Time: 2 Hours<\/span> General Instructions to candidates:<\/span><\/p>\n The given value of constants can be used wherever necessary.<\/span><\/p>\n Velocity or light in vacuum, c = 3 \u00d7 108<\/sup> m\/s Questions 1 and 2 carry 1 Score each. Answer both. (2 \u00d7 1 = 2)<\/span><\/p>\n Question 1. Question 2. Answer any six questions from question number 3 to 10. Each question carries 2 Scores. (Scores: 6 \u00d7 2 = 12)<\/span><\/p>\n Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Answer any five questions from question number 11 to 16. Each question carries 3 Scores. (Scores: 5 \u00d7 3 = 15)<\/span><\/p>\n Question 11. Question 12. Question 13. Question 14. b) Consider two thin convex lenses of focal lengths f1<\/sub> and f2<\/sub> kept in contact. Let O be an object kept at a distance \u2018u\u2019 from the first lens L1<\/sub>, I1<\/sub> is the image formed by the first lens at a distance v1<\/sub>. Question 15. b) Speed decreses Question 16. b) According to Law of Radioactive decay, Answer any four questions from question number 17 to 21. Each question carries 4 Scores. (4 \u00d7 4 = 16)<\/span><\/p>\n Question 17. b) Question 18. Question 19. b) when magnetic flux in a metal block changes, induced currents are produced. This current is called eddy current. Question 20. Question 21. Answer any three questions from question number 22 to 25. Each question carries 5 Scores. (Scores: 3 \u00d7 5 = 15)<\/span><\/p>\n Question 22. Question 23. Question 24.
\nCool off time : 15 Minutes
\nMaximum : 60 Score<\/span><\/p>\n\n
\nPlank\u2019s constant, h = 6.64 \u00d7 10-34<\/sup>\u00a0Js
\nCharge of electron, e = 1.6 \u00d7 10-19<\/sup>C
\nMass of the electron, m = 9.1 \u00d7 10-31\u00a0<\/sup>kg
\nAvogadro number = 6.025 \u00d7 1023<\/sup> atom\/mol
\n\u03b50<\/sub> = 8.854\u00a0 \u00d7 10-12\u00a0<\/sup>C2\u00a0<\/sup>N-1\u00a0<\/sup>m-2<\/sup>
\n\u03bc0<\/sub> = 4\u03c0 \u00d7 10-7\u00a0<\/sup>Tm\/A<\/p>\n
\nHow will you represent a resistance of 3700 \u03a9 \u00b1 5% using colour code?
\nAnswer:
\nOrange, Violet, Red and Gold.<\/p>\n
\nDraw the symbol of NAND gate.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-1.png\")
<\/p>\n
\nTwo equal and opposite charges placed in air as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-2.png\")
\na) Redraw the figure and show the direction of dipole moment (P), direction of resultant electric field (E) at P.
\nb) Write an equation to find out the electric field at P.
\nAnswer:
\na)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-3.png\")
\nb) \\(\\overline{\\mathrm{E}}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\overline{\\mathrm{P}}}{\\mathrm{r}^{3}}\\)<\/p>\n
\nCalculate the electrical capacitance of earth. The radius of earth is 6400 km.
\nAnswer:
\nCapacitance of earth C = \\(4 \\pi \\varepsilon_{0} \\mathrm{R}\\)
\nC = \\(4 \\pi \\varepsilon_{0}\\) \u00d7 6400 \u00d7 103<\/sup>
\n= 710.8 \u00b5F<\/p>\n
\nWhat do you mean by drift velocity? Write the relation between drift velocity and electric current.
\nAnswer:
\nThe average velocity acquired by an electron under the applied electric field is called drift velocity.
\nI = neAVd<\/sub><\/p>\n
\nA galvanometer is connected as shown in the figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-4.png\")
\na) This combination can be used as ………..
\n(voltmeter\/rheostat\/ammeter)
\nb) Derive an expession to find the value of resistance S.
\nAnswer:
\na) Ammeter
\nb)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-5.png\")
\nLet G be the resistance of the galvanometer, giving full deflection for a current Ig<\/sub>.
\nTo convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement Ig current flows through Galvanometer and remaining (I – Ig) current flows through shunt resistance.
\nSince G and S are parallel
\nP.d Across G = p.d across S
\nIg \u00d7 G = (I – Ig)s
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-35.png\")
<\/p>\n
\nAn unpolarised light incident on a medium as shown in figure.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-6.png\")
\nShow that the direction of reflected and refracted waves are mutually perpendicular.
\nAnswer:
\nAccording to Brewster\u2019s law
\nn = tan p = \\(\\frac{\\sin p}{\\cos p}\\) ………(1)
\nAccording to Snell\u2019s law
\nn = \\(\\frac{\\sin p}{\\sin x}\\) …….(2)
\nComparing (1) and (2), we get
\n\\(\\frac{\\sin p}{\\sin x}=\\frac{\\sin p}{\\sin x}\\)
\nsin x = cos p = sin(90 – p)
\nx = 90 – p, but p = r
\nx = 90 – r
\nx + r =90\u00b0
\nIf x + r is 90\u00b0, angle between reflected and refracted ray will be 90\u00b0.<\/p>\n
\nHow many electrons, protons and neutrons are there in 16g of 8<\/sub>O16<\/sup>?
\nAnswer:
\nNumber of atoms in 16g of 8<\/sub>O16<\/sup> = 6.023 \u00d7 1023<\/sup>
\nNumber of protons = 8 \u00d7 6.023 \u00d7 1023<\/sup>
\nNumberof neutrons = 8 \u00d7 6.023 \u00d7 1023<\/sup><\/p>\n
\nThe given block diagram shows general form of a communication system.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-34.png\")
\na) Identify the blocks X and Y.
\nb) What is the difference between attenuation and amplification?
\nAnswer:
\na) X – transmitter
\nY – Receiver
\nb) Attenuation: The loss of strength of a signal while propagating through a medium is known as attenuation.
\nAmplification: It is the process of increasing the amplitude of a signal.<\/p>\n
\nWhat is the population covered by the transmission, if the average population density around the tower is 1200 km-2<\/sup>? The antenna height is 160 m.
\nAnswer:
\nDistance travelled by the wave d = \\(\\sqrt{2 R h}\\)
\nArea = \u03c0d2<\/sup>
\n= \u03c0(\\(\\sqrt{2 R h}\\))2<\/sup>
\n= \u03c02Rh
\n= 3.14 \u00d7 2 \u00d7 6400 \u00d7 0.160 Km2<\/sup>
\nPopulation covered = Area \u00d7 population density
\n= 3.14 \u00d7 2 \u00d7 6400 \u00d7 0.16 \u00d7 1200
\n= 7.72 \u00d7 106<\/sup><\/p>\n
\nTwo spheres encloses charges as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-7.png\")
\na) Derive and expression for electric field intensity at any point on the surface S2.
\nb) What is the ratio of electric flux through S1<\/sub> and S2<\/sub>?
\nAnswer:
\na) According to Guess law
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-8.png\")
\nb) Flux a charge
\nfor first sphere \u03a61<\/sub> \u03b1 Q
\nfor second sphere \u03a62<\/sub> \u03b1 3Q
\n\u2234 \\(\\frac{\\phi_{1}}{\\phi_{2}}=\\frac{Q}{3 Q}=\\frac{1}{3}\\)
\n= 1 : 3<\/p>\n
\nThe electric field lines of a positive charge is as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-9.png\")
\na) Give the sign of potential difference Vp – Vq.
\nb) Give the sign of work done by the field in moving a small positive charge from \u2018Q\u2019 to \u2018P\u2019.
\nc) What is the shape of equipotential surface near the charge?
\nAnswer:
\na) positive.
\nb) negative.
\nc) spherical.<\/p>\n
\na) Name the part of electromagnetic spectrum:
\ni) used in radar systems.
\nii) produced by bombarding a metal target by a high speed electrons.
\nb) Electromagnetic waves are produced by ………..
\n(charges at rest\/charges in uniform motion\/ charges in accelerated motion)
\nc) Why only microwaves are used in microwave ovens?
\nAnswer:
\na) i) Micro waves
\nii) X – rays
\nb) Charges in accelerated motion.
\nc) The frequency of micro waves matches with the rotational frequency of water molecule.<\/p>\n
\nTwo thin convex lenses of focal length f1<\/sub> and f2<\/sub> are placed in contact:
\na) If the object is at principal axis, draw ray diagram of the image formation by this combination of lenses.
\nb) Obtain a general expression for effective focal length of the combination in terms of f1<\/sub> and f2<\/sub>.
\nAnswer:
\na)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-10.png\")
<\/p>\n
\nThen from the lens formula, we can write,
\n\\(\\frac{1}{f_{1}}=\\frac{1}{v_{1}}-\\frac{1}{u}\\) ……..(1)
\nThis image will act as the virtual object for the second lens and the final image is formed at I (at a distance v). Then
\n\\(\\frac{1}{f_{2}}=\\frac{1}{v}-\\frac{1}{v_{1}}\\) ……..(2)
\nAdding eq (1) and eq (2)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-11.png\")
\nIf the two lenses are replaced by a single lens of focal length \u2018F\u2019 the image is formed at \u2018v\u2019. Then we can write,
\n\\(\\frac{1}{F}=\\frac{1}{v}-\\frac{1}{u}\\) …….(4)
\nfrom eq(3) and (4),
\n\\(\\frac{1}{F}=\\frac{1}{f_{1}}+\\frac{1}{f_{2}}\\) …….(3)<\/p>\n
\nA plane wave-front AB incident on a surface XY and undergoes refraction as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-12.png\")
\na) Prove Snell\u2019s law of refraction by using this figure.
\nb) What is the change in speed of propagation and frequency of the refracted wave?
\nAnswer:
\na) AB is the incident wavefront and c1<\/sub> is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c2<\/sub> is the velocity of the wavefront in the second medium. AC is a plane separating the two media.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-13.png\")
\nThe time taken for the ray to travel from P to R is
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-14.png\")
\nO is an arbitrary point. Hence AO is a variable.
\nBut the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq (2) should be zero.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-15.png\")
\nwhere 1<\/sup>n2<\/sub> is the refractive index of the second medium w.r.t. the first. This is the law of refraction.<\/p>\n
\nfrequency remains same.<\/p>\n
\na) State radioactive decay law.
\nb) Prove the relation N(t) = N0<\/sub>e-\u03bbt<\/sup>.
\nAnswer:
\na) The number of nuclei undergoing decay per unit time is proportional to number of nuclei in the sample at that time.<\/p>\n
\n\\(\\frac{\\mathrm{d} \\mathrm{N}}{\\mathrm{d} \\mathrm{t}}\\) = -\u03bbN
\n\\(\\frac{\\mathrm{d} \\mathrm{N}}{\\mathrm{d} \\mathrm{t}}\\) = -\u03bbdt
\nIntegrating
\nIn N = -\u03bbt + C …….. (1)
\nC is the constant of integration. To get value of C, let us assume that initially (t = 0) the number of nuclei be N0<\/sub>.
\n\u2234 C = In N0<\/sub>
\nSubstituting for C in equation (1) we get,
\nIn N – In N0<\/sub> = -\u03bbt
\nIn \\(\\frac{\\mathrm{N}}{\\mathrm{N}_{0}}\\) = -\u03bbt
\n\\(\\frac{\\mathrm{N}}{\\mathrm{N}_{0}}\\) e-\u03bbt
\nN = N0<\/sub>e-\u03bbt<\/sup><\/p>\n
\nTwo charged partjeles q1 and q2 are moving through a uniform magnetic field (B) as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-16.png\")
\na) What is the shape of path of q1<\/sub> and q2<\/sub>.
\nb) Derive an expression for cyclotron frequency with the help of a neat diagram.
\nAnswer:
\na) q1<\/sub> charges undergoes helical motion and q2<\/sub> moves in a circular path.<\/p>\n
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-17.png\")
\nLet ‘v’ be the velocity of ion, q the charge of the ion and B the magnetic flux density. If the ion moves along a semicircular path of radius ‘r\u2019, then we can write
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-18.png\")
<\/p>\n
\nThe B-H curve of a ferromagnetic material is as shown in figure:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-19.png\")
\na) What do you mean by ferromagnetic material? Write any two properties of it.
\nb) Identify the value of retentivity and coercivity from the figure.
\nc) A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate magnetic intensity (H) and magnetic field (B).
\nAnswer:
\na) Ferromagnetic substances are those which gets strongly magnetized in an external magnetic field.
\nProperties: High susceptibility and permeability.
\nb) Ob represents retentivity and Oc coercivity.
\nc) H = nI
\n= 1000 \u00d7 2
\n= 2 \u00d7 103<\/sup> A\/m
\nB = \u00b5s<\/sub>\u00b5r<\/sub>H
\n= \u00b50<\/sub> \u00d7 400 \u00d7 2 \u00d7 103<\/sup>
\nB = IT<\/p>\n
\na) Derive an expression for self inductance of a solenoid.
\nb) What do you mean by eddy current? Write any two applications of it.
\nAnswer:
\na) Consider a solenoid (air core) of length\/.number of turns N and area cross section A. let \u2018n\u2019 be the no. of turns per unit length (n = N\/I)
\nThe magnetic flux linked with the solenoid,
\n\u03a6 = BAN
\n\u03a6 = \u03bc0<\/sub>nIAN (since B = \u03bc0<\/sub> ni)
\nbut \u03a6 = LI
\n\u2234 LI = \u03bc0<\/sub>nIAN
\nL = \u03bc0<\/sub>nAN
\nIf solenoid contains a core of relative permeability
\n\u03bcr
\nThen L = \u03bc0<\/sub>\u03bcr<\/sub>nAN<\/p>\n
\nApplications:<\/p>\n\n
\nThe given graph shows the variation of KE with frequency of incident radiations for two surfaces A and B.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-20.png\")
\na) Which of the two metals has greater work function? For which of the metals will stopping potential be more for the same frequency of incident radiation?
\nb) The work function of caesium metal is 2.14 eV. When light of frequency 6 \u00d7 1014<\/sup> Hz incident on the metal surface, what is the maximum KE of the photoelectrons and stopping potential?
\nAnswer:
\na) A
\nb) KE = h\u03c5 – \u03a60<\/sub>
\n= 6.6 \u00d7 10-34<\/sup> \u00d7 6 \u00d7 1014<\/sup> – 2.14 \u00d7 1.6 \u00d7 10-19<\/sup>
\n= 0.551 \u00d7 10-19\u00a0<\/sup>J
\n= 0.345 eV
\n\u2234 Stopping potential = 0.345 V<\/p>\n
\na) Name different series of lines observed in hydrogen spectrum.
\nb) Draw energy level diagram of hydrogen atom.
\nAnswer:
\na) Laymen, Balmer, Paschen, Bracket, Pfund
\nb)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-21.png\")
<\/p>\n
\nThe circuit diagram of a potentiometer for determining the emf \u2018E’ of a cell of negligible internal resistance is as shown in figure.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-22.png\")
\na) State the principle of working of a potentiometer.
\nb) How the balancing length AJ changes when the value of R1<\/sub> decreases?
\nc) Derive an expression to find out internal resistance of a cell.
\nAnswer:
\na) Potential difference between two points of current carrying conductor is directly proportional to the length of the wire between two points.
\nb) Balancing length decreases.
\nc)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-23.png\")
\nThe key (K1<\/sub>) in the primary circuit is closed and the key is the secondary (K2<\/sub>) is open. Jockey is moved to get zero deflection in galvanometer. The balancing length l1<\/sub> (from A) is found out.
\nThen we can write.
\nE1<\/sub> \u221d l1<\/sub> …….. (1)
\nKey K2<\/sub> is put in the circuit, corresponding balancing length (l2<\/sub>) is found out. Let V be the applied voltage, then
\nV1<\/sub> \u221d l1<\/sub> ……..(2)
\n\u2018V\u2019 is the voltage across resistance box.
\nCurrent through resistance box
\ni.e., voltage across resistance,
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-24.png\")
<\/p>\n
\nThe phasor diagram of the alternating voltage across an unknown device X and current flowing through it are shown below.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-25.png\")
\na) Identify the device X.
\nb) Draw the graphical variation of current and voltage with \u03c9t through this device.
\nc) Draw the phasor diagram which shows the relation among VR, VL, VC and I in a series LCR circuit.
\nAnswer:
\na) Resistor
\nb)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-26.png\")
\nc)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-27.png\")
<\/p>\n
\nA ray of light passing through a prism. If the refracted ray QR is parallel to the base BC, as shown in figure.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-28.png\")
\na) Derive an expression to find out deviation (d).
\nb) Draw a graph which shows the variation of \u2018d\u2019 with \u2018i\u2019.
\nc) Calculate the angle of minimum deviation if A = 60\u00b0 and refractive index of the material of prism is 1.62.
\nAnswer:
\na)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-29.png\")
\nThe angle between the emergent ray and incident ray is the deviation \u2018d\u2019.
\nIn the quadrilateral AQMR,
\n\u2220Q + \u2220R = 180\u00b0
\n[since N1M and NM are normal]
\ni.e., \u2220A + \u2220M = 180\u00b0 …….(1)
\nIn the \u0394 QMR,
\n\u2234 r1<\/sub> + r2<\/sub> + \u2220M = 180\u00b0 …….(2)
\nComparing eq (1) and eq (2)
\nr1<\/sub> + r2<\/sub> = \u2220A …….(3)
\nFrom the \u0394 QRT,
\n(i1<\/sub> – r1<\/sub>) + (i2<\/sub> – r2<\/sub>) = d
\n(i1<\/sub> + i2<\/sub>) – (r1<\/sub> + r2<\/sub>) = d
\nbut, r1<\/sub> + r2<\/sub> = A
\n\u2234 (i1<\/sub> + i2<\/sub>) – A = d<\/p>\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-30.png\")
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-31.png\")
<\/p>\n
![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-32.png\")
![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/01\/Plus-Two-Physics-Previous-Year-Question-Paper-March-2018-33.png\")