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# Part Test - 9 (JEE Advanced 2021)

## 72 Questions MCQ Test National Level Test Series for JEE Advanced 2020 | Part Test - 9 (JEE Advanced 2021)

Description
This mock test of Part Test - 9 (JEE Advanced 2021) for JEE helps you for every JEE entrance exam. This contains 72 Multiple Choice Questions for JEE Part Test - 9 (JEE Advanced 2021) (mcq) to study with solutions a complete question bank. The solved questions answers in this Part Test - 9 (JEE Advanced 2021) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Part Test - 9 (JEE Advanced 2021) exercise for a better result in the exam. You can find other Part Test - 9 (JEE Advanced 2021) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Solution:

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) :

Solution:

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 :

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 :

Solution:

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 :

Solution:

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

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 :

Solution:

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) :

Solution:

V are power, pressure amplitude and velocity respectively.

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 :

Solution:

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 :

Solution:

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

*Multiple options can be correct
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 :

Solution:

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
QUESTION: 10

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

Solution:

*Multiple options can be correct
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.

Solution:

*Multiple options can be correct
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

Solution:

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.

Solution:

The mean position of the particle in statement-1 is x = -(b/a). and the force is always proportional to displacement

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.

Solution:

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.

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.

Solution:

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.

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.

Solution:

The principle of superposition of waves is always valid.

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 :

Solution:

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.

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 :

Solution:

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

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  :

Solution:

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 :

Solution:

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 :

Solution:

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) :

Solution:

QUESTION: 23

Solution:

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.

Solution:

(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 ..

QUESTION: 25

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

Solution:

QUESTION: 26

The correct IUPAC name of following compound is

Solution:

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).

Solution:

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.

Solution:

S1: Most stable resonating structure is

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

QUESTION: 29

Choose the strongest base among the following :

Solution:

Only in (D) the l.p of N atom is not involved in resonance with benzene ring.

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 -

Solution:

QUESTION: 31

The feasible reaction is :

Solution:

Salicylic acid  is more acidic than p-hydroxy benzoic acid.

QUESTION: 32

The incorrect information about the given reaction is :

Solution:

The compound (I) does not racemise during enolisation, because the carbon atom with α-H is not asysmmetric carbon atom.

*Multiple options can be correct
QUESTION: 33

For Cyclooctatetraene which is/are correct :

Solution:

*Multiple options can be correct
QUESTION: 34

The compound in which resonance is/are possible :

Solution:

There is conjugation in A,B,D but not in C.

*Multiple options can be correct
QUESTION: 35

Which of the following compounds will show hyperconjugation ?

Solution:

Except carbanion all the above compounds have α-H which can show hyperconjugation.

*Multiple options can be correct
QUESTION: 36

Which of the following statement/s is/are true about the following compounds.

Solution:

(I)  is cyclodec-2-en-1-one whereas (II) is cyclodec-3-en-1-one so they are structural isomers. (III) is just a trans-isomer of (I) cis-isomer

QUESTION: 37

Statement-1 :  is stronger acid than

Statement-2 : ‘F’ exerts stronger –I effect and weaker + m effect than –OH group so the conjugate base of F–COOH is more stable anion.

Solution:

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

QUESTION: 38

Statement-1 : Carbon–oxygen bonds are of equal length in acetate ion.

Statement-2 : Bond length decreases with the multiplicity of bond between two atoms.

Solution:

Acetate ion shows resonance and thus both the bonds have identical bond length.

QUESTION: 39

Statement-1 : Resonance energy of phenanthrene is more than anthracene.

Statement-2 : Phenanthrene has more aromatic character and more delocalisation than anthrance

Solution:

Phenanthrene has five resonating structures while anthracene has four resonating structures only.

QUESTION: 40

Statement-1 : In gas phase oxalic acid has zero dipole moment.

Statement-2  : It is due to intramolecular H bonding in the following way.

Solution:

In gas phase, zero dipole moment is achieved dut to intramolecular H-bonding as shown in the above figure.

QUESTION: 41

Broadly speaking, there are four types of stereoisomers, namely conformational isomers, geometrical isomers, enantiomers and diastereomers. Although conformational isomers, have same configuration, geometrical isomers have different atoms attached to each of the doubly bonded carbon atom, and enantiomers are due to chirality in the molecule. Enantiomers are also known as optical isomers and diastereomers are those stereoisomers which are not enantiomers. In simple compounds, number of stereoisomers in achiral molecule is given by the relation 2n, where n is the number of dissimilar chiral carbon.

Observe the following structures and answer the questions given below :

Q. How many stereoisomers are possible in pent-3-en-2-ol, CH3CH=CHCHOHCH3

Solution:

CH3CH=CHCH(OH)CH3 has two stereocentres. Hence 4 stereoisomers.

QUESTION: 42

Broadly speaking, there are four types of stereoisomers, namely conformational isomers, geometrical isomers, enantiomers and diastereomers. Although conformational isomers, have same configuration, geometrical isomers have different atoms attached to each of the doubly bonded carbon atom, and enantiomers are due to chirality in the molecule. Enantiomers are also known as optical isomers and diastereomers are those stereoisomers which are not enantiomers. In simple compounds, number of stereoisomers in achiral molecule is given by the relation 2n, where n is the number of dissimilar chiral carbon.

Observe the following structures and answer the questions given below :

Q. The correct names for the above structure is :

Solution:

In naming of the above compound, follow the rules for priority order (E,Z) and (R,S) system.

QUESTION: 43

Broadly speaking, there are four types of stereoisomers, namely conformational isomers, geometrical isomers, enantiomers and diastereomers. Although conformational isomers, have same configuration, geometrical isomers have different atoms attached to each of the doubly bonded carbon atom, and enantiomers are due to chirality in the molecule. Enantiomers are also known as optical isomers and diastereomers are those stereoisomers which are not enantiomers. In simple compounds, number of stereoisomers in achiral molecule is given by the relation 2n, where n is the number of dissimilar chiral carbon.

Observe the following structures and answer the questions given below :

Q. The above structure can have how many other diastereomers ?

Solution:

Other stereisomers  of above (2R,3E) stereoisomer are (2R,3Z), (2S,3E), (2S,3Z) out of this (2S,3Z) is enantiomers. Hence no of diastereomers = 2.

QUESTION: 44

Ortho effect is a special type of effect that is shown by o-substituents, but it is not necessarily just a steric effect. This ortho-effect operates with  the benzoic acids. Irrespective of the polar type nearly all o-substituted benzoic acid are stronger than unsubstituted benzoic acid. Benzoic acid is a resonance hybrid and so the carboxyl group is coplanar with the ring. An o-substitueut tends to prevent this coplanarity. Ortho effect also operates in substituted anilines where as ortho substituent has base weakening effect.

Q. What is the order of acidity of following compounds :

Solution:

(i) Is strongest due to intramolecular bond formation in conjugate base.
(iv) Comes next due to intramolecular H bonding wih OH group.
(iii) Comes next due to ortho effect.

QUESTION: 45

Ortho effect is a special type of effect that is shown by o-substituents, but it is not necessarily just a steric effect. This ortho-effect operates with  the benzoic acids. Irrespective of the polar type nearly all o-substituted benzoic acid are stronger than unsubstituted benzoic acid. Benzoic acid is a resonance hybrid and so the carboxyl group is coplanar with the ring. An o-substitueut tends to prevent this coplanarity. Ortho effect also operates in substituted anilines where as ortho substituent has base weakening effect.

Q. What is the acidity order of the following compounds ?

Solution:

(V) Is most acidic due to SIR effect.
(III) Comes next due to ortho effect.
In (II) (IV) CH3 group decreases acidic strength due to +I effect and +H effect respectively.

QUESTION: 46

Ortho effect is a special type of effect that is shown by o-substituents, but it is not necessarily just a steric effect. This ortho-effect operates with  the benzoic acids. Irrespective of the polar type nearly all o-substituted benzoic acid are stronger than unsubstituted benzoic acid. Benzoic acid is a resonance hybrid and so the carboxyl group is coplanar with the ring. An o-substitueut tends to prevent this coplanarity. Ortho effect also operates in substituted anilines where as ortho substituent has base weakening effect.

Q. What is the order of basicity of following compounds?

Solution:

Due to electronic effects.

QUESTION: 47

Match the type of isomers mentined in column - I with the specific notations used in differant kinds of isomerism in column - II
Column-I                                     Column-II
(A)    Geometrical isomers            (p)    E,Z
(B)    Enantiomers                        (q)    anti-gauche
(C)    Diastereomers                     (r)     anti-syn
(D)    Conformational isomers        (s)    R,S
(t)     D, L

Solution:

Geometrical isomers are named as (E,Z) or (anti, syn) or (cis,trans). Enantiomers are named according to (R,S) nomenclature.Diasteriomers existsboth ingeometricalandopticalisomers Hence, allfour kinds of nomenculature, namely (p), (r), (s) and (t) . Conformational isomers are named as gauche form, anti form eclipsed form and partially eclipsed form.

QUESTION: 48

Solution:

(A) t-Butyl alcohol will not give bicarbonate test, iodoform test, tollen's test and 2,4 DNP test.

is ketone. It will not give bicarbonate test (test for acids), Lucas test (test of alcohols) and Tollen's test (test for aldehydes) but it will give 2,4 DNP test.
(C) Picric acid gives bicarbonate test but it will not give test of other functional groups.
(D) Ozonolysis product is 3 moles of CHO–CHO being aldehyde it will give Tollen's test and 2,4 DNP test but not
other tests.

QUESTION: 49

In a triangle PQR as shown in figure given that x : y : z = 2 : 3 : 6.
Then the value of ÐQPR is –

Solution:

QUESTION: 50

In a triangle ABC,

Solution:

QUESTION: 51

If in an equilateral triangle, inradius is a rational number then which of the following is not true.

Solution:

QUESTION: 52

In a ΔABC, tangent of half of difference of two angles is  1/3 the tangent of half of the sum of two angles, Ratio of the sides opposite to these angles is –

Solution:

QUESTION: 53

Solution:

QUESTION: 54

Solution:

QUESTION: 55

The value of the expression

Solution:

QUESTION: 56

Base angles of triangle are then height of the triangle is –

Solution:

*Multiple options can be correct
QUESTION: 57

If in ΔABC, A = 90°, c, cos B, and sin B are rational numbers then –

Solution:

By sine rule a & b are rational

*Multiple options can be correct
QUESTION: 58

In a triangle length of two longer sides are 10 and 9 and its angles are in A.P. then length of smaller side may be –

Solution:

*Multiple options can be correct
QUESTION: 59

For function f(x) = ln (sin–1 (logπ (cos–1x)))

Solution:

*Multiple options can be correct
QUESTION: 60

Solution:

QUESTION: 61

Statement - 1: In a Δ ABC a, b, c denotes length of the sides and  then triangle is quilateral.

Statement - 2 : Sum of non-negative numbers = 0 ⇒ each number is zero.

Solution:

QUESTION: 62

Statement -1 : In ΔABC if r1 = 2r2 = 3r3 then a : b = 5 : 4

Statement -2 : In ΔABC if xr1 = yr2 = zr3 = (x + y + z) r, then a : b : c = y + z : x + z : x + y

Solution:

QUESTION: 63

Statement -1 : If angle A and B satisfy the equation tan + p tanθ – 1 = 0 then triangle ABC is right angled

Solution:

QUESTION: 64

Statement -1 : In ΔABC, (a + b + c) (b + c – a) = λ be possible only if 0 < λ < 4.

Statement -2 : – 1 < cos θ < 1 where 'θ' is an angle of triangle

Solution:

QUESTION: 65

Let the inverse of the function f : f-1(x) = sin-1x. Here it should be noted that the function sin-1x does not have the conventional range.
The graph of f-1 is
Also for x, y > 0, we have

Q. The solution of the inequality (sin–1x)2 – 3 sin–1 x + 2 > 0 is

Solution:

QUESTION: 66

Let the inverse of the function f : f-1(x) = sin-1x. Here it should be noted that the function sin-1x does not have the conventional range.
The graph of f-1 is
Also for x, y > 0, we have

Q. The complete solution set of the equation  is

Solution:

QUESTION: 67

Let the inverse of the function f : f-1(x) = sin-1x. Here it should be noted that the function sin-1x does not have the conventional range.
The graph of f-1 is
Also for x, y > 0, we have

Q. The value of   is

Solution:

QUESTION: 68

DEF is the triangle formed by joining the points of contact of the incircle with the sides of the ΔABC ;

Q. Sides of triangle DEF are

Solution:

QUESTION: 69

DEF is the triangle formed by joining the points of contact of the incircle with the sides of the ΔABC ;

Q. Angles of triangle DEF are

Solution:

QUESTION: 70

DEF is the triangle formed by joining the points of contact of the incircle with the sides of the ΔABC ;

Q. Area of triangle DEF is

Solution:

QUESTION: 71

Solution:

QUESTION: 72

Solution:

equilatral triangle

(C) a cos B = b cos A
By projection formula triangle is isoseles
(D) a cos A = b cos B
By projection formula triangle is equilateral, so isoseles also.