CREATIVE LEARNING CLASSES, KARKALA NEET KEY ANSWERS WITH DETAILED SOLUTION  2022 VERSION CODE β R6 SECTION A 1) Two resistors of resistance100 π΄ and 200 π΄ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 π΄ to that in 200π΄ in a give time is: 1) 4:1 2) 1:2 3) 2:1 4) 1:4 π£2 We know that π = π 1 Therefore, ππΌ π Given π 1 =100Ξ© and π 2 =200Ξ© π π So, π1 = π 2 2 1 π1 200 = π2 100 π1 : π2 = 2: 1 Option (3) 2) Two hollow conducting spheres of radii R1 and R2 (R1 >> R2) have equal charges. The potential would be: 1) Dependent on the material property of the sphere 2) More on bigger sphere 3) More on smaller sphere 4) Equal on both the spheres 1 π Potential (V) = 4ππ (π ) 0 Charge is same 1 Therefore, π = π π 1 > π 2 π1 < π2 Option (3) π 3) When two monochromatic lights of frequency, v and π are incident on a photoelectric metal, their ππ stopping potential becomes and Vs respectively. The threshold frequency for this metal is: π (Note: The information given is practically incorrect) π π 1) π π 2) 2 v 3) 3 v 4) π π From Einstein Photoelectric equation βπ = π + β π0 ππ Case 1: π1 = π π0 = 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI ππ βπ = π + β 2 2βπ = 2π + β ππ 2βπ β 2π = β ππ β¦β¦β¦β¦β¦..(1) π Case 2: π2 = 2 π0 = ππ βπ = π + β ππ 2 βπ β π = β ππ β¦β¦β¦β¦β¦.(2) 2 From (1) and (2) βπ 2βπ β 2π = βπ 2 βπ 2βπ β = 2π β π 2 3 π = βπ 2 We Know that π = βπ0 3 So π0 = 2 π Option (1) 4) As the temperature increases, the electrical resistance: 1) Decreases for conductors but increases for semiconductors 2) Increases for both conductors and semiconductors 3) Decreases for both conductors and semiconductors 4) Increases for conductors but decreases for semiconductors Resistance increases for conductor but decreases for semiconductor with temperature. Option (4) 5) If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is: 1) 1:2 2) 1:1 3) βπ: π 4) π: βπ HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI π Speed of transverse Wave π£ = βπ Here βmβ is constant π£πΌβπ π£1 π1 =β π£2 π2 Given π1 = π and π2 = 2π π£1 π =β π£2 2π π£1 : π£2 = 1: β2 Option (4) 6) Match list I with list II List I List II (Electromagnetic wave) (Wavelength) (a) AM radio waves (i) 1010 m (b) Microwaves (ii) 102 m (c) Infrared radiations (iii) 102 m (d) Xrays (iv) 104 m Choose the correct answer from the options given below: 1) (a) β(ii), (b) β (iii), (c) β (iv), (d) β (i) 2) (a) β (iv), (b) β (iii), (c) (ii) , (d) β (i) 3) (a) β(iii), (b) β (ii), (c) β (i), (d) β (iv) 4) (a) β (iii), (b) β (iv), (c) β (ii), (d) β (i) AM Radio wave: 102m Microwave: 102m Infrared radiation: 104m Xray: 1010m Option (1) 7) The ratio of the radius of gyration of a thin uniforms disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is: 1)π: βπ 2) 2:1 3) βπ: π 4) 4:1 ππ 2 π Case 1: ICM = and π1 = 2 β2 ππ 2 π Case 2: ICM = and π2 = 4 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI π π1 β2 So = π π2 2 π1 : π2 = β2 : 1 Option (3) 8) A biconvex lens has radii of curvature. 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is: 1) Infinity 2) +2D 3) +20 D 4) +5D 1 1 1 By Lens Makers Formula = (π β 1) (π β π ) π 1 2 1 1 1 = (1.5 β 1) (20 β (β20)) π 1 2 = (0.5) ( ) π 20 π = 20ππ 100 P= π 100 P= 20 P = +5D Option (4) 9) The graph which shows the variation of the de Broglie wavelength (π)of a particle and its associated momentum (p) is: β Debroglie wavelength π = π 1 ππΌ π So rectangular hyperbola. Option (1) 10) When light propagates through a material medium of relative permittivity ππ and relative permeability ππ the velocity of light, πis given by: (e = velocity of light in vacuum) π π π 1) π = 2) π = π 3) π = β π π 4) π = βππ βππ ππ π π π Dielectric constant ππ = π 0 π = ππ π0β¦β¦β¦β¦..(1) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Β΅ Relative Permeability Β΅π = Β΅ 0 Β΅ = Β΅π Β΅0 β¦β¦....(2) 1 Velocity of light in air π = β¦β¦β¦β¦β¦(3) β π0 π 0 1 But velocity of light in medium π£ = βΒ΅π 1 π£= βΒ΅π Β΅0 ππ π0 1 1 π£= βΒ΅π ππ βΒ΅0 π0 π π£= (from equation (3)) β Β΅ π ππ Option (1) 11) In the given nuclear reaction, the element X is: 22 11 Na β X + e + + v 22 23 23 22 1) 12 Mg 2) 11 Na 3) 10 Ne 4) 10 Ne 22 11ππ β π₯ + β + + π It has under go π½ + decay Z is increased by 1, mass number remains the same 22 10π is 22 10πβ Option (4) 12) An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is: 1) 4 2) 1 3) 2 4) 3 1β isochoric 2βadiabatic 3β isothermal 4β isobaric Option (3) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 13) The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second. 1) 1:1:1:1 2) 1:2:3:4 3) 1:4:9:16 4) 1:3:5:7 2πβ1 We have π π = π( ) 2 2(1) β 1 π 2(2) β 1 3π 5π 7π π 1 = π ( ) = ; π 2 = π ( )= ; π 3 = ; π 4 = 2 2 2 2 2 2 π 1 : π 2 : π 3 : π 4 = 1: 2: 5: 7 Option (4) 14) Two bodies of mass 10 kg and 30 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is: 10 20 1) 5m 2) m 3) m 4) 10 m 3 3 π2 π 20π₯10 200 20 π1 = = = = π π1 + π2 10 + 20 30 3 Option (3) 15) The energy that will be ideally radiated by a 10kW transmitted in 1 hour is: 1) 1 ο΄ 105 J 2) 36 ο΄ 207 J 3) 36 ο΄ 104 J 4) 36 ο΄ 105 J Energy radiated = power x time = (100kW)(1hr) = (102x103W)3600s = 36x107J Option (2) 16) An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is (g = 10 ms2) 1) 23500 2) 23000 3) 20000 4) 34500 Fm is force exerted by motor HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Fnet = 0 Fm β 20000 β 3000 = 0 Fm = 23000N Power delivered = Fm x v = 23000 x 1.5 = 34500W Option (4) 17) A copper wire of length 10 m and radius (102/βπ )m has electrical resistance of 10 ο . The current density in the wire for an electric field strength of 10 (V/m) is: 1) 105 A/m2 2) 104 A/m2 3) 106 A/m2 4) 105A/m2 10β2 L = 10m π= π R = 10β¦ E = 10V/m βπ πΌ π πΈ. π 10π10 10 π΄ π½= = = = 2 = 10β4 = 105 2 π΄ π π΄ π π΄ 10(ππ ) π( ) π π Option (1) 18) If a soap bubble expands, the pressure inside the bubble: 1) Is equal to the atmospheric pressure 2) Decreases 3) Increases 4) Remains the same 2π Pinside = Poutside + π 2π If radius increases, decreases π pressure inside decreases. Option (2) 19) Given below are two statements Statement I: BiotSavartβs law gives us the expression for the magnetic field strength of an infinitesimal current element (IdI) of a current carrying conductor only. Statement II: Biot Savartβs law is analogous to Coulombβs inverse square law of charge with the former being related to the filed produced by a scalar source. IdI while the latter being produced by a vector source, q. In light of above statements choose the most appropriate answer from the options given below: 1) Statement I is incorrect and Statement II is correct 2) Both Statement I and Statement II are correct 3) Both Statement I and Statement II ate incorrect 4) Statement I is correct and Statement II is incorrect βqβ charge is not a vector Statement I is correct. Option (4) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 20) A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2:2:1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragments is: 1) πβππ 2) π 3) βππ 4) πβππ 2π 2π 1π Masses of pieces are , , 5 5 5 From law of conservation of linear momentum ππ£β² 2ππ£ 2 = β( ) π₯2 5 5 π£ β² = 2β2 π£ Option (4 ) 21) A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represented the speed of the ball (ο΅ ) as a function of time (t) is: 1) D 2) A 3) B 4) C Initially the ball accelerates in the liquid. As the speed increases, viscous drag force increases decreasing the net acceleration. At a certain point, the drag force becomes equal to weight (neglecting force of buoyancy) and ball moves with constant speed. Option (3) 22) The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s2 is: 1) 104 π 2) 2π 3) 4π 4) 12π HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 1200 f0 = 1200 rpm = rps = 20 rps 60 3120 f = 3120 rpm = rps = 52 rps 60 πβπ0 angular acceleration πΌ = π‘ 2π(πβπ0 ) = π‘ 2π(52β20) = 16 = 4π rads2 Option (3) 23) The peak voltage of the source is equal to 1) 1/ 2 times the rms value of the ac source 2) The value of voltage supplied to the circuit 3) The rms value of the ac source 4) 2 times the rms value of the ac source π0 ππππ = β2 π0 = β2 ππππ β2 times the rms value of the ac source. Option (4) 24) The dimension [MLT2A2] belong to the 1) Electric permittivity 2) Magnetic flux 3) Selfinductance 4) Magnetic permeability π΅ πΉ Magnetic permeability (π0 ) = πβ = πβ 2 π = [ππΏ2 π β2 π΄β2 ] Option (4) 25) The displacement time graph of two moving particles makes angles of 30 0 and 450 with the x axis as shown in the figure. The ratio of their respective velocity is: 1) 1: βπ 2) βπ:1 3) 1:1 4) 1:2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Velocity = slope of xt graph 1 π£1 = tan 30 = β3 π£2 = tan 45 = 1 π£ 1 Ratio of velocities = π£1 = 2 β3 Option (1) 26) In the given circuits (a), (b) and (c), the potential drop across the two pn junctions are equal in: 1) Both circuits (a) and (c) 2) Circuit (a) only 3) Circuit (b) only 4) Circuit (c) only Circuit (a) and (c) are identical. Therefore, potential drop across pn junctions are equal. Option (1) 27) The angle between the electric lines of force and the equipotential surface is: 1) 1800 2) 00 3) 450 4) 900 Electric field lines are always normal to the equipotential surface. Angle between them is 900. Option (4) 28) Plane angle and solid angle have: 1) Both units and dimensions 2) Units but no dimensions 3) Dimensions but no units 4) No units and no dimension Solid angle and plane angles have units (steradian and radian respectively) but no dimensions. Option (2) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 29) In a Youngβs double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is: 1) 12 2) 6 3) 8 4) 9 π = 600 ππ πβ² = 400 ππ Width of the screen is 8π½ ππ· π½= π Let π½ β² be the new fringe width for wavelength πβ² = 400 ππ πβ² π· π½β² = π Let n be the number of fringes on the screen. Width = ππ½ β² 8π½ = ππ½ β² π· 8π π π= π· πβ² π 8π π= πβ² 8 Γ 600 π= 400 π = 12 Option (1) 30) A light ray falls on a glass surface of refractive index βπ, at an angle 600. The angle between the refracted and reflected rays would be: 1) 1200 2) 300 3) 600 4) 900 π = β3 π = 600 Using Snellβs law sin π =π sin π sin π sin π = π β3 ( ) 2 sin π = β3 1 sin π = 2 π = 300 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Angle between reflected ray and refracted ray = 180 β π β π = 180 β 60 β 30 = 900 Option (4) 31) In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be: 1) 120 HZ 2) Zero 3) 30 Hz 4) 60 Hz Same as input, 60 Hz Option (4) 32) A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational filed intensity at that point is: 1) 180 N/kg 2) 0.05 N/kg 3) 50 N/kg 4) 20 N/kg πΉ 3 π π = π = 60Γ10β3 = 50 ππ Option (3) 33) A square loop of side 1 m and resistance 1 ο is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is: 1) Zero weber2) 2 weber 3) 0.5 weber 4) 1 weber ΙΈ = π΅π΄ cos π = 0.5 Γ 12 Γ πππ 0 = 0.5 π€β πβ π Option (3) 34) Let T1 and T2 be the energy of an electron in the first and second excited state of hydrogen atom respectively. According to the Bohrβs model of an atom, the ratio T1: T2 is: 1) 9:4 2) 1:4 3) 4:1 4) 4:9 β13.6 πΈπ = π2 1 πΈπ β π2 π1 = 2, π2 = 3 πΈ1 : πΈ2 = 9: 4 Option (1) 35) A long solenoid of a radius 1mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the center of solenoid is: 1) 6.28 ο΄ 104 T 2) 6.28 ο΄ 102 T 3) 12.56 ο΄ 102 T 4) 12.56 ο΄ 104T π 100 π΅ = π0 ππΌ = π0 πΌ = 4π Γ 10β7 Γ β3 Γ 1 = 12.56 Γ 10β2 π πΏ 10 Option (3) 36) Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 1) 8 2) 11 3) 9 4) 10 π π = 2πβ π π β βπ π1 121 11 =β = π2 100 10 10π1 = 11π2 β΄ They will be in phase after 10 vibrations of longer pendulum Or they be in phase after 11 vibrations of shorter pendulum Option (2) 37) The volume occupied by the molecules contained in 4.5 kg water at STP, if the inter molecular forces vanish away is 1) 5.6 m3 2) 5.6 ο΄ 106 m3 3) 5.6 ο΄ 103 m3 4) 5.6 ο΄ 103 m3 ππ = ππ π ππ π ππ π 4.5 Γ 103 Γ 8.3 Γ 273 π= = = = 5.66 π3 π ππ 18 Γ 105 Option (1) 38) Two transparent media A and B are separated by a plane perpendicular. The speed of light in those media are 1.5 ο΄ 108 m/s and 2.0 ο΄ 108 m/s, respectively. The critical angle for a ray of light these two media is: 1) tan1 (0.750) 2) sin1 (0.500) 3) sin1 (0.750) 4) tan1 (0.500) Using Snellβs Law, ππ΄ sin ππ = ππ΅ sin 90 π π sin ππ = Γ1 π£π΄ π£π΅ π£π΄ 1.5 Γ 108 sin ππ = = = 0.75 π£π΅ 2.0 Γ 108 ππ = sinβ1(0.75) Option (3) 39) match list I with List II List β I List  II (a) Gravitational constant (G) (i) [L2T2] (b) Gravitational potential energy (ii) [M1L3T2] (c) Gravitational potential (iii) [LT2] (d) Gravitational intensity (iv) [ML2T2] Choose the correct answer from the options given below: HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 1) (a) β (iv), (b) (ii), (c) β (i), (d) β (iii) 2) (a) β (ii), (b) (i), (c) β (iv), (d) β (iii) 3) (a) β (ii), (b) (iv), (c) β (i), (d) β (iii) 4) (a) β (ii), (b) (iv), (c) β (iii), (d) β (i) πΊπ1π2 a) Using πΉ = π2 [G]= [ML3T2] b) All energies = [π1 πΏβ2 π β2 ] ππ c) π = ππ [ππ] = [πΏ2 π β2 ] πΉ d) π = π [π] = [πΏ1 π β2 ] Option (3) 40) A series LCR circuit with inductance 10 H, capacitance 10 ο F , resistance 50 ο is connected to an ac source of voltage, V = 200sin (100t) volt. If the resonant frequency of the LCR circuit is v0 and the frequency of the sources is v, then πππ 1) π = ππππ―π, ππ = π―π 2) ππ = π = πππ―π π ππ ππ 3) ππ = π = π―π 4) ππ = π―π, π = πππ―π π π π 100 50 Frequency of the source π = 2π = = π»π§ 2π π 1 1 50 Resonant frequency π0 = = = π»π§ 2πβπΏπΆ 2πβ10Γ10Γ10β6 π Option (3) 41) A ball is projected with a velocity,0 ms1, at an angle of 600 with the vertical direction. Its speed at the highest of its trajectory will be: 1) 10 ms1 2) Zero 3) 5 3ms β1 4) 5 ms1 β3 Speed at highest point = vcosΞΈ= 10x cos 30= 10x = 5β3π/π 2 Option (3) 42) From Ampereβs circuital law for a long straight wire of circular crosssection carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is: 1) A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region. 2) Uniform and remains constant for both the regions. 3) A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region. HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 4) A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1/r dependance for the outside region. A linearly increasing function of distance r up to the boundary of the wire and then decreasing one with 1/r dependence for the out side region. Option (4) 43) Given below are two statements: One is labelling. Assertion (A) and the other is labelled as Reason (R). Assertion (A): The stretching of a spring is determined by the should modulus of the material of the spring. Reason (R): A coil spring of copper ha smore tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the more appropriate answer from the options given below: 1) (A) is false but (R) is true. 2) Both (A) and (R) are tru and (R) is the correct explanation of (A) 3) Both (A) and (R) are true and (R) u=is not be correct explanation of (A) 4) (A) is true but (R) is false After stretching shape will change So, assertion is correct But tensile strength of steel is more than copper So, reason is wrong. Option (4) 44) The truth table for the given circuit is 1) A B C 0 0 0 0 1 1 1 0 0 1 1 1 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 2) A B C 0 0 0 0 1 1 1 0 1 1 1 0 3) A B C 0 0 1 0 1 0 1 0 0 1 1 1 4) A B C 0 0 1 0 1 0 1 0 1 1 1 0 A B C 0 0 1 0 1 0 1 0 1 1 1 0 Option (4) 45) The nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is: 1) 25:16 2) 1:1 3) 4: 5 4) 5: 4 Radius of nucleus R= R0 A1/3 1 π 1 125 3 5 = [ ] = π 2 64 4 Option (4) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 46) A Wheatstone bridge is used to determine the value of unknown resistance X by adjusting the variable resistance Y as shown in the figure. For the most precise measurement of X, the resistance P and Q 1) Do not play any significant role 2) Should be approximately equal to 2X 3) Should be approximately equal and are small 4) Should be very large and unequal π π =π π They do not play any significant role. Option (3s) 47) The area of a rectangular field (in m2) of length 55.3m and breadth 25 m after rounding of value for correct significant digit is: 1) 14 x 102 2) 138 x 101 3) 1382 4) 1382.5 Area = l x b = 55.3x 25 = 1382.5 = 14x 102 m2 Option (1) 48) The big circular coil of 1000 turns and average radius 10 m is rotated about its horizontal diameter at 2 rad s1. If the vertical component of earthβs magnetic field at that place is 2 x 10 5 T and electrical resistance of the coil is 12.56 ο , then the maximum induced current in the coil will be: 1) 2A 2) 0.25 A 3) 1.5 A 4) 1 A πβ induced current = ππ‘ π₯ π π π π₯ π΄ π₯ π΅πππ π ππ΄π΅π 1000 π₯ ππ 2 π₯ 2π₯ 10β5 π₯2 = ( )= = = 1π΄ ππ‘ π π 12.56 Option (4) 49) Twopoint charges q and +q are placed at a distance of L, as shown in the figure, The magnitude of electric filed intensity at a distance R(R >>L) varies as: HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI π π π π 1)πΉπ 2) πΉπ 3) πΉπ 4) πΉπ It acts as a dipole Option (3) 50) The capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connect to another uncharged capacitor of capacitance C = 900pF as shown in figure (b) 1) 1.5 x 106 J 2) 4.5 x 106 J 3) 3.25 x 106 J 4) 2.25 x 106 J Q = CV = 900X 100 =9nC after connecting to second capacitor Qβ +Qβ = 45nC 2 πβ² (45π₯10β9 ) 2025π₯ 10β18 Energy stored = 2( 2πΆ)2 = = = 2.25π₯ 10β6 π½ 900π₯ 10β12 900π₯10β12 Option (4) DEPARTMENT OF PHYSICS β Dr. Gananath Shetty B β Mr. Amithav Guha β Mr. Abhishek Kumar β Mr. Karthik M B β Mr. Thirumala Reddy β Mr. Mohammed Fanu β Mrs. Nidhi B Shetty β Mr. Joel M F β Mr. Lakshman Bajila β Mr. Ranjith β Mr. Suhas Gore P β Mr. Anand Kumar Sharma β Mr. Sharath Rai A N β Mr. Pawan Kumar HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI CREATIVE EDUCATION FOUNDATION MOODBIDRI (R) Website : www.creativeedu.in Phone No. : +91 901984449
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