JEE Main Mock Test - 2 - JEE MCQ

  (orbital speed υ₀ of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

in horizontal direction  

in vertical direction

Consider the equilibrium of wooden block.
Forces acting in the downward direction are

(a) Weight of wooden cylinder

(b) Force due to pressure (P₁) created by liquid of height h₁ above the wooden block is
= P₁ × π (2r)² = [P₀ + h₁rg] × π (2r)²
Force acting on the upward direction due to pressure P₂ exerted from below the wooden block and atmospheric pressure is

At the verge of rising

So heat is given by 

H = I² Rt

time is fixed, unit time, Resistance of wire will remain the same. 

H ∝ I² 

So,

H = k I²    ---(1)

if the current becomes twice'

H' = k (2I)²    ---(2)

(2) divided by (1) we get

So, H' : H = 4 : 1

So we see, the heat generated in unit time becomes 4 times of its initial value.

Value of the acceleration due to gravity at any latitude φ,
g' = g - ω²Rcos²φ; where φ is latitude.
There will be no change in gravity at poles as φ = 90°.
At all other points, as ω increases, 'g' will decrease.

Moment of inertia of the whole disc about the axis passing through the centre of mass and perpendicular to the plane of disc,

Moment of inertia of the smaller part of the disc about the centre of the original disc,

Here,

Length of wire l = 2.5m

Area of cross-section, A = 1×10^–6 m²

Hanging mass, m = 15 kg

Elongation, e = ?

Energy stored in the wire, E = ?

Young’s modulus of steel, Y = 2×10¹¹ Nm^–2.

We have,

Young’s modulus of elasticity,

e = 1.875 ×10^-3 m

Again, 

Energy stored in the stretched wire,

      E = 1/2​ .F .e

Or, E = 1/2​ × mg × e 

Or, E = 1/2​ × 15×10 ×1.875 ×10^-3

∴∴ E = 0.141 J

Here,

The time period for geostationary satellite,

T = 24 hrs = 24×60×60

= 86400 sec

Radius of earth, R = 6370km = 6.37×10⁶m

Height of satellite, h = ?

We know, time period,

∴ h = 36122818.44 m = 36122.82 Km

Concept of Reducing Agent:
A reducing agent is a substance that loses or "donates" an electron to another substance in a redox chemical reaction. Therefore, a good reducing agent is the one that gets oxidized easily, or in other words, the one that can easily lose electrons.

Characteristics of a Good Reducing Agent:

• Electron Loss: A better reducing agent is the one that loses more electrons. This is because by losing electrons, the reducing agent gets oxidized and in turn reduces the other substance. This is the basic principle of a redox reaction.
• Reactivity: The reactivity of the metal also determines its capacity as a reducing agent. Metals that are high in the reactivity series are good reducing agents. This is because they can easily lose electrons and get oxidized.
• Stability: Metals that are less stable are better reducing agents because they can easily lose electrons to attain a stable state.

Hence, Option A is the correct answer - a better reducing agent is the one that loses more electrons.

Fe²⁺ is added asFeSO₄
Fe⁺ is formed by charge transfer from NO to Fe²⁺ 



Fe⁺ has three unpaired electrons (N).

• 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

Carbonic acid is a weak acid it dissociates as follows 

Since both acids are weak cannot be 0.034 M as there is no complete dissociation. 

Total hydrogen ion concentration is approximately equal to concentration of hydrogen ion in first reaction as K₁ is larger. This is approximately equal to concentration of 

The concentration of  depends on  and that of  on . But . So The concentration of  is not double that of 

Zn(s) + Cu²⁺(aq) ⇌ Cu(s) + Zn²⁺(aq),

E^o = +1.10V

∴ E^o = 0.0591/n log₁₀K_eq

because at equilibrium, 

Ecell = 0

(n = number of electrons exchanged = 2)

1.10 = 0.0591/2 log₁₀K_eq
2.20/0.0591 = log₁₀K_eq

K_eq = antilog37.225

H₂SO₄ = 2 ( + 1) + x + 4 (- 2) = 0
x = 8 – 2 = +6 

Given :

ΔH⁰_f CO2 = −393.5 kJ mol⁻¹

ΔH⁰_f H2O = −285.8 kJ mol⁻¹

ΔH⁰_f C3H8 = −2220.2 kJ mol⁻¹

Combustion of propane

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

ΔH⁰_c = ∑ ΔH⁰_{f p} − ∑ ΔH⁰_{f r}

−2220.2 = −1180.5 − 1143.2 − ΔH⁰_f C3H8

ΔH⁰_f C3H8 = −103.5 kJ mol⁻¹

ΔH⁰_f C3H8 = −103.5 KJ

Given :

ΔH = 400 J mol⁻¹

ΔS = 0.2 J K⁻¹ mol⁻¹

T = 298 K

We know that ΔG = ΔH − TΔS

At equilibrium, ΔG = 0

∴ TΔS = ΔH

T = ΔH / ΔS = ( 400 J mol⁻¹ ) / ( 0.2 Jk⁻¹ mol⁻¹)

T = 2000K

ΔG = 400 − (2000 × 0.2)

 = 0

if T > 2000K ΔG will be negative

The reaction would be spontaneous only beyond 2000K

 (1+x)ⁿ= nC0+ nC1x+ nC2x²+...+ nC0xⁿ ....(1)
(1−x)ⁿ= nC0 − nC1x+ nC2x²−...+ nC0 xⁿ....(2)
⇒(1+x)ⁿ−(1−x)ⁿ = nC0+ nC1x+ nC2x²+...+ nC0xⁿ− nC0+ nC1x− nC2x²+...+ nC0xⁿ
 =2(nC1x+nC3x³+....+xⁿ)

Given: n=60 and put x=1
⇒(1+1)⁶⁰−(1−1)⁶⁰
 =2⁶⁰
⇒2⁶⁰=2(nC1x+nC3x³+....+xⁿ)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x³+....+xⁿ =2⁶⁰−1 =2⁵⁹

Sum of the present ages of husband, wife and child

= (23 × 2 + 5 × 2) + 1 = 57 years.

∴ Required average = (57/3) = 19 years .

Average age is 19 years which is a prime number, so it has only two divisors {1, 19}.

General term of the given expression is,

Here, q + 2r = 4

For p = 6, q = 4, r = 0, coefficient = 10!/[6! x 4!] = 210

For p = 7, q = 2, r = 1, coefficient = 10! /[7! x 2! x 1!] = 360

For p = 8, q = 0, r = 2, coefficient = 10!/[ 8! x 2!] = 45

Therefore, sum = 615