JEE Main Practice Mock Test - 1 Free Online Test 2026

Using the equation of trajectory:
y = x tan α ( 1 - x / R )

For points (3,4):
4 = 3 tan α ( 1 - 3 / R ) ...(i)

For points (4,3):
3 = 4 tan α ( 1 - 4 / R ) ...(ii)

Dividing (i) and (ii)...

Pitch = 3/6 = 0.5 mm
L.C. = 0.5 mm  / 50 = 1 / 100 mm = 0.01 mm
=0.001 cm

When an ideal gas is compressed adiabatically, its temperature and the average kinetic energy of the gas molecule increases because of collision of molecules with wall. Hence, Both A and R are true and R is the correct explanation of A.

Given that

Force 

Initial point (0, 0, 0)

Final point (2, 4, 0)

Work done by the variable force is,

y = x²

dy = 2x dx

dw = Fx dx + Fy dy
= 0 + (3xy) dy
= 3x × x² × 2x dx

= 192 / 5 J

Wavelength of the photon in terms of energy will be given by,

λ₂ = hc / E

De-Broglie of the proton will be,

λ₁ = h / p = h / √(2mₚE)

(where p is the momentum of the proton and mₚ is the mass of the proton)

From the above two equations,

Magnetic moment / Angular momentum = (q / 2m)
∴ Magnetic moment = (angular momentum) * (q / 2m)
= (Iω)(q / 2m)
= (1/2mR²) * (ω) * (q / 2m)
= (1/4)qωR²

The limiting force of static friction depends on both the normal force and the coefficient of static friction, but it does not depend on the area of contact.
The force of kinetic friction is indeed dependent on the nature of the materials in contact and the normal force, but not on the area of contact.
Hence, Statement I is incorrect but Statement II is correct.

Initially 

After charge Q is given

Charge distribution is shown above in the figure.

Then,

Taking motion in vertical direction,


(b) Taking motion in the horizontal direction,

The extra path difference due to the insertion of a sheet of refractive index μ and thickness t is,
Δx = (μ - 1) t
The shift in the interference pattern of Young's double-slit experiment is,
y = (D/d) Δx
⇒ y = (μ - 1) t (D/d)
Hence, the fringe pattern will be displaced by (D/d) (μ - 1) t.

The moment of inertia of a hollow sphere shell in terms of mass and radius of the shell is, I₀ = 2 / 3mr². Here, for sphere A:

Also, for sphere B:

The ratio, I_B / I_A = 6 / 10.

The formula to calculate the fundamental frequency in a closed pipe is given by

And, that in an open pipe is given by

Given that

From equations (1), (2) and (3), it follows that

⇒ v = 42 × 7

= 294 m s⁻¹

The circuit given in the problem can be redrawn as shown below. 

The three branches are connected between the same points P and Q and hence they are in parallel. For such a circuit the potential difference between the points P and Q is

Where E₁, E₂ and E₃ are the emf of the batteries present in the first, second and the third branch.
Similarly, R₁, R₂ and R₃ are the values of resistance in each of the respective branches.

The value of current through the 1 Ω resistance is

It can be observed from figure that P and Q shall collide if the initial component of velocity of P and Q on inclined plane i.e along incline should be equal. That is particle is projected perpendicular to incline.

∴∴ Time of flight 

Cyclic C₅H₁₀

For 3rd structure 2 cis-trans and 1 optical isomer are possible. Total 7 isomers.

When 2−bromobutane is treated with NaI in the presence of dry acetone then, iodine is substituted in the place of the bromine atom and formation of 2−iodobutane (B) takes place. This reaction is known as the Finkelstein reaction.

2−Iodobutane reacts with moist Ag2O to oxidise into corresponding alcohol, i.e., butan−2−ol (C).

Butan−2−ol is unsymmetrical and has one chiral carbon, therefore, the number of optical isomers = 2ⁿ, where n = the number of chiral centres.

So, the number of optical isomers  = 2¹ = 2.

In the same nucleophilic centre, nucleophlilicity is parallel to basic character. Stronger the acid, weaker is the conjugated base.
Acidic character order: CH₃SO₃H > CH₃COOH > CH₃OH > C₂H₅OH
Basic and nucelophilicity order:

Correct option: D

(A) is incorrect. Atomic radius does not decrease down the group; it generally increases on moving down because extra shells are added. The trend is not regular because d- and f-block contraction cause anomalies (for example, the radius of Ga is not much larger than that of Al).

(B) is correct. There is an overall tendency for electronegativity to decrease down the group as atomic size and shielding increase; however, small anomalies occur (notably for Ga) due to d-block contraction, which can make Ga's electronegativity slightly higher than that of Al.

(C) is incorrect. Although third-period atoms have vacant 3d orbitals, these orbitals are too high in energy to be effectively used for forming normal covalent bonds in aluminium compounds. The observation of six-coordinate Al in complexes is explained by coordinate bonding, ionic interactions and ligand bridging, not by effective participation of 3d orbitals to give covalency 6.

(D) is correct. Boron compounds show significant pπ-pπ interaction between boron's empty p-orbital and lone pairs of more electronegative atoms (for example oxygen), giving partial multiple-bond character in B-X bonds.

(E) is correct. Boron and silicon display a diagonal relationship that leads to similar chemical behaviour such as predominantly covalent bonding and formation of acidic oxides (examples: B2O3, SiO2).

Therefore the true statements are (B), (D) and (E); the correct choice is option D.

Acidic oxides → Mn₂O₇, CrO₃, V₂O₅
(due to higher oxidation state of central metal)

Basic oxides → Na₂O, CaO, BaO
(due to lower oxidation state of central metal)

Alkaline and alkali metal oxides are always basic.

Neutral oxides → CO, NO, N₂O

Amphoteric oxides → ZnO, PbO, BeO
(acidic as well as basic)

Hence, option (d) is correct.

Molecular forces of attraction get stronger as molecules get bigger in size. Further, as the branching increases, the surface area of the molecule decreases. Because of this, the Van der Waal's force of attraction between the molecule decreases and consequently boiling point decreases.
In the above three isomers given Structure (b) is a straight-chain with no branching, structure (c) is branched but has less surface area compared to the structure (a). Hence, the increasing order of their boiling points is, (c) < (a)<(b)

Since the product on ozonolysis is propanaldehyde and formaldehyde, it implies that the the double bond is shared between C₁ and C₂.
If the compound were option 1, the result of such elimination would have been 2 butene whose ozonolysis would give ethanals.
1−butene on ozonolysis gives propanaldehyde and formaldehyde thus, the hydrocarbon is

1-butene can be obtained by,

(i) When SO₂ gas reacts with acidified K₂Cr₂O₇ solution, sulphur dioxide is oxidized to sulphuric acid, and potassium dichromate is reduced to chromic sulphate. The orange color of potassium dichromate changes to green, indicating the reaction. Thus, SO₂ decolorizes K₂Cr₂O₇.

Reaction:
3SO₂ + K₂Cr₂O₇ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

(ii) When sulphur dioxide is boiled with H₂O₂ and then cooled, sulphuric acid is formed.

Reaction:
SO₂ + H₂O₂ → H₂SO₄

When BaCl₂ is added to sulphuric acid, a white precipitate of BaSO₄ is formed, which is insoluble in dilute HCl.

Reaction:
BaCl₂ + H₂SO₄ → BaSO₄ ↓ + 2HCl

(iii) When H₂S is passed into the solution, white turbidity is formed due to the formation of sulphur.

Reaction:
2H₂S + SO₂ → 2H₂O + 3S ↓

Thus, the gas (X) is SO₂.

Molecular Mass of SO₂:
= 32 + (2 × 16)
= 64 g/mol

Let x²e = t

When x = 1, y = 0

⇒ 1 log 1 = 0 + C

⇒ C = 0

Hence, x² log x = y sin y

The centroid on a non-equilateral triangle divides the line joining orthocentre and circumcentre in 2:1, so let the coordinates of circumcentre is (h,k), then

And image of orthocentre (5,8) with respect to side BC,x−y=0 is (8,5)
So, the radius of circumcircle of △ABC is

Then diameter = 4√10.

As, a̅ lies in the plane of the vectors b̅ and c̅, So, 

Compare with  a̅ = αî + 2ĵ + βk̂

Not in option
So, now consider


Compare with a̅ = αî + 2ĵ + βk̂

Given, 2x² (dy/dx) - 2xy + 3y² = 0
Dividing by 2x²y², we get

General solution will be

 is the particular solution.

when x = 1;  y(1) = 2/3

∵y = e^x,  y = e^−x will meet, when e^x = e^−x
⇒ e^2x = 1,
∴ x = 0,y = 1
∴ A and B meet on (0, 1), 
∴ A ∩ B ≠ ϕ

Let A = {a₁, a₂, a₃, ......, aₙ}
P and Q are subsets of A. Elements in P and Q can be chosen in the following ways:
Case I: aᵢ ∈ P, aᵢ ∈ Q for i = 1, 2, ..., n
Case II: aᵢ ∈ P, aᵢ ∉ Q for i = 1, 2, ..., n
Case III: aᵢ ∉ P, aᵢ ∈ Q for i = 1, 2, ..., n
Case IV: aᵢ ∉ P, aᵢ ∉ Q for i = 1, 2, ..., n
Since P and Q are non-intersecting sets, every element can be chosen as in Case II, Case III, or Case IV, i.e., each element can be chosen in 3 ways.
Total number of ways = 3 × 3 × ... (n times) = 3ⁿ
However, one case needs to be excluded when P = ∅ and Q = ∅.
∴ The number of valid ways = 3ⁿ − 1