Category: CBSE

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2014 

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1. 1.  All questions are compulsory. There are 26
      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 five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
   4.  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 the three questions of five marks weight age. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:

SET I

SECTION – A

Question.1. Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.
Answer : The amount of current flowing in two parallel wire in same direction through two infinitely long parallel wire separated by one metre which produces an attractive force of 2 x 10^-7 N/m, provided the wires, have negligible circular cross section and placed in vacuum, this flowing current can be defined as “One Ampere of Current.

Question.2. To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019 Hz belong ?
Answer : The frequency 5 x 10¹⁹ Hz lies in the gamma region of the. electromagnetic spectrum.

Question.3. What is the force between two small charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in air?
Answer : Taking all the units of physical quantities in S.I. units, we have,

Question.4. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.
Answer : The intensity of radiation can be defined as the energy related with photons emitted from a unit surface area in unit time. Its S.I units is joule/metre² second (J/m²s).

Question.5. The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown.

Question.6. Why is it found experimentally difficult to detect neutrinos % in nuclear β -decay?
Answer : Neutrinos are difficult to detect experimentally in β decay because they do not have any charge with almost zero mass and also they do not interact easily with any matter.

Question.7. Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons.
Answer : The use of A.C. voltage is preferred over the use of D.C. voltage because of the following reasons :
(i) The energy losses while transmission of A.C. voltage are very less as compared to D.C. voltage.
(ii) A.C. voltage can be controlled as required by using a transformer (i. e., stepped up or stepped down).

Question.8. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer : In this case the biconvex lens will behave as a diverging lens because the refractive index of water (1.33) is more than that of the material (1.25) of the lens.

Question.9. Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
OR
Using Bohr’s postulates of the atomic model, derive the expression for radius of n^th electron orbit. Hence obtain the expression for Bohr’s radius.
Answer : The magnitude of force according to Rutherford’s model of the atom is,


Here, negative sign indicates that the revolving electron is bound to the positive nucleus.
OR
Suppose m be the mass of an electron and v_nbe its speed in n^th orbit of radius r_n . From Rutherford model, the centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.

Question.10. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Answer : Let q be the charge on the charged capacitor. Energy stored in it is,

Question.11. Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s Circuital law to include the term due to displacement current.

The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and displacement current.

Question.12. A cell of emf “E’ and internal resistance V is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined.

Question.13. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier.
Answer : The circuit diagram for a p-n junction diode as a half wave rectifier is shown below :

Working : During the positive half cycle of the input a.c., the p-n junction is forward biased i.e., the forward current flows from p to n. In the forward biasing, the diode provides a very low resistance and allows the current to flow. Thus we get output across-load i.e. a.c. input will be obtained as d.c. output.
During the negative half cycle of the input a.c., the p-n junction is reversed biased i.e., the reverse current flows from n to p. In the reversed biasing, the diode provides a high resistance and hence a very small amount of current will flow through the diode which is of negligible amount. Thus no output is obtained across the load. The wave form of input and output is shown below :

Question.14. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10^-7 m² carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 10²⁸ m^-3.

Question.15. 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.

Ray ‘1’ and ‘2’ will fall on the side AC at an angle of incidence (i) if 45°. Critical angle of ray T ’ 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.16. Write the functions of the following in communication systems:
(i) Transducer
(ii) Repeater
Answer : (i) Transducer : A transducer is used to convert one form of energy to another form.
(ii) Repeater : A repeater is used to receive a signal from a transmitter, amplifies it and retransmits it to the receiver.

Question.17. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature.
Answer : (i) The behaviour of magnetic field lines in the presence of a paramagnetic substance is shown below :

(ii) The behaviour of magnetic field lines in the presence of a diamagnetic substance is shown below :

This distinguishing feature is because the relative permeability of the diamagnetic substance is less than 1, so the magnetic lines of force do not prefer passing through the substance and the relative permeability of a paramagnetic substance is greater than 1, so the magnetic lines of force prefer passing through the substance.

Question.18. Draw a circuit diagram n-p-n transistor amplifier, CE configuration. Under what condition does the transistor act as an amplifier?
Answer : The circuit diagram of an n-p-n transistor .amplifier in CE configuration is given below :

Condition : For a transistor to act as a amplifier, it must be operated close be to the centre of its active region.
The transistor behaves as an amplifier when the input circuit (emitter —base) is forward biased having low voltage VBB and the output circuit (collector-base) is reverse biased having high voltage Vcc.

Question.19. (a) Using the phenomenon of polarization, show how transverse nature of light can be demonstrated. (b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity IQ is incident on P₁. A third Polaroid P₃  is kept in between P₁  and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃ .
Answer : (a) Suppose that an ordinary light is incident normally on a pair of crystals C₁  and C₂ . When the incident ray of light passes through crystal C₁ , it gets plane polarised in the direction perpendicular-to the length of crystal. Now, we see that when the axes of two crystals are parallel, the intensity of the emerging light will be maximum. When the second crystal is placed perpendicular with respect to the first crystal, the intensity of light observed is zero. This is due to the electric field of the plane polarised light obtained from C₁  can vibrate only in one direction. Hence, when the axis of the crystal is perpendicular to its direction of vibration of electric field, it gets blocked. This shows the transverse nature of light.

(b) As given in the question, the polaroids P₁  and P₂ are placed with their axes perpendicular to each other. Also, polaroid P₃ . is placed at an angle of 30° with respect to P₁ .

Question.20. Define the term ‘mutual inductance’ between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r₁  and r₂ (r₁  >> r₂ ). Total number of turns in the two solenoids are N₁  and N₂ respectively.
Answer : The ratio of magnetic flux passing through one coil to the current passing through the other is known as coefficient of mutual inductance between the two coils.

Question.21. Answer the following:
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bride wire?
(c) Which material is used for the meter bridge wire and why?
OR
A resistance of R Ω draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance Ro Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Answer : (a) The connection between the resistors in a meter bridge is made of thick copper strips because the resistivity of a copper wire is very low. As, the connections are thick, so the area becomes large and the resistance of the wires becomes almost negligible.
(b) It is preferred to obtain the balance point in the middle of the meter bridge wire because it improves the sensitivity of the meter bridge.
(c) Constant an nichrome/Manganine is used for meter bridge wire because its temperature coefficient of resistance is almost negligible due to which the resistance of the wire does not get affected on increasing temperature of the wire during flow of current.
OR
While the slide is in the middle of the potentiometer only half of its resistance (Ro/2) will be between the points A and B. Hence, the total resistance between A and B, say. R1 will be given by the following expression :

Question.22. A convex lens of focal length 20 cm is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
Answer : Let us first locate the image of the point object S formed by the convex lens.
Here : u = -60 cm and f= 20 cm
From the lens formula, we have :

The image I1 is formed behind the mirror and acts as a virtual source for the mirror. The convex mirror forms the image H, whose distance from the mirror can be determined as :

Question.23. A voltage V = V₀  sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit ?

The average power over a cycle is average of the two terms on the R.H.S. of the above equation. The second term is time dependent, so, its average is zero.

Question.24. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
Answer: Conductors :
(i) In case of conductors, the valence band is completely filled and the conduction band can have two cases-either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other as shown below :

(ii) Even when a small current is applied, conductors can conduct electricity.
Insulators :
(i) In case of insulators, the energy gap between the conduction and valence bands is very large. And the conduction band is practically empty.

(ii) When an electric field is applied to such kind of material, the electrons find hard to receive such a large amount of energy to attain the conduction band. Thus, the conduction band remains to be empty. That is why no current flows through insulators.
Semiconductors :
(i) In case of semiconductor, the energy band structure of semiconductors is similar to insulators, but in this case, the size of forbidden energy gap is quite smaller than that of the insulators.

(ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it relatively easier to jump to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.

Question.25. For the past some time, Aarti had been observing some erratic body movement, unsteadiness and lack of coordination in the activities of her sister Radha, who also used,to complain of severe headache occasionally. Aarti suggested to her parents to get a (a) What, according to you, are the values displayed by Aarti?
(b) How can radioisotopes help a doctor to diagnose brain tumour?
Answer : (a) Aarti has showed awareness and responsibility towards her sister.
(b) A little amount radio-isotope like radioiodine is inserted into the body along with organic dyes which are absorbed strongly by the tumour tissue than the normal tissues. By detecting the emitted radiation, the radiologist get information about the size and location of the tumour.

Question.26. Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.
Answer : The two basic modes of communication are :
(1) Point-to-point communication.
(2) Broadcast communication
Amplitude Modulation: Amplitude modulation is obtained by changing the amplitude of the carrier waves according to the amplitude of the modulating wave.
Amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.

Question.27. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope Compare with what of an optical microscope which uses yellow light ?
Answer : The de-Broglie wavelength of the electrons is given by:

This formula shows that to enhance resolution, we have to use minimized wavelength and media with large indices of refraction.
For an electron microscope, p is equal to 1 (vacuum).
For an electron microscope, the electrons are accelerated through a 60,000 V potential difference.

As, λ is very little (roughly 10^-5 times smaller) for electron microscope than an optical microscope which use yellow light of wavelength (5700 A to 5900 A). So, the resolving power of electron microscope is about 10⁵ greater than that of optical microscope.

Question.28. Draw a labelled diagram of Van de Graff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations.
OR

(b) Consider two hollow concentric spheres, S₁  and S₂ , enclosing charges 2Q and 4Q respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere Si change if a medium of dielectric constant ‘ε_r ’ ‘ε_r ’ is introduced in the space inside S₁ in place of air? Deduce the necessary expression.

Answer : Van de Graff generator is a device used for building up high potential differences of the order of a few million volts. Principle : The working of Vande Graff Graff generator is based on the following two electrostatic phenomena.
(i) Discharging action at sharp points (corona discharge i.e., electric discharge takes place in air or gases readily at the pointed ends of conductors.
(ii) If a charged conductor is brought into internal contact with a hollow conductor, all of its charge transfers to the hollow conductor, howsoever high the potential of the latter may be.
Construction : It has a big spherical conducting shell (S) kept over insulating pillars. A long narrow insulating belt is wound around 2 pulleys’ P₁  and P₂ . B₁ and B₂ are 2 metal combs with sharp points. B₁ is known as spray comb and B₂ collecting comb.
Working : The spray comb provides positive potential by high tension source. The positive charge is sprayed on belt. As belt moves and touches the sphere, a negative charge is induced on the sharp ends of collecting comb B₂ and similar positive charge is induced on the further end of B₂ . This positive charge moves immediately to the outer surface of S. because of discharging action of sharp points of B₂ , the positive charge on the belt is neutralized. The uncharged belt moves downwards and collects the positive charge from B₁ , which is then collected by B₂ . This process is repeated and the positive charge on S goes on accumulating. In this way, voltage differences of as much as 6 or 8 million volts (with respect to the ground) can be created.
Use : Van de Graff generator used to create high potential differences that are used to accelerate charged particles such as electrons, protons, ions, etc. used for nuclear breakdown.
Limitations :
1. It is a series of combination that allows only one way for moving charge.
2. It can accelerate only the charged particles and not the uncharged particles.

Question.29. (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Youngs double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
OR
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 10^-16 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.
Answer : (a) In Youngs double slit experiment, the wave fronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light is maximum at the centre C of the screen and it is called central maxima.

Let S₁  and S₂ be two slits separated by a distance d. GG’ is the screen at a distance D from the slits S₁ and S₂ . Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits S₁  and S₂ is S₂ P — S₁P.

OR
(a) The phenomenon of bending of light round the sharp corners of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.

Consider a parallel light beam from a lens is incident on slit AB. As diffraction happens, the pattern is focussed on screen XY with help of lens L₂ . We will get a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes known as secondary maxima and minima.
Central Maximum: Each point on the plane wave front AB sends secondary wavelets in all directions: The waves from points equidistant from the centre C kept on the upper and lower half reach point O with zero path difference and so, reinforce each other, making maximum intensity at point O.
Positions and Widths of Secondary Maxima and Minima
Consider a point P on screen at which wavelets moving in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference similar to BN :

Question.30. (a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.
OR
(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.
(b) Answer the following :
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer.
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.



Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.
(b) The working principle of a cyclotron is that a charge particle can be accelerated to high energy by an oscillating electric field. A cyclotron uses an electric field to accelerate charge particles across the gap between the two D-shaped magnetic field regions. The magnetic field is perpendicular to the paths of the charged particles that makes them follow in circular paths within the two Ds. Each time the charged particles cross the Ds, it is accelerated by an alternating voltage. As its speed increases the radius of path of each particle also increases. So, the accelerated particles move in a spiral path to the other wall of the cyclotron.

Square wave electric fields are used to accelerate the charged particles in a cyclotron.

At the time the charge particle finishes its half circle, the accelerating electric field reverse so that it gets accelerated across the gap between the Ds.
The particle gets accelerated again and again, and its velocity increases. Therefore, a high kinetic energy is achieved.
OR
(a) Moving coil galvanometer : It is a device used for detecting and measuring small electric current.

Principle : The working is based upon the principle, when a current carrying coil suspended in a magnetic field experiences a torque.
Construction : It is consists of a coil with a large number of turns of insulated copper wire wounded on a metallic frame.
The coil is suspended by means of a phosphor bronze strip and a horse shoe magnet NS surrounds it. The lower end of the coil is attached with a hair spring. The scale of the pointer is attached to the other end of the spring.
Working : When current is passed, the couple acts on it. Since the plane remains parallel to the magnetic field in all position of the coil, the force on the vertical arms always remains perpendicular to the place of the coil.

(b) (i) When the cylindrical soft iron core placed inside the coil of a galvanometer, the magnetic field gets stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil, the magnetic field is always parallel to its plane.
(ii) Current Sensitivity : It is defined as the small deflection produced in the galvanometer when a unit current is passed through it.

SET II

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. A conducting loop is held above a current carrying wire ‘PQ’ as shown in the figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing.

Question.4. Why do the electrostatic field lines not form closed loops ?
Answer : Electrostatic field lines do not form closed loops because electric field is conservative in nature.

Question.5. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer: The refractive index of material of lens (1.5) is greater than the refractive index of water (1.33). So, it will behave as a converging lens.

Question.7. To which part of the electromagnetic spectrum does a wave of frequency 3 x 10¹³ Hz belong ?
Answer : The frequency 3 x 10¹³ Hz belongs to infrared, region of electromagnetic spectrum.

Question.9. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10^-17m² carrying a current of 1.8. A. Assume the density of conduction electrons to be 9 x 10²⁸m^-3.

Question.13. Write the functions of the following in communication systems:
(i) Transmitter
(ii) Modulator
Answer: (i) Transmitter: A transmitter converts the message signal from the source of information into a form suitable for transmission through a channel.
(ii) Modulator: Modulator : is a device which changes some characteristics like amplitude/frequency/phase angle of a high frequency carrier wave in accordance with the instantaneous value of the low frequency message signal through the appropriate superposition.

Question.21. (a) Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering.

Explanation : The dots in Diagram show Vibrations Perpendicularte to the plane of paper and double arrow show vibrations in the plane of paper. The elctrons in the molecule define to vibrate in both of these directions. The electrons vibrating parallel to the double arrows cannot send energy toward an observer looking at 90° to the direction of sun because their acceleration has no transverse component. The light scattered by the molecules in this direction has only dots. It is polarised perpendicular to the plane of paper.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P₁. A third Polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 45° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃.
Answer : (a) Unpolarized light scattering from air molecules shakes their electrons perpendicular to the direction of the original ray. The scattered light therefore, has a polarization perpendicular to the original direction and none parallel to the original direction.
(b) As given in the question, the polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Also, P₃ is placed at an angle of 45° with respect of P₁.
Now, we have :
Intensity of light after falling on P₁, I’ = I₀ /2

Question.22.Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.
Answer : The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by

Energy stored in an inductor : When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t. The current rises at the rate dl/dt.
So the induced e.m.f. is

Question.24. A convex lens of focal length 20 cm is placed co axially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
Answer : The beam incident on lens L is parallel to principal axis. Hence the lens forms an image I₁ at its focus, i. e., at a distance OI₁ (= 20 cm) from the lens.

The image I₁ is formed in front of the mirror and hence, acts as a real source for the mirror. The concave mirror forms the image I₂, whose distance from the mirror can be calculated as;

SET III

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. A conducting loop is held below a current carrying wire PQ as shown. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing.

Question.2. The graph shows variation of stopping potential V₀ versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why ?

Answer : Metal A has higher threshold energy because from the graph it is clear that the minimum frequency required to start photo emission is more in A than that of B.

Question.5. Why do the electric field lines never cross each other ?
Answer : At any point, if electric field lines cross each other it means at that point there are two directions of electric field, which is impossible.

Question.6. To which part of the electromagnetic spectrum does a wave of frequency 5 x 10¹¹ Hz belong ?
Answer : A wave of frequency 5 x 10¹¹ Hz will belong to the microwaves of electromagnetic spectrum.

Question.10. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10^-17 m² carrying a current of 2.7 A. Assume the density of conduction electrons to be 9 x 10¹⁸m^-3.

Question.18. Write the functions of the following in communication systems:
(i) Receiver
(ii) Demodulator
Answer:
(i) Receiver : The receiver functions on the received signal collected from transmitted signal at the channel output and reproduce the original message signal for sending.
(ii) Demodulator : Demodulator is a device which recovers the original information signal from the modulated wave at the receiver end.

Question.19. A convex lens of focal length 20 cm is placed co axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw a ray diagram to show the formation.

The positive sign describes that the image is. formed to the right of the lens.

The image I₁ is formed behind the mirror and thus acts as a virtual source for the mirror. The convex mirror forms the image I₂, whose distance from the mirror is given by :

Question.25. (a) A rod of length l is moved horizontally with a uniform velocity V in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.


After this, resultant force on the free electrons of the wire PQ becomes zero. The potential difference between the ends Q and P is given by, ‘
V = El = vBl
Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf, e = Bvl
(b) Lorentz force acting on a charge q which is moving with a speed v in a (normal) uniform magnetic field B, is Bqv.
All the charges will experience the same force.

Question.26. (a) Show, giving via suitable diagram, how unpolarized light can be polarised by reflection.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each  other. Unpolarised light of intensity Io is incident on P₁. A third Polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 60° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃.
Answer: (a) On reflection from a transparent medium a normal light beam becomes partially polarised. As the angle of incidence gets higher, the degree of polarization gets higher.

The reflected light beam becomes fully polarised at a certain  value. This angle of incidence is known as polarizing angle (p). At the interface of a refracting medium when light is incident at polarizing angle, the refractive index of the medium is similar to the tangent of the polarizing angle.

(b) As given, the Polaroid P₁ and P₂ are placed with their axes perpendicular to each other and Polaroid P₃ placed at 60° with respect to P₁.

CBSE Class 9 Maths Lab Manual – Equal Division of a Line Segment

Objective
To divide a line segment into equal parts.

Prerequisite Knowledge
Intercept Theorem: If there are three (or more) parallel lines and intercepts made by them on any transversal are equal, then intercepts made by them on any other transversal are also equal.

Materials Required
A sheet of ruled paper in which all the lines are parallel and equidistant, tracing paper or transparent sheet, geometry box.

Procedure

1. Take a piece of tracing paper sheet and draw a line segment of length 10 cm or 9 cm (let line segment is to be divided into 7 equal parts). Name it as PQ.
   
2. Now, take a ruled sheet, and mark the lines as 0, 1, 2, 3, 4…10 from top to bottom.
3. Place the paper strip (tracing paper) on the ruled sheet such that the point P coincides as with the line (ruled sheet) marked 0, and other point Q coincides with the line marked 7 as shown in fig.(i).
4. Mark the points on PQ which are the points of intersection with lines of ruled sheet as A₁, A₂, A₃,…A₆.
5. Now remove the ruled sheet.

Observation
7 equal parts will be obtained.
PA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆Q.
Here, marked axis PQ is a transversal on parallel lines of ruled sheet with equal intercept. Therefore, PQ also has equal intercepts (By intercept theorem).

Result
Line PQ is divided into 7 equal parts.

Learning Outcome
By paper activity, a line segment of any measurement can be divided into equal parts.

Activity Time

1. Draw a line segment of 7 cm, and divide it into 6 equal parts and draw perpendicular on the 4th point of line segment, by paper folding.
2. Take any rectangular sheet of 5 cm by 9 cm and divide the edges of sheet in 8 equal parts.
3. Take any rectangular sheet of 10 cm by 7 cm and divide along length in 10 equal parts and along the breadth in 7 equal parts (By paper folding).

Viva Voce

Question 1.
What do you mean by parallel lines ?
Answer:
Two distinct lines which are not intersecting are called as parallel lines.

Question 2.
What is transversal line ?
Answer:
A line intersecting two or more than two distinct lines at distinct points is called as transversal line.

Question 3.
What is intercept?
Answer:
If a transversal line intersects two parallel lines, the line segment intersected between the parallel lines on the transversal is called intercept.

Question 4.
How will you state intercept theorem ?
Answer:
If there are three (or more) parallel lines and the intercepts made by them on one transversal are equal, the intercepts on any other transversal are also equal.

Question 5.
What is line segment ?
Answer:
Line segment is a part of a line.

Question 6.
What do you mean by fractions ?
Answer:
Equal parts of a whole are called fractions.

Question 7.
What do you mean by equivalent fractions ?
Answer:
Two or more fractions representing the same part of the whole are known as equivalent fractions.

Question 8.
How will you bisect a rope without measuring it ?
Answer:
By folding rope and joining its end points, two equal parts of a rope will be obtained.

Question 9.
To divide a thread in 8 equal parts, how many successive folds will you give ?
Answer:
7 folds.

Question 10.
Explain the trisection of a strip.
Answer:
A strip is divided into three equal parts.

Question 11.
Explain the six equal divisions of a rope.
Answer:
If a rope is divided into six equal parts then each part is one-sixth of rope.

Question 12.
If the intercept made by a transversal between three parallel lines is in ratio 1:2, what intercept made by another transversal?
Answer:
Same ratio (1:2).

Multiple Choice Questions

Question 1.
If AB = 10 cm, is divided into 7 equal parts then what will be the value of each part ?
(i) \(\frac { 1 }{ 7 }\) cm
(ii) \(\frac { 10 }{ 7 }\) cm
(iii) \(\frac { 3 }{ 7 }\) cm
(iv) none of these

Question 2.
If ratio of intercepts made by a transversal is \(\frac { 2 }{ 3 }\) then what will be the ratio of intercepts made by another transversal ?
(i) \(\frac { 3 }{ 2 }\)
(ii) \(\frac { 2 }{ 3 }\)
(iii) \(\frac { 1 }{ 3 }\)
(iv) none of these

Question 3.
The length of a line segment is 12 cm, what is the length of each part, if it is divided into 10 equal parts ?
(i) 1.2 cm
(ii) 2.2 cm
(iii) \(\frac { 10 }{ 12 }\) cm
(iv) none of these

Question 4.
If PQ = 9 cm, is divided into three equal parts, what is ratio of first two parts to the remaining part ?
(i) 2:1
(ii) 1:2
(iii) 3 :2
(iv) none of these

Question 5.
Which theorem is used to get the equal division of a line segment ?
(i) Pythagoras theorem
(ii) mid-point theorem
(iii) intercept theorem
(iv) none of these.

Question 6.
What do you mean by transversal line ? A line which intersects:
(i) one line
(ii) two or more lines
(iii) angle
(iv) none of these

Question 7.
Write first two equivalent fractions of \(\frac { 5 }{ 2 }\)
(i) \(\frac { 5 }{ 2 } ,\frac { 10 }{ 2 }\)
(ii) \(\frac { 10 }{ 4 } ,\frac { 15 }{ 6 }\)
(iii) \(\frac { 2 }{ 5 } ,\frac { 4 }{ 10 }\)
(iv) none of these

Question 8.
If \(\frac { 5 }{ 7 } =\frac { x }{ 35 }\) what is the value of x?
(i) 25
(ii) 10
(iii) 1
(iv) none of these

Question 9.
Find x, if \(\frac { 21 }{ 9 } =\frac { 7 }{ x }\)
(i) 3
(ii) \(\frac { 1 }{ 3 }\)
(iii) 21
(iv) none of these

Question 10.
In the given figure, AB = BC = 4 cm, find DE, if EF = 5 cm

(i) 5 cm
(ii) 6 cm
(iii) 2 cm
(iv) none of these

Answers

1. (i)
2. (ii)
3. (i)
4. (i)
5. (iii)
6. (ii)
7. (ii)
8. (i)
9. (i)
10. (i)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Previous Year Solved  Papers  Class 12 Physical Education Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  The question paper consists of 26 questions.
2. All question are compulsory.
3. Answer to questions carrying 1 mark should be in approximately 10-20 words.
4. Answer to questions carrying 3 marks should be in approximately 30-50 words ,
5.  Answer to questions carrying 5 marks should be in approximately 75-100 words.

Question.1. Suggest any four ways through which women participation in sport across age group can be enhanced.
Answer :

1. Modification in Legislation.
2. Better coverage of women sports
3. Improvement in fitness and wellness movements –
4. Educating Women

Question.2. Trekking is a long adventurous journey undertaken on foot in areas where common means of transport are generally not available. Name any four important materials required that should be carried along.
Answer :

1. First Aid Box
2.  Sleeping Bags
3.  Pair of Shoes And Socks
4.  Rope

Question.3. Enlist two objectives of Intramurals.
Answer :

1. Develops moral and ethical values of the students.
2.  Provide opportunity to the maximum number of students to participate in sports.

Question.4. Enlist two sources for calcium and iron separately
Answer : Calcium sources : Cheese, Milk, Orange Juice, Eggs, Yogurt. Iron sources : Liver, meat, orange juice, egg.

Question.5. Explain correct sitting posture.
Answer : When we sit in a chair, our hips should be as far as back in the chair as possible. Head, spinal column, Shoulder and hips should be in straight line and erect. Legs should touch the ground and not in hanged position. Thighs should be in horizontal position.

Question.6. Calculate the physical fitness index using short formula for a 12 year old boy having completed Harvard Step Test for a duration of 3 minutes and a pulse rate of 54 beats for 1 to 1.5 minute.
Answer : The athlete’s fitness index score is calculated with the help of following formula. ‘
Fitness index score = (100 x test duration in seconds) divided by (2 x sum of heart beats in recovery period). (100 x 180 Sec)/ 2 x 54 = 500/3 = 166.66

Question.7. Your grandmother feels she has reduced her upper body flexibility and therefore she wants to test herself. Which test would you suggest her?
Answer : Back scratch test for upper body flexibility.

Question.8. Explain the term Hypertrophy of muscles.
Answer : Increase in size of the muscle fiber due to regular exercises or Hypertrophyis enlargement of heart due to regular exercises which is called “Athletic heart”.

Question.9. What do you understand by linear movement?
Answer : Linear movement refers to any movement along a straight line in one direction.

Question.10. Explain the term “realistic” in goal setting principles.
Answer : Realistic goads are achievable goals.

Question.11. Suggest any two isometric exercises for shoulder region.
Answer :

1.  Pushing against the wall.
2.  Holding push up position.

Question.12. What safety measures children should be taught while participating in trekking?
Answer :

1.  Avoid trekking during bad weather conditions.
2. To prevent insect bite do wear full sleeves shirts and full pants.
3.  Wear proper footwear so that you don’t slip while trekking.
4.  Don’t eat leaves, flowers etc. while trekking, they may be poisonous.

Question.13. Briefly explain the functions and resources of three fat soluble vitamins.
Answer : Fat Soluble vitamins are A, D, E, K.
Functions :

1.  Vitamin ‘D’ : The main function of vitamin D is to regulate the absorption of calcium and phosphorus in our bones and aid in cell-to-cell communication throughout the body.
2. Vitamin ‘E’ : Vitamin E (tocopherol) is a powerful, fat-
   soluble antioxidant that helps to protect cell membranes against the damage caused by free radicals, prevents the oxidation of LDL cholesterol and is also important in the formation of red blood cells (RBC), thus essential for blood coagulation.
3. Vitamin ‘K’ : Vitamin K occurs in two forms-Vitamin K] and Vitamin K2. Vitamin K is known as the clotting vitamin, because without it, blood would not clot. It also helps in prevention of haemorrhage.
   Sources of Vitamin A : Ghee, milk, curd, egg yolk, fish, tomato, papaya, spinach, carrot, pumpkin etc.
   Sources of Vitamin D : Egg yolk, fish, sunlight. Vegetables, cod liver oil, milk, cream, butter, tomato,
   carrot etc.
   Sources of Vitamin E : Green vegetables, kidney, liver, heart cotton seed, sprouts seeds, coconut oil.
   Sources of Vitamin K : Cauliflower, spinach, cabbage, tomatoes, wheat, egg and meat etc.

Question.14. Neeti along with her father was regular at district park in early morning. She realized that most of the children are obese. She along with her few classmates wanted to help those children. She discussed with her physical education teacher and the principal of the school. School decided to organize awareness rally for the neighbourhood.

1.  How obesity can be prevented? Give two ways.
2. Give any two disadvantages of obesity.
3.  What values are shown by Neeti and her
   classmates?

Answer :

1. Obesity can be prevented by :
   1. Avoid fast food, fatty food and over eating
   2. Regular exercise/ physical activity
2. Disadvantages of obesity :
   1. More chances of injury
   2. More disease/ physical health problems
3. Values shown:
   1. Good moral character
   2. Self-discipline
   3. Decisiveness
   4. Logical and decision maker

Question.15. Briefly explain the six physical benefits of exercise to children.
Answer :

1.  Exercise helps in healthy growth and development : Encouraging healthy lifestyles in children and adolescents is important for when they grow older. Participating in organized sports and games is not only of great fun but is very essential for healthy growth and development.
2.  Exercises improve self-esteem : Self-esteem can play a great role in how children feel about themselves and also how much they enjoy things or worry about things. Exercise reduces depressive symptoms and improves self-esteem in children.
3.  Exercise reduces blood sugar level : Lowering blood sugar level is not only the health issue of adults but also for children as they are also becoming trapped with high sugar level. Exercise is helpful in preventing sugar from accumulating in the blood by triggering muscles to receive more glucose from blood and utilize it for energy.
4.  Exercise helps in motor development : Exercise helps in increasing the motor development of infants and children at a very rapid pace which in due course improves in building fine movements later on in life. The movements of muscles become efficient and smooth by doing exercise in premature age; their movements turn out to be attractive.
5.  Exercise strengthens the heart : Exercise helps improve heart health, and can even reverse some heart disease risk factors. A child who exercises often, has the lowest risk for heart disease, but any amount of exercise is beneficial.
6.  Exercise makes stronger bones, muscles and joints : Exercise is vital for strong muscles, bones and joints. Exercise may help to children lower their risk of chronic pain related to muscles, bones and joints in the future. Exercise increases bone density which helps to make bones stronger.

Question.16. Explain the procedure for conducting Kraus- Weber test for measuring minimum muscular strength.
Answer : Kraus -Weber Test. This test consists of six items. It is commonly known as the Kraus-Weber Tests. These tests are supposed to measure the minimum muscular fitness of an individual. Infact, they measure a level of strength and flexibility of certain key muscle groups below which the functioning of whole body as a healthy individual seems to be endangered. These tests are graded on a pass-fail basis. But partial movements on each test can be scored from 0 to 10.
Six tests measures minimum muscular fitness of an individual
Test No.1. The subject lies down in supine position i.e., flat on his back and hands behind his neck. The examiner holds his feet to keep him on the ground. The subject is asked to perform on sit- up. If he performs one sit- up, he passes this test. If he cannot raise his shoulders from Ehe table or ground, his score remains zero.
Test No.2. The lying position for this test remains same i.e., in supine position except that his knees are bent and ankles
remains in touch with his buttocks. He is asked to perform one sit- up. If he is able to perform full sit- up, he passes this test. If he is unable to raise his shoulders from the table or ground, he gets zero.
Test No. 3. Subject lies in supine’position i.e., lies flat,on his back with his hands behind the neck. He is asked to raise his feet 10 inches from the ground. His knees should be straight. The examiner counts to 10 seconds. He passes this test if he holds that position for ten seconds. Scoring from 0-10 depends on the number of seconds he holds the appropriate position.
Test no. 4. Subject lies in prone position i.e., on his stomach with a pillow under his lower abdomen and his hands behind his neck. The examiner holds his feet down. The subject is asked to raise his chest, head and shoulders, while the examiner counts to 10 seconds. He passes the test if he is able to hold the exact position up to 10 seconds. Scoring from 0-10 depends on the number of seconds he holds the exact position.
Test No. 5. The subject’s position remains the same, but the examiner holds his chest down. The subject is asked to raise his feet. His knees should be straight. The examiner counts to 10 seconds. Scoring from 0-10 depends on the number of seconds he holds the position.
Test No.6. It is also known as floor- touch test. It measures the flexibility of trunk. The subject stands erect, bare footed, hands at sides and feet together. He is asked to lean down slowly to touch the floor with fingertips for 3 seconds. In this test bouncing or jerking is not allowed. The examiner holds his knees in order to prevent any bend; if it occurs. Scoring from 0-10 depends on the number of seconds he holds the position.

Question.17. Maintaining physical activities for a longer period, brings desirous changes in circulatory system. Justify your answer by highlighting three benefits of exercise.
Answer :

1. Heart size increases : The size of heart and strength of the cardiac muscles increases due to regular exercises as to the maximum extent the left ventricle adapts. The walls of the heart develop into stronger and thicker as shown in recent studies and the thickness of myocardial wall increases as well.
2.  Resting heart rate decreases : The resting heart rate decreases due to regular exercises. After duration of 10 week training programme, the resting heart rate may reduce up to 10 beats per minute from the normal of 72 beats per minute. The heart becomes more efficient due to regular exercises. In highly conditioned athletes the resting heart rate decreases to 30 beats/minute..
3.  Stroke volume increases at resting conditions : The stroke volume increases at resting conditions due to regular exercises. The stroke volume at rest remains up to 50-70 ml/ beat in untrained individuals; in trained individuals it ranges from 70-90 ml/beat and in the elite endurance athletes it ranges from 90-110 ml/beat.

Question.18. What is the difference between linear and angular ‘ motion? Explain through example.
Answer : Linear motion is any motion that moves along a straight line in one direction. The direction can either be horizontal, vertical or inclined direction. Example, approach run.
Angular motion is rotatory motion, it occurs when all points on a body or object move in a circular path about the same fixed central line or axis. A child swings and rotations in i hammer throw are the best example.

Question.19. What do you understand by relative strength? Explain ? the importance of body weight in determining relative
strength.
Answer: Relative strength is strength in relation to your body
weight. Relative strength have a determining importance in sports in which the athlete shifts his body in space without any additional external weight. (H/J and L/J) as well as in sports in which he has to restrict his own weight within the framework of weight division (e.g. boxing, wrestling, weight f lifting etc.) e.g. if 1 RM (repetition maximum) is 50 kgs and body wt. is 50 kg and if 1 RM is 70 kg and body wt. is 50 kg, in second category the relative strength is more.

Question.21.Being sports captain of the school, prepare five important committees with their responsibilities to conduct one day run for health race.
Answer :

1.  Publicity committee : The committee for publicity informs and announces the sports events, dates and venues in advance to the institutions, departments and public through newspapers, media, e-mail and website etc. Their main responsibility is to advertise the sports events,
2.  Boarding and lodging committee : The main responsibility of this committee is ensuring and making : needful arrangements for providing accommodation and
   serving food to all the officials and sports persons.
3.  Transportation committee : This committee is mainly responsible for providing the facilities regarding transportation of various teams to the venue of sports events or to the place of boarding and lodging. Their main duty is to make arrangements of transportation.
4.  Grounds and equipment’s committee : This committee comes under the technical committee which mainly deals with the technical area of ground marking and layouts of track arid field etc. This committee also makes necessary arrangements of equipment’s related to the game and athletic meet.
5. Official committee : This committee selects various officials such as referees, judges, and various qualified officials as per requirements of the particular sport.

Question.22. What are the important functions of our skeletal system?
Answer :
The adult human skeletal system consists of 206 bones, as well as a network of tendons, ligaments and cartilage that connects them. The skeletal system performs vital functions — support, movement, protection, blood cell production, calcium storage and endocrine regulation — that enable us to survive.

Question.23. Explain Sheldon’s classification of personality and explain
its importance in physical education and sports.
Answer :

1.  Endomorphic : An Endomorphic somatotype is also known as a viscerotonic. The characteristic traits of this somatotype usually includes being relaxed, tolerarit, comfortable, and sociable. Psychologically, they are also fun- loving, good humored, even-tempered, and they love food and affection. The Endomorph is physically “round”. They have wide hips and narrow shoulders that give a pear-shape They tend to have a lot of extra fat on their body and on their arms and thighs. They have skinny ankles and wrists
   that make the rest of their body look even bigger.
2. Ectomorphic : An ectomorph is the complete opposite of the Endomorph. Physically, they have narrow shoulders, thin legs and arms, little fat on the body, a narrow face and a narrow chest. They may eat just as much as the endomorph but never seem to gain any weight. They always stay skinny. Personality wise, they tend to be self- conscious, socially anxious, artistic, thoughtful, quiet, and private. They always keep to themselves and are a afraid to branch out.
3.  Mesomorphic : The mesomorph is in between the endomorph and thin ectomorph. They  have an attractive and desirable body. Physically, they tend to have a large head and broad shoulders with a narrow waist. They have a strong muscular body and strong arms and legs and little fat on the body. They work for the body they have so that they could have an attractive body. Psychologically, the mesomorph is adventurous and courageous. They are not afraid to break out and do new things with new people. They are assertive.

Question.24. What is movement speed? Explain the methods to develop speed endurance.
Answer : Movement speed is the time taken between the initiation of movement and the completion of the movement. It depends upon techniques, explosive strength, flexibility and coordinative abilities. It plays a vital role in boxing, gymnastics, swimming; throws and jumps etc. Where the minimum time is taken to complete the movement.
To develop the speed endurance we will have to work more on pace races because pace races means running the whole distance at a constant speed.Generally, 800 meters and above races are included in pace races. As a matter of fact, an athlete’s can run a distance of 300 meters at full speed but, in longer races such as 800 meters or above races he must conserve his energy by reducing his speed. For example, if there is a runner of 800 meter race his best time is 1 minute 40 second, so, he should run first 400 m in 49 seconds and next 400m in 51 seconds.

Question.25. Diet for sportspersons are important. What should be the aims of preparing diet for sportsperson?
Answer : Aims of preparing diet for sports person :

1. Maintaining body weight and body composition desired for that specific sport
2. Maintaining adequate pool of nutrient levels in the body
3. Adopting healthy nutritional practices during training and competition.
4. Carrying on with healthy nutritional practices during off season as well i.e when competition are not taking place.

Question.26. “Involvement in physical activities for longer period of time with moderate intensity can improve the quality of life.” Justify your answer.
Answer :

1.  Exercise helps in healthy growth and develop-ment : Exercise is an important part of keeping children healthy. Encouraging- healthy lifestyles in children and adolescents is important for when they grow older. Participating in organized sports and games is not only of great fun but is very essential for healthy growth and’ development.
2.  Exercises improve self-esteem : Exercise is necessary for your physical and mental health. Self-esteem can play a great role in how children feel about themselves and also how much they enjoy things or worry about things. Exercise reduces depressive symptSms and improves self-esteem in children.
3. Enhances flexibility : The stiffness of joints decreases due to exercising, in a way improving the flexibility. The elasticity of tendons, figments and joint capsules improves due to regular exercise.
4. Lessens stress and tension : Regular exercise has a distinctive capability to slow down the depression process by-reducing stress and tension. Actually, regular exercise lessens the levels of bodys stress hormones like adrenaline! and cortisol. The body’s natural painkillers and mood elevators i.e. the endorphins are produced due to regular exercise. These benefits of work out facilitate in delaying the process of ageing.
5. Connect with others : Spend time with positive people who enhance your life. A strong support system will buffer you from the negative effects of stress.
6. Keep your sense of humor : This includes the ability to laugh at you. The act of laughing helps your body fight stress in a number of ways.

CBSE Class 9 Maths Lab Manual – Angle at Centre is Double the Angle Subtended by Same Arc at Any Point on Circumference of Circle

Objective
To verify that the angle subtended by an arc at the centre of circle is double the angle subtended at any point on the remaining part of the circle, experimentally.

Prerequisite Knowledge

1. Basic terms related to circle.
2. Concept of an angle subtended by an arc at the centre and at the circumference of the circle.

Materials Required
Glazed papers, white sheet, pencil, a pair of scissors, gluestick.

Procedure

1. Cut one circle of radius 2.5 cm with centre O from red coloured glazed paper.
2. Cut three more circles from different coloured glazed papers of same radius.
3. Keep all four circles one on the other. Now fold along any part and press them to make a crease. On unfolding, we get chords of same length on each circle.
4. Name the chords, AB in first circle with centre O.
5. Join OA and OB with pencil [fig. (i)].
6. Take two other circles of yellow and green colour and put one on the other and fold it such as to form an angle on the circumference with the same chord AB [fig. (ii) and (iii)].
7. Name these angles as ACB where AB is a chord.
8. Cut angles from fig. (ii) and (iii), ∠ACB from yellow circle and ∠ACB from green circle.
9. Cut the small portion of ∠ACB from both the circles [fig (ii) and (iii)].
   
   
   
   
10. Paste these two cut outs of fig. (ii) and (iii) on the another (blue) circle [fig. (iv)] at centre O such that their arms lie on the radius OA and OB of circle.

Observation
We observe that two cut outs of angles fully cover ∠AOB in fig.(iv).
∠AOB = ∠ECF + ∠GCH (as ∠ECF = ∠GCH = ∠ACB).
= 2 ∠ACB

Result
Hence we verified that the angle subtended by an arc at the centre of circle is double the angle subtended by the same arc at any point on the remaining part of the circle.

Learning Outcome
Verification of above theorem can be done for arc AB as major arc or semicircular arc. For semicircle, angle on the diameter is of 90°.

Activity Time

1. Prove this theorem by taking different situations on the circle such as chord, a diameter, chord in minor segment, chord on major segment.
2. What do you observe if you take chord as a diameter?

Viva Voce

Question 1.
What is the segment of a circle ?
Answer:
A chord divides the circle into two parts. Each part is known as segment.

Question 2.
What divides a circle into two equal segments ?
Answer:
Diameter.

Question 3.
Two diameters of a circle are perpendicular to each other, how many equal sectors will they form in circle ?
Answer:
4 sectors.

Question 4.
Two chords AB and CD in a circle form ∠AOB and ∠COD at the centre O of the circle. If AB and CD are not equal in length what can you say about the angles ∠AOB and ∠COD ?
Answer:
∠AOB ≠ ∠COD.

Question 5.
If the angle at the centre is 60°, what will be the angle on the remaining part of circle subtended by an arc ?
Answer:
30°.

Question 6.
If the angles subtended by the chords of a circle at the centre are equal, then what will be the length of chords ?
Answer:
Length of the chords will be equal.

Question 7.
What will be the distance of the two equal chords from the centre ?
Answer:
Equal distance.

Question 8.
If the two chords are equal, then what will be the length of their corresponding arcs ?
Answer:
Equal length.

Multiple Choice Questions

Question 1.
In a circle, arc BC subtends 40° at the centre, the value of angle subtended by the arc BC at the remaining part of the circle is:
(i) 20°
(ii) 40°
(iii) 80°
(iv) none of these

Question 2.
In a circle, chord AB subtends 32° at circle, the value of angle subtended by it at the centre is:
(i) 16°
(it) 64°
(iii) 30°
(iv) none of these.

Question 3.
Angle subtended by the diameter of the semicircle is:
(i) 90°
(ii) 45°
(iii) 180°
(iv) none of these

Question 4.
Triangle formed by the chord and two radii in a circle is:
(i) right angled triangle
(ii) isosceles triangle
(iii) equilateral triangle
(iv) none of these

Question 5.
In the given figure, O is the centre of the circle, ∠BAC = 40°. Find x:

(i) 50°
(ii) 100°
(iii) 80°
(iv) none of these

Question 6.
In the given figure, find x, if ∠ACB = 30°:

(i) 60°
(ii) 15°
(iii) 120°
(iv) none of these

Question 7.
The angle formed by a chord is 110° at the centre of the circle, the value of angle formed by it at the remaining part of circle is:
(i) 55°
(ii) 220°
(iii) 100°
(iv) none of these

Question 8.
The angle formed by a chord in minor segment is half of the angle formed by it at centre is:
(i) straight angle
(ii) reflex
(iii) acute angle
(iv) none of these

Question 9.
Find the value of x, in the given figure, if ∠BOC = 42°:

(i) 21°
(ii) 84°
(iii) 40°
(iv) none of these

Question 10.
The angle at the centre of the circle is:
(i) 260°
(ii) 180°
(iii) 360°
(iv) none of these

Answers

1. (i)
2. (ii)
3. (i)
4. (ii)
5. (i)
6. (i)
7. (i)
8. (ii)
9. (i)
10. (iii)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Previous Year Solved  Papers  Class 12 Computer Science Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET I

SECTION B

Question.1. (a) Name the function/method required to
(i) check if a string contains only upper case letters
(ii) give the total length of the list.
(b) Rewrite the following Code in python after removing all syntax error(s). Underline each correction done in the code.

(c) Find and Write the output of the following python code :

(d) Find and write the output of the following python code :


(e) What are the possible outcome(s) executed from the following code ? Also specify the maximum and minimum values that can be assigned to variable NUMBER.

Question.2. (a) Illustrate the concept of inheritance with the help of a python code.
(b) What will be the output of the following python code ? Explain the try and except used in the code.

(c) Write a class PICTURE in Python with following specifications:

(d) What is operator overloading with methods ? Illustrate with the help of an example using a python code.
(e) Write a method in python to display the elements of list thrice, if it is a number and display the element terminated with ‘#’ if it is not a number.
For example, if the content of fist is as follows :


Answer: (a) Inheritance: Python supports inheritance, it even supports multiple inheritance. Classes can inherit from other classes. A class can inherit attributes and behaviour methods from another class, called the super class. A class which inherits from a super class is called .a subclass, also called heir class or child class. Super classes are sometimes called ancestors as well. There exists a hierarchy relationship between classes, we define a class Calendar Clock, which inherits both from the class Calendar and from the class Clock.





Here try and its except statement form a unit. The scope of the except clause is restricted to those statements specified by the immediately preceding try statement. A except statement cannot catch an exception thrown by another try statement.

(d) Operator overloading is a specific case of polymorphism in which’ some or all operators like +, = or == are treated as polymorphic functions and as such have different behaviours depending on the types of its arguments. here we overload the + sign, in the class Point.

Question.3. (a) What will be the status of the following list after fourth pass of bubble sort and fourth pass of selection sort used for arranging the following elements in descending order ?
14, 10, -12, 9, 15, 35
(b) Write a method in python to Search for a value in a given list (assuming that the elements in list are in ascending order) with the help of Binary Search method. The method sould return -1, if the value not present else it should return position of the value present in the list.
(c) Write PUSH (Books) and POP (Books) methods in python to add Books and remove Books considering them to act as Push and Pop operations of Stack.
(d) Write a method in python to find and display the prime numbers between 2 to N. Pass N as argument to the method.
(e) Evaluate the following post fix notation of expression.
Show status of stack after every operation.
84, 62, -, 14, 3, *, +

Question.4. (a) Differentiate between the following :
(i) f = open (‘diary.txt’, V)
(ii) f = open (‘diary, text’ , ‘w’)
(b) Write a method in python to read the content from a text file diary.txt line by fine and display the same on screen.
(c) Consider the following definition of class Member, write a method in python to write the content in a pickled file member.dat.

SECTION C

Question.5. (a) Observe the following table carefully and write the names of the most appropriate columns, which can be considered as (i) candidate keys and (ii) primary key:

(b) Consider the following DEPT and WORKER tables. Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii).

Note: DOJ refers to date of joining and DOB refers to date of Birth of workers. in ascending order of Wno.
(ii) To display the Name of all the FEMALE workers from the table WORKER.
(iii) To display the Wno and Name of those workers from the table WORKER who are born between ‘1987 01, 01 and ‘1991-12-01’.
(iv) To count and display MALE workers who have joined after T986-01-01’.
(v) SELECT COUNT (*), DCODE FROM WORKER GROUP BY DCODE HAVING COUNT (*1 > 1;
(vi) SELECT DISTINCT DEPARTMENTf ROM DEPT;
(vii) SELECT NAME, DEPARTMENT, CITY FROM WORKER W, DEPT D WHERE W. DCODE=D. DCODE AND WNO< 1003;

Question.6. (a) Verily the following using Boolean Laws :
X+Y’ = X.Y+X.Y’+ X’.Y‘
(b) Draw the Logic Circuit for the following Boolean Expression :
(U+V>.W + Z
(c) Derive a Canonical SOP expression for a Boolean function F represented by the following truth table.

Question.7. (b) What is a spam mail ?
(c) Differentiate between ftp and http.
(e) What is Worm ? How is it removed ?
(f) Out of the following, which all comes under cyber crime ?
(i) Stealing away a brand new computer from a showroom.
(ii) Getting in someone’s social networking account without his consent and posting pictures on his behalf to harass him.
(iii) Secretly copying files from server of a call center and selling it to the other organization.
(iv) Vewing sites on a internet browser.
(e) Perfect Edu Services Ltd. is an educational organization. – It is planning to set up its India campus at Chennai with its head office at Delhi. The Chennai campus has 4 main buddings – ADMIN, ENGINEERING, BUSINESS and MEDIA.
you as a network expert have to suggest the best network related solutions for their problems raised in (i) to (iv), keeping in mind the distances between the buildings and other given parameters.

(i) Suggest the most appropriate location of the server inside the CHENNAI campus (out of the 4 buildings), to get the best connectivity for maximum number of computers. Justify your answer.
(ii) Suggest and draw the cable layout to efficiently connect various’ buildings within the CHENNAI campus for connecting the computers.
(iii) Which hardware device will you suggest to be procured by the company to be installed to protect and control the internet uses within the campus ?
(iv) Which of the following will you suggest to establish the online face-to-face communication between the people in the Admin Office of CHENNAI campus and DELHI-I Head Office ?
(i) Cable TV
(ii) Email
(iii) Video Conferencing
(iv) Text Chat .
Answer : (b) spam Email, also known as junk email or * unsolicited bulk email (UBE), is a subset of electronic spam involving nearly identical messages sent to numerous
recipients by email. Clicking on links in spam email may send users to phishing web sites or sites that are hosting malware,
(c) As far as file transfer speeds, FTP can be a bit faster for large files, while HTTP can be faster for multiple smaller files. HTTP is’ easier to work with, and it was developed specifically to address some of the inefficiencies of FTP.
(e) A program that when run, has the ability to spread to other computers on its own using either mass-mailing techniques to email addresses found on your computer or by using the Internet to infect a remote computer using known security holes.
There are several steps that should be taken for computer worm removal.

• Check that all antivirus signatures are up-to-date.
• Scan the computer with antivirus software.
• If the scan detects a computer worm or other malware, use the software to remove malware and clean or delete infected files. A scan that detects no malware is usually indicative that symptoms are being caused by hardware or software problems.
• Check that the computer’s operating system is up-to-date and all software and applications have current patches installed.
• If a worm is difficult to remove, check online for specific computer worm removal utilities.

(f) Out of the following, following comes under cyber crime Getting in someone’s social networking account without his consent and posting pictures on his behalf to Secretly copying files from server of a call center and selling it to the other organization.
(e) (i) The most appropriate location of the server inside the CHENNAI campus (out of the 4 buildings), is ADMIN block because this block have maximum number of computers.

(iii) Firewall device will be produced by the company to be installed to protect and control the internet uses within the campus.
(iv) Video Conferencing will be best to establish the online face-to-face communication between the people in the Admin Office of CHENNAI campus and DELHI Head Office.

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2009

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26
   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 five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  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 the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Question.1. What is the electrostatics potential due to an electric dipole at an equatorial point?
Answer : The electric potential due to an electric dipole at an equatorial point is zero.

Question.2. Name the EM waves used for studying crystal structure of solids. What is its frequency range?
Answer : X-rays are used to study crystal structure of solids. The frequency range is 1 nm to 10^-3 nm.

Question.3. An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is direction of the magnetic field?
Answer : The magnetic field is given to be in the direction in which the electron moves. The force is given by

Question.4. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled?

Question.5. Two thin lenses of power + 6D and -2D are in contact. What is the focal length of the combination.

Question.6. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Question.7. Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii?

Question.8. Give the logic symbol of NOR gate.

Question.9. Draw 3 equipotential surfaces corresponding to a field that uniformly increase in magnitude but remains constant along z-direction. How are these surfaces different from that of a constant electric field along z-direction?
Answer : For constant electric field in z-axis equipotential surfaces will be plane parallel to xy- planes .

If field that increases in magnitude, the equipotential surfaces will be planes parallel to XY plane But, as the field increases such planes will get closer.

Question.10. Define electric flux. Write its S.I. unit A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?
Answer: Electric flux is the total number of electric field lines crossing an area. Its SI units in Nm /C. The electric flux through a spherical surface of radius R for a charge q enclosed by the surface is If radius is reduced to half, the electric flux remains the same.

Question.11. Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.
Answer : Refractive Index : The ratio of velocity of light in vacuum to the velocity of light in medium is called absolute refractive index of the medium.
Graph : The plot of angle of deviation versus angle of 0 incidence for a triangular prism is shown below :

Question.12. Calculate the current drawn from the battery in the given network.

Question.13. Answer the following questions:
(a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?
(b)The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer :
(a) X-rays from celestial objects cannot reach the earth’s surface. So for X-ray astronomy to be possible the satellite has to be present in space. However, visible light and lower wavelengths of light emitted by celestial objects reach the surface of the earth. So these objects can be observed with optical and radio telescopes.
(b) The ozone layer is crucial for human survival as they help to block UV radiations and other high frequency harmful radiations and prevent them from reaching the earth’s surface.

Question.15. Define the term ‘linearly polarised light’.
When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids?
Answer: ‘Linearly polarized light’ is an electromagnetic wave, in which the vibrations of electric field are restricted to a single plane. ‘
The intensity of transmitted light is given by I =I_m cos² θ
The intensity of transmitted light is maximum, when θ =0° or 180°, or the polarizing axis of the two polaroids is parallel to each other.

Question.16. A wire of 15 Ω resistance is gradually stretched to double
its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery.
Answer : When any resistor is stretched to double its original length, the new resistance becomes four times of original resistance.

Question.17. (a)a nucleus in its ground state is always less than the total mass of its constituents — neutrons and protons. Explain.
(b) Plot of graph showing the variation of potential energy of a pair of nucleons as a function of their separation.
Answer :
(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents because some mass is converted into energy, in accordance with the equation E = mc². This difference in mass is called the mass defect and the energy corresponding to the mass defect is the binding energy. This
is the energy that has to be supplied*to the nucleus to break it up into its constituents..
(b) The graph is

Question.18. Write the function of (i) Transducer and (ii) Repeater in the context of communication system.
OR
Write two factors justifying the need of modulation for transmission of a signal.
Answer :

1. Transducer converts energy from one form of another either at the input or at the output. For example sound signals are converted to electrical signals so that they can be transmitted through the communication channel.
2. Repeater stations receive the signal, amplify it and then transmit it. So they are a combination of a receiver and , transmitter. This helps to increase the range of transmission of signals.

OR
Two factors justifying the need of modulation for transmission of signal are :

1. Modulation helps to increase the frequency of the signal. This helps to transmit it over larger distances. This is because the power radiated by an antenna is proportional to 1/λ₂ The power radiated by the antenna increase for high frequencies.
2.  The size of the antenna required is proportional to (wavelength/4). So if waves of large wavelength are transmitted then the size of the antenna required is impractical. By increasing the frequency of the signal by modulation, its wavelength is decreased. So now shorter and more practical size antennas can be built.

Question.19. A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate. Derive the expression for the electric field at the surface of a charged conductor.
OR
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.
Answer : Representation of electric field :
The electric field due to a positive charge (+q) is represented as :

Electric field due to a point charge : Consider a point charge +q placed at the origin O of the coordinate frame. Let qo be any test charge placed at P.

Question.20. (i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l1 t with resistance R.and S as shown in the figure.

Question.21.

1. State Faradays law of electromagnetic induction.
2.  A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 x 10^-4 T and the dip angle is 30°?

Answer:

1.  Faraday’s law of electromagnetic induction states that the induced emf is proportional to the rate of change of magnetic flux.
2.  The induced emf will be given by E = Blv where the magnetic field B is perpendicular to the length l. In the question the earth’s magnetic field is given and the angle of dip is 30  degrees, so the magnetic field perpendicular to the direction of the plane is B sin 30°. ’
   

Question.22. In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on. the interference fringes if the’ monochromatic source is replaced by a source of white light?
Answer:

When the monochromatic source is replaced by a source of white light, the fringe width is changed.

Question.23. Draw a schematic arrangement of the Geiger-Marsden experiment. How did the scattering of a-particles by a-thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly.
Answer: Diagram of the Geiger Marsden experiment.

The alpha particles that are incident head-on with the gold nucleus experience a very large force of repulsion and undergo maximum deflection. Equating the kinetic energy of the incident alpha particle with the potential energy of the  alpha particle and gold nucleus, the sum of the approximate radius of the gold nucleus and alpha particles, can be found.

Question.24. Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated.
Answer : The sky waves are reflected from the ionosphere and received by a receiver. Space waves penetrate the ionosphere  and are intercepted by a satellite. They can also be used for line of sight communication.
This mode of propagation is used by short-wave broadcast service.
The space waves are the radiowaves of very high frequency (i.e. between 54 MHz to 4.2 MHz). The space waves can travel through atmosphere from transmitter antenna to- receiver antenna either directly or after reflection from ground in the earth’s troposphere region. That is why space wave propagation in also known as tropbspherical propagation. The space waves travel in straight line from transmitting
antenna to receiving antenna. Therefore, the space waves are used for line of sight communication such as television
broadcast, microwave link and satellite communication.

Question.26. Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw a input and out waveforms of the single. Write the expression for its voltage gain.
Answer. Circuit diagrame of a common emitter amplofire using an n-p-n- transisstor.

Question.27. Draw a plot showing the variation of binding energy per nucleon versus the mass number A. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
Answer: Binding energy curve : The variation of average B.E per nucleon with mass number A, shown in figure below.

A graph between the average binding energy per nucleon and the mass number A of different nuclei is a curve, called ‘binding energy curve’.

1.  For nuclei having mass number A=50 to A=80 are almost stable.
2.  For nucler having mass number above 80, the average BE/nucleon decreases slowly and drops to about 7.6 MeV for uranium (A = 238). This lower value of binding energy per nucleon fails to overcome the Coulombian repulsion.
3.  For nuclei having mass number below 50 also, the average BE/nucleon decreases and below 20, it decreases sharply, e.g. : For heavy hydrogen (H²), it in only about 1.1 MeV. This shows that the nuclei having mass number below 20 are comparatively less stable.
4. Below A = 50, the curve does not fall continuously, but has subsidiary peaks at 8o¹⁶, 6C¹² and  2He⁴. This shows that these nuclei are more stable than their immediate neighbours.
5.  Binding energy per nucleon is small for both light and heavy nuclei. When light nuclei fuse to form a heavy nucleus, high value of B.E is released in Nuclear Fusion. When a heavy nucleus splits into light nuclei, high value of B.E is released in Nuclear fission.

Question.28. Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.

1.  Show that time period of ions in a cyclotron is
   independent of both the speed and radius of circular path. .
2. What is resonance condition? How is it used to accelerate the charged particles?

OR
(a) Two straight long parallel conductors carry currents Ii and I2 in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them,
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.

1.  What is the direction ,of the magnetic moment of the current loop?
2.  When is the torque acting on the loop (A) maximum, (B) zero?

Answer : Diagram of Cyclotron

Charged particles are introduced between the dees. An alternating voltage applied between the dees accelerates, these particles by an electric field. A magnetic field that is perpendicular to the plane of the dees exerts a force of the particles that is given by F = q (V x B). This cause the particles to follow a circular trajectory. As’ they reach the dees the polarity is reversed and the particles are once more accelerated. This continues and highly energetic beams of charged particles are obtained. Magnetic field is perpendicular to dees.

Hence, the time period is independent of speed and radius of circular path.
(ii) Resonance condition : When the angular frequency of the rotating charged particle and the angular frequency of the
alternating voltage applied across the dees of the cyclotron match. The charged particles are only accelerated by the electric field. The magnetic field only keeps it moving along a circular tra.ck. The acceleration happens when the charged particle crosses the gap between the two dees. At this instant the field between the dees has to be reversed so that the electric field can accelerate the charged particle.

Question.29. (a) What are eddy currents ? Write their two applications.
(b) Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed V.

Deduce an expression for
(i) The emf iriduced across the arm ‘RS’,
(ii) The external force required to move the arm, and
(iii) The power dissipated as heat.
OR
(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy”. Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.
Answer : (a) When a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.
Applications of eddy currents :

1.  Magnetic brakes in trains.
2. Electromagnetic damping.
3.  Induction furnaces.
4. Electric power meter.

(b) (i) The emf induced across the arm B lv
On account of the presence of the magnetic field, there will be a force on the arm RS. This force I (lx B), is directed outwards in the direction opposite to the velocity of the arm RS. The magnitude of this force is,

(a) Lenz’s Law : An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.
Lenz’s law is shown with the negative sign in Faraday’s law of induction :

Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux or to exert a mechanical force opposing the motion.
Explanation of conservation of energy : When a magnet is moved near a current carrying coil, the direction of the induced current opposes the motion of the magnet. When the north pole of the magnet is moved towards the coil, the induced current flows in a direction so that near face of the coil acts as a magnetic north pole. The repulsion between two poles opposes the motion of the magnet towards the coil. Similarly, when the north pole of the magnet is from the coil, the direction of the induced current is such as to make the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil. In either case, therefore, work has to be done in moving the magnet. This mechanical work appears as electrical energy in the coil. Thus Lenz’s law is in accordance with the principle of conservation of energy.
(b) Mutual Inductance of two long coaxial solenoids : Consider the following fig. which shows two long co-axial solenoids each of length l. Let the radius of the inner solenoid S₁be r₁ and the number of turns per unit length be n₁.

Question.30. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object, (ii) Write three distinct advantages of a reflecting type telescope
(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.
OR
(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
(b) The near point of a hypermetropic person is 50 cm
from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

(ii)Advantage of reflecting type telescope over a refracting
type telescope are :

1.  Lenses suffer from chromatic aberrations that are not there in mirrors.
2.  Lenses also have spherical aberration; a parabolic mirror will be free of spherical aberration.
3.  It is easier to support large mirrors as the back surface is nonreflecting, but a lens needs support around its rim.

(a) Mirror forumula for concave Mirror :


(b) The near point of the eye of this person is at 50 cm. The required lens must have a focal length such that the virtual image of the book placed at 25 cm is formed at 50 cm. Then image can be focussed by the eye.

SET II

Note: Except for the following questions, all the remaining questions have been asked in Set-I, 2009.

Question.1. What is work done in moving a test charge q through a distance of 1 cm along the equatorial axis of electric dipole?
Answer : Since potential difference for equipotential surface
ΔV = 0
Work done in moving a positive test charge q through a distance 1 cm is
w= qΔV = q x 0= 0

Question.5. Two thin lenses of power +4 D and — 2D are in contact. What is focal length of combination?

Question.6. Give the logic symbol of NAND gate?

Question.7.Two nuclei have mass number in the ratio 8 the ratio of their nuclear radii?

Question.8.The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?

Question.9.

1. State the principle on which the working of an optical fiber is based.
2.  What are the necessary conditions for this phenomenon
   to occur?

Answer :

1. The phenomenon of reflection of light when the light travelling in a denser medium strike* the interface separating the denser medium and the rarer medium at an angle greater than the critical angle is called total internal reflection.
2.  Conditions :
   (a) Light must travel from a denser medium to rarer medium.
   (b) The angle of incidence in the denser medium must be greater than the critical angle for the two media in contact.

Question.10. (i) State the law that gives the polarity of the induced emf. (ii) A 15.0μF capacitor is connected to 220 V, 50 Hz source.
Find the capacitive reactance and the rms current.
Answer :
(i) Lenz’s law : According to this law direction of induced emf Or current in a circuit is such that it opposes the cause or change that produces it.

Question.23. Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively.
OR
(a) A charge +Q is placed on a huge spherical conducting shell
of radius R. Another small conducting sphere of radius r carrying charge is introduced inside the laige shell and is placed at its centre. Find the potential difference between two points, one lying on the sphere and the other on the shell.
(b) How would the charge between the two flow if they are
connected by a conducting wire? Name the device which works on this fact.
Answer : Electric field due to a uniformly charged infinite plane sheet: Suppose a thin non-conducting uniform surface

Electric field intensity E on the either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.
Let P be any point at a distance r from the sheet. Let the small area elements are perpendicular on the surface of the imagined cylinder, so electric flux is zero. are parallel on the two cylindrical edges P and Q, which contributes electric flux.
.^.. Electric flux over the edges P and Q of the cylinder is




(b) When both spheres are connected by a conducting wire then charge q on smaller sphere wifi flow onto the large sphere having charge Q. ,
Van de Graff generator works on this fact that charge to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II, 2009.

1.Define the term ‘potential energy’ of charge q’ at a distance ‘r’ in an external electric field.
Answer: Potential energy of a single charge ‘q’ ara distance Y in an external field = q.V (r)

Question.4.The stopping potential in an experiment on photoelectric effect is 2V What is the maximum kinetic energy of the photoelectrons emitted?

Question.5.Two thin lenses of power + 5D and —2.5 D are in contact: What is the focal length of the combination?
Answer :

Question.6.How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen in halved?

Question.7. Give the logic symbol of AND gate.
:

Question.8.Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of the their nuclear radii?

Question.11. (i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.

Question.16. A wire of 20 Ω resistance is gradually stretched” to double its original length. It is then cut into two equal parts. These parts are then connected in. parallel across a 4.0 volt battery. Find the current drawn from the battery.
Answer: When any resistor is stretched to double its original length. The new resistance becomes four times of its original resistance.

Question.25. Explain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. Draw its I-V characteristics.
Answer : Zener diode 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 the host atoms to create a large number of electrons. This results a large value of current inside the circuit.

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

Question.27. Define the activity of a radionuclide. Write its S.I. unit. Give a plot of the activity of a radioactive species versus time.
How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8th of its initial value?
Answer : The total decay rate R of one or more radionuclide is called the activity of that sample. The S.I. unit for activity is becquerel.
1 becquerel = 1 Bq = 1 decay per second
Graph: The graph between the activity of a radioactive species and time is :

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2016

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1. 1.  All questions are compulsory. There are 26
      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 five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
   4.  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 the three questions of five marks weight age. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:

SET I

SECTION-A

Question.1. A point charge +Q is placed at point O as show in the figure. Is the potential difference V_A — V_B positive, negative or zero ?

Question.2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ?
Answer : According to Gauss’s law

Flux depends only on the charge enclosed.
Hence, the electric flux remains constant.

Motion in Straight Line Class 11 NCERT Solutions are very helpful in scoring more marks in class 11 exams.

 

Question.3. Write the underlying principle of a moving coil galvanometer.
Answer : When a current carrying coil is placed in magnetic field then it experiences a torque.

Question.4. Why are microwaves considered suitable for radar systems ,used in aircraft navigation ?
Answer : Microwaves of frequency 1 GHz to 300 GHz bounces from even the smallest aircraft so that they are suitable to avoid getting bombed. Microwaves can penetrate through clouds also.

Question.5. Define ‘quality factor ‘ of resonance in series LCR circuit. What is its SI unit ?
Answer : The Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the impressed voltage, which is the voltage across R.

It is dimensionless hence, it has no units.

SECTION -B

Question.6. Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
Answer: (i) Attenuation: The loss of strength of a signal while . propagating through a medium is known as attenuation.
(ii) Demodulation : The process of retrieval of information from the carrier wave at the receiver is termed demodulation. This is the reverse process of modulation.

Question.7.

Question.8. 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 Calculated the released energy.

Answer : Gain in binding energy for nucleon is about 0.9 MeV.

Question.9. Two cell 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.

Question.10. State Brewster’s law.
The value of Brewster angle for a transparent medium is different for light of different colour. Give reason.
Answer : Brewster’s law : The law states that the tangent of the polarising angle of incidence of a transparent medium is equal to its refractive index. The light incident at this angle. When reflects back is perfectly polarised.
M = tan i_p
The refractive index of a material depends on the colour or wavelength of light. As the polarising angle depends on refractive index (M = tan ip), so it also depends on wavelength of light.

SECTION – C

Question.11. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.
Answer : Suppose we have a ring of radius a that carries a uniformly distributed positive charge q.

(i) The axial components dE cos θ and
(ii) The perpendicular component dE sin θ.
Since the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field.


This expression is similar to electric field due to point charge.

Question.12. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.
Answer: (i) Existence of threshold frequency: According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of Threshold frequency.
(ii) Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it depends on frequency and not on intensity.
(iii) Instantaneous emission of electrons : According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.

Question.13. (a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your

Question.14. (i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.3 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ?
Answer : (i) Mutual induction is the phenomenon of production of induced emf in one coil due to change of current in the neighbouring coil. The coil in which the current changes is called primary coil and the coil in which emf is induced is called the secondary coil.

Question.15. Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of ε_r= 4.

(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 µF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y

Question.16. Two long straight parallel conductors carry steady current I₁ and I₂ separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.

Question.17. How are e.m. waves produced by oscillating charges ?
Draw a sketch of linearly polarized e.m. waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.

Answer: These waves are constituted by varying or oscillating electric and magnetic fields. The electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave E is the envelope of electric intensity vector and B is the envelope of magnetic intensity vector.

OR
Correction in Amperes Circuital law (Modified Ampere’s law) : Maxwell removed the problem of current continuity and inconsistency observed in Ampere’s Circuital law by introducing the concept of displacement current, Displacement current arises due to change in electric flux

Conduction current is because of How of charges but displacement current is not because of How of charges but because of change in electric flux.

Question.18. (a) Explain any two factors which justify the need of modulating a low frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation.

Question.19. (i) Write the functions of three segments of a transistor.
(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.
Answer : (i) Three segments of transistor are :
(i) Emitter (ii) Base
(iii) Collector
Emitter : It is of moderate size and heavily doped, it supplies a large number of majority carriers which flow through the transistor.
Base : It is very thin and lightly doped and it separates emitter and collector region of transistor and controls the flow of charge carriers.
Collector : This segment is moderately doped and larger in size as compared to emitter. It collects a major portion of majority carriers supplied by the emitter.

Question.20. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also,
(b) Using mirror formula, explain why does a convex mirror always produce a virtual image.

Question.21. (i) State Bohr’s quantization condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits ?
(ii) Find the relation between the three wavelengths λ₁, λ₂ and λ₃ from the energy level diagram shown below.

Question.22. Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope.
Answer : Reflecting Telescope : The reflecting telescope makes us of a concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflective). After reflection the rays of light meet at a point where another convex mirror is placed. This mirror focusses light inside the telescope tube. The final image is seen through the eye piece. The images produced by the reflecting telescope is very bright and its resolving power is high.

Advantages:
(i) The resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.
(ii) There is no chromatic aberration as the objective is a mirror.

SECTION – D

Question.23. Meeta’s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.
Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father ?
(ii) What answer did Meeta’s father give ?
(iii) What are the tiny lights in traffic signals called and how do these operate ?
Answer : (i) Awareness for energy conservation, power saving and knowledge about traffic rules.
(ii) Meeta’s father said that these are LED light which consume less power and high reliability.
(iii) The tiny lights in traffic signals are Light Emitting Diode. These are operated by connecting the P-N junction diode in forward biased condition.

SECTION – E

Question.24. (i) An a. c. source of voltage V = V₀ sin tor is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ? (ii) In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L= X_C is put in series, the power factor becomes P₂. Calculate P₁/P₂.
OR
(i) Write the function of a transformer. State is principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100
turns and transformation ratio is also 100. The input voltage and power are respectively 220V and 1100 W Calculate:
(a) number of turns in secondary.
(b) current in primary.
(c) voltage across secondary.
(d) current in secondary.
(e) power in secondary.
Answer : (i) Let a, series LCR circuit is connected to an ac source V (Fig). We take the voltage of the source to be V = V₀ sin ωt.



OR
(i) A transformer is an electrical device for converting an alternating current at low voltage into that at high voltage or vice-versa.

• If it increases the input ac voltage, it is called step up transformer.
•  If it decreases the input ac voltage, it is called step down transformer.

Principle : It works on the principle of mutual induction i.e., When a changing current is passed through one of the tow inductively coupled coils, an induced emf is set up in the other coil.


Working Theory : As the AC flows through the primary, it generate an alternating Magnetic flux in the core which passes through the secondary coil.
Let N1 = No. of turns in primary coils
N2 = No. of turns in secondary coils
This changing flux set up an induced emf in the secondary,/ also a self induced emf in the primary.
If there is no leakage of magnetic flux, then flux linked with each turn of the primary will be equal to that linked with each of the secondary. According to Faraday’s law of induction.

• Copper loss : Some energy is lost due to the heating of copper wires used in the primary and secondary winding’s. This power less (P = 12R) can be minimised by using thick copper wires of low resistance.
• Eddy current loss : The alternating magnetic flux induces eddy current in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.
• Hysteresis loss : The alternating current carries the iron core through cycles of magnetisation and demagnetisation. Work done in each of these cycles and is lost as heat. This is called hysteresis loss and can be magnetised by using c.ore material having narrow hysteresis loop.
• Flux leakage: The magnetic flux produced by the primary may not fully pass through the secondary. Some of the flux may leak into air. This loss can be minimised by winding the primary and secondary coils over one another.

Question.25. (i)In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position V on the screen.
(ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.
OR
(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive ah expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light ? What is its cause ?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.




(ii) Compose of interference pattern observed in Young’s double slits and the single slits diffraction :

OR
(i) It is a transparent medium having two planes and non-parallel refracting surface inclined to each other and three surfaces are not participating in refraction.




(ii) Dispersion of light : These colours are often observed as light passes Through a triangular prism upon passing through the prism, the white light is separated into its component colours : red, orange, yellow, green, blue, and violet. The separation of visible light into its different colours is known as dispersion. Dispersion occurs because for different colour of light a transparent medium will have different refractive indices (µ).

Question.26. (i) Define the term drift velocity.
(ii) 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 ?
(iii) Why alloys like constantan and manganin are used for making standard resistors ?
OR
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (=l).

Answer : (i) Drift velocity is defined as the average velocity which the 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.
(iii) They are used to make standard resistors because :
(a) They have high value of resistivity
(b) Temperature coefficient of resistance is less.
(c) They are least affected by temperature.
OR
(i) Principle of potentiometer : The basic principle of potentio meter is that when a constant current flows through a wire of uniform cross-section area and the composition of the potential drop across any length of the wire is directly proportional to that length.
A potentiometer is a device used to measure an unknown emf or potential difference and internal resistance of a cell accurately.
Construciton :

1. A potentiometer consists of a long uniform cross-section of wire generally made of manganin or constantan.
2. Usually, 1 m long separate pieces of wire are fixed on a wooden board parallel to each other.
3. The wire are joined in series by thick copper strips.
4. The ends A and B are connected to a battery (called driving all), a plug key and rheostat.
5. A jockey J is provided with the help of which contact can be made at any point on the wire.
6. This circuit sends a constant current I through the wire AB.

Principle : When a constant current flows through a wire of uniform cross sectional area and composition the potential drop across any length of the wire is directly proportional to that length.
Let V be the potential difference across the portion of the wire of length l whose resistance is R
By Ohm’s law,

Looking for an easy way to Learning of new elementary english grammar and composition for class 7 answers, Solutions. You have to learn basic English Grammar topics like Tenses Verbs, Nouns, etc… In this article, we will review the best English Grammer Topics and compare them against each other.

Noun Phrase And Noun Clause Exercises For Class 7 CBSE Pdf

“Main (or independent) clause is a clause that expresses a complete thought and , can stand as a sentence. But subordinate (or dependent) clauses cannot stand by themselves and are further divided into three types,

1. Noun Phrase
2. Adjective Phrase
3. Adverb Phrase

Noun Phrase:
A phrase that acts as a noun in a sentence is called a noun phrase. It consists of a noun and other words that give information about the noun. It answers the question what or who. The noun phrase can be replaced by a pronoun.
Examples:

• The coffee – table books looked beautiful. (noun: the coffee – table books)
• I like going for a swim. (noun: going for a swim)
• Mona met the newly appointed chief. (noun: the newly appointed chief)

Remember, a noun phrase must have a noun.

Noun Clause:
The clause that tells you what someone says or does. It can also be the object of a verb.
A noun clause is a dependent clause that acts as a noun. To identify the noun clause in a sentence, we can ask the question ‘What’ after the verb. Noun clauses begin with words such as how, that, what, whatever, when, where, whether, which, whichever, who, whoever, whom, whomever, and why.
Noun clauses can act as subjects, direct objects, indirect objects.

Noun clauses as direct objects:
Examples:

• They know what my name is.
• I noticed where they were going.

Noun clauses as a subject:
Examples

• What she is doing is not known.
• What you said is not clear.
• Where she lives is not considered.

Noun clauses as an indirect object:
Examples:

• I will do whatever it takes to score good marks.
• Nancy will whoever finds the purse for her.

Noun Phrase Vs Noun Clause
Examples

• No one knows when he will come. (Noun-Clause)
• No one knows the time of his coming. (Noun-Phrase)
• She heard that he had succeeded. (Noun-Clause)
• She heard of his success. (Noun-Phrase)
• They will never know why he succeeded. (Noun-Clause)
• They will never know the reason for his success. (Noun-Phrase)
• The law will punish whosoever is guilty. (Noun-Clause)
• The law will punish the guilty. (Noun)
• Her friends want to know where she is living. (Noun-Clause)
• Her friends want to know her residence. (Noun)

Focus:
A noun clause can be a complement after a ‘be’ verb.
E.g.

• Why she was working so late is something we need to find out

A noun clause may sometimes be placed first in a sentence for emphasis.
E.g.

• How we’re going to deal with the issue is the question.

Presentation:
Read the picture story.

Noun Phrase And Noun Clause Worksheet for Class 7 CBSE

A. Underline the noun phrases in the following sentences.
1. He wished to talk to his manager.
2. The wicked man loves getting poor people into trouble,
3. The poor man wanted to pay back every penny he owed.
4. He hates to punish his children.
5. Horses prefer living in dark stables.
6. I will never do such a thing.
7. Have you ever tried to climb a tree?
8. He refused to answer my question.
9. He promised to get me something.
10. Why do you want to meet him?

B. In each of the following sentences replace the words in italics by suitable noun clauses.

1. No one can guess the time of his arrival.
___________________________________________
2. I predict a change in the weather
___________________________________________
3. I heard of his success.
___________________________________________
4. Nobody knows the reason of his failure.
___________________________________________
5. The jury believed the man to be guilty.
___________________________________________
6. The doctors expect an improvement in his health.
___________________________________________
7. I know him to be trustworthy.
___________________________________________
8. He confessed his guilt.
___________________________________________
9. I hope to be there in time.
___________________________________________
10. He is confident of winning the match.
___________________________________________
11. He seems to be a spy.
___________________________________________
12. His arrival was totally unexpected.
___________________________________________

C. Combine the following sentences using noun clauses.

1. I don’t know. What kind of a battery is this?
___________________________________________
2. My brother doesn’t know. How tall is he?
___________________________________________
3. I am not sure. Which chapter are we on?
___________________________________________
4. Scientists claim. They will cure cancer.
___________________________________________
5. Who is she? No one knows.
___________________________________________
6. Could you tell me? Where is the bookstore?
___________________________________________
7. Do you remember? How long did we stay in Qatar?
___________________________________________
8. Please tell us… What did you hear?
___________________________________________

D. Complete the following sentences by adding suitable noun clauses. The first one has been done for you.

1. I cannot understand why she is angry with me.
2. They said _______________________.
3. I think _______________________.
4. He told me _______________________.
5. hurts me. _______________________.
6. You can eat _______________________.
7. Can you guess _______________________.
8. I feel certain _______________________.
9. I do not know _______________________.
10. No one knows _______________________.