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JEE Advanced Mock Test - 4 (Paper I) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 4 (Paper I)

JEE Advanced Mock Test - 4 (Paper I) for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Advanced Mock Test - 4 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 4 (Paper I) MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 4 (Paper I) below.
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JEE Advanced Mock Test - 4 (Paper I) - Question 1
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When liquid medicine of density p is to be put in the eye it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r << R)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 1
The vertical force due to the surface tension on the drop is,

JEE Advanced Mock Test - 4 (Paper I) - Question 2
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The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

For a = 0, the value of d (maximum value of p as shown in the figure) is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 2
As the charge density is

The charge inside the nucleus is

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JEE Advanced Mock Test - 4 (Paper I) - Question 3
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Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x << l. Find the rate dq/dt with which the charge leaks off each sphere if their approach velocity varies as v = a/√x , where a is a constant.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 3

JEE Advanced Mock Test - 4 (Paper I) - Question 4
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A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m which is stretched under a load by 0.7 mm. Calculate the load, given that the Young's modulus of elasticity for copper is 1.1 × 1011 N m–2 and that for steel is 2 × 1011 N m–2.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 4
Here, r = 3.0/2 = 1.5mm = 1.5 × 10−3m

LC = 2.2 m, Ls = 1.6 m

ΔLc + ΔLs = 0.7 × 10–3m ...........(1)

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 5
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Consider the situation shown in figure and select the correct statement from the following.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 5

Since, all pulleys and threads are light and pulleys are frictionless, the tensions in threads passing over B and C are equal i.e., TB = TC. Suppose pulley C moves down with acceleration a, then acceleration of B is a upward.

Writing newton's 2nd law equation for various blocks.

It means C moves up and B moves down with acceleration a = g/11

T = 2TB = 2TC

Solving the various equation, we get T = 12.46 g

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 6
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It is observed that only 0.39% of the original radioactive sample remains un decayed after eight hours. Hence
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 6
From the law of radioactivity,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 7
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In the figure shown AB is rod of length 30 cm and area of cross-section 1 cm2 and thermal conductivity 336 S.I. units. The ends A and B are maintained at temperature 20°C and 40°C respectively. A point C of this rod is connected to a box D, containing ice at 0°C, through a highly conducting wire of negligible heat capacity. Find the rate at which ice melts in the box. [Assume latent heat of fusion for ice Lice = 80 cal g−1]
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 7
Thermal resistance of

Temperature of C

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 8
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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 = 1,120 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 (in Hz) as observed by the passengers in train B is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 9
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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 speed (in ms-1) of the block when it reaches the point Q is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 9

Using work energy theorem,

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 10
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A ray of light travelling in the direction  is incident on a plane mirror. After reflection, it travels along the direction  What is the angle of incidence (in degrees)?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 10

The angle of incidence is given by:

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 11
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A binary star consists of two stars A (mass 2.2Ms) and B (mass 11MS), where Ms is the mass of the Sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 11
Let stars A and B be rotating about their centre of mass with angular velocity ω.

Let the distance of stars A and B from the centre of mass be rA and r,, respectively, as shown in the figure.

Total angular momentum of the binary stars about the centre of mass is

Angular momentum of star B about the centre of mass is

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 12
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When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 12
In series,

Rate of heat produced in R is

In parallel,

Rate of heat produced in R is

According to the given problem, J1 = 2.25J2

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 13
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A large glass slab (μ = 5/4) of thickness 6 cm is placed over a point source of light on a plane surface. There is a bright circular patch of light on the top surface of the slab with radius R cm. What is the value of R?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 13
In the figure, C represents the critical angle.

JEE Advanced Mock Test - 4 (Paper I) - Question 14
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Answer the following by appropriately matching the lists based on the information given:

List I includes physical quantities and List II includes units for the physical quantities given in List I.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 14


Therefore, the SI unit of G is Nm2 kg-2.
∴ SI unit of GMeMs = (Nm2kg-2) × kg2 = Nm2 = kg ms-2 × m2 = k gm3s-2

= Nm kg-1 = kg ms-2 × m × kg-1 = m2 s-2

JEE Advanced Mock Test - 4 (Paper I) - Question 15
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If the given rolling body is a solid sphere and force F is acting at a distance of R/2 above the centre, calculate acm, αcm and friction.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 15

Let f be the value of the force of friction acting at the point of contact and it is directed backward as shown:

For translational motion, we have
F - f = Macm --- 1
For rotational motion, torque about the centre of mass
Fx + fR = I αcm --- 2
For the perfect rolling,
acm = αcm R --- 3
From 1, 2 and 3,
Angular acceleration of the solid sphere,

Given x = R/2
Acceleration of the centre of mass,

The negative sign indicates that the force of friction is acting in the forward direction.

JEE Advanced Mock Test - 4 (Paper I) - Question 16
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If the given rolling body is a ring and force F is passing through the centre of mass, calculate acm, αcm and friction.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 16

Let f be the value of the force of friction acting at the point of contact and it is directed backward as shown:

For translational motion, we have
F - f = Macm --- 1
For rotational motion, torque about the centre of mass
Fx + fR = I αcm --- 2
For the perfect rolling,
acm = αcm R --- 3
From 1, 2 and 3,
Angular acceleration of the solid sphere,

Given x = 0
Acceleration of the centre of mass,

The positive sign indicates that the force of the friction is acting in the backward direction.

JEE Advanced Mock Test - 4 (Paper I) - Question 17
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A solid sphere having mass 'M' and radius 'R' is placed on a rough inclined plane having inclination 30°. It moves down rolling without slipping. What will be the values of acm, αcm and friction?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 17


For translational motion, we have

Mgsin 30° - f = M acm
Mg/2 - f = Macm --- 1
For rotational motion, torque about the centre of mass,

JEE Advanced Mock Test - 4 (Paper I) - Question 18
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Directions: The following question is based on the paragraph given below.

The noble gases have closed-shell electronic configuration and are monatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to give XeO3. The compound can also be prepared using XeF6 as the starting compound. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

The chemical nature of the compounds XeF4 and XeF6 is expected to be
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 18
The oxidation state of Xe in XeF4 is +4 and in XeF6 is +6. These oxidation states of Xe are displayed only with elements with high electronegativity, like oxygen and fluorine. Since, Xe will have a tendency to get reduced, the nature of fluorides of xenon is expected to be oxidising.
JEE Advanced Mock Test - 4 (Paper I) - Question 19
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(1R, 3S)-Cis-1-Bromo-3-methyl cyclohexane. The product formed in the reaction is
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 19
In the presence of a polar aprotic solvent, like acetone, the mechanism followed will be SN2.

Walden inversion takes place at C1, where -Br is substituted by -OH.

Hence, the product formed will be (1S, 3S)-Trans-3-methyl cyclohexanol

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 20
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The total number of cyclic structural as well as stereoisomer possible for a compound with the molecular formula C5H10 is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 20

For a compound with molecular formula C5H10, the isomers are as follows:

Structures (vi) and (vii) are trans-isomers and are same. It can exist in d, I form.
So, the total number of cyclic as well as stereoisomer possible is 7.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 21
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Consider a reaction: aG + bH → products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 21

Since the rate doubles when [G] is doubled, the order of the reaction with respect to the reactant G is one.
Since the rate becomes 8 times when both [G] and [H] are doubled, the order of the reaction with respect to the reactant H is two.
Rate of reaction = k[G][H]2
Overall order of the reaction = 1 + 2 = 3

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 22
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The sum of the number of lone pairs of electrons on each central atom in the following species is
[TeBr6]2–, [BrF2]+, SNF3, and [XeF3]–
(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 22

Te in [TeBr6]2– is sp3d3 hybridised with 1 lone pair of electrons.
Br in [BrF2]+ is sp3 hybridised with 2 lone pairs of electrons.
S in SNF3 is sp3 hybridised with 0 lone pair of electrons.
Xe in [XeF3]– is sp3d2 hybridised with 3 lone pairs of electrons.
Total number of lone pairs = 6

JEE Advanced Mock Test - 4 (Paper I) - Question 23
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Match the reagents in List - 1 with their properties in List - 2, and choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 23


JEE Advanced Mock Test - 4 (Paper I) - Question 24
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Match the organic name reactions in List - 1 with appropriate reagents for the reactions in List - 2, and choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 24



JEE Advanced Mock Test - 4 (Paper I) - Question 25
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Answer by appropriately matching the information given in the three columns of the following table:

From the three columns, select the correct combination for the metal that is the weakest reducing agent.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 25

Beryllium is a reducing agent. Alkali metals are stronger reducing agents than alkaline earth metals due to their low ionisation energies and high negative values of their standard electrode potential.
Down the group (both groups 1 and 2), the reduction potential becomes more negative and the reducing power increases.
Carbonates of alkaline earth metals (and also LiCO3) are thermally unstable and decompose on heating to form their oxides and CO2.
BeCO3 is the least stable and is preserved in an atmosphere of CO2 to prevent decomposition.
BeH2 is a covalent hydride and has a polymeric structure.

Note: BeO is amphoteric, but option (1) cannot be selected because R is not a correct match.
Hence, (III)-(iv)-(P) is the correct option.

JEE Advanced Mock Test - 4 (Paper I) - Question 26
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Answer by appropriately matching the information given in the three columns of the following table:

From the three columns, select the correct combination for the formation of a Diphenyl ketone.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 26

Benzene undergoes acylation with acid halides in the presence of a Lewis acid catalyst, such as AlCl3 to form ketones.

JEE Advanced Mock Test - 4 (Paper I) - Question 27
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In the following question, [x] denotes the greatest integer less than or equal to x.
Which of the following options correctly matches the functions in Column I with their properties listed in Column II?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 27

For (A)
The function: f(x) = x|x|
The graph of the above function is shown as:

Since the function is continuous at every point, it is differentiable and strictly increasing in (-1, 1).
So, (A)  (p) and(q)

For (B)
Given function:

The graph of the above function is shown as:

 

The graph shows that the function has one non-differentiable point in (-1, 1). So, the function is not differentiable at one point in (-1, 1).
So, (B) → (r)
For (C)
f(x) = x + [x]

The graph of the function is shown as:

 

From the above, it is clear that the function is strictly increasing in (-1, 1) and has one non-differentiable point in (-1, 1).
So, (C)  (q) and (r)
For (D)
f(x) = |x - 1| + |x + 1|

From the graph, the function is continuous at every point in (-1, 1).
Hence, it is differentiable also in (-1, 1).
So, (D) → (p)

JEE Advanced Mock Test - 4 (Paper I) - Question 28
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Match the statements/expressions given in Column I with the values given in Column II.


Which of the following is the only correct option?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 28

For (A):
Given: 2sin2θ + sin22θ = 2
⇒ 2sin2θ + 4sin2θcos2θ = 2
⇒ sin2θ + 2sin2θ(1 - sin2θ) = 1

⇒ 3sin2θ - 2sin4θ - 1 = 0
⇒ sinθ = ± 1/√2, ± 1
⇒ θ = π/4, π/2
Since it lies between (-π/2, π/2),
θ = π/4
(A) - (q)
(B)
Let y = 3x/π
Then 1/2 ≤ y ≤ 3 ∀ x ∈ [π/6, π]
Now, f(y) = [2y] cos[y]
Critical points are: y = 1/2, y = 1, y = 3/2, y = 3
So, points of discontinuity = [π/6, π/3, π/2, π]
Since we have to find the point of discontinuity which lies between (π/6, π/2).
So, the required point of discontinuity is π/3.
(B) - (r)
(C) The matrix (A) obtained =

Volume = |A| = 1(2π - 0) - 1(π - 0) + 0(1 - 2) = 2π - π = π
Hence, volume of the parallelepiped = π
(C) - (t)

The half of the angle between

JEE Advanced Mock Test - 4 (Paper I) - Question 29
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Directions: Answer the following question by appropriately matching the information given in the three columns in the following table.

Here,
Column I contains equation of line
Column II contains any points in plane
Column III contains coordinates of foot of perpendicular when drawn from the points in column II

Which of the following combinations is incorrect?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 29

Here, we have to find the coordinates of the foot of the perpendicular for each of the points to each line in terms of k.
We can eliminate the particular option by substituting respective values.

Given equation of straight line is,

And for points (-3,2,3)
So,

x = 2k
y = 4k + 3
z = 3k - 2
As we know that,
a1a2 + b1b2 + c1c2 = 0
(-3) 2k + (4k + 3) x 2 + (3k - 2) x 3 = 0
-6k + 8k + 6 + 9k - 6 = 0
11k = 0
k = 0
So,
x = 2 x 0 = 0
x = 0
y = 4 x 0 + 3 = 3
y = 3
z = 3 x 0 - 2 = -2
z = -2
According to these values, coordinate of foot of perpendicular = (0,3,-2)
But in column III, this type of point is not given in any of the options.
And,

Then, (-3,2,3) does not satisfy the equation

Hence, (IV) (ii) (b) is the incorrect combination.

JEE Advanced Mock Test - 4 (Paper I) - Question 30
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Directions: Answer by appropriately matching the information given in the three columns of the following table.
Let

* Column 1 contains information about functional value of f(x), f'(x) and f ''(x).
* Column 2 contains information about limiting behaviour of f(x), f'(x) and f ''(x) at different points.
* Column 3 contains information about increasing/decreasing nature of f(x) and f'(x).

Which of the following option is the only incorrect combination?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 30


∴ f'(x) is strictly decreasing function for x ∈ (-∞, -2) and increasing function for x ∈ (-2, ∞)


∴ f(x) is increasing for all positive value of x.
f(x) is decreasing in x ∈ (-4, 0)



∴ From Column 1: II and III are correct.
From Column 2: b, and d are correct.
From Column 3: q, r and s are correct.

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