Practice Questions

Q19.An amplitude modulated waves is represented by expression vm = 5(1 + 0. 6 cos 6280t) sin(211 × 104t) V. The minimum and maximum amplitudes of the amplitudes modulated wave are, respectively: (1) 2 3 V, 5V (2) 52 V, 8V (3) 5V, 8V (4) 3V, 5V

Q19.Find the Binding energy per nucleon for 12050Sn. Mass of proton mp = 1. 00783 U , mass of neutron mn = 1. 00867 U and mass of tin nucleus msn = 119. 902199 U . (take 1U = 931 MeV ) (1) 7 .5 MeV (2) 9. 0MeV (3) 8.0MeV (4) 8 .5 MeV

Q19.The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited stale to the ground state? (1) 24. 2nm (2) 11.4nm (3) 35.8nm (4) 8.6nm

Q19.A particle moving with kinetic energy E has de Broglie wavelength λ . If energy ΔE is added to its energy, the wavelength become λ . Value of ΔE, is: 2 (1) E (2) 4E (3) 3E (4) 2E

Q19.An electron (mass m ) with initial velocity →v= v0ˆi + v0ˆj is in an electric filed E = −E0ˆk. If λ0 is initial de- Broglie wavelength of electron, its de-Broglie wave length at time t is given by: (1) λ0√2 (2) λ0 m2v2 0 √1+ e2E2t2m2v20 √1+ e2E2t2 (3) λ0 (4) λ0 m2v2 √1+ e2E2t22m2v20 √2+ e2E2t2 0

Q19.The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10−16s. The frequency of revolution of the electron in its first excited state (in s−1 ) is: (1) 1.6 × 1014 (2) 7.8 × 1014 (3) 6.2 × 1015 (4) 5.6 × 1012

Q19.An electron (of mass m ) and a photon have the same energy E in the range of a few eV . The ratio of the de- Broglie wavelength associated with the electron and the wavelength of the photon is ( c = speed of light in vacuum) (1) 1 c ( 2Em ) 1/2 (2) c(2mE)1/2 (3) 1 E 1/2 (4) E 1/2 c ( 2m ) ( 2m )

Q19.Given the masses of various atomic particles mP = 1, 0072 u , mn = 1, 0087 u , me = 0 .000548 u , mv = 0 , ¯¯md = 2 .0141 u , where p =proton, n ≡neutron, e ≡electron, v ≡antineutrino and d ≡deuteron. Which of the following process is allowed by momentum and energy conservation : ¯(1) n + n → deuterium atom (electron bound to the (2) p →n + e+ + v nucleus) (3) n + p →d + γ (4) e+ + e−→γ

Q19.A radioactive nucleus decays by two different processes. The half-life for the first process is 10 s and that for the second is 100 s. The effective half-life of the nucleus is close to: (1) 9 s (2) 6 s (3) 55 s (4) 12 s

Q20.Identify the operation performed by the circuit given below : (1) NAND (2) OR (3) AND (4) NOT

Q20.The current i in the network is (1) 0.2 A (2) 0.6 A (3) 0.3 A (4) 0 A

Q20.The least count of the main scale of a vernier calipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of the vernier scale coincides with a division of the main scale and the zero of vernier scale is lying right side of the zero of the main scale. When this vernier is used to measure the length of the cylinder the zero of the vernier scale between 3. 1 cm and 3. 2 cm and 4th VSD coincides with the main scale division. The length of the cylinder is (VSD is vernier scale division) (1) 3. 2 cm (2) 3. 21 cm (3) 3. 07 cm (4) 2. 99 cm

Q20.In the following, digital circuit, what will be the output a ' Z', when the input (A, B) are (1, 0), (0, 0), (1, 1), (0, 1) (1) 0, 0, 1, 0 (2) 1, 0, 1, 1 (3) 1, 1, 0, 1 (4) 0, 1, 0, 0

Q20.Using screw gauge of pitch 0. 1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as, (1) 2. 121 cm (2) 2. 124 cm (3) 2. 125 cm (4) 2. 123 cm

Q20.Boolean relation at the output stage- Y for the following circuit is: (1) − − (2) A + B A + B (3) A ∙B (4) − − A ∙ B

Q20.The activity of a radioactive sample falls from 700s−1 to 500s−1 in 30 minutes. Its half life is close to: (1) 72min (2) 62min (3) 66min (4) 52min → → → → → → → →

Q20.A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0. 5 mm is noticed on the pitch scale. The nature of zero error involved and the lest count of the screw gauge, are respectively: (1) Negative, 2μm (2) Positive 10μm (3) Positive 0. 1mm (4) Positive, 0. 1μm

Q20.A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5. 50 mm, 5. 55 mm, 5. 34 mm , 5. 65 mm . The average of these four reading is 5. 5375 mm and the standard deviation of the data is 0. 07395 mm. The average diameter of the pencil should therefore be recorded as : (1) (5 .5375 ±0 .0739) mm (2) (5 .5375 ±0 .0740) mm (3) (5 .538 ±0 .074) mm (4) (5 .54 ±0 .07) mm

Q20.If a semiconductor photo diode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is: Planck's constant h = 6. 63 × 10−34 J. s Speed of light c = 3 × 108 m s−1 (1) 1. 1 eV (2) 2. 0 eV (3) 1. 5 eV (4) 3. 1 eV

Q20.In the given circuit, value of Y is: (1) 0 (2) toggles between 0 and 1 (3) will not execute (4) 1 JEE Main 2020 (08 Jan Shift 2) JEE Main Previous Year Paper

Q20.If the screw on a screw-gauge is given six rotations, it moves by 3mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is: (1) 0.001cm (2) 0.02m (3) 0.01cm (4) 0.001mm

Q26.The region in the electromagnetic spectrum where the Balmer series lines appear is: (1) Visible (2) Microwave (3) Infrared (4) Ultraviolet

Q26.The shortest wavelength of H atom in the Lyman series is λ1 . The longest wavelength in the Balmer series of He+ is : (1) 36λ1 (2) 5λ1 5 9 (3) 9λ1 (4) 27λ1 5 5

Q26.For the Balmer series, in the spectrum of H atom, −v = RH{ n211 2 }, the correct statements among (IV) are, (I) As wavelength decreases, the lines in the series converge. (II) The integer n1 is equal to 2 / (III) The lines of the longest wavelength correspond to n2 = 3 . (IV) The ionization energy of hydrogen can be calculated from the wave number of these lines. (1) (I), (III), (IV) (2) (I), (II), (III) (3) (I), (II), (IV) (4) (II), (III), (IV)

Q26.The difference between the radii of 3rd and 4th orbits of L2+i is ΔR1 . The difference between the radii of 3rd and 4th orbits of He+ ΔR2. Ratio ΔR1 : ΔR2 is : (1) 8 : 3 (2) 3 : 8 (3) 2 : 3 (4) 3 : 2