Physics Classes

Category: 12th Assignments

  • Case study based questions on Photoelectric effect

    Photoelectric effect

    Photoelectric effect is the phenomenon of emission of electrons from a metal surface , when radiation of suitable frequency fall on them . The emitted electrons are called photoelectrons and the current so produced is called photoelectric current.

    i. With the increase of intensity of incident radiation on photoelectrons emitted by a photo tube , the number of photoelectrons emitted pre unit time is (a) increases (b) decreases (c) remain the same  (d) none of these .

    ii. It is Observed that photoelectrons emission stops at a certain time t after the light source is switched on . The stopping potential can be represented as (a) 2(K.E.)max/e (b) (K.E.)max/e  (c) (K.E.)max/3e (c) (K.E.)max/2e .

    iii. A point source of light of power 3.2×10-3 W mono energetic photons of energy 5.0eV and work function 3.0eV . The efficiency of photoelectrons emission is 1 for every 106 incident photons . Assume that the photoelectrons are instantaneously swept away after emission . The max kinetic energy of photon is (a) 4eV (b) 5eV (c) 2eV (d) Zero .

    iv. Which of the following is the application of photoelectric effect ?   (a)L.E.D  (b)Diode (c) Photocell     (d) Transistor .

    v. If the frequency of incident light falling on the photosensitive metal is doubled , the kinetic energy of the emitted photoelectrons is (a) Unchanged (b) halved (c) doubled (d) more than twice its initial value.

    To solve the test based on the chapter Dual nature of matter . Atoms and nuclei, and Semiconductor devices click here-

    Test Dual nature, Atoms Nuclei and semiconductor

  • Hots questions based on semiconductor

    Hots questions based on Semi conductors.

    These questions are important for class 12th board Examination.

    Q. 1. What are different types of compound semiconductors? Give two examples of each.

    Ans- The compound semiconductors are of following types

    • Inorganic semiconductors: Cds, GaAs, CdSe
    • Organic semiconductors: Anthracene, pthalocynines
    • Organic polymers: Polyparrole, polyaniline, polythiophene.

    Q. 2. Which characteristic of a semiconductors make them useful in fabrication of electronic devices?

    Ans- The number of charge carriers and their direction of flow can be controlled within the solid semiconductor.

    Q. 3. Give three advantages of semiconductor devices over conventional vacuum tube devices.

    Ans- 1. The charge carriers can be manipulated within the semiconductor unlike the vacuum tubes where the electrons were manipulated in vacuum.

    2. The size of semiconductors devices is very small because they do not require vacuum for manipulation of the charge carriers.

    3. The semiconductor devices consume lesser energy because unlike vacuum tube, they do not require heating for electron emission.

    Q. 4. What are the characteristics to be taken care of while doping a semiconductor? Justify your answer.

    Ans-(i) The number of dopant atoms should be small and restricted to just a few parts per million (ppm) of the pure semiconductor atoms.

    (ii) The size of the dopant atom used should be nearly equal to that of the semiconductor atom.

    The above characteristics are essential so that the doping process does not cause any major distortion in the original pure semiconductor lattice. The dopant atom just substitute some of the semiconductor atoms in their lattice.

    Q. 5. Give example of the dopants used to make

           (i) n-type semiconductor                       (ii) p-type semiconductor.

    Ans- n-type semiconductor are formed by using pentavalent impurity e.g. Arsenic (As);  Antimony (Sb), Phosphorous (P) p-type semiconductors are formed by adding trivalent impurity to pure semiconductor e.g., Indium (In), Boron (B), Aluminium (Al).

    Q. 6. What is the energy required to make electrons free for conduction in pure (i) germanium (ii) silicon? How does the energy value change in a doped semiconductor?

    Ans- In a pure semiconductor, the energy required to make electrons available for conduction is large. It is about 0.72 eV for germanium and about 1.1 eV for silicon. In doped germanium, the energy required becomes nearly 0.01 eV and in doped silicon it is about 0.05 eV.

    Q. 7. What is the net charge on a p-type extrinsic semiconductor? Justify.

    Ans- Zero, the crystal maintains its overall neutrality because the charge on additional charge carriers is just equal and opposite to that of the ionised cores in the lattice.

    Q.8. What should be the value of the reverse breakdown voltage of the diode used in a half-wave rectifier and why?

    Ans- The reverse breakdown voltage of the pn-junction diode used in half-wave rectifier should be higher than the peak voltage of the a.c. voltage at the secondary of the transformer. The condition is necessary to protect the pn-junction diode from reverse breakdown.

    Q.9. How is a sample of n-type semiconductor electrically neutral though it has an excess of negative charge carriers?

    Ans- n-type semiconductors are fabricated by adding group 15 impurity atom which are neutral and having one extra proton for every conduction electron. So the sample in neutral.

    Q.10. A student has to study the forward and reverse characteristics of a pn-junction. What kind of circuit arrangement should he use?

    Draw the typical shape of input and output characteristic curves likely to be obtained.

    Mark the knee voltage and reverse breakdown voltage on the curve. How can the static and the dynamic resistance be determined from the curves?

    In which of the regions does the pn-junction remain when used as (i) full-wave rectifier (ii) stabiliser?

    Q.11. What do you mean by host atoms in semiconductors?

    Ans- In impure semiconductors, the semiconductors atom (e.g. Ge or Si) are called host atoms.

    Q.12. What are optoelectronic devices? Give any two example of such devices.

    Ans- The semiconductor junction diodes in which the charge carriers are generated by photons are called photoelectronic devices. The examples of such devices are

    • Photodiodes
    • Light emitting diodes
    • Photovoltaic or solar cells.

    Q.13. Two semiconductor pieces X and Y of same length and equal area or cross-section are provided. One of the materials is pure semiconductor and the other is doped.

    (i) How will you identify the impure semiconductor from the two give pieces X and Y?

    (ii) For a given rise in temperature what will be the relative charge in the behaviour of the two materials?

    Ans- (i) Measure the electrical resistance offered by X and Y. The piece offering lower resistance will be impure semiconductor.

    (ii) Both X and Y will have negative temperature coefficient of resistance. However, the resistance of pure semiconductor will decrease sharply with rise in temperature as compared to the impure semiconductor.

    Q.14. A photodiode is operated is reverse bias region only although the forward current is known to be more than the current in reverse bias. Give reason.

    Ans- A photodiode is sensitive to light and is used to measure/detect intensity of light incident on it. In a pn-junction, photodiode; consider n-type region. The electrons in this region are majority carriers and the holes are minority charge carriers

    ؞                              ne >> nh.

    When the pn-junction is illuminated the bonds between same atoms breaks up due to photo-excitation leading to increase in ne and nh say by ∆ne and ∆nh

    We have                       ∆ne = ∆nh

    ؞                                    ∆nh/nh >>  ∆ne/ne

    i.e. The fractional increase in minority charge carriers is very large as compared to that in minority charge carriers.

    The current through pn-junction in reverse bias is due to drift of the minority charge carriers. The number density of minority carriers being sensitive to light intensity; the reverse current show a rapid change with intensity of incident light.

    So a photodiode is always operated in reverse bias.

    Q.15. What is the minimum frequency that can be detected by a photodiode?

    Ans- The minimum frequency (vm) that can be detected by a photodiode depends on the energy gap Eg of the semiconductor and is given by

                                H vm = Eg

        ؞                         vm = Eg/h.

    Q.16.What are the factors taken into consideration while selecting material suitable for fabrication of solar cell?

    Ans-The following factors are taken into consideration while fabricating solar cell:

    • Band gap is 1.0 eV to 1.8 eV
    • High optical absorption
    • Reasonable electrical conductivity
    • Availability of raw material
    • Cost.

  • Case study question on Compound microscope

    Compound microscope-

    A compound microscope is an optical instrument used for observing high magnified image of tiny objects. Magnifying power of compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended by the object , when both the final image and object at the least distance of distinct vision from the eye . It can be given that:m=mexm0, where me is the magnifying power of eyepiece and m0 is the magnifying power of objective lens. Suppose a lens of focal length 2.0cm and another of 6.25 cm separated by distance 15 cm .

    (i) The object distance of eye piece , so that final image is formed at the least distance of distinct vision will be (a) 3.45cm  (b)5cm (c)1.29cm (d) 2.59cm.

    (ii)How far from the objective should an object be placed in order to obtain the condition given in (i). (a)4.5cm (b)2.5cm (c) 1.5cm (d) 3.0cm .

    (iii) What  is the magnifying power of the microscope in case of list distinct of vision? (a)20 (b)30 (c)40 (d)10 .

    (iv) The intermediate image formed by the objective of a compound microscope is (a) real, inverted and magnified (b) real erect and magnified (c) virtual , erect and magnified (d) virtual, inverted and magnified .

    (v) The magnifying power of compound microscope is increases with .(a) f0 increases and fe decreased (b)fe increased and f0 decreased (c) both increased (d) both decreased .

  • Case study question on astronomical telescope

    Astronomical telescope-

    An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies libe stars, planets etc. It consists of two lenses . In normal adjustment of telescope , the final image is formed at infinity . Magnifying power of astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the angle subtended at the eye by the final image to the angle subtended at the eye , by the object directly , when the final image and the object both lie at the infinite distance from the eye . It is given by m = f0/fe. To increase magnifying power of an astronomical telescope in normal adjustment , focal length of objective lens should be small.

    (i)An astronomical telescope of magnifying power 7 consists of two thin lenses 40cm apart , in normal adjustment . The focal length of the lenses are (a) 5cm,35cm   (b) 7cm , 35cm (c)17cm , 35cm (d)5cm, 30cm.

    (ii)  An astronomical telescope has a magnifying power of 10. In normal adjustment , distance between the objective and eye piece is 22cm . the focal length of the objective lens is ,(a)25cm (b)10cm(c)15cm (d)20cm.

    (iii) In astronomical telescope compare to eye piece , objective lens has (a)negative focal length (b)zero focal length (c)small focal length (d)large focal length .

    (iv)To see stars use (a)simple microscope (b)compound microscope (c)endoscope (d)astronomical telescope.

    (v)For large magnifying power of astronomical telescope (a) f0<<fe (b)f0=fe (c)f0>>fe (d)none of these  .

  • Case study questions on ray optics

    Before to solve case study based questions of ray optics , students must complete these questions , which are basic questions based on ray optics , To get the questions click on the link given below-

    Very important topics for ray optics

    Case study based questions-

    1. Total internal reflection –

    Total internal reflection is the phenomenon of reflection of light into denser medium at the interface of denser medium with rarer medium . For this phenomenon to occur necessary condition is that light must move from denser to rarer and angle of incidence in denser medium must be greater than critical angle for the pair of media in contact. Critical angle depends on nature of medium and wavelength of light . We can show that µ=1/sinC (where C is the critical angle).

    (i) critical angle for glass air interface , where µ for glass is 3/2 , is (a)41.80 (b) 600 (c)300 (d)150 .

    (ii)Critical angle for water air interface is 48.60 , then what is the refractive index of water/(a)1 (b)3/2 (c)4/3 (d)3/4

    (iii) critical angle for air water interface for violet colour is 490 its value for red colour will be (a)490(b)500 (c)480 (d)can’t say.

    (iv) Which of the following is not due to total internal reflection?(a) working of optical fibre (b)difference between real depth and apparent depth of a pound.(c)mirage on hot summer day (d) brilliance of diamond.

    (v) critical angle of glass is θ1 and that of the water is θ2 critical angle for water and glass surface would be (µg > µw) (a) less than θ2 (b) between θ1 and θ2 (c)greater then θ2 (d)less than θ1 .

    2. Power of a lens-

    Power (P) of a lens is given as the reciprocal of focal length (P=1/f)where f should be in meter and P is in Diopter . for convex power is positive and concave power is –ve. When two or more lenses are kept in contact then power of the combined lens is given as P= P1 + P2+P3 …….

    (I) A convex and a concave lens is separated by distance d are then put in contact then the focal length of the combination  (a) becomes 0 (b) remain the same (iii) decreases (iv)increases.

    (ii)The two lenses of power +1.5D and +1.0D are placed in contact then the effective power of the combination will be (a)2.5D (b)1.5 D (c) 0.5D (d)3.25D .

    (iii) If the power of the lens is 5D then what is the focal length of the lens?(a)10cm (b) 20cm (c)15cm (d) 5cm

    (iv)Two thin lens of focal length +10cm and -5cm are kept in contact , the power of the combination is ? (a)-10D (b)-20D (c)10D (d)15D .

    (v)A convex lens of focal length 25cm is placed coaxially in contact with a concave lens of focal length 20cm the system will be ; (a) converging in nature (ii) diverging in nature (c) can be converging or diverging (d) None of the above.

    3.

    4. Refraction through a prism

    A prism is a portion of a transparent medium bounded by two planes face inclined to each other at a suitable angle . A ray of light suffers two refraction on passing through a prism and hence deviates through a certain angle from its original path . The angle of deviation of a prism is ɗ = (µ-1 )A , through which a ray deviates on passing through a thin prism  of small refracting angle A . If µ is the refractive index of the material of the prism then the prism formula is µ = sin(A+ɗm/2)/sin(A/2).

    (i) For which colour , angle of deviation is minimum ?(a) Red (b)yellow (c) violet (d) blue.

    (ii) When white light moves through vacuum (a) all colours have same speed (b)different colours have different speed (c) violet has more speed then red (d)red has more speed then violet.

    (iii)The deviation through a prism is maximum when angle of incidence is (a)450 (b) 700 (c)900(d)600 .

    (iv)What is the deviation produced by a prism of angle 60 ? (µ=1.644) (a)3.840 (b)4.595(c)7.259(d)1.252.

    (v) A ray of light falling at an angle of 500 is refracted through a prism and suffers minimum deviation . If the angle of prism is 600, then the angle of minimum deviation is (a) 450 (b) 750 (c)500 (d)400 .

    5.

    6.

    Very important topics for ray optics

  • Very important Short answer type questions of ray optics

    Short answer type questions- (2 marks),

    Before to solve this assignment students have to solve these questions given in the link below-

    Very important topics for ray optics

    This assignment provides Very important short answer type questions of ray optics which is very important for the class 12th board examinations.

    1.What are the necessary conditions for the phenomenon of total internal reflection to occur . Also write the relation between refractive index and critical angle for a given pair of optical media.

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

    3. Define critical angle . Obtain a relation between refractive index and critical angle for a pair of media . Does critical angle depends on the colour of light ? Explain.

    4. Draw ray diagrams to show how specially designed prism make use of total internal reflection to obtain the inverted image of the object by deviating rays (i) through 900 (ii) through 1800.

    5. Find the relation between real depth and apparent depth with the refractive index , when light moves from denser medium to rarer medium.

    6. A fish in the water tank sees the outside world as if fish at the vertex of cone such that the circular base of the cone coincide with the surface of water . Given the depth of water were fish is located being h and critical angle for water-air interface being ic , find out by drawing the ray diagram the relationship between the radius of the cone and height h.

    7. Draw the ray diagram showing the path of a ray of light entering through a triangular glass prism . Deduce the expression for the refractive index of glass prism in terms of the angle of minimum deviation and angle of the prism.

    8. A convex lens of focal length f1 is kept in contact with another lens of focal length f2 . Find an expression for the focal length of the combination.

    Watch the video of related topic combination of lenses click here-

    9. Why does the sun appear reddish at sunset or sunrise? For which colour the refractive index of prism material is maximum and minimum?

    10. Draw the ray diagram showing formation of image by a magnifying glass when the image is formed at infinity also find the angular magnification of image.

    11. Define the magnifying power of compound microscope when the final image is formed at infinity. Why does objective and the eyepiece of a compound microscope has short focal lengths ? explain.

    12. Draw  a schematic diagram of reflecting telescope (Cassegrain). Write its two advantages .or, Modern telescope prefer using suitable mirrors over using suitable lens , give two reasons for this preferences.

    13. Use a lens formula and show that (i) Object placed at the focal length of the convex lens for image at infinity.(ii) Object is placed between O(optical center) and F (focus) the image will be virtual.

    14. Draw a plot showing the variation of power of a lens with the wavelength of the incident light.

    15 A double convex lens made from a material of refractive index µ1 is immersed in liquid of refractive index µ2 where µ2>µ1 . What change if any, would occur in the nature of the lens?

    Very important topics for ray optics

  • Magnus effect and dynamic lift

    Magnus effect and dynamic lift-

    The upward force experience by a body when it moves through a fluid is called dynamic lift . Such a lift is felt by airfoil , aircraft win etc. Couse behind it can be explained by Bernoulli’s theorem . It also explain why a spinning ball is deviated from its parabolic path .This deviation is usually observed in many games like cricket, football golf, tennis etc.

    As a ball moves through air without spinning , it divides the streamline  above and below it symmetrically . Thus there is no pressure difference on the two sides of the ball.

    But when a ball spins as it moves through air , it drags along a layer of air with itself. When the ball moves forward , the air ahead of the ball rushes backward to fill the space left vacant by the ball. Thus the streamlines of air around this ball is shown in the fig. given below.

    The velocity of air above the ball opposite to the direction of spin and it is in the direction below the ball. Thus the relative velocity of air  with respect to the ball is larger at the upper part than that is in the region below  the ball . The streamline thus gets crowded above and rarified below .This cause the pressure above the to be smaller and that below it to be greater . This pressure difference causes an upward force on the ball. Hence the ball experiences a dynamic lift and is deviated from its path . This effect arising due to spinning of the ball is called ‘Magnus effect’.

    Airfoil and lift on aircraft wing:-  An airfoil is streamlined shaped , Solid body that is capable of generating a dynamic lift as it moves through a fluid .  streamline shape is special shape given to bodies moving in fluids, So as to produce minimum drag .

    The cross-section of air wings is also shaped like that  .Its upper portion is more curved then the lower surface .Its front edge is broader and the rear edge is narrow . When the airfoil moves through air , streamlines of air moves backward . The air above the foil has to travel a longer distance then that  below in the same time . Thus the flow speed on top becomes greater then that below it . The higher speed on reduces air pressure there. This pressure difference causes to lift the foil. Such dynamic lift also acts on the wings of an air craft and balances the weight of the plane.     

  • Total internal reflection

    In this topic we will discuss , what do you mean by the term Total internal reflection, critical angle . And we will practice some questions based on these topics.

    before to know this topic students must know Real depth and apparent depth. To know this topic click here-

    Total internal reflection and critical angle.

    Before to know about the total internal reflection we must know critical angle

    Critical angle –  It is the value of angle of incidence for which angle of refraction becomes 900, when light enters from denser to rarer medium .

    As we know ,

    1µ2 = sin i /sin r

    Since light is moving from denser medium to rarer medium

    So,  1µ2 = sin r /sin I;

    i.e. µ = sin 900 / sin c  [ here r= 900 then I is written by C (critical angle).

    So, sin C = 1/µ

    Then critical angle C = sin -1(1/µ).

    Total internal reflection-   (TIR) When the light in the denser medium strikes an interface separating the denser medium and a rarer medium at an angle of incidence greater then critical angle total light is reflected without any refraction through the rarer medium . This phenomenon is called “total internal reflection”.

    As shown in fig.

    I2< C ( critical angle) in this case refraction of light occurs .

    I1> C  in this condition T.I.R. occurs.

    Essential conditions for T.I.R is

    (i) Light must move from denser to rarer and (ii) angle of incidence should be greater then critical angle.

    Questions based on refraction of light ( apparent depth and real depth, total internal reflection and  

    Q1. A ray of light is incident on a glass plate as an angle of 600. What would be the refractive index of glass , if reflected and refracted rays are perpendicular to each other? (a)√3 . (b)3/2 (c) √3/2 (d) ½ . ans-a

    Q2. A glass of thickness 12mm is placed on a table . The lower surface of the slab has a black spot .At what depth from the upper surface will the sport appear when viewed from above ? (µg=1.50) . (a) 2mm (b) 4mm (c) 6mm (d) 8mm. ans-d

    Q3. Light enters at an angle of incidence in a transparent rod of refractive index of the material of the rod the light once entered into it will not leave it through its lateral surface whatsoever be the value of angle of incidence? (a) µ>√2 (b) µ = 1 (c) µ=1.1 (d) µ=1.3. ans-a

    Q4. A transparent cube contains a small air bubble . Its apparent distance is 2cm when seen through one face and 5cm when seen through other face . If the refractive index of the of the material of the cube is 1.5 the real length of the edge of the cube must be (a) 7cm (b) 7.5 cm (c) 10.5 cm (d) 14/3 cm. ans-c

    Q5. For the given incident ray as shown in figure , The condition of T.I.R of the ray will be satisfied , if the refractive index of the block will be  ,

    (a) (√3+1)/2      (b)(√2+1)/2     (c)√3/√2     (d)√7/√6. ans-c

    Q6. A ray of light travels from optical denser medium to rarer medium. The critical angle for the two media is C . The maximum possible deviation of the ray will be (a)2C (b) π/2 – C (c) π-C  (d) π-2C . ans-b

    Q7. The apparent depth of water in the cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/min. when water is being drained out at a constant rate . The amount of water drained in cc/minute is : (µ1 refractive index of air and µ2 refractive index of water) (a) xπR2µ1/µ2  (b) xπR2µ2/µ1          (c)  2πRµ1/µ2  (d) xπR2. ans-b

    Q8. Refractive index of glass with respect to medium is 4/3  .If the difference between the velocities of light in medium and glass is 6.25 x 107m/sec , then velocity of light in medium is (a)2.5 x 108m/s  (b) 0.125 x108m/s (c) 1.5 x 107m/s (d) 3 x 107m/s . ans-a

  • Integration

    UNITS AND DIMENSIONS ASSIGNMENT

    In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well.

    But for big addition problems, where the limits could reach to even infinity, integration methods are used. Integration and differentiation both are important parts of calculus. The concept level of these topics is very high. Hence, it is introduced to us at higher secondary classes and then in engineering or higher education.

    Ho can you increase the studding strength of physics watch the video given below–

    We can say, integration is the calculation of an integral. Integrals in mathematics are used to find many useful quantities such as areas, volumes, displacement, etc. When we speak about integrals, it is related to usually definite integrals. The indefinite integrals are used for anti-derivatives. Integration is one of the two major calculus topics in Mathematics, apart from differentiation(which measure the rate of change of any function with respect to its variables). It’s a vast topic which is discussed at higher level classes like in Class 11 and 12

    To download the important questions of indefinite integration click below.

    Assignment-Integraton1Download

    To download the important questions of indefinite integration click below.

    Assignment- Definite integration

    Assignment integration(2)Download

    These assignments are provided by Akash jha , Director of Ambika institute, situated in south-west Delhi . Contact NO.-8285114787.

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  • current objective assignment -1

                                             Current objective assignment – (1)

                                       Electric conduction ohm’s law and resistance.

    Answers of all the questions are given at the end of all questions of the assignment.

    1. for which if the following depends of drift velocity v on electric field E is Ohm’s law obeyed ?          (a) v α E    (b) v= constant    (c) v α E ­½      (d)  v α  E 2   .

    2. A potential difference is applied across the end of the metallic wire . If the potential difference is doubled , the drift velocity ;     (a) will be doubled,  (b) will be halved,   (c) will be quadrupled,   (d) no change .

    3. The maximum current that flows in the fuse wire , before it blows out , varies with the radius r as :(a) r3/2    (b)  r     (c) r2/3   (d) r ½

    4. The mass of three wires of copper are in the ratio of 1:3:5 and their length are in the ratio 5:3:1 . T he ratio of their electric resistance is ;   (a) 1:3:5  (b) 5:3:1 (c) 1:15:125 (d) 125:15:1

    5. The temperature (T) dependence of resistivity  of a semiconductor represented by :

    6. A wire of resistance R b= is stretched till its radius is half of its original value . Then the radius of the stretched wire is ; (a) 2R   (b) 4R   (c) 8R    (d) 16R   :

    7. A wire till its length is doubled of its original length , then the resistance of the stretched wire is; (a) 2R   (b) 4R   (c) 8R    (d) 16R   :

    8. If a copper wire is stretched to make it 0.1% longer, then the percentage change in its resistance is approximately (a) 0.1% (b) 0.2 %  (c) 0.4% (d) 0.8% .

    9. If a copper wire is stretched to make its radius decrease by 0.1% , then the % change in its resistance is approximately ; (a) 0.1% (b) 0.2 %  (c) 0.4% (d) 0.8% .

    10. A current of 5 amp exist in the 10 ohm resistance, for 4 minutes . How many coulombs pass through any cross-section of the resistor in this time ? (a) 12 (b) 120 (c) 1200 (d) 12000.

    11. The temperature coefficient of resistance of a wire is 0.00125/0C at 300K is one Ohm. The resistance of the wire will be 2 ohm at (a)1154K (b) 1100k (c) 1400k (d) 1127k .

    12. Drift speed of electrons is of the order of ; (a) zero (b) 10-2 (c) 105 (d) 100m/s ;

    13. Which represents the negative dynamic resistance?  

     (a) AB   (b) BC  (c)CD  (d) DA

    14. When a current flows in a wire , there exist in electric field in the direction of ;                               (a) flow of current, (b) opposite to the flow of current (c) perpendicular to the flow of current (d)at the angle of 450 to the flow of current.

    15. The drift velocity of free electron in a conductor is v when a current I is flow in it . If both the radius and current are doubled , then drift velocity will be;   (a) v (b) v/2 (c) v/4 (d) v/8 .

    16. The sides of rectangular block are 2,3 and 4cm . the ratio of maximum to minimum resistance between its parallel face is;    (a) 4    (b)3   (c) 2   (d) 1 ;

    17. The electric resistance of certain wire is R, if its length and radius both are doubled , then  ;        (a) resistance will be doubled and the specific resistance will be halved  (b) resistance will be halved and the specific resistance will be remain same  (c) resistance will be halved and the specific resistance will be doubled  (d) resistance will be unchanged and  also the specific resistance will be unchanged.

    18. Total no. of electrons present in 4 amp current flowing for 1 sec is ;  (a) 2.5 x 109 (b)2.5x 1020 (c) 1.25×1013 (d) 6x 108 .

    19. If the electric current through a bulb is 3.2 A, the no of electrons flowing through it in 1 sec is;    (a) 2 x 109  (b) 2 x 1019 (c) 3.2 x 109 (d) 1.6 x 1018 .

    20. The resistance of a wire is 5 ohm at 500c and 6 ohm at 1000c , the resistance of the wire at 00c will be;    (a) 3 ohm (b) 2 ohm (c) 1 ohm (d) 4 ohm .

    21. Two wires have length ,diameter and resistivity  all are in the ratio 1:2 the resistance of the first wire is 10 ohm, the resistance of the second wire is     ; (a) 5 (b) 10 (c) 20 (d) infinity.

    22. Two copper wire have their masses in the ratio 2:3 and the length in the ratio 3:4 , the ratio of their  resistance is  ;     (a) 4:9 (b) 27:32 (c) 16:9 (d) 27:128 (e) 1:2 ,

    23.On increasing the temperature of the conductor  its resistance increases because the  ;                 (a) relaxation time increases (b) mass of electron increases (c) electron density decreases (d) relaxation time decreases .

    24 .The resistance of a wire is R ohm . It is melted and stretched to n times its original length , the new resistance will be ; (a) R/n (b) n2R (c)R/ n2 (d) nR .

    25. The current flowing through a wire depends on time as I = 3t2 +2t +5 . The charge flowing through the cross-section of the wire in time from t=0 to t=2 sec is; (a) 22c (b) 20c (c) 18c (d) 5c.

    ans- Ans- 1-a   ;2- a   ; 3-a  ;4- d  ;5-b  ;6-d  ; 7-b ; 8-b ; 9-c  ;10-c ;  11-d; 12- b ; 13 -c;  14-a  ;15-b  ;16- a;  17-b ; 18-a ; 19-b  ; 20-d ; 21-b ; 22-b  ; 23-d   ;24-b  ;25-a