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JEE Advanced Mock Test - 6 (Paper II) - JEE MCQ


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30 Questions MCQ Test - JEE Advanced Mock Test - 6 (Paper II)

JEE Advanced Mock Test - 6 (Paper II) for JEE 2025 is part of JEE preparation. The JEE Advanced Mock Test - 6 (Paper II) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 6 (Paper II) MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 6 (Paper II) below.
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*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 1

A particle is projected from a horizontal floor with speed 10 m s−1 at an angle 30° with the floor, which strikes the floor after some time. Which of the following is correct?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 1

Minimum speed of a projectile is at a point when it reaches its maximum height as there, vertical velocity becomes zero.

⇒ vmin = 10cos30o = 5√3 ms−1

Time of flight of the given projectile is

Maximum height attained is

Range of the projectile

Displacement of the projectile at half second

=

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 2

An electron is moving along the positive x-axis. You want to apply a magnetic field for a short time, so that the electron may reverse its direction and move parallel to the negative x-axis. This can be done by applying the magnetic field along

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 2
For the magnetic field to be able to deflect the electron and push it along a circular path, the magnetic field needs to be applied in a direction perpendicular to the direction of motion.

Hence, the magnetic field can be applied along y-axis or z-axis.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 3

Refer to figure . Let ΔU1 and ΔU2 be the change in internal energy in processes A and B, respectively, ΔQ be the net heat given to the system in process A+B and ΔW be the net work done by the system in the process A+B.

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 3

Since, it is a cyclic process and change in internal energy for process A and B is ΔU1 and ΔU2, respectively.

Since, initial and final points are same, so change in internal energy will be zero for combined process AB.

⇒ ΔUA + ΔUB = 0

⇒ ΔU1 + ΔU2 = 0

Applying first law of thermodynamics for cyclic process,

ΔQ = ΔU + ΔW

But, ΔU = 0

⇒ ΔQ − ΔW = 0

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 4

An infinitely long solid cylinder of radius R has a uniform volume charge density ρρ. It has a spherical cavity of radius R2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression . The value of k is:


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 4

Let us assume that the spherical cavity is filled with charge of same density but opposite in sign then

On comparing this with given value in the question,

we get:

⇒ k = 6

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 5

A cylindrical cavity of diameter aa exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density JJ flows along the length. If the magnitude of the magnetic field at the point P is given by N/12μ0aJ, then the value of N is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 5

So, N = 5

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 6

A sphere of mass MM and radius R rests on two supports of the same height. One support is stationary while the other is moving with a velocity v = √2 m s−1 (see in the diagram). Assume the stationary support is rough and the moving support is frictionless so that the sphere is slipping at B. At the instant shown, AB = R√2=R2. Find the speed (in m s−1) of the centre of the sphere at the given instant.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 6

The sphere is rotating about point A, so the net velocity of point O will be along OB(assume u).

Applying wedge constraint at point of contact B between sphere and block,

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 7

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at least at A. Assume that collisions between the block and the incline are totally inelastic. (g = 10 m/s2)

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline, is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 7

In case of elastic collision, component of v1 parallel to BC will remain unchanged, while component perpendicular to BC will remain unchanged in magnitude but reverse in its direction.

Now, the vertical component of velocity of the block is given by:

Now the vertical component of velocity of the block is,

v' = vT cos 30° - v2 cos 60°

= 0

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 8

A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 8
Suppose the string becomes slack at point P.

v2 = gL cosθ ...(ii)

Using energy conservation,

v2 = v02 - 2gL(1 + cos θ) ...(iii)

From equations (i), (ii) and (iii):

v02 = gL(2 + 3 cos θ) ...(iv)

Using equations of motion,

L sin θ = v cos θ × t ...(v)

-L cos θ = v sin θ × t1/2gt2 ...(vi)

tanθ = √2 ...(vii)

So,

= g × 20 × [2 + 1.372}1/2

= (200 × 3.732)1/2

= 27.32 m/s

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 9

A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi , the value of a is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 9

For an adiabatic process, = constant

Ti

Substituting the given values, we get

For diatomic gas, y = 7/5

a = 327/5 - 1 = 322/5 = 22 = 4

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 10

A barometre contains two uniform capillaries of radii 1.44 x 10-3 m and 7.2 x 10-4 m. If the height of the liquid in narrow tube is 0.2 m more than that in the wide tube, then the true pressure difference is N x 103 N/m2. Find 'N'.
Density of liquid = 103 kg/m3, surface tension = 72 x 10-3 N/m and g = 9.8 m/s2


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 10

Let the pressure in wide and narrow capillaries of radii r1 and r2, respectively be P1 and P2.
Pressure below meniscus in wide is

Pressure below meniscus in narrow tube is

The pressure difference

True pressure difference is

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 11

Two identical conducting rods are first connected independently to two vessels, one containing water at 100oC and the other containing ice at 0oC. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 and q2 (g/s) be the rate of melting of ice in the two cases, respectively. How many times is q1 to q2?


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 11

As we have Q = mL hence 

If R is the thermal resistance of the given material, then heat flow is

Where dT is the temperature difference across the two points.
From 1 and 2
q ∝ 1/R
When rods are connected in parallel, the thermal resistance is R/2.
When rods are connected in series, the thermal resistance is 2R.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 12

A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of gas A with γ (= Cp/Cv) = 5/3 and other gas B with γ = 7/5 at a certain temperature T. The gram molecular weights of the gases A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation pV19/13= constant, in adiabatic process. Find the number of gram moles of the gas B in the gaseous mixture.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 12

Since the gases do not react with each other and are ideal, hence for the mixture
(nA + nB)CvT = nA(Cv)AT + nB(Cv)BT
nA and nB are the number of moles of gases A and B
As we have 

So we have 

Number of moles of the gas B is nB = n

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 13

A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to final state f along two different paths iaf and ibf as schematically shown in the figure. The work done by the system along the paths af, ib and bf is Waf = 200 J, Wib = 50 J and Wbf = 100 J, respectively. The values of heat supplied to the system along paths iaf, ib and bf are Qinf, Qib and Qbf, respectively. If the internal energy of the system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio 


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 13

Waf = 200 j
Wbf = 100 j
Wib = 50 j
Wia = 0 j
Qiaf, Qib, Qbf
Ub = 200 j, Uiaf = 500 j
ΔQiaf = ΔUiaf + Wiaf
500 = ΔUiaf + 200 j
ΔUiaf = 300 j
∴ ΔUibf = 300 j
Wibf = 150 j
∴ ΔQibf = 450 j = Qib + Qbf
Ui = 100 j, Ub = 200 j
ΔUjb = 100 j, Wib = 50 j
∴ Qib = 150 j
∴ 450 = Qib + Qbf
450 = 150 j + Qbf
Qbf = 300 j

JEE Advanced Mock Test - 6 (Paper II) - Question 14

In the circuit shown, the coil has inductance and resistance. When X is joined to Y, the time constant is ττ during growth of current. When the steady-state is reached, rate of production of heat in the coil is P joule per second. X is now joined to Z, and after long time of joining X to Z -

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 14
Let L and R be the inductance and the resistance of the coil respectively.

Let E = e.m.f. of the cellE = e.m.f. of the cell

τ = Time constant, I0 = E/R

Energy stored in the coil

=

=

= total heat produced in the coil

JEE Advanced Mock Test - 6 (Paper II) - Question 15

Calculate the velocity with which the liquid gushes out of the 4 cm2 outlet, if the liquid flowing in the tube is water and liquid in U tube has a specific gravity12 . Velocity of liquid at point A is √20.2 m s−1

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 15

⇒ A3.vB = A1v1 + A2v2

⇒ 30 = 4v1 + 8

⇒ 4v1 = 22 ⇒ v1 = 224 = 5.5 m/s

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 16

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

A particular class of polymers is mostly derived from naturally occurring polymers by chemical modifications. Which of the following polymers belong(s) to this class?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 16
Gun cotton is another name of cellulose nitrate which is obtained by the chemical modification of cellulose, which is a naturally occurring polymer.

Vulcanised rubber is obtained by the chemical modification of natural rubber in which natural rubber is heated with sulphur to improve its properties.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 17

Directions: The following question has four choices, out of which one or more is/are correct.

Select correct statement(s) about the chemical behaviour of oxo ions of halogens.

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 17

Both chlorine monoxide (Cl2O) as well as chlorine dioxide (ClO2) are used as bleaching agents to bleach wood pulp as well as to disinfect drinking water.

Hence, options (a), (b) and (c) are correct.

However, the reaction of chlorous acid with KI and HCl proceeds as follows.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper II) - Question 18

Which of the following cations cannot be separated by adding NH4Cl, NH4OH and (NH4)2CO3 to their solutions?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 18

NH4Cl + NH4OH + (NH4)2CO3 is the group reagent for 5th group. Ions present in the 5th group cannot be separated by this reagent.

Ca2+, Sr2+ - 5th group

NH4Cl + NH4OH + (NH4)2CO3 is the group reagent for 5th group. Ions present in the 5th group cannot be separated by this reagent.

Ba+2, Sr+2 - 5th group

NH4Cl + NH4OH + (NH4)2CO3 is the group reagent for 5th group. Ions present in the 5th group cannot be separated by this reagent.

Ca+2, Ba+2 - 5th group

So, (a), (b) and (d) are from the the same group.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 19

The value of log10 K(in integer) for a reaction A ⇌ B is (Given Δr H298k0= -54.07 kJ mol-1, Δr S298k010 JK-1 mol-1 and R = 8.314 JK-1 mol-1; 2.303 × 8.314 × 298 = 5705)


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 19

ΔG = -2.303 RT log K

ΔG = ΔH - TΔS = -54070 - 2980

= -57050 J

-57050 = -2.303 × 8.314 × 298 log K

log K = 9.9985 ≈ 10

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 20

The succeeding operations that enable this transformation of states are

1. heating, cooling, heating, cooling
2. cooling, heating, cooling, heating
3. heating, cooling, cooling, heating
4. cooling, heating, heating, cooling
(Mark the correct succeeding operations as per their respective numbering given in the question.)


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 20


K – L is an isobaric process and increase in the temperature will cause expansion of the gas.
L – M is an isochoric process and decrease in the temperature will cause decrease in the pressure of the gas.
M – N is an isobaric process and a further decrease in the temperature will cause contraction of the gas.
N – K is an isochoric process and increase in the temperature will cause increase in the pressure of the gas.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 21

The possible equation of L is x - √3y = k.  Find the sum of all possible values of k.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 21

Equation of L is
x - y√3 + c = 0
Length of perpendicular dropped from centre = radius of circle

Equating with x - √3y = k,we get, k = 1 and 5.
Hence, sum = 1 + 5 = 6.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 22

A common tangent to the two circles is Find the value of m.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 22

Equation of tangent at (√3, 1)
√3x + y = 4

B divides C₁C₂ in 2:1 externally
∴ B(6, 0)
Hence let equation of common tangent is
y - 0 = m(x - 6)
mx - y - 6m = 0
length of ⊥ dropped from center (0, 0) = radius

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 23

If chord PQ subtends an angle θ at the vertex of y2 = 4ax, and  Find the value of k.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 23

Angle made by chord PQ at vertex (0, 0) is given by:

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper II) - Question 24

Length of chord PQ is ma. Find the value of m.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 24

Let  as PQ is focal chord.

Point of intersection of tangents at P and Q

as point of intersection lies on y = 2x + a

length of focal chord 

JEE Advanced Mock Test - 6 (Paper II) - Question 25

Suppose, two heat engines are connected in series, such that the heat exhaust of the first engine is used as the heat input of the second engine. The efficiencies of the engines are η1 and η2, respectively. The net efficiency of the combination is given by

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 25


 

Q3 = Q2(1 - η2)
Now,

JEE Advanced Mock Test - 6 (Paper II) - Question 26

Two non-conducting solid spheres of radii R and 2R having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R from centre of the smaller sphere is zero. The ratio ρ12 can be

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 26


At point P, if resultant electric field is zero, then

At point Q, if resultant electric field is zero, then

JEE Advanced Mock Test - 6 (Paper II) - Question 27

During the dehydration of alcohols, the hydrogen is removed from the carbon atom having a lesser number of hydrogen atoms to yield the most stable alkene as a major product. Identify the most stable alkene formed during the following conversion:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 27


The reaction proceeds through the formation of a carbocation intermediate, which re-arranges by a hydride shift followed by ring expansion in order to reduce the Bayer strain.
Thus, the major product formed is 1,2-dimethyl cyclohexene.

JEE Advanced Mock Test - 6 (Paper II) - Question 28

Which chloro derivative of benzene among the following would undergo hydrolysis most readily with aqueous sodium hydroxide to furnish the corresponding hydroxyl derivative?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 28

The polarity of the C-Cl bond in 2, 4, 6-trinitrochlorobenzene is enhanced by three NO2 groups (EWG) at the o- and p-positions, which activate the molecule towards nucleophilic substitution.
Hence, it undergoes hydrolysis most readily

JEE Advanced Mock Test - 6 (Paper II) - Question 29

Consider the given series.

The sum of the series is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 29

Here,

Hence, this is the required solution.

JEE Advanced Mock Test - 6 (Paper II) - Question 30

The value of  

Detailed Solution for JEE Advanced Mock Test - 6 (Paper II) - Question 30


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