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CREATIVE LEARNING CLASSES, KARKALA NEET KEY ANSWERS WITH DETAILED SOLUTION - 2022 VERSION CODE – R6 SECTION A 1) Two resistors of resistance100 𝜴 and 200 𝜴 are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 𝜴 to that in 200𝜴 in a give time is: 1) 4:1 2) 1:2 3) 2:1 4) 1:4 𝑣2 We know that 𝑃 = 𝑅 1 Therefore, 𝑃𝛼 𝑅 Given 𝑅1 =100Ω and 𝑅2 =200Ω 𝑃 𝑅 So, 𝑃1 = 𝑅2 2 1 𝑃1 200 = 𝑃2 100 𝑃1 : 𝑃2 = 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 𝑞 Potential (V) = 4𝜋𝜀 (𝑅) 0 Charge is same 1 Therefore, 𝑉 = 𝑅 𝑅1 > 𝑅2 𝑉1 < 𝑉2 Option (3) 𝒗 3) When two monochromatic lights of frequency, v and 𝟐 are incident on a photoelectric metal, their 𝒗𝒔 stopping potential becomes and Vs respectively. The threshold frequency for this metal is: 𝟐 (Note: The information given is practically incorrect) 𝟑 𝟐 1) 𝟐 𝒗 2) 2 v 3) 3 v 4) 𝟑 𝒗 From Einstein Photoelectric equation ℎ𝜈 = 𝜙 + ⅇ𝑉0 𝑉𝑠 Case 1: 𝜈1 = 𝜈 𝑉0 = 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 𝑉𝑠 ℎ𝜈 = 𝜙 + ⅇ 2 2ℎ𝜈 = 2𝜙 + ⅇ𝑉𝑠 2ℎ𝜈 − 2𝜙 = ⅇ𝑉𝑠 ……………..(1) 𝜈 Case 2: 𝜈2 = 2 𝑉0 = 𝑉𝑠 ℎ𝜈 = 𝜙 + ⅇ𝑉𝑠 2 ℎ𝜈 − 𝜙 = ⅇ𝑉𝑠 …………….(2) 2 From (1) and (2) ℎ𝜈 2ℎ𝜈 − 2𝜙 = −𝜙 2 ℎ𝜈 2ℎ𝜈 − = 2𝜙 − 𝜙 2 3 𝜙 = ℎ𝜈 2 We Know that 𝜙 = ℎ𝜈0 3 So 𝜈0 = 2 𝜈 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) √𝟐: 𝟏 4) 𝟏: √𝟐 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 𝑇 Speed of transverse Wave 𝑣 = √𝑚 Here “m” is constant 𝑣𝛼√𝑇 𝑣1 𝑇1 =√ 𝑣2 𝑇2 Given 𝑇1 = 𝑇 and 𝑇2 = 2𝑇 𝑣1 𝑇 =√ 𝑣2 2𝑇 𝑣1 : 𝑣2 = 1: √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) 102 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) –(ii), (b) – (iii), (c) – (iv), (d) – (i) 2) (a) – (iv), (b) – (iii), (c)- (ii) , (d) – (i) 3) (a) –(iii), (b) – (ii), (c) – (i), (d) – (iv) 4) (a) – (iii), (b) – (iv), (c) – (ii), (d) – (i) AM Radio wave: 102m Microwave: 10-2m Infrared radiation: 10-4m X-ray: 10-10m 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)𝟏: √𝟐 2) 2:1 3) √𝟐: 𝟏 4) 4:1 𝑀𝑅 2 𝑅 Case 1: ICM = and 𝑘1 = 2 √2 𝑀𝑅 2 𝑅 Case 2: ICM = and 𝑘2 = 4 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 𝑅 𝑘1 √2 So = 𝑅 𝑘2 2 𝑘1 : 𝑘2 = √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 1 1 1 By Lens Makers Formula = (𝜂 − 1) (𝑅 − 𝑅 ) 𝑓 1 2 1 1 1 = (1.5 − 1) (20 − (−20)) 𝑓 1 2 = (0.5) ( ) 𝑓 20 𝑓 = 20𝑐𝑚 100 P= 𝑓 100 P= 20 P = +5D Option (4) 9) The graph which shows the variation of the de Broglie wavelength (𝝀)of a particle and its associated momentum (p) is: ℎ De-broglie wavelength 𝜆 = 𝑝 1 𝜆𝛼 𝑝 So rectangular hyperbola. Option (1) 10) When light propagates through a material medium of relative permittivity 𝒆𝟏 and relative permeability 𝝁𝒕 the velocity of light, 𝝊is given by: (e = velocity of light in vacuum) 𝒆 𝝁 𝒆 1) 𝝊 = 2) 𝝊 = 𝒆 3) 𝝊 = √ 𝒆 𝟏 4) 𝝊 = √𝝁𝟏 √𝒆𝟏 𝝁𝟏 𝟏 𝟏 𝜀 Dielectric constant 𝜀𝑟 = 𝜀 0 𝜀 = 𝜀𝑟 𝜀0…………..(1) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI µ Relative Permeability µ𝑟 = µ 0 µ = µ𝑟 µ0 ……....(2) 1 Velocity of light in air 𝑐 = ……………(3) √ 𝜇0 𝜀 0 1 But velocity of light in medium 𝑣 = √µ𝜀 1 𝑣= √µ𝑟 µ0 𝜀𝑟 𝜀0 1 1 𝑣= √µ𝑟 𝜀𝑟 √µ0 𝜀0 𝑐 𝑣= (from equation (3)) √ µ 𝑟 𝜀𝑟 Option (1) 11) In the given nuclear reaction, the element X is: 22 11 Na → X + e + + v 22 23 23 22 1) 12 Mg 2) 11 Na 3) 10 Ne 4) 10 Ne 22 11𝑁𝑎 → 𝑥 + ⅇ+ + 𝜗 It has under go 𝛽 + decay Z is increased by 1, mass number remains the same 22 10𝑋 is 22 10𝑁ⅇ 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 1→ isochoric 2→adiabatic 3→ isothermal 4→ 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 2𝑛−1 We have 𝑠𝑛 = 𝑔( ) 2 2(1) − 1 𝑔 2(2) − 1 3𝑔 5𝑔 7𝑔 𝑠1 = 𝑔 ( ) = ; 𝑠2 = 𝑔 ( )= ; 𝑠3 = ; 𝑠4 = 2 2 2 2 2 2 𝑠1 : 𝑠2 : 𝑠3 : 𝑠4 = 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 𝑚2 𝑑 20𝑥10 200 20 𝑑1 = = = = 𝑚 𝑚1 + 𝑚2 10 + 20 30 3 Option (3) 15) The energy that will be ideally radiated by a 10-kW transmitted in 1 hour is: 1) 1  105 J 2) 36  207 J 3) 36  104 J 4) 36  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 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 Fm is force exerted by motor HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Fnet = 0 Fm – 20000 – 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 (10-2/√𝝅)m has electrical resistance of 10  . 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) 10-5A/m2 10−2 L = 10m 𝑟= 𝑚 R = 10Ω E = 10V/m √𝜋 𝐼 𝑉 𝐸. 𝑙 10𝑋10 10 𝐴 𝐽= = = = 2 = 10−4 = 105 2 𝐴 𝑅𝐴 𝑅𝐴 10(𝜋𝑟 ) 𝜋( ) 𝑚 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 2𝑠 Pinside = Poutside + 𝑅 2𝑠 If radius increases, decreases 𝑅 pressure inside decreases. Option (2) 19) Given below are two statements Statement I: Biot-Savart’s 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- Savart’s law is analogous to Coulomb’s 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 ‘q’ 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) 𝟑√𝟐𝝊 2) 𝝊 3) √𝟐𝝊 4) 𝟐√𝟐𝝊 2𝑚 2𝑚 1𝑚 Masses of pieces are , , 5 5 5 From law of conservation of linear momentum 𝑚𝑣′ 2𝑚𝑣 2 = √( ) 𝑥2 5 5 𝑣 ′ = 2√2 𝑣 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 ( ) 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 𝜔−𝜔0 angular acceleration 𝛼 = 𝑡 2𝜋(𝑓−𝑓0 ) = 𝑡 2𝜋(52−20) = 16 = 4𝜋 rads-2 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 𝑉0 𝑉𝑟𝑚𝑠 = √2 𝑉0 = √2 𝑉𝑟𝑚𝑠 √2 times the rms value of the ac source. Option (4) 24) The dimension [MLT-2A-2] belong to the 1) Electric permittivity 2) Magnetic flux 3) Self-inductance 4) Magnetic permeability 𝐵 𝐹 Magnetic permeability (𝜇0 ) = 𝑛ⅈ = 𝑛ⅈ 2 𝑙 = [𝑀𝐿2 𝑇 −2 𝐴−2 ] 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: √𝟑 2) √𝟑:1 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 1 𝑣1 = tan 30 = √3 𝑣2 = tan 45 = 1 𝑣 1 Ratio of velocities = 𝑣1 = 2 √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) 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 Young’s 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 𝜆 = 600 𝑛𝑚 𝜆′ = 400 𝑛𝑚 Width of the screen is 8𝛽 𝜆𝐷 𝛽= 𝑑 Let 𝛽 ′ be the new fringe width for wavelength 𝜆′ = 400 𝑛𝑚 𝜆′ 𝐷 𝛽′ = 𝑑 Let n be the number of fringes on the screen. Width = 𝑛𝛽 ′ 8𝛽 = 𝑛𝛽 ′ 𝐷 8𝜆 𝑑 𝑛= 𝐷 𝜆′ 𝑑 8𝜆 𝑛= 𝜆′ 8 × 600 𝑛= 400 𝑛 = 12 Option (1) 30) A light ray falls on a glass surface of refractive index √𝟑, at an angle 600. The angle between the refracted and reflected rays would be: 1) 1200 2) 300 3) 600 4) 900 𝑛 = √3 𝑖 = 600 Using Snell’s law sin 𝑖 =𝑛 sin 𝑟 sin 𝑖 sin 𝑟 = 𝑛 √3 ( ) 2 sin 𝑟 = √3 1 sin 𝑟 = 2 𝑟 = 300 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Angle between reflected ray and refracted ray = 180 − 𝑖 − 𝑟 = 180 − 60 − 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 𝐹 3 𝑁 𝑔 = 𝑚 = 60×10−3 = 50 𝑘𝑔 Option (3) 33) A square loop of side 1 m and resistance 1  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 ɸ = 𝐵𝐴 cos 𝜃 = 0.5 × 12 × 𝑐𝑜𝑠0 = 0.5 𝑤ⅇ𝑏ⅇ𝑟 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 Bohr’s model of an atom, the ratio T1: T2 is: 1) 9:4 2) 1:4 3) 4:1 4) 4:9 −13.6 𝐸𝑛 = 𝑛2 1 𝐸𝑛 ∝ 𝑛2 𝑛1 = 2, 𝑛2 = 3 𝐸1 : 𝐸2 = 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  10-4 T 2) 6.28  10-2 T 3) 12.56  10-2 T 4) 12.56  10-4T 𝑁 100 𝐵 = 𝜇0 𝑛𝐼 = 𝜇0 𝐼 = 4𝜋 × 10−7 × −3 × 1 = 12.56 × 10−2 𝑇 𝐿 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 𝑙 𝑇 = 2𝜋√ 𝑔 𝑇 ∝ √𝑙 𝑇1 121 11 =√ = 𝑇2 100 10 10𝑇1 = 11𝑇2 ∴ 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  106 m3 3) 5.6  103 m3 4) 5.6  10-3 m3 𝑃𝑉 = 𝑛𝑅𝑇 𝑛𝑅𝑇 𝑚𝑅𝑇 4.5 × 103 × 8.3 × 273 𝑉= = = = 5.66 𝑚3 𝑃 𝑀𝑃 18 × 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  108 m/s and 2.0  108 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 Snell’s Law, 𝑛𝐴 sin 𝑖𝑐 = 𝑛𝐵 sin 90 𝑐 𝑐 sin 𝑖𝑐 = ×1 𝑣𝐴 𝑣𝐵 𝑣𝐴 1.5 × 108 sin 𝑖𝑐 = = = 0.75 𝑣𝐵 2.0 × 108 𝑖𝑐 = sin−1(0.75) Option (3) 39) match list I with List II List – I List - II (a) Gravitational constant (G) (i) [L2T-2] (b) Gravitational potential energy (ii) [M-1L3T-2] (c) Gravitational potential (iii) [LT-2] (d) Gravitational intensity (iv) [ML2T-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) – (iv), (b)- (ii), (c) – (i), (d) – (iii) 2) (a) – (ii), (b)- (i), (c) – (iv), (d) – (iii) 3) (a) – (ii), (b)- (iv), (c) – (i), (d) – (iii) 4) (a) – (ii), (b)- (iv), (c) – (iii), (d) – (i) 𝐺𝑚1𝑚2 a) Using 𝐹 = 𝑟2 [G]= [ML3T-2] b) All energies = [𝑀1 𝐿−2 𝑇 −2 ] 𝑑𝑉 c) 𝑔 = 𝑑𝑟 [𝑑𝑉] = [𝐿2 𝑇 −2 ] 𝐹 d) 𝑔 = 𝑚 [𝑔] = [𝐿1 𝑇 −2 ] Option (3) 40) A series LCR circuit with inductance 10 H, capacitance 10  F , resistance 50  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 𝟏𝟎𝟎 1) 𝒗 = 𝟏𝟎𝟎𝑯𝒛, 𝒗𝟎 = 𝑯𝒛 2) 𝒗𝟎 = 𝒗 = 𝟓𝟎𝑯𝒛 𝝅 𝟓𝟎 𝟓𝟎 3) 𝒗𝟎 = 𝒗 = 𝑯𝒛 4) 𝒗𝟎 = 𝑯𝒛, 𝒗 = 𝟓𝟎𝑯𝒛 𝝅 𝝅 𝜔 100 50 Frequency of the source 𝑓 = 2𝜋 = = 𝐻𝑧 2𝜋 𝜋 1 1 50 Resonant frequency 𝑓0 = = = 𝐻𝑧 2𝜋√𝐿𝐶 2𝜋√10×10×10−6 𝜋 Option (3) 41) A ball is projected with a velocity,0 ms-1, at an angle of 600 with the vertical direction. Its speed at the highest of its trajectory will be: 1) 10 ms-1 2) Zero 3) 5 3ms −1 4) 5 ms-1 √3 Speed at highest point = vcosθ= 10x cos 30= 10x = 5√3𝑚/𝑠 2 Option (3) 42) From Ampere’s 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 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 𝑅1 125 3 5 = [ ] = 𝑅2 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 𝑋 𝑃 =𝑄 𝑌 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 s-1. If the vertical component of earth’s magnetic field at that place is 2 x 10 -5 T and electrical resistance of the coil is 12.56  , then the maximum induced current in the coil will be: 1) 2A 2) 0.25 A 3) 1.5 A 4) 1 A 𝑑∅ induced current = 𝑑𝑡 𝑥 𝑅 𝑑 𝑁 𝑥 𝐴 𝑥 𝐵𝑐𝑜𝑠𝜃 𝑁𝐴𝐵𝑊 1000 𝑥 𝜋𝑟 2 𝑥 2𝑥 10−5 𝑥2 = ( )= = = 1𝐴 𝑑𝑡 𝑅 𝑅 12.56 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 𝟏 𝟏 𝟏 1)𝑹𝟔 2) 𝑹𝟐 3) 𝑹𝟑 4) 𝑹𝟒 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 106 J 3) 3.25 x 106 J 4) 2.25 x 106 J Q = CV = 900X 100 =9nC after connecting to second capacitor Q’ +Q’ = 45nC 2 𝑄′ (45𝑥10−9 ) 2025𝑥 10−18 Energy stored = 2( 2𝐶)2 = = = 2.25𝑥 10−6 𝐽 900𝑥 10−12 900𝑥10−12 Option (4) DEPARTMENT OF PHYSICS ❖ Dr. Gananath Shetty B ❖ Mr. Amithav Guha ❖ Mr. Abhishek Kumar ❖ Mr. Karthik M B ❖ Mr. Thirumala Reddy ❖ Mr. Mohammed Fanu ❖ Mrs. Nidhi B Shetty ❖ Mr. Joel M F ❖ Mr. Lakshman Bajila ❖ Mr. Ranjith ❖ Mr. Suhas Gore P ❖ Mr. Anand Kumar Sharma ❖ Mr. Sharath Rai A N ❖ 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