Practice Questions

Q15.An inductance coil has a reactance of 100 Ω . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is (1) 1. 1 × 10−2 H (2) 1. 1 × 10−1 H (3) 5. 5 × 10−5 H (4) 6. 7 × 10−7 H JEE Main 2020 (02 Sep Shift 2) JEE Main Previous Year Paper

Q15.A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dB /dt = 0. 032 Ts−1. The induced current in the loopis close to (Resistivity of the metal wire is 1. 23 × 10−8 Ωm ) (1) 0.43 A (2) 0.61 A (3) 0.34 A (4) 0.53 A

Q15.A point like object is placed at distance of 1m in front of a convex lens of focal length 0. 5m. A plane mirror is placed at a distance of 2m behind the lens. The position and nature of the image formed by the system is (1) 2. 6m from the mirror, real (2) 1m from the mirror, virtual (3) 1m from the mirror, real (4) 2. 6m from the mirror, virtual

Q15.Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following, which property best matches for the type of magnet required? (1) T : Large retentivity, small coercivity (2) P : small retentivity, large coercivity (3) T : Large retentivity, large coercivity (4) P : large retentivity, large coercivity

Q15.A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations will be seen on the galvanometer G attached across the coil? Three positions shown describe: (a) the magnet's entry (b) magnet is completely inside and (c) magnet's exit. JEE Main 2020 (04 Sep Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q15.At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5s, then induced JEE Main 2020 (08 Jan Shift 1) JEE Main Previous Year Paper EMF in the loop is: (1) 56μV (2) 28μV (3) 48μV (4) 36μV

Q15. As shown in the figure, a battery of emf ∈ is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc ( tc is the time constant of the circuit) is: (1) ∈L (2) ∈L eR2 R2 (1 −1e ) (3) ∈L (4) ∈R R2 eL2

Q15.Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm) (1) QA = VeC , QB = CV2 (2) QA = V C, QB = CV (3) QA = V C, QB = VeC (4) QA = CV2 , QB = VeC

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.A series L −R circuit is connected to a battery of emf V . If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches ( n1 ) times of its maximum value, is : (1) L √n (2) L √n+1 R ln( √n−1 ) R ln( √n−1 ) (3) L √n (4) L √n−1 R ln( √n+1 ) R ln( √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)

Q16.An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown in the figure. The connector has length l. and resistance R. It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is (1) μ0 Iv ℓ (2) μ0 Iv ℓ 4π Rr π Rr (3) 2μ0 Iv ℓ (4) μ0 Iv ℓ π Rr 2π Rr

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.If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eye-piece, should be close to: (1) 22mm (2) 2mm (3) 4mm (4) 33mm

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.Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum: (1) λvisible < λmicro waves < λradio waves < λx-rays (2) λradio waves > λmicro waves > λvisible > λx-rays (3) λx-rays < λmicro waves < λradio waves < λvisible (4) λvisible > λx-rays > λradio waves > λmicro waves

Q16.In a Young's double slit experiment, light of 500 nm is used to produce and interference pattern. When the distance between the slits is 0. 05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to : (1) 0. 17° (2) 0. 57° (3) 1. 7° (4) 0. 07°

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.A plane electromagnetic wave is propagating along the direction ˆi+ˆj , with its polarization along the direction √2 ˆk. The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant): (1) ˆi−ˆj ˆi+ˆj (2) ˆj−ˆi ˆi+ˆj B0 −k B0 + k √2 cos(ωt √2 ) √2 cos(ωt √2 ) ˆi+ˆj ˆi+ˆj ˆi+ˆj −k −k B0 (3) B0ˆkcos(ωt √2 ) (4) √2 cos(ωt √2 )

Q16.A plane electromagnetic wave, has frequency of 2. 0 × 1010 Hz and its energy density is 1. 02 × 10−8 J m−3 in vacuum. The amplitude of the magnetic field of the wave is close to ( 4πε01 = 9 × 109 Nm2C2 ) and speed of light = 3 × 108 m s−1 . (1) 150 nT (2) 160 nT (3) 180 nT (4) 190 nT

Q16.In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20m. The separation between P and Q is 5m and the phase of P is ahead of that of Q by 90°. A, B and C are three distinct point of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio : (1) 0 : 1 : 4 (2) 2 : 1 : 0 (3) 0 : 1 : 2 (4) 4 : 1 : 0

Q16.The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 4 for this wavelength, will be: 3 (1) 15o (2) 30o (3) 45o (4) 60o

Q16.A plane electromagnetic wave of frequency 25GHz is propagating in vacuum along the z-direction. At a → particular point in space and time, the magnetic filed is given by B = 5 × 10−8 ˆj T. The corresponding electric → field E is (speed of light = 3 × 108 m s−1 ) (1) 1.66 × 10−16ˆi mV (2) −1.66 × 10−16ˆi mV (3) −15ˆi mV (4) 15ˆi mV

Q16.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 z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be : (1) 2μ0I2a2 (2) 2μ0I2a2b πb π(a2+b2) (3) μ0I2a2b (4) μ0I2a2 2π(a2+b2) 2πb

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