Physics 2009 Set 1

8/7/2019 Physics 2009 Set 1

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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)

What is sky wave propagation?

Solution:

They type of propagation in which radio waves are transmitted towards the sky and are reflected by the ionosphere towards the desired location on earth is called sky wave propagation.


Write the following radiations in ascending order with respect to their frequencies:

X-rays, microwaves, UV rays and radio waves.

Solution:

The given radiations can be arranged in ascending order with respect to their frequencies as:

Radio waves < Microwaves < UV rays < X-rays


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Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight

solenoid. Why?

Solution:

Magnetic field lines form closed loops around a current-carrying wire. The geometry of astraight solenoid is such that magnetic field lines cannot loop around circular wires without

spilling over to the outside of the solenoid. The geometry of a toroid is such that magnetic field

lines can loop around electric wires without spilling over to the outside. Hence, magnetic fieldlines can be entirely confined within the core of a toroid, but not within a straight solenoid.


You are given following three lenses. Which two lenses will you use as an eyepiece and as an

objective to construct an astronomical telescope?

Lens Power (P) Aperture (A)

L1 3D 8 cm

L2 6D 1 cm

L3 10D 1 cm

Solution:

For constructing an astronomical telescope, the objective should have the maximum diameter. Of

the three lenses given, L1 has the maximum diameter.

The eyepiece should have the highest power for better magnification. Therefore, we use lens L3.


If the angle between the pass axis of polarizer and the analyser is 45º, write the ratio of the

intensities of original light and the transmitted light after passing through the analyser.

Solution:

I = Im cos

2

θ

Where,

I is the transmitted intensity

I mis the maximum value of the transmitted intensity

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θ is the angle between the two polarising directions


The figure shows a plot of three curves a, b, c, showing the variation of photocurrent vs collector

plate potential for three different intensities I1, I2 and I3 having frequencies v1, v2 and v3

respectively incident of a photosensitive surface.

Point out the two curves for which the incident radiations have same frequency but different

intensities.

Solution:

Curves a and b have the same frequency but different intensities.


What type of wavefront will emerge from a (i) point source, and (ii) distance light source?

Solution:

(i) For point source, wavefront will be spherical.

(ii) For a distannt light source, the wavefronts will be plane wavefronts.


Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?

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Solution:

Nuclear density is independent of mass number. Hence, both the atoms have the same nuclear

density.


A cell of emf ‘ E ’ and internal resistance ‘r ’ is connected across a variable resistor ‘ R’. Plot agraph showing the variation of terminal potential ‘V ’ with resistance R. Predict from the graph

the condition under which ‘V ’ becomes equal to ‘ E ’.

Solution:

V becomes equal to E when no current flows through the circuit.

The condition under which V will be equal to E is when R = ∞


(i) Can two equi-potential surfaces intersect each other? Give reasons.

(ii) Two charges −q and +q are located at points A (0, 0, −a) and B (0, 0, +a) respectively. Howmuch work is done in moving a test charge from point P (7, 0, 0) to Q (−3, 0, 0)?

Solution:

(i) Two equipotential surfaces cannot intersect each other because when they will intersect, the

electric field will have two directions, which is impossible.

(ii) Charge P moves on the perpendicular bisector of the line joining +q and −q. Hence, this perpendicular bisector is equidistant from both the charges. Thus, the potential will be same

everywhere on this line. Therefore, work done will be zero.


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By what percentage will the transmission ranges of TV tower be affected when the height of the

tower is increased by 21%?

Solution:

Let the transmission of the tower before transmission be =

Range after increase in height =

Ratio =

Height increase = 21%

Thus,


Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation

time.

Solution:

If there are N electrons and the velocity of the ith electron at a given time is vi where, i = (1, 2, 3,…N), then

(If there is no external field)

When an external electric field is present, the electrons will be accelerated due to this field by

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Where,

− e = Negative charge of the electron

E = External field

m = Mass of an electron

Let vi be the velocity immediately after the last collision after which external field was

experienced by the electron.

If vi is the velocity at any time t , then from the equation V = u + at , we obtain

For all the electrons in the conductor, average value of vi is zero.

The average of vi is vd or drift velocity.

This is the average velocity experienced by an electron in an external electric field.

There is no fixed time after which each collision occurs. Therefore, we take the average time

after which one collision takes place by an electron.

Let this time, also known as relaxation time, beτ. Substituting this in equation (i)

Then,

Negative sign shows that electrons drift opposite to the applied field.


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How does a charge q oscillating at certain frequency produce electromagnetic waves?

Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave

propagating along the Z-direction.

Solution:

An oscillating charge is an example of accelerated charge. We know from Maxwell’s theory thataccelerated charge radiates electromagnetic waves. These electromagnetic waves are produced

because oscillating charge produces oscillating magnetic field, which in turn produces an

oscillating electric field. This process goes on, giving rise to an electromagnetic wave.


A charge ‘q’ moving along the X- axis with a velocity is subjected to a uniform magnetic fieldB along the Z-axis as it crosses the origin O.

(i) Trace its trajectory.

(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

Solution:

(i)

The direction of magnetic field is along the negative X-direction. Hence, the magnetic force willact in such a way that this particle describes a circular motion as shown below.

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(ii) No, the charge does not gain kinetic energy because the force and velocity are perpendicular to each other.

Thus, force does not bring out any change in the velocity.


The following figure shows the input waveforms (A, B) and the output waveform (Y) of a gate.

Identify the gate, write its truth table and draw its logic symbol.

Solution:

The gate is the NAND gate.


State Biot-Savart law.

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A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the

direction of the magnetic field due to a small element d at point P situated at distance from the

element as shown in the figure.

Solution:

Biot-Savart’s law states that the magnitude of the magnetic field dB is proportional to the current

I , the element , and inversely proportional to the square of the distance r .

The direction of magnetic field is along the negative X-direction.


Why are high frequency carrier waves used for transmission?

OR

What is meant by term ‘modulation’? Draw a block diagram of a simple modulator for obtainingan AM signal.

Solution:

For transmitting a signal, the antenna should have a size comparable to the wavelength of the

signal (at least in dimension), where λ is the wavelength.

If the frequency of the signal is small, then its wavelength becomes very large and it is

impractical to make that large antennas for the corresponding large wavelengths. For higher

frequencies, wavelength is smaller, which is the reason why high frequency carrier waves areused for transmission.

OR

The process of superimposing information contained in a low frequency signal on a high

frequency signal is called modulation.

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A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme:

The mass number and atomic number of A are 180 and 72 respectively. What are these numbers

for A4?

Solution:

A has mass number as 180 and atomic number 172.

Formation of A1 by α-decay:

Formation of A2 by β decay:

Formation of A4:

In r -decay, mass number and atomic number remain the same.

Thus,

Mass number of A4 = 172

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Atomic number of A4 = 169


A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface.

Using Gauss’s law, derive an expression for an electric field at a point outside the shell.

Draw a graph of electric field E (r ) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.

Solution:

According to Gauss law,

Where,

q is the point charge

E is electric field due to the point charge

dA is a small area on the Gaussian surface at any distance and

is the proportionality constant

For a spherical shell at distancer

from the point charge, the integral is merely the sum of all differential of dA on the sphere.

Therefore,

Therefore, for a thin conducting spherical shell of radius R and charge Q, spread uniformly over

its surface, the electric field at any point outside the shell is

Where r is the distance of the point from the centre of the shell.

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The graph of electric field E (r ) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.


Three identical capacitors C1, C2 and C3 of capacitance 6 μF each are connected to a 12 V batteryas shown.

Find

(i) charge on each capacitor

(ii) equivalent capacitance of the network

(iii) energy stored in the network of capacitors

Solution:

The 12 V battery is in parallel with C 1 , C 2, and C 3. C 1 and C 2are in series with each other whileC 3 is in parallel with the combination formed by C 1 and C 2.

Total voltage drop across C 3 = 12 V

q3 = CV

Where, q = Charge on the capacitor

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C 1 , C 2 , C 3 = 6 μF (Given in the question)

q3 = 6 × 12 = 72 μC

Voltage drop across C 1 and C 2 combined will be 12 V.

Let the voltage drop at C 1 = V 1

Let the voltage drop at C 2 = V 2

Then,

V = V 1 + V 2

As both the capacitors are in series,

Then,

Or,

q = 36 micro coulombs

Thus, charge on each of C 1 and C 2 is 36 coulombs.


(a) The energy levels of an atom are as shown below. Which of them will result in the transition

of a photon of wavelength 275 nm?

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(b) Which transition corresponds to emission of radiation of maximum wavelength?

Solution:

Energy transitions for A,B,C, and D are:

A = 2 eV

B = 4.5 eV

C = 2.5 eV

D = 8 eV

Where,

E = Energy transition

λ = Wavelength

h = 6.63 × 10−34 Js

C = 3 × 108 m/s

For B, we have

Thus, B will result in transition of a photon of wavelength of 275 nm.

(b)

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For maximum wavelength, energy transition should be minimum.

A undergoes minimum energy transition.

A = 2 eV

Thus, photon in A will have the maximum wavelength.


A proton and an alpha particle are accelerated through the same potential. Which one of the twohas (i) greater value of de Broglie wavelength associated with it, and (ii) less kinetic energy?Justify your answers.

Solution:

Kinetic energy gained by the proton = eV

Kinetic energy gained by the alpha particle = 2eV

Let v1 be the velocity of the proton and v2 be the velocity of the alpha particle.

Mass of proton = m

Mass of alpha particle = 4m

Then their kinetic energies are .

Dividing (i) and (ii), we obtain

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It shows that velocity of proton is greater than that of the alpha particle

(i) de Broglie wavelength =

de Broglie wavelength of proton

Hence, the de Broglie wavelength of proton will be greater than that of alpha particle.

(ii) For kinetic energy:

K.E. of proton =

K.E. of alpha particle =

By substitution, we obtain

K.E of proton =

As , therefore, K.E of alpha particle will be larger.

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Where,

N is total number of turns

l is the length of the solenoid

Inductance,

Substituting, we obtain

Substituting the value of B, we obtain

Inductance L of a solenoid is:


The figure shows experimental set up of a meter bridge. When the two unknown resistances Xand Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω

is connected in series with X, the null point shifts by 10 cm. Find the position of the null pointwhen the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values

of the resistances X and Y.

Solution:

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For a metre bridge:

… (1)

Where, it is given that l 1 = 40 cm

… (2)

When 10 Ω resistance is added in series to X, null point shifts by 10 cm.

Substituting the value of X from equation (2), we obtain

Substituting the value of Y in equation (3), we obtain

X + 10 = 30

X = 20 Ω

Position of the null point when 10 Ω resistance is put in series with Y ,

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Derive the expression for force per unit length between two long straight parallel current

carrying conductors. Hence define one ampere.

OR

Explain the principle and working of a cyclotron with the help of a schematic diagram. Write the

expression for cyclotron frequency.

Solution:

Two long parallel conductors a and b separated by a distance l and carrying currents I a and I brespectively are shown below.

By Ampere’s circuital law, we have

Conductor b will experience a sideways force because of conductor a. Let this force be F ba.

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By symmetry,

F ba = − F ab

1 ampere is the value of that steady current which when maintained in each of the two very long,

straight, parallel conductors of negligible cross section and placed one metre apart in vacuum,would produce on each of these conductors a force equal to 2 × 10−7 Newton per metre of length.

OR

Cyclotron is a machine used to accelerate charged particles or ions to high energies. It uses both

electrical and magnetic fields in combination to increase the speed of the charged particles.

The particles move in two semi-circular containers D1 and D2, called Dees. Inside the metal box,the charged particle is shielded from external electric fields.

When the particle moves from one dee to another, electric field is acted on the particle.

The sign of the electric field is changed alternately, in tune with the circular motion of the

particle. Hence, the particle is always accelerated by the electric field. As the energy of the particle increases, the radius of the circular path increases.

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Time taken for a particle for one complete revolution = T

T =

Where is the cyclotron frequency

Then,

The above expression is the expression for cyclotron frequency.

The oscillator applies an ac voltage across the Ds and this voltage must have a frequency equalto that of cyclotron frequency.


Three light rays red (R), green (G) and blue (B) are incident on a right angled prism ‘abc’ at face

‘ab’. The refractive indices of the material of the prism for red, green and blue wavelengths are1.39, 1.44 and 1.47 respectively. Out of the three which colour ray will emerge out of face ‘ac’?

Justify your answer. Trace the path of these rays after passing through face ‘ab’.

Solution:

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The blue ray will emerge out of the face ‘ac’.

The three light rays will go through ‘ab’ as the three are perpendicular to ‘ab’. However, whenthey will hit ‘ac’, they will refract with an angle of incidence of 45°.

Refractive indices for three light rays are given.

Red ( R) = 1.39

Green (G) = 1.44

Blue ( B) = 1.47

Total internal reflection takes place if the angle of incidence is such that

The refractive indices are from air to prism. To convert them from prism to air, we take their

reciprocal.

For red:

sin r = 0.9828, which is less than 1

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Thus, red light will pass through the face ‘ac’.

Similarly, for green:

This is greater than 1. Therefore, it will not pass through, but reflect back in the same medium.

For blue:

Even this will reflect back because of total internal reflection.


(a) Derive an expression for the average power consumed in a series LCR circuit connected to

a.c. source in which the phase difference between the voltage and the current in the circuit is Φ.

(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in

receiving circuits? Name the factors on which it depends.

OR

(a) Derive the relationship between the peak and the rms value of current in an a.c. circuit.

(b) Describe briefly, with the help of labelled diagram, working of a step-up transformer.

A step-up transformer converts a low voltage into high voltage. Does it not violate the principle

of conservation of energy? Explain.

Solution:

(a) Power in ac circuit

Voltage v in an ac circuit is:

which drives through the circuit a current i

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are constants. Therefore,

By trigonometric identity,

The average value of cos 2 ωt is zero.

We have:

Thus,

The rms value in the ac power is expressed in the same form as dc power root mean square or

effective current and is denoted by Irms.

Peak current is

Therefore,

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(b)

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In a transformer with N s secondary turns and N p primary turns, induced emf or voltage E s is:

Back emf = E p =

E P = V P

E s = V s

Thus, Vs = … (i)

Dividing equations (i) and (ii), we obtain

If the transformer is 100% efficient, then

Thus, combining the above equations,

If N s > N p, then the transformer is said to be step-up transformer because the voltage is steppedup in the secondary coil.

No, the transformer does not violate the principal of conservation of energies. This can be easily

observed by the following equation:

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Power consumed in both the coils is the same as even if the voltage increases or current

increases, their product at any instant remains the same.


(i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor inits common emitter configuration. Draw the typical input and output characteristics.

(ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common

emitter amplifier.

OR

How is a zener diode fabricated so as to make it a special purpose diode? Draw I-Vcharacteristics of zener diode and explain the significance of breakdown voltage.

Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half waverectifier.

Solution:

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OR

Zener is fabricated such that both the p-type and the n-type are highly doped. This makes the

depletion region thin. When an electric field is applied, a high electric field appears across the

thin depletion region. When the electric field becomes very high, it knocks off electrons from thehost atoms to create a large number of electrons. This results in a large value of current inside the

circuit.

Zener has a sharp breakdown voltage and this property of zener is used for voltage regulation.

An ac current has a positive half cycle and a negative half cycle. A pn junction allows current to

pass only in one direction and that is when it is forward biased.

When a positive half-cycle occurs, the p-side has a lower potential. Therefore, the diode is nowforward biased and therefore conducts and this positive cycle is available for the load.

When a negative half cycle occurs, the n-side has a higher potential than the p-side. Hence, the

diode is now reverse biased and thus, does not conduct. As a result, this positive half cycle also

does not conduct. Therefore, it does not appear at the load and is cut-off.

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We obtain a waveform, which has only positive half cycles and therefore it is called half-wave

rectifier.


Trace the rays of light showing the formation of an image due to a point object placed on the axisof a spherical surface separating the two media of refractive indices n1 and n2. Establish the

relation between the distances of the object, the image and the radius of curvature from the

central point of the spherical surface.

Hence derive the expression of the lens maker’s formula.

OR

Draw the labelled ray diagram for the formation of image by a compound microscope.

Derive the expression for the total magnification of a compound microscope. Explain why boththe objective and the eyepiece of a compound microscope must have short focal lengths.

Solution:

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In the given figure, image is I and object is denoted as O.

The centre of curvature is C.

The rays are incident from a medium of refractive index to another of refractive index .

We consider NM to be perpendicular to the principal axis.

For ΔNOC, i is the exterior angle.

Therefore, i =∠ NOM +∠ NCM

Similarly,

r =∠ NCM −∠ NIM

i.e.,

According to Snell’s law,

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For small angles,

Substituting i and r , we obtain

Where, OM, MI, and MC are the distances

OM = −u

MC = + R

MI = v

Substituting these, we obtain

Applying equation (i) to lens ABCD, we obtain for surface ABC,

For surface ADC, we obtain

For a thin lens,

BI1 = DI1

Adding (ii) and (iii), we obtain

Suppose object is at infinity and DI = f , then

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8/7/2019 Physics 2009 Set 1


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Physics 2009 Set 1