Practice Questions

Q19.Consider a situation in which reverse biased current of a particular P −N junction increases when it is exposed to a light of wavelength ≤621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly. (1) 2 eV (2) 4 eV (3) 1 eV (4) 0. 5 eV

Q19.A radioactive sample is undergoing α decay, At any time t1, its activity is A and another time t2, the activity is A . What is the average life time for the sample ? 5 (1) ln 5 (2) ln(t2+t1) t2−t1 2 (3) t2−t1 (4) t1−t2 ln 5 ln 5

Q19.If f denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0, (N0) then for a collection of radioactive nuclei, the rate of change of f with respect to time is given as: [λ is the radioactive decay constant] (1) −λ(1 −e−λt) (2) λ(1 −e−λt) (3) λe−λt (4) −λe−λt

Q19.A radioactive sample disintegrates via two independent decay processes having half lives T (1) and T (2) 1/2 1/2 respectively. The effective half- life T1/2 of the nuclei is : (1) (2) T 1/2+T 1/2 (1) T (2) (2) T1/2 = T 1/2(1) + 1/2 T1/2 = (1) (2) T 1/2−T 1/2 (3) T 1/2T(1) 1/2(2) (4) None of the above T1/2 = (1) (2) T 1/2+T 1/2

Q19.Which one of the following will be the output of the given circuit? (1) NOR Gate (2) NAND Gate (3) AND Gate (4) XOR Gate

Q19.Four NOR gates are connected as shown in figure. The truth table for the given figure is : (1) A B Y (2) A B Y 0 0 1 0 0 0 0 1 0 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 (3) A B Y (4) A B Y 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 1 1

Q19.If λ1 and λ2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of λ1 : λ2 is : (1) 7 : 135 (2) 1 : 9 (3) 1 : 3 (4) 7 : 108

Q19.The following logic gate is equivalent to : (1) NOR Gate (2) OR Gate (3) AND Gate (4) NAND Gate

Q19.The half life period of a radioactive element 𝑥 is same as the mean life time of another radioactive element 𝑦. Initially they have the same number of atoms. Then: (1) 𝑥 and 𝑦 decay at the same rate always. (2) 𝑥-will decay faster than 𝑦. (3) 𝑦-will decay faster than 𝑥. (4) 𝑥 and 𝑦 have same decay rate initially and later on different decay rate. JEE Main 2021 (01 Sep Shift 2) JEE Main Previous Year Paper

Q19.The output of the given combination gates represents: JEE Main 2021 (17 Mar Shift 1) JEE Main Previous Year Paper (1) XOR Gate (2) NAND Gate (3) AND Gate (4) NOR Gate

Q19.There are 1010 radioactive nuclei in a given radioactive element. Its half-life time is 1 min. How many nuclei = 1. will remain after 30 s? (√2 414) (1) 7 × 109 (2) 105 (3) 2 × 1010 (4) 4 × 1010

Q20.LED is constructed from Ga −As −P semiconducting material. The energy gap of this LED is 1. 9 eV . Calculate the wavelength of light emitted and its colour. h = 6. 63 × 10−34 J −s and c = 3 × 108 m s−1 (1) 654 nm and red colour (2) 654 nm and orange colour (3) 1046 nm and red colour (4) 1046 nm and blue colour → +

Q20.When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4. 8 V . If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1. 6 V . The threshold wavelength of the metal is: (1) 2λ (2) 4λ (3) 8λ (4) 6λ + N acts on a particle. The work done by this force when the particle is moved from

Q20.Draw the output signal Y in the given combination of gates. JEE Main 2021 (26 Feb Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)

Q20.One main scale division of a vernier callipers is a cm and nth division of the vernier scale coincide with (n −1)th division of the main scale. The least count of the callipers in mm is : (1) 10na (2) 10a (n−1) (n−1) (3) (n−1) (4) 10a 10n n

Q20.If an emitter current is changed by 4 mA, the collector current changes by 3 . 5 mA. The value of 𝛽 will be: (1) 3.5 (2) 0.5 (3) 0.875 (4) 7

Q20.A carrier signal C(t) = 25 sin(2. 512 × 1010t) is amplitude modulated by a message signal m(t) = 5 sin(1. 57 × 108t) and transmitted through an antenna. What will be the bandwidth of the modulated signal ? (1) 8 GHz (2) 2. 01 GHz (3) 1987. 5 MHz (4) 50 MHz

Q20.An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km ? (1) 605 m (2) 302 m (3) 37. 8 m (4) 75. 6 m JEE Main 2021 (27 Aug Shift 2) JEE Main Previous Year Paper

Q20.In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be : (1) 0, 1, 0, 1 (2) 0, 0, 1, 1 (3) 1, 1, 1, 0 (4) 1, 0, 1, 0

Q20.In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current 𝐼1 will be : (1) 2 . 0 A (2) 3 . 75 A (3) 2 . 73 A (4) 2 . 35 A

Q20.Match List –I with List – II. List –I List – II (a) 10 km height over earth's surface (i) Thermosphere (b) 70 km height over earth's surface (ii) Mesosphere (c) 180 km height over earth's surface (iii) Stratosphere (d) 270 km height over earth's surface (iv) Troposphere (1) (a )-( iv ), ( b )-( iii ), ( c )-( ii ), ( d )-( i) (2) (a )-( i ), ( b )-( iv ), ( c )-( iii ), ( d )-( ii) (3) (a )-( iii ), ( b )-( ii ), ( c )-( i ), ( d )-( iv) (4) (a )-( ii ), ( b )-( i ), ( c )-( iv ), ( d )-( iii)

Q20.The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is (1) 1. 64 mm (2) 0. 82 mm (3) 0. 90 mm (4) 1. 80 mm

Q20.If a message signal of frequency fm is amplitude modulated with a carrier signal of frequency fc and radiated through an antenna, the wavelength of the corresponding signal in air is (1) c (2) c fc−fm fm (3) c (4) c fc fc+fm → → → → → →

Q20.What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of 150 km? The average population density around the tower is 2000 km−2 and the value of Re = 6. 5 × 106 m. (1) Height = 1731 m (2) Height = 1241 m Population Covered = 1413 × 105 Population Covered = 7 × 105 (3) Height = 1600 m (4) Height = 1800 m Population Covered = 2 × 105 Population Covered = 1413 × 108 → → →

Q20.The correct relation between α (ratio of collector current to emitter current) and β (ratio of collector current to base current) of a transistor is : (1) β = 1+αα (2) α = 1−αβ (3) β = 1−α1 (4) α = 1+ββ JEE Main 2021 (18 Mar Shift 2) JEE Main Previous Year Paper