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JEE Main 2021 August 26, Shift 2 (Physics)
Q1. In the shown Zener diode, power consumed will be:
(A) 0.15 π
(B) 0.12 π
(C) 2 π
(D) 1 π
Correct option: (B)
Solution:
The Zener diode offers constant potential difference. So, potential difference across 5γ is 10V and across 1γ is
14V.
Current through 5γ, π3 =10
5= 2ππ΄
Current through 1γ, π1 =14
1= 14ππ΄
Current through Diode, π2 = π1 β π3 = 12ππ΄
Power consumed in Zener diode, π = ππΌ = 0.12π
Q2. In a pendulum clock, the fractional error on the length of the string is 0.001. Calculate the error by the clock in a
day.
(A) 15.6π ππ
(B) 19.4π ππ
(C) 43.2π ππ
(D) 23.2π ππ
Correct option: (C)
Solution:
Time period of the simple pendulum is π β βπ
Therefore, the fractional error in calculated time, π₯π‘
π‘=
1
2
π₯π
π
For one day, π₯π‘ =1
2Γ 0.001 Γ (24 Γ 3600)
=> π₯π‘ = 43.2π ππ
Q3. The initial temperatures of liquids π΄, π΅ and πΆare 10ΒΊπΆ, 20ΒΊπΆ and 30ΒΊπΆ respectively. When π΄ & π΅ are mixed, the
temperature of the mixture is 16ΒΊπΆ. When π΅ & πΆ are mixed, the temperature of mixture is 26ΒΊπΆ. Find the
temperature of the mixture, when π΄ & πΆ are mixed.
(A) 160
13ΒΊπΆ
(B) 310
13ΒΊπΆ
(C) 180
13ΒΊπΆ
(D) 190
13ΒΊπΆ
Correct option: (B)
Solution:
Let the heat capacities of liquids A, B and C are π»1, π»2 and π»3 respectively.
When π΄ & π΅ are mixed,
π»1(16 β 10) = π»2(20 β 16)
=> 6π»1 = 4π»2 β¦ (1)
When π΅ & πΆ are mixed
π»2(26 β 20) = π»3(30 β 26)
=> 6π»2 = 4π»3 β¦ (2)
From equation (1) and (2)
4π»3 = 9π»1
When π΄ & πΆ are mixed, let the temperature of the mixture be T,
π»1(π β 10) = π»3(30 β π)
π =30π»3 + 10π»1
π»1 + π»3
π =10(3π»3 + π»1)
π»3 + π»1
π =310
13ΒΊπΆ
Q4. If π is the wavelength for which energy is πΈ. If π decreases to 75% of initial value, then find the energy gain.
(A) πΈ
3
(B) πΈ
2
(C) 4πΈ
3
(D) 3πΈ
4
Correct option: (A)
Solution: πβ² = 75% of π
= (75
100Γ π)
=3π
4
πΈ =βπ
π
β΄ πΈ1 =βπ
π
β΄ πΈ2 =βπ
πβ²=
βπ
(3π4 )
=4βπ
3π
β΄ πΈ2 =4πΈ
3
Therefore,
Energy gain = find energy β initial energy
=4βπ
3πβ
βπ
π
=4πΈ
3β πΈ =
πΈ
3
Q5. Angle between two vector π΄ and οΏ½ββοΏ½ is 60ΒΊ, then find angle between π΄ β οΏ½ββοΏ½ & π΄
(A) π‘ππβ1 (π΄π ππ 60β
π΅βπ΄πππ 60β)
(B) π‘ππβ1 (π΅π ππ 60β
π΄+π΅πππ 60β)
(C) π‘ππβ1 (π΅πππ 60β
π΄βπ΅π ππ 60β)
(D) π‘ππβ1 (π΅π ππ 60β
π΄βπ΅πππ 60β)
Correct option: (D)
Solution:
β΄ tanπ =π΅sin60Β°
π΄ β π΅cos60Β°
β΄ π = tanβ1 (π΅sin60Β°
π΄ β π΅cos60Β°)
Q6. A conical pendulum of length π moving in a circle of radius π
β2 Find the velocity of Pendulum:
(A) π
β2π
(B) βπ
2β2π
(C) βπ
β2π
(D) βπ
2π
Correct option: (C)
Given,
π =π
β2
600
B cos600
B sin600
sinΞΈ =π
π=
1
β2
Therefore, π = 45Β°
πcosΞΈ = ππ β¦ . (1)
πsinΞΈ =ππ£2
π β¦ . (2)
From equation (1) and (2),
We can write
πsinΞΈ
πcosΞΈ=
ππ£2
πππ
=π£2
ππ
β΄ tanΞΈ =
ππ£2
ππ
β2π
β΄ tan45Β° =π£2β2
ππ
β΄ πg = β2π£2
β΄ v = βππ
β2
Q7. In an Atwood machine the maximum stress that a string can tolerate without break is 24
πΓ 10β2. Find the radius
of string:
(A) 12.5
(B) 16.5
(C) 20.5
(D) 24.5
Correct option: (A)
Solution:
π1 = 5 kg
π2 = 3 kg
π =2π1π2π
π1 + π2
=2 Γ 5 Γ 3 Γ 10
5 + 3=
300
8=
75
2
Stress developed in rope =π
π΄
24
πΓ 102 =
752β
2
π2 = 75
48Γ 100 = 156.25
β΄ π = 12.5
Q8. Electric field of an electromagnetic wave is πΈ = 200π ππ (π (π‘ βπ₯
π))
π£
π . An electron is moving with speed
3 Γ 107π/π . Find magnitude of magnetic force on electron.
(A) 2.2 Γ 10β18 πππ€π‘ππ
(B) 3.2 Γ 10β18 πππ€π‘ππ
(C) 1.2 Γ 10β18 πππ€π‘ππ
(D) 4.2 Γ 10β18 πππ€π‘ππ
Correct option: (B)
Solution:
General equation of wave.
πΈ = πΈπ sin [π€ (π‘ βπ₯
π)]
β΄ πΈπ = 200
Now velocity of light is given by
πΆ =πΈπ
π΅π
Or
3 Γ 108 =200
π΅π
β΄ π΅π =200
3 Γ 108
Force experienced by electron
= ππ£π΅π
= (1.6 Γ 10β19) Γ (3 Γ 107) Γ (200
3 Γ 108)
= 3.2 Γ 10β18 N
Q9 Equation of SHM for two particles is π₯1 = 5π ππ (ππ‘ +π
4) and π₯2 = 5β2[π ππ2ππ‘ + πππ 2ππ‘] Then how many times
amplitude of second particle is greater than first.
Correct option: (2)
Solution:
Standard equation of SHM
π₯ = π΄ sin(ππ‘ + π)
where, π΄ = amplitude
Now,
π₯1 = 5 sin (ππ‘ +π
4)
β΄ π΄1 = 5
And π₯2 = 5β2[sin 2ππ‘ + cos 2ππ‘]
= 5β2 Γ β2 [(sin 2ππ‘) Γ1
β2+
1
β2cos(2ππ‘)]
= 10 [sin (2ππ‘ +π
4)]
β΄ π΄2 = 10
Clearly, π΄2 = 2π΄1
Q10. Find equivalent capacitance.
(A) π΄π0
π(
1
3+
πΎ1πΎ2
πΎ1+πΎ2)
(B) π΄π0
π(
1
2β
πΎ1πΎ2
πΎ1+πΎ2)
(C) π΄π0
π(
1
2+
πΎ1πΎ2
πΎ1+πΎ2)
(D) π΄π0
π(
1
2+
2πΎ1πΎ2
πΎ1+πΎ2)
Correct option: (C)
Solution:
The given configuration can be treated as series and parallel combination of the capacitors.
Capacitance, πΆ =πΎπ΄π0
π
πΆππ = πΆ1 +πΆ2πΆ3
πΆ2+πΆ3
πΆππ =π΄π0
2π+
πΎ1π΄π0π
ΓπΎ2π΄π0
ππΎ1π΄π0
π+
πΎ2π΄π0π
πΆππ =π΄π0
2π+
π΄π0
π(
πΎ1πΎ2
πΎ1+πΎ2) =
π΄π0
π(
1
2+
πΎ1πΎ2
πΎ1+πΎ2)
Q11. Find truth table for given logic gates
(A)
(B)
(C)
(D)
Correct option: (B)
π¦ = π΄ + (π΄ + π΅Μ Μ Μ Μ Μ Μ Μ Μ )Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ + π΅ + (π΄ + π΅Μ Μ Μ Μ Μ Μ Μ Μ )Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ
= (π΄ + (π΄ + π΅Μ Μ Μ Μ Μ Μ Μ Μ )) β (π΅ + π΄ + π΅Μ Μ Μ Μ Μ Μ Μ Μ )
π΄ π΅ π 0 0 1 1 1 1 0 1 0 1 0 0
Q12. A container has 1 mole of gas 'A' and 2 mole of gas 'B' confined in a space of volume 1 π3 at 127βπΆ
temperature. Find the pressure exerted on the wall of container in πππ.
Correct Answer: (10)
Solution:
Ideal gas equation
ππ = ππ π
Net pressure = ππ΄ + ππ΅
ππ΄ = pressure due to gas π΄
ππ΅ = pressure due to gas π΅
π = 1m3, π = 127 Β°C = 127 + 273
= 400 K
ππ΄π = ππ΄π π
β΄ ππ΄ =ππ΄π π
π=
1 Γ π Γ 400
1
= 8.314 Γ 400
= 3325.6 ππ
ππ΅π = ππ΅π π
ππ΅ =ππ΅π π
π=
2 Γ 8.314 Γ 400
1
= 6651.2 Pa
β΄ π = ππ΄ + ππ΅ = 9976.8
ππ = 9.976 kPa
= 10 kPa
Q13. Two identical circular rings of radius R are placed at R distance apart as shown with charges +Q and -Q.
Calculate potential difference between their centres.
(A) β3ππ
π
(B) ππ
π(2 β β2)
(C) β2ππ
π
(D) 2ππ
π
Correct option: (B)
Solution: Potential at A due to both rings, ππ΄ =ππ
π β
ππ
β2π
Potential at B due to both rings,ππ΅ =ππ
β2π β
ππ
π
ππ΄ β ππ΅ =ππ
π (2 β β2)
Q14. Height of transmission and receiver tower are 80π and 50π respectively. If radius of earth is 6400ππ. Then
find the range of LOS communication.
(A) 60ππ
(B) 116ππ
(C) 200ππ
(D) 245ππ
Correct option: (π΄)
Solution: Range of transmission = β2π β1 + β2π β2 = β2 Γ 6400 Γ 0.08 + β2 Γ 6400 Γ 0.05
Putting values range = 60ππ
Q15. In given circuit, find the value for resistance ' R ' So that bulb operate at the rated power of 500 πππ‘π‘.
(A) 20πΊ
(B) 30πΊ
(C) 40πΊ
(D) 50πΊ
Correct option: (A)
Solution:
R
i
200V
500W, 100V
For given rated power of bulb,
ππ =π2
ππ
β΄ π π =π2
ππ=
(100)2
500= 20 Ξ©
Current through bulb for rated power,
π =ππ
π=
500
100= 5 A
β΄ potential different across
Bulb = 100 V
Then, 100 = π Γ π
β΄ π =100
5= 20 Ξ©
Q16. A circular coil is rotating in uniform magnetic field shown in figure. Find the maximum induced emf?
(A) 2π΅ππ(ππ 2)
(B) π΅π(ππ 2)
(C) π΅ππ(ππ 2)
(D) β2π΅ππ(ππ 2)
Correct option: (3)
Solution:
π = ππ΄π΅ π sin ππ‘
= [π(ππ 2) π΅] π sin πππ‘
Hence, max value of emf
= [π(ππ 2) π΅] π
= π΅ππ(ππ 2)
Q17. Unpolarized light passes through polarizer π΄ and π΅, intensity of unpolarized light is I, then find out angle of
rotation of polarizer π΅, so final intensity becomes 3πΌ/8.
(A) 30Β°
(B) 40Β°
(C) 50Β°
(D) 60Β°
Correct option: (A)
Solution:
When unpolarized light passes through a polarizer, its intensity becomes half and the light becomes polarized with a
plane of polarization as along the axis of polarizer.
So, after passing through polarizer A, intensity will be πΌ
2.
From Malus law,
Intensity after passing through polarizer B will be,
πΌβ² =πΌ
2πππ 2π =
3πΌ
8
On solving, π = 30β
Q18. A Carnot engine working between β10βπΆ and 25βπΆ temperature limit. It produces a power of 35π then what
is the heat input rate in the engine.
(A) 190π½
(B) 290π½
(C) 298π½
(D) 250π½
Correct option: (C)
Efficiency of car not engine is given by
π = 1 βππΏ
ππ»
ππΏ = temperature of sink
= (β10Β°C)
= (β10 + 273)K = 263 K
ππ» = temperature of source
= 25Β°C
= 25 + 273 = 298 K
β΄ π = 1 β263
298=
35
298
β΄ πinput =πouput
π= (
298
35Γ 35)
= 298 J
Q19. A convex lens is placed in front of a concave mirror as shown. The radius of curvature of the mirror is 15 cm. If
location of image of the object is same as location of object, then calculate separation between object and new
image when mirror is removed.
(A) 10
(B) 15
(C) 25
(D) 35
Correct option: (D)
Solution:
In case-1, the rays will retrace the path.
In case-2, the final Image coincide with the centre of curvature of the mirror. The distance between object and final
image= 12 + 23 = 35 ππ
Q20. An equilateral triangular coil placed in a uniform magnetic field π΅ = 2 Γ 10β2π exist in horizontal direction. A
current of 0.2π΄ is flowing in the coil. If torque acting on coil is equal to π¦ Γ 10β5π β π then find π¦ =?
Solution:
|πβ| = πΌ |
π΄β|
= 0.2 Γ [β3
4π2]
= 0.2 Γβ3
4(0.1)2 =
β3
2Γ 103
|π¦β| = |
πβ Γ
π΅β|
= |πβ| |
π΅β| sin90
= (β3
2Γ 10β3) (2 Γ 10β3)
= β3 Γ 10β5 = 1.732 Γ 10β5
β΄ π¦ = 1.732
Q21. Four resistances 2πΊ, 6πΊ, 4πΊ, 8πΊ are arranged to find equivalent resistance of 46
3πΊ. Then which
arrangement is correct.
(A)
(B)
(C)
(D)
Correct option: (A)
Solution:
For parallel combination
π ππ =π 1π 2
π 1 + π 2
For series combination
π ππ = π 1 + π 2
A)
π parallel =2 Γ 4
4 + 2=
8
6=
4
3
β΄ π ππ =4
3+ 14 =
46
3 Ξ©
B)
π parallel =4 Γ 6
4 + 6=
24
10= 2.4
π ππ = 2.4 + 10 = 12.4 Ξ©
πΆ)
π parallel =6 Γ 2
6 + 2=
12
8=
4
3
π eq =4
3+ 12 =
40
3
D)
π parallel =8 Γ 6
8 + 6=
48
14=
24
7
π eq =24
7+ 6 =
66
7 Ξ©
Q22. A fighter jet plane is flying horizontally drops a bomb, find the nature of the path of the bomb as seen by the
pilot?
(A) hyperbola
(B) straight line
(C) parabola
(D) None of these
Correct option: (B)
Solution: Initially the bomb and fighter plane both are moving horizontally. After the bomb detaches from the plane,
its horizontal velocity will be the same as the plane. Due to gravity, the bomb falls vertically downward also.
Since they have identical horizontal velocities, hence the bomb will appear below the plane as seen from the plane.
Q23. For a simple pendulum if percentage error in gravitational acceleration is 1% and percentage error in time
period T is 1 % then find maximum percentage error in energy.
Correct answer: 2
Solution:
π = 2πββ
π
β΄ β =ππ2
4π2
Hence, relation in terms of fractional change will be (for small change)
ββ
β=
βπ
π+ 2 (
βπ
π)
β΄ % change in β
= (1 + 2(1)) = 3%
Expression of energy is given by
=1
2 ππ2π΄2 =
1
2π (
π
π) π΄2
=ππ΄2
2(
π
π)
=ππ΄2
2ππβ1
β΄ Fractional change in energy
=βπ
π+
ββ
β
β΄ % percentage change in energy
= (1 + 3)% = 4%
Q24. Two blocks of mass 2 ππ & 1 ππ are resting on a smooth surface as shown in figure. There is friction between
2 ππ & 1 ππ block with coefficient of friction π = 0.5. Find maximum force to be applied on 1 ππ for the two blocks
to move together.
(A) 5 π
(B) 10 π
(C) 15 π
(D) 20 π
Correct option: (C)
Solution: For maximum force on lower block πΉπππ₯, the friction between them will be limiting friction which causes
acceleration in the upper block. with relative rest, ππ between 2ππ & 1ππ must be maximum.
ππ πππ₯ = ππ = 0.5 Γ 2 Γ 10 = 10 π
Maximum possible common acceleration of blocks, ππππ₯ =ππ πππ₯
2=5 m /π 2
Since blocks are moving together, for maximum force,
πΉπππ₯ = ππ = 3 Γ 5 = 15 π
Q25. A Solid sphere of radius R and charge Q, what is the ratio of electric potential at a point inside and outside the
sphere with distance π /2 from surface:
(A) 33 / 16
(B) 35 / 16
(C) 37 / 16
(D) 40/16
Correct option: (A)
Electric potential at any point inside a solid sphere
=πΎπ
2π [3 β
π2
π 2]
In equation,
π =π
2
β΄ π1 =πΎπ
2π [3 β
(π 2β )2
π 2]
=πΎπ
2π [3 β
1
4]
=11 πΎπ
8π
Electric potential at any point outside the solid sphere
=πΎπ
π
In question, π = 3π 2β
β΄ π2 =πΎπ
(3π
2)
=2πΎπ
3π
β΄ π1
π2=
33
16
Q26. Equation of two travelling waves is given as:
π1 = π΄1π ππ[π(π₯ β π£π‘)]
π2 = π΄2π ππ[π(π₯ β π£π‘ + π₯0)]
Given π΄1 = 12, π΄2 = 7, π = 6.28, π₯0 = 3.5
Calculate the resultant amplitude after superposition of the given waves.
(A) 10
(B) 5
(C) 8
(D) 12
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