JEE Advanced Practice Test- 9 - JEE MCQ

JEE Advanced Practice Test- 9 - JEE MCQ

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54 Questions MCQ Test - JEE Advanced Practice Test- 9

JEE Advanced Practice Test- 9 for JEE 2024 is part of JEE preparation. The JEE Advanced Practice Test- 9 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Practice Test- 9 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Practice Test- 9 below.
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JEE Advanced Practice Test- 9 - Question 1

An engine whistling at a constant frequency n0 and moving with a constant velocity goes past a stationary observer. As the engine crosses him, the frequency of the sound heard by him changes by a factor f. The actual difference in the frequency of the sound heard by him before and after the engine across him is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 1

Simply use the above relation

JEE Advanced Practice Test- 9 - Question 2

A ball of density ρ0 falls from rest from point P onto the surface of a liquid of density ρ in time T. It enters the liquid stops moves up and returns to P in a total time 3T. Neglect viscosity, surface tension and splashing. The ratio p/p0 is equal to

Detailed Solution for JEE Advanced Practice Test- 9 - Question 2

JEE Advanced Practice Test- 9 - Question 3

When an object is placed in front of a concave mirror of focal length f, a virtual image is produced with a magnification of 2. To obtain a real image with a magnification of 2. The object has to be moved by a distance equal to

Detailed Solution for JEE Advanced Practice Test- 9 - Question 3

Calculate the initial and final distance of object from the concave mirror.

JEE Advanced Practice Test- 9 - Question 4

A solid cube is placed on a horizontal surface. The coefficient of friction between them is μ, where μ < 1/2 . A variable horizontal force is applied on the cube’s upper face, perpendicular to one edge and passing through the mid-point of that edge. The maximum acceleration with which it can move without toppling is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 4

To avoid t toppling

JEE Advanced Practice Test- 9 - Question 5

A spherical body of mass m, radius r and moment of inertia I about its centre moves along the x-axis. Its centre of mass moves with velocity = v0 and it rotates about its centre of mass with angular velocity ω0.LO = lω0 + mvor . The angular momentum of the body about the origin O is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 5

As we know
L0 = L1 + L2
L1 = due to rotational motion
L2 = due to linear motion.

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 6

DIRECTION ;- This section contains 2 multiple choice questions from 8 to 9. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONE OR MORE is/are correct

The temperature of a solid object is observed to be constant during a period. In this period

Detailed Solution for JEE Advanced Practice Test- 9 - Question 6

If temperature of a body is constant that means amount of hear energy received and amount of hear energy emitted are equal.

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 7

The electric potential decreases uniformly from 120 V to 80 V as one moves on the axis from x = –1 cm to x = 1 cm. The electric field at the origin

Detailed Solution for JEE Advanced Practice Test- 9 - Question 7

If direction of field is along the x-axis then its magnitude will be 20 V/cm other wise greatest then hat

JEE Advanced Practice Test- 9 - Question 8

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C) and D) out of WHICH ONLY ONE IS CORRECT.

paragraph for  question no.12 and 13

The base of a hollow right cone of semi vertical angle 30o is fixed to a horizontal plane. Two particle each of mass m are connected by a light inextensible string which passes through a small hole in the vertex of the cone. One particle A hangs at rest inside the cone. The other particle B moves on the outer smooth surface of the cone at a distance ℓ from vertex in a horizontal circle with centre at A. Neglecting friction, now answer the following.

The tention in the string is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 8

Now apply newton’s law of motion for the particle B.

JEE Advanced Practice Test- 9 - Question 9

The base of a hollow right cone of semi vertical angle 30o is fixed to a horizontal plane. Two particle each of mass m are connected by a light inextensible string which passes through a small hole in the vertex of the cone. One particle A hangs at rest inside the cone. The other particle B moves on the outer smooth surface of the cone at a distance ℓ from vertex in a horizontal circle with centre at A. Neglecting friction, now answer the following.

The angular velocity of B is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 9

Now apply newton’s law of motion for the particle B.

JEE Advanced Practice Test- 9 - Question 10

Paragraph For Question No.14 and 16

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C= 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 10

Initially current will pass through the resistance only.

JEE Advanced Practice Test- 9 - Question 11

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C2 = 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

The charge on capacitance C1 and C2 are respectively

Detailed Solution for JEE Advanced Practice Test- 9 - Question 11

Initially current will pass through the resistance only.

JEE Advanced Practice Test- 9 - Question 12

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C2 = 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

Long time after switch is opened the charge on C1 is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 12

Initially current will pass through the resistance only.

JEE Advanced Practice Test- 9 - Question 13

Column – I gives certain situation involving two thin conducting shells connected by a conducting wire via a key K. In all situation one sphere has net charge +q and other sphere has no net charge. After the key K is pressed

column – II gives some resulting effect. Match the figures in column with the statements in column – II.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 13

In such type of situation, we have to use these basic facts.
(a) charge will flow between the two body if there is potential difference between the two.
(b) during this process energy will lose due to sparking

JEE Advanced Practice Test- 9 - Question 14

A particle of mass 2 kg is moving on straight line under the action of force F = (8 – 2x)N. The
particle is released at rest from x = 6m. For the sub-sequent motion match the following (All the
values in the right column are in there SI units).

Detailed Solution for JEE Advanced Practice Test- 9 - Question 14

At equilibrium Fnet will be zero and at the extreme point, particle will be in state of rest.

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 15

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

In the arrangement shown in figure, pulley are light and frictionless, threads are inextensible and mass of blocks A, B and C are m1 = 5 kg, m2 = 4 kg and m3 = 2.5 kg respectively and co-efficient of friction for both the planes is μ = 0.50. Calculate the acceleration of block A (in m/sec), when the system is released from rest. (g = 10 m/sec)

Detailed Solution for JEE Advanced Practice Test- 9 - Question 15

In this situation block B will not move at all.

Now apply newton's laws of motion.

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 16

The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 16

Where R is radius of curvature.

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 17

A disc ‘A’ of mass M is placed at rest on the smooth inclined surface of inclination θ. A ball B of mass m is suspended vertically from the centre of the disc A by a light inextensible string of light ℓ as shown in the figure. If the acceleration of the disc B immediately after the system is released from rest

Detailed Solution for JEE Advanced Practice Test- 9 - Question 17

apply newton’s laws of motion for the body A and body B

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 18

A uniform rod AB of mass M and length R √2 is moving in a vertical plane inside a hollow sphere of radius R. The sphere is rolling on a fixed horizontal surface without slipping with velocity of its centre of mass 2v, when the end B is at the lowest position, its speed is found to be v as shown in the figure. If the kinetic energy of rod at this instant is 4/k mv2. Find k.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 18

length of rod is R √2

Velocity of point P
2v - ωR = v
KE of rod = rotational KE t translatory KE

JEE Advanced Practice Test- 9 - Question 19

Sulphur reacts with chlorine in 1 : 2 ratio and forms X. Hydrolysis of X gives a sulphur compoundY. What is the hybridization state of central atom in the compound Y?

Detailed Solution for JEE Advanced Practice Test- 9 - Question 19

Hybridisation of S in H2SO3 = 1/2 (6+2+0) = 4 = sp3 (3σ + 1π)

JEE Advanced Practice Test- 9 - Question 20

Find out the correct representation of trans - decaline

Detailed Solution for JEE Advanced Practice Test- 9 - Question 20

JEE Advanced Practice Test- 9 - Question 21

50 ml of a solution containing 10-3 mol of Ag+ is mixed with 50 ml of a 0.1 M HCl solution. Howmuch [Ag+] remains in solution? Given: Ksp of AgCl = 10-10

Detailed Solution for JEE Advanced Practice Test- 9 - Question 21

JEE Advanced Practice Test- 9 - Question 22

Where X is a compound that forms azodye with benzene diazonium chloride in faintly basic medium.
Hence the products P, X and Y are respectively.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 22

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 23

Which of the following solutions show lowering of vopour pressure on mixing?

Detailed Solution for JEE Advanced Practice Test- 9 - Question 23

In B and D compound after mixing undergoes H – bonding and hence boiling point will be raised and hence vapour pressure lowered.

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 24

Which of the following will produce aromatic compound as a product?

Detailed Solution for JEE Advanced Practice Test- 9 - Question 24

JEE Advanced Practice Test- 9 - Question 25

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C) and D) out of WHICH ONLY ONE IS CORRECT.

Paragraph For Question No.12 and 13

A white compound (A) on strong heating decomposes to produce two products (B) and (C). (B) on reaction with white phosphorus produces (D), which is a strong dehydrating agent. (D) on reaction with perchloric acid converts it to its anhydride.

The compound (A) is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 25

JEE Advanced Practice Test- 9 - Question 26

Paragraph For Question No.12 and 13

A white compound (A) on strong heating decomposes to produce two products (B) and (C). (B) on reaction with white phosphorus produces (D), which is a strong dehydrating agent. (D) on reaction with perchloric acid converts it to its anhydride.
The product/s obtained on hydrolysis of (D) is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 26

JEE Advanced Practice Test- 9 - Question 27

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The percent distribution of C in the product mixture B, C and D at any time is equal to

Detailed Solution for JEE Advanced Practice Test- 9 - Question 27

If k1, k2 and k3 be the rate constants of the reaction along path I, II and III respectively, then overall rate constant of consumption of A will be k1 + k2 + k3. So,

JEE Advanced Practice Test- 9 - Question 28

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The initial rate of consumption of A and the sum of the initial rate of formation of B, C and D arerespectively, taking [A] = 0.25 M, equal to

Detailed Solution for JEE Advanced Practice Test- 9 - Question 28

JEE Advanced Practice Test- 9 - Question 29

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The overall energy of activation of A along all the three parallel path is equal to

Detailed Solution for JEE Advanced Practice Test- 9 - Question 29

JEE Advanced Practice Test- 9 - Question 30

Match the following ;

Detailed Solution for JEE Advanced Practice Test- 9 - Question 30

JEE Advanced Practice Test- 9 - Question 31

Among the following, the molecule with the highest dipole moment is

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 32

5.6 L an unknown gas at NTP requires 12.5 calories to rise its temperature by 10oC at constant volume. What is the atomicity of the gas?

Detailed Solution for JEE Advanced Practice Test- 9 - Question 32

The gas is thus diatomic

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 33

After electrolysis of a NaCl solution with inert electrodes for a certain period of time, 60 ml of the solution was left which was found to be 1 N in NaOH. During the same time 31.75 g of Cu was deposited in a Cu voltameter in series with the electrolytic cell. Calculate the percentage of the theoretical yield of the sodium hydroxide obtained.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 33

No. of eq. of NaOH that can be produced theoretically for 100% current efficiency = no of equivalents of NaCl decomposed = no of eq. of Cu deposited

No. of eq. of NaOH produced experimentally =

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 34

Of the following amines how many can be separated by Hoffmann’s mustard oil reaction?

Detailed Solution for JEE Advanced Practice Test- 9 - Question 34

Hoffmann’s mustard oil reaction is a test of 1o amine.

JEE Advanced Practice Test- 9 - Question 35

If the equation x4 – 4x3 + ax2 + bx + 1 = 0 has four positive roots then

Detailed Solution for JEE Advanced Practice Test- 9 - Question 35

JEE Advanced Practice Test- 9 - Question 36

Find the point p on the line 2x + 3y + = 0 such that |PA - PB| is maximum where A is (2,0) and B is (0,2)

Detailed Solution for JEE Advanced Practice Test- 9 - Question 36

Thus the max value of |PA – PB| is AB
This is possible only when P lies on AB but P lies on AB
∴ P is the point of intersection of x + y = 2 and 2x + 3y + 1 = 0.

JEE Advanced Practice Test- 9 - Question 37

Find the coefficient of x5 in the expansion of (2 – x + 3x2)6

Detailed Solution for JEE Advanced Practice Test- 9 - Question 37

\

JEE Advanced Practice Test- 9 - Question 38

If tan3θ + tanθ = 2tan2θ then θ is equal to (n ∈ z)

Detailed Solution for JEE Advanced Practice Test- 9 - Question 38

θ = nπ,n ∈ z or 2θ = nπ ∵  θ = nπ/2  is not for possible as n is odd tanθ is not define.
Hence θ = nπ, n ∈ z is the only solution.

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 39

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct

If x, y, z are respectively perpendiculars from the circumcenter to the sides of ΔABC, (a, b, c are usual meanings) then

Detailed Solution for JEE Advanced Practice Test- 9 - Question 39

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 40

Which of the following is/are correct

Detailed Solution for JEE Advanced Practice Test- 9 - Question 40

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 41

The angle between the lines whose direction cosine are connected by the relations ℓ –5m+3n=0 and 7 ℓ2 + 5m2 – 3n2 = 0

Detailed Solution for JEE Advanced Practice Test- 9 - Question 41

*Multiple options can be correct
JEE Advanced Practice Test- 9 - Question 42

If  are three mutually perpendicular unit vectors and   is a unit vector which makes  equal angle with   and   then the value of

Detailed Solution for JEE Advanced Practice Test- 9 - Question 42

JEE Advanced Practice Test- 9 - Question 43

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions and based on the other paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct

Paragraph for Question Nos. 12 to 13

PQ is the double ordinate of the parabola y2 = 4x which passes through the focus S. ΔPQA is an isosceles right angle triangle, where A is on the axis of the parabola. Line PA meets the parabola at C and QA meets the parabola at B.

The area of trapezium PBCQ is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 43

end point of latus rectum P(1, 2) & Q(1, -2) ΔPAQ is isosceles right angled

∴ slope of PA is – 1
In equation y – 2 = - (x – 1)
x + y – 3 = 0

similarly equation of line QB
x – y – 3 = 0
solving x + y – 3 = 0 with parabola y2 = 4x

(3-x)2 = 4x

x = 1,9
∴ co-ordinate of B & C are (9, -6) & (9, 6) respectively
Area of trapezium

= 1/2 x (12 + 4) x 8
= 64 sq units

JEE Advanced Practice Test- 9 - Question 44

PQ is the double ordinate of the parabola y2 = 4x which passes through the focus S. ΔPQA is an isosceles right angle triangle, where A is on the axis of the parabola. Line PA meets the parabola at C and QA meets the parabola at B.

The circumradius of trapezium PBCQ is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 44

let the circumcenter of trapezium be T(n, 0)
Then PT = PB

or n = 7

JEE Advanced Practice Test- 9 - Question 45

Let z be a complex number satisfying z2 + 2αz + 1 = 0 where α is a parameter which can take any real value.

The roots of this equation lie on a certain circle if

Detailed Solution for JEE Advanced Practice Test- 9 - Question 45

one root lies inside the unit circle and other will outside the unit circle case||| where α is very large then

JEE Advanced Practice Test- 9 - Question 46

Amongst the following, the most basic compound is

JEE Advanced Practice Test- 9 - Question 47

PQ is the double ordinate of the parabola y2 = 4x which passes through the focus S. ΔPQA is an isosceles right angle triangle, where A is on the axis of the parabola. Line PA meets the parabola at C and QA meets the parabola at B.

For every large value of α the roots are approximately

Detailed Solution for JEE Advanced Practice Test- 9 - Question 47

one root lies inside the unit circle and other will outside the unit circle

case||| where α is very large then

JEE Advanced Practice Test- 9 - Question 48

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as
illustrated in the following example:
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

A function is defined as f; { a1,a2,a3,a4,a,,a6 } → { b1,b2,b3 }

Detailed Solution for JEE Advanced Practice Test- 9 - Question 48

total number of function = 36 = 729
(a) total number of onto  function

(b) since f(ai) ≠ bi
It means that a1, a2, a3 cannot be assigned images b1, b2, b3
Number of function = 2333 = 216
(c) number of invertible function = 0
as function is not one – one
(d) total many one function = 729 – 0
= 729

JEE Advanced Practice Test- 9 - Question 49

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as
illustrated in the following example:
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following:

Match the following:

Detailed Solution for JEE Advanced Practice Test- 9 - Question 49

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 50

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9(both inclusive).
Two friends have equal number of sons. There are 3 tickets for a cricket match which are to be distributed among the sons. The probability that two tickets go to the sons of one and one tickets goes to the sons of other is 6/7 . Then total number of boys is equal to (sum of son of each friend).

Detailed Solution for JEE Advanced Practice Test- 9 - Question 50

let each friend has n sons.
∴ 3 tickets can be distributed among 2n sons in 2nC3 ways.
The number of ways distributing 3 tickets such that two tickets go to the sons of one and one tickets goes to sons of the other.
nC2 x nC1 + nC1 x nC2 = 2 x nC1 x nC2  probability that two tickets go to the sons of one and one tickets goes the sons of the other

But from the question

hence total number of boys = 8

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 51

A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.
Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to the
hyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .

Detailed Solution for JEE Advanced Practice Test- 9 - Question 51

the equation of the tangents to the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q are

Where y = mx is intersecting line through (2, 0)
Let (h, k) be the point of intersection of these two lines

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 52

In a ΔABC,    then the value of  |Δ|,where |•| denote absolute value.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 52

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 53

then p+q is

Detailed Solution for JEE Advanced Practice Test- 9 - Question 53

*Answer can only contain numeric values
JEE Advanced Practice Test- 9 - Question 54

The area of region in first quadrant in which points are nearer to the origin then to the line x = 3.

Detailed Solution for JEE Advanced Practice Test- 9 - Question 54