KIITEE Mock Test - 3 - JEE MCQ

As g' = g − ω²Rcos²λ

The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of the earth makes with equatorial plane.

It is denoted by, λ.

For the poles λ = 90° and for equator λ = 0°.
(i) Substituting λ = 90° in the above expression,

we get,

g_pole = g − ω²Rcos²90°

∴ g_pole = g

i.e., there is no effect of rotational motion of the earth on the value of g at the poles.
(ii) Substituting λ = 0° in the above expression,

we get,

g_equator = g − ω²Rcos²0°

∴ g_equator = g − ω²R
i.e., the effect of rotation of the earth on the value of g at the equator is maximum.

= 27 + 0 – (8 + 12) = 27 – 20 = +7 J
According to work energy theorem,
Change in the kinetic energy
= Work done, W
= +7 J

Both the diodes are in reverse biased.

Given,

Number of turns in the coil, N = 400

The inductance of the coil, L = 8 mH

Current in the coil, i = 5 mA

Let the magnetic flux through each coil be ϕ, then

Viscosity is the property of fluids by the virtue of which it opposes any relative motion between the layers of fluid or between any object and fluid surrounding it. It is analogous to friction for fluids.
It depends strongly on temperature. In liquids, it usually decreases with increasing temperature, whereas, in gases, viscosity increases with increasing temperature.

= 0.03 sin(2πt - 0.01πx) m

Comparing it with general equation, we get y = asin(ωt-kx)

where, k = 2π/λ⇒λ = 200m

The phase deference between two particles is given by Δ∅ = kx = 2π/λ x x …(i)

Here, x = 25m

Substituting the values of x and λ in Eq. (i). we get Δ∅ = (2π/200) x 25 = (π/4)rad

Hydrogen combines with nitrogen to produce ammonia in Haber's process.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
In this process, iron oxide is used with small amounts of K₂O and Al₂O₃ to increase the rate of attainment of equilibrium.
Optimum conditions for the production of ammonia are a pressure of 200 atm and a temperature of 700 K.
Earlier, iron was used as a catalyst with molybdenum as a promoter in this reaction.

• ClO₃^- : A very reactive inorganic anion.
• The term chlorate can also be used to describe any compound containing the chlorate ion, normally chlorate salts. 
• Example: Potassium chlorate, KClO₃

In the titration of oxalic acid (weak acid) with a strong base like NaOH, oxalic acid is taken in a conical flask and NaOH is taken in a burette.

Hence, 'A' is bromo benzene.

f(x) = (sin(tan^-1x) + sin(cot^-1x))² - 1

= 1 + 2sin(cot^-1x) cos(cot^-1x) - 1
f(x) = sin(2cot^-1x) = sin(2tan^-1x) = sin 2θ
As |x| > 1

Hence, sin^-1(f(x) = sin^-1(sin(2tan^-1x)) = π - 2tan^-1x or - π - 2tan^-1x


Hence, option (2) is the answer.

Given: x * y = x² + y³ and (x * 1) * 1 = x * (1 * 1)
So, (x² + 1) * 1 = x * 2
⇒ (x² + 1)² + 1 = x² + 8
⇒ x⁴ + 2x² + 2 = x² + 8
⇒ (x²)² + x² – 6 = 0
⇒ (x² + 3)(x² – 2) = 0
∴ (x² + 3)(x² – 2) = 0
∴ x² = 2

L₁ : 4x - 3y + K₁ = 0
L₂ : 4x - 3y + K₂ = 0
Now,
-4 - 6 + K₁ = 0
⇒ K1 = 10
12 + 18 + K₂ = 0
⇒ K₂ = -30
⇒ Tangents to the circle are:
4x - 3y + 10 = 0
4x - 3y - 30 = 0
Length of diameter, 2r =  = 8
⇒ r = 4
Now, centre is the mid-point of A and B.
x = 1, y = -2
Equation of the circle,
(x - 1)² + (y + 2)² = 16