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JEE Advance 2020 with Solution: Paper - 1 - JEE MCQ


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30 Questions MCQ Test - JEE Advance 2020 with Solution: Paper - 1

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JEE Advance 2020 with Solution: Paper - 1 - Question 1

A football of radius R is kept on a hole of radius r (r < R) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle θ from the horizontal as shown in the figure below. The maximum value of θ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) -

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



For θmax, the football is about to roll, then N2 = 0 and all the forces (Mg and N1) must pass through contact point
∴ cos(90° – θmax) = 

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

A light disc made of aluminium (a nonmagnetic material) is kept horizontally and is free to rotate about its axis as shown in the figure. A strong magnet is held vertically at a point above the disc away from its axis. On revolving the magnet about the axis of the disc, the disc will (figure is schematic and not drawn to scale)- 

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

When the magnet is moved, it creates a state where the plate moves through the magnetic flux, due to which an electromotive force is generated in the plate and eddy currents are induced. These currents are such that it opposes the relative motion ⇒ disc will rotate in the direction of rotation of magnet.
Note : This apparatus is called Arago's disk.

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JEE Advance 2020 with Solution: Paper - 1 - Question 3

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to  scale)-

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 3

For no slipping at the ground,
Vcentre = ωR (R is radius of roller) 
∴ Velocity of scale = (Vcenter + ωr) [r is radius of axle] 
Given, Vcenter · t = 50 cm 
∴ Distance moved by scale = (Vcenter + ωr)t = 
Therefore relative displacement (with respect to centre of roller) is (75 – 50) cm = 25 cm.

JEE Advance 2020 with Solution: Paper - 1 - Question 4

A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 4

Torque experienced by circular loop = 
where  is magnetic moment 
 is magnetic field 
∴ τ = iπR2N B0 [at the instant shown θ = π/2] 

JEE Advance 2020 with Solution: Paper - 1 - Question 5

A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below. Correct shape of the emergent wavefront will be (figures are schematic and not drawn to scale)- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 5


JEE Advance 2020 with Solution: Paper - 1 - Question 6

An open-ended U-tube of uniform cross-sectional area contains water (density 103kg m−3). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density 800 kg m−3 is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio (h1/h2) of the heights of the liquid in the two arms is- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 6

h1 + h2 = 0.29 × 2 + 0.1
h1 + h2 = 0.68 ……(1)
⇒ P0 + ρkg(0.1) + ρwg(h1 - 0.1) [ρk = density of kerosene & ρw = density of water] -ρwgh2 = P0
⇒ ρkg(0.1) + ρwgh1 - ρwg x (0.1) = ρwgh2
⇒ 800 × 10 × 0.1 + 1000 × 10 × h1
– 1000 × 10 × 0.1 = 1000 × 10 × h2
⇒ 10000 (h1 – h2) = 200
⇒ h1 – h2 = 0.02 ……(2)

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 7

A particle of mass m moves in circular orbits with potential energy V(r)=Fr, where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E, respectively, then for the nth orbit (here h is the Planck’s constant)-

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 7

U = Fr 
[Using U = Potential energy and v = velocity, to avoid confusion between their symbols] 
⇒ Force = -dU/dr = -F
⇒ Magnitude of force = Constant = F
⇒ F = mv2/R ...(1)
⇒ mvR = nh/2π ...(2)


(B) is correct 
⇒ E = (1/2)mv2 + U
= (1/2)mv2 + FR

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 8

The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then (Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 8

A = 64 mm2,  T = 2500 K (A = surface area of filament, T = temperature of filament, d is distance of bulb from observer, Re = radius of pupil of eye) 
Point source  d = 100 m 
Re = 3mm 
(A) P = σAeT4
= 5.67 × 10–8 × 64 × 10–6 × 1 × (2500)4 (e = 1 black body) 
= 141.75 w 
Option (A) is wrong 
(B) Power reaching to the eye 

(C)  λm T = b  
λm × 2500 = 2.9 × 10-3
⇒ λm = 1.16 × 10-6
= 1160 nm 
Option (C) is correct 
(D) Power received by one eye of observer = 
 Number of photons entering into eye per second 

Option (D) is correct

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 9

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follows: [position] = [Xα]; [speed] = [Xβ]; [acceleration] =[Xp]; [linear momentum] = [Xq]; [force] = [Xr]. Then - 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 9

Given L = xα ……(1)
LT–1 = xβ ……(2)
LT–2 = xp ……(3)  
MLT–1 = xq ……(4)  
MLT–2 = xr ……(5)

From (3) 

From (4)  
M = xq–β
From (5)  ⇒ xq = xr  xα-β
⇒ α + r – q = β ……(6)
Replacing value 'α' in equation (6) from (A) 
2β – p + r – q = β
⇒ p + q – r = β (B) 
Replacing value of 'β' in equation (6) from (A) 
2α + 2r – 2q = α + p 
α = p + 2q – 2r

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 10

A uniform electric field, is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of . This particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure. Take q/m = 1010 Ckg-1. Then- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 10




2θ = 60°, 120   ⇒ θ = 30°, 60° 
Time of flight 
Time of flight 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 11

Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ. Its inner radius is R1 and outer radius is R2. If a voltage V0 is applied between its two ends, a current I flows in it. In addition, it is observed that a transverse voltage ΔV develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale)- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 11


All the elements are in parallel

Resistance 
 (A)
 will be inward direction in order to provide centripetal acceleration. Therefore electric field will be radially outward 
Vouter < Vinner  (C) 

 (I = neAVd ⇒ Vd ∝ i)

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 12

As shown schematically in the figure, two vessels contain water solutions (at temperature T) of potassium permanganate (KMnO4) of different concentrations n1 and n2 (n1 > n2) molecules per unit volume with Δn = (n1 - n2) << n1. When they are connected by a tube of small length l and cross-sectional area S, KMnO4 starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed v of the molecules is limited by the viscous force -βv on each molecule, where β is a constant. Neglecting all terms of the order (Δn)2, which of the following is/are correct? (kB is the Boltzmann constant)- 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 12

n1 >> (n1 - n2) = Δn
 
F = (n1 – n2)kBTS = ΔnkBTS (A) 

Force balance ⇒ Pressure × Area = Total number of molecules × βv 

Total number of molecules/sec = 

As Δn will decrease with time therefore rate of molecules getting transfer decreases with time.

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 13

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm  and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance ��R from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of ��R (in cm) is ______.-


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 13


Initially 

Suppose  xL = distance of left finger from centre when right finger starts moving 


Now  xR = distance when right finger stops and left finger starts moving 


*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 14

When water is filled carefully in a glass, one can fill it to a height h above the rim of the glass due to the surface tension of water. To calculate h just before water starts flowing, model the shape of the water above the rim as a disc of thickness h having semicircular edges, as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension, the water surface breaks near the rim and water starts flowing from there. If the density of water, its surface tension and the acceleration due to gravity are 103kg m−3, 0.07 Nm−1 and 10 ms−2, respectively, the value of h (in mm) is _________.-


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 14


r = h/2
Pressure at the bottom of disc = pressure due to surface tension 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 15

One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0). A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole pointing towards the charge q is fixed at the origin, the spring gets stretched to a length l and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by Δl << l from its equilibrium position and released, it is found to oscillate at frequency  The value of δ is ______.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 15

Δl → x 
At l : Fe = FSP


Fnet = Fsp – Fe = k(l + x) 


keq = 4k 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 16

Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It expands isothermally to volume 4V1. After this, the gas expands adiabatically and its volume becomes 32V1. The work done by the gas during isothermal and adiabatic expansion processes are Wiso and Wadia, respectively. If the ratio  ln 2, then �� is ________.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 16



Wadi

Wiso 

f = 16/9 = 1.7778 ≈ 1.78

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 17

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 ms−1 in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320 ms−1, the smallest value of the percentage change required in the length of the pipe is ____________.


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 17

 ...(1)
(l1 ⇒ initial length of pipe) 
 {VT Speed of tuning fork, l2 → new length of pipe} ...(2)


Therefore smallest value of percentage change required in the length of pipe is 0.625 

*Answer can only contain numeric values
JEE Advance 2020 with Solution: Paper - 1 - Question 18

A circular disc of radius R carries surface charge density  where σ0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is φ0. Electric flux through another spherical surface of radius R/4 and concentric with the disc is φ. Then the ratio  is ......


Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 18



∴ 

= 6.40 

JEE Advance 2020 with Solution: Paper - 1 - Question 19

If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the average and the roots mean square speeds, respectively, is 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 19

Graph represents symmetrical distribution of speed and hence, the most probable and the average speed should be same. But the root mean square speed must be greater than the average speed.

JEE Advance 2020 with Solution: Paper - 1 - Question 20

Which of the following liberates O2 upon hydrolysis?

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 20

(A) Pb3O4 is insoluble in water or do not react with water.
(B) 2KO2 + 2H2O → 2KOH + H2O2 + O2(g) ­↑ 
(C) Na2O2 + 2H2O → 2NaOH + H2O2
(D) Li2O2 + 2H2O → 2LiOH + H2O2

JEE Advance 2020 with Solution: Paper - 1 - Question 21

A colorless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous solution of NaCl, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue P and a solution Q. The residue P was soluble in aq. NH3 and also in excess sodium thiosulfate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively, are

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 21



JEE Advance 2020 with Solution: Paper - 1 - Question 22

Newman projections P, Q, R and S are shown below :

Which one of the following options represents identical molecules?

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 22

 2, 3, 3-trimethyl pentan-2-ol
 3-ethyl-2-methyl pentan-2-ol
 3-ethyl-2-methyl pentan-2-ol
 3-ethyl-2-methyl pentan-3-ol
Q and R is same.

JEE Advance 2020 with Solution: Paper - 1 - Question 23

Which one of the following structures has the IUPAC name
3-ethynyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid?

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 23


3-ethynyl-2-hydroxy-4-methyl-hex-3-en-5-ynoic acid.

JEE Advance 2020 with Solution: Paper - 1 - Question 24

The Fischer projection of D-erythrose is shown below.

D-Erythrose
D-Erythrose and its isomers are listed as P, Q, R, and S in Column-I. Choose the correct relationship of P, Q, R, and S with D-erythrose from Column II. 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 24


D-Erythrose

P-2, Q-1, R-1, S-3

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 25

In thermodynamics the P-V work done is given by 

For a system undergoing a particular process, the work done is ,

This equation is applicable to a

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 25

For 1 mole Vander Waal's gas

If Pext = P, means process is reversible. For Vanderwaal gas, expression is correct for all reversible 
process.

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 26

With respect to the compounds I-V, choose the correct statement(s).

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 26



(C) –NO2 is –I group (electron withdrawing group). It increases acid strength.
(D) Acid strength order : IV > V > I > II > III

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 27

In the reaction scheme shown below Q, R and S are the major products.

The correct structure of 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 27


*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 28

Choose the correct statement(s) among the following :

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 28

(A) [FeCl4]

[FeCl4] is sp3 hybridised and has tetrahedral geometry with 5 unpaired electrons.
[Co(en)(NH3)2Cl2]+ has three geometrical isomers.

(C) [FeCl4]

Number of unpaired electrons (n) = 5
Spin only magnetic moment = 
= 5.92 B.M.
[Co(en)(NH3)2Cl2]+

Number of unpaired electrons (n) = 0
Spin only magnetic moment = 
= 0
(D) [Co(en)(NH3)2Cl2]+

[Co(en)(NH3)2Cl2]+ is d2sp3 hybridised and has octahedral geometry with 0 unpaired electron.

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 29

With respect to hypochlorite, chlorate and perchlorate ions, choose the correct statement(s).

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 29

Hypochlorite ion : ClOΘ
Chlorate ion : ClOΘ3
Per chlorate ion : ClOΘ4
(A) Acidic order : 
Conjugate base order : ClO > ClO3 > ClO4
(B) Hypochlorite ion (ClOΘ):  Linear shape
Chlorate ion (ClO3Θ) :  Trigonal pyramidal shape
Perchlorate ion (ClOΘ4) Perfect tetrahedral shape due to resonance
In chlorate ion bond angle changes due to presence of lone pair on chlorine atom. While hypochlorite ion is linear and perchlorate ion is tetrahedral and there is no effect of lone pair on hypochlorite ion.
(C) Disproportionation reaction of 
(i) hypochlorite ion  : 3ClOΘ → 2Cl- + ClO3Θ
(ii) Chlorate ion : 
(D) 

*Multiple options can be correct
JEE Advance 2020 with Solution: Paper - 1 - Question 30

The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s). 

Detailed Solution for JEE Advance 2020 with Solution: Paper - 1 - Question 30

(A) ZM = 2 x (1/2) = 1
ZX = 4 x (1/4) = 1
∴ Empirical formula is MX
(B) Coordinate numbers of both M and X is 8.
(C) Bond length of M – X bond      


(D) 

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