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JEE Advance 2021 Question Paper with Solutions - 1 - JEE MCQ


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30 Questions MCQ Test - JEE Advance 2021 Question Paper with Solutions - 1

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JEE Advance 2021 Question Paper with Solutions - 1 - Question 1

The smallest division on the main scale of Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this caliper with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is;

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 1

Least count of Vernier calipers = 0.01 cm
Error in scale = 4 LC = 0.04 cm
Reading = 3.1 cm + 1 L.C = 3.1 cm + 0.01 = 3.11 cm
So correct diameter of the sphere = (3.11 + 0.04) cm = 3.15 cm

JEE Advance 2021 Question Paper with Solutions - 1 - Question 2

An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During this cycle, heat is absorbed by the gas in;

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 2


Given P – V diagram
For process (1)
ΔQ= nCPΔT
As P = constant and V increases so T will increase
So ΔQ> 0
For process (2)
ΔQ2 = nCVΔT
V = constant, P↓, So T↓
For process (3), ΔQ3 = nCPΔT < 0
For process (4), ΔQ4 = nCPΔT
As ΔT > 0
So ΔQ4 > 0

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JEE Advance 2021 Question Paper with Solutions - 1 - Question 3

An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens syStatement is;

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 3


So,

So, v = -20 cm
and 
So, v'= -12 cm
So total magnification = 
= 0.8

JEE Advance 2021 Question Paper with Solutions - 1 - Question 4

A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha-decay of Q in the first one hour is

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 4

Total no. of decays in 60 minutes = 1000 - 1000(1/2)3= 875
So, number of α-decay = 875 × 0.6 = 525

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 5

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x metres from the point O. The acceleration due to gravity g = 10 m/s2.

The value of t is ____.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 5

After splitting first part takes 0.5 sec to reach ground.
Initial velocity is same for both mass at the highest point in vertical direction. Displacement and acceleration in vertical direction is also same.
So, second part will also take 0.5 sec to reach the ground.

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 6

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x metres from the point O. The acceleration due to gravity g = 10 m/s2.
Q. The value of x is ____.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 6


Velocity of projectile at the highest point = 5 m/s 
Since there is no external force in horizontal direction, by conservation of momentum,

Distance covered by the second mass before landing =  Range /2 + 10(t)
= 7.5 m

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 7

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μC.

Q. The magnitude of q1 is ____.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 7


Switch connected to position P

Potential drop across  capacitor ΔV = 4/3μC
∴ Charge on Capacitor

∴ q1 = 1.33μC

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 8

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μC.

Q. The magnitude of q2 is ____.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 8

Switch at position Q:


Potential Difference  across capacitor ΔV = 2/3V
∴ Charge of capacitor q2 = CΔV = 1 x 2/3 = 0.67μC

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 9

Two point charges -Q and +Q/√3 are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its centre at (b, 0). All lengths are measured in metres.

Q. The value of R is _______ metres.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 9


Let's take two points (a, 0) and (C, 0) on equipotential circle.
Net potential at (C, 0) = 0

Net Potential at (a, 0) = 0

So, Radius



Radius = √3
= 1.73

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 10

Two point charges -Q and +Q/√3 are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its centre at (b, 0). All lengths are measured in metres.

Q. The value of b is _______ metres.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 10

b = a + radius

Centre = (3, 0)

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 11

A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is. The centre of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 11


For a solid cylinder,

⇒ a = 2F/3m
For hollow cylinder,

⇒ a = F/3m
For solid cylinder
f = F – m x (2F/3m) = (F/3) ≤ μmg
⇒ F ≤ 3μmg
Therefore,
A ≤ (2/3m) x (3μmg)
⇒ amax = 2μg

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 12

A wide slab consisting of two media of refractive indices n1 and n2 is placed in the air as shown in the figure. A ray of light is incident from medium n1 to n2 at an angle, where sin is slightly larger than 1/n1. Take the refractive index of air as 1. Which of the following statement(s) is(are) correct?

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 12

The ray diagram for the given conditions is:



The light finally must reflected back in the medium of refractive index n1 for all values of n2.

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 13

A particle of mass M = 0.2 kg is initially at rest in the xy-plane at a point (x = -l, y = -h), where l = 10 m and h = 1 m . The particle is accelerated at time t = 0 with a constant acceleration a = 10 m/s2 along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by  and , respectively.  and  are unit vectors along the positive x, y and z directions, respectively. If , then which of the following statements is/are correct?

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 13


*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 14

Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 14

For Balmer series:

n = 3,4,5 ---

For Layman Series

n = 2,3,4 ----
1/λmin = R
⇒ 

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 15

15:A long straight wire carries a current, l = 2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3.0 m/s (see the figure).
A resistor R = 1.4 and a capacitor C0 = 5.0 F are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct? [μ0 = 4 × 10–7 SI units. Take ln 2 = 0.7]

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 15

The equivalent circuit of the given arrangement is,


At t = 0, imax = ε/R = 1.68 x 10-6/1.4
= 1.2 x 10-6 A
At t = ∞, qmax = C0ε = 8.4 x 10-12 C

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 16

A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration along a fixed inclined plane with angle θ = 45°. P1 and P2 are pressures at points 1 and 2, respectively located at the base of the tube. Let β = (P1 - P2)/(ρgd), where ρ is density of water, d is the inner diameter of the tube and g is the acceleration due to gravity. Which of the following statements is/are correct?

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 16


When,

When,

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 17

An α -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of a uniform magnetic field which is normal to the velocities of the particles. Within this region, the -particle and the sulfur ion move in circular orbits of radii rα and rs respectively. The ratio rs/rα is ________.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 17

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 18

A thin rod of mass M and length a is free to rotate in a horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a/4 is pivoted on this rod with its centre at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity and the disc rotating about its vertical axis with angular velocity 4Ω. The total angular momentum of the syStatement about the point O is 

Q. The value of n is _________.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 18

Angular momentum of disc about O is:

Angular momentum of rod about O is:

So, n = 49

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 19

A small object is placed at the centre of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t = t1 and 50 K at t = t2. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio (t2/t1) is ____,


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 19


Similarly,



= 9

JEE Advance 2021 Question Paper with Solutions - 1 - Question 20

The major product formed in the following reaction is

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 20


Na in liquid NH3 reduces non-terminal alkyne into trans alkene.

JEE Advance 2021 Question Paper with Solutions - 1 - Question 21

Among the following, the conformation that corresponds to the most stable conformation of meso-butane-2,3-diol is

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 21

In (B), the given configuration represents meso-butane-2,3-diol and due to intramolecular hydrogen bonding, the gauche form is more stable.
(C) and (D) do not represent meso-isomer.

JEE Advance 2021 Question Paper with Solutions - 1 - Question 22

For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one face are shown for clarity), the packing fraction is approximately

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 22

Packing Fraction (f) = 

Now 2r-= a
∴ a / r- = 2
Also,


So, f = 
= 0.634 ≈ 0.63

JEE Advance 2021 Question Paper with Solutions - 1 - Question 23

The calculated spin only magnetic moments of [Cr(NH3)6]3+ and [CuF6]3- in BM respectively are:
(Atomic numbers of Cr and Cu are 24 and 29, respectively)

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 23

[Cr(NH3)6]3+ has d3 configuration.
So, as per CFT,
N = 3 and = 3.87 BM
[CuF6]3- has d8 configuration and is a weak field ligand.
So, N = 2 and = 2.84 BM

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 24

For the following reaction scheme, percentage yields are given along the arrow:

x g and y g are masses of R and U, respectively.
(Use molar mass (in g mol-1) of H, C and O as 1, 12 and 16, respectively)
Q. The value of x is ___.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 24




Molar mass of P = 40

So, number of moles of R = 0.01 mole
Molar mass of (R) = 162
So, x = 0.01 × 162 = 1.62 g

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 25

For the following reaction scheme, percentage yields are given along the arrow:

x g and y g are masses of R and U, respectively.
(Use molar mass (in g mol-1) of H, C and O as 1, 12 and 16, respectively)
Q. The value of y is ___.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 25

Molar mass of 'U' = 122 g or 100 g

Mass of 'U' = 0.1/2 x 0.8 x 0.8 x 122 = 3.90 gm

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 26

For the reaction: X(s)  Y(s) + Z(g), the plot of In  versus  is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and po = 1 bar.

(Given: , where the equilibrium constant, K =  and the gas constant, R = 8.314 J K-1 mol-1)
The value of standard enthalpy,  (in kJ mol-1), for the given reaction is ______.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 26

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 27

or the reaction: X(s)  Y(s) + Z(g), the plot of In  versus  is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and po = 1 bar.

(Given: , where the equilibrium constant, K =  and the gas constant, R = 8.314 J K-1 mol-1)
The value of  (in  J K-1 mol-1) for the given reaction at 1000 K is _______.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 27

Put the value of Ink &  ΔHo


= 17 x 8.314
= 141.338J / mol
= 141.34 J / mol

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 28

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x°C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y × 10-2 °C.
(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), Kb = 0.5 K kg mol-1; Boiling point of pure water is 100°C.)
Q. The value of x is ___.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 28

For solution A (AgNO3), i = 2
ΔTb = i Kb × m = 2 × 0.5 × 0.1 = 0.1
The B.P. of solution A is 100 + ΔTb
= 100.10°C

*Answer can only contain numeric values
JEE Advance 2021 Question Paper with Solutions - 1 - Question 29

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x°C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y × 10-2 °C.
(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), Kb = 0.5 K kg mol-1; Boiling point of pure water is 100°C.)
Q. The value of |y| is ___.


Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 29


Total moles = 0.05 x 3 + 0.05 x 3 = 0.3 moles
molality = 0.3 / 2 = 0.15
ΔTb = 0.15  x 0.5 = 0.075

B.P. of solution B = 100.75
Difference in B.P. of solution A & Solution B = 100.1  - 100.075
= 0.025
= 2.5 x 10-2
= Y = 2.5

*Multiple options can be correct
JEE Advance 2021 Question Paper with Solutions - 1 - Question 30

Given:

The compound(s) which on reaction with HNO3 will give the product having degree of rotation [α]= -52.7° is/are

Detailed Solution for JEE Advance 2021 Question Paper with Solutions - 1 - Question 30


We have to get the product (x) of [α]= -52.70, i.e. the enantiomer of the above product, which is only possible from (C) and (D).


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