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JEE Advanced Mock Test - 2 (Paper II) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 2 (Paper II)

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*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper II) - Question 1

If in a hydrogen atom, radius of nth Bohr orbit is rn, frequency of revolution of electron in nth orbit is fn, and area enclosed by the nth orbit is An , then which of the following graphs is/are correct?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 1
(i) Since in hydrogen atom rn ∝ n2, therefore graph between rn and n will be a parabola through origin and having increasing slope.

Since, rn ∝ n2, therefore rn/r1 = n2Hence, log(rn/r1) = 2 log n

(ii) It means, graph between log(rn/r1) and log n will be a straight line passing through and having positive slope (tanθ = 2)tanθ = 2.

(iii) If radius of an orbit is equal to r, then area enclosed by it will be equal to A = πr2.

Since rn ∝ n2, therefore An ∝ n4

It means, graph between log(An/A1) and log n will be a straight line passing through origin and having positive slope (tanθ = 4).

(iv) If frequency of revolution of electron is f, then its angular velocity will be equal to ω = 2π. Hence, its angular momentum will be equal to Iω = mr2. But according to Bohr's theory, it is equal to nh/2π, therefore,

It means, graph between log (fn/f1) and log n will be a straight line passing through origin and having negative slope, tan θ = −3. Hence, it will be as shown in diagram.

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper II) - Question 2

Which of the following is/are conservative force(s) ?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 2

Clearly for forces (1) and (2) the integration do not require any information of the path taken.

Taking : x2 + y2 = t

2x dx + 2y dy = dt

Which is solvable.

Hence, (1), (2) and (3) are conservative force.

But (4) requires some more information on path. Hence non-conservative.

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper II) - Question 3

When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then,

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 3
Work done = ΔKinetic energy

As the net work done is positive, the kinetic energy will increase. Due to air resistance torque, the angular momentum will decrease.

T2 ∝ r3, T will decrease.

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 4

Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.

The spread of frequency as observed by the passengers in train B is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 4

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 5

A resistance of 2Ω is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 5
Given, R = 2 Ω

From the first condition,

Now, from the second condition,

Put the value of x from equation (i) in (ii).

⇒ (100 - l)(80 - l) = l(l + 20)

⇒ 8000 - 100l - 80l + l2 = l2 + 20l

⇒ 8000 = 20l + 180l

Put this value in equation (ii).

⇒ x = 3 Ω

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 6

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s-2).

The magnitude of the normal reaction that acts on the block at the point Q is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 6

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 7

Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.

Value of E ( in eV) is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 7

WA = ionization energy of electron ion 2nd orbit of hydrogen atom = 3.4 eV

WB = ionization energy of electron in 2nd orbit of He+ ion

= 13.6 eV.

Now, given that

KA = 2KB

or (E − WA) = 2(E − WB)

∴ E = 2WB − WA = 23.8 eV

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 8

When you throw a ball in air with some velocity at some angle with horizontal, vertical component of velocity at highest point is zero and horizontal component of velocity remains unchanged.

Velocity of a projectile at height 15 m from ground is Here, is in horizontal direction and vertically upwards. Then

Angle of projectile with ground is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 8
ux = vx = 20 m/s [constant]

uy = 20 m/s

θ = 45o

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 9

A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 9
When 'S' is open

After the capacitors are fully charged, current flows through resistors only.

Let the charge on each capacitor be q μC.

So, we have

Total charge at junction X = 0 μC

When 'S' is closed:

Potential drop across the 3 W resistor = 1 × 3 = 3 V

Charge on the 3 μC capacitor is q1 = 3 × 3 μC = 9 μC

Potential drop across the 6 W resistor = 6 × 1 V = 6 V

Charge on the 6 μC capacitor is q2 = 6 × 6 μC = 36 μC

Total charge at X = (36 - 9) μC = 27 μC

Total charge flowing from Y to X = 27 μC

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 10

1 kg of 1H2 is used in fusion reactor to produce electric power.
The nuclear reaction is 1H2 + 1H2 → 2He4
It is given that mass of 1H2 is 2.0141 u and mass of 2He4 is 4.0026 u
The energy liberated is  P × 1014 J
To convert this nuclear energy into electrical energy. The water at 300 K is heated to convert into steam at temperature 800 K. The left over steam is at 400 K. The energy supplied to electric grid is  Q × 1014 J.
Specific heat capacity of water = 4.2 × 103 J/kg-k,
Specific heat capacity of steam = 2 × 103 J/kg-k,
Latent heat of vaporization of water = 2.25 × 106 J/kg
1amu = 931MeV/c2  

Find the value of P. 


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 10

Energy released in one nuclear reaction = (2×2.0141- 4.0026) × 931 MeV = 23.83 MeV
Total energy produced by 1 kg of 
Now, 5.74 × 1014 = m × 4.2 × 103 (100 - 27) + m × 2.25 × 106 + m × 2 × 103 (800 - 373)
⇒ m = 1.68 × 108 kg
∴ supplied energy = 1.68 × 108 × 2 × 103 × (800 - 400) = 1.34 × 1014 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 11

1 kg of 1H2 is used in fusion reactor to produce electric power.
The nuclear reaction is 1H2 + 1H2 → 2He4
It is given that mass of 1H2 is 2.0141 u and mass of 2He4 is 4.0026 u
The energy liberated is  P × 1014 J
To convert this nuclear energy into electrical energy. The water at 300 K is heated to convert into steam at temperature 800 K. The left over steam is at 400 K. The energy supplied to electric grid is  Q × 1014 J.
Specific heat capacity of water = 4.2 × 103 J/kg-k,
Specific heat capacity of steam = 2 × 103 J/kg-k,
Latent heat of vaporization of water = 2.25 × 106 J/kg
1amu = 931MeV/c2  

Find the value of Q. 


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 11

Energy released in one nuclear reaction = (2×2.0141- 4.0026) × 931 MeV = 23.83 MeV
Total energy produced by 1 kg of 
Now, 5.74 × 1014 = m × 4.2 × 103 (100 - 27) + m × 2.25 × 106 + m × 2 × 103 (800 - 373)
⇒ m = 1.68 × 108 kg
∴ supplied energy = 1.68 × 108 × 2 × 103 × (800 - 400) = 1.34 × 1014 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 12

A solid cylinder and hemisphere are joined as shown in the figure. Combined mass of the cylinder and the hemisphere is 1/3kg. The cylinder is painted shiny white and the hemisphere is painted dark black. The initial temperature of the body is Ti = 300 K and it is released form rest in gravity free space where the temperature of surrounding is 0K. The body attains terminal velocity v0. Its specific heat capacity (S) is given in the shown graph.

Find the total energy (in KJ) radiated by the body.


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 12

Q = mò sdT = 70 m KJ

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 13

A solid cylinder and hemisphere are joined as shown in the figure. Combined mass of the cylinder and the hemisphere is 1/3kg. The cylinder is painted shiny white and the hemisphere is painted dark black. The initial temperature of the body is Ti = 300 K and it is released form rest in gravity free space where the temperature of surrounding is 0K. The body attains terminal velocity v0. Its specific heat capacity (S) is given in the shown graph.

Find the terminal velocity v0 (in mm/s) of the body.


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 13


Since energy of radiation is uniformly spreaded in 2p solid angle. 

JEE Advanced Mock Test - 2 (Paper II) - Question 14

A.DC circuit consisting of two cells of emf V and 2V having no internal resistance are connected with two capacitors of capacity C and 2C and four resistors R, R, 2R and 2R as shown in figure. The ammeter and voltmeter used in the circuit are ideal.

The reading of the ammeter as soon as the switch is closed is

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 14
Capacitors will offer no resistance and hence the circuit becomes

Clearly reading of ammeter will be zero.

JEE Advanced Mock Test - 2 (Paper II) - Question 15

An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.

The molar specific heat of the gas in this process is given by C whose value is

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 15
It is given that there is decrease of the internal energy while the gas expands absorbing some heat, which numerically equal to the decrease in internal energy.

Hence, dQ = −dU

Thus, dQ = dW + dU = −dU

⇒ dW = −2dU (remembering that both dW and dU are negative)

⇒ PdV = 2CdT [dU = −CdT] ...... (1)

where C is the molar specific heat of the gas

Since the gas is ideal, PV = RT ....... (2)

From (2) P = RT/v

since the gas is diatomic.

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper II) - Question 16

Which of the following statements are correct?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 16
In an isocyanide, an electrophile attacks first followed by a nucleophile at the same carbon atom bearing negative charge. on partial hydrolysis in acidic medium, will give N-methylmethanamide while on complete hydrolysis in acidic medium it gives CH3NH2 and HCOOH.

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 17

If total entropy change during steps-1,3 and 5 be ’x’ J/K, then the value of  is .....


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 17

ΔS1 + ΔS2 + ΔS3 + ΔS4 + ΔS5 + ΔS6 = 0 (because the process is cyclic).
Also, ΔS2 = ΔS4 = ΔS6 = 0 (rev ersible adiabatic process).
So, ΔS1 + ΔS3 + ΔS5 = 0 = x

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 18

If during step-5, the system exchanges heat from a reserv oir at temperature ‘T5 K’, then value of T1/T2 is:


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 18

qtotal + wtotal = ΔU = 0 (cyclic process)
qtotal + (-700) = 0 
∴ qtotal = 700 J
Now, Q1 + Q3 + Q5 + Q2 + Q4 + Q6 = 700
500 + 800 + Q5 + 0 + 0 + 0 = 700
Q5 = -600 J
Also, ΔS1 + ΔS3 + ΔS5 + ΔS2 + ΔS4 + ΔS6 = 0 (cyclic process)

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 19

The ratio of the molar mass of ‘X’ to the total number of ozonides that ‘X’ can form, is…….


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 19

On analyzing the abov e paragraph, one can conclude that:

Also, ‘X’ can form a total of 10 ozonides including cross-ozonides.

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 20

In the compounds, X, Y, Z, A, B and C.
If the sum of the number of pi-bonds in all of the above product = m and if degree of unsaturation of (C) be n, then find the value of 2m/n.


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 20

m = 21; n = 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 21

If a·b·c = 0, then angle between the lines is q, then 1/2 tan2θ is


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 21

∠θ = π/3

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 22

Total number of ordered triplets (a, b, c) are


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 22

∠θ = π/3

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 23

The number of points P on ellipse is three, when the area of the , then the 
value of  is equal to 


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 23


Length of AB 
Let h → distance of point P from AB

Maximum area of DPAB = Area of ΔP2AB
Three if area of 

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper II) - Question 24

The number of point P on ellipse is one, when the area of  then the v alue of  is equal to


Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 24


Length of AB 
Let h → distance of point P from AB

Maximum area of DPAB = Area of ΔP2AB

JEE Advanced Mock Test - 2 (Paper II) - Question 25

A circular disc is allowed to roll on the surface of a rough horizontal fixed cylinder of radius R = /20 m. The centres of the cylinder and the disc are connected by a massless rgid rod. The ends of the massless rigid rod are hinged at the centre of the disc and centre of the cylinder so that the disc can roll over the outer surface of the cylinder without slipping at any point during motion. The disc is released from rest from the top of the cylinder as shown in the figure. Radius of the disc is r = 3/20 m.

Hinged points are v ery smooth so that no loss of energy takes place. Consider the situation when the rod becomes horizontal the magnitude of acceleration of point A (which is on the disc and is in contact with the cylinder) is aA and speed of point B is vB. Then choose the correct option. (take g = 10 m/s2)

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 25


JEE Advanced Mock Test - 2 (Paper II) - Question 26

An inclined plane ABCD is shown in the figure. It is inclined at an angle of 45° from horizontal. A block is placed on the inclined plane. The plane is accelerating along x-axis with an acceleration, a = 5 m/s2. The minimum value of coefficient of friction between the block and the plane required, so that the block does not slip on the plane, is (g = 10 m/s2)

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 26

JEE Advanced Mock Test - 2 (Paper II) - Question 27

Which of the following is an example of a redox reaction?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 27

JEE Advanced Mock Test - 2 (Paper II) - Question 28

A disaccharide does NOT give a positive test for Tollen’s reagent. Upon acidic hydrolysis, it gives an equimolar mixture of two different monosaccharides, both of which can be oxidized by bromine water. The correct structure of disaccharide is:

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 28

JEE Advanced Mock Test - 2 (Paper II) - Question 29

Let F : R → R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = -4 and F'(x) < 0 for all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R.
if  then which of the following expressions is correct?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 29


⇒ 40 = 27 F'(3) - F'(1) + 36
f'(x) = F(x) + x F'(x)
f'(3) = F(3) + 3 F'(3)
f'(1) = F(1) + F'(1)
9 f'(3) - f'(1) + 32 = 0

JEE Advanced Mock Test - 2 (Paper II) - Question 30

Consider the following function:
, where k is a positive integer.
Which of the following statements are true?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper II) - Question 30

Consider the expression:



Hence, this is the required solution.

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