JEE Main Mock Test - 5 Free Online Test 2026

 or 2sinθ cosθ = sin²θ or tanθ = 2

Method I: Considering equilibrium of each part of system The whole system is in equilibrium; therefore, for each part . From the free-body diagram of block C, T_C = 300 N.

Now consider the equilibrium of point O

Method II: Using Lami's theorem

By Lami's theorem, we have

R + P sin 60 = Mg
R = Mg − P sin 60°
Frictional force
F = μR
= μ[Mg − Psin60^∘  

The body just moves when
P cos 60^∘ = F

Or P = 5.36 N

⇒ R_α = a₀ + a ...(i)
Other equations are
T = ma...(ii)
F + T = ma₀...(iii)

Let the volume of the block be V m³.
Volume of block under liquid = 4V/5 m³
∴ Volume of liquid displaced =4V/5 m³
Now let the density of the liquid be ρkgm⁻³.
Mass of liquid displaced
= (volume of liquid displaced) × (density of liquid)
= 4V/5 ρkg
Weight of liquid displaced = 4V/5 × ρ × g newton
Relative density of wood = 0.8
∴ Density of wood = 0.8 × 1000
= 800 kg m⁻³
∴ Mass of the block = 800 × V kg
From the law of flotation,
Weight of block = weight of liquid displaced
or 800 × V × g = 4V/5 × ρ × g

Or = 1000 kg m⁻³  
Hence the correct choice is (b).

For small oscillations, sin2x = 2x i.e.,
a = −16x
Since aμ − x, the oscillations are simple harmonic in nature.

Conclude from the vector triangle (equilateral).

Binding energy of = 2.2MeV. Therefore, the binding energy of two  = 4.4 MeV. Now, the binding energy of  nucleus = 4 × 7.1 = 28.4MeV. Hence Q = 28.4−4.4 = 24MeV which is choice (c).

For ground wave propagation, the maximum range is

which is less than 100 km. Hence choice (a) is wrong. The maximum frequency which can be propagated via sky waves is v_max ≃ 9MHz which is more than 5 MHz. Hence a 5 MHz signal can be propagated via sky waves and not via ground waves. Thus the correct choice is (b).

In graphite, each carbon is sp²-hybridized and the single occupied unhybridized porbitals of C-atoms overlap side wise to give π-electron cloud which is delocalized and thus the electrons are spread out between the structure.

For a zero-order reaction, the half-life is directly proportional to the initial concentration of A. Hence, the reaction is of zero order.

Some complex ions of nitrogen show similarities with halide ions. These are called pseudohalide ions e.g., NCO⁻,CN⁻etc.
Halogen, among themselves, form complex ions which are called polyhalide ions. e.g., I⁻₃,BrI⁻₂ etc.
Similarly, interhalogens are the compounds of halogen in which one halogen is cation and other halogen is anion. E.g., IF₇, ICl₅, BrF₃,IF₅ etc .

In oxymercuration - demercuration, more stable carbocation is formed and water attacks on more substituted carbon. The reaction is as follows:

(p° - pₛ) / p° = X₂

(p° - pₛ) / 0.80 = 0.25

p° - pₛ = 0.25 × 0.80 = 0.20 atm

f(x) = (ax + b)(x + 1)

f(1) + f(2) = 0

⇒ (a + b)2 + (2a + b)3 = 0

⇒ 8a + 5b = 0

f(x) = (ax - 8a/5)(x + 1)

= a(x - 8/5)(x + 1)

∴ From the above diagram of f(x), f(x) is surjective but not injective.

B's - 1
A's - 2
R's - 2
C's - 1
K's - 1

The number of four-lettered words can be formed from (2 identical, 2 identical) or (2 identical, 2 different) or 4 different letters.

So, number of ways

²C₂ × (4! / 2!2!) + ²C₁ × ⁴C₂ × (4! / 2!) + ⁵C₄ × 4!

= 6 + 24 × (24 / 4) + 5 × 24

= 6 + 11 × 24

= 270

Let

By using roots of unity, there are 2021 complex number that satisfied the equation z²⁰²¹=1
∴ Total number of solutions = 2021 + 1 = 2022

0 < sin⁸θ ≤ sin²θ ...(i)
and 0 < cos¹⁴θ ≤ cos²θ ...(ii)
Adding (i) and (ii) ⇒ 0 < A ≤ 1

Let equation of hyperbola and ellipse be

⇒ a₁e₁ = a₂e₂ (since, same pair of foci)
e₂ = 1/2 (given)
Both intersect at (2,2)

The given equation can be written as

which is L.D. equation in y.

∴ Multiplying the given equation by I.F. = x sec x, and integrating, we get

Multiplying by cosx, this result can also be written as
xy = sin x + c cos x which is given in (a).
Hence the correct answers is (a): xy = sin x + c cos x

Let:
p = 1/6 (probability of getting a 6)
q = 5/6 (probability of not getting a 6)
Since A starts, A wins if:
He gets 6 on the 1st try: p
Or both A and B fail once, and A gets 6 on 3rd throw: q·q·p = q²p
Or both fail twice, and A gets 6 on 5th throw: q⁴p, and so on.
So: P(A wins) = p + q²p + q⁴p + ... = p(1 + q² + q⁴ + ...)
This is a geometric series with first term 1 and ratio q².
Sum of infinite GP = 1 / (1 − q²)
So: P(A wins) = p / (1 − q²)
⇒ (1/6) / (1 − (25/36)) = (1/6) / (11/36) = 6/11
Hence, P(B wins) = 1 − 6/11 = 5/11
Final Answer: Option A: 5/11 is correct.

If n³ + 2n + 1 = 3n² + 7 ⇒ n³ − 3n² + 2n − 6 = 0 ⇒ (n − 3)(n² + 2) = 0 ⇒ n = 3 as n ∈ N
⇒ n = 3 as n ∈ N So, x =3 × 3² + 7 = 34 ∈ X ∩ Y.
In (a) and (b) x ≠ 34, for any n ∈ N.

(b) 8/4 = 2, 64/4 = 16 ; but 4 is not prime.
Hence P∧Q → R, false
(c) (6)²/12 = 36/12 = 3 ; but 12 is not prime
Hence Q→R, false
(d) (4)²8 = 168 = 2;4/8 is not an integer
Hence Q → P, false

So 1 + 2 + 6 + x₄ + x₅ = 22
x₄ + x₅ = 13 (1)

= 5(8.24+19.36)
= 138

Hence, the other two numbers are 4 and 9.