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JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

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Q.1. Which of the following gives a reversible operation?    (2020)
(i)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(ii)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(iii)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(iv)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEAns.
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEThis is NOR gate, it is not reversible
Option (2)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEThis is OR gate, it is not reversible
Option (3)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEThis is NAND gate, it is not reversible
Option (4)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEThis is NOT gate, it is reversible.

Q.2. In the figure, potential difference between A and B is    (2020)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(1) 10 V
(2) 5 V
(3) 15 V
(4) zero
Ans. 
(1)
Solution.

We have
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEESince, 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEPotential difference between A and B isJEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Now,JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Therefore,JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.3. Boolean relation at the output stage-Y for the following circuit is    (2020)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(1)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(2) A + B
(3) A.B
(4)JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Ans. (4)
Solution.

Truth table of input output relation is
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEFrom truth table, we haveJEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.4. In the given circuit, value of Y is    (2020)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEESo, output of the circuit is given by
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Given that A = 1, B = 0
So, Y = 0+0 = 0

Q.5. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEAns. 
(12)
Solution.

From given circuit we have
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEFirst diode is forward biased and second diode is reversed biased. So, voltage at point A is given by JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.6. The current I in the network is    (2020)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEESo, current I in the circuit is given by JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Here Reff = 5 + 10 + 5 + 10 = 30Ω
Given that V = 9V
Therefore,JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.7. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEEAns.
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
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.8. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
We know that
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.9. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
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.10. To get output 1 at R, for the given logic gate circuit the input values must be    [2019]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
P + Q must be 0
Therefore, Y = 0, X = 1

Q.11. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
⇒ N = 1011 
Therefore, number of emitted electrons per second JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.12. For the circuit shown below, the current through the Zener diode is    [2019]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
= 14 mA
V1 = 50 V
V1 = I1R
50 = 10 × 103 × I1
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Current through diode is
I2 = I – I1
= 14 – 5= 9 mA

Q.13. In the given circuit the current through Zener Diode is close to    [2019]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE   [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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
= 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.14. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(1) 0.036
(2) 0.020
(3) 0.027
(4) 0.030

Ans. (2)
Solution. 
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
The second diode is reverse biased. Thus, current only flow through first diode and its value is JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.15. The output of the given logic circuit is    [2019]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Ans.
(3)
Solution.
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
We have,
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.16. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Now,
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
At input side in BE circuit
VBB = IBRB + VBE
= 25 × 10−6 × 100 × 103 + 1
⇒ VBB = 3.5 V

Q.17. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
= 3.75 × 1014 Hz
Usable frequency = 1% of f
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Therefore, required number of channel JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
= 6.25 × 10

Q.18. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.19. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

- wherein,

JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE is carrier frequency,

JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE are side band frequency.

JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

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.20. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.21. An amplitude modulated signal is plotted below:
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.22. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.23. To double the covering range of a TV transmission tower, its height should be multiplied by     [2019]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(2) 2
(3) 4
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Ans.
(3)
Solution.
Let d be the cover range of TV tower
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.24.  The reading of the ammeter for a silicon diode in the given circuit is:    [2018]
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(1) 0
(2) 11.5 mA
(3) 15 mA
(4) 13.5 mA
Ans:
(2)
Solution:
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.25.  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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.26. 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 (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(2)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(3)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
(4)
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Ans:
(3)
Solution:
Voltage gain
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Current gain
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE
Power gain = voltage gain x current gain
JEE Main Previous year questions (2016-20): Modern Physics- 1 - Notes | Study Physics For JEE - JEE

Q.27. 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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