The experimental result differs from the assumption. The result shows that the ring oscillator frequencies can be grouped into two groups, where the first group is the high frequency group for the test cases with a frequency at around 247 Mhz, and the second group is the low frequency group for the test cases with a frequency at around 235 Mhz. The difference between the average frequency of these two group is 11.64 MHz, which is about 5% of the average frequency. The standard deviations of the fast and slow group are 0.193 and 0.192 respectively. These results also show that this observation is not only true for a single CLB but all the CLB ring oscillators. When we observe the frequency distribution of these 32 test cases, we can see that actually it is of some pattern, as seen in Figure 6.6. The observation result for the first nine test cases is shown in Table 6.5. The frequency change seems to depend on the parity of the 5-bit test case control signal. When the parity of the test case is even, this test case belonged to the low frequency group. In contrast, the odd parity test cases have a high frequency. This can not be explained by the LUT structure which we assumed, as the change of the test cases is only choosing a different path in the LUTs. The path lengths are the same when the test configuration is not changed. Considering the test cases 0 and 1, the paths are very slightly changed. These kinds of the small difference may not cause such a significant change in ring oscillator frequency up to 5%. The current understanding of the LUT structure seems to be inaccurate. But still, as the LUTs are implemented with the 6-input XOR functions and all the input combinations are 57 6 Results Figure 6.6: Frequency comparison of 32 different test cases for a single CLB ring oscillator in one test configuration Table 6.5: Parity dependence of the frequency Test Case Binary Frequency 0 0000 low 1 0001 high 2 0010 high 3 0011 low 4 0100 high 5 0101 low 6 0110 low 7 0111 high 8 1000 high covered, every path that accesses the SRAM bit entries is tested and characterized. No matter what the hidden LUT structure is, the function and the input pins of the LUTs are identical for all the ring oscillators in a same test configuration and test case. As the ring oscillators in the array are using the same hard macro, it is ensured that they have an identical logic function and routing, so the delays of each CLB in different test cases are compared. Hence, the ring oscillator frequencies reflect the delay variation from CLB to CLB. 58 6.4 Temperature Dependency of Delay 6.4 Temperature Dependency of Delay The following experiments investigate the temperature influence on the ring oscillator frequency and the measurement operations influence on the temperature. Two types of experiments are presented in this section, one is using an external heating equipment to control the temperature, the other is measuring the self-heating in the FPGA caused by the proposed delay characterization method. 6.4.1 External Heating The setup of the external heating experiment is shown in Figure 6.7. The heating equipment is refitted from a hot air gun which is normally used for desolering. The fan of the FPGA board is removed and the muzzle of the hot air gun is attached to the heat sink of the FPGA board. We can adjust the heating speed by controlling the temperature and the air pressure of the hot air gun. Figure 6.7: The experiment setup for external heating In the external heat up experiment using the hot air gun, two rows of the ring oscillators in the array mentioned in Section 6.3 are chosen. In both of these two cases, hot air gun temperature is set to 100 , used the same heating air pressure. After each experiment, the hot air gun is turned off, the FPGA is put in the idle mode until it cools down and reaches the temperature equilibrium and then we start another experiment. The heating experiment lasts for ten minutes and results are sampled every second. As we would expect when the FPGA is heated up, the temperature raises and the ring oscillator frequency drops down, shown in Figure 6.9 (a). From Figure 6.9 (b) we can observe 59 6 Results that the temperature and frequency have an approximately linear correlation. Moreover, as the lines which depict the frequency decline of the ring oscillators are parallel, it shows that the frequency-temperature correlation is not depending on the ring oscillator speed. That

means in a reasonable operating temperature range (below 100 ), the frequency differences between every two ring oscillators are constant. Even though the absolute path delay will increase as temperature rise, the delay variation among the CLBs can be well measured by the ring oscillator array. (a) (b) Figure 6.8: Temperature and frequency correlation of CLB row 5 for array at coordinate X48Y94 (a) (b) Figure 6.9: Temperature and frequency correlation of CLB row 9 for array at coordinate X48Y94 Linear regression analysis is run for the two rows of ring oscillators. T

Coulomb's law states that the electrostatic force of attraction or repulsion acting between two stationary point charges is given by F=140q1q2r2F=140q1q2r2 where F denotes the force between two charges q₁and q₂separated by a distance rin free space, E_ois a constant known as permittivity of free space. Free space is vacuum and may be taken to be air practically. If free space is replaced by a medium, then E_ois replaced by (E_ok) or (E_oEr)where k is known as dielectric constant or relative permittivity. (i) In coulomb's law, F =kq1q2r2kq1q2r2,then on which of the following factors does theproportionality constant k depends?

(a) Electrostatic force acting between the two charges

(b) Nature of the medium between the two charges

(c) Magnitude of the two charges

(d) Distance between the two charges

(ii) Dimensional formula for the permittivity constant E_oof free space is

(a)[ML3T4A2](a)[ML3T4A2]	(b)[M1L3T2A2](b)[M1L3T2A2]
(c)[M1L3T4A2](c)[M1L3T4A2]	(d)[ML3T4A2](d)[ML3T4A2]

(iii) The force of repulsion between two charges of 1 C each, kept 1 m apart in vaccum is

(a)19109N(a)19109N	(b)[M1L3T2A2](b)[M1L3T2A2]
(c)9107N(c)9107N	(d)191012N(d)191012N

(iv) Two identical charges repel each other with a force equal to 10 mgwt when they are 0.6 m apart in air. (g = 10 ms^-2). The value of each charge is

(a) 2 mC

(b) 2 x10-7mC

(c) 2 nC

(d) 2C

(v) Coulomb's law for the force between electric charges most closely resembles with

(a) law of conservation of energy	(b) Newton's law of gravitation
(c) Newton's 2ndlaw of motion	(d) law of conservation of charge

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Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electrons which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron. [1.6 x 10^-19C] i.e. q = ne where n = 1,2,3,4, .... Hence no body can have a charge represented as1.1e,2.7e,35e,etc.1.1e,2.7e,35e,etc. Recently, it has been discovered that elementary particles such as protons or neutrons are composed of more elemental units called quarks. (i) Which of the following properties is not satisfied by an electric charge?

(a) Total charge conservation	(b) Quantization of charge
(c) Two types of charge	(d) Circular line of force

(ii) Which one of the following charges is possible?

(a) 5.8 x 10-18C	(b) 3.2 x 10-18C
(c) 4.5 x 10-19C	(d) 8.6 x 10-19C

(iii) If a charge on a body is1 nC, then how many electrons are present on the body?

(a) 6.25 x 1027	(b) 1.6 x 1019
(c) 6.25 x 1028	(d) 6.25 x 109

(iv)If a body gives out 10⁹electrons every second, how much time is required to get a total charge of 1 C from it?

(a) 190.19 years

(b) 150.12 years

(c) 198.19 years

(d) 188.21 years

(v) A polythene piece rubbed with wool is found to have a negative charge of3.2 x 1O^-7C.Calculate the numberof electrons transferred.

(a) 2 x 1012

(b) 3 x 1012

(c) 2 x 1014

(d) 3 x 1014

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Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure,having sides oflength L = 10.0 cm.The electric field is uniform, has a magnitude E = 4.00 x 10³N C^-Iand isparallel to the xy plane at an angle of 37 measured from the +x-axis towards the +y-axis. (i) Electric flux passing through surface S₆is

(a) -24 N m2C-1

(b) 24 N m2C-1

(c) 32 Nm2C-1

(d) -32 N m2C-1

(ii) Electric flux passing through surface s₁is

(a) -24 N m2C-1

(b) 24 N m2C-1

(c) 32 N m2C-1

(d) -32 N m2C-1

(iii) The surfaces that have zero flux are

(a) S1and S3

(b) S5and S6

(c) S2and S4

(d) S1and S2

(iv) The total net electric flux through all faces of the cube is

(a) 8 N m2C-1

(b) -8 N m2C-1

(c) 24 N m2C-1

(d) zero

(v) The dimensional formula of surface integralEdSEdSof an electric field is

(a) [M L2T-2A-1]	(b) [M L3T-3A-1]
(c) [M-1L3T-3A]	(d) [M L-3T-3A-1]

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When a charged particle is placed in an electric field, it experiences an electrical force. If this is the only force on the particle, it must be the net force. The net force will cause the particle to accelerate according to Newton's second law. So Fe=qE=maFe=qE=ma IfEEis uniform, thenaais constant anda=qE/ma=qE/m.If the particle has a positive charge, its acceleration is inthe direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite to theelectric field. Since the acceleration is constant, the kinematic equations can be used. (i) An electron of mass m, charge e falls through a distance h metre in a uniform electric field E. Then time of fall,

(a)t=2hmeE(a)t=2hmeE

(b)t=2hmeE(b)t=2hmeE

(c)t=2eEhm(c)t=2eEhm

(d)t=2eEhm(d)t=2eEhm

(ii) An electron moving with a constant velocity v along X-axis enters a uniform electric field applied along Y-axis. Then the electron moves.

(a) with uniform acceleration along Y-axis	(b) without any acceleration along Y-axis
(c) in a trajectory represented as y = ax2	(d) in a trajectory represented as y = ax

(iii) Two equal and opposite charges of masses m_land m₂are accelerated in an uniform electric field throughthe same distance. What is the ratio of their accelerations if their ratio of masses ism1m2=0.5?m1m2=0.5?

(a)a1a2=2(a)a1a2=2

(b)a1a2=0.5(b)a1a2=0.5

(c)a1a2=3(c)a1a2=3

(d)a1a2=1(d)a1a2=1

(iv) A particle of mass m carrying charge q is kept at rest in a uniform electric field E and then released. The kinetic energy gained by the particle, when it moves through a distance y is

(a)12qEy2(a)12qEy2

(b)qEy(b)qEy

(c)qEy2(c)qEy2

(d)qE2y(d)qE2y

(v) A charged particle is free to move in an electric field. It will travel

(a) always along a line of force

(b) along a line of force, if its initial velocity is zero

(c) along a line of force, if it has some initial velocity in the direction of an acute angle with the line of force

(d) none of these.

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In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled Mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of 1.6 x 10^-19C the charge of the electron. For this, he won the Nobel prize. (i) If a drop of mass 1.08 x 10^-14kg remains stationary in an electric field of 1.68 x 10⁵N C^-I, then the charge of this drop is

(a) 6.40 x 10-19C	(b) 3.2 x 10-19C
(c) 1.6 X 10-19C	(d) 4.8 x 10-19C

(ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are

(a) 4

(b) 3

(c) 5

(d) 8

(iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m^-1.If the mass of the drop is 1.6 X 10^-3g, the number of electrons carried by the drop is (g= 10 m s^-2)

(a) 1018

(b) 1015

(c) 1012

(d) 109

(iv) The important conclusion given by Millikan's experiment about the charge is

(a) charge is never quantized	(b) charge has no definite value
(c) charge is quantized	(d) charge on oil drop always increases.

(v) If in Millikan's oil drop experiment, charges on drops are found to be8C,12C,20C8C,12C,20Cthen quanta ofcharge is

(a)8C(a)8C

(b)20C(b)20C

(c)12C(c)12C

(d)4C(d)4C

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Gauss's law and Coulomb's law, although expressed in different forms, are equivalent ways of describing the relation between charge and electric field in static conditions. Gauss's law is0=qencl0=qencl,when q_enclis the netcharge inside an imaginary closed surface called Gaussian surface.=EdA=EdAgives the electric flux through the Gaussian surface. The two equations hold only when the net charge is in vacuum or air. (I) If there is only one type of charge in the universe, then(EElectric field,dsArea vector)(EElectric field,dsArea vector)

(a)Eds0Eds0on any surface

(b)EdsEdscould not be defined

(c)Eds=Eds=if charge is inside

(d)Eds=0Eds=0if charge is outside,Eds=q0Eds=q0if charge is inside

(ii) What is the nature of Gaussian surface involved in Gauss law of electrostatic?

(a) Magnetic

(b) Scalar

(c) Vector

(d) Electrical

(iii) A charge 10C is placed at the centre of a hemisphere of radius R = 10 cmas shown. The electric flux through the hemisphere (in MKS units) is

(a) 20 x 105

(b) 10 x 105

(c) 6 x 105

(d) 2 x 105

(iv) The electric flux through a closed surface area S enclosing charge Q is.If the surface area is doubled, thenthe flux is

(a)2(a)2

(b)/2(b)/2

(c)/4(c)/4

(d)(d)

(v) A Gaussian surface encloses a dipole. 'The electric flux through this surface is

(a)q0(a)q0

(b)2q0(b)2q0

(c)q20(c)q20

(d) zero

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Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in the given figure, the electric field at P is stronger that at Q. (i) Electric lines of force about a positive point charge are

(a) radially outwards	(b) circular clockwise
(c) radially inwards	(d) parallel straight lines

(ii) Which of the following is false for electric lines of force?

(a) They always start from positive charges and terminate on negative charges

(b) They are always perpendicular to the surface of a charged conductor

(c) They always form closed loops

(d) They are parallel and equally spaced in a region of uniform electric field

(iii) Which one of the following pattern of electric line of force in not possible in filed due to stationary charges?

(iv) Electric lines of force are curved

(a) in the field of a single positive or negative charge	(b) in the field of two equal and opposite charges
(c) in the field of two like charges	(d) both (b) and (c).

(v) The figure below shows the electric field lines due to two positive charges. The magnitudes E_A,E_Band E_cof the electric fields at points A, Band C respectively are related as

(a)EA>EB>EC(a)EA>EB>EC

(b)EB>EA>EC(b)EB>EA>EC

(c)EA=EB>EC(c)EA=EB>EC

(d)EA>EB=EC(d)EA>EB=EC

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When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole inuniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However some work is done in rotating the dipole against the torque acting on it. (i)The dipole moment of a dipole in a uniform external fieldEEisPPThen the torqueacting on the dipole is

(a)=PE(a)=PE

(b)=PE(b)=PE

(c)=2(P+E)(c)=2(P+E)

(d)=(P+E)(d)=(P+E)

(ii) An electric dipole consists of two opposite charges, each of magnitude 1.0Cseparated by a distance of2.0 cm. The dipole is placed in an external field of 10⁵NC^-1. The maximum torque on the dipole is

(a)0.2103Nm(a)0.2103Nm

(b)1103Nm(b)1103Nm

(c)2103Nm(c)2103Nm

(d)4103Nm(d)4103Nm

(iii) Torque on a dipole in uniform electric field is minimum whenis equal to

(a) 0

(b) 90

(c) 180

(d) Both (a) and (c)

(iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torqueon thedipole are

(a)F=0,=0(a)F=0,=0

(b)F0,0(b)F0,0

(c)F=0,0(c)F=0,0

(d)F0,=0(d)F0,=0

(v) An electric dipole of momentp is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an anglewith the direction of the field. Assumingthat the potential energyof the dipole to be zero when=90,the torque and the potential energy of the dipole will respectively be

(a)pEsin,pEcos(a)pEsin,pEcos

(b)pEsin,2pEcos(b)pEsin,2pEcos

(c)pEsin,2pEcos(c)pEsin,2pEcos

(d)pEcos,pEsin(d)pEcos,pEsin

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In practice, we deal with charges much greater in magnitude than the charge on an electron, so we can ignore the quantum nature of charges and imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as continuous charge distribution. There are three types of continuous charge distribution : (i) Line charge distribution (ii) Surface charge distribution (iii) Volume charge distribution as shown in figure. (I)Statement 1: Gauss's law can't be used to calculate electric field near an electric dipole. Statement 2: Electric dipole don't have symmetrical charge distribution.

(a) Statement 1 and statement 2 are true	(b) Statement 1 is false but statement 2 is true
(c) Statement 1 is true but statement 2 is false	(d) Both statements are false

(ii) An electric charge of8.85 X 10^-13C is placed at the centre of a sphere of radius 1 m. The electric fluxthrough the sphere is

(a) 0.2 N C-1m2

(b) 0.1 N C-1m2

(c) 0.3 N C-1m2

(d) 0.01 N C-1m2

(iii) The electric field within the nucleus is generally observed to be linearly dependent on r. So,

(a) a=O

(b)a=R2(b)a=R2

(c) a=-R

(d)a=2R3(d)a=2R3

(iv) What charge would be required to electrify a sphere of radius 25 cm so as to get a surface charge density of3Cm2?3Cm2?

(a) 0.75 C

(b) 7.5 C

(c) 75 C

(d) zero

(v) The SI unit of linear charge density is

(a) Cm

(b) Cm-1

(c) C m-2

(d) C m-3

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Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e.,=dqdS=dqdSTwolarge. thin metal plates are parallel and close to each other. On their inner faces, the plates have surface chargedensities of opposite signs having magnitude of17.0 x 10^-22C m^-2as shown. The intensity of electric field at apoint isE=0E=0where00=permittivity of free space. (i) E in the outer region of the first plate is

(a)171022N/C(a)171022N/C

(b)1.51025N/C(b)1.51025N/C

(c)1.91010N/C(c)1.91010N/C

(d) zero

(ii) E in the outer region of the second plate is

(a)171022N/C(a)171022N/C

(b)1.51015N/C(b)1.51015N/C

(c)1.91010N/C(c)1.91010N/C

(d) zero

(iii) E between the plates is

(a)171022N/C(a)171022N/C

(b)1.51015N/C(b)1.51015N/C

(c)1.91010N/C(c)1.91010N/C

(d) zero

(iv) The ratio of E from right side of B at distances 2 cm and 4 ern, respectively is

(a) 1: 2

(b) 2: 1

(c) 1: 1

(d)1:2(d)1:2

(v) In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is

(a) spherical

(b) cylindrical

(c) straight line

(d) none of these

5

Two point charges q₁and q₂of unequal magnitude are placed as shown below (i) Determine the ratio q₁:q₂ (ii)If one !lull point is at infinity, then where is another null point? (iii)If q₁and q₂are separated by a distance of 10 em, then find the position of a null point. (iv) Will a positive charge follow the electric lines of force if free to move?

5

An electric dipole is a system consisting of the two equal and opposite point charges seperated by a small and finite distance. If dipole moment of thissystem isppand it is placed in a uniform electricfieldEE. (i) Write the expression of torque experienced by a dipole. (ii) Identify two pairs of perpendicular vectors in the expression. (iii) Show diagrammatically the orientation of the dipole in the field for which the torque is (a) Maximum. (b) Half the maximum value. (c) Zero

5

A thin conducting shell contains a charge +Q distributed uniformly all overit. Now a point charge +q₁is placed at the centre ofthe shell, and another charge +q₂is pLacedoutside the shell. What is the net force on (i) charge q₁ (ii) charge q₂ (iii) spherical shell (iv) also determine charge density on the shell if radius of shell is R. (v) What is the net electric flux through the sphere?

A rectangular coil of n turns each of area A, carrying current I, when suspended in a uniform magnetic field B, experiences a torque =nIBAsin=nIBAsin Where is the angle which a normal drawn on the plane of coil makes with the direction of magnetic field. This torque tends to rotate the coil and bring it in an equilibrium position. In the stable equilibrium state, the resultant force on the coil is zero. The torque on the coil is also zero and the coil has minimum potential energy. Read the above passage and answer the following questions: (i) In which position, a current carrying coil suspended in uniform magnetic field experiences (a) minimum torque and (b) maximum torque? (ii) a circular coil of 200 turns, radius 5 cm carries a current of 2.0 A. It is suspended vertically in a uniform horizontal magnetic field of 0.20 T, with the plane of the coil making an angle with6060 the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning. (iii) what is the basic value displayed by the above study?

5

That night Vaikunth was preparing for his physics exam. Suddenly, the light in his room went off and he could not continue his studies. His cousin brother Vasu who had come to visit him was quick to react. Vasu using the torch (an android application) installed in his mobile phone found that the fuse had blown out. He checked the wiring and located a short circuit. He checked the wiring and located a short circuit. He rectified it and put a fuse wire. The light came to life again. Vaikunth had a sign of releif and continued his studies. Read the above passage and answer the following question. (i) What are the values projected by Vaikunth and Vasu? (ii) Why did Vasu have to check the wiring? (iii) What is an electric fuse? What characteristics you would prefer for a fuse wire?

5

Two linear parallel conductors carrying currents in the same direction attract each other and two linear parallel conductors carrying in opposite directions repel each other. The force acting per unit length due to currents I1andI2I1andI2in two linear parallel conductors held distance r apart in vacuum in SI unit is F=022I1I2rF=022I1I2r Read the above passage and answer the following questions: (i) What is the basic reason for the force between two linear parallel conductors currents? (ii) Two straight wires A and B of lengths 2 cm and 20 cm, carrying currents 2.0 A and 5.0 A respectively in opposite directions are lying parallel to each other 4.0 cm apart. The wire A is held near the middle of wire B. What is the force on 20 cm long wire B? (iii) What does this study imply in day to day life?

5

Self- inductance is the property of a coil by virtue of which the coil oppose any change in the strength of current flowing through it by inducing an emf in itself. The induced emf is also called back emf. Self-inductance represents electric inertia which is measured in terms of coefficient of self-inductance(L). We can show that L = L=IL=I=eI/teI/t Read the above passage and answer the following questions: (i) How does the self-inductance of a coil represent its electric inertia? (ii) An emf of 100V is induced in a coil when the current in it changes from 5A to 1A in0.4s. Find the self-inductance of the coil.

5

Raj is in XII standard. His Physics teacher demonstrated an experiment to explain Faraday's laws of electromagnetic induction. Raj interrupted his lecture and asked, "Is there any possibility of induced emf due to the earth's magnetism"? The teacher was stunned for a moment and gave this question for group discussion. Finally, the students came out with correct answer. (i) Write the values that you learnt from this incident. (ii) What can be reason for Raj's question?

5

Mr.Kamal and Jayant were playing football in the ground. In the middle of the match, Kamal fell down as there was a cramp in his right leg. Jayant rushed towards him and picked him up and told him not to move. They went to the nearby hospital.Doctor examined his legs and advised for an X-ray test for the confirmation of fracture. (i) What are the values Jayant show here? (ii) How are X-rays produced? (iii)Can you give one other application of X-ray?

5

Mr.Vishwanathan, a retired professor of physics was walking with his grandson. It was last week of December and it was drk aruond 5.30 pm. The streetlights were on and the yellow light flooded the area around. The boy asked professor why yellow lights were used when light was beighter. The professor answered that during foggy days the tiny droplets act as prisms splitting white light light into its constituwnt colours and thus reducing the clarity. Read the above passage and give the answer of the following questions: (i) What phenomenon was the professor referring to? Why does it happen? (ii) Give one application of prism. (iii) What values of the boy are being reflected from the conversation?

5

Work function of a metal is the minimum amount of energy required by an electron to just escape from the metal surface without any kinetic energy.The expression for work function of metal is 0=hv0=hc/00=hv0=hc/0 where h is Plank's constant;v₀is the threshold wavelength and c is the velocity of light in vacuum. (i) Why different metals has different work function? (ii) What is the threshold wavelength of the incident radiation for a metal surface whose work function is 1.2 eV.Givenh=6.631034Js;h=6.631034Js;1eV=1.61019J1eV=1.61019J (iii) What do you learn from this study?

5

Nuclear fission is the phenomenon of splitting of a heavy nucleus into two or more lighter nuclei. Nuclear fussion is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. In both these processes, a certain mass (m)(m) disappears, which appears in the form of nuclear energy, E=(m)c2.E=(m)c2. The release of energy is so sudden that it cannot be controlled. This causes havoc. Nuclear fission is the basis of an atom bomb and nuclear fusion is the basis of a nydrogen bomb. A powerful device, called Nuclear Reactor has been developed, where in nuclear energy produced is utilised for constructive purposes. Read the above passage and answer the following questions: (i) How much energy is released in the fission of one nucleus of U235U235 and in how much time? (ii) Give an estimate of devastation potential of an atomic explosion. (iii) Should we ban nuclear research in this field? Give your views briefly.

5

A nuclear reactor is a powerful device, wherein nuclear energy is utilised for peaceful purposes. It is based upon controlled nuclear chain reaction. The nuclear chain reaction is controlled by the use of control rods (of boron or cadmium) and moderators like heavy water, graphite, etc. The whole reactor is protected with concrete walls 2 to 2.5 metre thick, so that radiations emitted during nuclear reactions may not produce harmful effects. Read the above passage and answer the following questions: (i) Give any two merits of nuclear reactors. (ii) What is radiactivewaste? (iii) Why do people often oppose the location site of a nuclear reactor? What do you suggest?

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Shyamsawhis younger brother wondering with a question which deals with emission of light from a vapour lamp. He was anxious to know how different colours wrebeing emitted by different lights. He also saw mercury and sodium vapour lamps in the Physics lab and was curious to know what is inside the lamps. On seeing his anxiety to know more about it, Shyam explained about absorption of energy and re-emission of photons in the visible region. He also advised him not to touch or break any items in the lab for the knowledge. Read the above passage and answer the following questions: (i) What is the moral you derivefrom Shyam's behaviour? (ii) Which series in the hydrogen spectrum is in the visible region? (iii) Write the quality displayed by Shyam's brother.

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A pure semiconductor germaniumor silicon,free of every impurity is called intrinsic semiconductor, At room temperature, a pure semiconductor has a very small number of current carriers (electrons and holes).Hence,its conductivity is low. When the impurity atoms of valence five or three are doped in a pure semiconductor,we get respectivelynn type orpp type extrinsic semiconductor.In case of a doped semiconductor.nenh=n2i;nenh=ni2;whenneneandnhnhare the number density of electrons and holes respectively andniniis the number density of intrinsic charge carriers in a pure semiconductor.The conductivity of extrinsic semiconductor is much higher than that that of intrinsic semiconductor. Read the above passage and answer the following question: (i) Name two materials to be doped in pure semiconductor of silicon to get (a)pptype semiconductor (b)nntype semiconductor (ii)What do you learn from the above study?

5

Khusbhu Mowar is an electronic engineer. She was working in a project for college exhibition. Her younger sister asked Khushubu about a small black coloured device. which Khusbhu was working on. She clearly explained the working of integrated circuit(I_C) to her sister. Read the above passage and answer the following questions: (i) What according to you, are the values displayed by Khusbhu? (ii) What is an integrated circuit? What are the uses of the integrated circuit? (iii) What do you mean by microprocessor?

5

Mohan went to the market to purchase a TV set. He got confused with so many features and functionsof electronic appliances. He took the help of his friend Sohan, a science student.Sohan explained him about the communication system, digital and analog signals. This knowledge proved of great help to Mohan in purchasing a colour TV. Read the above passage and the answer the following questions: (i) What do you mean by the term communication? (ii)Which type of signals are better? (iii) What type of nature Sohan has? (iv)What is the minimum number of communication satellites required for global communication coverage?

5

Group discussion was arranged in class XII on the topic of atmosphere. Three groups were made. Teacher asked the question "why can Moonbe not used as acommunication satellite?" Answers were given by all the three groups. Each, groupcan give only one reason. Teacher told them that reason given by each group is correct. The groups collected all the three, reasons and came to correct conclusion," (i) Give the correct reason for the above question. (ii) What values were showed by all the three groups?