Practice Questions

Q16.The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is → → E = E0 The magnetic field B, at the moment t = 0 is: jcos(ωt −kx). → → (1) E0 (2) B = B = E0√μ0ϵ0 √μ0ϵ0 cos(kx)ˆk cos(kx)ˆj (3) → (4) → E0 B = E0√μ0ε0 B = √μ0ϵ0 cos(kx)ˆk cos(kx)ˆj

Q16.The electric fields of two plane electromagnetic plane waves in vacuum are given by E1→ = E0ˆj cos(ωt −kx) −→ and E2 = E0ˆk cos(ωt −ky) At t = 0, a particle of charge q is at origin with a velocity →v= 08cˆj ( c is the speed of light in vaccum). The instantaneous force experienced by the particle is: + + (1) E0q(0.8ˆi −ˆj 0.4ˆk) (2) E0q(0.4ˆi −3ˆj 0.8ˆk) (3) E0q(−0.8ˆi + ˆj + ˆk) (4) E0q(0.8ˆi + ˆj + 0.2ˆk)

Q16.An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 Ω resistor. The ratio of the currents at time t = ∞ and at t = 40 s is close to: (Take e2 = 7.389 ) (1) 1.06 (2) 1.15 (3) 1.46 (4) 0.84

Q16.An electron is constrained to move along the y-axis with a speed of 0. 1 c (c is the speed of light) in the → presence of electromagnetic wave, whose electric field is E = 30ˆj sin(1. 5 × 107t −5 × 10−2x) V m−1. where t in in seconds and x is im meters.The maximum magnetic force experienced by the electron will be: (given c = 3 × 108 m s−1 and electron charge = 1. 6 × 10−19 Coloumbs (1) 3. 2 × 10−18 N (2) 2. 4 × 10−18 N (3) 4. 8 × 10−19 N (4) 1. 6 × 10−19 N

Q16.In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by ˆk and 2ˆi −2ˆj, respectively. What is the unit vector along direction of propagation of the wave. (1) + k) √2 1 (ˆi +ˆj) (2) √21 (ˆj (3) + 1 (ˆi √5 2ˆj) (4) √51 (2ˆi +ˆj)

Q17.A square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about z-axis is given by. (1) μ0I 2a2 (2) μ0I 2a3 2πb 2πb2 (3) 2μ0I2a2 (4) 2μ0I2a3 πb πb2

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.An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification (m) versus distance of the object from the mirror (x) is correctly given by (Graphs are drawn schematically and are not to scale) JEE Main 2020 (08 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)

Q17.In a Young's double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be : (1) 24 (2) 30 (3) 18 (4) 28

Q17.A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 36% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which JEE Main 2020 (07 Jan Shift 1) JEE Main Previous Year Paper the analyser needs to be rotated further, to reduce the output intensity to zero, is (sin−1( 35 ) = 37°) (1) 53o (2) 37o (3) 90o (4) 45o

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 spherical mirror is obtained as shown in the figure from a hollow glass sphere, if an object is positioned in front of the mirror, what will be the nature and magnification of the image of the object? (Figure down as schematic and not to scale) JEE Main 2020 (02 Sep Shift 1) JEE Main Previous Year Paper (1) Inverted, real and magnified (2) Erect, virtual and magnified (3) Erect, virutual and unmagnified (4) Inverted, real and unmagnified

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.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.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 thin lens made of glass (refractive index = 1.5 ) of focal length f = 16cm is immersed in a liquid of refractive index 1.42 . If its focal length in liquid is fl , then the ratio fl/f is closest to the integer: (1) 1 (2) 9 (3) 5 (4) 17

Q17.When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to: JEE Main 2020 (03 Sep Shift 1) JEE Main Previous Year Paper (1) 0. 81 eV (2) 1. 02 eV (3) 0. 52 eV (4) 0. 61 eV

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.For a concave lens of focal length f , the relation between object and image distance u and v, respectively, from its pole can best be represented by ( u = v is the reference line): (1) (2) (3) (4)

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.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.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.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.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.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