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Grade XII DELHI SET 3

Physics (Theory)

Time allowed: 3 hours] [Maximum marks:70

General Instructions:

(i) All questions are compulsory.

(ii) There are 30 questions in total. Question Nos. 1 to 8 are very short answer type questions and

carry one mark each.

(iii) Question Nos. 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 in one question of

two marks; one question of three marks and all three questions of five marks each. You have to

attempt only one of the choice in such questions.

(iv) Use of calculators is not permitted.

(v) You may use the following values of physical constants wherever necessary:

c = 3 × 108 ms

–1

h = 6.626 × 10 –34

Js

e = 1.602 × 10–19

C

0 = 4 × 10–7

Tm A–1

9 2 2

0

19 10 Nm C

4

Mass of electron me = 9.1 × 10–31

kg

Q 19. (a) Why are coherent sources necessary to produce a sustained interference pattern?

(b) In Young’s double slit experiment using monochromatic light of wavelength , the intensity

of light at a point on the screen where path difference is , is K units. Find out the intensity

of light at a point where path difference is /3. 3

Solution:

(a) Coherent sources produce coherent light which have wavelength restricted to a very small

range. So the sources have almost same wavelength, hence produce stable and sustained

interference pattern.

(b) We know


2

0

2

0

0

0

2P h ase d ifferen ce p ath d ifferen ce

A t p a th d ifferen ce .

2P h ase d ifferen ce , = 2

In ten sity, I 4 I co s2

2o r, K 4 I co s g iven I K a t p a th d if feren ce

2

o r, K 4 I

Ko r, I

4

N o w , a t p a

2

0

2

th d ifferen ce3

2 2'

3 3

1 2In ten sity, I' 4 I co s

2 3

Ko r, I' = 4 co s

4 3

KI'

4

Q 20. Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the

photon picture of electromagnetic radiation.

Write the three salient features observed in photoelectric effect which can be explained using

this equation. 3

Solution:

Einstein’s photoelectric equation,

hv = Kmax +

Where

h = Planck’s constant

v = frequency

max2

max max 0

0

V maximum velocity of emitted photoelectron1K mv eV

V Stopping potential2

Work function

According to Planck’s quantum theory, light radiations consist of small packets of energy.

Einstein postulated that a photon of energy h is absorbed by the electron of the metal surface,

then the energy equal to is used to liberate electron from the surface and rest of the energy


h becomes the kinetic energy of the electron.

Energy of photon is,

E = hv

Where, h = Planck’s constant

v = frequency of light

The minimum energy required by the electron of a material to escape out of it is ‘’, work

function.

The additional energy acquired by the electron appears as the maximum kinetic energy ‘Kmax’ of

the electron.

i.e.,

Kmax = hv –

maxor, K Einstein's photoelectric equationhv

Kmax = eV0.

Salient features observed in photoelectric effect: —

The stopping potential and hence the maximum kinetic energy of emitted electrons varies

linearly with the frequency of incident radiation.

There exists a minimum cut – off frequency v0, for which the stopping potential is zero.

Photoelectric emission is instantaneous.

Q 21. Name the three different modes of propagation of electromagnetic waves. Explain, using a

proper diagram the mode of propagation used in the frequency range from a few MHz to 40

MHz. 3

Solution:

Three modes of propagation of electromagnetic waves,

(a) Ground waves,

(b) Sky waves,

(c) Space waves.

Sky wave propagation is used in the frequency range from a few MHz to 40 MHz.


In the ionosphere of the Earth’s atmosphere, there are a large number of charged particles (ions).

The ionosphere is situated about 65 km – 400 km above the surface of the Earth. The ionization

of molecules occurs due to the absorption of the ultraviolet rays and high energy radiation from

the Sun. The ionosphere acts as a reflecting layer for certain range of frequencies (3MHz-

30MHz). The transmitting antenna sends the EM signals of this frequency range towards the

ionosphere. When the EM waves strikes the ionosphere, it is reflected back to the Earth. A

receiving antenna at a remote location on the Earth receives these reflected signals.

Q 22. (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the

circumference of the electron in the nth

orbital state in hydrogen atom is n times the de Broglie

wavelength associated with it.

(b) The electron in hydrogen atom is initially in the third excited state. What is the maximum

number of spectral lines which can be emitted when it finally moves to the ground state? 3

Solution:

(a) According to Bohr’s second postulate of quantization, the electron can revolve round the

nucleus only in those circular orbits in which the angular momentum of the electron is integral

multiple of 2π

h where h is Planck’s constant (= 6.62 × 10

–34 Js) .

So, if m is the mass of electron an v is the velocity of electron in permitted quantized orbit with

radius r than

12π

hmvr n

Where n is the principle quantum number and can take integral values like

n = 1, 2, 3…

This is the Bohr’s quantization condition.


Now, de-Broglie wavelength is given as

h

mv

Where is wavelength associated with electron.

v is the velocity of electron.

h – is the velocity of electron.

m – mass of electron

2h

vm

Putting value of v from (2) in (1)

2π

π 2π

2π

h hm r n

m

rh nh

r n

Now circumference of the electron in the nth

orbital state of Hydrogen atom with radius r is 2r.

(b) In n is the quantum number of the highest energy level involved in the transitions, then the

total number of possible spectral lines emitted is

( 1)

N2

n n

Third excited state means fourth energy level i.e. n=4. Here, electron makes transition from n =

4 to n = 1 so highest n is n = 4

Thus, possible spectral lines

4(4 1)N

2

4 3

2

6

6 is the maximum possible number of spectral lines.

Q 23. A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A straight wire

carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, find

the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop

due to the current carrying wire. 3


Solution:

i M B MB sin

Here, M and B have the same direction

= 0°

MBsin 0

(ii) We know BF Bi l

On line AB, and CD magnetic forces are equal and opposite. So, they cancel out each other.

Magnetic force on line AD.


0

F B Attractive

B 4 cm = 0.04 m

IB

2πr

IF Attractive

2π

i l

i l l

i l

r

Magnetic force on line CB

0

0

F' B' Repulsive

F' F ' B'

IB'

2πr'

IF' Repulsive

2π '

i l

i l

i l

r

So net force

0

F F F'

I 1 1

2π '

n

i l

r r

Given

2A

I 1A

2cm 0.02 m

' 2 2.5 cm 4.5 cm

0.045 m

4 cm 0.04 m

i

r

r

l

Plug-in these values in above equation

7

7

7

7

7

4π×10 2 1 0.04 1 1F

2π 0.02 0.045

1 14 10

2 4.5

16 10 0.5 0.22

16 10 0.278

4.45 10 N

n

The direction of the force on the loop will be towards left.


Q 24. In the figure a long uniform potentiometer wire AB is having a constant potential gradient along

its length. The null points for the two primary cells of emfs 1 and 2 connected in the manner

shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) 1/2 and (ii)

position of null point for the cell 1. 3

How is the sensitivity of a potentiometer increased?

OR

Using Kirchoff’s rules determine the value of unknown resistance R in the circuit so that no

current flows through 4 resistance. Also find the potential difference between A and D.

Solution:


(i) Apply Kirchhoff’s law in loop ACFGA:-

1 2120

= potential drop per unit length.

Or, 1 2 120 ... 1

Loop AEHIA:-

2 1300

Or, 2 2 120 300 (By substituting value of 1 from equation (1))

Or, 22 = (300 - 120)

Or, 2 = 90 …(2)

Thus, 1 = 90 + 120

1 = 210 …(3)

Hence, 1

2

210 7

90 3

(ii) As we know,

= l

Thus from equation (2) and (3).

Null point for cell 2 is 90 cm.

And for cell 1, it is 210 cm.

Sensitivity of the potentiometer can be increased by :

(a) Increasing the length of the potentiometer wire

(b) Decreasing the resistance in the primary circuit

OR

Apply Kirchhoff’s In loop ABCFA:-

I + I + 4I1 = 9 – 6

2I + 4I1 = 3 …(1)


As there is no current flowing through the 4 resistance,

I1 = 0

Or, 2I = 3

Or, I = 1.5A

Thus the current through resistances R is 1.5A.

As there is no current through branch CF, thus equivalent circuit will be,

By applying Kirchhoff’s loop law we get,

1.5 + 1.5 + R (1.5) = 9 – 3

R = 2

Potential difference between A and D = (9 – 3) = 6V

Q 25. The figure shows a series LCR circuit with L = 10.0 H, C = 40 F, R = 60 connected to a

variable frequency 240 V source, calculate

(i) the angular frequency of the source which drives the circuit at resonance,

(ii) the current at the resonating frequency,

(iii) the rms potential drop across the inductor at resonance. 3

Solution:

(i) Resonant angular frequency


o6

36

1

1 1

LC 10 40 10

1 1

20 10400 10

1000

20

50rads

(ii) At resonant frequency, we know that the inductive reactance cancels out the capacitive

reactance.

The impedance = Z = 60 the value of resistance

The current amplitude at resonant frequency

o vo

E 2 E 2 ×240I = = =

Z R 60

339.36= =5.66A

60

(iii) The R.M.S. value of current

ov

I 5.66I = = = 4A

2 2

For R.M.S potential drop across inductor

L V L

V

V = I X

= I × L

= 4 × 5 0 × 1 0

= 2 0 0 × 1 0

= 2 0 0 0 V

Q 26. Use Huygens’s principle to explain the formation of diffraction pattern due to a single slit

illuminated by a monochromatic source of light.

When the width of the slit is made double the original width, how would this affect the size and

intensity of the central diffraction band? 3

Solution:

Consider a parallel beam of monochromatic light is incident normally in a slit of width b as

shown if figure. According to Huygens’s principle every point of slit acts as a source of

secondary wavelets spreading in all directions. Screen is placed at a larger distance.

Consider a particular point P on the screen receives waves from all the secondary sources. All

these waves start from different point of the slit and interface at point P to give resultant

intensity.


Point Po is at bisector plane of the slit. At Po, all waves are traveling equal optical path. So are in

phase the waves thus interface constructively with each other and maximum intensity is

observed. As we move from Po, the wave arrives with different phase and intensity is changed.

Intensity at point P is given by

2

0

sinI I

Where = sin θb

For central maxima 0 thus,

0I I

When the width of slit is made double the original width intensity will get four times of its

original value.

Width of central maximum is given by,

2D

a

Where, D = Distance between screen and slit,

= Wavelength of the light,

a = Size of slit.

So with the increase in size of slit the width of central maxima decreases. Hence, double the

size of the slit would result in half the width of the central maxima.

Q 27. (i) What characteristic property of nuclear force explains the constancy of binding energy per

nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?

(ii) Show that the density of nucleus over a wide range of nuclei is constant-independent of

mass number A. 3

Solution:

(i) The approximate constancy of BE/A over most of the range is saturation property of nuclear

force,

In heavy nuclei: nuclear size > range of nuclear force.


So, a nuclear sense approximately a constant number of neighbours and thus, the nuclear BE/A

levels off at high A. This is saturation of the nuclear force.

(ii) To find the density of nucleus of an atom, we have an atom with mass number lets say A.

(Here, we are neglecting mass of the orbital electrons)

Mass of the nucleus of the atom of the mass number A.

= A a.m.u

= A × 1.660565 × 10–27

kg

Let radius of nucleus is R.

Then, volume of nucleus, 34

πR3

31

30

3

0

4π R A

3

4πR A

3

Now, we know R0 = 1.1 × 10–15

m

Volume of nucleus 3

15 34π 1.1 10 A m

3

Density of the nucleus

27

315

17 3

Mass of nucleus

Volume of nucleus

A × 1.6605 × 10

4π 1.1 10 A

3

2.97 10 kgm

Thus, we can see the density of nuclei is independent of the mass number and is constant for all

nuclei.

physics grade12 set3 delhi 5 - meritnation.com · salient features observed in photoelectric...

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