Accelerate to win

Exclusive Exclusive HOTS HOTS

Get Finetuned With Boards Bahubali Bahubali 10 10

SAMPLE SAMPLE PAPERS PAPERS (designed (designedfor forcomplete complete revision revision to to score score

PHYSICS-XII PHYSICS-XII The Final Touch with

The Bahubali Series

Bahubali Series

Sample Papers Physics-XII 2021 Your search for Exclusive HOTS Questions Ends Here

By A team of subject experts

Accelerate Publications To win

Sample Papers Physics-XII First Edition Copyright

2021 by Accelerate Publications

271-B Mianwali Gurugram-122001 Call/Whatsapp : 7982680217 Price : ` 250/All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of publishers.

Limits Of Liability: While best possible efforts have been put in by the publishers and their experts in preparing the book, Accelerate Publications and their Experts make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and, specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. Beyond the description of this paragraph there are no warranties extended. No warranty may be created or extended by sales representatives. Disclaimer: Accuracy of the contents of this book has been checked. Since deviations cannot be prevented entirely, Accelerate Publications or its Experts cannot guarantee full agreement. As the book is intended for educational purpose, Accelerate Publications or its experts shall not be legally responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This Publication is designed to provide accurate and authoritative information with respect to the subject matter covered. It is sold on the understanding that the publisher is not engaged in rendering professional services.

Typeset by Harish Kumar.

Preface We are pleased to bring forward the Physics portion of Bahubali Series. It contains 10 sample papers based on the latest pattern of 30% reduced syllabus for session 2020-21. Conscious efforts have been made for framing the questions so that these papers serve as almost a complete revision of syllabus. CBSE has already announced that most of the questions asked will be based on understanding of concepts and its applications . The questions of these papers have been designed keeping this point in mind. As a result you will find most of the questions based on understanding and applications. All the papers are completely solved. The presentation of answers is based on reasoning. All possible efforts have been taken to answer the questions in a simple but precise manner. Once you go through these papers you will be ready to handle the boards paper with ease. Your mindset will be to expect tricky questions with the background to answers such questions with confidence. These papers will end the search of a serious student to find a set of sample papers which can enable him to score 100% marks confidently. All the best for boards! Experts of Accelerate Publications

Your search for Exclusive HOTS Questions Ends Here

Contents 1. Sample Paper 1 Solutions to Sample Paper 1 2. Sample Paper 2 Solutions to Sample Paper 2 3. Sample Paper 3 Solutions to Sample Paper 3 4. Sample Paper 4 Solutions to Sample Paper 4 5. Sample Paper 5 Solutions to Sample Paper 5 6. Sample Paper 6 Solutions to Sample Paper 6 7. Sample Paper 7 Solutions to Sample Paper 7 8. Sample Paper 8 Solutions to Sample Paper 8 9. Sample Paper 9 Solutions to Sample Paper 9 10. Sample Paper 10 Solutions to Sample Paper 10

1-7 8-14 15-20 21-26 27-33 34-39 40-46 47-53 54-60 61-67 68-75 76-81 82-87 88-94 95-101 102-107 108-114 115-120 121-127 128-134

1

Accelerate Publications

SAMPLE PAPER - 1 Maximum Marks: 70 Marks

Time Allowed: 3 hours

General Instructions: (1) All questions are compulsory. There are 33 questions in all. (2) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (3) Section A contains ten very short answer questions and four assertion reasoning MCQs of 1 mark each, Section B has two case based questions of 4 marks each, Section C contains nine short answer questions of 2 marks each, Section D contains five short answer questions of 3 marks each and Section E contains three long answer questions of 5 marks each. (4) There is no overall choice. However internal choice is provided. You have to attempt only one of the choices in such questions. SECTION – A All questions are compulsory. In case of internal choices, attempt any one of them. 1. 2.

3.

Name the physical quantity having unit A-m2. Give one use of an electromagnetic wave having wavelength 0.1 pm. or Name the electromagnetic wave produced by bombarding an energetic electron beam on a metal target. Give its wavelength range. Y A charged particle of charge +q and mass m is projected along +X direction with a velocity ‘v’ in a region where a uniform magnetic field is present in +Z direction. Find the direction of force acting on the charged particle. V

+q

X

B Z

4.

Is coefficient of self inductance a scalar or a vector quantity ? What are the dimensions of coefficient of self inductance in terms of M, L, T and A. or I (DECREASES) A coil is placed vertically below a straight current carrying wire in which the current decreases. What is the direction of induced current in the coil. COIL

5. 6. 7.

8.

An electron in higher orbit of a hydrogen atom has less negative total energy. Comment on the statement. The work function of a metal is hc/o. Light of wavelength  falls on the metal surface. What is the condition for photoelectric effect to take place. Is the mass of proton equal to / greater / smaller than the mass of neutron. or Two isotopes of chlorine have approximate masses of 35 u and 37 u and approximate relative abundances of 75% and 25% respectively. Find the average mass of chlorine atom. The movement of holes from p-side to n-side of a p-n junction gives rise to a current. Name the current. Why does this movement takes place as soon as a p-n junction is formed ? or

Accelerate Publications

9. 10. 11. 12.

13.

14.

2

The p-side of a junction diode is connected to – 2 V and n-side to – 4 V. Is the junction forward or reverse biased ? Give reason for your answer. Which p-n junction (photodiode / LED / solar cell) is used in reverse bias and why ? Why is the junction area of a solar cell much larger ? Assertion : Electrostatic force are the strongest forces in nature. Reason : Electrostatic force is weaker than gravitational force. Assertion : When an electric dipole is placed in a non-uniform electric field, in general, it experiences a force as well as torque. Reason : The forces acting on the two charges of dipole are unequal and their line of action is different. Assertion : A concave lens always forms a virtual and diminished image for a real objects and has positive focal length when nlens < nsurrounding medium. Reason : A convex lens always forms an inverted image for a real object and has a positive focal length when nlens > nsurrounding medium. Assertion : When the apparatus of Young’s double slit experiment is immersed in a liquid, the fringe width decrease. Reason :  liquid 

 nliquid

.

SECTION – B Questions 15 and 16 are Case Study based questions and are compulsory. Attempt any 4 sub parts from each question. Each question carries 1 mark. 15.

The concept of electric field lines was introduced by Michael Faraday. It is used to depict the nature of electric field in a given region. The properties of electric field lines are

The field linesoriginate from positive charge and terminate at the negative charge The direction of electric field strength at a point is given by the tangent drawn on the electric field line at that point  The field lines are continuous  The field lines never intersect  In a region of the strong electric field, lines are very close to each other whereas in the region of weak electric field lines are far  In the region of uniform electric field line, there are equidistant parallel lines having the same direction The field lines are always normal to the surface of the conductor (1) A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that Q1 (a) |Q1| > |Q2| Q2 (b) |Q1| < |Q2| (c) at a finite distance to the left of Q1 the electric field is zero. (d) None (2) Refer the above question and choose the correct statement (a) | Q1 | = | Q2 | (b) Q1 is negative and Q2 is positive  

3

Accelerate Publications

(3)

(c) The electric field will be zero some where at a point lying on the line joining Q1 and Q2 between Q1 and Q2. (d) At a finite distance to the right of Q2, the electric field will be zero. A metallic solid sphere is placed in a uniform electric field. The 1 1 line of force follow the path shown in figure as 2 3 4

(4)

(a) 1 (b) 2 (c) 3 (d) 4 A metallic shell has a point charge ‘q’ kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces ?

(b)

(a)

(5)

(d)

( c)

A point positive charge is brought near an isolated conducting sphere (see figure). The electric field is best given by

+q

+q (a)

(b)

+q

+q (c)

16.

2 3 4

(d)

(a) Fig (a) (b) Fig (c) (c) Fig (b) (d) Fig (d) The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The line ab and cd represent wavefronts of two consecutive crests of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of two consecutive crests of the b d light wave in medium-2 after refraction. (1) Light travels as a MEDIUM-1 (a) parallel beam in each medium a c X Y f h (b) convergent beam in each medium (c) divergent beam in each medium MEDIUM-2 e g (d) divergent beam in one medium and convergent beam in the other medium. (2) The phases of the light wave at c, d, e and f are c, d, e and f respectively. It is given that c  f. (a) c cannot be equal to d (b) d can be equal to e (c) (d - f) is equal to (c - e) (d) (d - c) is not equal to (f - e) (3) Speed of light is (a) the same in medium-1 and medium-2 (b) larger in medium-1 than in medium-2 (c) large in medium-2 than in medium-1 (d) different at b and d. (4) If 1 and 2 are wave lengths in medium-1 and medium-2 respectively the

4

Accelerate Publications

(5)

(a) 1 = 2 (b) 1 > 2 (c) 1 < 2 (d) depends on nature of medium-1 and medium-2 If v1 and v2 are the frequencies of light in medium-1 and medium-2 respectively then (a) v1 = v2 (b) v1 > v2 (c) v1 < v2 (d)depends on various factors

SECTION – C All questions are compulsory. In case of internal choices, attempt anyone. 17.

18.

19.

20. 21.

22.

Two long conductors, separated by a distance d carry current I1 and I2 in the same direction. They exert a force 𝐹 on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. Find the new value of force between them Define a wavefront. Using Huygens’ principle, draw the shape of a refracted wavefront, when a plane wave is incident on a convex lens. or (a) When a wave is propagating from a rarer to a denser medium, which characteristic of the wave does not change and why ? (b) What is the ratio of the velocity of the wave in the two media of refractive indices 1 and 2 ? Deduce an expression for torque acting on a dipole of dipole moment 𝑝 when 𝑝 is directed at an angle of ‘’ with a uniform electric field. or (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic, i.e., large scale charges ? Draw the circuit diagram of a full wave rectifier and state how it works. A square loop of side 5 cm is placed vertically in the east-west plane. A uniform magnetic field of 0.1 T. is set-up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.5 s at a steady rate. Find the magnitude of induced emf. Figure shows a double slit experimental set-up for observing interference fringes due to different component colours of white light. What would be the predominant colour of fringes observed at the point (i) O, (the central point) (ii)

P, where S2P – S1P =

b ? 2

Here  b is the wavelength of the blue colour. 23.

24.

What is the name given to an intrinsic semiconductor when it is doped with pentavalent atoms. What are the majority charge carriers in such type of a semiconductor considering the number of dopant atoms per unit volume to be much much greater than the electron-hole pairs in intrinsic semiconductor. Write the formula for electrical conductivity of such a doped conductor in terms of elementary charge ‘e’, the number of dopant atom per unit volume Nd and the mobility of the majority charge carriers. (a) The earth’s magnetic field varies from point to point in space. Does it also change with time ? If so, on what time scale does it change appreciably ? (b) The earth’s core is known to contain iron. Yet, geologists do not regard this as a source of the earth’s magnetism. Why ? or In a hydrogen atom, the electrons moves in an orbit of radius 0.5 Å making 1016 revolution per second. Find the magnetic moment associated with the orbital motion of electron.

5

Accelerate Publications

25.

Calculate the amount of charge flown in 2 second through a wire in which a variable current I = 3 t is flowing in the wire. Current is measured in Coulomb.

SECTION – D All questions are compulsory. In case of internal choices, attempt any one. 26.

I

(a) If resistance R in the circuit ‘a’ is decreased, what will be the direction of induced current in circuit b ? R

a

b

(b) Predict the direction of induced current in a metal ring when the ring is moved towards straight conductor with a constant speed v. The conductor is carrying current I in the direction shown in the figure.

(c) A jet plane is travelling towards west at a speed of 1800 kmh-1. What is the voltage difference developed between the ends of a wing having a span of 25 m, if the earth’s magnetic field at the location is of the 27.

magnitude 5 10-4 T and the dip angle is 30° ? (a) A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38  as shown in the figure. Find the value of current in the circuit.

10 V

200 V

28.

38 

(b) In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9  is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell. or Answer the following : (a) Why are the connections between resistors in a meter bridge made up of thick copper strips ? (b) Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire? (c) Which material is used for the meter bridge wire and why ? (i) The figure shows the plot of binding energy (BE) per nucleon as a function of mass number A. The letters A, B, C D C, D and E represent the positions of typical nuclei on the E curve. Point out, giving reasons, the two processes (in BE B A terms of A, B, C, D and E), one of which can occur due to A nuclear fission and the other due to nuclear fusion. MASS NUMBER A

29.

(ii) In the reaction C  A + B, is energy released or absorbed ? Give reason. or (a) State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ? (b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. (a) Deduce electrical conductivity of a doped semiconductor in terms of ne (electron concentration), nh (hole concentration), e (electron mobility), h (hole mobility). Given electron charge is e.

6

Accelerate Publications

(b) The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. Find the band gap of the semiconductor. Given h = 6.63 1034 Js 30.

and 1 eV = 1.6 1019 J. Using Bohr’s postulates, obtain expressions for (i) Kinetic energy (ii) Potential energy of an electron in hydrogen like atom. Draw energy level diagram showing energies of various orbits and series of spectral lines.

SECTION – E All questions are compulsory. In case of internal choices, attempt any one. 31.

(a) Three circuits, each consisting of a switch ‘S’ and two capacitors, are initially charged, as shown in the figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor (i) increase, (ii) decrease and (iii) remain same ? Give reasons. S 6Q

+

S

2C

C

+

3Q

6Q

+

S

C

(a)

(b) Find the resultant electric force on a charge Q.

C

+

6Q

3Q

+

(b)

C

+

3Q

( c) q

Q

q

3C

a

Q

or (a) Find the equivalent capacitance of the network. For a 300 V supply, determine the charge and voltage across each capacitor.

100 pF C1 200 pF C2

200 pF 300 V

C3

C4 100 pF

32.

(b) Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2. (a) Three students X, Y and Z performed an experiment for studying variation of a.c. with angular frequency in a series LCR circuit and obtained the graph. (see in figure) They all used a.c. sources of same rms value and inductance of the same value. What can we conclude about. (i) Capacitance value (ii) Resistance value used by them ?

7

Accelerate Publications

33.

In which case the quality factor will be maximum ? What can we concluded about the nature of impedance of set-up at frequency 0 ? (b) A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, find the charge flowing through the galvanometer. or L Consider tan electrical circuit consisting a charged capacitor ‘C’ connected to an inductor L through a key K which is opened. At time t = 0, the key is closed and L – C oscillations start. Draw the circuit diagram showing current in the circuit, charge on the capacitor and K the magnetic field lines in the inductor at t = T/8, t = 2T/8, t = 3T/8, t + – + – = 5T/8. L-C oscillations can be compared with oscillations of a + – spring mass system. Find the analogy of m (mass), spring constant + – C (K) of spring – mass system with L – C oscillation. (a) Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen. (b) The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is

   , and . 6 4 3

or (a) Three rays of light-red (R), green (G) and blue (B) are incident on the face AB of a right angled prism ABC in figure. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Trace the path of the rays through the prism.

A

B G R 45 C

(b) On what factors does angle of deviation of a ray passing through a prism depends. (c) On what factors does dispersive power depends.

8

Accelerate Publications

SOLUTION OF SAMPLE PAPER – 1 1. 2.

Magnetic dipole moment. –rays are used in treatment of cancer / tumour. or X-rays. Wave length range is 1 nm to 10-3 nm.

3.

𝐹=𝑞 𝑣×𝐵

4.

= 𝑞 𝑣𝑖 × 𝐵𝑘 = 𝑞𝑣𝐵 𝑖 × 𝑘 = 𝑞𝑣𝐵 −𝑗 . The force is directed towards negative Y-axis. ‘L’ is a scalar quantity Coefficient of self inductance =

13. 14. 15.

 BA [ MLT 2 ][L2 ] FA = =  I I qvI [ AT ][LT 1 ][ A]

= [ ML2T 2 A2 ]

5.

or The induced current is in anticlockwise direction. This is in accordance to Lenz’s laws. The statement is correct. E= 

e2 8 o r

Greater the value of ‘r’ lesser is the value of E in negative. 6.

  o

7.

Mass of proton (1.00785 u) is slightly smaller than the mass of neutron (1.008665 u) or 35  75  37  25 Average mass =  35.5u 100 Diffusion current. This is because of difference in the hole concentration in p-side and n-side. or Forward biased because p-side is connected to a higher potential and n-side to lower potential. Photodiode is used in reverse bias as it is easy to detect the fractional change in current when light is allowed to fall on it. Larger surface area enables greater amount of incident light to fall on junction which converts into higher power output. D. Nuclear force is the strongest force in nature. Electrostatic force is stronger than gravitational force. A

8.

9.

10.

11.

12.

16.

When forces are unequal a net force will act on the dipole. When the line of action of forces are different, a torque is produced. C A (1) (a) The electric field lines are originating from Q1 and terminating on Q2. Therefore Q1 is positive and Q2 is negative. As the number of lines associated with Q1 is greater than that associated with Q2, therefore |Q1| > Q2|. (2) (d) At a finite distance on the left of Q1, the electric field intensity cannot be zero because the electric field created by Q1 will be greater than Q2. This is because the magnitude of Q1 is greater and the distance Q  smaller  E  2  . At a finite distance to the r   right of Q2, the electric field can be zero. Here, the electric field created by Q2 at a particular point will cancel out the electric field created by Q1. (3) (d) The electric lines of force cannot enter the metallic sphere as electric field inside the solid metallic sphere is zero. Also, the origination and termination of the electric lines of force from the metallic surface is normal to the surface. (4) (c) Electric field everywhere inside the metallic portion of shell is zero. Hence options (a) and (d) are incorrect. Electric field lines are always normal to a surface. Hence option (b) is incorrect. Only option (c) represents the correct answer. (5) (a) Due to charge +q, induced charges will be developed in the isolated conducting sphere. (1) (a) For plane wave fronts the beam of light is parallel.

9

Accelerate Publications

(2)

(c) Since points c and d are on the same wavefront, d  c Similarly e   f

19.

Consider an electric dipole placed in a uniform electric field at an angle  as shown in the figure. B+

  d   f  c  e

2l

(3)

17.

18.

(b) The gap between consecutive wavefronts in medium 2 is less than that is medium-1. Therefore, wavelength of light in medium-2 is less than that in medium-1. Therefore, speed of light is more in medium 1 and less in medium-2. (4) (b) The gap between consecutive wavefronts in medium 2 is less than that is medium-1. Therefore, wavelength of light in medium-2 is less than that in medium-1. (5) (a) Frequency of light does not change during refraction. For between two long conductor carrying current,  2I I F = 0 1 2 l 4 d  2(2 I1 ) I 2 F' 2 2F F' 0 l    F' 4 3d F 3 3 The direction of F ' is opposite to F. Locus of points in the same state of vibration i.e. same phase is called a wavefront.

(a)

(b)

or When a wave is propagating from rarer to denser medium, frequency of wave does not change. This is because as light enters the denser medium its velocity and wavelength falls proportionately such that frequency v remains constant   . Thus frequency is a  characteristics of source and has nothing to do with the medium.

 2 v1  1 v 2

qE

2l son  –  A

C E

Torque = Force  (perpendicular distance between the line of action of force) q

 E

t

 t = F  (BC)  t = qE  2l sin   t = (q  2l) E sin  t = pE sin [ p  q  2l ] 𝜏= 𝑝× 𝐸 (a)

20.

or Electric charge of a body is always an integral multiple of elementary charge (e)

q = ne where e = 1.6 10 19 C. In other words, electric charge can take any value but it can have value only in integral jumps of the elementary charge. (b) The magnitude of elementary charge is extremely small as compared to the charge on bodies that we practically deal with. Therefore we can ignore quantisation of charge at macroscopic level. During first half cycle when A is at a higher potential with respect to B, D1 is forward biased and D2 is reverse biased. D1 becomes conducting and current flows in load resistance RL.

10

Accelerate Publications

During the next half cycle when A is at a lower potential with respect to B, D1 is reverse biased and D2 is forward biased. D2 becomes conducting and current flows in RL in the same direction as the first half cycle. In this way a.c. converts into d.c. 21.

e

 ( Bl 2 cos 45)  t t

22.

23. 24.

m

2 2

19

= 1.6  10

16

10 2

 10  3.14  (0.5  10

)

 1.251023 Am2

25.

I=

dq dt

2



2



 q  Idt  3tdt 0

3

26.

(a) (b)



1  0.1  0   e  5  10   3.5  104 V 2  0.5  (i) At O, white colour is obtained as all the colours component have zero path difference and therefore undergo constructive interference. (ii) A path difference of b/2 indicate that the blue colour undergo destructive interference and therefore blue colour will not be present at P. From white light if blue colour is removed then the remaining colour is yellow. Therefore, yellow colour of fringe will be obtained at P. n-type semiconductor. The majority charge carriers are electrons.  = e Nde (a) Yes, the magnetic field of earth changes with time. An appreciable change is observed in roughly a few hundred years. (b) The temperature ( 6000°C) in the core of earth is too high for any magnetism to be retained. It is the presence of radioactive substance in the core of earth. or e m = IA = r 2  er 2 T

0

t2  q = 3  = 6C  2 0 (a) The current in circuit b will be in anticlockwise direction. (b) In clockwise direction.

The voltage difference developed between the ends of wing are

e  BV 1v

27.

B   e   (l 2 cos 45)  t  



(c)

(a)

(b)

(c) 28.

(i)

(ii)

(a)

5   5  104  sin 30  25  1800  18   = 3.125 V 200 10 I= = 5A 38 E  350 V  300 E 350 35    V 300 30 E   35  r  R   1  9  1 = 1.5  V   30  or The connection between the resistors in a meter bridge is made up of thick copper strips to minimise the resistance. This is to minimise errors due to contact resistance developed at the points of contact where the wire is connected to the metal strips. Magnanin or constantan due to high specific resistance. ‘E’ can undergo nuclear fission. By doing so, ‘E’ converts into ‘C’ and ‘D’. We see that B.E/A for’C’ and ‘D’ is greater than ‘E’ due to which fission is a favourable process and energy will be released. Energy will be absorbed as A and B have less binding energy per nucleon as compared to C. or According to Bohr’s postulate, electron in a hydrogen atom can revolve only in certain permissible orbits for which its angular momentum is an integral multiple of (h/2). h mvr  n 2 While an electron revolves in an orbit, it neither gains nor loses energy. Therefore each orbit is characterised by a fixed energy and is called a stable orbit. Explanation of de-Broglie for the stability of orbits.

11

Accelerate Publications

According to de Broglie



h mv

 Eg =

...(i)

For the electron to revolve in phase around the nucleus, the circumference of the orbit should be integral multiple of de-Broglie wavelength. That is ...(ii) 2r  n From (i) and (ii) h  h  2r  n mvr  n   2  mv  (b)

30.

V

–

e

 2.18 1018 (1) 2   2.18 1018 (1) 2    42 12  

v=

Also

 Kinetic Energy =

I  e(neve  nhvh ) A

Potential Energy =

Semi conductor

Ve

Vh e

J  e(neve  nhvh )

... (ii)

Ze 2 nh Ze 2  v   2nh 2mv 40 mv2

= 0.3081016  3.081015 Hz. I  I h  Ie

h

Ze 2 40 mv2

From (i) and (ii)

2.18  1018  15 16  6.63  1034

A

mv2 1 ( Ze)(e)  r 40 r 2

 r

I  Ae(neve  nhvh )

...(iii)

1 2 me4 Z 2 mv  (from iii) 2 2 8 0 n 2 h 2 me4 Z 2 1 ( Ze)(e) =  2 40 r 4 0 n 2 h 2

E = 0 eV

n= 

E5 = 0.54 eV

Pfund

E4 = 0.85 eV

Ih

Brackett

E3 = 1.5 eV

Ie

Paschen

E2 = 3.4 eV

I   J  A   

E  e(neve  nhvh )

E1 = 13.6 eV

Balmer

The band gap Eg equals the energy carried by a photon of wavelength 2840 nm. Hence hc Eg  hv  

n= 5 n= 4 n= 3 n= 2

n= 1

Lyman Energy level diagram for for hydrogen atom

31.

(a)

(i)

v   v   e ne e  n h  E E    e(nee nhh )

(b)

m

r

ze

 1  6.63  1034 v = 2.18 1018 1    16 

(a)

= 8.02  102 J  0.5 eV nh nh  r ... (i) mvr  2 2mv

E  h  E4  E1

=

29.

6.631034  3 108 2480109

(ii)

Initial potential difference across the 6Q 3Q  plates of capacitor (2C) = 2C C Initially potential difference across the 3Q plates of capacitor (C) = C As the potential is same, no change flow take place. Potential difference on the left 6Q capacitor = C

Accelerate

1. In accordance with Board 2021 (After 30% syllabus deletion)

to win

Exclusive HOTS Exclusive HOTS

Get Finetuned With Boards Bahubali 10 Bahubali 10

SAMPLEPAPERS PAPERS SAMPLE (designed (designedfor forcomplete complete revision revision to to score score 90%+)

PHYSICS-XII PHYSICS-XII

2. In accordance with Board 2021 pattern that

The Final Touch with

The Bahubali Series

includes Ø Case study/ paragraph based questions

Accelerate to win

Exclusive HOTS Exclusive HOTS

Ø Assertion and reasoning questions

Get Finetuned With Board Bahubali 10 Bahubali 10

SAMPLEPAPERS PAPERS SAMPLE (designed (designedfor forcomplete complete revision revision to to score score 90%+)

PHYSICS-XII Mathematics-XII

3. Provides complete revision of syllabus

The Final Touch with

The Bahubali Series

4. Special emphasis given on

Accelerate to win

Ø Critical thinking based questions

Exclusive Exclusive HOTS HOTS

Get Finetuned With Board

Ø Reasoning based questions

Bahubali 10 Bahubali 10

SAMPLE SAMPLEPAPERS PAPERS (designed (designedfor forcomplete complete revision revision to to score score 90%+)

Ø Application based questions

Chemistry-XII PHYSICS-XII The Final Touch with

The Bahubali Series

5. All papers solved with detailed and accurate solutions.

` 250/-