Part Test - 10 (JEE Advanced 2020)


72 Questions MCQ Test National Level Test Series for JEE Advanced 2020 | Part Test - 10 (JEE Advanced 2020)


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This mock test of Part Test - 10 (JEE Advanced 2020) for JEE helps you for every JEE entrance exam. This contains 72 Multiple Choice Questions for JEE Part Test - 10 (JEE Advanced 2020) (mcq) to study with solutions a complete question bank. The solved questions answers in this Part Test - 10 (JEE Advanced 2020) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Part Test - 10 (JEE Advanced 2020) exercise for a better result in the exam. You can find other Part Test - 10 (JEE Advanced 2020) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

An interference is observed due to two coherent sources 'A' & 'B' having phase constant zero, separated by a distance 4 l along the y - axis where l is the wavelength of the sources. A detector D is moved on the positive x - axis. The number of points on the x - axis excluding the 
points, x = 0 & x = ¥ at which maximum will be observed are

Solution:

QUESTION: 2

The observer 'O' sees the distance AB as infinitely large. If refractive index of liquid is m1 and that of glass is m2, then   is :

Solution:

Using formula of spherical surface taking 'B' as object 

QUESTION: 3

An object of height 1 cm is kept perpendicular to the principal axis of a convex mirror of radius of curvature 20 cm. If the distance of the object from the mirror is 20 cm then the distance between heads  of the image and the object will be:

Solution:

QUESTION: 4

A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The object is at 15 cm from the lens. The length of the image is:

Solution:

QUESTION: 5

An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :

Solution:

QUESTION: 6

A ray of light falls on a plane mirror. When the mirror is turned, about an axis which is at right angle to the plane of the mirror through 20º the angle between the incident ray and new reflected ray is 45º. The angle between the incident ray and original reflected ray was therefore : 

Solution:

If mirror is turned, about an axis perpendicular to plane of mirror, then there will be no change in incident angle and reflected angle so angle between incident & reflected rays after rotation will be same as before

QUESTION: 7

The focal length of a lens of refractive index 3/2 is 10 cm in air. The focal length of that lens in a medium of refractive index 7/5 is:

Solution:

QUESTION: 8

A converging lens of focal length f is placed just above a water surface parallel to the surface without touching it. A point source of light S is placed inside the water vertically before the lens at a depth f from it. This arrangement will produce:

Solution:

Due to refraction at water-air interface, the object is at a distance less than the focal length of the lense, thus, a virtual image is created inside water

*Multiple options can be correct
QUESTION: 9

A small air bubble is trapped inside a transparent cube of size 12 cm. When viewed from one of the vertical faces, the bubble appears to be at 5 cm from it. When viewed from opposite face, it appears at 3 cm from it.

Solution:

*Multiple options can be correct
QUESTION: 10

A parallel beam of light (l =5000 Å) is incident at an angle a = 30° with the normal to the slit plane in a young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is I0. Point O is equidistant from S1 & S2.The distance between slits is 1mm.    

Solution:

*Multiple options can be correct
QUESTION: 11

Figure shows two thin slabs of glass, one is rectangular and the other is triangular wedge shaped. A monochromatic light incident nearly normally on the slabs as shown in figure.

Solution:

In case x no fringes are formed because light passes the slab normaly & in case y fringes are obtained.

Suppose for point A ‘t’ is such that it satisfies the condition for bright interference. The same ‘t’ will be
present throughout the line A A’ & therefore the line AA’ will be bright & a bright line will be seen. The same
applies for dark lines.Hence fringes are straight line

*Multiple options can be correct
QUESTION: 12

A Young’s double slit experiment is conducted in water (m1) as shown in the figure and a glass plate of thickness t and refractive index m2 is placed in the path of S2. The magnitude of the phase difference at O is (Assume that ‘l’ is the wavelength of light in air). O is symmetrical with respect to S1 and S2 .

Solution:

QUESTION: 13

Statement-1 : Two coherent point sources of light having nonzero phase difference are separated by small distance. Then on the perpendicular bisector of line segment joining both the point sources,  constructive interference cannot be obtained.

Statement-2 : For two waves from coherent point sources to interfere constructively at a point, the magnitude of their phase difference at that point must be 2mp ( where m is a nonnegative integer) .

Solution:

Statement 1 is false because constructive interference can be obtained if phase difference between cohrent
sources is 2π, 4π , 6π, etc.

QUESTION: 14

Statement-1 : A spherical surface of radius of curvature R separates two media of refractive index n1 and n2 as shown. If an object O (a thin small rod) is placed upright on principal axis at a distance R from pole (i.e, placed at centre of curvature), then the size of image is same as size of object.

Statement-2 : If a point object is placed at centre of curvature of spherical surface separating two media of different refractive index, then the image is also formed at centre of curvature, i.e., image distance is equal to object distance.

Solution:

In the situation of statement-1, the magnitude of image (v) and object (u) distance is same.

Hence statement-1 is false and statement-2 is true.

QUESTION: 15

Statement-1 : A parallel beam of light is incident on a thin convex lens and is also parallel to the principal axis of convex lens as shown. The magnitude of deviation of each ray of this beam produced by given convex lens is different.

Statement-2 : A thin convex lens can be assumed to be made of prisms of small angles. The magnitude of deviation produced by prism of small angle for small angles of incidence depends on angle of prism. Each ray of the beam in situation of statement-1 is incident on a prism of different angles, hence the magnitude of deviation for each ray is different.

Solution:

The lens can be assumed to be made of prisms of small but different angles. The lowest ray of beam is incident on
prism of smaller angle while the topmost ray of beam is incident on prism of larger angle. Hence all rays of beam
suffer different deviations. Therefore both statements are correct and statement-2 is the correct explanation.

QUESTION: 16

Statement-1 : A ray is incident from outside on a glass sphere surrounded by air as shown. This ray may suffer total internal reflection at second interface.

Statement 2 : For a ray going from denser to rarer medium, the ray may suffer total internal reflection.

Solution:

From symmetry the ray shall not suffer TIR at second interface, because the angle of incidence at first interface equals to angle of emergence at second interface. Hence statement 1 is false

QUESTION: 17

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n1) is incident at an angle 'q' as shown. Distance S1O = S2O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n1 and n2 respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n1 = 4/3, n2 = ,   10/9

answer the following :

Q. The y-coordinate of the point where the total phase difference between the interefering waves is zero, is 

Solution:

It is negative because upper path in medium n2 is longer than lower path in the same medium.

QUESTION: 18

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n1) is incident at an angle 'q' as shown. Distance S1O = S2O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n1 and n2 respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n1 = 4/3, n2 = 10/9,     
answer the following :

Q. If the intensity due to each light wave at point 'O' is I0 then the resultant intensity at point 'O' will be - 

Solution:

Path lengths in medium 2 are equal for point O. Therefore path difference = d sinθ

QUESTION: 19

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n1) is incident at an angle 'q' as shown. Distance S1O = S2O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n1 and n2 respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n1 = 4/3, n2 = 10/9,     
answer the following :

Q. y-coordinate of the nearest maxima above 'O' will be - 

Solution:

As we go up from point O, path difference will increase. At O, phase difference is 3π + π/3
and when it becomes 4π, there will be maximum. Extra path difference created in medium 2 must lead to 2π/3
phase difference.

QUESTION: 20

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. Minimum focal length of eye lens of a normal person is

Solution:

QUESTION: 21

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. Maximum focal length of eye lens of normal person is

Solution:

QUESTION: 22

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be.

Solution:

QUESTION: 23

Match the following :         
             Column-I                                                    Column-II
    (A)     Object is between optic center and 1st        (p) Image is inverted
             principle focus in a diverging lens
    (B)    Object is between optic center and 1st         (q) Image is Erect
             principle focus of a converging lens
    (C)    Object is between optic center and 2nd         (r) Image is of greater size
             principle focus of a diverging lens                    than the object
    (D)     Object is between optic center and 2nd        (s) Image is of smaller size 
             principle focus of a converging lens                  than the object
                                                                              (t) Image distance is more than |f|

Solution:

QUESTION: 24

A double slit interference pattern is produced on a screen, as shown in the figure, using monochromatic light of wavelength 500 nm. Point P is the location of the central bright fringe, that is produced when light waves arrive in phase without any path difference. A choice of three strips A, B and C of transparent materials with different thicknesses and refractive indices is available, as shown in the table. These are placed over one or both of the slits, singularly or in conjunction, causing the interference pattern to be shifted across the screen from the original pattern. In the column-I, how the strips have been placed, is mentioned whereas in the column-II, order of the fringe at point P on the screen that will be produced due to the placement of the strip(s), is shown. Correctly match both the columns.

Solution:

By using (μ – 1)t = nλ , we can find value of n, that is order of the fringe produced at P, if that particular strip has been placed over any of the slit. If two strips are used in conjuction (over each other), path
difference due to each is added to get net path difference created. If two strips are used over different slits, their path differences are subtracted to get net path difference.

QUESTION: 25

The compound 'X' is :

Solution:

QUESTION: 26

Observe the following reaction and identify the end product.

Solution:

QUESTION: 27

Solution:

QUESTION: 28

The major product of following sequence of reactions is

Solution:

In hofmann elimination the more acidic β-hydrogen is removed.

QUESTION: 29

Solution:

QUESTION: 30

The final product of the following reaction is 

Solution:

QUESTION: 31

Give the product (P) of the above reaction sequence.

Solution:

QUESTION: 32

Solution:

*Multiple options can be correct
QUESTION: 33

Which of the following is/are correctly matched?

Solution:

Terylene contains ester linkage not ketone.

*Multiple options can be correct
QUESTION: 34

Which statements is/are correct about the following reaction :

   

Solution:

The chain propagation step involves 

the most stable R* is formed most readily.

*Multiple options can be correct
QUESTION: 35

An organic compound boiled with HNO3 cooled and then treated with AgNO3 , a white ppt is obtained. The compound may be :

Solution:

The compound with ionisable Cl give white ppt with AgNO3.

*Multiple options can be correct
QUESTION: 36

The alkene limonene has the followning structure.

A product (P) results from the reaction of limonene and chlorine water. Which of the following cannot be 'P' ?

Solution:

QUESTION: 37

Statement-1 Alkylation of benzene by Friedel -Crafts reaction gives polyalkylated benzene.

Statement-2 The ring gets activated for further substitution after the introduction of one alkyl group.

Solution:

An electron donor group increases the reactivity towards alkylation so further alkylation occurs.

QUESTION: 38

Statement-1 : 

 

Statement-2 : When acetone is treated with a base (like –OCH3) then it forms a carbanion which is resonance
stablised

Solution:

Both statements are true.

QUESTION: 39

Statement-1 : Aryl halides are extremely less reactive towards nucleophilic substitution reactions.

Statement-2 : In haloarenes the electron pairs of halogen atom are in conjugation with π electrons of the ring. More over due to more 's' character of sp2 carbon of ring, the C–X bond strength increases and cleavage becomes difficult.

Solution:

S-2 correctely explains S-1.

QUESTION: 40

Statement-1 : In the acidic medium tetrahydro furfuryl alcohol is transformed in 

Statement-2 : Acid catalysed dehydration of 3º and 2° alcohols takes place by E1 (carbocation)
mechanism.

Solution:

QUESTION: 41

Both the Cβ–H and Cα–X bonds are breaking in the transition state of E2 reactions. The rate of E2 reaction is of second order. The rate shows primary isotopic (deuterium) effect, i.e. if Cβ–H is replaced by Cβ–D, the rate of reaction decreases sharply. The rate is also corelated with nucleofugality i.e. the ability of leaving group to leave. With a better nucleofuge, the rate of reaction increases. This is called the element effect.

The element effect is also observed in E1cB reactions in which the second step is rate determining and elimination of first order w.r.t. conjugate base is observed. For E1cB reaction KH/KD ~ 1, therefore proton abstraction is not involved in the rate determing step. In D2O the incorporation of Deuterium at Cβ–H in the substrate is found many times faster than rate of elimination, in E1cB reaction.

In E1 reaction only nucleofuge departs in slow step and a carbocation intermediate is formed.

Q. For the given compounds I and II the rate of elimination by  What is true about this reaction

Solution:

The breaking of Cβ–H is faster than Cβ–D and both Cβ–H and Cβ–Br are breaking in same step. It is an E2 reaction.

QUESTION: 42

Both the Cβ–H and Cα–X bonds are breaking in the transition state of E2 reactions. The rate of E2 reaction is of second order. The rate shows primary isotopic (deuterium) effect, i.e. if Cβ–H is replaced by Cβ–D, the rate of reaction decreases sharply. The rate is also corelated with nucleofugality i.e. the ability of leaving group to leave. With a better nucleofuge, the rate of reaction increases. This is called the element effect.

The element effect is also observed in E1cB reactions in which the second step is rate determining and elimination of first order w.r.t. conjugate base is observed. For E1cB reaction KH/KD ~ 1, therefore proton abstraction is not involved in the rate determing step. In D2O the incorporation of Deuterium at Cβ–H in the substrate is found many times faster than rate of elimination, in E1cB reaction.

In E1 reaction only nucleofuge departs in slow step and a carbocation intermediate is formed.

Q. Observe the given reaction. In this reaction  What is not true about this reaction

 

Solution:

Both EtOH or EtOD give same base  so rate of reaction will be same this is an E1cb reaction.

QUESTION: 43

Both the Cβ–H and Cα–X bonds are breaking in the transition state of E2 reactions. The rate of E2 reaction is of second order. The rate shows primary isotopic (deuterium) effect, i.e. if Cβ–H is replaced by Cβ–D, the rate of reaction decreases sharply. The rate is also corelated with nucleofugality i.e. the ability of leaving group to leave. With a better nucleofuge, the rate of reaction increases. This is called the element effect.

The element effect is also observed in E1cB reactions in which the second step is rate determining and elimination of first order w.r.t. conjugate base is observed. For E1cB reaction KH/KD ~ 1, therefore proton abstraction is not involved in the rate determing step. In D2O the incorporation of Deuterium at Cβ–H in the substrate is found many times faster than rate of elimination, in E1cB reaction.

In E1 reaction only nucleofuge departs in slow step and a carbocation intermediate is formed.

Q. Consider the following statement, “In 80% aqueous ethanol, t-Butyl iodide solvolyses 100 times as rapidly as t-Butylchloride, but the ratio of elimination to substitution products is the same for chloride and iodide.” This statement indicates that

Solution:

Rate of solvolysis is rate of formation of carbocation with the elimination of leaving group.

QUESTION: 44

Hydrolysis of cumene hydroperoxide is an important reaction in organic chemistry. This reaction involves migration of phenyl group to electron-deficient oxygen.

Another reaction which involves group migration is the Schmidt reaction, in which isocyanide intermediate is formed.

Now consider the following scheme of reaction and answer the following questions.

Q. What are X and X1

Solution:

QUESTION: 45

Hydrolysis of cumene hydroperoxide is an important reaction in organic chemistry. This reaction involves migration of phenyl group to electron-deficient oxygen.

Another reaction which involves group migration is the Schmidt reaction, in which isocyanide intermediate is formed.

Now consider the following scheme of reaction and answer the following questions.

Q. The product P should be 

Solution:

QUESTION: 46

Hydrolysis of cumene hydroperoxide is an important reaction in organic chemistry. This reaction involves migration of phenyl group to electron-deficient oxygen.

Another reaction which involves group migration is the Schmidt reaction, in which isocyanide intermediate is formed.

Now consider the following scheme of reaction and answer the following questions.

Q. Conversion of C6H5CON3  to C6H5NCO is an example of rearrangement where an hydrocarbon group migrates to :

Solution:

QUESTION: 47

Match the column
Match the reagents/reaction of Column-II with the appropriate reactants of Column-I
(more than one option of column-II may match with one option of column-I)
        Column I                                                 Column II
       Reactants                                             Reagents / Reaction

Solution:

(A), (B), (C) all homoatomic multiple bond undergo electrophilic addition reaction.

(D) all alkanes undergo free radical substitution.

QUESTION: 48

Solution:

(A) It is an alkene and it can undergo oxidation, hydrogenation, electrophilicaddition and hydration.
(B) It is an ester with aromatic ring and it can undergo all the mentioned reactions.
(C) It is a nitro compound which can only be hydrogenated to amine.
(D) It is a primary alcohol which can be oxidised to aldehyde or carboxylic acid.

QUESTION: 49

Solution:

QUESTION: 50

let f(t) be a continous function for

Solution:

QUESTION: 51

Solution:

QUESTION: 52

Solution:

QUESTION: 53

Solution:

Degree of Diff. equation is not defined

QUESTION: 54

Solution:

QUESTION: 55

The area of the region of the plane bounded by 

Solution:

QUESTION: 56

Let f(x) = [x] + {x}2 ;and
x = 5 is (where [ . ] and { . } represent the greatest integer and fractional part functions respectively)

Solution:

*Multiple options can be correct
QUESTION: 57

Solution:

*Multiple options can be correct
QUESTION: 58

Solution:

*Multiple options can be correct
QUESTION: 59

A function y = f(x) satisfying the differential equation  is such that,
y → 0 as x →∞ then :

Solution:

*Multiple options can be correct
QUESTION: 60

Solution:

QUESTION: 61

Statement-1 : Let f be real valued function such that f(2) = 2 and f'(2) = 1, then 

Statement-2 : Let f(x) =

Solution:

Statement - 1 is true.
Statement - 2 is Lebnitz formula so it is true and statement -2 explains Statement - 1.

QUESTION: 62

 Statement-1 : 
Statement-2 : If f(x) is continuous in [a, b] and m and  are greatest and least value of f(x) in [a, b], then 

Solution:

Hence statement 1 is true and Statement 2 is correct explanation of 1
Hence (A)

QUESTION: 63

Statement-1 : Let f(x) be an even function which is periodic, then  
is also periodic.
Statement-2 : If α(x) is a differentiable and periodic function, then α'(x) is also periodic.

Solution:

Statement-2 : 

Statement-1 :

QUESTION: 64

Statement-1 : If then the value of the definite integral  where { } denotes the fractional part function, is 6.

Statement-2 : where a, b  N such that b > a and T is period of f(x).

Solution:

QUESTION: 65

Comprehension # 1

 

This is one of the important property for the integrable function f(x)

Q. 

Solution:

QUESTION: 66

Comprehension # 1

 

This is one of the important property for the integrable function f(x)

 

Q.  then the
value of k is

Solution:

QUESTION: 67

Comprehension # 1

 

This is one of the important property for the integrable function f(x)

Q. 

Solution:

QUESTION: 68

Comprehension # 2
Let f(x) = x – [x] , g(x) = ex where [ . ] denotes greatest integer function

Q.   

Solution:

QUESTION: 69

Comprehension # 2
Let f(x) = x – [x] , g(x) = ex where [ . ] denotes greatest integer function

Q. Area bounded by the curve y = g(f(x)), x-axis and the lines x = –3 and x = 4 is 

Solution:

QUESTION: 70

Comprehension # 2
Let f(x) = x – [x] , g(x) = ex where [ . ] denotes greatest integer function

Q. Orthogonal trajectory of the curve y = k g(x), where k is any arbitrary constant is 

Solution:

QUESTION: 71

Solution:

QUESTION: 72

Solution:

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