SRMJEE Mock Test - 6 (Engineering) - JEE MCQ

Let initial height of the body from surface = R
So, initial PE, u₁ = -
Final height of the body from surface, R + 3R = 4R = h₂
Where (GM_e = gR²)
So, final gravitational PE, u₂ = = =
Change in gravitational PE = u₂ – u₁
=

Magnetic flux ^φ = BA ...(i)
and also F = ilB or B = F/il...(ii)
Now, from Eqs. (i) and (ii)
=
Dimensions of
=
= [ML² T^-2 I^-1]

The initial momentum of parent nucleus is zero.
Hence, the momenta of the emitted α-particle and of the daughter nucleus will be equal (and opposite) to each other (momentum-conservation), i.e., p_α = -p_α

Kinetic energy of α–particle
= 6.7 MeV (Given)

Kinetic energy or recoil energy of daughter nucleus 

∴ Recoil energy of daughter nucleus = 0.123 MeV

A photocell employs photoelectric effect to convert change in the intensity of illumination into a change in photoelectric current.
When light of suitable frequency illuminates a metal surface, electrons are emitted from the metal surface. These photo(light)-generated electrons are called photoelectrons. The emission of electrons causes flow of electric current in the circuit.
In the photo-cell, the collector is maintained at a positive potential with respect to emitter, so that electrons ejected from it are attracted towards the collector. Keeping the frequency of the incident radiation and the potential fixed, the intensity of light is varied and the resulting photoelectric current is measured each time. It is found that the photocurrent increases linearly with intensity of incident light. The photocurrent is directly proportional to the number of photoelectrons emitted per second.

Since the shape of the fringes depends on the source of light (shape) or that of the slits. Therefore, the shape of the interference fringes will be circular as the source is a point size.

Let the temperature of common interface be T℃ Rate of heat flow

In steady state, the rate of heat flow should be same in whole system ie,

Hence, heat flow from composite slab is

[from Eq. (i)]

Accordingly, 

By comparing Eqs. (ii) and (iii), we get

⇒ f = 1/3

For an LR circuit, we know:

tan ϕ = X_L / R

where:

• ϕ = Phase angle between current and emf
• X_L = Inductive reactance
• R = Resistance

According to the question:

ϕ = 45° ⇒ tan 45° = 1

We know that X_L = ωL, so:

X_L = R ⇒ ωL = R

Since ω = 2πf, we have:

R = 2πνL = 2 × π × 50 × 0.2 = 62.8 Ω

HBr reacts the fastest with 2-methyl propan-2-ol due to the formation of stable tertiary carbocation.

^{_{The ligands having vacant π type orbitals have a tendency to receive back donated π electrons. Thus, these are called π acid ligands or π acceptor ligands. For example, CO and CN have lone pairs as well as π orbitals which take part in the formation of π bond with central metal atom as observed in carbonyls.}}

Cell reaction is Mg(s) + S_n²⁺ → Mg²⁺ + S_n(s)

= 2.23 V

All other are water soluble.

Phenol ca be converted salicylaldehyde in the presence of chloroform and KOH. This reaction is known as Reimer-Tiemann reaction. In this reaction, the mixture chloroform and KOH gives dichlorocarbene, which acts as an electrophile in this reaction.

We have . = . = . = 0
Let = = = a
= a² + b² + c² + 2 ( . + . + . )
= a² + b² + c² = 3a²
( + + ) . = cosθ
a² + . + . =a² cos θ
a² = a² cosθ
cosθ =
θ = cos^-1

Required area =
=
=
= ^79/24

The two given curves are y = x + 2 and y = x².

To find their points of intersection, we equate y:

x² = x + 2

⇒ x² - x - 2 = 0

⇒ (x - 2)(x + 1) = 0

⇒ x = 2 or x = -1.

The graphs of the curves are given below.

Required area is the area of the shaded portion.

Given:

In question it is given,

3A = 2B

From the equation (1):
3(x + 4y) = 2(y − 2x + 2)
⇒ 3x + 12y = 2y − 4x + 4
⇒ 3x + 4x + 12y − 2y = 4
⇒ 7x + 10y = 4
⇒ x = (4 − 10y) / 7

From the equation (2):
3(2x + y + 1) = 2(2x − 3y − 1)
⇒ 6x + 3y + 3 = 4x − 6y − 2
⇒ 6x − 4x + 3y + 6y = −2 − 3
⇒ 2x + 9y = −5

Substituting x = (4 − 10y) / 7 in 2x + 9y = −5:
2(4 − 10y) / 7 + 9y = −5
⇒ (8 − 20y) / 7 + 9y = −5
⇒ 8 − 20y + 63y = −35
⇒ 43y = −43
⇒ y = −1

Now, x = (4 − 10(−1)) / 7 = (4 + 10) / 7 = 2

Hence, x = 2, y = −1.

Let the point whose locus is required be P(h, k).

Then, the equation of the normal to the hyperbola can be given as,

Also, the equation of the normal at any point R(asecϕ, btanϕ) on the hyperbola can be given as,

On comparing the equations (i) and (ii) we get,

After eliminating the parameter ϕ, we get,

Since α and β are the roots of the equation 2x² - 3x - 6 = 0, then:

α + β = 3/2 and αβ = -3

We know that:

(α − β)² = (α + β)² − 4αβ

Substituting the values:

(α − β)² = 57/4

Given: AP:PB = 2:1

Let's assume AP = 2r and BP = r

Let the chord AB make an angle θ with the positive x-axis

Using the parametric equation of a line, we get:

A = (1 - 2r cosθ, 1 - 2r sinθ)
B = (1 + r cosθ, 1 + r sinθ)

Since A lies on the circle x² + y² = 4,
(1 - 2r cosθ)² + (1 - 2r sinθ)² = 4 ...(i)

Similarly, since B lies on the circle x² + y² = 4,
(1 + r cosθ)² + (1 + r sinθ)² = 4 ...(ii)

Solving equations (i) and (ii), we get:
|r| = 1

Let cos⁻¹(1/8) = θ, where 0 < θ < π/2.

⇒ (1/2) cos⁻¹(1/8) = θ/2

⇒ cos((1/2) cos⁻¹(1/8)) = cos(θ/2)

Now, cos⁻¹(1/8) = θ ⇒ cosθ = 1/8

⇒ 2cos²(θ/2) - 1 = 1/8

⇒ cos²(θ/2) = 9/16 ⇒ cos(θ/2) = 3/4

[∵ 0 < θ/2 < π/4, so cos(θ/2) ≠ -3/4]

Using parametric coordinates of the parabola, A (0, 2), B = (t₁² - 4, t₁), and C = (t₂ - 4, t), we have ∠CBA = π/2. Therefore, the product of the slopes of AB and BC = -1.

This leads to the equation:
(2 - t₁)/(4 - t₁²) * (t₁ - t)/(t₁² - t₂) = -1

Simplifying this, we get:
12 + t₁ * (t + t₁) = -1
t₁² + (2 + t) t₁ + (2t + 1) = 0
We get a quadratic equation in t₁.

For real roots, the discriminant D ≥ 0:
(2 + t)² - 4(2t + 1) ≥ 0
t² - 4t ≥ 0
t ∈ (-∞, 0] ∪ [4, ∞)

In ΔBDC,
BD = DC
So, ∠BDC = 180° - (26° + 26°) = 128°
In cyclic quadrilateral ABDC,
∠BAC = 180° - 128° = 52°
In ΔABC,
AB = AC
So, ∠ACB = = 64°
So, ∠ABE = ∠ACD = ∠ACB + ∠BCD
∠ABE = 64° + 26° = 90°

According to the given information in the question, diagram can be drawn as,

B is the mother of H. C is the nephew of E. A is the husband of H. A is third to the left of H. Both the neighbours of C are females. F and G are daughters of H.
Thus, option D is the correct answer.

Let f(x) = (ax² + bx + c)(ax² - dx - c)

⇒ D₁ = b² - 4ac
⇒ D₂ = d² + 4ac

⇒ D₁ + D₂ = b² - 4ac + d² + 4ac

∴ D₁ + D₂ ≥ 0

Therefore, at least one of D₁ or D₂ is positive.

Hence, the polynomial has at least two real roots.

According to the passage, inflation will directly affect the plans which had been postponed, as the cost will be more for the same goods in future. Thus, 'postponed' is the most suitable option.

According to the passage, 'ramp up' is used to talk about the increase in production. Thus, the opposite will be to decrease gradually. Thus, 'dampen', which means making less, will be the most suitable answer.
'Dole out' means to give or deliver in small portions.
Other options are rather synonymous to 'ramp up'.