Practice Questions

Q28.An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ . The common emitter current gain β is: (1) 6 × 102 (2) 102 (3) 104 (4) 60

Q28.A He+ ion is in its first excited state. Its ionization energy is: (1) 13.60 eV (2) 48.36 eV (3) 54.40 eV (4) 6.04 eV

Q28.To get output '1' at R, for the given logic gate circuit the input values must be: (1) X = 1, Y = 1 (2) X = 0, Y = 0 (3) X = 0, Y = 1 (4) X = 1, Y = 0

Q28.Consider the nuclear fission, Ne20 →2 He4 + C12 . Given that the binding energy/nucleon of Ne20, He4 and C12 are 8.03 MeV, 7.86 MeV , respectively. Identify the correct statement: (1) Energy of 12. 4 MeV will be supplied. (2) Energy of 9. 72 MeV has to be supplied. (3) Energy of 3. 6 MeV will be released. (4) 8. 3 MeV energy will be released.

Q28. Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by: (assume that the Ge diode has large breakdown voltage) (1) 0.8 V (2) 0.4 V (3) 0.2 V (4) 0.6 V

Q28.In Li+ + ,electron in first Bohr orbit is excited to a level by a radiation of wavelength λ.When the ion gets de excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ ? (Given: ℎ= 6.63 × 10-34 J s ; 𝑐= 3 × 108 m s-1 ) (1) 10.8 nm (2) 9.4 nm (3) 11.4 nm (4) 12.3 nm

Q28.The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6 V, the current through the 100Ω resistance (in Amperes) is: (1) 0.036 (2) 0.02 (3) 0.027 (4) 0.03

Q28.Radiation coming from transitions 𝑛= 2 to 𝑛= 1 of hydrogen atoms fall on He+ ions in 𝑛= 1 and 𝑛= 2 states. The possible transition of helium ions as they absorb energy from the radiation is: (1) 𝑛= 2 ⟶𝑛= 3 (2) 𝑛= 2 ⟶𝑛= 4 (3) 𝑛= 2 ⟶𝑛= 5 (4) 𝑛= 1 ⟶𝑛= 4

Q28.Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6 V . If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current? (1) 1.5 mA (2) 3.5 mA (3) 7.5 mA (4) 2.5 mA

Q28.An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is 1240 eV (for a photon of wavelength λ, energy E = ) λ ( in nm ) (1) n = 6 (2) n = 5 (3) n = 7 (4) n = 4

Q28.Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be : (1) 889.2 nm (2) 488.9 nm (3) 388.9 nm (4) 642.7 nm

Q28.Consider a tank made of glass (refractive index 1.5 ) with a thick bottom. It is filled with a liquid of refractive index 𝜇. A student finds that, irrespective of what the incident angle 𝑖 (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of 𝜇 is: 3 4 (1) (2) √5 3 5 5 (3) (4) √3 √ 3

Q28.The ratio of mass densities of nuclei of 40Ca and 16O is close to: (1) 1 (2) 0.1 (3) 2 (4) 5

Q28.In the given circuit the current through Zener Diode is close to: (1) 0.0 mA (2) 6.7 mA (3) 4.0 mA (4) 6.0 mA

Q28.A 100V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? (1) 0.3 (2) 0.6 (3) 0.4 (4) 0.5

Q28.In the figure, given that VBB supply can vary from 0 to 5.0 V , VCC = 5 V , βdc = 200, RB = 100 kΩ, RC = 1 kΩ and VBE = 1.0V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively: (1) 25 μA and 2.8 V (2) 20μA and 2.8V (3) 25μA and 3.5V (4) 20μA and 3.5V

Q29.A message signal of frequency 100 MHz and peak voltage 100 V is used execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V . The modulation index and difference between the two side band frequencies are: (1) 4; 2 × 108 Hz (2) 0.25; 2 × 108 Hz (3) 4; 1 × 108 Hz (4) 0.25; 1 × 108 Hz

Q29.Two radioactive substances A and B have decay constants 5λ and λ respectively. At t = 0, a sample has the 1 2 same number of the two nuclei. The time taken for the ratio of the number of nuclei to become will be e 1 1 (1) (2) λ 2λ 2 1 (3) (4) λ 4λ

Q29.The reverse break down voltage of a Zener diode is 5.6 V in the given circuit. The current I𝑧 through the Zener is: (1) 10 mA (2) 7 mA (3) 17 mA (4) 15 mA

Q29.For the circuit shown below, the current through the Zener diode is (1) zero. (2) 5 mA . (3) 9 mA . (4) 14 mA . JEE Main 2019 (10 Jan Shift 2) JEE Main Previous Year Paper

Q29.The logic gate equivalent to the given logic circuit is: (1) NAND (2) AND (3) NOR (4) OR

Q29. An amplitude modulated signal is plotted below: Which one of the following best describes the above signal? (1) (9 + sin (2.5π × 105t)) sin (2π × 104t)V (2) (1 + 9 sin (2π × 104t)) sin (2.5π × 105t)V (3) (9 + sin (2π × 104t)) sin (2.5π × 105t)V (4) (9 + sin (4π × 104t)) sin (5π × 105t)V

Q29.In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108 m/s, h = 6.6 × 10−34 J-s ) (1) 6.25 × 105 (2) 4.87 × 105 (3) 3.75 × 106 (4) 3.86 × 106

Q29.In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V, if the base resistance is 1 kΩ and the current amplification of the transistor is 100 then the input signal voltage is (1) 10 mV (2) 1 mV (3) 0 .1 V (4) 1 V

Q29.A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given: radius of earth = 6.4 × 106 m). (1) 48 km (2) 40 km (3) 80 km (4) 65 km