SRMJEE Mock Test - 3 (Engineering) - JEE MCQ

 (as ω will be same in both cases)



= 6
The correct answer is 6.

It is given that, 
We calculate the vector cross product by solving determinant,

By taking magnitude, we get,

As the magnitude of the vector is

Second order reactions can be defined as chemical reactions wherein the sum of the exponents in the corresponding rate law of the chemical reaction is equal to two. The rate of such a reaction can be written as r = k[A]².
The half life of the reaction is 1/k[A].

Manganate (MnO₄²⁻) and permanganate (MnO₄⁻) are tetrahedral due to sp³ hybridization of Mn, with π-bonding from p-orbitals of oxygen overlapping with d-orbitals of manganese, stabilizing the structure.

Iron oxide is basic impurity which is removed by adding acidic flux silica.
FeO + SiO₂ → FeSiO₃ (slag)

The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics. The rate of reaction depends upon the concentration of only one reactant, which is tertiary butyl bromide.
Rate ∝ [Reactant]¹

A: Reaction occurs through the SN1 mechanism.
True. The hydrolysis of t-butyl bromide (a tertiary alkyl halide) occurs via the SN1 mechanism. The tertiary carbocation formed in this mechanism is stable due to the inductive and hyperconjugative effects from the surrounding alkyl groups, making the SN1 pathway favored.

B: The intermediate formed is a carbocation.
True. In the SN1 mechanism, the first step is the loss of the leaving group (Br⁻), which results in the formation of a tertiary carbocation as the intermediate. This carbocation then reacts with the hydroxide ion (OH⁻) to form t-butanol.

C: Rate of the reaction doubles when the concentration of alkali is doubled.
False. In the SN1 mechanism, the rate-determining step is the formation of the carbocation, which depends only on the concentration of the alkyl halide (t-butyl bromide), not the hydroxide ion concentration. Therefore, doubling the concentration of NaOH will not affect the rate of the reaction significantly, as the reaction is independent of the concentration of the nucleophile (OH⁻).

D: Rate of the reaction doubles when the concentration of t-butyl bromide is doubled.
True. In an SN1 mechanism, the rate is directly proportional to the concentration of the substrate (t-butyl bromide). So, if the concentration of t-butyl bromide is doubled, the rate of the reaction will also double.

Conclusion: The incorrect statement is: C: Rate of the reaction doubles when the concentration of alkali is doubled.

Reason is false. Fully filled orbitals are more stable due to exchange energy. Cr ([Ar] 3d⁵ 4s¹) has 6 unpaired electrons.

Faraday's second law of electrolysis states that when the same quantity of electricity is passed through different electrolytes, the mass of different ions liberated is directly proportional to their equivalent weights.

Equivalent weight = Atomic mass / Valence factor

In the first cell at the cathode:
Fe²⁺ + 2e⁻ ⟶ Fe
Equivalent weight of Fe²⁺ (E_Fe)_₁ = 56 / 2

In the second cell at the cathode:
Fe³⁺ + 3e⁻ ⟶ Fe
Equivalent weight of Fe³⁺ (E_Fe)_₂ = 56 / 3

Hence, the ratio of the mass of Fe deposited in the first cell at the cathode to the mass of Fe deposited in the second cell at the cathode is:
Mass of Fe (first cell) / Mass of Fe (second cell) = (E_Fe)_₁ / (E_Fe)_₂ = 3/2

A metal complex consists of a central metal atom or ion that is bonded to one or more ligands, which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal.
Higher the charge and smaller the size of the ligand, more stable is the complex formed, as the ligand can approach the metal more closely. In case of neutral monodentate ligands, high dipole moment and small size favours the more stable complex.
A metal ion with a high charge and a small size forms more stable complexes as it can attract the ligands more closely. As the size will increase, the ionisation potential will decrease, which means that a greater ionisation potential favours the more stable complex.

Let r be the radius of the circle.
r² = (1 - 4)² + (2 - 6)²
r² = 25
Area = πr² = 25 π sq. units

Let the line be ax + by + c = 0.
According to the given condition
= 0
⇒ a + b + c = 0 (Same as when (x, y) is (1, 1) in the equation ax + by + c = 0)
We can say that the line passes through (1, 1).

Let E₁ be the event that a screw is drawn from box A, E₂ be the event that a screw is drawn from box B, and E₃ be the event that a screw is drawn from box C.
Then, P(E₁) = P(E₂) = P(E₃) = 1/3
Let D be the event that a defective screw is drawn.
Then, , and
Then, the probability that a defective screw came from box A is:

AB = √2 units

Similarly,  BC = √2 units and CA = √2 units

Hence, ΔABC  is an equilateral triangle.

Hence, the distance of the orthocentre O from the sides is equal to the distance of the incentre from the sides, that is, the inradius of the triangle.

Since, in an equilateral triangle, the orthocentre and incentre coincides.

According to question,
Given, Standard deviation = 8 and Mean = 10
Then, the coefficient of variation is
 = S.D.A.M. ×100
= 8/10 × 100
= 80
Hence, the correct answer is A.

Given, 

Also,  (Using Newton Leibnitz Theorem)

Now Let, 

Substituting 1 + x³ = t², then 3x²dx = 2tdt

Let OP = h

According to the diagram:
AO = BC, AB = OC = 100

PC = PO − OC = h − 100

In ΔAPO,
cotα = AO / PO ⇒ AO = h cotα

In ΔBPC,
cotβ = BC / PC ⇒ BC = (h − 100) cotβ

Since AO = BC,
h cotα = (h − 100) cotβ

Mean of observations = Mean of (4, 6, 8, 15, 2) + x
∴ Mean of (4, 6, 8, 15, 2) = = 35/5 = 7
Mean of observations = 7 + x
But, mean is given as 10.
∴ 7 + x = 10
x = 3

C.P. = Rs. 200

Profit = S.P. - C.P. = 280 - 200 = Rs. 80
Profit% = = 40%

Total number of cases = 20
Number divisible by 3 and 5 both = 15 (one only)
Therefore, Probability of card with a number divisible by 3 and 5 = 1/20
So, option 4 is correct.

In the passage, “sustenance” refers to the requirement for maintaining a continuous market for listed securities, as in: “A large number of shares ... is essential for the sustenance of a continuous market.” ‘Maintenance’ (preserving a condition) fits best, as public float ensures the market’s ongoing operation. ‘Subsistence’ implies bare minimum survival, ‘presence’ means existence, and ‘evidence’ means proof, none of which fit the context.

In the passage, “averse” describes promoters’ reluctance to share capital: “promoters ... are generally averse to giving a large share in the capital/benefits to the public.” ‘Hesitant’ (unwilling or reluctant) matches this meaning. ‘Agreeable’ is an antonym, ‘disputable’ means debatable, and ‘indebted’ means owing gratitude, none of which fit.