SRMJEE Mock Test - 2 (Engineering) - JEE MCQ

Since, the frequency n of the light does not change as light travels from air into glass, we have
v_a = nλ_a and v_g = nλ​​​​​​​_g
Therefore,

When the proton enters the region of the magnetic field, it will experience a force F given by: F = q (u B); where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it does not do any work. Hence, the magnitude of the velocity of the particle will remain unchanged; only the direction of the velocity changes. Hence, v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along positive x-axis and B is directed out of the page; the proton will move in a circle in the x-y plane in the clockwise direction. Hence, its y-coordinate will be negative, when it leaves the region. Thus, the correct choice is (4).

Ampere is the unit of current, so we can write it as Q/t
Hour is the unit of time, so we can write it as t.
Ampere-hour is = Q. Hence, ampere-hour is the unit of electric charge.

Initially the radius of the bubble is r. After reaching surface it becomes 2r. The atmospheric pressure is given as,
P_atm = P cm of water
So, the volume is changing in the air bubble but the temperature remains unchanged.
For isothermal process,
P₁V₁ = P₂V₂
Let the height of water surface be h.

⇒ h + P = 8P
⇒ h = 7P

The fringe width in Young’s double-slit experiment

β = λD / d

where λ is the wavelength of light used
D is the distance between slit and screen.
d is the distance between the slit.

∴ β' = λ(2D) / (d/2) = 4λD / d = 4β

According to Faraday's law,

We know, if there is a varying current in a coil, then the magnetic field associated with the loop would also be varying, thus an eddy current would be produced.
Again equation (1) clearly depicts that a change of magnetic flux through an area would cause eddy current.
Hence both assertion and reason are true and the reason is the correct explanation of the assertion.

A ball is dropped from a height H.

As we know, the change in momentum is impulse:
J = ΔP,
where J is impulse and ΔP is the change in momentum when a ball is dropped from a height H.

Then, its velocity just before it reaches the ground is:
Vi = √(2gH) = √(10 × 10) = 10 m/s.

Here, it loses 75% of its initial kinetic energy.
Therefore,
Kf = (1/4) K_i.

Using the kinetic energy formula KE = (1/2)m V², we get:
(1/2) m V_f² = (1/8) m V_i².

Substituting Vi = √(2gH), we get:
(1/2) m V_f² = (1/8) m (√(2gH))².

Given that:
H = 10 m,
g = 10 m/s²,

we get:
V_f = √5 m/s ≈ 2.24 m/s.

Change in momentum (ΔP) = P_f - P_i
= m(V_f) - m(-V_i)
= m(V_f + V_i).

Thus, impulse J = 1.05 N·s.

Let γ be the coefficient of volume expansion of the metal, and the coefficient of linear expansion is given by α = 2 × 10⁻⁵ K⁻¹.

The relation between the coefficient of volume expansion and the coefficient of linear expansion is given by:

γ = 3α = 3 × 2 × 10⁻⁵ = 6 × 10⁻⁵ K⁻¹.

Change in volume is given by:

ΔV/V = γΔT = 6 × 10⁻⁵ × (473 - 273) = 1.2 × 10⁻².

Percent change in volume is given by:

(ΔV/V) × 100 = 1.2 × 10⁻² × 100 = 1.2%.

Since the amine (X) reacts with benzenesulphonyl chloride and forms a product which is soluble in KOH, so amine (X) must be a primary amine.
Secondary amine forms a product which is insoluble in KOH.
Tertiary amine does not react with benzenesulphonyl chloride.

The first step is the Kolbe–Schmitt reaction or Kolbe process which is a carboxylation chemical reaction that proceeds by heating sodium phenoxide with carbon dioxide under pressure, then treating the product with sulfuric acid. The final product is an aromatic hydroxy acid which is also known as salicylic acid.
The second step involves formation of aspirin by reaction between salicylic acid and acetic anhydride in the acidic medium.

The catalytic hydrogenation of acid chloride to yield aldehyde is called Rosenmund reduction.

Clemmensen reduction: It involves reduction of carbonyl group in the presence of Zn/Hg/conc. HCI to form alkane.

Wolff-Kishner reduction:

Birch reduction:

Primary amine reacts with CS₂ and then HgCl₂ to form isothiocynate having pungent smell of mustard oil. Hence, it is called mustard oil reaction.2RNH₂ + S = C = S → S = C(SH)NHR

Here, R is alkyl part.
The pungent smell of the mustard oil is due to a sulphur containing compound.

Beryllium (Be) has a higher ΔiH (ionization enthalpy) than boron (B). In both cases, the electron to be removed belongs to the same principal shell.

In Be (Z = 4): 1s², 2s², the electron to be removed is a 2s electron.
In B (Z = 5): 1s², 2s², 2p¹, the electron to be removed is a 2p electron.
The 2s electron penetrates closer to the nucleus compared to the 2p electron. This means 2s electrons experience a stronger attraction from the nucleus than 2p electrons. As a result, a higher amount of energy is required to remove a 2s electron compared to a 2p electron. Thus, Be has a higher ionization enthalpy than B.

Ionization Enthalpy of Oxygen vs. Nitrogen and Fluorine
Oxygen (O) has a lower ionization enthalpy than nitrogen (N) and fluorine (F).

Electronic configurations:
Nitrogen (N, Z = 7): 1s², 2s², 2pₓ¹, 2pᵧ¹, 2p��¹
Oxygen (O, Z = 8): 1s², 2s², 2pₓ², 2pᵧ¹, 2p��¹
Fluorine (F, Z = 9): 1s², 2s², 2pₓ², 2pᵧ², 2p��¹
Across a period, ionization enthalpy generally increases from left to right due to the decrease in atomic size and increase in nuclear charge.

However, the ionization enthalpy of nitrogen is greater than that of oxygen. This is because nitrogen has a more stable half-filled p-orbital configuration (2p³), which provides extra stability. Removing an electron from this stable configuration requires more energy.

Thus, oxygen has a lower ionization enthalpy than nitrogen and fluorine.

Rate of S_N1 reaction depends on the stability of carbocation (intermediate)
in molecule P after losing Cl⁻ carbocation will form and that will be stabilized by resonance with a double bond and electron-donating group −OCH₃
in molecule Q also carbocation will be stabilized by resonance with a double bond and −OCH₃ group but resonance with both is not continuous hence Q will be less reactive then P.
In R molecule there will not be any substitution reaction because of partial double bond character in C-Cl bond due to resonance with lone pair of Cl and a double bond, hence least reactive.
Order of reactivity is P > Q > R

⇒ dy = xlogx dx
The above equation is of variable separable form.
So, integrating,

We have f(x) = .
Therefore, fof(x) = x ⇔ f (f (x)) = x

Obviously, d = -a satisfies this relation.

The first year students can be arranged among themselves in 3! ways, the second year students can be arranged among themselves in 2! ways.
For each arrangement of the students in their respective groups, the three groups can be arranged in 3! ways.
∴ Number of ways of arrangement = 3! × 2! × 1 × 3! = 72.

In equilateral triangle, 

Circumradius  = a/√ 3

Inradius of equilateral triangle = a/2√ 3

Ratio of circumradius and inradius = (a/√ 3)/(a/2√ 3) = 2/1

∴ Ratio of circumradius and inradius of equilateral triangle is 2 : 1

The given equation can be written as

but given  α/β + β/α = 4/5

or 

It is a quadratic equation in λ, let roots be λ₁ & λ₂⁡_.

then λ₁ + λ₂ = 16, λ₁λ₂ = 1

∴ 

= 256 - 2 = 254.

We have,

sin⁻¹(x) + sin⁻¹(y) = 2π/3.

Now, we know that

sin⁻¹(z) + cos⁻¹(z) = π/2, where z ∈ [−1, 1].

Therefore,

(π/2 − cos⁻¹(x)) + (π/2 − cos⁻¹(y)) = 2π/3.

⇒ π − (cos⁻¹(x) + cos⁻¹(y)) = 2π/3.

⇒ cos⁻¹(x) + cos⁻¹(y) = π − 2π/3 = π/3.

We know, that the sum of squares of the first n natural number is 

Given series is 2² + 4² + 6² + ⋯ + 20²

Hence, The correct answer is 1540.

The profit percentage is calculated using the formula:

Profit % = (SP - CP) / CP × 100

where SP is the total selling price and CP is the total cost price for the TV set as well as the DVD player.

Given Data:
Selling Price (SP) = ₹26,400
Cost Price (CP) = ₹18,000 + ₹4,000 = ₹22,000
Calculation:
Profit % = (26,400 - 22,000) / 22,000 × 100
= 4,400 / 22,000 × 100
= 20%

Conclusion: The profit percentage is 20%.

Kamini, Komal, Kavita, Kamya, Kashish, Kumkum, Kaveri, Kokila and Kamna are sitting at a round table, facing the center.

Komal is fourth to the right of Kaveri and second to the left of Kamini. Kashish is fourth to the left of Kamini and third to the right of Kavita. Kumkum is third to the right of Kamna. Kokila is second to the left of Kamya.

Here nine persons are sitting around a circle facing the centre.

While facing centre-right direction refers to the anticlockwise direction and left refers to Clockwise direction.

So, the final arrangement will be,

So, Kavita is the second to the right of Kamini.

Hence, this is the correct answer.

Out arrows are representing person facing outside and in arrows representing person facing inwards.

The person sitting between L and Z in anticlockwise direction with respect to Z. Anticlockwise direction is opposite to clockwise.

Therefore, Four persons are sitting between L and Z- W, Y, S, X.