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HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI CREATIVE EDUCATION FOUNDATION, KARKALA SECOND PU ANNUAL EXAMINATION MARCH– 2023 PHYSICS DETAILED SOLUTION PART - A I. Select the correct option from the given choices : 15X 1 = 15 1) Physical quantity measured in terms of “ Coulomb” is A) electric charge B) electric current C) electric flux D) electric field Ans: A) 2) The electric field inside the cavity of a charged conductor is zero, this is known as A) charging B) grounding C) electrostatic shielding D) electrostatic induction Ans: C) 3) An example for polar molecule is A) oxygen molecule B) nitrogen molecule C) water molecule D) hydrogen molecule Ans: C) 4) The magnitude of the drift velocity per unit electric field is A) mobility B) drift velocity C) relaxation time D) resistivity Ans: A) 5) The device used to accelerate charged particle is A) electroscope B) cyclotron C) galvanometer D) ammeter Ans: B) 6) The net magnetic flux through any closed surface is zero is in accordance with A) Gauss’s law in magnetism B) Gauss’s law in electrostatics C) Ampere’s circuital law D) Biot-Savart’s law Ans: A) 7) S.I unit of mutual inductance of pair of coils is A) Henry B) Ohm C) Farad D) Ohm-metre Ans: A) 8) If the number of turns of a solenoid is doubled, the self inductance of the solenoid will A) remains unchanged B) be doubled C) be halved D) becomes four times Ans: D) 9) The relation between peak value of current ( ) m i and rms value of current (I) is A) 2 m i I  B) 2 m I i  C) 2 m I i  D) 2 m i I  Ans: A) 10) The ultraviolet region of the electromagnetic spectrum lies between A) X-ray region and visible region B) Microwave region and Radiowave region C)  - rays region and X-rays region D) Visible region and microwave region Ans: A) 11) Snell’s law of refraction invalid at an angle of incidence (i) is A) i = 30 0 B) i = 60 0 C) i = 0 0 D) i = 90 0 Ans: C) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 12) When a point source of light is placed at the principal focus of a thin convex lens, the shape of the emergent wave front is A) Spherical convergent wave front B) Spherical divergent wave front C) Plane wave front D) Cylindrical wave front Ans: C) 13) C.J. Davison – L.H. Germer experiment proved A) Wave nature of electrons B) Particle nature of electrons C) Wave nature of light D) Particle nature of light Ans: A) 14) Function of moderator in a nuclear reactor is A) to slow down fast neutrons B) to absorb the neutrons C) to reduce heat energy D) to control the chain reaction Ans: A) 15) Energy gap (E g ) between the valence band and the conduction band for conductor is A) E g = 0 B) E g < 3eV C) E g > 3eV D) E g = 3eV Ans: A) II. Fill in the blanks by choosing appropriate answer given in the bracket for all the following question: ( curie temperature, electric dipole, transverse, isotopes, Zener diode) 16) A pair of equal and opposite point charges q and -q separated by a distance 2a is an ________ Ans: Electric dipole 17) Temperature of transition from ferro magnetism to paramagnetism is called _________ Ans: Curie temperature 18) Phenomenon of polarization proves the _____ nature of light waves. Ans: Transverse 19) Nuclei having same atomic number and different mass number are called ______ Ans: Isotopes 20) _____ is used as voltage regulator. Ans: Zener diode PART-B III. Answer ANY FIVE of the following questions: 21) On what factors does the capacitance of a parallel plate capacitor depends? Ans: The size and shape of the conductor. The dielectric medium surrounding the conductor. 22) State and explain Ampere’s Circuital Law. Statement: The line integral of magnetic field around any closed path in air or vacuum is equal to 𝜇 ଴ times the total current enclosed by that path. ර 𝐵 ሬ ⃗ 𝑑𝑙 ሬ ሬ ሬ ⃗ = 𝜇 ଴ 𝐼 ර 𝐵 𝑐𝑜𝑠 𝜃 𝑑𝑙 = 𝜇 ଴ 𝐼 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI Where, ∮ 𝐵 ሬ ⃗ 𝑑𝑙 ሬ ሬ ሬ ሬ ⃗ is the line integral of 𝐵 ሬ ⃗ around the closed path and I is the current enclosed by that path. dl = elemental length of the closed path and θ= angle between 𝐵 ሬ ⃗ 𝑎𝑛𝑑 𝑑𝑙 ሬ ሬ ሬ ⃗ 23) Define magnetic dip and declination at a place. Magnetic dip: Dip at a place is the angle between the earth’s total magnetic field and the horizontal drawn in the magnetic meridian. Magnetic declination: The angle between the magnetic meridian and the geographic meridian at a place is known as the declination. 24) What are eddy current? Mention any one use of it. Current induced in a thick conductor, when the conductor is placed in a changing magnetic field is called eddy current. It is used in Electromagnetic brakes: Eddy current braking can be used to control the speed of trains. 25) Write two sources of energy Loss in a transformer. Loss due to heating of coils Loss due to flux leakage 26) What is displacement current? Give the expression for it. Displacement current is that current which appears in the region in which the electric field and hence the electric flux is changing with time. It is equal to ε 0 times the rate of change of electric flux through a given surface. 𝐼 ௗ = 𝜀 ଴ 𝑑 ∅ ா 𝑑𝑡 27) Mention the expression for limit of resolution at a telescope and explain the terms. Resolving limit dθ = ଵ ଶଶ஛ ୟ where λ is the wavelength of the light entering the telescope a is the aperture of the objective 28) Name the spectral series of hydrogen atom lies in a) Ultraviolet region and b) Visible region of electromagnetic spectrum.  Lyman series  Balmer series HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 29) Give any two difference between nuclear fission and nuclear fusion. Nuclear fission Nuclear fusion The process of splitting of a heavy nucleus into two lighter nuclei of comparable masses together with the liberation of few neutrons and energy is called nuclear fission. The process of combination of two or more lighter nuclei to form a heavy nucleus together with the liberation of energy is known as nuclear fusion. Energy released per fission reaction is large Energy released per fusion reaction is less. PART C IV. Answer ANY FIVE of the following question. 30) Write any three properties of electric field lines.  Electric field lines are straight, when they represent the electric field due to an isolated charge, and are curved when they represent the field due to two or more charges placed nearby.  The electric field lines are directed away from a positive charge and directed towards negative charge.  The tangent drawn to an electric field line at any point gives the direction of electric field at that point. 31) Draw a labeled Wheatstone’s bridge and hence write the balancing condition in terms of resistance. 3 1 4 2 R R R R  HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 32) How would you convert a galvanometer into an ammeter? explain. An ammeter is an instrument used for measurement of current in amperes. A galvanometer is very sensitive device and it can be used to measure very low currents. A galvanometer can be converted into an ammeter by connecting a low resistance called shunt resistance ‘S’ in parallel with the coil of the galvanometer. The shunt resistance reduces the resistance of the galvanometer. Let R g be the resistance of the galvanometer. It requires a current I g for full scale deflection. Let S be the shunt resistance to be connected to convert it into an ammeter of range 0 to 1A. The value of S should be such that, when a current I enter the instrument, only a small portion I g flows through galvanometer. PD across the galvanometer = PD across shunt resistance I ୥ R ୥ = ൫ I − I ୥ ൯ S S = I ୥ R ୥ I − I ୥ This is the expression for shunt required to convert galvanometer into ammeter 33) Write three differences between diamagnetic and paramagnetic materials. Properties of diamagnetic substance:  Diamagnetic substance is repelled by a strong magnet.  When the diamagnetic substance is placed in an external magnetic field, the field lines are expelled and field inside the material is reduced.  When a rod of diamagnetic substance is suspended in a uniform magnetic field, the rod comes to rest with its longest axis at right angles to the direction of the field. Properties of paramagnetic substance:  They get weakly attracted to a magnet.  They have tendency to move from a region of weak magnetic field to strong magnetic field.  When a paramagnetic substance is placed in a magnetic field, the magnetic lines of force prefer to pass through the substance rather than through air. Therefore, the resultant field B inside the substance is more than the external field 𝐵 ଴ HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 34) Derive an expression for motional e.m.f induced in a conductor moving perpendicular to the uniform magnetic field. Let us consider a straight conductor PQ moving in a uniform magnetic field B, acting into the plane of the paper. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field. The Lorentz force on this charge is F = qvB in magnitude and its direction is towards Q. All charge irrespective of their position experience the same force in magnitude and direction. The work done in moving the charge from P to Q is; 𝑊 = 𝑞𝑣𝐵𝑙 Emf induced is the work done per unit charge. ∈ = ௐ ௤ ∈ = 𝐵𝑙𝑣 35) Arrive the relation between focal length and radius of curvature of a spherical concave mirror. Consider a concave mirror of radius of curvature R. Let C be the center of curvature. Consider a ray of light LM travelling parallel to the principal axis be incident on the mirror at M. Let CM be the normal at M. Let θ be the angle of incidence. Draw MD perpendicular to the principal axis. From the fig: ∠𝐿𝑀𝐶 = ∠𝑀𝐶𝑃 = 𝜃 [Since alternating angles] Using law of reflection; ∠𝐿𝑀𝐶 = ∠𝐶𝑀𝐹 =  Therefore, ∠𝐿𝑀𝐹 = ∠𝑀𝐹𝑃 = 2𝜃 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI From the ΔMCD, 𝑡𝑎𝑛𝜃 = ெ஽ ஼஽ ............... (1) From the ΔMFD, 𝑡𝑎𝑛 2 𝜃 = ெ஽ ி஽ .............. (2) For small value of  ; 𝑡𝑎𝑛𝜃 ≈ 𝜃 and tan2θ ≈ 2 𝜃 Therefore, equation (1) and (2) can be rewritten as: 𝜃 = ெ஽ ஼஽ ............. (3) 2 𝜃 = ெ஽ ி஽ ............ (4) Dividing equation (3) by equation (4) we get; 𝜃 2 𝜃 = 𝑀𝐷 𝐶𝐷 𝐹𝐷 𝑀𝐷 ଵ ଶ = ி஽ ஼஽ 𝐹𝐷 = ஼஽ ଶ ............. (5) Since D is very close to point P Then FD = FP = f and CD = CP = R Equation (5) becomes: 𝑓 = ோ ଶ 36) Give the three postulates of Bohr’s atomic model. Postulate 1: Electrons revolve round the nucleus only in certain stable orbits (called stationary orbits) without the emission of radiant energy. Postulate 2: Electrons revolve around the nucleus only in those orbits for which the angular momentum of the electron is integral multiple of h 2 π where h is Planck’s constant. i.e., Angular momentum, mvr = 𝑛 ℎ ଶగ where n = 1, 2, 3 .... ∞ (Bohr’s quantization rule) where n is called principal quantum number. Postulate 3: An electron might make a transition from one of its (specified non-radiating) orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the inner and the outer orbits. i.e., hv = E o – E i (Bohr’s frequency condition) The frequency of the emitted photon is given by v = ୉ ౥ ି ୉ ౟ ୦ where E i and E o are the energies of the electron in inner and outer orbits. HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 37) Calculate the mass defect and binding energy of helium nucleus ( 2 He 4 ) using the following date in MeV Mass of proton = 1.00727 u Mass of neutron = 1.00866 u Mas of helium nucleus = 4.00260 u Ans: Mass defect = 2 M p + 2M n – M He = 2(1.00727) + 2(1.00866) – 4.00260 = 2.01454 + 2.01732 – 4.00260 = 0.02926 u Binding energy = mass defect (u)  931 (Me V/u) = 0.02926  931 = 27.24106 MeV 38) Write the logical symbol and truth table of NAND gate. Circuit symbol: Truth table: Input A Input B Output 𝑌 = 𝐴 𝐵 _ _ _ _ 0 0 1 0 1 1 1 0 1 1 1 0 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI PART – D V. Answer any three of the following questions 39) State Gauss’s law in electrostatics. Derive an expression for the electric field at a point due to infinitely long thin charged straight wire using Gauss’s law. It states that, the total electric flux passing through any closed surface is ଵ ఌ బ times the total charge enclosed by the surface. ∅ ா = 𝑞 𝜀 ଴ Consider an infinitely long thin straight wire with uniform linear charge density λ. Let P be a point at a distance r from the wire. To find the electric field at point P, draw a cylindrical Gaussian surface of radius r and length l Let ∆𝑆 ଵ , ∆𝑆 ଶ and ∆𝑆 ଷ , be the three small elemental areas on the Gaussian cylinder. The flux though ∆𝑆 ଶ and ∆𝑆 ଷ is zero. Since electric field E is perpendicular to normal drawn to the surface. ∆∅ ଶ = ∆∅ ଷ = 𝐸∆𝑆𝑐𝑜𝑠90° = 0 Therefore, flux through the circular caps of the cylinder ∅ ଶ = ∅ ଷ = ∑ 𝐸 ሬ ⃗ . ∆𝑆 ሬ ሬ ሬ ሬ ⃗ = 0 ௔௟௟ Flux through the curved part of the cylinder ∆∅ ଵ = 𝐸 ሬ ⃗ . ∆𝑆 ሬ ሬ ሬ ሬ ⃗ 𝑐𝑜𝑠0 Then total flux ∅ = ∅ ଵ = ∑ 𝐸∆𝑆 ௔௟௟ ∅ = 𝐸 × 2𝜋𝑟𝑙 ---------- (1) Using Gauss’s law ∅ ா = ௤ ఌ బ = ఒ௟ ఌ బ ------------- (2) Comparing equation (1) and (2) 𝐸 × 2𝜋𝑟𝑙 = ఒ௟ ఌ బ 𝐸 ሬ ⃗ = ఒ ଶగఌ బ ௥ 𝑛 ො Where, 𝑛 ෝ is the unit vector in the plane normal to the wire passing through the point. HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 40) Derive the expression for conductivity of a material 2 ne m    , where the terms have their usual meaning. Let us consider a conductor of length x  and a planar area A. Let the potential difference V is applied across the ends of the conductor and electric field 𝐸 ሬ ⃗ is set up in the conductor. Let n be the number of free electrons per unit volume in the conductor. Total number of electrons in the conductor is; n x Volume of the conductor n x A x V d Δt Since each electron carries a charge –e, then the total charge transported across the area A in the direction opposite to E, in a time t  is; q= - n e A V d t  The total charge transported along 𝐸 ሬ ⃗ across the area is; q= n e A V d t  By definition the amount of charge crossing the area A in time t  is; q = I t  𝐼𝛥𝑡 = 𝑛𝐴𝑒𝑉 ௗ 𝛥𝑡 𝐼 = 𝑛𝐴𝑒𝑉 ௗ This is the relation between current I and drift velocity V d We know that drift velocity 𝑉 ௗ = ௘ா → ఛ ௠ Then I = 𝑛𝐴𝑒( ௘ா → ఛ ௠ ) I = ୬୅ୣ మ τ ୫ E → We know that Current density j → = ୍ ୅ j → A = ୬୅ୣ మ τ ୫ E → j → = ୬ୣ మ τ ୫ E → σ E → = ୬ୣ మ τ ୫ E → (Since j → = σ E → ) σ = ୬ୣ మ τ ୫ is conductivity of the material 41) Obtain the expression for the force between two straight long parallel conductors carrying current. Hence define. “ampere” . Consider two long straight parallel conductors ‘a’ and ‘b’ of length L separated by a distance ‘d’ and carrying a current I a and I b in the same direction. Each conductor is in the region of magnetic field produced by the other, Therefore, each conductor experiences a force. HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI The magnitude of the magnetic field at any point P on the conductor ‘b’, due to current I a in the conductor ‘a’ is; B ୟ = μ బ ଶ π ୍ ୟ ୢ ............. (1) Using right hand thumb rule, the direction of magnetic field is perpendicular to the plane of the paper directed inwards. Now the conductor ‘b’ will experience a mechanical force and is given by; F ୠୟ = I ୠ B ୟ L F ୠୟ = μ బ ଶ π ୍ ౗ ୍ ౘ ୢ L ............... (2) Using Fleming’s left-hand rule, the direction of 𝐹 ௕௔ towards the conductor ‘a’. Similarly magnetic field at any point on conductor ‘a’ due to current in conductor ‘b’ is 𝐵 ௕ = ஜ బ ଶ஠ ୍ ౘ ୢ (acting out wards) ............. (3) Then the force experience by the conductor ‘a’ is F ୟୠ = I ୟ B ୠ 𝐿 F ୠୟ = μ బ ଶ π ୍ ౗ ୍ ౘ ୢ L (towards the conductor b) ............ (4) from equation (2) and (4) F ୟୠ = − F ୠୟ = ఓ బ ூ ೌ ூ ್ ଶగௗ 𝐿 ... One ampere : It is defined as that current flowing in each of the two infinitely long parallel conductors of negligible cross section, separated 1m apart in vacuum, would produce a force of 2 × 10 -7 N per meter length of each conductor. 42) Arrive at the expression for refractive index of the material of the prism in terms of angle of prism and angle of minimum deviation. Let ABC be the principal section of a prism of refractive index n, A be its refractive angle and BC its base. A ray of monochromatic light PQ incident on the refracting face AB, is refracted along QR and emerges along RS. Let i and r 1 be the angle of incidence and refraction respectively on the face AB. Let r 2 and e be the angle of incidence and emergence respectively on the face AC. The normal drawn to the faces AB and AC meets at N. The angle between incident ray PQZ and the emergence ray MRS are called the angle of deviation (δ). HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI From the quadrilateral QARN: ∠𝐴 + ∠𝑄𝑁𝑅 = 180 0 -------- (1) From the ∆QNR r 1 + r 2 + ∠𝑄𝑁𝑅 = 180 0 -------- (2) Comparing equation (1) and (2) A = r 1 + r 2 -------- (3) From the diagram total deviation = deviation at the first face + deviation at the second face SMZ = MQR + MRQ δ = (i– r 1 ) + (e - r 2 ) δ = (i -e) - (r 1 + r 2 ) δ = ( i +e ) – A ----------------------(4) Thus, the angle of deviation depends on the angle of incidence. A plot between angle of deviation and angle of incidence is shown below As the angle of incidence is gradually increased from a small value, the angle of deviation first decreases, reaches a minimum and then increases. Hence, we see from the graph for any value of deviation δ, there are two values of the angle of incidence, i and e. At minimum the incident ray and emergent ray are symmetrical with respect to the refracting faces of the prism and the refracted ray travels parallel to the base of the prism. At minimum deviation, i = e = I, r 1 = r 2 = r and δ =D m Then equation (3) becomes. A = r + r = 2r or r = A/2 ------------ (5) And equation (4) becomes, D m = i+i-A 2i = D m +A 𝑖 = ஽ ೘ ା஺ ଶ ................ (6) From Snell’s law, the refractive index of the material of the prism is given by, 𝑛 ଶଵ = ௦௜௡ ௜ ௦௜௡ ௥ Using equation (5) and (6), 𝑛 ଶଵ = ௦௜௡ [ ಲశವ ೘ మ ] ௦௜௡ [ ಲ మ ] HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 43) a) What is meant by photo electric effect? b) Define work function. c) Write the three experimental observation of photo electric effect. a) The phenomenon of emission of electrons by a metal surface when a radiation of suitable frequency is incident on it is called Photoelectric Effect. b) The minimum amount of external energy that must be supplied to a free electron in order that it may be just emitted out of the metal surface is known as work function of that metal. c) The Photoemission is an instantaneous process. For every photo emissive material, there is a minimum frequency below which no photo emission takes place. This minimum frequency is called threshold frequency. For frequency greater than threshold frequency the photoelectric current increases linearly with increase in intensity. 44) a) What is rectification? b) Draw the circuit diagram and input-output waveforms of a full wave rectifier. c) explain the working of a full wave rectifier. a) The process of converting A.C into pulsating D.C is called rectification. b) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI c) During the positive half cycle of the input voltage i.e.., when potential at A is positive and at B is negative, the diode D 1 is forward biased and D 2 is reverse biased. So D 1 conducts & D 2 does not conduct. During the negative half cycles of the input voltage i.e.., when potential at A is negative and at B is positive the diode D 2 is forward biased and D 1 is reverse biased. So D 2 conducts and D 1 does not conduct. Thus, in each cycle the current flows through the load resistor R L in the same direction. Hence output obtained is unidirectional (D.C.) and pulsatory. Answer any of the following questions. 45) Charge 2 , 4 c c   and 6 c  are place at the three corners A, B and C respectively of a square ABCD of side X metre. Find the charge that must be placed at the fourth corner so that the total potential at the centre of the squares is zero. Let V 0 be me potential at centre, V 0 = V A + V B + V C + V P 2 (4 ) (6 ) 2 2 2 D K C K C K C O V x x x                          12 2 D K C V x         12 2 2 KQ K C x x                12 Q C    46) Three resistors 2 , 3 6 and    are combined in parallel. What is the total resistance of the combination? If this combination is connected to a battery of emf 2V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery. HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI 1 1 1 1 Re 2 3 6 ff    Reff = 1  The current drawn from the battery is 𝐼 = ா ோ௘ ௙௙ = ଶ ଵ = 2𝐴 1 2 1 2 I A   2 2 0.67 3 I A A   3 2 1 0.33 6 3 I A    47) A sinusoidal voltage of peak value 283 V and frequency 50 Hz, is applied to a series LCR circuit in which 3 , 25.48 796 R L mH and C F      . Calculate: a. impedance of the circuit. b. the phase difference between the voltage across the source and the current. Given 3 R   , L = 25.48 mH, C = 796 F  a) Impedance 2 2 ( ) L C Z R X X    3 2 2 2 3.14 50 25.48 10 L X L fL fL             8 L X   6 1 1 1 2 2 3.14 50 796 10 C X wC fc          4 L X   2 2 3 (8 4) 5 Z      HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI b) 1 8 4 4 4 tan tan 3 3 3 L C X X R           0 53   48) Two narrow slits in Young’s double slit experimental are 0.18 mm apart. When they are illuminated by a monochromatic light, fringes of width 2.7 mm are obtained on a screen 0.8 m away. find the wavelength of light used. If the source is replaced by another source of wavelength 450 nm, find the change in the fringe width. d = 0.18 mm 2.7 mm   D = 0.8 m ?   We know that, 1 1 D d    3 3 (0.8) 2.7 10 0.18 10       6 1 0.6075 10 m     1 607.5 nm   2 450 nm   9 6 2 3 450 10 0.8 2000 10 2 0.18 10 m mm            Change in fringe width 1 2       2.7 2 0.7 mm mm mm    DEPARTMENT OF PHYSICS HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE UDUPI CREATIVE EDUCATION FOUNDATION KARKALA  Dr. Gananath Shetty B  Mr. Joel Manoj Fernandes  Mrs. Nidhi B Shetty  Mr. Abhishek  Mr. Lakshmana  Mr. Vankadara Pavan Kumar  Mr. Y K Thirumala Reddy  Mr. Sharath Rai  Mr. Karthik M B  Mr. Fanu V P  Mr. Ananda Sharma  Mr. Chandra Bhushan Singh  Mr. Ranjith  Ms. Sharel fernandes www.creativeedu.in Phone No: 9019844492