Practice Questions

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.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.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.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

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.Consider the following statements: A . Atoms of each element emit characteristics spectrum. B. According to Bohr's Postulate, an electron in a hydrogen atom revolves in a certain stationary orbit. C. The density of nuclear matter depends on the size of the nucleus. D. A free neutron is stable but a free proton decay is possible. E. Radioactivity is an indication of the instability of nuclei. Choose the correct answer from the options given below. (1) A, B, C, D and E (2) A, B and E only. (3) B and D only (4) A, C and E only

Q19.The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation: (1) Phase (2) Intensity (3) Amplitude (4) Frequency

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

Q20. Assertion A : If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is 5 mm and there are 50 total divisions on a circular scale, then the least count is 0. 001 cm. Reason R : Least Count = Total divisionsPitchon circular scale In the light of the above statements, choose the most appropriate answer from the options given below. (1) A is not correct but R is correct. (2) Both A and R are correct and R is the correct explanation of A. (3) A is correct but R is not correct. (4) Both A and R are correct and R is NOT the correct explanation of A.

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.Statement I : To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load 𝑅𝐿. Statement II : To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with 𝑅𝐿. In the light of the above statements, choose the most appropriate answer from the options given below : (1) Both Statement I and Statement II are false (2) Both Statement I and Statement II are true (3) Statement I is false but Statement II is true (4) Statement I is true but Statement II is false

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

Q20.In amplitude modulation, the message signal 𝑉m ( 𝑡) = 10sin2𝜋× 105𝑡 volts and carrier signal 𝑉C𝑡= 20sin2𝜋× 107𝑡 volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is 𝛼 kHz. The value of 𝛼 is: (1) 200 kHz (2) 500 kHz (3) 100 kHz (4) 0

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 a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0. 5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading. (1) 5. 20 mm (2) 5. 00 mm (3) 5. 15 mm (4) 5. 25 mm

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.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.Given below are two statements: Statement I : p −n junction diodes can be used to function as a transistor, simply by connecting two diodes, back to back, which acts as the base terminal Statement II: In the study of transistors, the amplification factor β indicates ratio of the collector current to the base current. In the light of the above statements, choose the correct answer from the options given below. (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Statement I is false but Statement II is true (4) Both Statement I and Statement II are true

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.The vernier scale used for measurement has a positive zero error of 0. 2 mm. If while taking a measurement it was noted that 0 on the vernier scale lies between 8. 5 cm and 8. 6 cm. Vernier coincidence is 6, then the correct value of measurement is cm. (least count = 0. 01 cm ) (1) 8. 36 cm (2) 8. 54 cm (3) 8. 58 cm (4) 8. 56 cm

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.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.For the circuit shown below, calculate the value of Iz : JEE Main 2021 (20 Jul Shift 1) JEE Main Previous Year Paper (1) 25 mA (2) 0. 15 A (3) 0. 1 A (4) 0. 05 A

Q20.For a certain radioactive process, the graph between ln R and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately: (1) 9. 15 sec (2) 6. 93 sec (3) 2. 62 sec (4) 4. 62 sec

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 → → →