JEE Main Mock Test - 18 - JEE MCQ

If a be the acceleration of the body, then the distance traveled by the body in 2 s is given as,

x₁ = ut + (1/2) at²
= 0 × 2 + (1/2) a × (2)²

x₁ = 2a

Velocity of the body at the end of 2 s is given as,

v = u + at
= 0 + a × 2

v = 2a

Distance traveled by the body in the next 2 s is given as,

x₂ = vt + (1/2) at²

= 2a × 2 + (1/2) a × (2)² [from Eq. (ii)]

= 4a + 2a

x₂ = 6a = 3 × 2a

⇒ x₂ = 3 × x₁ [from Eq. (i)]

⇒ x₂ = 3x₁

According to the problem, the mass of gases is equal, so the number of moles will not be equal, i.e., μ_A ≠ μ_B.

From the ideal gas equation P_V = μRT,

⇒ (P_AV_A) / μ_A = (P_BV_B) / μ

[As the temperature of the container is equal]

From this relation, it is clear that if P_A = P_B, then

V_A / V_B = μ_A / μ_B ≠ 1, i.e., V_A ≠ V_B

Similarly, if V_A = V_B, then P_A / P_B = μ_A / μ_B ≠ 1, i.e., P_A ≠ P_B

Dimension of torque = [M¹ L² T⁻²]
Dimension of angular momentum = [M¹ L² T⁻¹]
∴ Relation A−4, B−3 is right.

Given, v₁ = 14 m/s and v₂ = 28 m/s

As we know that, tanθ = v² / rg ⇒ r = v² / g tanθ

In the first case, r₁ = v₁² / g tanθ = (14)² / g tanθ  ... (i)

In the second case,

r₂ = (28)² / g tanθ  [for similar ramp remains same]

r₂ = (14 × 2)² / g tanθ = 4 (14)² / g tanθ ⇒ r₂ = 4r₁  [using Eq. (i)]

Hence, radius should be increased by factor 4.

Given Work Functions:
W_Na = 2.75 eV
W_Cu = 4.65 eV
W_Au = 51 eV
The frequency range of visible light is 4 × 10¹⁴ Hz to 8 × 10¹⁴ Hz, so we calculate the maximum and minimum energy of photons in visible light.

Using E = hν,

Maximum Photon Energy:
E_max = (6.63 × 10⁻³⁴ × 8 × 10¹⁴) / (1.6 × 10⁻¹⁹)
= 331 eV

Minimum Photon Energy:
Similarly,
E_min = (6.63 × 10⁻³⁴ × 4 × 10¹⁴) / (1.6 × 10⁻¹⁹)

Now, comparing the work functions of the mentioned metals with E_max, we see that only sodium can emit photoelectrons.

Further, a sodium device can also work with photons of energy higher than 2.75 eV, such as X-rays and UV rays.

[OA→ circular path]

[AB→ circular path]

[BC→ parabolic path]

From B to C,

Co-ordinate of pt. C

∴ x = 2 & y = 2 So 2 + 2 = 4

u_rel = 0 and a_rel = g + g / 4 = 5g / 4

(v_b / e)² = 2 * (5g / 4) * h₁ (just before collision)

v_b' / e = e * v_b (just after collision)

⇒ e² * (v_b / e)² = 2 * (5g / 4) * h₂
⇒ h₂ = e² * h₁

Meter Bridge works on the principle of balanced wheatstone bridge network.
Since J is mentioned as the Null point,

where,
R_AJ→Resistance of the part of wire of length(AJ+end correction),
R_JB→Resistance of the part of wire of length(JB+end correction),
X and R are the resistances of the resistors mentioned in the question,
As Resistance of a wire is directly proportional to the length of the wire, we can write,

From equations (i) and (ii), we can write,

Refer figure, A and B are the extreme position and O is the mean position.
For SHM, acceleration and force is always directed towards the mean position. It is given that the direction A to B is positive.

Given, power P = 2 × 10⁻³ W
Energy of a photon,
E = hν
= 6.6 × 10⁻³⁴ × 6 × 10¹⁴ J
Where h is Planck's constant.
Number of photons emitted per second is,
n = P / E
= (2 × 10⁻³) / (6.6 × 10⁻³⁴ × 6 × 10¹⁴)
≈ 5 × 10¹⁵

Acceleration
a = F/m = 4/20 = 1/5ms⁻²
(Displacement in n^th second is given by S_nth = u + a/2(2n − 1)
Distance covered by body in 3rd second
= 1/2 × 1/5 × (2 × 3 − 1) = 5/10 = 1/2m
∴  W = 4 × 1/2 = 2J

The constituents of nucleic acids are nitrogenous bases, sugar and phosphoric acid. The sugar present in DNA is D(−)−2-deoxyribose and the sugar present in RNA is D(−) ribose. Due to these D(−)-sugar components, DNA and RNA molecules are chiral molecules.

This product is highly stable conjugate di-ene system with benzene ring also so it is formed here as major product.

Br₂ undergoes disproportionation reaction. It participates in oxidation as well as reduction reaction in alkaline medium to give Br⁻ and BrO₃^-

Reduction half reaction:
Step 1: Balancing the charge and bromine.
Br₂  + 2e⁻  →  2Br⁻
Oxidation half reaction:
Br₂ → 2BrO₃^-
Step 1: Balancing O atoms by adding OH⁻ ions
Br₂ + 6OH⁻ → 2BrO₃^-
Step 2: Balancing H atoms by adding H₂O and balance oxygen atoms accordingly.

Step 3: Balancing the charge on both sides by adding electrons.
Br₂ + 12 OH⁻ →   2BrO₃^- + 6H₂O + 10e⁻
For this, the reduction half equation needs to be multiplied by 5, then added to the oxidation half. So that the electrons cancel out in both the reactions and the net reaction will be:

Dividing the equation by 2 to convert into a simple ratio:

Hydroboration - oxidation reaction of alkenes leads to anti-Markovnikov's hydration. Further addition of water adds in syn-manner, i.e., H and OH are added to the same face of the double bond leading to formation of trans-product. In short, hydroboration-oxidation of alkenes is regioselective as well as stereoselective.

P is sp³ hybridised in P₄. so, p character is 75% in it.

Star marked atoms are chiral centers.

Taking the abscissa of P as a, we get,

P(a, 2a/3)

Similarly, taking the abscissa of Q as b, we get,

Q(b, -2b/3)

It is given that the area of △OPQ = 5

Area of △OPQ = (1/2) × Absolute value of determinant

Solving for the determinant, we get:

Area of △OPQ = 5

4ab / 3 = ±10

So,

4ab = ±30 ...(i)

Now, let the midpoint of PQ be (h, k)

2h = a + b ...(ii)

And,

2k = (2a - 2b) / 3

a - b = 3k ...(iii)

Now,

4ab = (a + b)² - (a - b)²

Substituting the values of 4ab, a + b, and a - b from equations (i), (ii), and (iii), we get:

±30 = 4h² - 9k²

Hence, the required locus is given as:

4x² - 9y² = ±30

Given function:
f(x) = P e^(2x) + Q e^x + R x

Differentiation:
f'(x) = 2P e^(2x) + Q e^x + R

Given:
31 = 2P e^{² ln 2} + Q e ^{ln 2} + R
8P + 2Q + R = 31 ...(i)

Also, -1 = P + Q ...(ii)

15P + 6Q = 39 ...(iii)

Solving (i), (ii), and (iii), we get P = 5, Q = -6, R = 3.

General term of (a + b/x²)¹¹ is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ (ax)¹¹⁻ʳ × (b/x²)ʳ

⇒ T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ a¹¹⁻ʳ bʳ x¹¹⁻³ʳ

For coefficient of x⁻⁷,

11 - 3r = -7
⇒ 3r = 18
⇒ r = 6

So,

T₍₆₊₁₎ = C₆¹¹ a¹¹⁻⁶ b⁶ x⁻⁷

T₍ᵣ₊₁₎ = (C₆¹¹ a⁵ b⁶) x⁻⁷

So, L = C₆¹¹ a⁵ b⁶

Similarly, the general term of (bx² + a/x)¹¹ is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ (bx²)¹¹⁻ʳ × (a/x)ʳ

⇒ T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ b¹¹⁻ʳ × aʳ × x⁽²²⁻³ʳ⁾

So, for the coefficient of x⁷,

22 - 3r = 7 ⇒ r = 5

Hence,

M = C₅¹¹ b⁶ a⁵

So, L + M = 2 × C₆¹¹ a⁵ b⁶ = 924a⁵b⁶

And, the general term of (ax² + b/x)¹² is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹² (ax²)¹²⁻ʳ × (b/x)ʳ

T₍ᵣ₊₁₎ = C₍ᵣ₎¹² a¹²⁻ʳ bʳ x⁽²⁴⁻³ʳ⁾

For the coefficient of x⁶,

24 - 3r = 6 ⇒ r = 6

So, the coefficient of x⁶ is

= C₆¹² a⁶ b⁶ = 924 a⁶ b⁶

Hence,

L + M = (1/a) × [Coefficient of x⁶ in (ax² + b/x)¹²]

Setting x = π/4 − t in the first integral and x = π/4 + t in the second integral,

We have,

Given vectors:

ā = 2î + k̂

b̄ = î + ĵ + k̂

c̄ = 4î - 3ĵ + 7k̂

Calculations:

ā · c̄ = (2î + k̂) · (4î - 3ĵ + 7k̂) = 15

ā · b̄ = (2î + k̂) · (î + ĵ + k̂) = 3

Solving for r̄:

r̄ × b̄ = c̄ × b̄

⇒ ā × (r̄ × b̄) = ā × (c̄ × b̄)

⇒ (ā · b̄) r̄ - (ā · r̄) b̄ = (ā · b̄) c̄ - (ā · c̄) b̄

⇒ (ā · b̄) r̄ = (ā · b̄) c̄ - (ā · c̄) b̄

[Since (ā · r̄) = 0]

⇒ (ā · b̄) r̄ = (ā · b̄) c̄ - (ā · c̄) b̄

⇒ 3r̄ = 3c̄ - 15b̄

⇒ r̄ = c̄ - 5b̄

⇒ r̄ = (4î - 3ĵ + 7k̂) - 5(î + ĵ + k̂)

⇒ r̄ = (-î - 8ĵ + 2k̂)

Then,

r̄ · b̄ = (-î - 8ĵ + 2k̂) · (î + ĵ + k̂)

= -1 - 8 + 2 = -7

f'(x) < 0

∴ f(x³ + f(x)) ≤ f(-f(x) - x³)

⇒ x³ + f(x) ≥ -f(x) - x³

⇒ f(x) + x³ ≥ 0

⇒ 3x² + 6x - 1 ≤ 0

As x ∈ Z, x ∈ {-2, -1, 0}

The number of ways they can be arranged in a row is 9 / !2!3!4! = 1260
The three colour blocks of balls can be arranged in 3! = 6 ways.
∴ The desired number of ways is 1260 − 6 = 1254.

Let A(0, -3), B(-2√3, 3), and C(2√3, 3)
Let, D, E & F are the mid-point of sides AB, BC and CA respectively.

∴ ΔDEF is an equilateral triangle.

The equation of line passing through the point (−3,2,3) and parallel to a line with direction ratios 3,3,−1 will be,

Now let any point on the line will be, M(3λ −3, 3λ + 2,3 − λ)
We know the direction ratios of the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by
(x₂ − x₁, y₂ − y₁, z₂ − z₁)
Therefore, the direction ratios of line PM,
D.R of PM(3λ − 7, 3λ − 4, 5−λ)
Since, PM is perpendicular to the line
⇒  3(3λ − 7) + 3(3λ − 4) − 1(5 − λ) = 0
⇒  λ = 2
⇒  M(3, 8, 1)
Now distance of point P from point M,

Given equation of hyperbola is

Here, a² = cos²α and b² = sin²α
(i.e., comparing with standard equation )
We know, foci = (±ae, 0) where

⇒Foci = (±1, 0)
whereas vertices are (±cos α, 0)
⇒ae = 1
⇒Eccentricity, e = 1/cosα
Hence, foci remains constant with change in 'α'.

Here is the corrected text without LaTeX:
Set S = {1, 2, 3, ..., 12}
Given that A ∪ B ∪ C = S and A ∩ B = B ∩ C = A ∩ C = ϕ
Set S is partitioned into 3 equal parts A, B, C, each containing 4 elements.
Thus, the number of ways to partition is:
¹²C₄ × ⁸C₄ × ⁴C₄ = (12! / (4! 8!)) × (8! / (4! 4!)) × (4! / (4! 0!)) = 12! / (4!)³.

⇒ 4λ = 1