JEE Main Part Test - 2 - JEE MCQ

# JEE Main Part Test - 2 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Part Test - 2

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JEE Main Part Test - 2 - Question 1

### A planet of mass M is revolving around sun in an elliptical orbit. If dA is the area swept in a time dt, angular momentum can be expressed as

Detailed Solution for JEE Main Part Test - 2 - Question 1

JEE Main Part Test - 2 - Question 2

### A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is

Detailed Solution for JEE Main Part Test - 2 - Question 2

(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

JEE Main Part Test - 2 - Question 3

### A spherical hole is made in a solid sphere of radius R. The mass of the original sphere was M.The gravitational field at the centre of the hole due to the remaining mass is

Detailed Solution for JEE Main Part Test - 2 - Question 3

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

JEE Main Part Test - 2 - Question 4

Three particles P, Q and R placed as per given figure. Masses of P, Q and R are √3 m, √3 m and m respectively. The gravitational force on a fourth particle ‘S’ of mass m is equal to

Detailed Solution for JEE Main Part Test - 2 - Question 4

in horizontal direction

in vertical direction

JEE Main Part Test - 2 - Question 5

A uniform ring of mass M and radius R is placed directly above uniform sphere of mass 8M and of same radius R. The centre of the ring is at a distance of d = √3R from the centre of sphere. The gravitational attraction between the sphere and the ring is

Detailed Solution for JEE Main Part Test - 2 - Question 5

Gravitational field due to the ring  at a distance d = √3R on its axis is:

force on sphere

JEE Main Part Test - 2 - Question 6

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered wooden cylindrical block of diameter 4r, height h and density ρ/3.

Situation I : Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II : The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure.
The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q. Find the minimum value of height h1 (in situation 1), for which the block just starts to move up?

Detailed Solution for JEE Main Part Test - 2 - Question 6

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

JEE Main Part Test - 2 - Question 7

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container.
As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

Q. If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of

Detailed Solution for JEE Main Part Test - 2 - Question 7

From principle of continuity,

JEE Main Part Test - 2 - Question 8

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Take Young's modulus of copper as 42 × 109Pa

Detailed Solution for JEE Main Part Test - 2 - Question 8

Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10-4 m2
Modulus of elasticity of copper from standard list, η = 42× 109 N/m2
By definition, Modulus of elasticity, η = stress/strain

⇒ Strain = F/Aη

⇒ Strain = 3.65 × 10-3
Hence, the resulting strain is 3.65 × 10-3

JEE Main Part Test - 2 - Question 9

If the elastic limit of copper is 1.5 × 108 N/ m2, determine the minimum diameter a copper wire can have under a load of 10.0 kg if its elastic limit is not to be exceeded.

Detailed Solution for JEE Main Part Test - 2 - Question 9

JEE Main Part Test - 2 - Question 10

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?

Detailed Solution for JEE Main Part Test - 2 - Question 10

JEE Main Part Test - 2 - Question 11

Water is flowing continuously from a tap having an internal diameter 8 × 10-3m. The water velocity as it leaves the tap is 0.4 ms-1. The diameter of the water stream at a distance 2 × 10-1 m below the tap is close to

[AIEEE 2011]

Detailed Solution for JEE Main Part Test - 2 - Question 11

JEE Main Part Test - 2 - Question 12

A jar is filled with two non-mixing liquids 1 and 2 having densities respectively. A solid ball, made of a material of density , is dropped in the jar. It comes to equillibrium in the position shown in the figure.

Which of the following is true for

[AIEEE 2008]

Detailed Solution for JEE Main Part Test - 2 - Question 12

The density of ball should be greater than liquid 1 because it sinks in it.
Also density of ball should be less than density of liquid 2 as it floats on it.

JEE Main Part Test - 2 - Question 13

There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is h. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

Detailed Solution for JEE Main Part Test - 2 - Question 13

Let ρ = density of the liquid.

Let α = area of a cross-section of each hole.

Volume of liquid discharged per second at a hole = αv.

Mass of liquid discharged per second = αvρ.

Momentum of liquid discharged per second = αv2ρ.

∴ the force exerted at the upper hole (to the right) = αρv22

and  the force exerted at the lower hole (to the left) = αρv12

Net force on the tank = 2αρgh.

JEE Main Part Test - 2 - Question 14

Two soap bubbles with radii   come in contact. Their common surface has a radius of curvature r. Then

Detailed Solution for JEE Main Part Test - 2 - Question 14

Let p0 = atmospheric pressure,

p1 and p2 = pressures inside the two bubbles

Let r = radius of curvature of the common surface

JEE Main Part Test - 2 - Question 15

In an adiabatic process internal energy of gas

Detailed Solution for JEE Main Part Test - 2 - Question 15

From the first law of thermodynamics,
we know, dU = dQ - dW ; (work done BY the system is considered +ve)
For an adiabatic process, dQ = 0, and hence, dU = -dW
For an ideal gas expansion, we see that work done
BY the system is +ve (recall the sign convention for work done), i.e., dW > 0.
Therefore, dU is less than 0, and thus, the internal energy decreases.

*Multiple options can be correct
JEE Main Part Test - 2 - Question 16

At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C?

Detailed Solution for JEE Main Part Test - 2 - Question 16

*Multiple options can be correct
JEE Main Part Test - 2 - Question 17

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 >T1) . The rate of heat transfer through the slab, in a steady state is with f equal to

Detailed Solution for JEE Main Part Test - 2 - Question 17

The thermal resistance

JEE Main Part Test - 2 - Question 18

Heat generated through a resistive wire will increase _______ times in unit time, if current in the wire becomes twice.

Detailed Solution for JEE Main Part Test - 2 - Question 18

So heat is given by

H = I2 Rt

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

H ∝ I

So,

H = k I2    ---(1)

if the current becomes twice'

H' = k (2I)   ---(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.

JEE Main Part Test - 2 - Question 19

A sphere of mass m and diameter D is Heated by temperature ΔT , if Coefficient of linear expansion is α what will be the change in the Surface Area

Detailed Solution for JEE Main Part Test - 2 - Question 19

Initial Surface Area = 4πR2
= 4 π (D/2)2
= 4 π D2 / 4 = π D2
Surface Area at ΔT
= 4π(D'/2)2
Now
D'=D(1+α ΔT)
So Surface Area at ΔT
= 4π(D2/4 )(1+α ΔT)2
= π D2 (1+ α2ΔT2 + 2α ΔT)
Change in surface Area
= π D2 (α2ΔT2 + 2α ΔT)
= π D2α ΔT (α ΔT+ 2)

JEE Main Part Test - 2 - Question 20

For any material, density ρ , mass m and volume V are related by ρ = m/V and B is coefficient of volume expansion then which one is true

Detailed Solution for JEE Main Part Test - 2 - Question 20

ρ = m/V
ρV = m
differentiating at w.r.t T
d(ρV)/dT = 0
ρ(dV/dT) = - V(dρ/dT)
dV/VdT= - (1 /ρ) (dρ/ dT)
Now since B = dV/VdT
So B = -(1/ρ) (dρ/ dT)

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 21

A wire of length 2.5 m and area of cross section 1×10–6 m2 has a mass of 15 kg hanging on it. What is the extension produced? How much is the energy stored in the standard wire if Young’s modulus of wire is 2×1011 Nm–2.

Detailed Solution for JEE Main Part Test - 2 - Question 21

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

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 22

A man can jump 1.5m on earth. Calculate the approximate height he might be able to jump on a planet whose density is one quarter of the earth and where radius is one third that of the earth.

Detailed Solution for JEE Main Part Test - 2 - Question 22

Here,

Height on earth, hE = 1.5m

Height on the planet, hp = ?

If ρP and ρE be the density and  RP and RE be the radius of planet and earth respectively then according to question,

For a given man, Initial K.E. of man at planet = Initial K.E. of man at earth.

and according to Work energy theorem, we can write

Potential Energy at planet = Potential energy at earth

Or, mgph= mgEhE

Or, hP =1.5 × 3 × 4 = 18m

∴∴ The man can jump upto a height of 18m.

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 23

Find the height of the geostationary satellite above the earth assuming earth as a sphere of radius 6370 km.

Detailed Solution for JEE Main Part Test - 2 - Question 23

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

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 24

A circular hole of diameter 2.00 cm is made in an aluminium plate at 0 0 C .what will be the diameter at 1000 C?
Linear expansion for aluminium = 2.3 x 10-3 / 0 C

Detailed Solution for JEE Main Part Test - 2 - Question 24

Diameter of circular hole in aluminium plate at 00 C = 2.0 cm
With increase in temperature from 00 C to 1000 C diameter of ring increases
using
L = L0(1 + αΔT)
where L0 = 2.0 cm
α = 2.3 x  10-3 / 0 C
ΔT = (100 -0) = 100 0 C
we can find diameter at 1000 C
L = 2(1 + 2.3 x 10-3 x 100)
= 2.46 cm

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 25

A gas is contained in a cylinder with a moveable piston on which a heavy block is placed. Suppose the region outside the chamber is evacuated and the total mass of the block and the movable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gas increases by 1580 J. What is the distance s through which the piston rises? (Round off answer to 2 decimal places)

Detailed Solution for JEE Main Part Test - 2 - Question 25

Total heat supplied =Work done + Change in internal energy
So work done = 2140 - 1580 = 560 J
Let s be the distance moved then
the work done is given by = Fs
Fs = 560
s = 560/F
= 560/102 x 10
s = .55 m

JEE Main Part Test - 2 - Question 26

A metal in a compound can be displaced by another metal in the uncombined state. Which metal is a better reducing agent in such a case?

Detailed Solution for JEE Main Part Test - 2 - Question 26

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.

JEE Main Part Test - 2 - Question 27

In which case change of oxidation number of V is maximum?

Detailed Solution for JEE Main Part Test - 2 - Question 27

JEE Main Part Test - 2 - Question 28

The oxidation number of P in Ba(H2PO2)2, Ba(H2PO3)2 and Ba(H2PO4)2 are respectively

Detailed Solution for JEE Main Part Test - 2 - Question 28

*Multiple options can be correct
JEE Main Part Test - 2 - Question 29

The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO(nitrosyl).

Q. Magnetic moment  of Fe in the ring is

Detailed Solution for JEE Main Part Test - 2 - Question 29

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

Fe+ has three unpaired electrons (N).

JEE Main Part Test - 2 - Question 30

Which is chlorate (I) ion?

Detailed Solution for JEE Main Part Test - 2 - Question 30
• 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
JEE Main Part Test - 2 - Question 31

How much heat will be required at constant pressure to form 1.28 kg of CaC2 from CaO(s) & C(s) ?

Given :

ΔfH°(CaO, s) = -152 kcal/mol

ΔfH°(CaC2, s) = -14 kcal/mol

ΔfH°(CO, g) = -26 kcal/mol

Detailed Solution for JEE Main Part Test - 2 - Question 31

JEE Main Part Test - 2 - Question 32

The enthalpy of neutralisation of a weak acid in 1 M solution with a strong base is -56.1 kJ mol-1. If the enthalpy of ionization of the acid is 1.5 kJ mol-1 and enthalpy of neutralization of the strong acid with a strong base is -57.3 kJ equiv-1, what is % ionization of the weak acid in molar solution (assume the acid to be monobasic) ?

Detailed Solution for JEE Main Part Test - 2 - Question 32

The correct answer is option C
Ideally, the enthalpy of neutralization should be - 57.3 K.J + 1.5 K.J = - 55.8 KJ.
But it is - 56.1 KJ.
∴ Energy used for neutralization
= 57.3 - 56.1 = 1.2 KJ
∴ Percent ionization of weak acid

∴ % of weak acid in solution = 20%

JEE Main Part Test - 2 - Question 33

From the following data, the heat of formation of Ca(OH)2(s) at 18°C is ………..kcal:

Detailed Solution for JEE Main Part Test - 2 - Question 33

The correct answer is Option B.

Ca(s) + O2(g) + H2(g) → Ca(OH)2 , ΔHf = ?

Desired equation = eq (iii) + eq(i) - eq (ii)

ΔHf = (−151.80)+(−15.26)−(−68.37)
ΔHf = (-151.80)+(-15.26)-(-68.37)
ΔHf = −235.43KCalmol−1

JEE Main Part Test - 2 - Question 34

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be -

[AIEEE-2008]

Detailed Solution for JEE Main Part Test - 2 - Question 34

JEE Main Part Test - 2 - Question 35

In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11 Selection the correct statement for a saturated 0.034 M solution of the carbonic acid.

[AIEEE-2010]

Detailed Solution for JEE Main Part Test - 2 - Question 35

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

JEE Main Part Test - 2 - Question 36

How many litres of water must be added to litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

[AIEEE-2013]

Detailed Solution for JEE Main Part Test - 2 - Question 36

Volume of the original solution = 1 L

pH of the new solution = 2

Volume of the new solution = ?
As per volumetric principle

That means the volume of water added to the original solution of 1 L = 10 −1 = 9 L

JEE Main Part Test - 2 - Question 37

For the following electrochemical cell reaction at 298 K,

E°cell = 1.10 V

Detailed Solution for JEE Main Part Test - 2 - Question 37

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq),

E= +1.10V

∴ Eo = 0.0591/n log10Keq

because at equilibrium,

Ecell = 0

(n = number of electrons exchanged = 2)

1.10 = 0.0591/2 log10Keq
2.20/0.0591 = log10Keq

Keq = antilog37.225

JEE Main Part Test - 2 - Question 38

We know that the relationship between Kc and Kp is Kp = Kc (RT)Δn
What would be the value of Δn for the reaction NH4Cl (s) ⇔ NH3 (g) + HCl (g)

Detailed Solution for JEE Main Part Test - 2 - Question 38

JEE Main Part Test - 2 - Question 39

For the equilibrium,

at 1000 K. If at equilibrium pCO = 10 then total pressure at equilibrium is

Detailed Solution for JEE Main Part Test - 2 - Question 39

C(s) + CO2(g) <=========> 2CO(g)
Kp = pCO2/pCO2
GIven Kp = 63 and pCO = 10pCO2
Putting the value of pCO in above equation,
63 = 100(pCO2)2/pCO2
Or pCO2 = 0.63
pCO = 6.3
Therefore, total pressure = 6.3+0.63 = 6.93 atm

JEE Main Part Test - 2 - Question 40

Ammonium carbamate dissociates as,

In a closed vessel containing ammonium carbamate in equilibrium with its vapour, ammonia is added such that partial pressure of NH3 now equals the original total pressure. Thus, ratio of the total pressure to the original pressure is

Detailed Solution for JEE Main Part Test - 2 - Question 40

JEE Main Part Test - 2 - Question 41

A 0.10 M solution of a weak acid, HX, is 0.059% ionized. Evaluate Ka for the acid.

Detailed Solution for JEE Main Part Test - 2 - Question 41

Since the acid is only 0.059% ionized, therefore the concentration of ions in solution = 0.1 x 0.059 / 100 = 0.000059

Ka = [H+] [X] / [HX] = (0.000059)2 / 0.1 = 3.5 x 10-8

JEE Main Part Test - 2 - Question 42

AgCI(s)is sparingly soluble salt,

AgCl (s)  Ag+(aq) + Cl-(aq)

There is

Detailed Solution for JEE Main Part Test - 2 - Question 42

When ammonia is added, solubility of AgCl increases due to formation of complex salt which decreases the concentration of radicals in the product side and tus drives the reaction in forward direction.
When we add KCl common ion effect is applied in presence of common ion solubility decreases and reaction goes in backward direction.

JEE Main Part Test - 2 - Question 43

Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

[AIEEE-2010]

Detailed Solution for JEE Main Part Test - 2 - Question 43

For precipitation

[Ag+][Br-] > Ksp[AgBr]

[Br-]min = (5 x 10-13)/0.05 = 10-11 M

Mass of potassium bromide needed = 10-11 x 120

= 1.2 x 10-9 g

JEE Main Part Test - 2 - Question 44

Which one of the following is the approximate pH of 0.01 M solution of NaOH at 298 k?

Detailed Solution for JEE Main Part Test - 2 - Question 44

First off, since NaOH is a strong base, it will dissociate completely into Na+ and OH-. Thus, we know that we have 0.01 M OH-.

However, we do not know anything about the concentration of H+. Fortunately, we do not need to, as pH + pOH = 14. So, if we find pOH, we can solve for pH. p is a mathematical function equivalent to -log. So, pH actually means -log[H+] (Note that brackets indicate concentration of).

pOH = -log 0.01M OH-

pOH = 2

pH + 2 = 14

pH = 12

This result makes sense, since a solution of strong base should have a high pH.

JEE Main Part Test - 2 - Question 45

If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

Detailed Solution for JEE Main Part Test - 2 - Question 45

We know that,
From the first law of thermodynamics
ΔQ=ΔU+ΔW
Also,
ΔU=−ΔW
Hence, option (A) is correct.

JEE Main Part Test - 2 - Question 46

What is the oxidation state of S in H2SO4 & H2SO3?

Detailed Solution for JEE Main Part Test - 2 - Question 46

H2SO4 = 2 ( + 1) + x + 4 (- 2) = 0
x = 8 – 2 = +6
H2SO3 = 2 ( + 1) + x + 3 ( – 2) = 0
x = 4

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 47

Calculate the standard heat of formation of propane, if its heat of combustion is −2220.2 KJmol−1 the heats of formation of CO2 (g) and H2O(1) are −393.5 and −285.8 kJ mol−1 respectively.

Detailed Solution for JEE Main Part Test - 2 - Question 47

Given :

ΔH0f CO2 = −393.5 kJ mol−1

ΔH0f H2O = −285.8 kJ mol−1

ΔH0f C3H8 = −2220.2 kJ mol−1

Combustion of propane

C3H8 + 5O2 → 3CO2 + 4H2O

ΔH0= ∑ ΔH0f p  ∑ ΔH0f r

−2220.2 = −1180.5 − 1143.2 − ΔH0f C3H8

ΔH0f C3H8 = −103.5 kJ mol−1

ΔH0f C3H8 = −103.5 KJ

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 48

For the reaction at 298 K: 2A +B → C
ΔH = 400 J mol−1; ΔS = 0.2 JK−1mol−1
Determine the temperature at which the reaction would be spontaneous.

Detailed Solution for JEE Main Part Test - 2 - Question 48

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

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 49

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

2SO2(g) + O2(g) ⇌ 2SO3(g)
A mixture of SO2 and O2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10−2 M SO3, 3.5 × 10−3 M O2, and 3.0 × 10−3 M SO2. Calculate K and Kp at this temperature.

Detailed Solution for JEE Main Part Test - 2 - Question 49

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

To solve for Kp, where Δn = 2 − 3 = −1:

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

H2(g) + I2(g )⇌ 2HI(g)

A mixture of H2 and I2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10−2 M HI, 6.47 × 10−3 M H2, and 5.94 × 10−4 M I2. Calculate K and Kp for this reaction.

K = 48.8; Kp = 48.8

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 50

Calculate the value of ∆U and ∆H on heating 128.0 g of oxygen from 0o C to 1000 C. CV and CP on an average are 21 and 29 J mol-1 K-1. (The difference is 8Jmol-1 K-1 which is approximately equal to R)

Detailed Solution for JEE Main Part Test - 2 - Question 50

We know
∆U   = n Cv (T2-T1)
∆H   = n CP (T2- T1)
Here
n = 128/32 4 moles ;
T2 = 100o
C = 373K;
T1 = 0o
C = 273K
∆U = n Cv (T- T1)
∆U =    4 x 21 x (373 - 273)
∆U =    8400 J
∆U =     8.4 kJ
∆H =  n Cp (T2- T1)
∆H =    4 × 29 × (373 - 273)
∆H =    11600 J
∆H =    11.6 kJ

JEE Main Part Test - 2 - Question 51

If the coefficients of (r +1)th term and (r + 3)th term in the expansion of (1+x)2n be equal then

Detailed Solution for JEE Main Part Test - 2 - Question 51

Tr+1 = 2nCr(x)r (1)2n-r
Tr+3 = 2nCr+2 (x)r+2 (1)2n-r-2
Tr+1 : Tr+3
= 2nCr = 2nCr+2
=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
=> (r+2)(r+1) = (2n-r)(2n-r-1)
=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
=> 4n(n-r-1) -2(n-r-1)
Therefore n - r - 1 = 0
=> n = r+1

JEE Main Part Test - 2 - Question 52

In the expansion of (1+x)60, the sum of coefficients of odd powers of x is

Detailed Solution for JEE Main Part Test - 2 - Question 52

(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

JEE Main Part Test - 2 - Question 53

The middle term in the expansion of (1+x)2n is

Detailed Solution for JEE Main Part Test - 2 - Question 53

Middle term in the expansion of (1+x)2n; is
=tn+1 = 2nCn.1(2n−n).xn
= {(2n)!.xn}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xn
= {{2n[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xn
= [[(2n−1)(2n−3)....3×1]/n!] 2nxn

JEE Main Part Test - 2 - Question 54

Detailed Solution for JEE Main Part Test - 2 - Question 54

JEE Main Part Test - 2 - Question 55

The circumcentre of the triangle with vertices (0, 0), (3, 0) and (0, 4) is

Detailed Solution for JEE Main Part Test - 2 - Question 55

Step-by-step explanation:

The given points when plotted show a right angled triangle

Since we know that the circumcentre of a right angled triangle lies on the midpoint of the hypotenuse, using section formula:

C=(3/2,4/2) => (3/2,2)

JEE Main Part Test - 2 - Question 56

The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12), then orthocentre of this triangle is

Detailed Solution for JEE Main Part Test - 2 - Question 56

JEE Main Part Test - 2 - Question 57

Area of a triangle whose vertices are (a cos q, b sinq), (–a sin q, b cos q) and (–a cos q, –b sin q) is

Detailed Solution for JEE Main Part Test - 2 - Question 57

JEE Main Part Test - 2 - Question 58

The point A divides the join of the points (–5, 1) and (3, 5) in the ratio k : 1 and coordinates of points B and C are (1, 5) and (7, –2) respectively. If the area of ΔABC be 2 units, then k equals

Detailed Solution for JEE Main Part Test - 2 - Question 58

JEE Main Part Test - 2 - Question 59

If A(cosa, sina), B(sina, – cosa), C(1, 2) are the vertices of a ΔABC, then as a varies, the locus of its centroid is

Detailed Solution for JEE Main Part Test - 2 - Question 59

JEE Main Part Test - 2 - Question 60

Two perpendicular tangents to the circle x2+y2 = r2 meet at P. The locus of P is

Detailed Solution for JEE Main Part Test - 2 - Question 60

JEE Main Part Test - 2 - Question 61

The locus of a variable point whose distance from the point (2, 0) is 2/3 times its distance from the line x = 9/2 is

Detailed Solution for JEE Main Part Test - 2 - Question 61

JEE Main Part Test - 2 - Question 62

The locus of a point such that two tangents drawn from it to the parabola y2 = 4ax are such that the slope of one is double the other is

Detailed Solution for JEE Main Part Test - 2 - Question 62

Let the point be (h, k)

Now equation of tangent to the parabola y= 4ax whose slope is m is

as it passes through (h, k)]

JEE Main Part Test - 2 - Question 63

T is a point on the tangent to a parabola y2 = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then

Detailed Solution for JEE Main Part Test - 2 - Question 63

JEE Main Part Test - 2 - Question 64

Tangents are drawn from the points on the line x – y + 3 = 0 to parabola y2 = 8x. Then the variable chords of contact pass through a fixed point whose coordinates are

Detailed Solution for JEE Main Part Test - 2 - Question 64

Let (k,k+3) be the point on the line x−y+3=0
Equation of chord of contact is S1=0
⇒yy1=4(x+x1)
⇒y(k+3)=4(x+k)
⇒4x−3y−k(y−4)=0
Therefore, straight line passes through fixed point (3,4)

JEE Main Part Test - 2 - Question 65

The focus of the parabola x2−8x+2y+7 = 0 is

Detailed Solution for JEE Main Part Test - 2 - Question 65

Parabola is x2 – 8x + 2y + 7 = 0
∴   (x – 4)2 = – 2y – 7 + 16
∴   (x – 4)2 = – 2[y – (9/2)]
∴   x2 = – 4ay
⇒ x = x – 4, y = y – (9/2) and 2 = 4a
i.e. a = (1/2)
Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/2) = -1/2
∴    x = 4    and y = 4
∴ focus [4, 4].

JEE Main Part Test - 2 - Question 66

From the focus of the parabola y2 = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is

Detailed Solution for JEE Main Part Test - 2 - Question 66

Focus of parabola y2 = 8x is (2,0). Equation of circle with centre (2,0) is (x−2)2 + y2 = r2
Let AB is common chord and Q is mid point i.e. (1,0)
AQ2 = y2 = 8x
= 8×1 = 8
∴ r2 = AQ2 + QS2
= 8 + 1 = 9
So required circle is (x−2)2 + y2 = 9

JEE Main Part Test - 2 - Question 67

In an examination, ten students scored the following marks: 60, 58, 90, 51, 47, 81, 70, 95, 87, 99. The range of this data is

JEE Main Part Test - 2 - Question 68

Two vertices of a triangle are (3,−2) and (−2, 3) and its orthocentre is (−6, 1). The coordinates of its third vertex are-

Detailed Solution for JEE Main Part Test - 2 - Question 68
Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

Solving Equations(i)i and (ii)ii, we get

α = −1, β = 6

∴ The third vertex is (−1, 6)

JEE Main Part Test - 2 - Question 69

is equal to

Detailed Solution for JEE Main Part Test - 2 - Question 69

JEE Main Part Test - 2 - Question 70

Let ,Q,R and S be four points on the ellipse 9x2 + 4y2 = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If , where p and q are coprime, then p + q is equal to :

Detailed Solution for JEE Main Part Test - 2 - Question 70

Given, points P and Q are on the ellipse defined by 9x2 + 4y= 36 which simplifies to . This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis a = 3 along the y-axis and semi-minor axis b = 2 along the x-axis.
OP is the distance from origin O to point P, which is given by :

1. Let's represent the given point  in polar coordinates. We can write

Simplifying this, we obtain :

2. Similarly, if we represent the point R as ,   (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or π/2  radians. In terms of sin and
it too should satisfy the equation of the ellipse. We have :

Simplifying this, we obtain :

3. From equations (1) and (2), we have :

4. Now, note that lines PQ and RS are perpendicular and pass through the origin, so PQ =2OP and RS = 2OR. Thus,

5. Substituting the values of r1 and r2, we obtain :

6. Hence, p= 13 and q = 144.

7. So, the final answer p + q = 13 + 144=157.

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 71

If one of the lines of my2 + (1 - m2)xy - mx2 = 0 is a bisector of the angle between the lines xy = 0, then m(m > 0) is

Detailed Solution for JEE Main Part Test - 2 - Question 71
Equation of bisectors of lines, xy = 0 are y = ± x

Putting y = ± x in my2 + (1 - m2)xy - mx2 = 0, we get

(1 - m2)x2 = 0

⇒ m = ±1

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 72

The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one-year old child. If the average age of the family now is λ, then the number of divisors of λ are

Detailed Solution for JEE Main Part Test - 2 - Question 72

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}.

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 73

If the sum of the coefficients of all even powers of x in the product (1 + x + x2 + … + x2n) (1 – x + x2 – x3 + … + x2n) is 61, then find the value of n.

Detailed Solution for JEE Main Part Test - 2 - Question 73

Let (1 + x + x2 + … + x2n) (1 – x + x2 – x3 + … + x2n) = a0 + a1x + a2x2 + a3x3+ ……

Put x = 1 => 2n + 1 = a0 + a1 + a2 + a3 + …… …(i)

Put x = -1 => 2n + 1 = a0 – a1 + a2 – a3 + …… …..(ii)

Add (i) and (ii), we get

2n + 1 = 61

Or n = 30

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 74

Find the coefficient of x4 in the expansion of (1 + x + x2)10 .

Detailed Solution for JEE Main Part Test - 2 - Question 74

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

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 75

If the sum of the series

is α/β, where α and β are co-prime, then α + 3β is equal to __________.

Detailed Solution for JEE Main Part Test - 2 - Question 75

We can rewrite the given series as follows :

The first few terms of the series are :

We can now see that each group of terms forms a geometric series with a common ratio of 12:

The series can be rewritten as :

Now, we can simplify and rewrite the series inside the parentheses as:

The series inside the parentheses is an infinite geometric series with the first term �=−13 and the common ratio �=−13:

Thus, the sum of the series is 12, and α = 1 and β = 2 are co-prime.
Therefore, α  + 3β = 1 + 3 x 2 = 7

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