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JEE Main Practice Test- 2 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Practice Test- 2

JEE Main Practice Test- 2 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main Practice Test- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 2 below.
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JEE Main Practice Test- 2 - Question 1
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The volumes of containers A and B, connected by a tube and a closed valve are V and 4 V, respectively. Both the containers A and B have the same ideal gas at pressures (temperatures) 5.0 ×105 Pa(300 K) and 1.0 ×105Pa (400 K), respectively. The valve is opened to allow the pressure to equalise, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers.

Detailed Solution for JEE Main Practice Test- 2 - Question 1

∴ p1 > p2 so, the equalise pressure of the p is reduced from p1

Given, T1 = 300

T2 = 400

V2 = 4V

p1 = 5 x 105

JEE Main Practice Test- 2 - Question 2
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A block of mass 10 kg is suspended through two light spring balances as  shown in figure

Detailed Solution for JEE Main Practice Test- 2 - Question 2

The FED of the spring balances and the block are as shown in figure.

T1=10g T1=T2 ⇒ T2=10g where, T1 and T2 are readings of spring balances as shown in figure.

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JEE Main Practice Test- 2 - Question 3
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If E = 100 sin (100t) volt and are the instantaneous values of voltage and current, then the R.M.S values of voltage and current are respectively.

Detailed Solution for JEE Main Practice Test- 2 - Question 3

The instantanous valur of voltae is 

E= 100sin (100 t) V ...(1)

 (compare it with

E = E0 sin (ωt) V

we get,

E0 = 100 V, ω = 100 rads-1

the r.m.s value of voltage is 

The instantaneous value of current is 

compare it with, I=I0sin(ω t + Φ)

we get

I0 = 100 mA, ω = 100 rads-1

The r.m.s value of current is

JEE Main Practice Test- 2 - Question 4
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Two spherical bodies of mass M and 5 M and radii R and 2 R, respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is
Detailed Solution for JEE Main Practice Test- 2 - Question 4

Initial separation between both centers = 12R
Final separation (when collision occurs) = 3R
Thus both the centers need to travel 9R combinely in such a way that their COM does not move as no external force is acting over the 2 mass system.
Thus let say if smaller mass travel x distance, bigger would eventually travel rest 9R - x
Thus we we write the equation of displacement of the COM, and taking direction of displacement of bigger mass as positive, we get
M (-x) + 5M (9R - x) / (M + 5M) = 0
Thus we get, -Mx + 45MR - 5Mx = 0
We get 6x = 45R
Thus x = 7.5R

JEE Main Practice Test- 2 - Question 5
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Mark the correct option.
Detailed Solution for JEE Main Practice Test- 2 - Question 5

A charge moving along a circle is equivalent to a current carrying coil, but with respect to magnetic field on the axis of circle. It is equivalent only for the average value of the magnetic field and not for instantaneous values. While if two charge particles are moving symmetrically along a circle at diametrically opposite points then the average as well as instantaneous magnetic field on its axis is same as due to a current carrying coil.

JEE Main Practice Test- 2 - Question 6
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The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be
Detailed Solution for JEE Main Practice Test- 2 - Question 6

Initial volume = final volume

JEE Main Practice Test- 2 - Question 7
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Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

If EI, EII and EIII are the percentage errors in g, i.e.,

for students I, II and III, respectively, then
Detailed Solution for JEE Main Practice Test- 2 - Question 7

The period of oscillation (T) of a simple pendulum of length ℓ is given by

T = 2π√(ℓ/g)

Therefore, g = 4π2 ℓ/T2 so that the fractional error in g is given by

∆g/g = (∆ℓ/ℓ) + 2(∆T/T)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the maximum possible error].

Here ∆ℓ = 0.1 cm and ∆T = 0.1 s

The percentage error is 100 times the fractional error so that

EI = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

EII = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

EIII = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus EI is minimum. 

JEE Main Practice Test- 2 - Question 8
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Write down the expression for capacitance of a spherical capacitor whose conductors radii are R1 and R2(R2>R1),when inner sphere is grounded.
Detailed Solution for JEE Main Practice Test- 2 - Question 8

JEE Main Practice Test- 2 - Question 9
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A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is
Detailed Solution for JEE Main Practice Test- 2 - Question 9

Here, position 1 and position 2 can be calculated by using 

work- energy Theorem


JEE Main Practice Test- 2 - Question 10
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Acceleration of each block is given as g/5√2. Find the magnitude and direction of force exerted by string on pulley. (μ = 0.4)
Detailed Solution for JEE Main Practice Test- 2 - Question 10

Let coefficient of friction be u. and and 3m block is moving down the incline, then Acceleration  

 

Force exerted by string/on pulley is √2T as shown in figure

∴ F = 6mg/5

JEE Main Practice Test- 2 - Question 11
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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
Detailed Solution for JEE Main Practice Test- 2 - Question 11

for closed pipes,

so in this case 6 possibilities

JEE Main Practice Test- 2 - Question 12
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Three identical spheres, each of mass 1 kg are kept as shown in the figure below, touching each other, with their centres on a straight line. If their centres are marked P, Q, R respectively, the distance of centre of mass of the system from P is
Detailed Solution for JEE Main Practice Test- 2 - Question 12

Radius

r1= 0

r2 = PQ

r3 = PR

the distance of centre of mass of the system from 'P' is given by

JEE Main Practice Test- 2 - Question 13
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A resistor R and 2μF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed (log10 2.5 = 0.4)
Detailed Solution for JEE Main Practice Test- 2 - Question 13

Wen Neon bulb is fulled with gas.

No, current frows through it because the resistance is infine.

JEE Main Practice Test- 2 - Question 14
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A α−particle passes through a potential difference of 2×106V and then it becomes incident on a silver foil. The charge number of silver is 47. The energy of incident particles will be: (in joule)
Detailed Solution for JEE Main Practice Test- 2 - Question 14

Energy of incident particles ie, alpha particles will be 

JEE Main Practice Test- 2 - Question 15
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Direction: Question is based on the following paragraph.

A wire of length L, mass m and carrying a current is suspended from point O as shown. An infinitely long wire carrying the same current I is at a distance L below the lower end of the wire. Given, I = 2A, L= 1m and m = 0.1 kg (ln 2 = 0.693)

 

What is angular acceleration of the wire just after it is released from the position shown?
Detailed Solution for JEE Main Practice Test- 2 - Question 15

 

JEE Main Practice Test- 2 - Question 16
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The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits according to the currents through the battery just after the switch is closed, greatest first
Detailed Solution for JEE Main Practice Test- 2 - Question 16

In circuit ( 1 ), on closing the switch, the current in the inductor is zero due to self induction, ie, i1=0.

In circuit (2), on closing the switch the current in the inductor is zero due to self-induction.
image


Therefore,

i2 = i′ = E / R1
In circuit (3), on closing the switch, the current in the inductor is again zero due to the same reason.
image


Therefore,

i2 = i′ = E / (R1 + R2) 

Thus, it is obvious that,

i2>i3>i1(=0)

JEE Main Practice Test- 2 - Question 17
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A bullet when fired into a fixed target loses half of its velocity after penetrating 20 cm.How much further it will penetrate before coming to rest?
Detailed Solution for JEE Main Practice Test- 2 - Question 17

 

JEE Main Practice Test- 2 - Question 18
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Directions: Question are Assertion - Reaction type each of these contains two statements: Statement I (Assertion), Statement II (Reason) Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes a, b, c and d given below:

Statement I : The isothermal curves intersect each other at a certain point

Statement II: The isothermal changes takes place slowly, so the isothermal curves have very little slope.
Detailed Solution for JEE Main Practice Test- 2 - Question 18

To carry out isothermal process, a perfect gas is compressed or allowed to expand very slowly. Isothermal curves never intersect each other as they have very little slope.

JEE Main Practice Test- 2 - Question 19
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A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be
Detailed Solution for JEE Main Practice Test- 2 - Question 19

∴ increase in capiacitance

JEE Main Practice Test- 2 - Question 20
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A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?  (Take g=10m/s2)
Detailed Solution for JEE Main Practice Test- 2 - Question 20

for second ball

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 21
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Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. A 40 cm diameter pipe branches into two pipes of diameters 10 cm and 20 cm each. If the average velocities of water that flows through 10 cm and 20 cm pipes are 6 m/s and 2 m/s, respectively, then the speed of the water that flows through the 40 cm pipe (in m/s) is 

(Answer up to nearest integer)

Detailed Solution for JEE Main Practice Test- 2 - Question 21

Let x be the velocity of water flowing through 40 cm diameter branch
V(40): Volume of water flowing through 40cm diameter
V(20): Volume of water flowing through 20cm diameter
V(10): Volume of water flowing through 10cm diameter
V(40) = V(20) + V(10) - - - (1)
Volume of water flowing = (Cross-sectional area) * (Velocity of water)
(1) → x*π (0.20)2 = 2*π (0.10)2 + 6*π (0.05)2
x*(0.04) = 2*(0.01) + 6*(0.0025)
x = (0.035) / (0.04) = 35/40 = 0.875
Nearest integral value is 1

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 22
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Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. Two point objects A and B are 40 cm apart. A convex lens L of focal length 15 cm should be placed in between them such that the images due to the two point objects coincide. If the least distance between L and A is x, then the value of x – 1 is

Detailed Solution for JEE Main Practice Test- 2 - Question 22

Let VA be the image distance due to the object A. X is the distance between the object A and the lens.

image due to the second object:

As one of the images should be virtual in order to satisfy the given condition.


the lesser value is 10. Hence, the value of x-1 = 9

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 23
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Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. An air-filled parallel plate capacitor having circular plates has a capacitance of 10 pF. When the radii of the plates are increased two times, the distance between them is halved and if a medium of dielectric constant k is introduced, the capacitance increases 16 times. The value of k is

Detailed Solution for JEE Main Practice Test- 2 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 24
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Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. Two parallel identical plates carry equal and opposite charges having a uniform charge of 88.9 C. Positive plate is fixed on the ceiling of a box and the negative plate has to be suspended. If the area of the plates is 6.35 sq. m and 'm' is the mass of the negative plate, then the value of [m] in kg, where [ ] stands for maximum integer value, is

Detailed Solution for JEE Main Practice Test- 2 - Question 24

force of attraction between the plates = weight of the negative plate for it to be suspended

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 25
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Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

A radioactive sample S1 having an activity of 5 μ Ci and half life of 20 years has twice the number of nuclei as another sample S2, which has an activity of 10 μCi. The half lives (in years) of S2 is

Detailed Solution for JEE Main Practice Test- 2 - Question 25

JEE Main Practice Test- 2 - Question 26
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Directions: Questions are based on the following paragraph.

When ammonium vanadate is heated with oxalic add solution, a compound Z is formed. A sample of Z was titrated with KMnO4 solution in hot acidic solution. The resulting liquid was reduced with SO2, the excess SO2 boiled off, and the liquid again titrated with KMnO4. The ratio of the volumes of KMnO4 used in the two titrations was 5 : 1. KMnO4 oxidises all oxidation state of vanadium to Vanadium (+V) and SO2 reduces vanadium (+V) to vanadium (+IV). Read the above experiment and answer the following questions. If vanadium exists as , reduced species by SO2 would be
Detailed Solution for JEE Main Practice Test- 2 - Question 26

 is reduced to +4 oxidation state which is 

JEE Main Practice Test- 2 - Question 27
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In a cubic dosed packed structure of mixed oxides, the lattice is made up of oxide ions, one eighth of tetrahedral/voids are occupied by divalent ions (A2+), while one half of the octahedral voids are occupied by trivalent ions(B3+)What is the formula of the oxide ?
Detailed Solution for JEE Main Practice Test- 2 - Question 27

Let number of oxides = x

Number of octahedral void = x

Number of tetrahedral void = 2x

Number of 

 

 Hence, formula of oxide is AB2O4.

JEE Main Practice Test- 2 - Question 28
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Which of the following sequence of reaction is the best means to furnish the conversion RCH2OH→RCH2NH2
Detailed Solution for JEE Main Practice Test- 2 - Question 28

JEE Main Practice Test- 2 - Question 29
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The standard heat of combustion of carbon(s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and - 1108.76 kJ/mol respectively. The standard heat of formation of carbon disulphide(l) is
Detailed Solution for JEE Main Practice Test- 2 - Question 29

on putting various enthalpy of formation in equation III 

(reactants) - 1108.76 = [-393.3 + 2(-293.72)] - 

=-1108.76 = -393.3 -2 x 293.72 - 

JEE Main Practice Test- 2 - Question 30
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The volume of 0.1 M oxalic acid that can be completely oxidised by 20 mL of 0.025 M KMnO4 solution is
Detailed Solution for JEE Main Practice Test- 2 - Question 30

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