Questions 1 – 7 Carry 1 Score each. Answer any six questions. (6 \u00d7 1 = 6)<\/span><\/p>\nQuestion 1.
\nWrite the expression for intensity of electric field near to the surface of a charged conductor with uniform charge density \u03c3.
\nAnswer:
\nE = \\(\\frac{\\sigma}{\\varepsilon_{0}}\\)<\/p>\n
Question 2.
\nThe unit of electrical resistivity is ……….
\nAnswer:
\n\u2126m (ohm metre)<\/p>\n
Question 3.
\nE.m.f can be induced in a coil placed in an external magnetic field by:
\na) Changing the intensity of magnetic field
\nb) Changing the area of coil
\nc) Changing the orientation of the coil
\nd) All of the above
\nAnswer:
\nd) All of the above<\/p>\n
Question 4.
\nWrite the energy and momentum associated with a moving photon.
\nAnswer:
\nE = h\u03bd, P = mc<\/p>\n
Question 5.
\nThe value of relative permeability of a diamagnetic material is;
\na) \u03bcr<\/sub> > 1
\nb) \u03bcr<\/sub> = 1
\nc) \u03bcr<\/sub> < 1
\nd) \u03bcr<\/sub> = 1
\nAnswer:
\nc) \u03bcr<\/sub> < 1<\/p>\nQuestion 6.
\nThe electric field amplitude of an electromagnetic wave is 15V\/m. Find the magnetic field amplitude of the wave.
\nAnswer:
\nB = \\(\\frac{E}{C}=\\frac{15}{3 \\times 10^{8}}\\) = 5 \u00d7 10-8<\/sup>T<\/p>\nQuestion 7.
\nA convex lens is placed in a medium having refractive index greater than that of the lens. The lens now behaves as;
\na) Converging lens
\nb) Diverging lens
\nc) Plane glass plate
\nd) None of the above
\nAnswer:
\nb) Diverging lens<\/p>\n
Questions 8 to 15 carry 2 score each. Answer any 7 questions. (7 \u00d7 2 = 14)<\/span><\/p>\nQuestion 8.
\na) ‘Electrostatic field is always normal to the surface of a charged conductor’. Justify the statement.
\nb) The value of elelctric potential at the surface of a charged conductor is 10V. Find the vlaue of intensity of elecric field and potential at a point interior to it.
\nAnswer:
\na) The surface of charged conductor is an equi- potential surface. In equipotential surface, potential is same and hence electric field is always normal to surface of a charged conductor.
\nb) E = 0<\/p>\n
Question 9.
\nA galvanometer with a coil of resistance 12\u2126 shows a full scale deflection for a current of 2.5 mA. How can it be convereted into an ammeter of range 7.5 A?
\nAnswer:
\nG= 12\u2126, Ig<\/sub>= 2.5 mA = 2.5 \u00d7 10-3<\/sup>A, I = 7.5 A
\nS = \\(\\frac{I_{g} \\times G}{I-I_{g}}=\\frac{2.5 \\times 10^{-3} \\times 12}{7.5-2.5 \\times 10^{-3}}=4 \\times 10^{-3} \\Omega\\)<\/p>\nQuestion 10.
\nThe curves shown in figure are drawn for different magnetic materials. Among the three curves, name the curve that,
\n
\na) Represent the material usually used for making permanent magnets.
\nb) Represent the material usually used for making electromagnetis.
\nAnswer:
\na) A
\nb) C<\/p>\n
Question 11.
\nA capacitor C, a variable resistance R and a bulb B are connected in series to a.c. mains in the circuit as shown. The bulb glows with some brightness. How the glow of the bulb change if.
\n
\na) A dielectric slab is introduced between the plates of the capacitor.
\nb) The resistance R is increased keeping the same capacitance.
\nAnswer:
\na) The brightness of bulb increases.
\n(When dielectric slab is introduced, capacitance increases hence capacitive reactance decreases, then bulb glows with more brightness.)
\nb) The brightness of bulb decreases.<\/p>\n
Question 12.
\nJames Clerk Maxwell modified Ampere\u2019s circuital theorem by introducing the concept of displacement current.
\na) What do you mean by displacement current?
\nb) Write down the equation for displacement current.
\nAnswer:
\na) The current due to time varying electric field is called displacement current.
\nb) id<\/sub> = \\(\\varepsilon_{0} \\frac{\\mathrm{d} \\varphi}{\\mathrm{dt}}\\)<\/p>\nQuestion 13.
\nAn object AB is kept in front of a concave mirror as shown in figure.
\n
\na) Complete the ray diagram showing the image formation of object.
\nb) How will the position and intensity of image be affected if the lower half of the mirror\u2019s reflecting surface is painted black?
\nAnswer:
\na)
\n
\nb) The position of image does not charge. But the intensity decreases.<\/p>\n
Question 14.
\nAssuming that the two diodes D1<\/sub> and D2<\/sub> used in the electric circuit as shown in figure are ideal. Find out the value of current flowing through 2.5\u2126 resistor.
\n
\nAnswer:
\nSince D2<\/sub> is reverse biased, no current flows through 3\u2126. We need to consider only the other two resistance 3\u2126 and 2.5\u2126, which are connected in series. Total resistance, R = 3\u2126 + 2.5\u2126 = 5.5\u2126
\nv = 10V
\nI = \\(\\frac{V}{R}=\\frac{10}{5.5}\\) = 1.81 A<\/p>\nQuestion 15.
\na) Mention the function of the following used in the communication system
\n(i) Transducer
\n(ii) Transmitter
\nb) Figure shows the block diagram of a AM transmitter. Identify the boxes X and Y.
\n
\nAnswer:
\na) i) Transducer: The device that converts one form of energy into another is called tranducer.
\nii) Transmitter – Atransmittertransmits the information<\/p>\n
b) x \u2192 Modulator
\ny \u2192 Power Amplifier<\/p>\n
Questions 16 to 22 carry 3 scores each. Answer any 6 questions. (6 \u00d7 3 = 18)<\/span><\/p>\nQuestion 16.
\nThree capacitors of capacitances 2\u00b5F, 3\u00b5F and 4\u00b5F are connected in series.
\na) Find the equivalent capacitance of the combination
\nb) The plates of a parallel plate capacitor have an area 20 cm2<\/sup> each are separated by a distance of 2.5 mm. The capacitor is charged by connecting it to a 400V supply. How much electrostatic energy is stored in the capacitor?
\nAnswer:
\na) C1<\/sub> = 2\u03bcF = 2 \u00d7 10-6<\/sup>F
\nC2<\/sub> = 3\u03bcF = 3 \u00d7 10-6<\/sup>F
\n
\nb) A = 20cm2<\/sup> = 20 \u00d7 10-4<\/sup>m2<\/sup>
\nd = 2.5 mm = 2.5 \u00d7 10-3<\/sup>m
\n\u03b50<\/sub> = 8.85 \u00d7 10-12<\/sup>C2<\/sup>\/Nm2<\/sup>
\nv = 400v
\n<\/p>\nQuestion 17.
\nA circuit using potentiometer and a battery of negligible internal resistance is set up as shown to develop a constant potential gradient along the wire PQ. Two cells of e.m.fs E1<\/sub> and E2<\/sub> are connected in sereis as in figure in combination 1 and 2. The balance points are obtained respectively at 400 cm and 240 cm from point P. Find
\n
\na) The ratio between E1<\/sub> and E2<\/sub>
\nb) Balancing length for the cell with emf E1<\/sub> only.
\nAnswer:
\na) From the connection (1), we get
\nE1<\/sub> + E2<\/sub> = k \u00d7 400 ………(1)
\nk = constant
\nFrom the connection (2), we get
\nE2<\/sub> – E1<\/sub> = 240 K ……….(2)
\nAdding (1) and (2), we get
\n2E2<\/sub> = 640 K
\nE2<\/sub> = 320 K
\nSubtituting for E2<\/sub> in equation (1)
\nE1<\/sub> + 320 K = 400 K
\nE1<\/sub> = 80 K
\nThe ratio \\(\\frac{E_{1}}{E_{2}}=\\frac{1}{4}\\)
\nb) l1<\/sub> be the balancing length of E1<\/sub>
\nE1<\/sub> = kl1<\/sub>
\n80k \u221d kl1<\/sub>
\nl1<\/sub> = 80 cm<\/p>\nQuestion 18.
\nA conducting rod PQ of length \u2018l\u2019 connected to a resistance \u2018R\u2019 is moved at a uniform speed \u2018V\u2019, normal to a uniform magnetic field \u2018B\u2019.
\n
\na) Deduce the expresssion for e.m.f induced in the conductor.
\nb) Find the magnitude and direction of current through the conductor.
\nAnswer:
\na) flux, \u03a6 = BA
\nHere, Area A = ldx
\ndx is small displacement rod in time \u0394 T
\n\u03a6 = B \u00d7 l \u00d7 dx
\n<\/p>\n
Question 19.
\nIn tuner circuits, we use the phenomenon of resonance.
\na) Write the condition of resonance in series LCR circuit.
\nb) A series LCR circuit uses L = 0.1 H, C = 10\u03bcF and R = 100\u2126. Find the value of frequency at which the amplitude of current is maximum.
\nAnswer:
\na) L\u03c9 = \\(\\frac{1}{\\mathrm{C} \\omega}\\)
\nb) L = 0.1 H, C = 10\u03bcF = 10 \u00d7 10-6<\/sup>F
\n<\/p>\nQuestion 20.
\nThe focal length of a lens has dependence on its radii of curvatures and refractive index. Derive Lens maker\u2019s formula.<\/p>\n
Question 21.
\na) Write the expression for the de Broglie wavelength associated with a charged particle having charge \u2018q\u2019 and mass \u2018m\u2019, when it is accelerated by a protential of \u2018V\u2019 volts.
\nb) A proton and an electron have same kinetic energy. Which one has greater value of de Broglie wavelength and why?
\nAnswer:
\n
\nFor electron, \u03bbe<\/sub> = \\(\\frac{h}{\\sqrt{2 m_{e} K E_{e}}}\\)
\nBut KEp<\/sub> = KEe<\/sub> and mp<\/sub> > Me<\/sub>, hence \u03bbe<\/sub> > \u03bbp<\/sub>
\nElectron has greater de Broglie wavelength.<\/p>\nQuestion 22.
\nFind the binding energy per nucleon of \\({ }_{20}^{40} \\mathrm{Ca}\\) nucleus.
\nGiven m\\({ }_{20}^{40} \\mathrm{Ca}\\) = 39.962589u.
\nmp = 1.00783u, mn = 1.00867u.
\nTake 1 amu = 931 MeV\/c2<\/sup>
\nAnswer:
\nMass detect, \u0394m = Zmp<\/sub> + (A – Z) mn<\/sub> – M
\n\u0394m = 20 mp<\/sub> + 20 mn<\/sub> – 39.962589
\n= 20 \u00d7 1.00783 + 20 \u00d7 1.00867 – 39.962589
\n= 0.367411 u
\nBinding energy = 0.367411 \u00d7 931 = 342.06 MeV<\/p>\nQuestions from 23 – 26 carry 4 scores. Answer any 3 questions. (3 \u00d7 4 = 12)<\/span><\/p>\nQuestion 23.
\nGauss\u2019s theorem is useful for finding the intensity of elelctric field.
\na) Write the Gauss\u2019s law in its mathematical form
\nb) Using the law, prove that intensity of electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
\nAnswer:
\n\\(\\phi=\\oint \\overrightarrow{\\mathrm{E}} \\cdot \\mathrm{d} \\overrightarrow{\\mathrm{s}}=\\frac{\\mathrm{q}}{\\varepsilon_{0}}\\)<\/p>\n
Question 24.
\na) Express Ohm\u2019s law in terms of current density, electrical resistivity and intensity of electric field.
\nb) Explain the variation of resistance of a semiconductor with temperature. Also draw the graph showing the variation of resistivity of silicon with temperature.
\nAnswer:
\na) J = \u03c3E
\nb) As temperature increases, the number density of electrons (\u03b7) of semiconductor increases.
\n\u03c1 = \\(\\frac{m}{h e^{2} \\tau}\\)
\nHence resistivity decreases with increase in temperature.
\n<\/p>\n
Question 25.
\nThe relation between magnetic field and current is given by Biot-Savart\u2019s law.
\na) Write the expression for the magnetic field at point along the axis of a circuit loop of radius \u2018R\u2019 carrying a current ‘I’.
\nb) From the above expression. Find the value of magnetic field at the centre of the loop.
\nc) Sketch the magnetic field lines for current carrying circular loop
\nAnswer:
\na) B = \\(\\frac{\\mu_{0} \\mathrm{NIR}^{2}}{2\\left(\\mathrm{x}^{2}+\\mathrm{R}^{2}\\right)^{3 \/ 2}}\\)
\nI \u2192 Current
\nN \u2192 Number of turns
\nX \u2192 distance tot he point from the center of loop
\nb) B = \\(\\frac{\\mu_{0} N I}{2 a}\\)
\nc)
\n<\/p>\n
Question 26.
\n
\na) Identify the diode in the circuit and write the use of the resistance Rs.
\nb) Explain how the diode helps as to stabilize the output voltage of the circuit.
\nc) Name the type of biasing used in this diode
\nAnswer:
\na) Zenerdiode, Voltage regulator
\nb) Solar Cell: Solar cell is junction diode used to convert solar energy into electrical energy.<\/p>\n
Circuit details:
\n
\nIts p-region is thin and transparent and is called emitter. The n-region is thick and is called base. Output is taken across RL<\/sub>.<\/p>\nWorking: When light falls on this layer, electrons from the n-region cross to the p-region and holes in the p-region cross into the n-region. Thus a voltage is developed across RL<\/sub>.
\nSolar cells are used to charge storage batteries during daytime.
\nc) Reverse bias<\/p>\nQuestions 27 to 29 carry 5 scores. Answer any 2 questions. (2 x 5 = 10)<\/span><\/p>\nQuestion 27.
\nIn the figure, PQ is the incident ray on the equilateral glass prism ABC.
\n
\na) Complete the ray diagram showing the passage of light and mark the angle of deviation.
\nb) Derive an expression for the refractive index of the material of the prism.
\nAnswer:
\n
\nb)
\n
\nABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. \u2220A is the angle of prism.<\/p>\n
A ray PQ incidents on the face AB at an angle i1<\/sub>. QR is the refracted ray inside the prism, which makes two angles r1<\/sub> and r2<\/sub> (inside the prism). RS is the emergent ray at angle \u2018i2<\/sub>\u2019<\/p>\nThe angle between the emergent ray and incident ray is the deviation ‘d\u2019.
\nIn the quadrilateral AQMR,
\n\u2220Q + \u2220R = 180\u00b0
\n[since N1<\/sub>M and NM are normall
\ni.e. \u2220A + \u2220M = 18O\u00b0 ………(1)
\nIn the \u0394QMR,
\n\u2234 r1<\/sub> + r2<\/sub> + \u2220M = 180\u00b0……..(2)
\nComparing eq (1) and eq (2)
\nr1<\/sub> + r2<\/sub> = \u2220A ……….(3)
\nFrom the \u0394QRT,
\n(i1<\/sub> – r1<\/sub>) + (i2<\/sub> – r2<\/sub>) = d
\n[since exterior angle equal sum of the opposite interior angles]
\n(i1<\/sub> + i2<\/sub>) – (r1<\/sub> + r2<\/sub>) = d
\nbut, r1<\/sub> + r2<\/sub> = A
\n\u2234 (i1<\/sub> + i2<\/sub>) – A = d
\n(i1<\/sub> + i2<\/sub>) = d + A ……..(4)
\nIt is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D.
\nAt the minimum deviation position,
\ni1<\/sub> = i2<\/sub> =i, r1<\/sub> = r2<\/sub> = r and d=D
\nHence eq (3) can be written as,
\n
\n
\nIt is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.<\/p>\nQuestion 28.
\nAccording to Christian Huygens wave theory, light emanating from a source as wave fronts.
\na) What is the shape of wave front emerging form a linear source?
\nb) Derive the mathematical expression for the bandwidth of interference bands obtained in young\u2019s double slit experiment with the help of suitable diagram.
\nAnswer:
\na) Cylinder
\nb)
\n
\nS1<\/sub> and S2<\/sub> are two coherent sources having wave length A. Let \u2018d\u2019 be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O\u2019 is a point on the screen equidistant from S1<\/sub> and S2<\/sub>.
\nHence the path difference, S1<\/sub>O – S2<\/sub>O = 0
\nSo at ‘O’ maximum brightness is obtained.
\nLet \u2018P\u2019 be the position of nm bright band at a distance xn<\/sub> from O. Draw S1<\/sub>A and S2<\/sub>B as shown in figure.
\nFrom the right angle \u0394S1<\/sub>AP
\n
\n
\n
\nThis is the width of the bright band. It is the same for the dark band also.<\/p>\nQuestion 29.
\na) The radius of nth stationary orbit of hydrogen atom is:
\nrn<\/sub> = \\(\\frac{\\mathrm{n}^{2} \\mathrm{~h}^{2} \\varepsilon_{0}}{\\pi \\mathrm{me}^{2}}\\)
\nUsing Bohr postulates, obtain an expression for the energy of electron in the stationary states of H-atom.
\nb) Draw the energy level diagram showing how the spectrallines corresponding to Balmerseries occur due to transition between the energy levels.
\nAnswer:
\na) Consider hydrogen atom which has one electron revolving around nucleus having one proton. The centripetal force for electron is provided by electrostatic force between electron and proton
\n
\nb) Energy level diagram of hydrogen atom
\n
\nNote: An electron can have any total energy above E=0ev. In such situations electron is free. Thus there is a continuum of energy states above E = 0ev.<\/p>\n