JEE Main 2020 Paper with Solutions Physics - Shift 1- 8th January is given here. Students are recommended to practise these questions so that they can improve their accuracy and problem-solving skills. The questions and solutions can be accessed from our page directly. The PDF format is also available which can be downloaded easily. Ultimately, students will be able to face the exam with a better confidence level.
1. A particle of mass π is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed π about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is
Solution:
The centripetal force is provided by the spring force.
mΟ2(β + x) = kx
kx = m(β + x)Ο2
x = mβΟ2 /(Β k - mΟ2)
2. Three charged particles A, B and, C with charge -4q, +2q and -2q are present on the circumference of a circle of radius π. The charges particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x- direction is:
Solution:
Applying superposition principle,
By symmetry, net electric field along the x-axis.
3. A thermodynamic cycle xyzx is shown on a V- T diagram .
The P-V diagram that best describes this cycle is : (Diagrams are schematic and not upto scale)
Solution:
For the given V - T graph
For the process x β y; V β T; π =constant.
For the process y β z; V =constant
Only β²πβ² satisfies these two conditions.
4. Find the co-ordinates of center of mass of the lamina shown in the figure below.
Solution:
The Lamina can be divided into two parts having equal mass π each.
5. The plot that depicts the behavior of the mean free time π (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graph are schematic and not drawn to scale)
Solution:
Ο β 1/βT
6. Effective capacitance of parallel combination of two capacitors πΆ1 and πΆ2 is 10 ΞΌF. When these capacitor are individually connects to a voltage source of 1 π, the energy stored in the capacitor πΆ2 is 4 times of that in πΆ1. If these capacitors are connected in series, their effective capacitance will be:
Solution:
Given that,
C1 + C2 = 10 ΞΌF β¦(i)
4( Β½ C1V2) = Β½ C2V2
4C1 = C2 β¦(ii)
From equations (i) and (ii)
C1 = 2 ΞΌF
C2 = 8 ΞΌF
If they are in series
Ceq = C1C2/(C1+ C2)
= 1.6 ΞΌF
7. Consider a uniform rod of mass 4π and length πΏ pivoted about its centre. A mass π is moving with a velocity π£ making angle ΞΈ=Ο/4 to the rodβs long axis collides with one end of the rod and stick to it. The angular speed of the rod-mass system just after collision is
Solution:
There is no external torque on the system about the hinge point. So,
= 6π£/7β2L = 3β2 π£/7L
8. When photons of energy 4 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy ππ΄ eV and de-Broglie wavelength ππ΄. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is ππ΅= (ππ΄ β 1.5) eV. If the de-Broglie wavelength of these photoelectrons ππ΅ = 2ππ΄, then the work function of metal B is
Solution:
TA = 2eV
KEB = 2 - 1.5 = 0.5 eV
β B = 4.5 - 0.5 = 4 eV
9. The length of a potentiometer wire of length 1200 ππ and it carries a current of 60 ππ΄. For a cell of emf 5 π and internal resistance of 20Ξ©, the null point on it is found to be at 1000 ππ. The resistance of whole wire is
Solution:
Let Resistance per unit length of potentiometer wire =Β Ξ»
β Ξ» Γ 1000 Γ 60 Γ 10<sup>-3</sup> = 5Β
β Ξ» 5 / 60
Resistance of potentiometer wire = 1200Β Γ 5 / 60 = 100Ξ©
10. The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eyepiece?
Solution:
m = f0/fe = 5
f0 = 5fe
f0 + fe = 5fe + fe = 6fe = length of the tube
6fe = 60 cm
fe = 10 cm
11. Consider two solid spheres of radii π
1= 1 m, π
2 = 2 m and masses M1 & M2, respectively. The gravitational field due to two spheres 1 and 2 are shown. The value of M1/M2 is
Solution:
Gravitation field will be maximum at the surface of a sphere. Therefore,
GM2/22 = 3 and GM1/12 = 2
(M2/M1)Γ ΒΌ = 3/2
M1/M2 = 1/6
12. Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6 Γ 10β27 kg)
Solution:
K. E = 1 Γ 106 eV = 1.6 Γ 10β13 J
= Β½ mev2
Where me is the mass of the electron = 1.6Γ10 -27
1.6 Γ 10-13 = Β½ Γ 1.6 Γ 10-27 Γ v2
v = β2Γ107 m/s
Bqv = mea
B = (1.6 Γ10 -27 Γ 1012 )/ (1.6Γ10-19Γβ2Γ 107)
= 0.71 Γ 10-3 T
= 0.71 mT
13. If finding the electric field around a surface is given by
Solution:
14. The dimension of stopping potential π0 in photoelectric effect in units of Planckβs constant (h), speed of light (c), and gravitational constant (G) and Ampere (A) is
Solution:
V = K(h)a(I)b(G)c(c)d
Unit of stopping potential is (V0) Volt.
We know [h] = ML2Tβ1
[I] = A
[G] = Mβ1L3Tβ2
[C] = LTβ1
[V] = ML2Tβ3Aβ1
ML2Tβ3Aβ1 = (ML2Tβ1)a(A)b (Mβ1L3Tβ2)c(LTβ1)d
ML2T-3A-1 = Ma-cL2a+3c+dTβaβ2cβdAb
a - c = 1
2a + 3c + d = 2
-a - 2c - d = -3 b = -1
On solving,
c = -1
a = 0
d = 5
b = -1
V = K(h)0(A)β1(G)β1(c)5
15. A leak proof cylinder of length 1 m, made of metal which has very low coefficient of expansion is floating in water at 0o C such that its height above the water surface is 20 cm. When the temperature of water is increases to 4o C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4o C relative to the density at T = 0o C is close to
Solution:
Since the cylinder is in equilibrium, itβs weight is balanced by the Buoyant force.
mg = A(80)(Ο0oc)g
mg = A(79)(Ο4oc)g
Ο4oc/ Ο0oc = 80/79 = 1.01
16. The graph which depicts the result of Rutherford gold foil experiment with Ξ±- particle is:
π: Scattering angle
π βΆ Number of scattered πΌ β particles is detected
(Plots are schematic and not to scale)
Solution:
N β 1/sin4(ΞΈ/2)
17. At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is:
Solution:
= 56Γ500Γ10-8/5
= 56Γ 10-6 V
18. Choose the correct Boolean expression for the given circuit diagram:
Solution:
First part of figure shown OR gate and second part of figure shown NOT gate.
So, Y =
19. Consider a solid sphere of density Ο(r) = Ο0 (1- r2/R2), 0< r β€ R. The minimum density of a liquid in which it float is just
Solution:
Let the mass of the sphere be m and the density of the liquid be ΟL
Ο= Ο0 ( 1 - r2/R2), 0< r β€ R
Since the sphere is floating in the liquid, buoyancy force (FB) due to liquid will balance the weight of the sphere.
πΉ = mg
ΟL (4Ο/3) R3g = β«Ο(4Οr2 dr)g
ΟL (4Ο/3) R3 = β« Ο0 ( 1 - r2/R2) 4Οr2dr
ΟL (4/3)Ο R3 =
ΟL = (2/5) Ο0
20. The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 4/3 for this wavelength, will be
Solution:
If the speed of light in the given medium is V then,
V = 1/β(ΞΌΖ)
We know that, n = c/v
n = β(ΞΌr Ζr) = 2
sin ΞΈc = Β½
ΞΈc = 300
21. A body of mass π=0.10 ππ has an initial velocity of
Solution:
Mass of each object, π1= π2 = 0.1 ππ
Initial velocity of 1st object, π’1=5 π/π
Initial velocity of 2nd object, π’2 =3 π/π
Final velocity of 1st object, π1=
For elastic collision, kinetic energy remains conserved
Initial kinetic energy (Ki) = final kinetic energy (Kf)
Β½ mu12 + Β½ mu22 = Β½ mV12 + Β½ mV22
Β½ m(5)2 + Β½ m(3)2 = Β½ m(16β2)2 + Β½ mV22
V2 = β2 m/s
Kinetic energy of second object = Β½ mV22
= Β½ Γ 0.1 Γ β22
= 0.1
= 1/10 J
x = 1
22. A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of lens material is 1.5, then the focal length of the lens (in cm) is
Solution:
Applying Lens makersβ formula,
1/f = (ΞΌ - 1)[(1/R1) - (1/R2)]
R1 =β
R2 = -30 cm
ΞΌ = 1.5
1/f = (1.5 - 1)[(1/β) - (1/-30)]
1/f = 0.5/30
f = 60 cm
23. A particle is moving along the x-axis with its coordinate with time t given by x(t) = -3t2 + 8t + 10 π. Another particle is moving along the y-axis with its coordinate as a function of time given by y = 5 - 8t3 m. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as βv . Then v (m/s) is
Solution:
x = β3t2 + 8t + 10
= 2
y = 5 β 8t3
v = β(242 + 22)
v = 580 m/s
24. A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is __Hz.
Solution:
V = β(B/Ο)
Vpipe/ Vair =
Vpipe = Vair/β2
fn = Vpipe(n+1)/2l
f1-f0 = Vpipe/ 2l
= 300/2β2
= 106.06 Hz ( if β2 = 1.414) β 106 Hz
25. Four resistors of resistance 15 Ξ©, 12 Ξ©, 4Ξ© and 10Ξ© respectively in cyclic order to form a wheatstoneβs network. The resistance that is to be connected in parallel with the resistance of 10 to balance the network is __.
Solution:
[(10R)/(10+R) ]Γ12 = 15 Γ4
R = 10Ξ©