Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Set 6 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Physics Set 6 with Solutions
Time : 3 Hours
Maximum Marks: 70
General Instructions :
- There are 33 questions in all. All questions are compulsory.
- This question paper has five sections: Section A, Section B, Section C. Section D and Section E.
- Air the sections are compulsory.
- Section A contains sixteen questions, twelve MCQ and four Assertion-Reasoning based questions of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study-based questions of four marks each and Section E contains three long answer questions of five marks each.
- There is no overall choice. However, an internal choice has been provided in one question in Section B, one question in Section C, one question in each CBQ in Section D and all three questions in Section E. You have to attempt only one of the choices in such questions.
- Use of calculators is not allowed.
- You may use the following values of physical constants wherever necessary
(i) c = 3 ×108 m/s
(ii) me = 9.1 × 10-31 kg
(iii) e = 1.6 × 10-19 C
(iv) μ0= 4π × 10-17 TmA-1
(v) h = 6.63 × 10-34 Js
(vi) ε0 = 8.854 ×10-12 C2N-1m-2
(vii) Avogadro’s number = 6.023 × 1023 per gram mole
Section – A
The following questions are multiple-choice questions with one correct answer. Each question carries 1 mark. There is no internal choice in this section.
Question 1.
Electric charges q1– q1,+ q2 and- q2 are placed in free space and S is a spherical Gaussian surface. The electric flux passing over the surface S is:
(a) due to + q2 only
(b) due to all charges
(c) zero
(d) due to positive charge only
Answer:
(c) zero
Explanation: Here total charge = q1 + q2 – q1 – q2 = 0.
So, total electric flux out of the surface is zero.
Question 2.
Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.
(a) The work done in figure (i) is the greatest
(b) The work done in figure (ii) is the least
(c) The work done is the same in figure (i), (ii) and (iii).
(d) The work done in figure (iii) is greater than figure (ii) but equal to that in figure (i).
Answer:
(c) The work done is the same in figure (i), (ii) and (iii).
Explanation: In all the three figures, VA = 20 V and VB = 40 V. Work done in carrying a charge q from A to B is W = q (VB – VB).
Question 3.
When a current I is set up in a wire of radius r, the drift velocity is Vd- If the same current is set up through a wire of radius 2r, the drift velocity will be:
(a) 4vd
(b) 2vd
(c) vd/2
(d) vd/4
Answer:
(d) vd/4
Question 4.
A charge of 1 C is moving in a magnetic field of 0.5 T with velocity of 10 m/s. Force experienced is:
(a) 0.5 N
(b) 5 N
(c) 10 N
(d) 0 N
Answer:
(b) 5 N
Explanation: Force on change due to magnetic field.
F=qvB
F =1 × 0.5 × 10
= 5 N
Question 5.
A charge particle after being accelerated through a potential difference V enters in a uniform magnetic field and moves in a circle of radius r. If V is doubled, the radius of the circle will become :
(a) 2r
(b) √2r
(c) 4r
(d)\(\frac{1}{\sqrt{2} r}\)
Answer:
(b) √2r
Explanation: Radius of circular path
Question 6.
A paramagnetic substance is placed in a magnetic field which is increased till the magnetisation of the substance becomes constant. Now, on lowering the temperature, the magnetisation :
(a) will increase
(b) will decrease
(c) will remain unchanged
(d) may increase or decrease
Answer:
(c) will remain unchanged
Explanation: Magnetisation becomes constant i.e., all the magnetic moments have got aligned in the direction of die applied field. So now, if the temperature is decreased, thermal vibration of the paramagnetic material will reduce. But as all the magnetic moments are already aligned in the direction of the field so no further alignment can take place due to reduced thermal motion. Thus, there will be no negative effect of decreasing the temperature on the magnetisation.
Question 7.
Lenz’s law gives:
(a) the direction of the induced current
(b) the magnitude of the induced emf
(c) the magnitude of the induced current both
(d) the magnitude and direction of the induced current
Answer:
(a) the direction of the induced current
Explanation: Lenz’s law states that an induced electric current flows in a direction such that the current opposes the change that induced it.
Question 8.
What is the velocity of electromagnetic waves in free space?
(a) \(c=\frac{1}{\mu_0 \varepsilon_0}\)
(b) c = μoε0
(c) \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
(d) None of these
Answer:
(c) \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Explanation: As the constant po and so do not depend on frequency or wavelength of e.m. waves so the value of c remains same. Fience, the velocity of e.m. waves in free space is : \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Question 9.
An AC supply gives Vn„s = 30V, which passes through 10Q resistance. The power dissipated in it is :
(a) 45 √2 W
(b) 90 √2 W
(c) 45 W
(d) 90 W
Answer:
(d) 90 W
Explanation: We know that,
Question 10.
A fish which is at a depth of 12 cm in water \(\left(\mu=\frac{4}{3}\right)\) is viewed by an observer on the bank of a lake. Its apparent depth as observed by the observer is:
(a) 3 cm
(b) 9 cm
(c) 12 cm
(d) 16 cm
Answer:
(b) 9 cm
Explanation: D = Real depth, d = apparent depth
Question 11.
In the a-particle scattering experiment, the shape of the trajectory of the scattered a-particles depend upon :
(a) only on impact parameter.
(b) only on the source of a-particles.
(c) both impact parameter and source of a-particles.
(d) impact parameter and the screen material of the detector.
Answer:
(a) only on impact parameter.
Explanation: In the a-particle scattering experiment, the shape of the trajectory of the scattered a-particles only depends upon impact parameters.
Question 12.
The forbidden energy band gap in conductors, semiconductors and insulators are Eg1, Eg2, and Eg3 respectively. The relation among them is:
(a) Eg1 = Eg2 = Eg3
(b) Eg1 < Eg2 < Eg3
(c) Eg1 > Eg2 > Eg3
(d) Eg1 < Eg2 > Eg3
Answer:
(b) Eg1 < Eg2 < Eg3
Explanation: Band gap of the insulator is largest as it restricts the flow of electrons through it.
So, Eg1 < Eg2 < Eg3.
(Direction: Question 13 to 16) Two statements are given-one labelled Assertion (A) and the other labelled
Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, and R is also false.
Question 13.
Assertion: Semiconductors do not obeys Ohm’s law.
Reason: Current can not be determined by the rate of flow of charge carriers.
Answer:
(d) A is false, and R is also false.
Explanation: The assertion is not true. In fact, semiconductor obeys Ohm’s law for low values of electric field (- 106 V/m). Above this, the current becomes almost independent of electric field. Hence, assertion is false and reason is also false.
Question 14.
Assertion: Optical fibres make use of total internal reflection.
Reason: Light undergoes successive total internal reflections as it moves through an optical fibre.
Answer:
(b) Both A and R are true, but R is not the correct explanation of A.
Explanation: Optical fibres are fabricated with high-quality composite glass or quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding. When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end.
Question 15.
Assertion: Kirchhoffs junction rule can be applied to a junction of several lines or a point in a line.
Reason: When steady current is flowing, there is no accumulation of charges at any junction or at any point in a line.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation: Junction rule or Kirchhoffs first lawor Kirchhoffs current law (KCL) states that the algebraic sum of the currents meeting at a junction (point) in an electrical circuit is always zero. Or, the sum of currents flowing towards the junction is equal to sum of currents leaving the junction.
Question 16.
Assertion: Bohr’s postulates states that the electrons in stationary orbits around the nucleus do not radiate.
Reason: According to classical physics, all moving electrons radiate.
Answer:
(b) Both A and R are true and R is not the correct explanation of A.
Explanation: According to classical physics, all moving charged particle radiates e.m. radiations. So, moving electrons will also radiate energy.
Section – B
Question 17.
Name the parts of the electromagnetic spectrum which is:
(i) suitable for radar systems used in aircraft navigation
(ii) used to treat muscular strain
(iii). used as a diagnostic tool in medicine
Write in brief, how these waves can be produced.
Answer:
(i) Microwaves are suitable for radar systems that are used in aircraft navigation.
These rays are produced by special vacuum tubes, namely klystrons, magnetrons and Gunn diodes.
(ii) Infrared waves are used to treat muscular strain. These rays are produced by hot bodies and molecules.
(iii) X-rays are used as a diagnostic tool in medicine. These rays are produced when high energy electrons are stopped suddenly on a metal of high atomic number.
Question 18.
Two charges 2 μC and- 2 μC are placed at points A and B 6 cm apart.
(i) Identify an equipotential surface of the system.
(ii) What is the direction of the electric field at every point on this surface?
Answer:
The situation is represented in the given figure.
(i) An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
(ii) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
Question 19.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.
Answer:
Ray ‘1’ and ‘2’ will fall on the side AC at an angle of incidence (i) of 45°. Critical angle of ray ‘1’ is greater than i, so it will get refracted from the prism. Critical angle of ray ‘2’ is less than that of i, so it will undergo total internal reflection.
Question 20.
(i) Explain the formation of energy bands in crystalline solids.
(ii) Draw the energy band diagrams of (a) a metal and (b) a semiconductor.
Answer:
(i) An enormously large number of energy levels closely spaced in a very small energy range constitute an energy band.
The allowed energy bands are separated by regions in which energy levels cannot exist. These forbidden regions are called band gaps or energy gaps.
The highest energy band occupied by the valence electrons is called the valence band and the next empty allowed band is called the conduction band.
Question 21.
A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.
OR
Distinguish between nuclear fission and nuclear fusion.
Answer:
Binding energy of the nucleus,
B1 = 7.6 × 240 = 1824 MeV.
Binding energy of each product nucleus,
B2 = 8.5 × 120 = 1020 MeV
Then, energy released as the nucleus breaks,
E = 2 B2 – B1 = 2 × 1020 -1824 = 216 MeV.
OR
Nuclear fission | Nuclear fusion | |
(i) | It is the process in which a heavy nucleus split up into two lighter nuclei of nearly equal masses. | It is the process in which two lighter nuclei combine together to form a heavy nucleus. |
(ii) | Comparatively less amount of energy is released. | Enormous amount of energy is released. |
(iii) | Nuclear fission may take place at ordinary temperature. | A very high temperature of the order of million of degree is required. |
(iv) | The sources of fissionable materials is limited. | The sources of fusion reaction i.e., hydrogen is more plentiful (air and water). |
(v) | The products of nuclear fission are in general radioactive and hence pose a radiation hazard. | The products of fusion are non-radioactive and pose no radiation hazard. |
Section – C
Question 22.
(a) Define the term drift velocity.
(b) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?
(c) Why alloys like constantan and manganin are used for making standard resistors?
OR
Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Answer:
(a) Drift velocity is defined as the average velocity with which the free electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.
OR
Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Q and 0.3 Q respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Answer:
(a) Drift velocity is defined as the average velocity with which the free electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.
Where p is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.
(c) They are used to make standard resistors because:
(i) They have high value of resistivity.
(ii) Temperature coefficient of resistance is less.
(iii) They are least affected by temperature.
OR
Question 23.
An observer to the left of a solenoid of N turns each of cross-section area A observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic momentum
M = NLA.
Answer:
Given that the current flows in the clockwise direction for an observer on the left side of the solenoid. This means that left face of the solenoid acts as south pole and right face acts as north pole. Inside a bar magnet the magnetic field lines are directed from south to north. Therefore, the magnetic field lines are directed from left to right in the solenoid.
Magnetic moment of single current carrying loop is given by m’ = IA
Where, I = Current flowing through the loop
A = Area of the loop
So, Magnetic moment of the whole solenoid is given by
M = Nm‘ = N (IA)
Question 24.
Four point charges Q, q, Q and q are placed at the comers of a square of side ‘a’ as shown in the figure. Find the:
(i) resultant electric force on a charge Q, and
(ii) potential energy of this system.
Answer:
(i) Force on charge Q at B due to charge q
Question 25.
Explain the formation of potential barrier and depletion region in a p-n junction diode. What is the effect of applying forward bias on the width of depletion region?
Answer:
Two important processes involved during the formation of p-n junction are diffusion and hole. Formation of depletion region and potential barrier: At the instant of p-n junction formation, the free electrons near the junction diffuse across the junction into the p-region and combines with holes. Thus, on combining with the hole, it makes a negative ion and leaves a positive ion on n-side. These two layers of immobile positive and immobile negative charges form the depletion charge.
Further, as electrons diffuse across the junction, a point is reached where the negative charge repels only further diffusion of electron. This depletion region now acts as a barrier. Now, the external energy is supplied to get electrons to move across the barrier of electric field. The potential difference required to move the electron through the electric field is called barrier potential. The effect of forward bias on p-n region is that the depletion layer will be reduced.
Question 26.
State Bohr’s quantisation condition ofangular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.
Answer:
According to Bohr’s quantisation of angular momentum, the stationary orbits are those in which angular momentum of electron is an integral multiple of \(\frac{h}{2 \pi}\) i.e., mvr = n \(\frac{h}{2 \pi}\) Where, n = 1, 2, 3
For Brackett series, n1 = 2 and n2 = ∞
Brackett series is found in infrared region of electromagnetic spectrum.
Question 27.
A small telescope has an objective lens of focal length 150 cm and eye piece of focal length 5 cm.
What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
Answer:
If the telescope is in normal adjustment, i.e., the final image is at infinity.
If tall tower is at distance 3 km from the objective lens of focal length 150 cm. It will form its image at distance v0. So,
Question 28.
Explain the properties of different waves in electromagnetic spectrum.
OR
Write down properties of e.m. wave.
Answer:
Electromagnetic spectrum: Since, electromagnetic waves carry energy and momentum from source to a receiver. In early time, the only electromagnetic waves recognized were radio waves and visible. light. It is now known that other forms of electromagnetic waves exist which are distinguished by their frequency and wavelength as given below :
(i) Gamma Rays: These are the waves emitted by radioactive nuclei and during certain nuclear reactions. Their wavelength varies as 10-14 m to 10-10 m. These waves are highly penetrating and produce serious change when absorbed by living tissues. Consequently those working near such dangerous radiations must be protected with heavily absorbing material such as thick layers of lead.
(ii) X-rays: These waves are commonly produced by the deceleration of high energy electron bombarding a metal target. Their wavelength varies as 10-13 m to 10-8 m. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. X-rays are used in surgery, radiotherapy, engineering, detective departments, industries and in scientific research.
(iii) Ultra-violet Rays: These are the waves constituting 10% of sun light. Their wavelength varies as 3.8 × 10-7 m to 6 × 10-8 m. Ultra violet rays are used for checking the minerals samples by making use of its property of causing fluorescence. These rays are also used in the study of molecular structures. These are used in sterlizing surgical instruments.
(iv) Visible Light: These are the electro-magnetic waves constituting 50% of sun light. Their wavelength varies as 4 × 10-7 m to 7 × 10-7 m. These waves make the objects visible to use. These are used in many ways in our day-to-day life such as drying, decaying, making food for plants, keeping atmosphere warm etc.
(v) Infra-red Rays: These waves are also called heat waves. Their wavelength varies 1 × 10-3 m to 7 × 10-7 m. These waves are produced by hot bodies and molecules and are readily absorbed by most materials. The infra-red energy absorbed by a substance appears as heat since the energy agitates the atoms of the body, increasing their vibrational or translational motion, which results in temperature rise. Infra-red radiation has many practical and scientific applications, including physical therapy, infra-red photography and vibrational spectroscopy.
(vi) Micro Waves: These are also called short wavelength radio waves. Their wavelength varies from 1 mm to 30 cm and are generated by electronic devices. Because of the short wavelength, they are well suited for the radar systems used in aircraft navigation and for studying the atomic and molecular properties of matter. Microwave ovens represent an interesting domestic application of these waves. In a recent proposal, it was suggested that solar energy could be harnessed by beaming microwaves down to earth from a solar collector in space.
(vii) Long Waves or Radio Waves : These waves are produced by oscillating electronic circuits. Their wavelength is of the order of 1 m. These waves are used as carrier waves in radio broadcasting and T.V. transmission.
OR
(i) Changes in electric and magnetic fields occur simultaneously. Electromagnetic wave attain their maxima and minima at the same place and at the same time.
(ii) The electric and magnetic fields directions are mutually perpendicular to each other and as well as to the direction of propagation of wave.
(iii) The electric field vector (E) and magnetic field vector (B) are related by c = \(\frac{E_0}{B_0}\) Eo here B0 are the amplitudes of the respective fields and c is speed of light.
(iv) The velocity of electromagnetic waves in free space, c = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
(v) The velocity of electromagnetic waves in a material medium = \(\frac{1}{\sqrt{\mu \varepsilon}}\) where, p and e are absolute permeability and absolute permittivity of the material medium respectively.
(vi) Electromagnetic waves follow the principle of superposition.
(vii) Electromagnetic waves transfer energy as they propagate through space. This energy is divided equally between electric and magnetic fields.
(viii) Electromagnetic waves can transfer energy as well as momentum to objects placed on their paths.
(ix) Electromagnetic waves do not require material medium to travel.
Section – D
Question 29.
Read the passage given below and answer the following questions:
Faraday Cage:
A Faraday cage or Faraday shield is an enclosure made of a conducting material. The fields within a conductor cancel out with any external fields so the electric field within the enclosure is zero. These Faraday cages act as a big hollow conductors, you can put things in it to shield them from electrical fields. Any electrical shocks the cage receives, pass harmlessly around the outside of the cage.
(i) Which of the following material can be used to make a Faraday cage?
(a) Plastic
(b) Glass
(c) Copper
(d) Wood
Answer:
(c) Copper, as Faraday cage is made from conducting material.
(ii) Example of a real-world Faraday cage is:
(a) Car
(b) Plastic box
(c) Lighting rod
(d) Metal rod
Answer:
(a) Car, as it can save us from lightning.
(iii) What is the electrical force inside a Faraday cage when it is struck by lightning?
(a) The same as the lightning
(b) Half that of the lightning
(c) Zero
(d) A quarter of the lightning
Answer:
(c) Zero, as electrical field inside the cage is zero.
OR
An isolated point charge +q is placed inside the Faraday cage. Its surface must have charge equal to:
(a) Zero
(b) +q
(c) -q
(d) +2q
Answer:
(c) -q, charge is +q inside it. So, -q will be induced on outer surface
(iv) A point charge of 2C is placed at centre of Faraday cage in the shape of cube with surface of 9 cm edge.
The number of electric field lines passing through the cube normally will be:
(a) 1.9105 × 1011 Nm²/C entering the surface
(b) 1.9105 × 1011 Nm²/C leaving the surface
(c) 2.0105 × 1011 Nm²/C leaving the surface
(d) 2.26 × 1011 Nm²/C leaving the surface
Answer:
(d) 2.26 × 1011 Nm²/C leaving the surface
Leaving the surface 2.26 × 1011 Nm²/C
Question 30.
Read the passage given below and answer the following questions:
Lenz’s Law
This law is mostly used to find the direction current induced in a circuit. According to this law the polarity of e.m.f. induced in the circuit is such that it opposes or checks the variation in magnetic flux responsible for it. It we move north pole of a bar magnet towards coil, then magnetic flux linked with the coil changes (i.e., increases). Due to this current is induced in the coil in anticlockwise direction. The magnetic moment related with the induced current has north polarity towards the north pole. If we move away the bar magnet then magnetic moment has south polarity.
Lenz’s law obeys the law of conservation of energy.
(i) Lenz’s law is used to find the direction of:
(a) Electric field
(b) Force
(c) Induced current
(d) Electrostatic force
Answer:
(c) Induced current
(ii) Lenz law follows the law of:
(a) Conservation of force
(b) Conservation of mass
(c) Conservation of momentum
(d) Conservation of energy
Answer:
(d) Conservation of energy
(iii) What will be the polarity of magnetic moment towards bar magnet if we move the north of magnet towards coil ?
(a) North
(b) South
(c) Both (a) and (b)
(d) None of these
Answer:
(a) North
(iv) If we move away the bar magnet from coil then what will be the direction of induced current in coil?
(a) Clockwise
(b) Anticlockwise
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Clockwise
OR
As we move the north pole of magnet towards coil, the magnetic flux linked with the coil:
(a) Increases
(b) Decreases
(c) Remain same
(d) None of these
Answer:
(a) Increases
Section – E
Question 31.
(a) (i) State Biot-Savart law and express this law in the vector form.
(ii) Two identical circular coils, P and Q each of radius R, carrying currents1 A and √3 A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.
(b) Write Ampere’s circuital law. State its one application.
OR
(a) Derive the expression for the force acting between two long parallel current carrying conductors. Hence, define1 A current.
(b) A bar magnet of dipole moment 3 Am² rests with its centre on a frictionless pivot. A force F is applied at right angles to the axis of the magnet, 10 cm from the pivot. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of 30° with the field.
Calculate the value of F.
How will the equilibrium be effected if F is withdrawn?
Answer:
(a) (i) Biot-Savart Law: It states that the magnetic field strength (dB) produced due to a current element I
and length dl at a point having position vector \(\vec{r}\) relative to current element is
(1) directly proportional to the current I, i.e., dB × I.
(2) directly proportional to the length dl of the element i.e., dB ∝ dl.
(3) directly proportional to sin θ, where θ is the angle between dl and r, i.e., dB ∝ sin θ.
(4) inversely proportional to the square of the distance r from the current element
(b) The line integral of the magnetic field along any closed path is equal to po times the current enclosed by the path.
Application of Ampere’s circuital law:
Magnetic field at a point due to a long straight wire— Assume a long straight wire carrying a current i, P is point at a distance ‘R’ from the wire. B is the magnetic field at point P. As per Ampere circuital law, we have
Amperian loop at P-Here it is a circle of radius R.
Since B is constant at any point on the loop,
B × 2πR = μoi
\(\mathrm{B}=\frac{\mu_0 i}{2 \pi \mathrm{R}}\)The direction of the magnetic field is found by right hand thumb rule.
Curl the fingers of the right hand with thumb extended and hold the wire such that the thumb is along the direction of the current, then the curled fingers give the direction of the magnetic field at any point.
OR
(a) Suppose two long thin straight long conductors (or wires) PQ and RS are place parallel to each other, carrying current I1 and I2 respectively. When the current flow in same direction they experience an attractive force and when they carry current, in opposite directions, they experience repulsive force.
Consider a current element ΔL of wire RS. The magnetic field produced by current carrying conductor PQ at the location of other wire RS.
\(\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \pi r}\)According to Maxwell’s right hand rule the direction of magnetic field will be perpendicular to the plane of the people.
ΔF = B1I2 ΔL sin 90°
\(=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi r} \Delta \mathrm{L}\)Thus, 1 ampere is defined as the current which when flow in each of the parallel conductors placed at separation of 1 m in vacuum exerts a force of 2 × 10-7 N on 1 m length of either wire.
Question 32.
(a) (i) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(ii) Discuss briefly how wave theory of light cannot explain these features.
(b) Using photon picture of light, show how Einstein’s photoelectric equation can be established. Write two features of photoelectric effect which cannot be explained by wave theory.
OR
(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.
Answer:
(a) (i) 1. The photoelectric effect will not occur when the frequency of the incident light is less than the threshold frequency. Different materials have different threshold frequencies and most elements have threshold frequencies in the ultraviolet region of the electromagnetic spectrum-.
2. The maximum kinetic energy of a stream of photoelectrons increases linearly with the frequency of the incident light above the threshold frequency.
3. The rate at which photoelectrons are emitted from a photosensitive surface is directly proportional to the intensity of the incident light when the frequency is constant.
(ii) Classical wave theory cannot explain:
1. The existence of threshold frequency because it predicts that electrons would absorb enough energy to escape and there would not be any threshold frequency.
2. The almost immediate emission of photoelectrons as, according to this theory, electrons require a period of time before sufficient energy is absorbed by it to escape from the metal; however, such a thing does not happen practically.
3. The independence of KE of photoelectrons on intensity and dependence on frequency because it cannot explain why maximum KE is dependent on the frequency and independent of intensity.
(b) Einstein’s photoelectric equation: Einstein explained the various laws of photoelectric emission on the basis of Planck’s quantum theory. According to Planck’s quantum theory, light radiations consist of tiny packets of energy called ‘quanta’. One quantum of light radiation is called a photon which travels with the speed of light.
The energy of a photon is given by,
E – hv
where, h is Planck’s constant and v is the frequency of light radiation.
Einstein assumed that one photoelectron is ejected from a metal surface, if one photon of suitable light radiation falls on it.
Consider a photon of light of frequency v, incident on a photosensitive metal surface. The energy of the photon (= hv) is spent in two ways :
(1) A part of the energy of the photon is used in liberating the electron from the metal surface which is equal to the work function Φo of the metal.
(2) The rest of the energy of the photon is used in imparting the maximum kinetic energy Kmax to the emitted photoelectron.
If vmax is the maximum velocity of the emitted photoelectron and m is its mass, then maximum kinetic energy of the photoelectron is,
This equation is called Einstein’s photoelectric equation.
Features of photoelectric effect which can not be explained by wave theory:
(a) The wave theory could not explain the instantaneous process of photoelectric effect.
(b) It could not explain, why the photoelectric emission is independent of intensity.
OR
(a) Wave nature of radiation cannot explain the following:
(1) The immediate ejection of photoelectrons.
(2) The presence of threshold frequency for a metal surface.
(3) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.
(b) Photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based on particle nature of light.
Its basic features are:
(1) In interaction with matter, radiation behaves as if it is made up of particles called photons.
(2) Each photon has energy (E = hv), momentum \(\left(p=\frac{h v}{c}\right)\) where c is the speed of light.
(3) All photons oflight ofa particular frequency v, or wavelength X, have the same energy \(\left(\mathrm{E}=h \mathrm{v}=\frac{h c}{\lambda}\right)\) and momentum \(\left(p=\frac{h v}{c}\right)\)
(4) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons emitted per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
(5) Photons are electrically neutral and are not deflected by electric and magnetic fields.
(6) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, number of photons may not be observed.
Question 33.
(a) “Two independent monochromatic sources of light cannot produce a sustained interference pattern.”
Give reasons.
(b) Two coherent light waves of intensity 5 × 10-2 Wm-2 each superimpose and produce the interference pattern on a screen. At a point where the path difference between the waves is λ/6, λ being wavelength of the wave, find the
(i) phase difference between the waves.
(ii) resultant intensity at the point.
(iii) resultant intensity in terms of the intensity at the maximum.
OR
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
Answer:
(a) The condition for the sustained interference is that both the sources must be coherent (i.e., they must have the same wavelength and the same frequency and they must have the same phase or constant phase difference).
Two sources are monochromatic if they have the same frequency and wavelength. Since, they are independent, i.e., they have different phases with irregular difference, they are not coherent sources.
OR
(a) The resultant intensity in Young’s experiment is given by
\(\mathrm{I}_{\mathrm{R}}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi\)
When slit is not covered, then Io is the intensity from each slit.
Maximum intensity (Imax) occurs when Φ = 0°.
Minimum intensity (Imax) occurs when Φ = 180°.
If one slit is covered with glass to reduce its intensity by 50%, then
(b) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpendicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Further, the shortest visible wave, blue, produces a bright fringe first.