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JEE Advance 2020 with Solution: Paper - 2 - JEE MCQ


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30 Questions MCQ Test - JEE Advance 2020 with Solution: Paper - 2

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*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 1

A large square container with thin transparent vertical walls and filled with water (refractive index 4/3) is kept on a horizontal table. A student holds a thin straight wire vertically inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the wire from this corner, another student sees two images of the wire, located symmetrically on each side of the line of sight as shown. The separation (in cm) between these images is ____________. 


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 1

We will assume that observer sees the image of object through edge ⇒ α = 45° 

By applying Snell's Law 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 2

A train with cross-sectional area St is moving with speed vt inside a long tunnel of cross-sectional area S0 (S0 = 4St). Assume that almost all the air (density ρ) in front of the train flows back between its sides and the walls of the tunnel. Also, the air flow with respect to the train is steady and laminar. Take the ambient pressure and that inside the train to be p0. If the pressure in the region between the sides of the train and the tunnel walls is p, then  The value of N is ________.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 2


Applying Bernoulli's equation 

From equation of continuity

From (i) and (ii) 

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*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 3

Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density 900 kg m−3 are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of 200 V across the discs. As a result, an oil drop of radius 8×10−7 m stops moving vertically and floats between the discs. The number of electrons present in this oil drop is ________. (neglect the buoyancy force, take acceleration due to gravity = 10 ms−2 and charge on an electron (e) = 1.6×10–19 C)


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 3


When terminal velocity is achieved  
qE = mg 

⇒ n ≈ 6

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 4

A hot air balloon is carrying some passengers, and a few sandbags of mass 1 kg each so that its total mass is 480 kg. Its effective volume giving the balloon its buoyancy is V. The balloon is floating at an equilibrium height of 100 m. When N number of sandbags are thrown out, the balloon rises to a new equilibrium height close to 150 m with its volume V remaining unchanged. If the variation of the density of air with height h from the ground is , where  ρ0 = 1.25 kg m-3 and h0 = 6000 m, the value of N is _________.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 4


480 x g = vρ1g
(480 - N) g = vρ2g

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 5

A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment  is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N×(mgh), where g is the acceleration due to gravity, then the value of N is _________ . (Note that for three coplanar forces keeping a point mass in equilibrium, F/sinθ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)
 


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 5


Now, from ΔOAB  
α + 90 - θ + 90 - θ = 180 
⇒ α = 2θ

Now charge is in equilibrium at point B.
So, using sine rule 


⇒ substituting this in equation (i)

⇒ Uf = 2mgh
W = ΔU = Nmgh = N = 2 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 6

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be _______ (take the acceleration due to gravity = 10 ms−2 and the universal gas constant = 8.3 J mol−1K−1).


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 6

Assuming temperature remains constant at 300 K
From P1V1 = P2V2 


6x = 16 – x2
x2 + 6x – 16 = 0
x = 2
distance = 4 + 2 = 6m 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 7

A beaker of radius r is filled with water (refractive index 4/3) up to a height H as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed ω. This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is h (h≪H, h≪r)), as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature R. Which of the following is/are correct? (g is the acceleration due to gravity) 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 7


In ΔOAB  
R2 = (R - h)2 + r2
R2 = R2 - 2hR + h2 + r2
⇒ 2hR = h2 + r2 

Now considering equation of surface 


*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 8

A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v0 and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 8

By the energy conservation (ME) between bottom point and point Y 

Now at point Y the centripetal force provided by the component of mg


At point x and z of circular path, the points are at same height but less then h. So the velocity more than a point y.
So required centripetal = mv2/r is more. 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 9

A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with angular speed ω about the pivot. The maximum angular speed ωM is achieved for x = xM. Then

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 9

by the angular momentum conservation about the suspension point. 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 10

In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode) at a distance d from the cathode. The target is kept at a potential V higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current I is decreased to I/2 ,  the potential difference V is increased to 2V, and the separation distance d is reduced to d/2 , then 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 10


Hence I decreases 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 11

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is α. The spheres are now immersed in a dielectric liquid of density 800 kg m−3 and dielectric constant 21. If the angle between the strings remains the same after the immersion, then  

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 11

The net electric force on any sphere is lesser but by Coulomb law the force due to one sphere to another remain the same. 

In equilibrium

After immersed is dielectric liquid. 
As given no change in angle α.

when ρ = 800 Kg/m3 


d = density of sphere 

d = 840

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 12

Starting at time t = 0 from the origin with speed 1 ms−1, a particle follows a two-dimensional trajectory in the x-y plane so that its coordinates are related by the equation y = x2/2. The x and y components of its acceleration are denoted by ax and ay, respectively. Then 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 12


⇒ vy = xvx 
difference wrt time

ay = vx2 + xax
Option (A)  
If ax = 1 and particle is at origin  
(x = 0, y = 0) 
ay = vx2
ay = 12 = 1  
At origin, at t = 0 sec
speed = 1 given 
(B)  Option
ay = vx2 + xax
given in option B, ax = 0
⇒ ay = vx2  
If ax = 0, vx = constant = 1, (all the time)
⇒ ay = I2 = 1 (all the time)
(C) at t = 0, x = 0  vy = xvx  
speed = 1  
vy = 0  
vx = 1
(D) ay = vx2 + xax  
vy = xvx
ax = 0 (given in D option)
⇒ ay = vx2
If ax = 0 ⇒ Vx = constant initially (vx = 1)
⇒ ay = 12 = 1  
at t = 1 sec  
vy = 0 + ay × t = 1 × 1 = 1


(θ → angle with x axis)  

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 13

A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R - a). For a≪R the magnitude of the work done in the process is given by (4πp0R2)X, where X is a constant and γ = Cp/Cv = 41/30. The value of X is________.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 13

= W = (ΔP)avg x 4πR2a

{for small change (ΔP)avg <P> arithmetic mean} 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 14

In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance R3 has temperature coefficient 0.0004°C−1. If the temperature of R3 is increased by 100°C, the voltage developed between S and T will be   _____ volt.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 14

R'3 = 300 (1 + αΔT)
= 312 Ω
Now


VS – VT = 312I1 – 500I2
= 41.94 – 41.67
= 0.27 V

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 15

Two capacitors with capacitance values C1 = 2000 ± 10 pF and C2 = 3000 ± 15 pF are connected in series. The voltage applied across this combination is V = 5.00 ± 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is _______. 


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 15


*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 16

A cubical solid aluminium (bulk modulus =  = 70 GPa) block has an edge length of 1 m on the surface of the earth. It is kept on the floor of a 5 km deep ocean. Taking the average density of water and the acceleration due to gravity to be 103 kg m−3 and 10 ms−2, respectively, the change in the edge length of the block in mm is _____.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 16

dV/V = 3da/a 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 17

The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is________ mJ. 


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 17

Mutal inductance is producing flux in same direction as self inductance. 

= 55 mJ

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 18

A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 18





Now from equ. (i) 



ΔT = 8.33

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 19

The 1st , 2nd and the 3rd ionization enthalpies I1, I2 and I3, of four atoms with atomic numbers n, n+1, n+2 and n+3 , where n < 10, are tabulated below. What is the value of n?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 19


By observing the values of I1, I2 & I3 for atomic number (n+2), it is observed that I2 >>I1. This indicates that number of valence shell electrons is 1 and atomic number (n+2) should be an alkali metal.
Also for atomic number (n+3), I3 >> I2.
This indicates that it will be an alkaline earth metal which suggests that atomic number (n+1) should be a noble gas & atomic number (n) should belong to Halogen family. Since n < 10; hence n = 9 (F atom)
Note : n = 1 (H atom) cannot be the answer because it does not have I2 & I3 values.

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 20

Consider the following compounds in the liquid form :
O2, HF, H2O, NH3, H2O2, CCl4, CHCl3, C6H6, C6H5Cl.
When a charged comb is brought near their flowing stream, how many of them show deflection as per the following figure?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 20

Here polar molecules in the liquid form will be attracted/deflected near charged comb.
Polar molecules : HF, H2O, NH3, H2O2, CHCl3, C6H5Cl  (6-polar molecules)
Nonpolar molecules : O2, CCl4, C6H6

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 21

In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed ?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 21

KMnO4 + KI → MnO2 + I2
Eq of KMnO4 = Eq of I2
4 × 3 = n × 2
n = 6

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 22

An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after the addition of 1 mL of 3% H2O2, a blue alcohol layer was obtained. The blue color is due to the formation of a chromium (VI) compound 'X' . What is the number of oxygen atoms bonded to chromium through only single bonds in a molecule of X?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 22


Here the structure of CrO5 is :-

Here, single bonded O-atoms with Cr is = 04

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 23

The structure of a peptide is given below 


If the absolute values of the net charge of the peptide at pH = 2, pH = 6, and pH = 11 are |z1| , |z2| and |z3| , respectively, then what is |z1| + |z2| + |z3| ?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 23

|z1| + |z2| + |z3| = 5

At pH = 2 
 of Tyrosine and Lysine is +ve charged (+1 each) +2 |z1| = 2

At pH = 6
NH2 of Lysine (+1), COOH (–1) of glutamic acid, so because of dipolar ion exist |z2| = 0

At pH = 11
COOH of Glutamic acid (–1)
COOH of Lysine (–1)
OH of phenol ( –1)

|z3| = 3

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 2 - Question 24

An organic compound (C8H10O2) rotates plane-polarized light. It produces pink color with neutral FeCl3 solution. What is the total number of all the possible isomers for this compound?


Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 24

C8H10O2 → Gives FeCl3 test means Phenol derivative

Rotate plane polarized light means optically active

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 25

In an experiment, m grams of a compound X (gas/liquid/solid) taken in a container is loaded in a balance as shown in figure I below. In the presence of a magnetic field, the pan with X is either deflected upwards (figure II), or deflected downwards (figure III), depending on the compound X. Identify the correct statement(s)

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 25

Paramagnetic compound (X) are attracted towards magnetic field and the pan is deflected 
downwards. 
While the Diamagnetic compound (X) are repelled by magnetic field and pan is deflected upward.
(A) X ⇒ H2O → Diamagnetic (correct)
(B) X ⇒ K4[Fe(CN)6](s) → Diamagnetic (correct)
Here Fe2+ + Strong field ligand → 3d6 ⇒ [t2g6, eg0]
(C) X ⇒ O2Paramagnetic (correct)
Here O2(g) is paramagnetic due to two-unpaired electrons present in π* (antibonding orbitals).
(D) X ⇒ C6H6 (ℓ) → Diamagnetic (Incorrect)
It is due to presence of 0 unpaired electrons.

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 26

Which of the following plots is(are) correct for the given reaction? ([P]0 is the initial concentration of P)

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 26



*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 27

Which among the following statements is(are) true for the extraction of aluminium from bauxite?

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 27


(B) Function of Na3AlF6 is to lower the melting point of electrolyte.
(C) During electrolysis of Al2O3, the reactions at anode are :

(D) The steel vessel with a lining of carbon acts as cathode.

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 28

Choose the correct statements among the following.

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 28

(A) SnCl2.2H2O is a reducing agent since Sn2+ tends to convert into Sn4+.

(C) First group cations (Pb2+) form insoluble chloride with HCl that is PbCl2 however it is slightly soluble in water and therefore lead +2 ion is never completely precipitated on adding hydrochloric acid in test sample of Pb2+, rest of the Pb2+ ions are quantitatively precipitated with H2S in acidic medium.
So that we can say that filtrate of first group contain solution of PbCl2 in HCl which contains Pb2+ and Cl
However in the presence of conc. HCl or excess HCl it can produce H2[PbCl4]
So, we can conclude A, B or A,B,C should be answers.

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 29

Consider the following four compounds I, II, III, and IV.

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 29


pKb different between I and II is 0.53 and that of III and IV is 4.6.
So option (B) is incorrect Correct Statement (C), (D)
The most basic compound in the given option is (II) and least basic compound is (III) In 2,4,6-trinitro aniline (III) due to strong –R effect of –NO2 groups, the ℓ.p. of -NH2 is more involved with benzene ring hence it has least basic strength.
Whereas (IV) N,N-Dimethyl 2,4,6-trinitro aniline, due to steric inhibition to resonance (SIR) effect; the lone pair of nitrogen is not in the plane of benzene, hence make it (ℓ.p.) more free to protonate  

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 2 - Question 30

Consider the following transformations of a compound P. 

Choose the correct options. 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 2 - Question 30

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