Practice Questions

Q17.Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistance are now connected in parallel combination to the same battery, the electric power consumed will be : (1) 60 W (2) 240 W (3) 120 W (4) 30 W

Q17. In the circuit shown, the potential difference between A and B is (1) 1 V (2) 2 V (3) 3V (4) 6 V

Q17.To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained: If Vo is almost zero, identify the correct statement: (1) The emf of the battery is 1.5 V and its internal (2) The value of the resistance R is 1.5 Ω resistance is 1.5 Ω (3) The potential difference across the battery is (4) The emf of the battery is 1.5 V and the value of R 1.5 V when it sends a current of 1000 mA is 1.5 Ω

Q17.A carbon resistance has a following colour code. What is the value of the resistance? (1) 6.4 MΩ ± 5% (2) 64 kΩ ± 10% (3) 530 kΩ ± 5% (4) 5.3 MΩ ± 5%

Q17.The correct figure that shows, schematically, the wave pattern produced by the superposition of two waves of frequencies 9 Hz and 11 Hz, is JEE Main 2019 (10 Apr Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)

Q17.In the given circuit the cells have zero internal resistance. The currents (in amperes) passing through resistance R1 and R2 respectively, are: (1) 0, 1 (2) 2, 2 (3) 0.5 , 0 (4) 1, 2

Q17.A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60o at the center, the equivalent resistance between these two points will be: (1) 5 3 Ω (2) 125 Ω (3) 7 2 Ω (4) 52 Ω

Q17.A capacitor with capacitance 5 μF is charged to 5 μC . If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done? (1) 6.25 × 10–6 J (2) 2.55 × 10–6 J (3) 2.16 × 10–6 J (4) 3.75 × 10–6 J

Q17.In the given circuit, the charge on 4 μF capacitor will be: (1) 9.6 μC (2) 5.4 μC (3) 24 μC (4) 13.4 μC

Q17.When the switch 𝑆, in the circuit shown, is closed, then the value of current 𝑖 will be: (1) 2 A (2) 5 A (3) 3 A (4) 4 A

Q17.Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, −2) and (0, −2) . The work required to put a fifth charge Q at the origin of the coordinate system will be: (1) Q2 (2) Q2 4πϵ0 2√2πϵ0 (3) Q2 + 1 (4) Q2 + 1 4πϵ0 (1 √5 ) 4πϵ0 (1 √3 )

Q17.For the circuit shown, with 𝑅1 = 1.0 Ω , 𝑅2 = 2.0 Ω , 𝐸1 = 2 V and 𝐸2 = 𝐸3 = 4 V, the potential difference between the points ‘a’ and ‘b’ is approximately ( in V ): (1) 3.3 (2) 2.3 (3) 2.7 (4) 3.7

Q17.An ideal battery of emf 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is: (1) 480 Ω (2) 495 Ω (3) 395 Ω (4) 490 Ω

Q17.A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10−4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V it should be connected to a resistance of: (1) 6250 ohm (2) 250 ohm (3) 6200 ohm (4) 200 ohm

Q17.A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is: (1) 1 V (2) 2 V (3) 3 V (4) 5 V

Q18.In a Wheatstone bridge (see fig.), Resistances P and Q are approximatelyequal. When R = 400Ω , the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405Ω . The value of Y is close to JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper (1) 401.5ohm (2) 404.5ohm (3) 403.5ohm (4) 402.5ohm

Q18.A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is (1) 560 pJ (2) 600 pJ (3) 508 pJ (4) 692 pJ

Q18.In the given circuit the internal resistance of the 18V cell is negligible. If R1 = 400 Ω, R3 = 100 Ω and R4 = 500 Ω and the reading of an ideal voltmeter across R4 is 5 V , then the value of R2 will be: JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper (1) 550 Ω (2) 300 Ω (3) 450 Ω (4) 230 Ω

Q18.A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is: (1) 2 Ω (2) 12 Ω (3) 8 Ω (4) 4 Ω

Q18.A 200 Ω resistor has certain colour code. If one replaced the red colour by green in the code, the new resistance will be: (1) 300 Ω (2) 100 Ω (3) 400 Ω (4) 500 Ω

Q18.A galvanometer having a resistance of 20Ω and 30 division on both sides has figure of merit 0.005 ampere/ division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is: (1) 100Ω (2) 120Ω (3) 80Ω (4) 125Ω

Q18.In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that 0 (2) R(T) = R0 (1) R(T) = R0eT2/T2 T2 (3) R(T) = R0e−T2/T2 0 (4) R(T) = R0e−T20/T2

Q18.Drift speed of electrons, when 1.5 A current flows in a copper wire of cross section 5 mm2 is 𝑣𝑑. If the electron density in copper is 9 × 1028 m-3 the value of 𝑣𝑑 in mm 𝑠-1 is close to (Take charge of an electron to be = 1.6 × 10-19 C) (1) 0.2 (2) 3 (3) 2 (4) 0.02

Q18.A moving coil galvanometer, having a resistance G, produces full scale deflection when a current IG flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to Io(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V(V = GI0) by connecting a series resistance RV to it. Then, JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper Ig (1) = and RA = 2 (2) RARV = G2 and RARV RARV = (I0−Ig) RV G2( I0−IgIg ) ( (I0−Ig)Ig ) (3) Ig RA I0−Ig 2 (4) RA Ig 2 RARV = G2( I0−Ig ) and RV = ( Ig ) RARV = G2 and RV = ( I0−Ig )

Q18.In the given circuit diagram, the currents, I1 = −0.3 A, I4 = 0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are: (1) 0.4A, 1.1A, 0.4A (2) 1.1A, −0.4A, 0.4A (3) 1.1A, 0.4A, 0.4A (4) −0. 4A, 0. 4A, 1. 1A