SRMJEE Mock Test - 1 (Engineering) - JEE MCQ

According to Ohm's law:
V = IR
R =

Dimensional formula of the potential difference is = [M¹L²T^-3A^-1]
Dimensional formula of = = [M¹L³T^-3A^-2]
Or we can say that dimensional formula of conductivity σ = is [M^-1L^-3T³A²]

Let the forces be F₁, F₂, F₃, F₄ and F₅.
All the forces are of equal magnitude and are at equal angle with each other.
Using polygon rule of vector addition
We are getting a regular pentagon which is closed. Hence, the resultant of all the forces will be zero

In classical mechanics, when any charged particle will be in acceleration, it will radiate energy in form of electromagnetic waves and will lose its energy.
In Rutherford's model, electron revolves around the nucleus and will be in accelerated motion. So it will lose kinetic energy in form of electromagnetic waves and will cause decrease in radius.

Similarly, 

⇒ 3r + 24 = 4r + 16 ⇒ r = 8Ω

cannot be selected for the formation of desired product because substitution does not take place at the bridgehead carbon.
Hence, the Williamson synthesis for the given compound should be done by reaction between sodium salt of Bicyclo [2,2,2] octan-1-ol and benzyl iodide ( I- is a good leaving group).

The reactivity series of metals is:

It is clear that hydrogen is present above Cu in the reactivity series i.e., it is more reactive than Cu. So, Cu cannot displace hydrogen from dilute acids.

Due to resonance, the C-Cl has double bond characteristic and thus, it becomes difficult to break that bond. Hence, vinyl chloride will not react with phenol to give ethers

ΔP = 5.1 torr ;
Let, x is atomicity of sulphur.
Now,

or 

Schizophrenia is a type of psychosis characterised by alterations in thought, perception, emotions, language, sense of self, and behaviour.

Electron affinity is defined as "the energy released when an extra electron is added to a neutral gaseous atom." The increasing order of electron affinity is:

2s² 2p⁴ < 3s² 3p⁴ < 2s² 2p⁵ < 3s² 3p⁵

O < S < F < Cl

General electron affinity decreases with the increase in the size of an atom, as the nuclear attraction decreases down a group. The value of electron affinity increases as we move along a period since the size of atoms decreases in a period. Electron affinity of O and F are less than S and Cl respectively due to their very small size.

2A + B → product
[B] is doubled, half-life did not change.
Half-life is independent of change in concentration of reactant i.e., the first order.
First order w.r.t. B:

When [A] is doubled, the rate increased by two times.
First order w.r.t. A:
Hence, net order of reaction = 1 + 1 = 2
Unit for the rate constant = concentration⁽¹⁻ⁿ⁾t⁻¹
=(mol.L⁻¹)⁻¹ s⁻¹
= L mol⁻¹ s⁻¹

In triangle ABC, ∠A = 60°
We know that cos A =
cos 60° =
Since, cos 60° = 1/2


⇒ 6c = 18 + 2c² - 32
⇒ 2c² - 6c - 14 = 0
⇒ c² - 3c - 7 = 0

So, required equation is c² - 3c - 7 = 0.

(i) Since f(2 + 3i) = f(2 - 3i), f is many-one.
(ii) Since the modulus of a complex number can never be negative, range(f) ≠ R. So, f is into.

We have x = –ay.
Therefore,
–a²y + z = 0 ⇒ z = a²y
Thus, az + y = 0 ⇒  (a³ + 1)z = 0
For the system of equations to have infinite number of solutions, a³ + 1 = 0
⇒ a = –1 [∴ a is real]

Given: f(x) =
⇒ log f(x) = log (3^{1 + logx}) - log (x^{log 3})
⇒ log f(x) = (1 + log x) log 3 - log 3 log x = log 3
⇒ f(x) = 3, which is a constant
f(1994) = 3, which will also be a constant.

f(x) will be continuous at a point x = 0, if f(0−) = f(0+) = f(0).

Now, f(0) = a[0] = 0.

Now,

∴ f(x) is continuous at x = 0.

Now, at positive integral points, f(x) is discontinuous because [x] is discontinuous for all integral points.

So, f(x) is discontinuous at an infinite number of points.

Let the numbers be x₁, x₂, …, x₁₀ (N = 10).

Given:
x₁ + x₂ + x₃ + x₄ = 4 × 11 = 44
x₅ + x₆ + … + x₁₀ = 6 × 16 = 96
∑xᵢ = 140

Variance formula:
σ² = (∑xᵢ² / N) - (∑xᵢ / N)²

Substituting the given values:
σ² = (2000 / 10) - (140 / 10)²
σ² = 200 - 196 = 4

Standard deviation:
σ = 2

Differentiate both sides w.r.t. x using Newton-Leibnitz Rule

Given,

adj(adj A) = |A|ⁿ⁻² ⋅ A

|A| = 3(−3 + 4) + (−20) + 4(−2 − 0)
|A| = 3 + 6 − 8
|A| = 1

So, we get:

adj(adj A) = A

Hence, the trace is 1 (the sum of the diagonal elements).

Let y = (x − a)(x − c) / (x − b) = (x² − (a + c)x + ac) / (x − b)

Rearranging, we get:
y(x − b) = x² − (a + c)x + ac

Expanding further:
x² − (a + c + y)x + ac + by = 0 ----(1)

Since x is real, the discriminant of equation (1) must be non-negative:

(a + c + y)² − 4(ac + by) ≥ 0

Expanding:
y² + 2(a + c)y + (a + c)² − 4ac − 4by ≥ 0

Rearranging:
y² + 2(a + c − 2b)y + (a − c)² ≥ 0 ----(2)

Since y takes all real values, inequality (2) must hold for all y. This is possible only if:

4(a + c − 2b)² − 4(a − c)² < 0 (since the coefficient of y² is positive)

Factorizing:
(a + c − 2b + a − c)(a + c − 2b − a + c) < 0

Simplifying:
(2a − 2b)(2c − 2b) < 0

Dividing by 4:
(a − b)(c − b) < 0

Thus, b lies between a and c. One possible ordering is a < b < c.

Draw line parallel PX
PX ∥ RS

PX = RS = 12 + 3 = 15 cm

QX = QS − PR = 12 − 3 = 9 cm

PQ = √(15² − 9²) = √(225 − 81) = √144 = 12 cm

STBM is a square,
MB = BT = ST = SM = 12 cm

In ΔAPR and ΔASQ,
AP/AQ = PR/SQ
AP + PQ = 3 × 12 + 12 = 14
4AP = 12 + 12
3AP = 12
AP = 4 cm

Therefore, option B will be the correct option.

The line 'Multinational corporations could take the top micro-finance institutions to the next level...' suggests the answer.
Nothing about the management about the micro-finance institutions or the involvement of banks has been mentioned, so options 1 and 2 are negated. The line 'The poorest of the world's poor, who are predominantly in Asia and Africa get left out" negates option 3 also.