Practice Questions

Q17.The electric field of a plane electromagnetic wave is given by E = E0(ˆx + ˆy) sin(kz −ωt). Its magnetic field will be given by (1) E0 c (−ˆx + ˆy) sin(kz −ωt) (2) E0c (ˆx + ˆy) sin(kz −ωt) (3) E0 c (ˆx −ˆy) sin(kz −ωt) (4) E0c (ˆx −ˆy) cos(kz −ωt)

Q17.A beam of plane polarized light of large cross-sectional area and uniform intensity of 3 .3 W m–2 falls normally on a polarizer (cross-sectional area 3 × 10–4 m2 ), which rotates about its axis with an angular speed of 31 .4 rad s−1 . The energy of light passing through the polarizer per revolution, is close to: (1) 1 .0 ×10−5 J (2) 1. 0 × 10−4 J (3) 1. 5 × 10−4 J (4) 5. 0 × 10−4 J

Q17.For a plane electromagnetic wave, the magnetic field at a point x and time t is : → B(x, t) = T. [1 .2 ×10−7 sin(0 .5 ×103x + 1 .5 ×1011t)ˆk] → → The instantaneous electric field E corresponding to B is : → → (1) (2)V V E(x, t) = E(x, t) = [−36 sin(0. 5 × 103x + 1. 5 × 1011t)ˆj] m [36 sin(1 × 103x + 0. 5 × 1011t)ˆj] m → → (3) (4)V V E(x, t) = E(x, t) = [36 sin(0. 5 × 103x + 1. 5 × 1011t)ˆk] m [36 sin(1 × 103x + 15 × 1011t)ˆi] m

Q17.There is a small source of light at some depth below the surface of water (refractive index = 43 ) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, JEE Main 2020 (09 Jan Shift 2) JEE Main Previous Year Paper percentage of light that emerges out of surface is (nearly): [Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2πrh ] (1) 21% (2) 34% (3) 17% (4) 50%

Q17.Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is: (1) 1 (2) 250 500 (3) 1 (4) 500 250

Q17.A vessel of depth 2h is half filled with a liquid of refractive index 2√2 and the upper half with another liquid of refractive index √2 . The liquids are immiscible. The apparent depth of the inner surface of the bottom of the vessel will be JEE Main 2020 (09 Jan Shift 1) JEE Main Previous Year Paper (1) h (2) h √2 2(√2+1) (3) h (4) 3√2h 3√2 4

Q17.The correct match between the entries in column I and column II are : I II Radiation Wavelength a Microwave i 100 m b Gamma rays ii 10−15m c A.M. radio iii 10−10m JEE Main 2020 (05 Sep Shift 2) JEE Main Previous Year Paper d X–rays iv 10−3m (1) (a)–(ii), (b) – (i), (c) – (iv), (d) – (iii) (2) (a)–(i), (b) – (iii), (c) – (iv), (d) – (ii) (3) (a)–(iii), (b) – (ii), (c) – (i), (d) – (iv) (4) (a)–(iv), (b) – (ii), (c) – (i), (d) – (iii)

Q18.Visible light of wavelength 6000 × 10−8cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at θ1, then θ1 is close to (1) 20o (2) 30o (3) 25o (4) 45o

Q18.A double convex lens has power P and same radii of curvature r of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1. 5 P is : JEE Main 2020 (06 Sep Shift 2) JEE Main Previous Year Paper (1) 2R (2) R2 (3) 3R (4) R 2 3 ¯

Q18.Particle A of mass mA = m2 moving along the x -axis with velocity v0 collides elastically with another particle B at rest having mass mB = m3 . If both the particles move along the x -axis after the collision, the change Δλ in the wavelength of the particle A , in terms of its de-Broglie wavelength (λ0) before the collision is: (1) Δλ = 32 λ0 (2) Δλ = 52 λ0 (3) Δλ = 2λ0 (4) Δλ = 4λ0

Q18.An electron of mass m and magnitude of charge e at rest, gets accelerated by a constant electric field E . The rate of change of de-Broglie wavelength of this electron at a time t is (ignore relativistic effects) (1) dλ dt = − eEth (2) dλdt = −2heEt (3) dλ dt = −2heEt2 (4) dλdt = − eEt2h

Q18.Two light waves having the same wavelength λ in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2 . After this the phase difference between the two waves is: (1) 2π λ ( L2n1 −L1n2 ) (2) 2πλ ( L1n1 −L2n2 ) (3) 2π λ (n1L1 −n2L2) (4) 2πλ (n2L1 −n1L2) JEE Main 2020 (03 Sep Shift 2) JEE Main Previous Year Paper

Q18.When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is TB = (TA −1.5)eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is: (1) 4eV (2) 2eV (3) 1.5eV (4) 3eV

Q18.You are given that 73Li = 7. 0160u,Mass of Mass of 42He = 4. 0026u and Mass of 11He = 1. 0079H When 20g of 73Li is converted into 42 He by proton capture, the energy liberated, (in kWh ), is : [Mass of nucleon = 1 GeV /c2 ] (1) 4. 5 × 105 (2) 8 × 106 (3) 6. 82 × 105 (4) 1. 33 × 106

Q18.Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632. 8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1. 27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to : (1) 1. 27μm (2) 2 . 87 nm (3) 2 nm (4) 2. 05 μm

Q18.A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1. 878 × 10–4 . The mass of the particle is close to : (1) 4. 8 × 10−27 kg (2) 9. 1 × 10−31 kg (3) 1. 2 × 10−28 kg (4) 9. 7 × 10−28 kg

Q18.In a radioactive material, a fraction of active material remaining after the time t is 9 . The fraction that 16 was remaining after the time t is : 2 (1) 4 (2) 3 5 5 (3) 3 (4) 7 4 8

Q18.Two coherent sources of sound, S1 and S2, produce sound waves of the same wavelength λ = 1 m are in phase. S1 and S2 are placed 1. 5 m apart (see fig). A listener, located at L, directly in front of S2, finds that the intensity is at a minimum when he is 2 m away from S2. The listener moves away from S1, keeping the distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then d is : (1) 12m (2) 5m (3) 2m (4) 3m

Q18.In a double – slit experiment, at a certain point on the screen the path difference between the two interfering waves is 1 8 th of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is: (1) 0.853 (2) 0.672 (3) 0.568 (4) 0.760 →

Q18.The aperture diameter of a telescope is 5m . The separation between the moon and the earth is 4 × 105km . With light of wavelength of 5500Å , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to: (1) 60m (2) 20m (3) 200m (4) 600m

Q18.In a Young’s double slit experiment, the separation between the slits is 0.15mm . In the experiment, a source of light of wavelength 589nm is used and the interference pattern is observed on a screen kept 1.5m away. The separation between the successive bright fringes on the screen is: (1) 6.9mm (2) 3.9mm (3) 5.9mm (4) 4.9mm

Q19.The radius R of a nucleus of mass number A can be estimated by the formula R = (1. 3 × 10−15)A1/3m . It follows that the mass density of n nucleus is of the order of: (Mprot ≅Mneut ≃1. 67 × 10−27 kg) (1) 103 kg m−3 (2) 1010 kg m−3 (3) 1024 kg m−3 (4) 1017 kg m−3

Q19.Given figure shows few data points in a photo-electric effect experiment for a certain metal. The minimum energy for ejection of electrons from its surface is: (Planck's constant h = 6.62 × 10–34 J.s) (1) 2.27 eV (2) 2.59 eV (3) 1.93 eV (4) 2.10 eV

Q19.When a diode is forward biased, it has a voltage drop of 0. 5 V . the safe limit of current through the diode is 10 mA . If a battery of emf 1. 5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is : (1) 300 Ω (2) 50 Ω (3) 100 Ω (4) 200 Ω

Q19.In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1 , the frequency of radiation emitted is proportional to : (1) 1 (2) 1 n n3 (3) 1 (4) 1 n2 n4