Practice Questions

Q18.A free neutron decays into a proton but a free proton does not decay into neutron. This is because (1) neutron is an uncharged particle (2) proton is a charged particle (3) neutron is a composite particle made of a proton (4) neutron has larger rest mass than proton and an electron

Q18.Equivalent resistance between the adjacent corners of a regular n -sided polygon of uniform wire of resistance R would be : (1) (n−1)R (2) (n−1)R n2 (2n−1) (3) n2R (4) (n−1)R n−1 n

Q18.Name the logic gate equivalent to the diagram attached (1) NAND (2) AND (3) NOR (4) OR

Q18.A student is provided with a variable voltage source V , a test resistor RT = 10Ω , two identical galvanometers G1 and G2 and two additional resistors, R1 = 10MΩ and R2 = 0. 001Ω . For conducting an experiment to verify ohm's law, the most suitable circuit is: (1) (2) (3) (4) JEE Main 2023 (06 Apr Shift 2) JEE Main Previous Year Paper

Q18.In this figure the resistance of the coil of galvanometer G is 2 Ω. The emf of the cell is 4 V. The ratio of potential difference across C1 and C2 is (1) 1 (2) 45 (3) 5 (4) 3 4 4

Q18.The angular momentum for the electron in Bohr’s orbit is 𝐿. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be (1) Zero (2) 2𝐿 (3) 𝐿 (4) 𝐿 2

Q19.The logic performed by the circuit shown in figure is equivalent to JEE Main 2023 (11 Apr Shift 1) JEE Main Previous Year Paper (1) AND (2) NOR (3) OR (4) NAND

Q19.Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 Ω resistor will be, (1) 1. 0 A (2) 0. 4 A (3) 0. 6 A (4) 0. 2 A

Q19.The resistance of a wire is 5 Ω. Its new resistance in ohm, if stretched to 5 times of its original length will be : JEE Main 2023 (25 Jan Shift 2) JEE Main Previous Year Paper (1) 625 (2) 125 (3) 5 (4) 25

Q19.Match the List I with List II List I List II AIntrinsic I Fermi-level near valence band Semiconductor Fermi-level at middle of valence and conduction B n-type semiconductor II band C p-type semiconductor III Fermi-level near conduction band DMetals IVFermi-level inside conduction band Choose the correct answer from the options given below: (1) (A) → I, (B) → II, (C) → III, (D) → IV (2) (A) → II, (B) → I, (C) → III, (D) → IV (3) (A) → II, (B) → III, (C) → I, (D) → IV (4) (A) → III, (B) → I, (C) → II, (D) → IV

Q19.Match List I with List II List – I List – II (Current configuration) (Magnetic field at point O ) + A I. B0 = μ0I4πr [π 2] JEE Main 2023 (25 Jan Shift 1) JEE Main Previous Year Paper B II. B0 = μ04 Ir C III. B0 = μ0I2πr [π −1] + D IV. B0 = μ0I4πr [π 1] Choose the correct answer from the option given below: (1) A-III, B-IV, C-I, D-II (2) A-I, B-III, C-IV, D-II (3) A-III, B-I, C-IV, D-II (4) A-II, B-I, C-IV, D-III

Q19.A zener diode of power rating 1 . 6 W is to be used as voltage regulator. If the zener diode has a breakdown of 8 V and it has to regulate voltage fluctuating between 3 V and 10 V The value of resistance 𝑅𝑠 for safe operation of diode will be (1) 10 Ω (2) 12 Ω (3) 13 . 3 Ω (4) 13 Ω

Q19.As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 4√3 cm . The magnetic field at the centroid O of the triangle is : (Neglect the effect of earth’s magnetic field) (1) 4√3 × 10−4 T (2) 4√3 × 10−5 T (3) √3 × 10−4 T (4) 3√3 × 10−5 T

Q19.The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be : (1) 8 T (2) 4 T (3) 2 T (4) 16 T

Q19.The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is : (1) 1 (2) 1 8 4 (3) 7 (4) 3 8 4

Q19.The magnitude of magnetic induction at mid-point O due to current arrangement as shown in figure will be . (1) μ0I (2) 0 2πa (3) μ0I (4) μ0I 4πa πa

Q19.A massage signal of frequency 3 kHz is used to modulate a carrier signal of frequency 1 . 5 MHz. The bandwidth of the amplitude modulated wave is (1) 6 MHz (2) 6 kHz (3) 3 MHz (4) 3 kHz

Q20.A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed: A. By changing the magnitude of the magnetic field within the coil. B. By changing the area of coil within the magnetic field. C. By changing the angle between the direction of magnetic field and the plane of the coil. D. By reversing the magnetic field direction abruptly without changing its magnitude. Choose the most appropriate answer from the options given below: (1) A and B only (2) A, B and C only (3) A, B and D only (4) A and C only

Q20.Find the mutual inductance in the arrangement, when a small circular loop of wire of radius R is placed inside a large square loop of wire of side L(L ≫R) . The loops are coplanar and their centres coincide : JEE Main 2023 (29 Jan Shift 1) JEE Main Previous Year Paper (1) M = √2μ0R2L (2) M = 2√2μ0RL2 (3) M = 2√2μ0R2L (4) M = √2μ0RL2

Q20.The amplitude of 15sin ( 1000π𝑡) is modulated by 10sin ( 4π𝑡) signal. The amplitude modulated signal contains frequency (ies) of ( 𝐴) 500 Hz ( 𝐵) 2 Hz ( 𝐶) 250 Hz ( 𝐷) 498 Hz ( 𝐸) 502 Hz Choose the correct answer from the options given below: (1) A only (2) A, D and E only (3) B only (4) A and B only

Q20.A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0. 2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is: (1) 2. 4 × 103 A m−1 (2) 1. 2 × 103 A m−1 (3) 1 A m−1 (4) 2. 4 × 10−3 A m−1

Q20.A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A . The loop is in a uniform magnetic field of magnitude 0. 75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be 130x N . The value of x is _______.

Q20.A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is: (1) N : n (2) n2 : N 2 (3) N 2 : n2 (4) n : N

Q20.Which of the following frequencies does not belong to FM broadcast. (1) 106 𝑀𝐻𝑧 (2) 64 𝑀𝐻𝑧 (3) 99 𝑀𝐻𝑧 (4) 89 𝑀𝐻𝑧

Q20.A square loop of area 25 cm2 has a resistance of 10 Ω . The loop is placed in uniform magnetic field of magnitude 40. 0 T . The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1. 0 sec, will be JEE Main 2023 (29 Jan Shift 2) JEE Main Previous Year Paper (1) 2. 5 × 10−3 J (2) 1. 0 × 10−3 J (3) 1. 0 × 10−4 J (4) 5 × 10−3 J