Practice Questions

Q26.In a Frank - Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV . The minimum wavelength of photons emitted by mercury atoms is close to: (1) 250 nm (2) 1700nm (3) 220nm (4) 2020nm

Q26.Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm-2 . If the surface has an area of 25 cm2 , the momentum transferred to the surface in 40 min time duration will be: (1) 6.3 × 10-4 N s (2) 5.0 × 10-3 N s (3) 3.5 × 10-6 N s (4) 1.4 × 10-6 N s

Q26.Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ? (1) λ (2) λ (5−√2) 2(√5−2) (3) λ (4) λ 2(5−√2) √5−2

Q26.A particle of mass m moves in a circular orbit in a central potential field U(r) = 12 kr2. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as: (1) rn ∝n2, En ∝ n21 (2) rn ∝√n, En ∝n (3) rn ∝n, En ∝n (4) rn ∝√n, En ∝1n

Q26.In a double slit experiment, when a thin film of thickness t having refractive index μ is introuduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( λ is the wavelength of the light used): (1) λ (2) λ 2 ( μ - 1 ) ( 2μ - 1 ) (3) 2λ (4) λ ( μ - 1 ) ( μ - 1 )

Q26.In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10−12 m, the minimum electron energy required is close to: (1) 100 keV (2) 25 keV (3) 1 keV (4) 500 keV

Q26.The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper 4 .7 eV, what will be the maximum kinetic energy of the photoelectrons? (c = 3 × 108 m s−1, h = 6 .6 ×10−34 J s) (1) 6 .82 eV (2) 7 .72 eV (3) 12 .5 eV (4) 8 .52 eV

Q27.The output of the given logic circuit is: (1) − (2) − A B A B ¯(3) AB + AB (4) − − A B + A B

Q27.The electric field of light wave is given as →E = 10–3 cos( 5×10–72πx −2π × 6 × 1014 t)ˆx NC . This light falls on a metal plate of work function 2 eV . The stopping potential of the photo-electrons is: Given, E (in eV ) = λ(in12375Å) (1) 0.72 V (2) 2.0 V (3) 2.48 V (4) 0.48 V

Q27.In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is L. If an electron jumps from N -shell to the L -shell, the wavelength of emitted radiation will be: (1) 27 λ (2) 16 λ 20 25 (3) 25 λ (4) 20 λ 16 27

Q27.A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λx' and 'λy' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is: (1) λx −λy (2) λxλy λx+λy (3) λxλy (4) λx + λy λx−λy JEE Main 2019 (09 Apr Shift 2) JEE Main Previous Year Paper

Q27.Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be: (1) 3 : 8 (2) 8 : 1 (3) 1 : 8 (4) 9 : 8

Q27.A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Ã …. The radius of the atom in the excited state, in terms of Bohr radius a0 , will be: (1) 25a0 (2) 9a0 (3) 16a0 (4) None of the above JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper

Q27.At a given instant, say t = 0, two radioactive substance A and B have equal activities. The ratio RB of their RA activities after time t itself decays with time t as e−3t. If the half-life of A is ln2, the half-life of B is: (1) 2ln2 (2) 4ln2 (3) ln2 (4) ln2 4 2

Q27.Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to: (1) 400 (2) 200 (3) 150 (4) 360 JEE Main 2019 (10 Jan Shift 1) JEE Main Previous Year Paper

Q27.Two particles move at right angle to each other. Their de Broglie wavelengths are 𝜆1 and 𝜆2 respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength 𝜆 of the final particle, is given by: 1 1 1 1 (1) 2 + = 1 (2) 𝜆= 𝜆1 + 𝜆2 (3) 𝜆= √𝜆1𝜆2 (4) 2 + 2 2 𝜆2 𝜆= 𝜆1 𝜆2 𝜆1 𝜆2

Q27.In a radioactive decay chain, the initial nucleus is 232 90 Th. At the end, there are 6 α-particles and 4β-particles which are emitted. If the end nucleus is AZX, A and Z are given by: (1) A = 208; Z = 82 (2) A = 208; Z = 80 (3) A = 200; Z = 81 (4) A = 202; Z = 80

Q27.Two radioactive materials A and B have decay constants 10λ and λ , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time: (1) 1 (2) 1 10λ 9λ (3) 1 (4) 11 11λ 10λ

Q27.Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16 . The intensity of the waves are in the ratio: (1) 25: 9 (2) 16: 9 (3) 5: 3 (4) 4: 1

Q27.A 2 mW laser operates at a wavelength of 500 nm . The number of photons that will be emitted per second is: [Given Planck's constant h = 6.6 × 10-34 J s, speed of light c = 3.0 × 108 m / s ] (1) 1.5 × 1016 (2) 5 × 1015 (3) 2 × 1016 (4) 1 × 1016 JEE Main 2019 (10 Apr Shift 2) JEE Main Previous Year Paper

Q27.A metal plate of area 1 × 10−4 m2 is illuminated by a radiation of intensity 16 millim2 W . The work function of the metal is 5 eV . The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: [1eV = 1.6 × 10−19J] (1) 1014 and 10 eV (2) 1012 and 5 eV (3) 1011 and 5 eV (4) 1010 and 5 eV

Q28.A He+ ion is in its first excited state. Its ionization energy is: (1) 13.60 eV (2) 48.36 eV (3) 54.40 eV (4) 6.04 eV

Q28.Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6 V . If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current? (1) 1.5 mA (2) 3.5 mA (3) 7.5 mA (4) 2.5 mA

Q28.Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be : (1) 889.2 nm (2) 488.9 nm (3) 388.9 nm (4) 642.7 nm

Q28.In the given circuit the current through Zener Diode is close to: (1) 0.0 mA (2) 6.7 mA (3) 4.0 mA (4) 6.0 mA