JEE Main Mock Test - 5 - JEE MCQ

An infinite wire, carrying current, is bent in the following way.
Two semi-infinite parts and a bent finite part. We have to find the induced magnetic field at an external point o, at a distance R. The corresponding figure is shown below.

Magnetic field induced by a finite current-carrying conductor

The magnetic field at a distance R is given by:

B = (μ₀ I / 4πR) (cos θ₁ - cos θ₂) ........(i)

where θ₁ and θ₂ are the angles made by the endpoints of the wire to the external point.
Using the right-hand thumb rule, we determine the direction of the magnetic field.

For the 1st semi-infinite part:

Given: θ₁ = 0 and θ₂ = 90° = π/2
Substituting these values in equation (i), we get:

B₁ = (μ₀ I / 4πR) ⊗

For the bent portion:

Given: θ₁ = 45° and θ₂ = 135°
The magnetic field is calculated as:

B₂ = (μ₀ I / 4πR) [ (1/√2) + (1/√2) ] ⊗
⇒ B₂ = (μ₀ I / 2√2πR) ⊗

For the second semi-infinite part:

Given: θ₁ = π/2 and θ₂ = 0
The magnetic field is:

B₃ = (μ₀ I / 4πR) ⊗

Total Magnetic Field:

Summing up the individual contributions, the total magnetic field at point O is:

B = B₁ + B₂ + B₃
⇒ B = (μ₀ I / 2√2πR) ⊗

 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:

The oxidation number of cobalt in the given complex is +2.

Co⁺² in the complex is sp³ hybridised and the complex has a tetrahedral geometry.
Number of unpaired electrons (n) = 3
Spin magnetic moment (m_s)

(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

u² −2u + 2 = 0
⇒ u = 1±i
So, α = 1+i and β = 1−i
Now given that, x = cotθ − 1

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 are (a).

Let S denote the success (getting a '6') and F denote the failure (not getting a '6').
Thus, P(S) = 1/6 = p,P(F) = 5/6 = q
P (A wins in first throw) = P(S) = p
P (A wins in third throw) = P(FFS) = qqp
P (A wins in fifth throw) = P(FFFFS) = qqqqp
So, P (A wins) = p + qqp + qqqqp +…

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