Question 5.
\nWrite any two characteristic properties of nuclear force. [1]
\nAnswer :
\nCharacteristic properties of nuclear forces are:<\/p>\n
- \n
- Nuclear forces are strongest forces in nature.<\/li>\n
- Nuclear forces are charge independent.<\/li>\n<\/ul>\n
Question 6.
\nTwo bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. [1]
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-1.png?resize=390%2C108&ssl=1\")
\nAnswer:
\nIn this situation, a will become positive with respect to b, as current induced is in clockwise direction.
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-2.png?resize=386%2C106&ssl=1\")
\nQuestion 7.
\nWhat happens to the width of depletion layer of a p-n junction when it is
\n(i) forward biased,
\n(ii) reverse biased ? [1]
\nAnswer :
\n(i) Forward biased : Potential drop across the junction decreases and diffusion of holes and electrons across the junction increases. It makes the width of the depletion layer smaller.
\n(ii) Reverse biased : Potential drop across the junction increases and diffusion of holes and electrons across the junction decreases. It makes the width of the depletion layer larger.<\/p>\nQuestion 8.
\nDefine the term ‘stopping potential’ in relation to photoelectric effect. [1]
\nAnswer :
\nStopping potential is the minimum negative (retarding) potential of anode for which photo current stops or becomes zero. It is denoted by Vs<\/sub>. The value of stopping potential is different for different metals but it is independent of the intensity of incident light and depends on the frequency of the incident light.<\/p>\nQuestion 9.
\nA thin straight infinitely long conducting wire having charge density \u03bb is enclosed by a cylindrical surface of radius r and length ‘l’ its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [2]
\nAnswer :
\nCharge enclosed by the cylindrical surface q = \u03bbl
\n\\(\\begin{aligned} \\mathrm{Flux}, & \\phi=\\frac{q}{\\varepsilon_{0}} \\\\ \\phi &=\\frac{\\lambda l}{\\varepsilon_{0}} \\end{aligned}\\)<\/p>\nQuestion 10.
\nPlot a graph showing the variation of Coulomb force (F) versus \\(\\left(\\frac{1}{r^{2}}\\right)\\), where r is the distance between the two charges of each pair of charges : (1 \u03bcC, 2 \u03bcC) and (2 \u03bcC, -3 \u03bcC). Interpret the graphs obtained. [2]
\nAnswer :
\nThe following graph shows the variation of Coulomb force (F) versus r.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-3.png?resize=367%2C502&ssl=1\")
\nQuestion 11.
\nWrite the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity \\(\\vec{v}\\) in a magnetic field \\(\\vec{B}\\) . Show that no work is done by this force on the charged particle. [2]
\nAnswer:
\nMagnetic Lorentz force is given by
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-4.png?resize=367%2C181&ssl=1\")
\nQuestion 12.
\nWhat are eddy currents ? Write any two applications of eddy currents. [2]
\nAnswer :
\nWhen a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.
\nApplications of eddy currents :<\/p>\n- \n
- Magnetic brakes in trains.<\/li>\n
- Electromagnetic damping.<\/li>\n
- Induction furnaces.<\/li>\n
- Electric power meter.<\/li>\n<\/ul>\n
Question 13.
\nWhat is sky wave communication ? Why is this mode of propagation restricted to the frequencies only up to few MHz ? [2]<\/p>\nQuestion 14.
\nIn the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-5.png?resize=375%2C479&ssl=1\")
\nQuestion 15.
\nA parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. [2]
\nAnswer:
\nGauss’ law states that the electric flux \u03a6E of a parallel plate capacitor having an area A, and a total charge Q is given by
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-6.png?resize=366%2C349&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-7.png?resize=375%2C327&ssl=1\")
<\/p>\nQuestion 16.
\nNet capacitance of three identical capacitors in series is 1 \u03bcF. What will be their net capacitance if connected in parallel ? Find the ratio of energy stored in the two configurations if they are both connected to the same source. [2]
\nAnswer:
\nNet capacitance in series = 1 \u03bcF
\nIf C1<\/sub> = C2<\/sub> = C3<\/sub> = C
\nLet C be the capacitance of each of three capacitors and Cs<\/sub> and Cp<\/sub> be the capacitance of series and parallel combination respectively.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-8.png?resize=318%2C422&ssl=1\")
\nQuestion 17.
\nUsing the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release in energy in the processes of nuclear fission and nuclear fusion can be explained.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-9.png?resize=363%2C193&ssl=1\")
\nAnswer:
\nThe above curve shows that:<\/p>\n- \n
- When a heavy nucleus breaks into two medium sized nuclei (in nuclear fission), the BE\/nucleon increases resulting in the release of energy.<\/li>\n
- When two small nuclei combine to form a relatively bigger nucleus in nuclear fusion, BE\/nucleon increases, resulting in the release of energy.<\/li>\n<\/ul>\n
Question 18.
\nIn the meter bridge experiment, balance point was observed at J with AJ = l. [2]
\n(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point ?
\n(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected ?
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-10.png?resize=388%2C541&ssl=1\")
\n(ii) By interchanging galvanometer and battery, there will be no change in the balance point position.<\/p>\nQuestion 19.
\nA convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
\n(i) a medium of refractive index 1.65,
\n(ii) a medium of refractive index 1.33. [3]
\n(a) Will it behave as a converging or a diverging lens in the two cases ?
\n(b) How will its focal length change in the two media ?
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-11.png?resize=371%2C398&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-12.png?resize=373%2C472&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-13.png?resize=368%2C114&ssl=1\")
\n(b) (i) As seen from equation (ii), focal length becomes negative and increases in magnitude.
\n(ii) As seen from equation (ii), focal length remains positive and increases in magnitude<\/p>\nQuestion 20.
\nDraw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v1<\/sub> > v2<\/sub>, of incident radiation having the same intensity. In which case will be stopping potential be higher ? Justify your answer. [3] Answer:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-14.png?resize=303%2C238&ssl=1\")
\nStopping potential is more for the curve corresponding to the frequency v2<\/sub>\u00a0(\u2235 v1<\/sub>> v2<\/sub>). This is due to the fact that with increase in the frequency, the kinetic energy of emitted photo electrons also increases. Therefore, we need more negative potential to stop these electrons.<\/p>\nQuestion 21.
\nWrite briefly any two factors which demonstrate the need for modulating a signal. Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as a modulating signal. [3]<\/p>\nQuestion 22.
\nUse the mirror equation to show that [3]
\n(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
\n(b) a convex mirror always produces a virtual image independent of the location of the object.
\n(c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
\nAnswer:
\nMirror equation is given as,
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-15.png?resize=368%2C476&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-16.png?resize=366%2C270&ssl=1\")
\nQuestion 23.
\nDraw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input-output wave forms. [3]
\nAnswer:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-17.png?resize=420%2C225&ssl=1\")
\nRectification : Rectification means conversion of ac into dc. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode conducts only when it is forward biased; it does not conduct when it is reverse biased.<\/p>\nWorking : The ac input voltage across secondary S1<\/sub> and S2<\/sub> changes polarity after each half cycle. Suppose during the first cycle of input ac signal, the terminal S1<\/sub> is positive relative to centre tap and S2<\/sub> is negative relative to it. Then diode D1<\/sub> is forward biased and D2<\/sub> is reverse biased. Therefore, diode D1<\/sub> conducts while D2<\/sub> does not. Thus, the current in load resistance RL<\/sub> is in the same direction for both half cycles of input ac signal and the output current is a continuous series of unidirectional pulses.
\nIn a full wave rectifier, if input frequency is f Hertz, then output frequency will be 2f Hertz because for each cycle of input, two positive half cycles of output are obtained.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-18.png?resize=344%2C297&ssl=1\")
\nQuestion 24.
\n(a) Using de-Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
\n(b) The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ? [3]
\nAnswer :
\n(a) The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized, i.e.,
\n\\(m v r=\\frac{n h}{2 \\pi}, \\text { where } n=1,2,3, \\ldots . .\\)
\nAccording to de-Broglie, a stationary orbit is that which contains an integral number of de-Broglie waves associated with the revolving electron.
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-19.png?resize=371%2C330&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-20.png?resize=369%2C552&ssl=1\")
\nQuestion 25.
\nYou are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. [3]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-21.png?resize=297%2C120&ssl=1\")
\nQuestion 26.
\nA compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope. [3]
\nOR
\nA giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.<\/p>\nIf this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.42 x 106<\/sup> m and the radius of the lunar orbit is 3.8 x 108<\/sup> m.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-22.png?resize=334%2C373&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-23.png?resize=372%2C341&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-24.png?resize=371%2C370&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-25.png?resize=348%2C344&ssl=1\")
\nQuestion 27.
\nTwo heating elements of resistances R1<\/sub> and R2<\/sub> when operated at a constant supply of voltage V, consume powers P1<\/sub> and P2<\/sub> respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in
\n(i) series and
\n(ii) parallel across the same voltage supply. [3]
\nAnswer:
\n(i) In series,
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-26.png?resize=372%2C213&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-27.png?resize=267%2C263&ssl=1\")
\nQuestion 28.
\n(a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.
\n(b) In what respect is a toroid different from a solenoid ? Draw and compare the pattern of the magnetic field lines in the two cases.
\n(c) How is the magnetic field inside a given solenoid made strong ? [5]
\nOR
\n(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
\n(b) “Increasing the current sensitivity of a galvonometer may not necessarily increase its voltage sensitivity.” Justify this statement.
\n(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range.
\nAnswer:
\n(a) Magnetic field inside a long solenoid is uniform every where and approximately zero outside it. Fig. shows a sectional view of long solenoid current coming out of the plane of the papers at points marked (\u2022) and current entering the plane of the paper at points marked (x). To find the magnetic field B at any point inside the solenoid, consider a rectangular loop abed as Amperian loop. According to Ampere’s circuital law,
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-28.png?resize=274%2C175&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-29.png?resize=396%2C555&ssl=1\")
\n(b) Difference: In a toroid, magnetic lines do not exist outside the body. Toroid is closed whereas the solenoid is open on both sides. Magnetic field is uniform inside a toroid whereas for solenoid, it is different at the two ends and center.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-30.png?resize=326%2C436&ssl=1\")
\n(c) Strength of magnetic field :
\n(1) By inserting the ferromagnetic substance inside the solenoid.
\n(2) By increasing the current through the solenoid.
\nOR
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-31.png?resize=402%2C251&ssl=1\")
<\/p>\nPrinciple : Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.
\nWorking : When current (I) is passed in the coil, torque t acts on the coil, given by
\nT = NIAB sin \u03b8
\nwhere, \u03b8 = Angle between normal to plane of coil
\nB = Magnetic field of strength
\nN = No. of turns in a coil.
\nFor equilibrium,
\ndeflecting torque = restoring torque
\nNIAB = C\u03b8
\n\\(\\Rightarrow \\quad \\theta=\\frac{\\mathrm{NAB}}{\\mathrm{C}} \\mathrm{I}\\)
\nwhere,\u00a0 C = Torsional rigidity of the wire
\n\u21d2\u00a0 \u03b8 \u221d I
\nThe deflection of coil is directly proportional to the current flowing in the coil.
\n(b) Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension wire.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-32.png?resize=367%2C223&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-33.png?resize=344%2C102&ssl=1\")
\nIt means voltage sensitivity is dependent on current sensitivity and resistance of galvanometer, R. If we increase current sensitivity and galvanometer resistance is high, then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity.
\n(c) Conversion of a galvanometer to ammeter :
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-34.png?resize=357%2C131&ssl=1\")
\nA galvanometer can be converted into ammeter by connecting a shunt (low resistance) in parallel with the galvanometer and its value is given by
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-35.png?resize=103%2C68&ssl=1\")
\nQuestion 29.
\nState the working of a.c. generator with the help of a labelled diagram. The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity \u03c9. Deduce the expression for the alternating e.m.f. generated in the coil. What is the source of energy generation in this device ? [5]
\nOR
\n(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by \u03c0\/2 in phase.
\n(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
\n(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.
\n(ii) What is the direction of the e.m.f. ?
\n(iii) Which end of the wire is at the higher potential ?
\nAnswer:
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-36.png?resize=267%2C230&ssl=1\")
\nWorking : When a coil (armature) rotates inside a uniform magnetic field, magnetic flux linked with the coil changes w.r.t. time. This produces an e.m.f. according to Faraday’s law.<\/p>\nFor first half of the rotation, the current will be from one end (first ring) to the other end (second ring). For second half of the rotation, it is in opposite sense.<\/p>\n
To calculate the magnitude of e.m.f. induced, Suppose
\nN = number of turns in the coil.
\nA = area enclosed by each turn of coil.
\n\\(\\vec{B}=\\) = strength of magnetic field.
\n\u03b8= angle which normal to the coil makes with \\(\\vec{B}=\\) at any instant t,
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-37.png?resize=376%2C488&ssl=1\")
\nPut in (ii), if we denote NAB\u03c9 as e0<\/sub>, then e = e0<\/sub> sin \u03c9t
\nSource of energy: Mechanical energy.
\nThe word generator is a misnomer, because nothing is generated by the machine, it is an alternator converting one form of energy to another.
\nOR
\n(a) Circuit containing inductance only : Let an alternating emf given by E = E0<\/sub> sin \u03c9t, …(i)
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-38.png?resize=213%2C113&ssl=1\")
<\/p>\nbe applied across a pure (zero resistance) coil of inductance L. As the current i in the coil grow continuously, an opposing emf is induced in the coil whose magnitude is \\(\\mathrm{L} \\frac{d i}{d t}\\) where \\(\\frac{d i}{d t}\\)\u00a0 is the rate of change of current. But this should be zero because there is no resistance in the current. Thus,
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-39.png?resize=367%2C512&ssl=1\")
\nFrom equations (i) and (ii), it is proved that voltage is ahead of current by \u03c0\/2.
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-40.png?resize=296%2C136&ssl=1\")
\n(b) (i) e = BLv
\n(ii) Direction of e.m.f is from west to east.
\n(iii) Wire 1 is at greater potential than wire 2.<\/p>\nQuestion 30.
\nState the importance of coherent sources in the phenomenon of interference. In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width. How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water ? [5]
\nAnswer :
\nTwo sources of light which continuously emit light waves of same frequency with a zero or constant phase difference between them are called coherent sources. They are necessary to produce sustained interference pattern.<\/p>\nA thin film of oil spread over water shows beautiful colors due to interference of light. If coherent sources are not taken, the phase difference between the two interfering waves will change continuously and a sustained interference pattern will not be obtained.
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-41.png?resize=375%2C320&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-42.png?resize=370%2C486&ssl=1\")
\nConstructive interference: The intensity of light will be maximum at those places where the path difference between the interfering light waves is zero or an integral multiple of \u03bb, i.e., \u03bb , 2\u03bb ,……….
\nHence for maximum intensity, we have
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-43.png?resize=205%2C81&ssl=1\")
\nDestructive interference : The intensity of light will be minimum at those places where the path difference between the interfering light-waves is
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-44.png?resize=366%2C257&ssl=1\")
\ndestructive interference.
\nSo fringe width is directly proportional to \u03bb On immersing the apparatus in water, the wavelength of light decreases (\u03bb\u03c9 = \u03bb\/n).
\nTherefore, fringe width will decrease in water.<\/p>\nCBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-II<\/h3>\n
Note : Except for the following questions, all the remaining questions have been asked in previous Set.<\/p>\n
Question 1.
\nA hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere ? [1]
\nAnswer :
\nPotential inside the charged sphere is constant and equal to potential on the surface of conductor. Therefore, potential at the center of sphere is 5 V.<\/p>\nQuestion 2.
\nHow are X-rays produced ? 11]
\nAnswer :
\nX-rays are produced when electron strike a metal target. The electrons are liberated from the heated filament and accelerated from the high voltage towards the metal target. X-rays are also produced when electrons collide with the atom and nuclei of metal target.<\/p>\nQuestion 3.
\nWhere on the surface of Earth is the angle of dip zero ? [1]
\nAnswer :
\nAt magnetic equator angle of dip is zero.<\/p>\nQuestion 4.
\nState the principle of working of a transformer. Can a transformer be used to step up or step down a d.c. voltage ? Justify your answer. [2]
\nAnswer :
\nTransformer principle : It is a device which converts high voltage a.c. into low voltage a.c. and vice-versa.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-45.png?resize=289%2C162&ssl=1\")
\nIt is based upon the principle of mutual induction. When alternating current is passed through a coil, an induced e.m.f. is set up in the neighboring coil.<\/p>\nWorking: When an alternating current is passed through the primary, the magnetic flux through the iron core changes which does two things. It produces e.m.f. in the primary and an induced e.m.f. is also set up in the secondary. If we assume that the resistance of primary is negligible, the back e.m.f. will be equal to the voltage applied to the primary.
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-46.png?resize=369%2C325&ssl=1\")
\nA transformer can not be used to step up or step down a d.c. voltage because d.c. can not produce a changing magnetic flux in the core of the transformer and no emf will be induced.<\/p>\nQuestion 5.
\nIn the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-47.png?resize=339%2C193&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-48.png?resize=367%2C197&ssl=1\")
<\/p>\nQuestion 6.
\nWhat is ground wave communication ? On what factors does the maximum range of propagation in this mode depend ? [2]<\/p>\nQuestion 7.
\nA convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
\n(i) a medium of refractive index 1.6,
\n(ii) a medium of refractive index 1.3. [3]
\n(a) Will it behave as a converging or a diverging lens in two cases ?
\n(b) How will its focal length change in the two media ?
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-49.png?resize=362%2C458&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-50.png?resize=369%2C174&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-51.png?resize=370%2C303&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-52.png?resize=367%2C287&ssl=1\")
<\/p>\nCBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-III<\/h3>\n
Note : Except for the following questions, all the remaining questions have been asked in previous Set.<\/p>\n
Question 1.
\nA hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V. What is the potential at the centre of sphere ? [1]
\nAnswer :
\nPotential inside the charged sphere is constant and equal to potential on the surface of conductor. So therefore, potential at the centre of sphere is 12 V.<\/p>\nQuestion 2.
\nHow are microwaves produced ? [1]
\nAnswer :
\nMicrowaves are electromagnetic waves with wavelength ranging from as long as meter to as short as one millimeter, or equivalently with frequencies between 300 MFz and 300 GHz. Microwaves are produced by vacuum tubes devices that operate on the ballistic motion of electron controlled by magnetic or electric fields. Some different kinds of microwaves emitters are the cavity magnetron, the klystron, the travelling wave tube (TWT), the gyrotron and all stars.<\/p>\nQuestion 3.
\nMention various energy losses in transformer. [2]
\nAnswer :
\nMagnetic core losses are exaggerated with higher frequencies, eddy currents in the iron core, resistance of windings or copper loss, hysteresis loss and flux leakage are energy losses in transformers. Transformers energy losses tend to worsen with increasing frequency.<\/p>\nQuestion 4.
\nIn the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-53.png?resize=361%2C472&ssl=1\")
\nQuestion 5.
\nA thin straight infinitely long conducting wire having charge density \u03bb is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [2]
\nAnswer:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-54.png?resize=249%2C309&ssl=1\")
\nThere will be no electric flux through the circular ends of the cylinder.
\nSo, according to Gauss’s law,
\n\\(\\mathrm{Flux}, \\quad \\phi=\\frac{q}{\\varepsilon_{0}}\\)
\nSince, charge enclosed by gaussian surface
\n\\(\\begin{aligned} \\text { i.e. } & q=\\lambda \\times l \\\\ \\therefore & \\phi=\\frac{\\lambda l}{\\varepsilon_{0}} \\end{aligned}\\)<\/p>\nQuestion 6.
\nA converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length. [3]
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-55.png?resize=372%2C405&ssl=1\")
<\/p>\nCBSE Previous Year Question Papers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"
CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-I Question 1. Define electric dipole moment. Write its S.I. unit. [1] Answer : Electric dipole moment : Dipole moment is a measure of strength of electric dipole. It is vector quantity whose …<\/p>\n","protected":false},"author":29,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\n
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It is vector quantity whose …\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.cbselabs.com\/cbse-previous-year-question-papers-class-12-physics-2011-outside-delhi\/\" \/>\n<meta property=\"og:site_name\" content=\"CBSE Labs\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/ncertsolutionsbooks\/\" \/>\n<meta property=\"article:published_time\" content=\"2019-10-25T09:35:35+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-09-18T09:45:58+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-1.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:creator\" content=\"@learncbse\" \/>\n<meta name=\"twitter:site\" content=\"@learncbse\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Bhagya\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"18 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.cbselabs.com\/#organization\",\"name\":\"LearnCBSE\",\"url\":\"https:\/\/www.cbselabs.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/ncertsolutionsbooks\/\",\"https:\/\/www.instagram.com\/learncbse.in\/\",\"https:\/\/www.youtube.com\/user\/CBSEPapers\",\"https:\/\/in.pinterest.com\/learncbsein\/\",\"https:\/\/twitter.com\/learncbse\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.cbselabs.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2017\/01\/LearnCBSE.in_.png?fit=210%2C198&ssl=1\",\"contentUrl\":\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2017\/01\/LearnCBSE.in_.png?fit=210%2C198&ssl=1\",\"width\":210,\"height\":198,\"caption\":\"LearnCBSE\"},\"image\":{\"@id\":\"https:\/\/www.cbselabs.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.cbselabs.com\/#website\",\"url\":\"https:\/\/www.cbselabs.com\/\",\"name\":\"CBSE Labs\",\"description\":\"NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12\",\"publisher\":{\"@id\":\"https:\/\/www.cbselabs.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.cbselabs.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.cbselabs.com\/cbse-previous-year-question-papers-class-12-physics-2011-outside-delhi\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-1.png?fit=390%2C108&ssl=1\",\"contentUrl\":\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/10\/CBSE-Previous-Year-Question-Papers-Class-12-Physics-2011-Outside-Delhi-1.png?fit=390%2C108&ssl=1\",\"width\":390,\"height\":108,\"caption\":\"CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi 1\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.cbselabs.com\/cbse-previous-year-question-papers-class-12-physics-2011-outside-delhi\/#webpage\",\"url\":\"https:\/\/www.cbselabs.com\/cbse-previous-year-question-papers-class-12-physics-2011-outside-delhi\/\",\"name\":\"CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi - 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