JEE Main Mock Test - 2 Free Online Test 2026

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

Elastic limit (allowable stress) of copper, σ_el =1.5 ×10⁸ N m⁻²
Load m =10.0 kg ⇒ F = mg =10.0 × 9.8 = 98 N

So, Option B is the correct answer.

The P − V equation is given as

putting P = P₀, we get


P attains the maximum value at V = 3V₀/2

Resultant amplitude

∴ I ∝ A²
⇒ I ∝ 4a² 

u = −15
f = 30 cm
from lens formula,

So distance from plane mirror is
u₂ = 30 + 15 = 45 cm
So u₂ = v₂ for plane mirror
So from the lens 45 + 15 = 60 cm to the right of it.

Liquid has a tendency to acquire minimum surface area due to surface tension. In gravity free space a drop of any size will be spherical.
In case of gravity, bigger drop will have larger force of gravity which tends to distort the spherical shape of the drop.
Hence,  both assertion and reason are correct and reason is the correct explanation of assertion

Here,

Length of wire l = 2.5m

Area of cross-section, A = 1×10^–6 m²

Hanging mass, m = 15 kg

Elongation, e = ?

Energy stored in the wire, E = ?

Young’s modulus of steel, Y = 2×10¹¹ Nm^–2.

We have,

Young’s modulus of elasticity,

e = 1.875 ×10^-3 m

Again, 

Energy stored in the stretched wire,

      E = 1/2​ .F .e

Or, E = 1/2​ × mg × e 

Or, E = 1/2​ × 15×10 ×1.875 ×10^-3

∴∴ E = 0.141 J

= 37.22

Hydrolysis of substituted chlorosilanes yield corresponding silanols which undergo polymerisation.

Polymerisation of dialkyl silandiol yields linear thermoplastic polymer.

Volume of solution = 1000 ml
Mass of the solution = V × d
= 1000 mL × 1.04 g/mL = 1040 g
Amount of solute in 1000 mL solution
= 1 M
= 1 M × Molecular Mass = 1M × 74.5 g/mol
= 74.5 g
Mass of water = Mass of solution − Mass of
KCl = 1040 g−74.5 g = 965.5 g
Molality of the solution = m =

= 1.0357
ΔT_bi.K_b⋅m = 2 × 0.52 K kg mol⁻¹ × 1.0357
= 1.078°C.
Boiling point of the solution
=100^∘C + 1.078^∘C = 101.078^∘C

The reaction is 4NH₃ + 3O₂ → 2 N₂ + 6H₂O


= 1.05 mol L⁻¹ s⁻¹   

The reaction follows first-order kinetics. Hence k = 0.693/_t1/2 = 0.693/(80 s). For the reaction to be 80%, [A]_t/[A]₀ = 0.2. Hence

The acidic strength of an acid is defined as the tendency of the acid to dissociate into its corresponding ions in order to release a proton and the anion. The acidic strength of a strong acid is high and that of a weak acid is low. This means that on dissociation, a strong acid releases its proton at a much faster rate than the weak acid.

Oxidation number: The greater the oxidation number of the central atom of the oxide, the greater the acidic strength will be.

n a periodic table, the acidic strength of an atom increases across a period and decreases down the group. In the above options, the correct order of acidic strength of the oxides is:
Cl₂O₇ > SO₃ > P₄O₁₀

The oxidation state of the oxides is as follows:
Let the oxidation number of the chlorine atom be 'x′.
Cl₂O₇ = 2x + 7(−2) → x = +7
Let the oxidation number of the sulfur atom be ′y′.
SO₃ = y + 3(−2) ⇒ y = +6
Let the oxidation number of the phosphorus atom be ′z′.
P₄O₁₀ − 4z + 10(−2) → z − +5
Thus, as we can see that the order of oxidation is in a decreasing manner, this means that this option is correct.
So, the correct answer is Option (a).

Nucleosides are the structural subunit of nucleic acids such as DNA and RNA. A nucleoside, composed of a nucleobase, is either a pyrimidine (cytosine, thymine or uracil) or a purine (adenine or guanine), a five carbon sugar which is either ribose or deoxyribose. The base connected to sugar at  position. Nucleotides are building blocks of nucleic acids DNA and RNA. Nucleotides are composed of a nitrogenous base, a five-carbon sugar (ribose or deoxyribose), and at least one phosphate group. Thus, a nucleoside plus a phosphate group yields a nucleotide.

- PEt₃ and AsPh₃ as ligands can form dπ−dπ bond with transition metals.
- The N−N single bond is weaker than the single P−P bond because of high inter-electronic repulsion of the non-bonding electrons.
- Nitrogen has unique ability to form pπ−pπ multiple bonds with itself, carbon and oxygen.
- Nitrogen cannot form dπ−pπ bond as other heavier elements of its group.

H₂SO₄ = 2 ( + 1) + x + 4 (- 2) = 0
x = 8 – 2 = +6 

k + |k + z²| = |z|²
Hence k and |z²| are collinear.
Now k ∈ R⁻
Since, k and z² is collinear, then z² has to be purely real.
Thus, arg(z²) = (2n − 1)π
Now z  =re^iθ
z² = r²e^2iθ = me^2iθ
Hence arg(z²) = 2arg(z)
Therefore, arg(z) = 1/2 arg(z²)

Putting n = 1, we get
arg(z) = π/2
Hence, option 'C' is correct.

Given, 2 sec2α = tanβ + cotβ 

Equation of directrix is x + y = 0


On simplifying we get-
(x − y)² = 8(x + y − 2)

We have,

On integrating both sides,

On integrating both sides,

Sum of the present ages of husband, wife and child

= (23 × 2 + 5 × 2) + 1 = 57 years.

∴ Required average = (57/3) = 19 years .

Average age is 19 years which is a prime number, so it has only two divisors {1, 19}.

General term of the given expression is,

Here, q + 2r = 4

For p = 6, q = 4, r = 0, coefficient = 10!/[6! x 4!] = 210

For p = 7, q = 2, r = 1, coefficient = 10! /[7! x 2! x 1!] = 360

For p = 8, q = 0, r = 2, coefficient = 10!/[ 8! x 2!] = 45

Therefore, sum = 615