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JEE Main 2020 Question Paper with Solution (7th January - Morning) - JEE MCQ


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30 Questions MCQ Test - JEE Main 2020 Question Paper with Solution (7th January - Morning)

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JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 1

A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley)

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 1


v = ωR (no slipping)



 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 2

Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 2

Take 1kg mass at origin


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JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 3

In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 3

For 2nd minima
d sinθ = 2λ
sinθ = (given)
......(1)
So for 1st minima is
d sinθ = λ
sinθ = (from equation (i))
θ = 25.65° (from sin table)
θ ≈ 25°

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 4

There are two infinite plane sheets each having uniform surface charge density +σ C/m2. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:
   

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 4


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 5

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by k = k0(1 + αx). Calculate capacitance of system: (given αd << 1)

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 5

Capacitance of element 

Capacitance of element 

Given αd << 1

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 6

A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placed coaxially near its centre. During the time interval 0 ≤ t ≤ 1, the induced current IR and the induced emf VR in the ring vary as:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 6


and 

 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 7

If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 7

So θ > 45° and 90 – θ < 45º so only one option is correct i.e. 18.4º
angle rotated should be = 90° – 71.6° = 18.4°

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 8

Three moles of ideal gas A with  is mixed with two moles of another ideal gas B with . The of mixture is (Assuming temperature is constant)

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 8


on rearranging we get




γmix = 1.42

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 9

Given magnetic field equation is B = 3 × 10–8 sin(ωt + kx + φ)then appropriate equation for electric field (E) will be :

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 9

 (speed of light in vacuum)
E0 = B0C = 3 × 10–8 × 3 × 108
= 9 N/C
So E = 9 sin (ωt + kx + φ)

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 10

There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constant k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 10

In damped oscillation 
ma + bv + kx = 0


In the circuit

Comparing equation (i) and (ii)
m = L, b = R, k = 1/c

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 11

A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 11

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 12

A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited state 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 12


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 13

Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 13

Case-I
If final image is at least distance of clear vision


Case-II
If final image is at infinity

fe = 22 mm 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 14

1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given γ = 1.40 and 31.4= 4.65

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 14


now work done 
Closest ans is 90.5 J

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 15

A string of length 60 cm, mass 6gm and area of cross section 1mm2 and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 1011 N/m2. Find extension in string. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 15


after substituting value of μ,v,l,A and Y we get
Δl = 0.3 mm

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 16

Which of the following gate is reversible 

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 16

A logic gate is reversible if we can recover input data from the output eg. NOT gate

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 17

A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing through a point at a distance of L/4 from centre and perpendicular to rod.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 17


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 18

A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height equal to radius of earth, it ejects a rocket of mass M/10 and itself starts orbiting the earth in circular path 
of radius 2R, find the kinetic energy of rocket.

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 18




Kinetic energy 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 19

The current 'i' in the given circuit is

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 19


JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 20

A current carrying circular loop is placed in an infinite plane if φ1 is the magnetic flux through the inner region and φ0 is magnitude of magnetic flux through the outer region, then  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 20

As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,
∴ φi = - φ0

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 21

Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) and F(0, 0, 5). Find magnetic flux through loop due to magnetic field 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 21



φ = (3 × 25) + (4 × 25)  = 175 weber

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 22

A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 22

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 23

A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10–5/°C along the x-axis and 5 × 10–6/°C along y-axis and z-axis. If coefficient of volume expansion of the solid is C × 10–6/°C then the value of C is 


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 23

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 24

A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces are frictionless)


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 24

As only consevative internal force acts upon the mass and earth system, thus we can say mechanical energy is conserved. Thus we get that net loss in PE = net gain in KE
Loss in PE = mg Δh = 1 x 10 x 1 = 10J
Thus gain in KE = 10J

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 25

On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x  


Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 25

Energy of photon. E == 4eV  >  2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule
No. of photons falling per second

No. of photoelectron emitted per second

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 26

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 26


2 × 0.34 = + 1 x 0.522
Eº1 = 0.68 – 0.522
Eº1 = 0.158

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 27

Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 28

Arrange the following in order of their pKb value
  (B)  CH3–NH–CH3    (C) CH3–CH=NH  

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 28

Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is aliphatic amine, here the 'N' is sp3 whereas in option 'C' the 'N' is sp2, hence B is more basic than C.

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 29

1-Methylethylene oxide Product 'X' will be –

Detailed Solution for JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 29

--excess HBr---> 


 

JEE Main 2020 Question Paper with Solution (7th January - Morning) - Question 30

Correct order of Intermolecular forces

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