JEE Main Mock Test - 19 - JEE MCQ

Statement (A): When the temperature increases the viscosity of gases increases and the viscosity of liquids decreases.
Statement (B): Water does not wet an oily glass because cohesive force of oil is less than that of water.
Statement (C) : A liquid will wet a surface of a solid if the angle of contact is greater than 90°.

N = m(g − a),N < mg if a(↓) and N > mg if a (↑)
Reading of spring balance is less than m if a(↓) and reading of spring balance is

greater than m if a(↑)

Excess pressure for a liquid drop is given by 

P_inside - P_outside = T / R₁ + T / R₂ 

i.e., Excess pressure = T / R₁ + T / R₂

where R₁ and R₂ are radii of curvatures of the drop in two mutually perpendicular directions. 
In this case, R₁ = R and R₂ = ∞

∴ Excess pressure = T / R + T / ∞ = T / R

Dimensions of E = [ML² T⁻²]
Dimensions of G = [M⁻¹ L³ T⁻²]
Dimensions of I = [MLT⁻¹]
And dimension of M = [M]
∴ Dimensions of GIM² / E² = [M⁻¹ L³ T⁻²] [MLT⁻¹] [M²] / [ML² T⁻²]²
= [T]
= Dimensions of time

Given, MI = 2.5 kg m⁻²

W = 40 rad s⁻¹

T = 10 Nm

As T = Iα

10 = 2.5α

α = 4 rad s⁻²

Now, ω = ω₀ + αt

60 = 40 + 4 × t

20 = 4t

t = 5 s

PR = d
∴ PO = d sec θ
and CO = PO cos 2θ = d sec θ cos 2θ
Path difference between the two rays is,
Δx = PO + OC = (d sec θ + d sec θ cos 2θ)
Phase difference between the two rays is
Δϕ = π (one is reflected, while another is direct)
Therefore, the condition for constructive interference should be,

The electric field at the point P due to the entire cylinder is

The electric field at the P due to the spherical cavity is

E_net = E₁ − E₂

∴  K = 16

Given, velocity, v = 6t − 3t²
As we know that,
v = dx / dt
Here, x is the displacement of the particle.
Now, dx = vdt
Integrate on the both sides, limit t = 0 to t = 2, we get

Average velocity, v_avg = Total displacement / Total time taken 

= 4/2 = 2 m/s

Hence, the average velocity of the object between t=0 to t=2 s is 2 m/s.

The forward biased resistance of a diode is

The major product of

In acidic medium KMnO₄ undergoes redox reaction with H₂C₂O₄ and HNO₂ and converted into Mn²⁺.

P_Total = P_HNO₃ + P_NO₂ + P_H₂O + P_O₂

∴ P_NO₂ = 4P_O₂ & P_H₂O = 2P_O₂

∴ P_Total= P_HNO₃ + 7P_O₂

⇒ 30 - 2 = P_O₂ × 7 ⇒ P_O₂ = 4

ΔT_f = K_f × (w₁ × 1000) / (w₂ × m₁)

w₁ = weight of solute  
w₂ = weight of solvent  
m₁ = molar mass of solute  
K_f = 4.3 K kg mol⁻¹  

Now,  
ΔT_f = 4.3 × (2 × 1000) / (57.3 × 500)  

= 17.2 / 57.3 = 0.30 K

CO₃²⁻ → sp² → %s = 33.33%

XeF₄ → sp³d² → %s = 16.67%

I₃⁻ → sp³d → %s = 20%

NCl₃ → sp³ → %s = 25%

BeCl₂ → sp → %s = 50%

The liquid phase separates that are vaporized a water vapour carries compounds, and they are placed in a condensation flask that is placed nearby. Now distillation takes place at a lower temperature. Steam distillation can be applied in case if the substance is very sensitive to heat. Vapours are condensed after distillation process.

A graph plotted between  log k versus 1/T for calculating activation energy is shown as:

From the Arrhenius equation

Comparing with the equation of straight line, y = mx + c, we get a straight line having positive intercept on y-axis and negative slope.

k = (2.303 / t) log (P₀ / (2P₀ - Pₜ))

k = (2.303 / 100) log (2 × 0.5 / 0.5 - 0.6)

k = 2.23 × 10⁻³ /s

Rate = kP_SO₂Cl₂

The total pressure is 0.65 atm.

P_SO₂Cl₂ = 2P₀ - Pₜ = 2 × 0.5 - 0.65 = 0.35 atm

Hence, rate = 2.23 × 10⁻³ × 0.35

Rate = 7.805 × 10⁻⁴ atm/s

The sum of partial pressures is given by:
Σ_{partial pressures} = p₀ − x + 2x + x + 28(Vapor pressure of moisture)
= p₀ + 2x + 28 = 188
⇒p₀ + 2x = 160 mm Hg   (i)
After a long time, when P_A = 0, we have:
P_B = 2P₀ and P_C = P₀
So,

From equation (i) & (ii), we get,
∴  x = 20 mm Hg

∴ adj(adjA) = |A|A (by property)
A adj(adj A) = |A|A²… (i)

= 2(5 − 12) −1(3 − 2) −1(18 − 5)

= −14 − 1 − 13 = −28

tr(A(adj(adjA))) = tr(|A|A²) = −28[6 + 40 + 12]

= −28 × 58 = −1624

Here, 

Hence, f(x) is not differentiable at x=−3,−1,0,1,3
⇒S = {−3,−1,0,1,3}

Point on the line = (2t, 2 + 3t, 3 + 4t)

Equating to B(q, 5, 7), we get t = 1 and hence q = 2

∴ B = (2, 5, 7)

D.R.s of AB are (p - 1, -6, 4)

AB is perpendicular to the line ⇒ (p - 1)(2) + (-6)(3) + (4)(4) = 0

∴ p = 2 and p - q = 0

∴ (x, y) ∈ R ⇒ xʸ = yˣ

∴ (x, x) ∈ R as xˣ = xˣ ∀ x ∈ I - {0}

∴ R is reflexive

Now, (x, y) ∈ R ⇒ xʸ = yˣ ⇒ yˣ = xʸ

⇒ (y, x) ∈ R ∴ R is symmetric

Now, (x, y) ∈ R ⇒ xʸ = yˣ and (y, z) ∈ R ⇒ yᶻ = zʸ

∴ xʸ = yˣ ⇒ y = xʸ / x and yᶻ = zʸ

⇒ yᶻ = zʸ ⇒ (xʸ / x) ᶻ = zʸ

⇒ xʸᶻ = z ʸᶻ / x = zˣ

⇒ xʸᶻ ≠ zˣ² ⇒ (x, z) ∉ R

R is not transitive.

If AB = BA  ⇒a = b
Hence AB = BA is possible for infinitely many B′s.

Given point A(27, −243, 81) in space, now images of point A in xyh plane, yzh plane and zx-planes are B (27, −243, −81), C(−27, −243, 81) and D(27, 243, 81) respectively.
Hence centroid of the triangle BCD is = (9, −81, 27) ≡ (α, β, γ)
hence, α + β + γ= −45

Given, liquid evaporates at a rate proportional to its surface are.

We know, volume of cone = 13πr²h

The surface area is given by:
S = πr²
The volume is given by:
V = (1/3) πr²h
Also, we have:
tanθ = R / H
and r / h = tanθ
From equations (2) and (3), we get:
V = (1/3) πr³ cotθ
and S = πr³
On substituting Eq. (4) into Eq. (1), we get:

∴ Required time after which the cone is empty, T = H/k

Since, f(x) is continuous at x = 0
Therefore, 

⇒ a + b = k 
Hence, k = a + b.

Let  be the unit vector  which is coplanar to vectors 2î + ĵ + k̂ and î − ĵ + k̂.

Since  is orthogonal to 5î + 2ĵ + 6k̂ so,

Using (1) and (2)

The shortest distance between two lines  is given by

Given equation of lines

Comparing with the standard form, we get 

The shortest distance is given by

Hence, the value of K is 7.

Let, x be the length of an edge, V be the volume and S be the surface area of the cube.

S = 6x²