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JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR PDF Download

Q.1. A modulating signal is a square wave, as shown in the figure.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRIf the carrier wave is given as c(t) = 2sin⁡(8𝜋t) volts, the modulation index is:      (JEE Main 2023)
(1) 1/4
(2) 1/2

(3) 1
(4) 1/3

Ans. (2)
Modulation index μ = Am/Ac
Am = 1 & Ac = 2
μ = 1/2


Q.2. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement.
Reason R: The current in the forward bias is more than the current in the reverse bias for a 𝑝 − 𝑛 junction diode.
In the light of the above statements, choose the correct answer from the options given below:       (JEE Main 2023)
(1) A is true but R is false
(2) A is false but R is true
(3) Both A and R are true and R is the correct explanation of A
(4) Both A and R are true but R is NOT the correct explanation of A

Ans. (4)
Photodiode works in reverse bias and its is used as a intensity detector. (True)
Forward bias current is more as compaired to reverse bias current (True)


Q.3. The logic gate equivalent to the given circuit diagram is :       (JEE Main 2023)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(1) NAND
(2) OR
(3) AND
(4) NOR

Ans. (1)
by truth table
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
NAND gate


Q.4. Which of the following gives a reversible operation?    (2020)
(i)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(ii)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(iii)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(iv)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRAns.
(4)
Solution.
A logic gate is reversible if it’s input can be recovered from it’s output
Option (1)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRThis is NOR gate, it is not reversible
Option (2)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRThis is OR gate, it is not reversible
Option (3)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRThis is NAND gate, it is not reversible
Option (4)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRThis is NOT gate, it is reversible.

Q.5. In the figure, potential difference between A and B is    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(1) 10 V
(2) 5 V
(3) 15 V
(4) zero
Ans. 
(1)
Solution.

We have
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRSince, the diode is forward bias so diode behave as short circuit and it will conduct so equivalent circuit will be
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRPotential difference between A and B isJEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Now,JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Therefore,JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.6. Boolean relation at the output stage-Y for the following circuit is    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(1)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(2) A + B
(3) A.B
(4)JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Ans. (4)
Solution.

Truth table of input output relation is
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRFrom truth table, we haveJEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.7. In the given circuit, value of Y is    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(1) 0
(2) toggles between 0 and 1
(3) will not execute

(4) 1
Ans. 
(1) 
Solution.
Let two input of the given circuit be A and B
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRSo, output of the circuit is given by
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Given that A = 1, B = 0
So, Y = 0+0 = 0

Q.8. Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. Built in potential in each diode is 0.7 V. For the input voltage shown in the figure, the voltage (in Volts) at point A is __________.    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRAns. 
(12)
Solution.

From given circuit we have
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRFirst diode is forward biased and second diode is reversed biased. So, voltage at point A is given by JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.9. The current I in the network is    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR(1) 0.2 A
(2) 0.6 A
(3) 0.3 A
(4) 0
Ans. 
(3) 
Solution
.
In the given circuit both diodes are reversed biased. So, they are open circuited.
Redraw the given circuit we have following new figure
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRSo, current I in the circuit is given by JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Here Reff = 5 + 10 + 5 + 10 = 30Ω
Given that V = 9V
Therefore,JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.10. The circuit shown below is working as an 8 V DC regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is (considering both zener diodes are identical) ______.    (2020)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSRAns.
(40)
Solution.
Given that Vinn = 12V, Vout = 8V
So, voltage across resistance is V = Vin - Vout = 12 - 8 = 4V
Current across resistance isJEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Total power dissipated across Zener diodes is
P = VI = 8 x 1 x 10-2
= 80 x 10-3 W = 80 mW
Power dissipated across each Zener diode is
P1 = P/2 = 80/2 = 40mW

Q.11. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m−3 and their mobility is 1.6 m2/(V·s), then the resistivity of the semiconductor (since it is an n-type semiconductor, contribution of holes is ignored) is close to    [2019]
(1) 2 Ω·m
(2) 4 Ω·m
(3) 0.4 Ω·m
(4) 0.2 Ω·m
Ans.
(3)
Solution.
Current in the semiconductor is,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
We know that
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.12. Ge and Si diodes start conducting at 0.3 V and 0.7 V, respectively. In the following figure if Ge diode connection is reversed, the value of V0 changes by (assume that the Ge diode has large breakdown voltage)    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 0.8 V
(2) 0.6 V
(3) 0.2 V
(4) 0.4 V
Ans.
(4)
Solution.
In the given circuit,
voltage, V = 12 V
Resistance, R = 5 kΩ
So,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Output voltage
V0 = RI
= (5 × 103) × (2.34 × 10−3)
= 11.7 V
When the connection of Ge diode are reversed then the current will be through Si. Current JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Therefore, output voltage
16 = 1R
= (2.26 × 10−3) × (5 × 103)
= 11.3
The value of V0 = (11.7 – 11.3) V
= 0.4 V

Q.13. To get output 1 at R, for the given logic gate circuit the input values must be    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) X = 0, Y = 1
(2) X = 1, Y = 1
(3) X = 1, Y = 0
(4) X = 0, Y = 0
Ans. 
(3)
Solution.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
P + Q must be 0
Therefore, Y = 0, X = 1

Q.14. A metal plate of area 1 × 10−4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photoelectrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be (1 eV = 1.6 × 10−19 J)    [2019]
(1) 1014 and 10 eV
(2) 1012 and 5 eV
(3) 1011 and 5 eV
(4) 1010 and 5 eV
Ans. 
(4)
Solution.
Energy incident on plate per second = IA
= 1.6 × 10−3 × 1 × 10−4
= 1.6 × 10−7 W
Kinetic energy
K = hv− ϕ
= 10 – 5 = 5 eV
Now,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
⇒ N = 1011 
Therefore, number of emitted electrons per second JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.15. For the circuit shown below, the current through the Zener diode is    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 9 mA
(2) 5 mA
(3) Zero
(4) 14 mA
Ans. 
(1)
Solution.
If Zener diode does not undergo breakdown
V1 = 120 – 50
= 70
⇒ V = IR
70 = I × 5 × 10
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
= 14 mA
V1 = 50 V
V1 = I1R
50 = 10 × 103 × I1
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Current through diode is
I2 = I – I1
= 14 – 5= 9 mA

Q.16. In the given circuit the current through Zener Diode is close to    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 0.0 mA
(2) 6.0 mA
(3) 6.7 mA
(4) 4.0 mA
Ans. 
(1)
Solution.
When potential drop across 1500 Ω is 10 V then current flowing through it is
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR   [V = IR]
= 6.61 mA
Now, 2 V will be the potential difference across 500 Ω.
Thus, electric current flowing through it is
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
= 4 mA
So, I2 > I1 this condition is not possible.
Therefore, voltage across Zener diode must be less than 10 V therefore it will not work in break down region and its resistance will be infinite and current through it is equal to 0.

Q.17. The circuit shown below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in amperes) is    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 0.036
(2) 0.020
(3) 0.027
(4) 0.030

Ans. (2)
Solution. 
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
The second diode is reverse biased. Thus, current only flow through first diode and its value is JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.18. The output of the given logic circuit is    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Ans.
(3)
Solution.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
We have,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.19. In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5 V, βdc = 200, RB = 100 kΩ, RC = l kΩ and VBE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively    [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 25 μA and 3.5 V.
(2) 20 μA and 3.5 V.
(3) 25 μA and 2.8 V.
(4) 20 μA and 2.8 V.
Ans.
(1)
Solution.
For output, at saturation, VCE = 0
In CE circuit we have
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Now,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
At input side in BE circuit
VBB = IBRB + VBE
= 25 × 10−6 × 100 × 103 + 1
⇒ VBB = 3.5 V

Q.20. 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)    [2019]
(1) 3.75 × 106
(2) 3.86 × 106
(3) 6.25 × 105
(4) 4.87 × 105
Ans.
(3)
Solution.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
= 3.75 × 1014 Hz
Usable frequency = 1% of f
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Therefore, required number of channel JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
= 6.25 × 10

Q.21. 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).    [2019]
(1) 65 km
(2) 48 km
(3) 80 km
(4) 40 km
Ans.
(1)
Solution.
Let hr and hR be the height of transmitter tower and height of receiver respectively.
Maximum distance up to which signal can be broadcasted is
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.22. The modulation frequency of an AM radio station is 250 Hz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?     [2019]
(1) 2750 kHz
(2) 2900 kHz
(3) 2250 kHz
(4) 2000 kHz
Ans.
(4)
Solution.
Side band frequency -
AM wave contains three frequency JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

- wherein,

fc is carrier frequency,

JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR are side band frequency.

JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Range of signal 2250KHz to 2750KHz from option for 2000KHz , from = 200 Hz

range = 1800KHz to 2200KHz.

So, only option (4) lies in the given range.


Q.23. 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)    [2019]
(1) 1785 and 1715.
(2) 892.5 and 857.5.
(3) 178.5 and 171.5.
(4) 89.25 and 85.75.
Ans. 
(4)
Solution.
We have,
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.24. An amplitude modulated signal is plotted below:
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Which one of the following best describes the above signal?    [2019]
(1) (9 + sin (2.5π × 105 t)) sin(2π × 104t) V
(2) (1 + 9sin (2π × 104 t)) sin(2.5π × 105t) V
(3) (9 + sin(2π × 104 t))sin(2.5π × 105t) V
(4) (9 + sin(4π × 104 t))sin(5π × 105t) V
Ans.
(3)
Solution.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.25. A 100V carrier wave is made to vary between 160V and 40V by a modulating signal. What is the modulation index?    [2019]
(1) 0.3
(2) 0.5
(3) 0.6
(4) 0.4
Ans.
(3)
Solution.
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.26. To double the covering range of a TV transmission tower, its height should be multiplied by     [2019]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(2) 2
(3) 4
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Ans.
(3)
Solution.
Let d be the cover range of TV tower
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.27.  The reading of the ammeter for a silicon diode in the given circuit is:    [2018]
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(1) 0
(2) 11.5 mA
(3) 15 mA
(4) 13.5 mA
Ans:
(2)
Solution:
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.28. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?    [2018]
(1) 2 x 103
(2) 2 x 104
(3) 2 x 105
(4) 2 x 106
Ans:
C
Solution:
No. of telephonic channels that can be transmitted simultaneously
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.29. In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given respectively by:  
(β is current gain, IB, IC and IE are respectively base, collector and emitter currents).    [2018]
(1)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(2)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(3)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
(4)
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Ans:
(3)
Solution:
Voltage gain
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Current gain
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR
Power gain = voltage gain x current gain
JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

Q.30. The number of amplitude modulated broadcast stations that can be accomodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be:     [2018]
(1) 10
(2) 15
(3) 20
(4) 8
Ans:
A
Solution:
The station will require a band width of 30 kHz
So,
No. of stations = 300/30
= 10

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