of the intensity transmitted from

the first Polaroid.

(b) It cannot be said with certainty whether the light beam is unpolarised or polarised. In case of

unpolarised light, we will observe no intensity variation because all directions are equivalent. Similarly, in

case of circularly polarised light, the electric field vector rotates with some angular velocity. Therefore,

even in this case, all directions are equivalent and hence no intensity variation will be observed. In case of

linearly polarised light, there will be an intensity variation because the polaroid tries to block the

components which are perpendicular to its pass axis.


(a) Using Gauss’ law, derive an expression for the electric field intensity at any point outside a uniformly

charged thin spherical shell of radius R and charge density σ C/m2. Draw the field lines when the charge

density of the sphere is (i) positive, (ii) negative.

(b) A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100

µC/m2. Calculate the

(i) Charge on the sphere

(ii) Total electric flux passing through the sphere

OR

(a) Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field.

(b) Calculate the work done to dissociate the system of three charges placed on the vertices of a triangle

as shown.

Here q = 1.6 × 10−10C

Solution:

(a)Electric field intensity at any point outside a uniformly charged spherical shell:

Consider a thin spherical shell of radius R and with centre O. Let charge + q be uniformly distributed over

the surface of the shell.

Let P be any point on the Gaussian sphere S1 with centre O and radius r, as shown in the following figure.

According to Gauss’s law, we can write the flux through ds as:



At any point on the surface of the shell, r = R

For charge density σ, q = 4πR2σ

When the charged density of the sphere is positive and negative, the respective electric field lines are

shown as follows:

(b)Diameter of the sphere = 2.5 m

Radius of the sphere,

Charge density,

(i)

(ii) Total electric flux,

OR

(a) Dipole in Uniform External Field:

Let an electric dipole of length 2a be placed in a uniform electric field at an angle θ, as shown in the

following figure.

Dipole moment of the dipole is:



Force on charge (+ q) = + q E along the direction of

Force on charge (− q) = − q along the opposite direction of

Since is uniform, the net force on the dipole is zero.

Being equal, unlike, and parallel, these forces form a couple, rotating the dipole in clockwise direction.

∴Magnitude of torque = Force × Arm length of couple

The direction of is given by the right hand screw rule and is normal to vectors and .

(b) Let q1 = q, q2 = − 4q, and q3 + 2q

And, r = 10 cm = 0.10 m

Total work done

The negative sign indicates that the system is bounded and of work will be required to

dissociate the system.



physics 2008 set 1.1 sol

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