Practice Questions

Q13.The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA For a cell of emf 5 V and internal resistance of 20 Ω the null point on it is found to be at 1000 cm The resistance of whole wire is: (1) 80Ω (2) 120Ω (3) 60Ω (4) 100Ω

Q14.A LCR circuit behaves like a clamped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be: (1) L ↔m, C ↔k, R ↔b (2) L ↔1b , C ↔ m1 , R ↔1k (3) L ↔k, C ↔b, R ↔m (4) L ↔m, C ↔1k , R ↔b

Q14.A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic → substance. The whole system is placed in a uniform magnetic field B . Then the field inside the paramagnetic substance is: → (1) (2) Zero B → → → → (3) (4) much large than B and parallel to B much large than B but opposite to B

Q14.A wire A , bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of an arc of a circle, carrying a current of 3 A and having radius of 4 cm , are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is: (1) 4 : 6 (2) 6 : 4 (3) 2 : 5 (4) 6 : 5

Q14.In LC circuit the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V (t) = 10sin(314t) is applied to the circuit, the current in the circuit is given as: (1) 0.52 cos(314t) (2) 10 cos(314t) (3) 5.2 cos(314t) (4) 0.52 sin(314t)

Q14.In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is: (1) +2V (2) −2V (3) –1V (4) +1V

Q14.An AC circuit has R = 100Ω, C = 2μF and L = 80 mH, connected in series. The quality factor of the circuit is : (1) 2 (2) 0.5 (3) 20 (4) 400

Q14.A long, straight Wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance a and 2a , respectively from the axis of the wire is: 3 (1) 2 (2) 2 3 (3) 1 (4) 3 2 2 −

Q14.A 750 Hz, 20 V(rms) source is connected to a resistance of 100 Ω, an inductance of 0. 1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/°C ) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to : (1) 418 s (2) 245 s (3) 365 s (4) 348 s →

Q14.The electric field of a plane electromagnetic wave is given by E = E0 cos(kz + ωt) . At t = 0 , a positively √2 charged particle is at the point (x, y, z) = (0, 0, πk ) . If its instantaneous velocity at (t = 0) is v0ˆk , the force acting on it due to the wave is: (1) parallel to ˆi+ˆj (2) zero √2 (3) antiparallel to ˆi+ˆj (4) parallel to ˆk √2

Q14.A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected corrected correctly to the resistor. In the circuit : (1) ammeter is always used in parallel and voltmeter (2) Both ammeter and voltmeter must be connected is series in parallel (3) ammeter is always connected in series and (4) Both ammeter and voltmeter must be connected voltmeter in parallel in series

Q14.A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0. 4 T at a temperature of 4 K . When the sample is placed in an external magnetic field of 0. 3 T at a temperature of 24 K , then the magnetisation will be : (1) 1 A/m (2) 4 A/m (3) 2. 25 A/m (4) 0. 75 A/m

Q14.A charged particle (mass m and charge q) moves along X axis with velocity V0 . When it passes through the → origin it enters a region having uniform electric field E = −Eˆj which extends upto x = d. Equation of path of electron in the region x > d is: qEd qEd (1) y = (x −d) (2) y = ( d2 −x) mv2 mv2 0 0 qEd (3) y = (4) y = qEd2 x mv2 0 mv20

Q14.A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale) (1) (2) (3) (4)

Q14.A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at JEE Main 2020 (08 Jan Shift 2) JEE Main Previous Year Paper N and D. Magnetic filed at the center of circle is: (1) + − 2πR μ0I (π 1 ) (2) 2πRμ0I (π 1 ) √2 √2 μ0I (3) 2πR (π + 1) (4) μ0I2R

Q14.Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6 × 10−27kg ) (1) 0.71mT (2) 7.1mT (3) 0.071mT (4) 71mT

Q14.A galvanometer of resistance G is converted into a voltameter of range 0 −1 V by connecting a resistance R in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0 −2V will be : (1) G (2) R1 (3) R1 −G (4) R1 + G

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.An iron rod of volume 10−3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns cm−1 . If a current of 0. 5 A is passed through the solenoid, then the magnetic moment of the rod will be : (1) 50 × 102 Am2 (2) 5 × 102 Am2 (3) 500 × 102 Am2 (4) 0. 5 × 102 Am2

Q15.A charged particle going around in a circle can be considered to be a current loop. A particle of a mass m → carrying charge q is moving in a plane with speed v under the influence of magnetic field B. The magnetic moment of this moving particle is : B (1) mv2→ B (2) −mv2→ 2B2 2πB2 B (3) −mv2→ B (4) −mv2→ B2 2B2

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.A charged particle of mass ‘ m ’ and charge ‘ q ’ moving under the influence of uniform electric field →E ˆi and a −→ uniform magnetic field B ˆk follows a trajectory from point P to Q as shown in figure. The velocities at P and Q → → are respectively, vi and −2vj . Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale) 3 mv2 (A) E = 2 ( qa ) (B) Rate of work done by the electric field at P is 32 ( mv3a ) (C) Rate of work done by both the fields at Q is zero (D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav . (1) (A), (C), (D) (2) (B), (C), (D) (3) (A), (B), (C) (4) (A), (B), (C), (D) −

Q15.An electrical power line, having a total resistance of 2Ω, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately : (1) 72% (2) 91% (3) 85% (4) 96%

Q15.A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0 , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at: (1) 2.5 s and 7.5 s (2) 2.5 s and 5.0 s (3) 5.0 s and 7.5 s (4) 5.0 s and 10. 0 s

Q15.If the magnetic field in a plane electromagnetic wave is given by → B = 3 × 10−8 sin(1.6 × 103x + 48 × 1010t)ˆjT, then what will be expression for electric field ? → → (1) V (2) V E = E = (60 sin(1.6 × 103x + 48 × 1010t)ˆk m ) (9 sin(1.6 × 103x + 48 × 1010t)ˆk m ) (3) → V (4) → V E = E = (3 × 10−8 sin(1.6 × 103x + 48 × 1010t)ˆj m ) (3 × 10−8 sin(1.6 × 103x + 48 × 1010t)ˆi m )