Practice Questions

Q17.A heating element has a resistance of 100 Ω at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500 oC more than the room temperature. The temperature coefficient of resistance of the heating element is (1) 5 × 10−4 oC−1 (2) 2 × 10−4 oC−1 (3) 1 × 10−4 oC−1 (4) 0.5 × 10−4 oC−1

Q17.Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistance of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between, (1) 11. 7 V and 11. 8 V (2) 11. 6 V and 11. 7 V (3) 11. 5 V and 11. 6 V (4) 11. 4 V and 11. 5 V

Q17.A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 × 10−6 N , the value of V is approximately: C2 2 = 8.85 × 10−12 (ε0 N.m ) ) (1) 150 V (2) 100 V (3) 250 V (4) 300 V

Q17.In a meter bridge, as shown in the figure, it is given that resistance Y = 12.5Ω and that the balance is obtained at a distance 39.5 cm from end A (by jockey J). After interchanging the resistances X and Y , a new balance point is found at a distance l2 from end A . What are the values of X and l2 ? (1) 19.15Ω and 39.5 cm (2) 8.16Ω and 60.5 cm (3) 19.15Ω and 60.5 cm (4) 8.16Ω and 39.5 cm

Q17.In the given circuit all resistance is of the value of R ohm each. The equivalent resistance between A and B is: (1) 5R (2) 3R 2 (3) 5R (4) 2R 3

Q18.A galvanometer with its coil resistance 25 Ω requires a current of 1 mA for its full deflection. In order to construct an ammeter to read up to a current of 2 A the approximate value of the shunt resistance should be: (1) 1.25 × 10−2 Ω (2) 2.5 × 10−3 Ω (3) 2.5 × 10−2 Ω (4) 1.25 × 10−3 Ω

Q18.On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ . How much was the resistance on the left slot before interchanging the resistances? (1) 910 Ω (2) 990 Ω (3) 505 Ω (4) 550 Ω JEE Main 2018 (08 Apr) JEE Main Previous Year Paper

Q18.The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns/cm.. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is: (1) 2 mA (2) 1 mA (3) 40μA (4) 20μA

Q18.The B −H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns cm−1 . The current that should be passed in the solenoid to demagnetize the ferromagnet JEE Main 2018 (15 Apr) JEE Main Previous Year Paper completely is, (1) 2 mA (2) 20 μA (3) 1 mA (4) 40 μA

Q18.A capacitor C1 is charged up to a voltage V = 60 V by connecting it to battery B through switch (1), Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors C2 = 3.0μF JEE Main 2018 (15 Apr Shift 2 Online) JEE Main Previous Year Paper and C3 = 6.0μF through a switch (2) as shown in the figure. The sum of final charges on C2 and C3 is: (1) 36μC (2) 20μC (3) 54μC (4) 40μC

Q19.In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA division−1 . In the absence of shunt resistance, the galvanometer produces a deflection of θ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ , is closest to: 2 (1) 550 Ω (2) 220 Ω (3) 55 Ω (4) 110 Ω

Q19.A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of the magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below]: (1) 8Nμ0I (2) 4Nμ0I 51/2R 53/2R (3) 4Nμ0I (4) 8Nμ0I 51/2R 53/2R

Q19.In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω ,a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell. (1) 2.5 Ω (2) 1 Ω (3) 1.5 Ω (4) 2 Ω

Q19.A Helmholtz coil has pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P , midway between the centres A and C , is given by (Refer to figure): (1) 4Nμ0I (2) 8Nμ0I 53/2R 53/2R (3) 4Nμ0I (4) 8Nμ0I 51/2R 51/2R

Q19.A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be: (1) Increased 8 times (2) Doubled (3) Halved (4) Unchanged

Q20.An ideal capacitor of capacitance 0 .2 μF is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0. 5 mH . The current at a time when the potential difference across the capacitor is 5 V is: (1) 0. 15 A (2) 0. 17 A (3) 0. 34 A (4) 0. 25 A

Q20.A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with respect to normal axis then magnetic moment of the loop is: JEE Main 2018 (16 Apr Online) JEE Main Previous Year Paper (1) 3 qωr2 (2) 1 qωr2 2 2 (3) qωr2 (4) 4 qωr2 3

Q20.An ideal capacitor of capacitance 0.2μF is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH . The current at a time when the potential difference across the capacitor is 5 V , is: (1) 0.17 A (2) 0.15 A (3) 0.34 A (4) 0.25 A JEE Main 2018 (15 Apr Shift 1 Online) JEE Main Previous Year Paper .

Q20.A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × 10−2 m . The magnetic field at the centre of the triangle will be: (1) 4 × 10−5 Wb/m2 (2) Zero (3) 2 × 10−5 Wb/m2 (4) 8 × 10−5 Wb/m2

Q20.The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1 . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2 . The ratio B1 is: B2 (1) 1 (2) 2 √2 (3) √3 (4) √2

Q21.A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be: (1) 3 nBAω (2) 3nBAω 2 (3) nBAω (4) 1 nBAω 2

Q21.An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, rα respectively in a uniform magnetic field B. The relation between re, rp, rα is: (1) re < rα < rp (2) re > rP = rα (3) re < rp = rα (4) re < rp < rα

Q21.A monochromatic beam of light has a frequency v = 2π3 × 1012 Hz and is propagating along the direction ˆi+ˆj√2 . It is polarized along the ˆk direction. The acceptance form the magnetic field is: (1) E0 ˆi+ˆj (ˆi+ˆj) (2) E0 (ˆi+ˆj) .→r+ (3 × C ( √2 ) √2 .→r −(3 × C ˆk √2 Cos[104 1012)t] Cos[104 1012)t] (3) E0 (ˆi−ˆj) (ˆi+ˆj) (4) E0 (ˆi+ˆj+ˆk) (ˆi+ˆj) + (3 × C √2 √2 .→r+ (3 × C √3 √2 .→r Cos[104 1012)t] Cos[104 1012)t] JEE Main 2018 (15 Apr) JEE Main Previous Year Paper

Q21.A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is: (1) mgR cos θ (2) mgR sin θ B2l2 B2l2 (3) mgR tan θ (4) mgR cot θ B2l2 B2l2

Q21.A monochromatic beam of light has a frequency v = 2π3 × 1012 Hz and is propagating along the direction ^i+^j√2 It is polarized along the ^k direction. The acceptable form for the magnetic field is: (1) E0 (2) E0 cos cos ⋅→r −(3 × ⋅→r −(3 × C C ( ^i−^j√2 ) [104 ( ^i−^j√2 ) 1012)t] ( ^i−^j√2 ) [104 ( ^i+^j√2 ) 1012)t] (^i+^j+^k) (3) E0 (4) E0 cos + (3 × + (3 × C ^k cos C √3 [104 ( ^i+^j√2 )→r 1012)t] [104 ( ^i+^j√2 )→r 1012)t]