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Part Test - 9 (JEE Advanced)


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Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 1

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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 :

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 3

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 4

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

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 5

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 6

V are power, pressure amplitude and velocity respectively.

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 7

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 8

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

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

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 10

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

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

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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.
 

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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.
 

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 16

The principle of superposition of waves is always valid.
 

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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.
 

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 18

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

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 19

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 20

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 21

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 22

Part Test - 9 (JEE Advanced) - Question 23

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 23

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - 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 ..

Part Test - 9 (JEE Advanced) - Question 25

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 25

Part Test - 9 (JEE Advanced) - Question 26

The correct IUPAC name of following compound is

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 26

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 27

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 28

S1: Most stable resonating structure is 

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

Part Test - 9 (JEE Advanced) - Question 29

Choose the strongest base among the following :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 29

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

Part Test - 9 (JEE Advanced) - 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 for Part Test - 9 (JEE Advanced) - Question 30

Part Test - 9 (JEE Advanced) - Question 31

The feasible reaction is :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 31

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

Part Test - 9 (JEE Advanced) - Question 32

The incorrect information about the given reaction is :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 32

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
Part Test - 9 (JEE Advanced) - Question 33

For Cyclooctatetraene which is/are correct :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 33

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 34

The compound in which resonance is/are possible :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 34

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

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 35

Which of the following compounds will show hyperconjugation ?

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 35

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

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 36

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 36

(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

Part Test - 9 (JEE Advanced) - 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.

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 37

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

Part Test - 9 (JEE Advanced) - 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.

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 38

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

Part Test - 9 (JEE Advanced) - Question 39

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

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 39

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

Part Test - 9 (JEE Advanced) - 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.

  

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 40

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

Part Test - 9 (JEE Advanced) - 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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 41

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

Part Test - 9 (JEE Advanced) - 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 : 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 42

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

Part Test - 9 (JEE Advanced) - 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 ? 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 43

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.
 

Part Test - 9 (JEE Advanced) - 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 :

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 44

(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.

Part Test - 9 (JEE Advanced) - 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 ?

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 45

(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.

Part Test - 9 (JEE Advanced) - 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?

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 46

Due to electronic effects.

Part Test - 9 (JEE Advanced) - 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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 47

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.

Part Test - 9 (JEE Advanced) - Question 48

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 48

(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.

Part Test - 9 (JEE Advanced) - Question 49

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 49

Part Test - 9 (JEE Advanced) - Question 50

In a triangle ABC, 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 50

Part Test - 9 (JEE Advanced) - Question 51

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 51

Part Test - 9 (JEE Advanced) - 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 – 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 52

Part Test - 9 (JEE Advanced) - Question 53

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 53

Part Test - 9 (JEE Advanced) - Question 54

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 54

Part Test - 9 (JEE Advanced) - Question 55

The value of the expression 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 55

Part Test - 9 (JEE Advanced) - Question 56

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 56

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 57

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 57

By sine rule a & b are rational

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - 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 –

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 58

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 59

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 59

*Multiple options can be correct
Part Test - 9 (JEE Advanced) - Question 60

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 60

Part Test - 9 (JEE Advanced) - 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.

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 61

Part Test - 9 (JEE Advanced) - 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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 62

Part Test - 9 (JEE Advanced) - Question 63

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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 63

Part Test - 9 (JEE Advanced) - 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 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 64

Part Test - 9 (JEE Advanced) - 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 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 65

Part Test - 9 (JEE Advanced) - 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 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 66

Part Test - 9 (JEE Advanced) - 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 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 67

Part Test - 9 (JEE Advanced) - 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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 68

Part Test - 9 (JEE Advanced) - 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

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 69

Part Test - 9 (JEE Advanced) - 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 

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 70

Part Test - 9 (JEE Advanced) - Question 71

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 71

Part Test - 9 (JEE Advanced) - Question 72

Detailed Solution for Part Test - 9 (JEE Advanced) - Question 72

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.

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