Practice Questions

Q19.Which of the following circuits is reverse - biased ? (1) (2) (3) (4)

Q19.The explosive in a Hydrogen bomb is a mixture of 1H2, 1H3 and 3Li6 in some condensed form. The chain reaction is given by 3Li6 + 0n1 →2He4 + 1H3; 1H2 + 1H3 →2He4 + 0n1 During the explosion the energy released is approximately [Given : M(Li) = 6. 01690 amu, M(1H2) = 2. 01471 amu, M(2He4) = 4. 00388 amu and 1 amu = 931. 5 MeV] (1) 28. 12 MeV (2) 12. 64 MeV (3) 16. 48 MeV (4) 22. 22 MeV

Q19.In the given circuit, the voltage across load resistance ( 𝑅𝐿) is: (1) 8 . 75 V (2) 9 . 00 V (3) 8 . 50 V (4) 14 . 00 V

Q19.From the statements given below : (A) The angular momentum of an electron in 𝑛𝑡ℎ orbit is an integral multiple of ℎ. (B) Nuclear forces do not obey inverse square law. JEE Main 2024 (01 Feb Shift 2) JEE Main Previous Year Paper (C) Nuclear forces are spin dependent. (D) Nuclear forces are central and charge independent. (E) Stability of nucleus is inversely proportional to the value of packing fraction. Choose the correct answer from the options given below : (1) (A), (B), (C), (D) only (2) (A), (C), (D), (E) only (3) (A), (B), (C), (E) only (4) (B), (C), (D), (E) only

Q19.An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is : (1) E = U (2) 2E = U (3) 2E = 3U (4) E = 2U

Q19.The I −V characteristics of an electronic device shown in the figure. The device is: (1) a diode which can be used as a rectifier (2) a zener diode which can be used as a voltage regulator (3) a transistor which can be used as an amplifier (4) a solar cell

Q19.Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below. (1) Both statement I and statement II are false (2) Statement I is false but statement II is true (3) Statement I is true but statement II is false (4) Both statement I and statement II are true

Q19.Which of the following nuclear fragments corresponding to nuclear fission between neutron (10n) and uranium isotope (23592 U) is correct : (1) 144 56 Ba + 8936Kr + 410n (2) 14456 Ba + 8936Kr + 310n (3) 140 56 Xe + 9438Sr + 310n (4) 15351 Sb + 9941Nb + 310n

Q19.The output (Y ) of logic circuit given below is 0 only when : (1) A = 1, B = 0 (2) A = 0, B = 1 (3) A = 0, B = 0 (4) A = 1, B = 1

Q19.A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is 1.42eV. The wavelength of light emitted from the LED is : (1) 1400 nm (2) 650 nm (3) 875 nm (4) 1243 nm

Q19.The truth table of the given circuit diagram is : JEE Main 2024 (27 Jan Shift 2) JEE Main Previous Year Paper (1) A B Y (2) A B Y 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 0 (3) A B Y (4) A B Y 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 0

Q20.There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm . With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is : (1) 3.35 mm (2) 4.65 mm (3) 4.55 mm (4) 4.60 mm

Q20.In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm then how many main scale divisions are there in 1 cm ? (1) 10 (2) 5 (3) 20 (4) 40

Q20.While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of the wire is x mm . The value of x is : 50 (1) 21 (2) 142 (3) 71 (4) 42

Q20.One main scale division of a vernier caliper is equal to m units. If nth division of main scale coincides with (n + 1)th division of vernier scale, the least count of the vernier caliper is : (1) n (2) 1 (n+1) (n+1) (3) m (4) m (n+1) n(n+1) 5 for |→a| = n|→b| The integer

Q20.Identify the logic operation performed by the given circuit. JEE Main 2024 (31 Jan Shift 1) JEE Main Previous Year Paper (1) NAND (2) NOR (3) OR (4) AND

Q20.If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0 . 5 mm the Vernier constant of travelling microscope is: (1) 0 . 1 mm (2) 0 . 1 cm (3) 0 . 01 cm (4) 0 . 01 mm → →

Q20.Which of the diode circuit shows correct biasing used for the measurement of dynamic resistance of p-n junction diode : (1) (2) (3) (4)

Q20.The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is 𝑁%. The value of 𝑁 is: (1) 4 (2) 8 (3) 6 (4) 5

Q20.A Zener diode of breakdown voltage 10 V is used as a voltage regulator as shown in the figure. The current through the Zener diode is (1) 50 mA (2) 0 (3) 30 mA (4) 20 mA

Q20.Identify the physical quantity that cannot be measured using spherometer : (1) Radius of curvature of concave surface (2) Specific rotation of liquids (3) Thickness of thin plates (4) Radius of curvature of convex surface

Q20.Given below are two statements: one is labelled as Assertion(A) and the other is labelled as Reason (R). Assertion (A) : In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. Reason (R) : The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are correct and (R) is the (2) Both (A) and (R) are correct but (R) is not the correct explanation of (A) correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true

Q20.A vernier callipers has 20 divisions on the vernier scale, which coincides with 19th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to _____ mm . (1) 0.5 (2) 2 (3) 5 (4) 1

Q20.The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm . The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is: (1) 2.0 g/cm3 (2) 1.7 g/cm3 (3) 2.2 g/cm3 (4) 2.5 g/cm3 JEE Main 2024 (08 Apr Shift 1) JEE Main Previous Year Paper −−−

Q20.Following gates section is connected in a complete suitable circuit. For which of the following combination, bulb will glow (ON) : (1) A = 0, B = 0, C = 0, D = 1 (2) A = 0, B = 1, C = 1, D = 1 (3) A = 1, B = 0, C = 0, D = 0 (4) A = 1, B = 1, C = 1, D = 0