HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 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 200 π΄ in a give time is: 1) 4:1 2) 1:2 3) 2:1 4) 1:4 We know that π = π£ 2 π
Therefore, ππΌ 1 π
Given π
1 =100 Ξ© and π
2 =200 Ξ© So, π 1 π 2 = π
2 π
1 π 1 π 2 = 200 100 π 1 : π 2 = 2 : 1 Option (3) 2) Two hollow conducting spheres of radii R 1 and R 2 (R 1 >> R 2 ) 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 Potential (V) = 1 4 π π 0 ( π π
) Charge is same Therefore, π = 1 π
π
1 > π
2 π 1 < π 2 Option (3) 3) When two monochromatic lights of frequency, v and π π are incident on a photoelectric metal, their stopping potential becomes π π π and V s respectively. The threshold frequency for this metal is: (Note: The information given is practically incorrect) 1) π π π 2) 2 v 3) 3 v 4) π π π From Einstein Photoelectric equation β π = π + β
π 0 Case 1: π 1 = π π 0 = π π 2 CREATIVE LEARNING CLASSES, KARKALA NEET KEY ANSWERS WITH DETAILED SOLUTION - 2022 VERSION CODE β R6 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI β π = π + β
π π 2 2β π = 2 π + β
π π 2β π β 2 π = β
π π .................(1) Case 2: π 2 = π 2 π 0 = π π β π 2 = π + β
π π β π 2 β π = β
π π ................(2) From (1) and (2) 2β π β 2 π = β π 2 β π 2β π β β π 2 = 2 π β π π = 3 2 β π We Know that π = β π 0 So π 0 = 3 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) β π : π 4) π : β π HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Speed of transverse Wave π£ = β π π Here βmβ is constant π£πΌ β π π£ 1 π£ 2 = β π 1 π 2 Given π 1 = π and π 2 = 2 π π£ 1 π£ 2 = β π 2 π π£ 1 : π£ 2 = 1 : β 2 Option ( 4 ) 6) Match list I with list II List I List II (Electromagnetic wave) (Wavelength) (a) AM radio waves (i) 10 - 10 m (b) Microwaves (ii) 10 2 m (c) Infrared radiations (iii) 10 - 2 m (d) X - rays (iv) 10 - 4 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 wa ve: 10 2 m Microwave: 10 - 2 m Infrared radiation: 10 - 4 m X - ray: 10 - 10 m 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 : β π 2) 2:1 3) β π : π 4) 4:1 Case 1: I CM = π π
2 2 and π 1 = π
β 2 Case 2: I CM = π π
2 4 and π 2 = π
2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI So π 1 π 2 = π
β 2 π
2 π 1 : π 2 = β 2 : 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 By Lens Makers Formula 1 π = ( π β 1 ) ( 1 π
1 β 1 π
2 ) 1 π = ( 1 5 β 1 ) ( 1 20 β 1 ( β 20 ) ) 1 π = ( 0 5 ) ( 2 20 ) π = 20 ππ P = 100 π P = 100 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: De - broglie wavelength π = β π ππΌ 1 π So rectangular hyperbola. Option (1) 10) When light propagates through a material medium of relative permittivity π π and relative permeability π π the velocity of light, π is given by: (e = velocity of light in vacuum) 1) π = π β π π π π 2) π = π 3) π = β π π π π 4) π = β π π π π Dielectric constant π π = π π 0 π = π π π 0 ..............(1) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Relative Permeability ΞΌ π = ΞΌ ΞΌ 0 ΞΌ = ΞΌ π ΞΌ 0 ..........(2) Velocity of light in air π = 1 β π 0 π 0 ...............(3) But velocity of light in medium π£ = 1 β ΞΌ π π£ = 1 β ΞΌ π ΞΌ 0 π π π 0 π£ = 1 β ΞΌ π π π 1 β ΞΌ 0 π 0 π£ = π β ΞΌ π π π (from equation (3)) Option (1) 11) In the given nuclear reaction, the element X is: 22 11 Na X e v + β + + 1) 22 12 Mg 2) 23 11 Na 3) 23 10 Ne 4) 22 10 Ne ππ β π₯ + β
+ + π 11 22 It has under go π½ + decay Z is increased by 1, mass number remains the same π 10 22 is πβ
10 22 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 curv e which represents the adiabatic process among 1, 2, 3 and 4 is: 1 ) 4 2) 1 3) 2 4) 3 1 β isochoric 2 β adiabatic 3 β isothermal 4 β 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 1 st , 2 nd , 3 rd and 4 th second. 1) 1:1:1:1 2) 1:2:3:4 3) 1:4:9:16 4) 1:3:5:7 We have π π = π ( 2 π β 1 2 ) π 1 = π ( 2 ( 1 ) β 1 2 ) = π 2 ; π 2 = π ( 2 ( 2 ) β 1 2 ) = 3 π 2 ; π 3 = 5 π 2 ; π 4 = 7 π 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: 1) 5m 2) 10 3 m 3) 20 3 m 4) 10 m π 1 = π 2 π π 1 + π 2 = 20 π₯ 10 10 + 20 = 200 30 = 20 3 π Option (3) 15) The energy that will be ideally radiated by a 10 - kW transmitted in 1 hour is: 1) 1 ο΄ 10 5 J 2) 36 ο΄ 20 7 J 3) 36 ο΄ 10 4 J 4) 36 ο΄ 10 5 J Energy radiated = power x time = (100kW)(1hr) = (10 2 x10 3 W)3600s = 36x10 7 J 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 ms - 1 . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is (g = 10 ms - 2 ) 1) 23500 2) 23000 3) 20000 4) 34500 F m is force exerted by m o tor HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI F net = 0 F m β 20000 β 3000 = 0 F m = 23000N Power delivered = F m x v = 23000 x 1.5 = 34500W O ption ( 4 ) 17) A copper wire of length 10 m and radius (10 - 2 / β π
)m has electrical resistance of 10 ο . The current density in the wire for an electric field strength of 10 (V/m) is: 1) 10 5 A/m 2 2) 10 4 A/m 2 3) 10 6 A/m 2 4) 10 - 5 A/m 2 L = 10m π = 10 β 2 β π π R = 10β¦ E = 10V/m π½ = πΌ π΄ = π π
π΄ = πΈ π π
π΄ = 10 π 10 10 ( π π 2 ) = 10 π ( 10 β 4 π ) = 10 5 π΄ π 2 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 P inside = P outside + 2 π π
If radius increases, 2 π π
decreases pressure inside decreases. Option (2) 19) Given below are two statements Statement I: Biot - Savartβs 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 approp riate 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 β qβ charge is not a vector Statement I is correct. O ption ( 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) π β π π 2) π 3) β π π 4) π β π π Masses of pieces are 2 π 5 , 2 π 5 , 1 π 5 From law of conservation of linear momentum ππ£ β² 5 = β ( 2 ππ£ 5 ) 2 π₯ 2 π£ β² = 2 β 2 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. O ption (3) 22) The an gular 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/s 2 is: 1) 104 π
2) 2 π
3) 4 π
4) 12 π
HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI f 0 = 1200 rpm = 1200 60 rps = 20 rps f = 3120 rpm = 3120 60 rps = 52 rps angular acceleration πΌ = π β π 0 π‘ = 2 π ( π β π 0 ) π‘ = 2 π ( 52 β 20 ) 16 = 4 π rads - 2 O ption ( 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 π πππ = π 0 β 2 π 0 = β 2 π πππ β 2 times the rms value of the ac source. Option (4) 24) The dimension [MLT - 2 A - 2 ] belong to the 1) Electric permittivity 2) Magnetic flux 3) Self - inductance 4) Magnetic permeability Magnetic permeability ( π 0 ) = π΅ π β
= πΉ π β
2 π = [ π πΏ 2 π β 2 π΄ β 2 ] Option ( 4 ) 25) The displacement time graph of two moving particles makes angles of 30 0 and 45 0 with the x - axis as shown in the figure. The ratio of their respective velocity is: 1 ) 1: β π 2) β π :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 x - t graph π£ 1 = tan 30 = 1 β 3 π£ 2 = tan 45 = 1 Ratio of velocities = π£ 1 π£ 2 = 1 β 3 Option (1) 26) In the given circuits (a), (b) and (c), the potential drop across the two p - n 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 p - n junctions are equal. Option (1) 27) The angle between the electric lines of force and the equipotential surface is: 1) 180 0 2) 0 0 3) 45 0 4) 90 0 Electric field lines are always normal to the equipotential surface. Angle between them is 90 0 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 sa me region of the screen is: 1) 12 2) 6 3) 8 4) 9 π = 600 ππ π β² = 400 ππ Width of the screen is 8 π½ π½ = ππ· π Let π½ β² be the new fringe width for wavelength π β² = 400 ππ π½ β² = π β² π· π Let n be the number of fringes on the screen. Width = π π½ β² 8 π½ = π π½ β² π = 8 π π· π π β² π· π π = 8 π π β² π = 8 Γ 600 400 π = 12 Option (1) 30) A light ray falls on a glass surface of refractive index β π , at an angle 60 0 . The angle between the refracted and reflected rays would be: 1) 120 0 2) 30 0 3) 60 0 4) 90 0 π = β 3 π = 60 0 Using Snellβs law sin π sin π = π sin π = sin π π sin π = ( β 3 2 ) β 3 sin π = 1 2 π = 30 0 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 = 90 0 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 gravitat ional filed intensity at that point is: 1) 180 N/kg 2) 0.05 N/kg 3) 50 N/kg 4) 20 N/kg π = πΉ π = 3 60 Γ 10 β 3 = 50 π 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 weber 2) 2 weber 3) 0.5 weber 4) 1 weber ΙΈ = π΅π΄ cos π = 0 5 Γ 1 2 Γ πππ 0 = 0 5 π€β
πβ
π Option (3) 34) Let T 1 and T 2 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 T 1 : T 2 is: 1) 9:4 2) 1:4 3) 4:1 4) 4:9 πΈ π = β 13 6 π 2 πΈ π β 1 π 2 π 1 = 2 , π 2 = 3 πΈ 1 : πΈ 2 = 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΄ 10 - 4 T 2) 6.28 ο΄ 10 - 2 T 3) 12.56 ο΄ 10 - 2 T 4) 12.56 ο΄ 10 - 4 T π΅ = π 0 ππΌ = π 0 π πΏ πΌ = 4 π Γ 10 β 7 Γ 100 10 β 3 Γ 1 = 12 56 Γ 10 β 2 π 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 = 2 π β π π π β β π π 1 π 2 = β 121 100 = 11 10 10 π 1 = 11 π 2 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 m 3 2) 5.6 ο΄ 10 6 m 3 3) 5.6 ο΄ 10 3 m 3 4) 5.6 ο΄ 10 - 3 m 3 ππ = ππ
π π = ππ
π π = ππ
π ππ = 4 5 Γ 10 3 Γ 8 3 Γ 273 18 Γ 10 5 = 5 66 π 3 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΄ 10 8 m/s and 2.0 ο΄ 10 8 m/s, respectively. The critical angle for a ray of light these two media is: 1) tan - 1 (0.750) 2) sin - 1 (0.500) 3) sin - 1 (0.750) 4) tan - 1 (0.500) Using Snellβs Law, π π΄ sin π π = π π΅ sin 90 π π£ π΄ sin π π = π π£ π΅ Γ 1 sin π π = π£ π΄ π£ π΅ = 1 5 Γ 10 8 2 0 Γ 10 8 = 0 75 π π = sin β 1 ( 0 75 ) Option (3) 39) match list I with List II List β I List - II (a) Gravitational constant (G) (i) [L 2 T - 2 ] (b) Gravitational potential energy (ii) [M - 1 L 3 T - 2 ] (c) Gravitational potential (iii) [LT - 2 ] (d) Gravitational intensity (iv) [ML 2 T - 2 ] 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) a) Using πΉ = πΊ π 1 π 2 π 2 [G]= [ML 3 T - 2 ] b) All energies = [ π 1 πΏ β 2 π β 2 ] c) π = ππ ππ [ ππ ] = [ πΏ 2 π β 2 ] d) π = πΉ π [ π ] = [ πΏ 1 π β 2 ] Option ( 3 ) 40) A series LCR circuit with inductance 10 H, capacitance 10 F ο , resistance 50 ο is connected to an ac source of voltage, V = 200sin (100t) volt. If the resonant frequency of the LCR circuit is 0 v and the frequency of the sources is v, then 1) π = ππππ―π , π π = πππ π
π―π 2) π π = π = πππ―π 3) π π = π = ππ π
π―π 4) π π = ππ π
π―π , π = πππ―π Frequency of the source π = π 2 π = 100 2 π = 50 π π»π§ Resonant frequency π 0 = 1 2 π β πΏπΆ = 1 2 π β 10 Γ 10 Γ 10 β 6 = 50 π π»π§ Option (3) 41) A ball is projected with a velocity,0 ms - 1 , at an angle of 60 0 with the vertical direction. Its speed at the highest of its trajectory will be: 1) 10 ms - 1 2) Zero 3) 1 5 3 ms β 4) 5 ms - 1 Speed at highest point = vcosΞΈ= 10x cos 30= 10x β 3 2 = 5 β 3 π / π Option ( 3 ) 42) From Ampereβs circuital law for a long straight wire of circular cross - section 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 in creasing 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 t he 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 s teel 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= R 0 A 1/3 π
1 π
2 = [ 125 64 ] 1 3 = 5 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 ( 3 s ) 47) The area of a rectangular field (in m 2 ) of length 55.3m and breadth 25 m after rounding of value for correct significant digit is: 1) 14 x 10 2 2) 138 x 10 1 3) 1382 4) 1382.5 Area = l x b = 55.3x 25 = 1382.5 = 14x 10 2 m 2 Option ( 1 ) 48) The big circular coil of 1000 turns and average radius 10 m is rotated about its horizontal diameter at 2 rad s - 1 . 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 induced current = π β
ππ‘ π₯ π
= π ππ‘ ( π π₯ π΄ π₯ π΅πππ π π
) = ππ΄π΅π π
= 1000 π₯ π π 2 π₯ 2 π₯ 10 β 5 π₯ 2 12 56 = 1 π΄ Option ( 4 ) 49) Two - point 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 1) π πΉ π 2) π πΉ π 3) π πΉ π 4) π πΉ π 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 10 - 6 J 2) 4.5 x 10 6 J 3) 3.25 x 10 6 J 4) 2.25 x 10 6 J Q = CV = 900X 100 =9nC after connecting to second capacitor Qβ +Qβ = 45nC Energy stored = 2 ( π β² 2 πΆ ) 2 = ( 45 π₯ 10 β 9 ) 2 900 π₯ 10 β 12 = 2025 π₯ 10 β 18 900 π₯ 10 β 12 = 2 25 π₯ 10 β 6 π½ Option ( 4 ) DEPARTMENT OF PHYSICS β Dr. Gananath Shetty B β Mr. Abhishek Kumar β Mr. Thirumala Reddy β Mrs. Nidhi B Shetty β Mr. Lakshman Bajila β Mr. Suhas Gore P β Mr. Sharath Rai A N β Mr. Amithav Guha β Mr. Karthik M B β Mr. Mohammed Fanu β Mr. Joel M F β Mr. Ranjith β Mr. Anand Kumar Sharma β Mr. Pawan Kumar CREATIVE EDUCATION FOUNDATION MOODBIDRI (R) Website : www.creativeedu. in Phone No. : +91 901984449