Physics 2005 Set 1 Sol

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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)

An electric dipole of dipole moment 20 × 10−6 Cm is enclosed by a closed surface. What is the net fluxcoming out of the surface?

Solution:

Net flux coming out of the closed surface is zero because the net charge on the electric dipole is zero.


An electron beam, projected along + X-axis, experiences a force due to a magnetic field along the + Y-axis. What is the direction of the magnetic field?

Solution:

The magnetic field is along the + Z-axis.


The power factor of an AC circuit is 0.5. What will be the phase difference between voltage and current inthis circuit?

Solution:

Power factor,

Phase difference is 60°.


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

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Electrons are emitted from a photosensitive surface when it is illuminated by

(i) red light (ii) blue light.

Solution:

(i) Electrons are not emitted with red light.

(ii) Electrons are emitted with blue light.


What should be the length of the dipole antenna for a carrier wave of frequency 3 × 108 Hz?

Solution:

Length of the dipole antenna =


Define ‘electric line of force’ and give its two important properties.

Solution:

An electric field line is a path, straight or curved, such that tangent to it at any point gives the direction ofelectric field intensity at that point.

Properties of field lines:

(i) Tangent to the electric field line at any point gives the direction of electric intensity at that point.

(ii) No two electric lines of force can intersect each other.


State Lenz’s law.

Solution:

Lenz’s law:

It states that the direction of induced current or emf in a circuit is always such that it opposes the causewhich produces it.


Why is convex mirror used as driver’s mirror?

Solution:

Convex mirror is used as driver’s mirror because field of view of a convex mirror is large.


(a) Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

(b) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be thecapacitance if the distance between the plates be reduced by half and the space between them is filled witha substance of dielectric constant K = 6?

OR



OR

Three point charges of + 2 µC, − 3 µC and − 3µC are kept at the vertices A, B and C respectively of anequilateral triangle of side 20 cm as shown in the figure. What should be the sign and magnitude of thecharge to be placed at the mid-point (M) of side BC so that the charge at A remains in equilibrium?

Solution:

(a) When a dielectric is placed in an external field, the field causes polarisation of the dielectric inducing afield in the opposite direction. Thus, the net electric field inside the dielectric decreases.

(b)

Capacitance with dielectric:

OR

Force exerted on + 2µC charge by the charge at B,

along AB

Force exerted on + 2µC charge by charge at C,

along AC

The resultant force of F1 and F2 is:

For the equilibrium of charge at A, the charge q placed at point M must be a positive charge so that itexerts force along MA.

Force between charges at A and M




Draw V − I graph for ohmic and non-ohmic materials. Give one example for each.

Solution:

(i) V − I graph for ohmic material

(ii) V − I graph for a non-ohmic material


Define the terms ‘Magnetic Dip’ and ‘Magnetic Declination’ with the help of relevant diagrams.

Solution:

(i) Angle of dip (S):

The angle made by the earth’s total magnetic field with the horizontal component of earth’s magnetic

field (BH) is called angle of dip at any place.

(ii) Magnetic declination (α):

The angle between the geographic meridian and magnetic meridian at a place is called magnetic declinationat that place.


In the figure given below, a bar magnet moving towards the right or left induces an e.m.f in the coils (1)and (2). Find, giving reason, the directions of the induced currents through the resistors AB and CD whenthe magnet is moving (a) towards the right, and (b) towards the left.

Solution:



(a) When the magnet is moved towards right, the right end of coil (1) develops S-polarity and the left endof coil (2) also develops S-polarity. In coil (1), current flows from A to B while in coil (2), current flowsfrom D to C.

(b) When the magnet is moved towards left, the right end of coil (1) develops N-polarity and left end ofcoil (2) also develops N-polarity.

In coil (1), current flows from B to A while in coil (2), current flows from C to D.


(i) Draw the graphs showing variation of inductive reactance and capacitive reactance with frequency ofapplied a.c. source.

(ii) Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than theapplied voltage of the a.c. source? Justify your answer.

Solution:

(i) Variation of inductive reactance with frequency is shown in figure given below.

Variation of capacitive reactance with frequency is shown in figure given below.

(ii) Yes, the voltage drop across the inductor or the capacitor in a series circuit can be greater than theapplied voltage. In this case, these two voltages are not in same phase. Hence, they cannot be added likeordinary numbers.


The image of a candle is formed by a convex lens on a screen. The lower half of the lens is painted blackto make it completely opaque. Draw the ray diagram to show the image formation. How will this image bedifferent from the one obtained when the lens is not painted black?

Solution:

When the lower half of the lens is painted black, the image formed will still be of full size but the intensity ofthe image will be lesser.




An electric dipole is held in a uniform electric field. (i) Using suitable diagram, show that it does notundergo any translatory motion, and (ii) derive an expression for the torque acting on it and specify itsdirection.

Solution:

(i) Consider an electric dipole consisting of two equal and opposite point charges, q at A and + q at B,separated by a small distance.

AB = 2a, having dipole moment

Let this dipole be held in a uniform external electric field at an angle θ with the direction of .

Force on charge −q at A

= + q , in a direction opposite to

Force on charge +q at B

= q , along the direction of

Net translator forces on the dipole = qE − qE = 0

Hence, the dipole does not undergo any translatory motion.

(ii) The equal and unlike forces acting at different points form a couple, which rotates the dipole in theanticlockwise direction. Thus, the couple tends to align the dipole axis along the direction of field .

Torque = Moment of the couple

τ = Force × Arm of couple

τ = F × BC

τ = F × 2asinθ = qE 2asinθ

Or, τ = pEsinθ

Where, P = q (2a) = Dipole moment


A galvanometer with a coil of resistance 120 ohm shows full scale deflection for a current of 2.5 mA. Howwill you convert the galvanometer into an ammeter of range 0 to 7.5 A. Determine the net resistance of theammeter.

When an ammeter is put in a circuit, does it read slightly less or more than the actual current in the originalcircuit? Justify your answer.

Solution:



Here,

Net resistance of the ammeter:


Give reasons for the following:

(i) Long distance radio broadcasts use short-wave bands.

(ii) The small ozone layer on the top of the stratosphere is crucial for human survival.

(iii) Satellites are used for long distance TV transmission.

Solution:

(i) Long distance radio broadcasts use shortwave bands because shortwave bands are easily reflectedback to the earth by the ionosphere.

(ii) The small ozone layer on the top of the stratosphere is crucial for human survival because ozone layerabsorbs the ultraviolet rays from the sun and prevents it from reaching the earth and causing damage tolife.

(iii) Television signals cannot be properly reflected back by the ionosphere. Therefore, their reflection isaffected by using satellites.


A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in thecapacitor?

Solution:

Here, C = 12 pF = 12 × 10−12 F

V = 50 Volt


Define the term ‘resistivity’ and write its S.I. unit. Derive the expression for the resistivity of a conductor in



Define the term ‘resistivity’ and write its S.I. unit. Derive the expression for the resistivity of a conductor interms of number density of free electrons and relaxation time.

OR

State the principle of potentiometer. Draw a circuit diagram used to compare the e.m.f. of two primarycells. Write the formula used. How can the sensitivity of a potentiometer be increased?

Solution:

Resistivity: Resistivity of the material of a conductor is defined as the resistance offered by unit length andunit area of cross section by a wire of the given material of conductor.

Expression for resistivity: Current through a conductor is given by the expression:

I = nAeVd

Drift velocity,

By Ohm’s law,

Comparing with the equation, , we obtain

This is the required expression for resistivity.

OR

Principle: The fall of potential across any portion of the wire is directly proportional to the length of thatportion provided the wire is of uniform area of cross-section and a constant current is flowing through it.

V ∝ l

Comparison of emfs of two primary cells:



The circuit diagram is shown in the figure.

When the key K is closed, a constant current flows through the potentiometer wire. On closing the keyK1, the cell E1is included in the circuit. The jockey is adjusted till galvanometer shows no deflection.

Suppose for cell E1, the balancing length is l1. Then:

E1 = kl1 … (i)

Where k is the potential gradient

Now the null point is obtained for cell E2 by closing key K2. Let l2 be the balancing length in this case.

Then:

E2 = kl2 … (2)

Dividing equation (i) by equation (ii), we obtain

The sensitivity of a potentiometer can be increased by increasing the length of its wire.


Explain, with the help of diagram, the principle and working of an a.c. generator. Write the expression forthe e.m.f. generated in the coil in terms of its speed of rotation.

Solution:

A.C. Generator

Principle: Whenever a closed coil is rotated in a uniform magnetic field about an axis perpendicular to thefield, the magnetic flux linked with coil changes and an induced emf is set up across its ends.

The essential parts of an a.c. generator are shown in the figure. Initially, the armature coil ABCD ishorizontal. As the coil is rotated clockwise, the arm AB moves up and CD moves down.

By Fleming’s right hand rule, the induced current flows along ABCD. In second half rotation, the arm CDmoves up and AB moves down. The induced current flows in the opposite direction i.e., along DCBA.Thus, an alternating current flows in the circuit.

The magnetic flux linked with the coil at any instant is

Φ = �BA cosωt



Φ = �BA cosωt

Induced emf will be

Where E0 = �BAω = Peak value of induced emf


A figure divided into squares, each of size 1 mm2, is being viewed at a distance of 9 cm through amagnifying lens of focal length 10 cm, held close to eye.

(a) Draw a ray diagram showing the formation of the image.

(b) What is the magnification produced by the lens? How much is the area of each square in the virtualimage?

(c) What is the angular magnification of the lens?

Solution:

(a) Ray diagram showing the formation of the image:

(b) Area of each square (i.e., object) = 1 mm2

u = −9 cm, f = +10 cm

For a thin lens,

or v = −90 cm

Magnification,

Area of each square in the virtual image

= (10)2 × 1 mm2 = 100 mm2


Ultraviolet light of wavelength 2271 Å from a 100 W mercury source radiates a photo cell made ofmolybdenum metal. If the stopping potential is 1.3 V, estimate the work function of the metal. How would

the photo cell respond to high intensity (105 Wm−2) red light of wavelength 6328 Å produced by a He−Ne laser?

Plot a graph showing the variation of photoelectric current with anode potential for two light beams ofsame wavelength but different intensity.

Solution:



Solution:

λ = 2271 Å = 2271 × 10−10 m

V0 = 1.3 V

W0 = ?

Work function,

Threshold frequency,

� 1.007 × 10+15 Hz

Frequency of red light

= 0.474 × 1015

= 4.74 × 1014 HZ

As the frequency of red light is less than that of threshold frequency, there will be no photoelectricemission with He-Ne laser.


(a) Draw a graph showing the variation of potential energy of a pair of nucleons as a function of theirseparation. Indicate the regions in which nuclear force is (i) attractive, and (ii) repulsive.

(b) Write two characteristic features of nuclear force which distinguish it from the Coulomb force.

Solution:

(a)



(i) The potential energy is minimum at a distance of r0 = 0.8 fm. At this distance, force between nucleons

is zero. For distances larger than 0.8 fm, negative P.E. goes on decreasing. The nuclear forces areattractive.

(ii) For distances less than 0.8 fm, negative P.E. decreases to zero and then becomes positive. The nuclearforces are repulsive.

(b) Properties of nuclear forces:

(i) Nuclear force is always attractive in nature.

(ii) Nuclear force is a short range force.


(a) Show that the decay rate ‘R’ of a sample of a radionuclide is related to the number of radioactivenuclei ‘�’ at the same instant by the expression R = λ�.

(b) The half-life of against α-decay is 1.5 × 1017 S. What is the activity of a sample of having

25 × 1020 atoms?

Solution:

(a) Rate of disintegration of a radioactive sample,

According to radioactive decay law,

(b) Activity of a sample

= 11550 disintegrations/second


Explain with the help of a circuit diagram, how the thickness of depletion layer in a p-n junction diodechanges when it is forward biased. In the following circuits which one of the two diodes is forward biasedand which is reverse biased?

(i)



(ii)

Solution:

(i) p-n junction is reverse biased.

(ii) p-n junction is forward biased.

In forward biasing, the applied voltage V of battery B mostly drops across the depletion region and thevoltage drop across the p-side and n-side of the p-n junction is negligibly small. It is due to the fact that theresistance of depletion region is very high as it has no free charge carriers. In forward biasing, the forwardvoltage opposes the potential barrier VB. As a result of it, the potential barrier height is reduced and width

of the depletion layer decreases.


Distinguish between analog and digital communication. Write any two modulation techniques employed forthe digital data. Describe briefly one of the techniques used.

Solution:

An analog communication system makes use of analog signals, which vary continuously with time. A digitalcommunication system makes use of a digital signal, which has only two values of voltage, either high orlow.

The modulation techniques for digital data are:

(a) Amplitude shift keying (ASK)

(b) Frequency shift keying (FSK)

(c) Phase shift keying (PSK)

In amplitude shift keying modulation, two different amplitudes of the carrier frequency represent twobinary values. Binary digit 0 represents the absence of carrier wave while the binary digit 1 represents thepresence of the carrier wave of constant amplitude.





Draw a schematic diagram of a single optical fibre structure. Explain briefly how an optical fibre isfabricated. Describe in brief, the mechanism of propagation of light signal through an optical fibre.

Solution:

Optical fibres consist of several thousands of very long fine quality fibres of glass or quartz. The diameter

of each fibre is of the order of 10−4 cm with refractive index of material being of the order of 1.5. Thefibres are coated with a thin layer of material of lower refractive index of the order of 1.48.

Propagation of light through an optical fibre:

Light incident on one end of the fibre at a small angle passes inside and undergoes repeated total internalreflections inside the fibre. It finally comes out of the other end, even if the fibre is bent or twisted in anyform.


(a) With the help of a labelled diagram, explain the principle and working of a moving coil galvanometer.

(b) Two parallel coaxial circular coils of equal radius ‘R’ and equal number of turns ‘�’, carry equalcurrents ‘I’ in the same direction and are separated by a distance ‘2R’. Find the magnitude and directionof the net magnetic field produced at the mid-point of the line joining their centres.

OR

(a) State Biot-savart’s law. Using this law, derive the expression for the magnetic field due to a currentcarrying circular loop of radius ‘R’ at a point which is at a distance ‘x’ from its centre along the axis of theloop.

(b) Two small identical circular loops, marked (1) and (2), carrying equal currents, are placed with thegeometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction ofthe net magnetic field produced at the point O.

Solution:

(a) Principle:

Its working is based on the fact that when a current carrying coil is placed in a magnetic field, itexperiences a torque.

Suppose the coil PQRS is suspended freely in the magnetic field.

Let, l = Length PQ or RS of the coil



Let, l = Length PQ or RS of the coil

b = Breadth QR or S, P of the coil

n = Number of turns in the coil

Area of each turn of the coil, A = l × b

Let B = Strength of the magnetic field in which coil is suspended

I = Current passing through the coil in the direction PQRS

Let, at any instant, α be the angle, which the normal drawn on the plane of the coil makes with thedirection of magnetic field.

The rectangular coil carrying current when placed in the magnetic field experiences a torque whosemagnitude is given by:

τ = nIBA sin α

Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up inthe suspension wire.

Let θ be the twist produced in the phosphor bronze strip due to rotation of the coil and K be the restoringtorque per unit twist of the phosphor bronze strip. Then:

Total restoring torque produced = Kθ

In equilibrium position of the coil,

Deflecting torque = Restoring torque

∴ �IBA = kθ

Or,

Where, [Constant for a galvanometer]

It is known as galvanometer constant.

(b) Magnetic field at the mid-point due to loop 1

, acting towards right

Magnetic field at the mid-point due to loop 2


∴ Total field at the mid-point


OR

(a) According to Biot-Savarts’ law, the magnitude of the magnetic field induction dB at a point P due tocurrent element depends upon the factors as stated below.

(i) dB ∝ I



(ii) dB ∝ dl

(iii) dB ∝ sinθ

(iv)

Combining these factors, we obtain

(b) Consider a circular coil of radius a with centre O. Let the plane of the coil be perpendicular to theplane of the paper and current I be flowing in the coil in the direction shown in figure. Suppose P is anypoint on the axis of the circular coil at a distance x from its centre O. Clearly, OP = x

Consider two small elements of the coil, each of length dl, at C and D, which are situated at diametricallyopposite edges. Then:

Let ∠CPO = Φ = ∠DPO

According to Biot Savart’s law, the magnitude of magnetic field induction (i.e., magnetic flux density) at Pdue to current element at C is given by:

( a is small, therefore θ = 90°)

The direction of is in the plane perpendicular to and and is directed as given by right-handed

screw rule that is acting along PE perpendicular to CP.

Similarly, the magnitude of magnetic filed induction at P due to current element of length dl at D is givenby:

Its direction is along PF perpendicular to DP.

From (i) and (ii),

Resolving and into two rectangular components, we obtain

(i) dB cosΦ acts along PY and dB sinΦ acts along PX



(ii) acts along and acts along PX

Since the components of the magnetic field induction acting along PY and are equal and opposite,they cancel each other and the components of the magnetic field acting along PX (i.e., along the axis of thecoil) being in the same direction are added up.

Thus, total magnetic field induction at P due to current through the whole circular coil is given by:

If there are n turns in the coil, then

(b)

Magnetic field at O due to loop1

, acting towards left

Magnetic field at O due to loop 2

, acting vertically upwards

Here, R is the radius of each loop.

Resultant field at O will be



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


(a) How is a wavefront different from a ray? Draw the geometrical shape of the wavefronts when (i) lightdiverges from a point source, and (ii) light emerges out of the convex lens when a point source is placed atits focus.

(b) State Huygen’s principle. With the help of a suitable diagram, prove Snell’s law of refraction usingHuygen’s principle.

OR

(a) In Young’s double slit experiment, deduce the conditions for (i) constructive and (ii) destructiveinterference at a point on the screen. Draw a graph showing variation of the resultant intensity in theinterference pattern against position ‘x’ on the screen.

(b) Compare and contrast the pattern which is seen with two coherently illuminated narrow slits inYoung’s experiment with that seen for a coherently illuminated single slit producing diffraction.

Solution:

(a) The locus of all the particles of the medium, which at any instant are vibrating in the same phase, iscalled the wavefront. An arrow drawn normal to the wavefront and pointing in the direction of propagationof disturbance represents a ray of light.

(i) Wavefronts of the light emerging from a point source are spherical in shape as shown below:

(ii) When a point source is placed at the focus of a convex lens, the wavefront of the emerging light areplane wavefronts as shown below:

(b) Huygen’s principle:

(i) Every point on a given wavefront acts as a fresh source of secondary wavelets, which travel in alldirections with the speed of light.

(ii) The forward envelope of these secondary wavelets gives the new wavefront at any instant.

Snell’s law of refraction:



Let the surface AB represent a surface of separation of two media in which velocities of light are C1 and

C2 respectively.

Let the wavefront PQ be incident at an angle i to the surface AB. By the time Q reaches surface AB at S,the disturbance at P reaches R such that RS is the refracted wavefront. The time taken by a ray to travelfrom T to U is

As the new wavefront is the forward envelop of all secondary wavelets, this implies that the time taken isindependent of the position of O on the surface. Therefore, for t to be independent of O,

This proves Snell’s law of refractions.

OR

(a) Young’s double slit experiment: Consider two narrow rectangular slits S1 and S2 placed perpendicular

to the plane of paper. Slit S is placed on the perpendicular bisector of S1S2 and is illuminated with

monochromatic light.

The slits are separated by a small distance d. A screen is placed at a distance D from S1, S2.

Consider a point P on the screen at distance x from O.

The path difference between the waves reaching P from S1 and S2 is:

P = S2P − S1P

Draw S1N perpendicular to S2P. Then,

P = S2P − S1P = S2P − NP = S2N

From right-angled ∆S1S2N,



∴ P = S2N = S2S1 sinθ = d sinθ

From ∆COP,

When θ is small,

For constructive interference,

Position of nth bright fringe,

When n = 0, xn = 0, central bright fringe is formed at O.

For destructive interference,

Thus, alternate bright and dark fringes are formed on the screen.

(b)

Interference Diffraction

1. All bright fringes are of sameintensity.

1. Intensity of bright fringes decreases with the increase indistance from the central bright fringe.

2. Widths of interferences mayor may not be equal.

2. Widths of diffraction fringes are never equal.


(a) Distinguish between metals, insulators and semiconductors on the basis of their energy bands.



(b) Why are photodiodes used preferably in reverse bias condition? A photodiode is fabricated from asemiconductor with band gap of 2.8 eV. Can it detect wavelength of 6000 nm? Justify. 3

Or

(a) Explain briefly, with the help of circuit diagram, how V = I characteristics of

p - n junction diode are obtained in (i) forward bias, and (ii) reverse bias. Draw the shape of the curvesobtained.

(b) A semiconductor has equal electron and hole concentration of 6 × 108 /m3.

On doping with certain impurity, electron concentration increases to 9 × 1012 /m3.

(i) Identify the new semiconductor obtained after doping.

(ii) Calculate the new hole concentration.

Solution:

(a) On the basis of energy band diagrams:

(i) Metals:

The energy band diagrams for metals are such that either the conduction band is partially filled [figure (a)]or the conduction and valence band partly overlap each other [figure (b)].

(ii) Insulators:

Here, the valence band is completely filled, the conduction band is empty, and forbidden gap is quite large.[Figure (c)]

(iii) Semiconductors: At OK, the conduction band is empty and the valence band is filled. The material isan insulator at low temperatures. However, the energy gap between valence and conduction bands issmall. [Eg = 1.17 eV for Si and 0.74 eV for Ge] At room temperature, some valence electrons acquire

enough thermal energy and jump to the conduction band where they can conduct electricity. [Figure (d)]

(b) If a photodiode is used in forward bias, it will always allow current to pass through it. Therefore, it willnot act as a switch. But if this diode is connected in reverse bias, normally it would not allow current topass through. It will pass current only if light falls on the junction. The depletion region does not have anycharge carriers but when light falls on this depletion region, charge carriers (electron-hole pairs) arecreated temporarily, which are pulled away from the junction under the influence of electric field. Hence,current flows inside the circuit as long as light falls on the junction.

A photon of 6000 nm wavelength possesses energy E given by:



The energy of this photon is much less than the band gap energy (2.8 eV). The photon does not possesssufficient energy to cross the band gap of the photodiode. Hence, it will not detect this light.

Or

(a) The below given figure shows a forward biased p-n junction and its voltage-current graph.

The below given figure shows a reverse biased p-n junction and its voltage-current graph.

In both the cases, the applied voltage can be changed by using a rheostat. For different values of voltages,the value of current is noted. A graph is plotted between V and I as shown above.

(b) ni = 6 × 108/m3

ne = 9 × 1012/m3

As the new electron concentration is greater than the new hole concentration, therefore, the new semi-conductor is of n-type.



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Physics 2005 Set 1 Sol