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JEE Advanced Part Test- 9 Free Online Test 2026


MCQ Practice Test & Solutions: JEE Advanced Part Test- 9 (72 Questions)

You can prepare effectively for JEE Mock Tests for JEE Main and Advanced 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "JEE Advanced Part Test- 9". These 72 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Number of Questions: 72

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JEE Advanced Part Test- 9 - Question 1

A particle performs SHM with a time period T and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/T from the extreme position is :

Detailed Solution: Question 1

JEE Advanced Part Test- 9 - Question 2

A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration √3 g m/s2. The period of small oscillations of the pendulum about its equilibrium position is (g = π2 m/s2) :    

Detailed Solution: Question 2

With respect to the cart, equilibrium position of the pendulum is shown.If displaced by small angle θ from this position, then it will execute SHM about this equilibrium position,
time period of which is given by :

JEE Advanced Part Test- 9 - Question 3

A particle is subjected to two simple harmonic motions along x and y directions according to, x = 3 sin 100 πt; y = 4 sin 100 πt :

Detailed Solution: Question 3

JEE Advanced Part Test- 9 - Question 4

When a wave pulse traveling in a string is reflected from a rigid wall to which string is tied as shown in figure. For this situation two statements are given below :                 

   

    (1) The reflected pulse will be in same orientation of incident pulse due to a phase change of p radians
    (2) During reflection the wall exert a force on string in upward direction
    For the above given two statements choose the correct option given below :

Detailed Solution: Question 4

Reflected pulse will be inverted as it is reflected by a denser medium. The wall exerts force in downward direction.
 

JEE Advanced Part Test- 9 - Question 5

The equation of displacement due to a sound wave is s = s0 sin2 (ω t - kx). If the bulk modulus of the medium is B, then the equation of pressure variation due to that sound is :

Detailed Solution: Question 5

JEE Advanced Part Test- 9 - Question 6

A point source of power 50π watts is producing sound waves of frequency 1875Hz. The velocity of sound is 330m/s, atmospheric pressure is 1.0 x 105 Nm-2, density of air is 1.0 kgm-3.  Then pressure amplitude at r =  m from the point source is (using π = 22/7) :      

Detailed Solution: Question 6

V are power, pressure amplitude and velocity respectively.

JEE Advanced Part Test- 9 - Question 7

An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are :

Detailed Solution: Question 7

JEE Advanced Part Test- 9 - Question 8

A wire having a linear mass density 5.0 ´ 10-3 kg/m is stretched between two rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 480 Hz. The length of the wire is :

Detailed Solution: Question 8

Two consecutive frequencies are 420 Hz & 480 Hz. So the fundamental frequency will be 60 Hz.

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

A particle is executing SHM between points -Xm and Xm, as shown in figure-I. The velocity V(t) of the particle is partially graphed and shown in figure-II. Two points A and B corresponding to time t1 and time t2 respectively are marked on the V(t) curve :

Detailed Solution: Question 9

At time t1, velocity of the particle is negative i.e. going towards –Xm. From the graph, at time t1, its speed is decreasing. Therefore particle lies in between –Xm and 0.
At time t2, velocity is positive and its magnitude is less than maximum i.e. it has yet not crossed O.
It lies in between –Xm and 0.
Phase of particle at time t1 is (180 + θ1).
Phase of particle at time t2 is (270 + θ2)
Phase difference is 90 + (θ2 – θ1)
θ2 – θ1 can be negative making  < 90° but can not be more than 90°.

*Multiple options can be correct
JEE Advanced Part Test- 9 - Question 10

In a standing wave on a string rigidly fixed at both ends.

Detailed Solution: Question 10

*Multiple options can be correct
JEE Advanced Part Test- 9 - Question 11

For a certain transverse standing wave on a long string, an antinode is formed at x = 0 and next to it, a node is formed at x = 0.10 m. the displacement y(t) of the string particle at x = 0 is shown in figure.

Detailed Solution: Question 11

*Multiple options can be correct
JEE Advanced Part Test- 9 - Question 12

A wave pulse moving in the positive x-direction along the x-axis is represented by the wave function  where x and y are in centimeters and t is in seconds. Then  

Detailed Solution: Question 12

JEE Advanced Part Test- 9 - Question 13

Statement-1 : A particle is moving along x-axis. The resultant force F acting on it is given by F = – ax – b. Where a and b are both positive constants. The motion of this particle is not SHM.

Statement-2 : In SHM resultant force must be proportional to the displacement from mean position.

Detailed Solution: Question 13

The mean position of the particle in statement-1 is x = -(b/a). and the force is always proportional to displacement
from this mean position. The particle executes SHM about this mean position. Hence statement-1 is false

JEE Advanced Part Test- 9 - Question 14

Statement-1 : Two waves moving in a uniform string having uniform tension cannot have different velocities.

Statement-2 : Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only.

Detailed Solution: Question 14

Two waves moving in uniform string with uniform tension shall have same speed and may be moving in opposite directions. Hence both waves may have velocities in opposite direction. Hence statement-1 is false.
 

JEE Advanced Part Test- 9 - Question 15

Statement-1 : In a small segment of string carrying sinusoidal wave, total energy is conserved.

Statement-2 :  Every small part moves in SHM and in SHM total energy is conserved.

Detailed Solution: Question 15

Every small segment is acted upon by forces from both sides of it hence energy is not conserved, rather it is transmitted by the element.
 

JEE Advanced Part Test- 9 - Question 16

Statement-1 : When two vibrating tuning forks having frequencies 256 Hz and 512 Hz  are held near each other, beats cannot be heard.

Statement-2 : The principle of superposition of waves is valid only if the frequencies of both the interfering waves are equal or nearly equal.

Detailed Solution: Question 16

The principle of superposition of waves is always valid.
 

JEE Advanced Part Test- 9 - Question 17

A small block of mass m is fixed at upper end of a massless vertical spring of spring constant K =   and natural length '10L'. The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring-block system is released from rest in the shown position.

   

Q. At the instant speed of block is maximum, the magnitude of force exerted by spring on the block is :

Detailed Solution: Question 17

When speed of block is maximum, net force on block is zero. Hence at that instant spring exerts a force of magnitude 'mg' on block.
 

JEE Advanced Part Test- 9 - Question 18

A small block of mass m is fixed at upper end of a massless vertical spring of spring constant K =  and natural length '10L'. The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring-block system is released from rest in the shown position.

Q. As the block is coming down, the maximum speed attained by the block is :

Detailed Solution: Question 18

At the instant block is in equilibrium position, its speed is maximum and compression in spring is x given by 

JEE Advanced Part Test- 9 - Question 19

A small block of mass m is fixed at upper end of a massless vertical spring of spring constant K =   and natural length '10L'. The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring-block system is released from rest in the shown position.

Q. Till the block reaches its lowest position for the first time, the time duration for which the spring remains compressed is  :

Detailed Solution: Question 19

JEE Advanced Part Test- 9 - Question 20

There is a point source of sound placed at (0, h) as shown in figure. Two detectors D1 and D2 are placed at positions (D,d/2) and (D,– d/2) respectively. Take h < < D. The source emitted a sound pulse at a certain time. Assuming velocity of sound in the surrounding medium is v. 

Q. The time gap between the recordings made by the detectors will approximately be :  

Detailed Solution: Question 20

JEE Advanced Part Test- 9 - Question 21

There is a point source of sound placed at (0, h) as shown in figure. Two detectors D1 and D2 are placed at positions (D,d/2) and (D,– d/2) respectively. Take h < < D. The source emitted a sound pulse at a certain time. Assuming velocity of sound in the surrounding medium is v. 

Q. If the source emits continuous waves, and the pressures recorded by the two detectors are superposed at every instant in detector D0 (which is equidistant from D1 & D2), the resultant pressure amplitude will be maximum if the minimum frequency of the source is :  

Detailed Solution: Question 21

JEE Advanced Part Test- 9 - Question 22

There is a point source of sound placed at (0, h) as shown in figure. Two detectors D1 and D2 are placed at positions (D,d/2) and (D,– d/2) respectively. Take h < < D. The source emitted a sound pulse at a certain time. Assuming velocity of sound in the surrounding medium is v. 

Q. If the source is shifted slightly towards positive X direction. The minimum frequency required for the super posed pressure amplitude (detected at D0) to be maximum will (as compared to the answer in above question) :  

Detailed Solution: Question 22

JEE Advanced Part Test- 9 - Question 23

Detailed Solution: Question 23

JEE Advanced Part Test- 9 - Question 24

Match the statements in column-I with the statements in column-II.
            Column-I                                                            Column-II 
  (A)  A tight string is fixed at both ends and         (p)   At the middle, antinode is formed 
        sustaining standing wave                                       in odd harmonic
  (B)  A tight string is fixed at one end and           (q)   At the middle, node is formed 
        free at the other end                                              in even harmonic
  (C)  Standing wave is formed in an open organ   (r)    At the middle, neither node nor 
        pipe. End correction is not negligible.                     antinode is formed
  (D)  Standing wave is formed in a closed           (s)    Phase difference between SHMs of any
        organ pipe. End correction is not negligible.            two particles will be either p or zero.  

                                                                        (t)    The displacement of the particle in the                                                                                                                                                        middle is always non zero.

Detailed Solution: Question 24

(A) Number of loops (of length λ/2) will be even or odd and node or antinode will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(C) Number of loops (of length λ/2) will be even or odd and antinode or node will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be π ..

JEE Advanced Part Test- 9 - Question 25

Acidic strength of marked hydrogen of following compound in decreasing order is

Detailed Solution: Question 25

JEE Advanced Part Test- 9 - Question 26

The correct IUPAC name of following compound is

Detailed Solution: Question 26

JEE Advanced Part Test- 9 - Question 27

A hydrocarbon (P) on ozonolysis in presence of zinc gives only one dicarbonyl compound, which gives both Tollen’s and iodoform test. Identify the structure of (P).

Detailed Solution: Question 27

JEE Advanced Part Test- 9 - Question 28

Consider the true/false of the following statements:
    S1: The most stable resonating structure of p-nitrophenol (not having aromatic ring) is .
    S2 : In  all C–O bonds are of equal length.
    S3 : CH3COONa is more resonance stabilised than the protonated acid CH3COOH.
    S4 : Benzene ring is more electron dense in phenol than phenoxide.

Detailed Solution: Question 28

S1: Most stable resonating structure is 

S4 : In  group has greater +m effect, so it makes phenoxide ion more electron dense.

JEE Advanced Part Test- 9 - Question 29

Choose the strongest base among the following :

Detailed Solution: Question 29

Option D is the strongest base.

The dominant factor controlling basicity here is how available the lone pair on nitrogen is to accept a proton. If the lone pair is delocalised by resonance with an aromatic ring or is part of an aromatic system, its availability for protonation is reduced.

In option A the nitrogen lone pair is on an sp2-hybridised nitrogen directly attached to the benzene ring and is delocalised by resonance with the ring; this decreases basicity compared with an aliphatic amine.

In options B and C the nitrogen lone pair is also either conjugated with the aromatic system or involved in stabilising interactions with the ring; as a result the lone pair is less available and these compounds are less basic than a non-conjugated amine.

In option D the nitrogen is sp3-hybridised and its lone pair is not conjugated with the benzene ring, so the lone pair is more localized and more available to accept a proton. Any adjacent alkyl-type framework gives an additional electron-donating inductive effect, further increasing basicity.

Therefore, when resonance delocalisation is compared with an isolated sp3 nitrogen, the isolated sp3 nitrogen (option D) is the strongest base.

JEE Advanced Part Test- 9 - Question 30

A hydrocarbon ‘X’ C7H10 is catalytically hydrogenated to C7H14 ‘Y’. 'Y' gives six monochloro products after photochemical chlorination. The structure of 'X'  is -

Detailed Solution: Question 30

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