Practice Questions
14,828 questions across 23 years of JEE Main — find and practise any topic!
Q11.A galvanometer 𝐺 of 2𝛺 resistance is connected in the given circuit. The ratio of charge stored in 𝐶1 and 𝐶2 is: 2 3 (1) (2) 3 2 1 (3) 1 (4) 2
Q12.A galvanmeter has a coil of resistance 200Ω with a full scale deflection at 20μA. The value of resistance to be added to use it as an ammeter of range (0 −20)mA is ; (1) 0.40Ω (2) 0.20Ω (3) 0.50Ω (4) 0.10Ω
Q12.An electric toaster has resistance of 60 Ω at room temperature (27 °C). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2. 75 A, the temperature attained by toaster is around: (if α = 2 × 10−4 ∘C−1 ) (1) 694°C (2) 1235°C (3) 1694°C (4) 1667°C
Q12.The value of unknown resistance (x) for which the potential difference between B and D will be zero in the arrangement shown, is : (1) 3Ω (2) 42Ω (3) 9Ω (4) 6Ω
Q12.A square loop of side 15 cm being moved towards right at a constant speed of 2 cm/s as shown in figure. The front edge enters the 50 cm wide magnetic field at t = 0 . The value of induced emf in the loop at t = 10 s will JEE Main 2024 (09 Apr Shift 2) JEE Main Previous Year Paper be : (1) 0.3mV (2) zero (3) 4.5mV (4) 3mV
Q12.Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _____ to _____ times of its initial length if the water is to be boiled in 15 minutes. (1) decreased, 3/4 (2) increased, 4/3 (3) decreased, 4/3 (4) increased, 3/4
Q12.The ratio of heat dissipated per second through the resistance 5Ω and 10Ω in the circuit given below is : (1) 1 : 2 (2) 2 : 1 (3) 4 : 1 (4) 1 : 1 −
Q12.In the given circuit, the terminal potential difference of the cell is : (1) 2 V (2) 3 V (3) 4 V (4) 1.5 V
Q12.An infinitely long positively charged straight thread has a linear charge density λCm−1 . An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation JEE Main 2024 (04 Apr Shift 1) JEE Main Previous Year Paper of the kinetic energy of electron as a function of radius of circular path from the wire is : (1) (2) (3) (4)
Q12.The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be : (1) 60 cm (2) 45 cm (3) 30 cm (4) 15 cm
Q12.The resistance per centimeter of a meter bridge wire is 𝑟, with 𝑋 𝛺 resistance in left gap. Balancing length from left end is at 40 cm with 25 𝛺 resistance in right gap. Now the wire is replaced by another wire of 2𝑟 resistance per centimeter. The new balancing length for same settings will be at (1) 20 cm (2) 10 cm (3) 80 cm (4) 40 cm
Q12.The reading in the ideal voltmeter 𝑉 shown in the given circuit diagram is: (1) 5 V (2) 10 V (3) 0 V (4) 3 V
Q12.A current of 200 μA deflects the coil of a moving coil galvanometer through 60°. The current to cause 𝜋 deflection through radian is 10 (1) 30 μA (2) 120 μA (3) 60 μA (4) 180 μA
Q12.If the collision frequency of hydrogen molecules in a closed chamber at 27∘C is Z, then the collision frequency of the same system at 127∘C is : (1) √3 2 z (2) √32 Z (3) 3 4 Z (4) 34 Z
Q12.In the given circuit, the current in resistance R3 is: (1) 1 A (2) 1. 5 A (3) 2 A (4) 2. 5 A
Q12. Match List-I with List-II : Choose the correct answer from the options given below : (1) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) (2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (3) (A)-(III), (B)-(IV), (C)- (I), (D)-(II) (4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Q12.A wire of length 10 cm and radius √7 × 10−4 m connected across the right gap of a meter bridge. When a resistance of 4. 5 Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R × 10−7 Ω m , then value of R is : (1) 63 (2) 70 (3) 66 (4) 35
Q12.An electric bulb rated 50 W −200 V is connected across a 100 V supply. The power dissipation of the bulb is: (1) 25 W (2) 12.5 W (3) 50 W (4) 100 W
Q12.In a metre-bridge when a resistance in the left gap is 2 Ω and unknown resistance in the right gap, the balance length is found to be 40 cm . On shunting the unknown resistance with 2 Ω, the balance length changes by: (1) 22 . 5 cm (2) 20 cm (3) 62 . 5 cm (4) 65 cm JEE Main 2024 (01 Feb Shift 2) JEE Main Previous Year Paper
Q12.An alternating voltage 𝑉( 𝑡) = 220sin100𝜋t volt is applied to a purely resistive load of 50 𝛺. The time taken for the current to rise from half of the peak value to the peak value is: (1) 5 ms (2) 3 . 3 ms (3) 7 . 2 ms (4) 2 . 2 ms
Q12.A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6. 4 mH inductor to produce oscillations. The maximum current in the circuit would be : (1) 3. 2 A (2) 1. 5 A (3) 2. 0 A (4) 1. 2 A
Q13.Match List I with List II List I List II 1 A. Gauss’s law of magnetostatics i. ∮ →𝐸· d→𝑎= ∫𝜌d𝑉 𝜀0 B. Faraday’s law of electro magnetic induction ii. ∮ →𝐵· d→𝑎= 0 → → -d → C. Ampere’s law iii. ∮ 𝐸· d 𝑙= 𝐵· d →𝑎 d𝑡∫ D. Gauss’s law of electrostatics iv. ∮ →𝐵· d →𝑙= 𝜇0𝐼 Choose the correct answer from the options given below: (1) A-I, B-III, C-IV, D-II (2) A-III, B-IV, C-I, D-II (3) A-IV, B-II, C-III, D-I (4) A-II, B-III, C-IV, D-I
Q13.An element Δl = Δx^i is placed at the origin and carries a large current I = 10 A. The magnetic field on the y- axis at a distance of 0.5 m from the elements Δx of 1 cm length is: JEE Main 2024 (06 Apr Shift 1) JEE Main Previous Year Paper (1) 4 × 10−8 T (2) 10 × 10−8 T (3) 8 × 10−8 T (4) 12 × 10−8 T
Q13.Two charges 𝑞 and 3𝑞 are separated by a distance ‘𝑟’ in air. At a distance 𝑥 from charge 𝑞, the resultant electric field is zero. The value of 𝑥 is : (1) ( 1 + √3 ) (2) 𝑟 𝑟 3 ( 1 + √3 ) 𝑟 (3) (4) 𝑟( 1 + √3 ) ( 1 + √3 ) JEE Main 2024 (31 Jan Shift 1) JEE Main Previous Year Paper
Q13.The electrostatic force → and magnetic force acting on a charge q moving with velocity v can be (→F2) (F1) written : (1) →F1 = q →E, →F2 = q(→V × →B) (2) →F1 = q →B, →F2 = q(→B × →V ) (3) →F1 = q →E, →F2 = q(→B × →V ) (4) →F1 = q →V ⋅→E, →F2 = q(→B ⋅→V )