VITEEE PCME Mock Test - 2 - JEE MCQ

Family of circle touching a given line at a given point
(x - 1)² + (y - 1)² + λ(x - y) = 0
This circle passes through point (1, -3)
So, 0 + 16 + λ(1 + 3) = 0 ⇒ λ = -4
So, required circle
x² + 1 - 2x + y² + 1 - 2y - 4x + 4y = 0
x² + y² - 6x + 2y + 2 = 0

Given, x = sin t, y = cos pt 




⇒ (1−x²) y₂ − xy₁ + p²y = 0

In the standard form of an ellipse x²/a² + y²/b_² = 1
Focal distances are,
PS = a(1 + ecosθ)
PS = a(1 − ecosθ)
Hence in the standard form of an ellipse sum of the focal distances of a point is 2a.

Given,
x = t² + 2t - 1 …(i)
y = 3t + 5 ⇒ t = (y - 5)/3 …(ii)
On putting the value of t in Eq. (i), we get

This is an equation of a parabola.

Let 
  exist only when 
So, the degree of numerator should be greater than or equal to the degree of denominator for the limit to be zero.
Therefore, a₀ = a₁ = 0

Here, limit should be of the form 0/0
So, 1 + a₂ = 0

⇒ a₃ = 2
Hence, f(x) = 2x³ − x² = −x² + 2x³

Given:

4y³ - 8a²yx² - 3ay²x + 8x³ = 0

This is a homogeneous equation of degree 3, hence it represents three lines passing through the origin.

Now, to find the slopes of these lines, divide the equation by x³:

⇒ 4(y/x)³ - 3a(y/x)² - 8a²(y/x) + 8 = 0
⇒ 4(m)³ - 3a(m)² - 8a²(m) + 8 = 0

Let the roots of the cubic equation be m₁, m₂, and m₃.

Therefore, m₁m₂m₃ = -8/4 = -2.

Given m₁m₂ = -1, we find m₃ = 2.

Now substitute m = 2 into the equation:

⇒ 4(2)³ - 3a(2)² - 8a²(2) + 8 = 0
⇒ 4a² + 3a - 10 = 0

Solving the quadratic equation:

⇒ 4a² + 3a - 10 = 0

The sum of the possible values of a is -3/4.

Given, B = A⁻¹A′

Post-multiplying both sides by A, we get:
BA = A⁻¹A′A

Since AA′ = A′A, we can write:
BA = A⁻¹AA′

Simplifying:
BA = A′

Taking the transpose on both sides:
(BA)′ = (A′)′

This gives:
A′B′ = A

Pre-multiplying by A⁻¹, we get:
A⁻¹A′B′ = A⁻¹A

Since A⁻¹A = I, it follows that:
BB′ = I

Thus, the correct answer is D) I.

∴ |z| = 1

In Linear Programming Problems (LPP), the objective function (like maximizing profit or minimizing cost) is subject to a set of linear constraints (inequalities).
The feasible region is the area (usually a polygon) that satisfies all constraints.
According to the Fundamental Theorem of Linear Programming, the maximum or minimum value of the objective function occurs at one of the vertices (corner points) of the feasible region.
This means:
You don't need to check all points, just evaluate the objective function at the vertices of the feasible region.

So, the value is maximum at:
A vertex of the feasible region (not necessarily the farthest or center).

If a quadratic equation has imaginary roots, it implies that the quadratic expression is always positive or always negative.

Given the quadratic function:
f(x) = ax² − bx + 1

At x = 0, we get f(0) = 1.

Since f(0) > 0, it means f(x) > 0 for all x ∈ R (i.e., the quadratic expression is always positive).

Now, evaluating at x = -1:
f(-1) > 0

This simplifies to:
a + b + 1 > 0

h₁ : h₂ : h₃ = (1 + k₁) : (1 + k₂) : (1 + k₃)
h₁ : h₂ : h₃ = (1 + 1) : (1 + 1/2) : (1 + 2/5)
h₁ : h₂ : h₃ = 2 : 3/2 : 7/5
h₁ : h₂ : h₃ = 20 : 15 : 14

Energy received from the sun
= 2 cal cm^–2 (min)^–1
= 8.4 J cm^–2 (min)^–1
Energy of 1 photon received from sun,
= 3.6 10^–19 J
∴ Number of photons reaching the earth per cm² per minute will be


= 2.33 × 10¹⁹ photons / cm² / min

Resistance R = ΔV/Δi
Negative resistance occurs when current decreases as voltage increases. Region AB shows this behavior.

When charged body is at rest or static position, it produces electric field and when it is in motion, it produces both electric and magnetic fields. Thus, any oscillating charged body will produce electromagnetic radiations.

Let us name the different points in the question as A, B, D, E, F, and G.

Now, we can clearly see that points A and E are connected through the conducting wire, so we can join them because they have the same potential. We will call the new point A only.

Similarly, points B and F have the same potential. So, we join them and call the new point B only (as shown in the diagram below).

Now, we observe that the two capacitors between points A and G are in parallel, so we find the equivalent capacitance of these two capacitors. Similarly, the two capacitors between points B and D are in parallel, so we find the equivalent capacitance of these two capacitors.

C_eq = C + C = 2C (between A and G)
C_eq = C + C = 2C (between B and D)

Now, in the limb AGB, there are two capacitors in series, so the equivalent capacitance can be written as:
1/C_eq = 1/C + 1/2C ⇒ C_eq = 2C/3

Similarly, in the limb ADB, the capacitors are in series, so we find the equivalent capacitance:
1/C_eq = 1/C + 1/2C ⇒ C_eq = 2C/3

Now, it is clearly seen that the two capacitors, each having a capacitance of 2C/3, are in parallel. The final equivalent capacitance between points A and B is:
C_eq = 2C/3 + 2C/3 = 4C/3.

Fuse wire : It is used in a circuit to control the maximum current flowing in circuit. It is a thin wire having high resistance and is made up of a material with low melting point.

n the presence of friction, acceleration, a = (gsinθ − μgcosθ)
∴ Time taken to slide down the incline,

In the absence of friction, time taken to slide down the incline, 
According to the question, t₁ = 2t₂

When a charged particle q is moving in a uniform magnetic field with velocity such that angle between and be θ,
Then the force experienced by the charge particle is
F = qvB sin θ
F = 0 for θ = 0° or 180°
Since force on charged particle is non-zero, so angle between and so θ van have any value other than zero and 180°.

In NaCl, Cl^- forms the FCC lattice and Na⁺ is present in the octahedral void.
Therefore, along each edge, there are Cl^- ions at the corners and a Na⁺ ion at the edge centre.
Hence, edge length (a) = 2(r of Na⁺ + r of Cl^-) = 2y.

1. Cr³⁺ + e⁻ → Cr²⁺, E° = –0.41 V

• Negative E° value ⇒ Cr³⁺ is not easily reduced to Cr²⁺
• So, Cr³⁺ is more stable than Cr²⁺

So, A: Cr³⁺ is less stable than Cr²⁺ is ❌ incorrect

2. Mn³⁺ + e⁻ → Mn²⁺, E° = +1.51 V

• High positive value ⇒ Mn³⁺ is easily reduced, so it readily gains electrons
• Therefore, Mn³⁺ acts as a strong oxidizing agent
  B: Mn³⁺ is a good oxidizing agent is correct

3. Cr²⁺ → Cr³⁺ + e⁻, E° = +0.41 V (reverse of reduction)

• Since Cr²⁺ is easily oxidized, it acts as a good reducing agent, not oxidizing
  So, C: Cr²⁺ is a good oxidizing agent is false

4. Mn³⁺ → Mn²⁺, as E° is positive, the reverse reaction is less favorable
⇒ Mn²⁺ is more stable than Mn³⁺
So, D: Mn³⁺ is more stable than Mn²⁺ is false

When phenol (C₆H₅OH) is heated with ammonia (NH₃) under high pressure and in the presence of ZnCl₂, a reduction and amination process occurs, where the –OH group is replaced by an –NH₂ group on the benzene ring.

This gives:
C₆H₅NH₂ → Aniline
This process is known as reductive amination of phenol.

Since the secondary alcohol on oxidation gives two different acids, acetic acid and propanoic acid, the alcohol is secondary alcohol and will contain 5 C atoms.

Transition elements due to similar (almost) sizes exhibit both vertical and horizontal similarities. Same electronic configuration can also be a reason for vertical similarities in these elements.

In given graph
 has slope > 0
Now, ΔH = −nRTlnk_eq


⇒ ΔH < 0
∴ exothermic reaction

Thus, they differ in magnetic moment

There are no unpaired electrons at nickel.
Thus, both A and B are correct

Aniline on reaction with sulphuric acid give sulphanilic acid as shown below.

Sulphanilic acid gives 2,4,6-tribromophenol with bromine water. The reaction is shown below,

^ΔU - state function
Q - Path function
^ΔU = Q + W
Since ΔU_a = ΔU_b, and W_A < W_B (more negative), Q_a ​< Q_b​.
So, ΔU_a​ = ΔU_b​,Q_a​ < Q_b

Let's suppose that Pran and Khan start from point A.

At origin (0,0): Pran walks west 10 km to P(-10, 0).

Khan walks east 10 km to (10, 0), turns right (south), walks 15 km to Q(10, -15).
Distance between them = = = = 25 km
Hence, they are 25 km away from each other.
So, 4^th option is the answer.
Correct answer: D) 25 km

The numbers containing exactly one 7 in between 200 to 400 are:
207, 217, 227, 237, 247, 257, 267, 270, 271, 272, 273, 274,
275, 276, 278, 279, 287, 297, 307, 317, 327, 337, 347, 357,
367, 370, 371, 372, 373, 374, 375, 376, 378, 379, 387 and 397.
For every 10 numbers, there would be 1 number containing exactly one 7, and in 270's and 370's each, there would be 9 such numbers.
Hence, the total numbers containing the digit 7 exactly once are 36.