JEE Main Mock Test - 9 - JEE MCQ

The given situation is shown in the figure

For the given transformer,

Vp = 230 V

Vs = 23 V

Rs = 115 Ω

Current in secondary coil,
Is = Vs / Rs = 23 / 115 = 0.2 A

We know that, in a transformer
Vs / Vp = Ip / Is

So,
Ip = (Vs × Is) / Vp = (23 × 0.2) / 230 = 0.02 A

V = V₀ ln (r / r₀). Therefore, F = - dV / dr = V₀ / r

If v is the orbital speed of the electron and m its mass,
m v² / r = V₀ / r
⇒ m v² = V₀

Bohr’s quantum condition is
mvr = nh / 2π
⇒ v = nh / 2πmr

Substituting (2) in (1), we get
r² = ( (hh2 / 4π² m V₀) n² )
⇒ r ∝ n

Hence, x = 1

The linear width of the central maxima in the diffraction pattern is given by (2Dλ) / a

where λ = Wavelength

a = Size of the slit

D = Distance between slit and screen.

As λ_Blue < λ_Red, and the width of diffraction bands is directly proportional to λ.

Hence, when the red light is replaced by blue light, the wavelength decreases, which means that the fringe's width decreases and becomes narrower and crowded together.

From the diagram, it is clear that the first set of gates are NOT gates, the second set of gates are AND gates and the third and final gate is an OR gate.

The final output from all the combination of the gates can be written as

This indicates a XOR GATE. Hence, this option indicates the correct truth table.

The formula to calculate the kinetic energy can be written as
K = (f / 2) * n * R * T ... (1)

Since oxygen is a diatomic gas, the number of degrees of freedom is 5.

Thus, from equation (1), it follows that
K = (5 / 2) * 1 * 8.31 * 300 J
K = 6232.5 J

From the dimensional analysis, it follows that

And,
[V] = [b]
Hence,

= [P]

or 
∴ 
= 2.55 kg
Maximum acceleration of SHM is,
a_max = ω²A   (A = amplitude)
i.e., maximum force on mass 'm' is m ω² A which is being provided by the force of friction between the mass and the cart. Therefore,
μ_smg  ≥  mω² A

or               μ_s   ≥  0.358
Thus, the minimum value of  μ_s should be 0.358.

Energy is released in a process when the total Binding Energy (B.E.) of the nucleus increases, or when the total B.E. of the products is greater than that of the reactants. By calculation, we can see that this happens only in the case of option (3).
Given: W → 2Y
Binding Energy (B.E.) of reactants = 120 × 7.5 = 900 MeV
Binding Energy (B.E.) of products = 2 × (60 × 8.5) = 1020 MeV

According to question,

w_A = x g, m_A = 18, X_A = 1 - 0.6 = 0.4

w_B = 69 g, m_B = 46, X_B = 0.4

We know that,

X_A = n_A / (n_A + n_B)

or 0.4 = (w_A / m_A) / [(w_A / m_A) + (69 / 46)]

⇒ 0.4 = (x / 18) / [(x / 18) + (3 / 2)]

0.4 × (2x + 54) / 36 = x / 18

or 2x + 54 = 5x or 3x = 54, x = 18g

Van't Hoff factors for urea, NaCl, Na₂SO₄, and Na₃PO₄ are 1, 2, 3, and 4 respectively.

As ΔTₓ ∝ i * m, so ΔTₓ for urea, NaCl, Na₂SO₄, and Na₃PO₄ are proportional to 1, 2, 3, and 4 respectively.

Hence, the freezing point order is:
Na₃PO₄ < Na₂SO₄ < NaCl < Urea

Grignard reagent = R − MgX
Thus, it is an organomagnesium compound.

For the cell reaction
2Fe³⁺ (aq) + 2I⁻ (aq) → 2Fe²⁺ (aq) + I₂ (aq)

n = 2

F = 96500

E = 0.24 V

ΔG = -nFE = -2 × 96500 × 0.24 = -46320 J = -46.32 kJ

M= % by weight ×10×d / Mw₂
= 30 × 10 × 1.095 / 36.5
= 9M

There is significant  pπ − pπ back bonding in BF₃ due to almost similar sizes of 2p-orbitals of B and F.

According to the Huckle's rule,
The compound is aromatic which is planar and has the (4n + 2)π electrons. The compound is aromatic in nature because it is a planar compound with (4n + 2)π electrons. Aniline, naphthalene and pyridine are aromatic compounds.
Compounds :-

Hence, the three compounds are aromatic in nature.

The complex [Cr(NH³)₆]Cl₃ have Cr³⁺ ion.

Here two inner d-orbitals are vacant so it involves d²sp³ hybridisation as it involves (n − 1)d orbitals for hybridisation. It is an inner orbital complex.

Each end of double bond must contain different atoms (or) groups to show geometrical isomerism. There should be restricted rotation and planarity in the molecule to show geometrical isomerism. Geometrical isomers are named as cis and trans isomers.

This is a case of Hofmann reaction:
The Hofmann bromamide is the organic reaction of conversion of a primary amide to a primary amine with one fewer carbon atom.
CH₃CONH₂ + Br₂ + 4 NaOH→CH₃NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

Stability of +5 state decreases from top to bottom but because of high electronegativity and smaller size of fluorine bismuth can exist in this form.

The beta elimination reaction takes place in the reaction between an alkyl halide and alcoholic KOH. Only one type of beta hydrogen is available in the given alkyl halide. Hence, the product is

x = 2

There are eight covalent bonds in C₃O₂, so x = 8 and y = 3 as 3 C-atoms are sp-hybridised.
So x + y = 8 + 3 = 11.

Since the distance of the vertex from the origin is √2 and the focus is 2√2, here a = √2.

∴ V(1,1) and F(2,2), i.e., the axis lies on y = x.

Length of latus rectum = 4a = 4√2

By the definition of a parabola, PM² = (4a)(PN)

where PN is the length of the perpendicular upon line NV.

The equation of the line:
x + y - 2 = 0

⇒ (x - y)² / 2 = 4√2 × [(x + y - 2) / √2]

∴ (x - y)² = 8(x + y - 2)

Given, curve is

2x² + y² = 2x

2x² − 2x + y² = 0

which represents an ellipse.

Here, a = 1/2, b = 1/√2, h = 1/2, k = 0

Consider a point P(h + a cosθ, k + b sinθ)

= P(1/2 + (1/2) cosθ, (1/√2) sinθ) on the ellipse from which the distance of the point (a, 0) is maximum. Let Q(a, 0).

Now,

For maxima and minima, put dy / dθ = 0

⇒ - (1/2 - a) sinθ + (1/4) · 2 sinθ cosθ = 0

⇒ sinθ (- 1/2 + a + 1/2 cosθ) = 0

⇒ sinθ = 0 or - 1/2 + a + 1/2 cosθ = 0

⇒ θ = 0 or cosθ = 1 - 2a

⇒ sin²θ = 1 - cos²θ

= 1 - (1 - 2a)²

= -4a² + 4a

Now, d²y/dθ² < 0 for cosθ = 1 - 2a

Thus, the distance PQ is maximum when cosθ = 1 - 2a and sin²θ = -4a² + 4a

Now, the required longest distance is

Understanding the Function

Since the given function involves a limit, we analyze its behavior for different values of x. The denominator contains sin(a−x), which suggests that we may use the small-angle approximation or L'Hôpital’s rule if necessary.

After simplification (detailed steps not shown for brevity), the function can be rewritten in a solvable form.

Finding the Roots of f(x)⋅∣x²−1∣ = 1

Rearrange the equation:

Since ∣x² − 1∣ represents a transformed quadratic function, we solve the equation for different values of x. Through algebraic manipulation, solving for xxx results in 5 distinct roots.

Since a, b, c are in A.P., therefore,
b - a = d and c - b = d,
where d is the common difference of the A.P.
Thus,
a = b - d and c = b + d
Given, abc = 64
⇒ (b - d) * b * (b + d) = 64
⇒ b(b² - d²) = 64
But, b(b² - d²) ≤ b * b² (as d² ≥ 0 always)
⇒ b(b² - d²) ≤ b³
⇒ b³ ≥ 64
⇒ b ≥ 4
Hence, the minimum value of b is 4.

We have,



Centre of smallest circle is A
Centre of largest circle is B

Hence,

Hence,
a = 3, b = 2
a + b = 5

But

Hence d = 6.

The constant term is

= 1 + 60 + 360 + 160 = 581.

Clearly guests will stay either in '✓'  or in 'x'

Therefore, number of required ways = 2 × 4! = 48