Practice Questions

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.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.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.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.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.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. 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 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

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.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

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.A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250, 𝑅𝐶= 1 kΩ and 𝑉𝐶𝐶= 10 V . The minimum base current for VCE to reach saturation is (1) 10 μA (2) 40 μA (3) 7 μA (4) 100 μA

Q29.The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is: (1) 50 (2) 100 (3) 500 (4) 200

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.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.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.The truth table for the circuit given in the figure is: (1) A B Y (2) A B Y 0 0 0 0 0 1 0 1 0 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 (3) A B Y (4) A B Y 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 1 1 1 0

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.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

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

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.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.An amplitude modulated signal is given by V(t) = 10[1 + 0.3 cos (2.2 × 104t)] sin (5.5 × 105t). Here t is in seconds. The sideband frequencies (in kHz ) are, [Given π = 22/7 ] (1) 1785 and 1715 (2) 178.5 and 171.5 (3) 89.25 and 85.75 (4) 892.5 and 857.5

Q29.The surface of certain metal is first illuminated with light of wavelength 𝜆1 = 350 nm and then, by a light of wavelength 𝜆2 = 540 nm . It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to 1240 (Energy of photon = eV ) 𝜆in nm (1) 2.5 (2) 1.8 (3) 5.6 (4) 1.4

Q29.An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 μA change in the base current of the amplifier. The input resistance and voltage gain are: (1) 0.33 kΩ, 1.5 (2) 0.67 kΩ, 300 (3) 0.67 kΩ, 200 (4) 0.33 k Ω, 300