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# Part Test - 4 (JEE Advanced)

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## 66 Questions MCQ Test JEE Main & Advanced Mock Test Series | Part Test - 4 (JEE Advanced)

Part Test - 4 (JEE Advanced) for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The Part Test - 4 (JEE Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Part Test - 4 (JEE Advanced) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Part Test - 4 (JEE Advanced) below.
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Part Test - 4 (JEE Advanced) - Question 1

### An opaque sphere of radius a is just immersed in a transparent liquid as shown in figure. A point source is placed on the vertical diameter of the sphere at a distance a/2 from the top of the sphere. One ray originating from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is 30º. The refractive index of the liquid is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 1

Part Test - 4 (JEE Advanced) - Question 2

### A thin oil film of refractive index 1.2 floats on the surface of water (m = 4/3)  When a light of wavelength l = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is:

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 2

Part Test - 4 (JEE Advanced) - Question 3

### In a YDSE experiment if a slab whose refractive index can be varied is placed in front of one of the slits then the variation of resultant intensity at mid-point of screen with 'µ' will be best represented by (µ ³ 1).[ Assume slits of equal width and there is no absorption by slab; mid point of screen is the point where waves interfere with zero phase difference in absence of slab]

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 3

In absence of film or for μ=0 intensity is maximum at screen. As the value of μ is increased, intensity shall decrease and then
increase alternately. Hence the correct variation is

Part Test - 4 (JEE Advanced) - Question 4

Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made reflecting. The acceleration of two images
formed in those two reflecting surfaces w.r.t. each other is:

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 4

Part Test - 4 (JEE Advanced) - Question 5

A ray is incident on the first prism at an angle of incidence 53º as shown in the figure. The angle between side CA and B'A' for the net deviation by both the prisms to be double of the deviation produced by the first prism, will be :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 5

Let be the angle of emergence from the first prism be ‘e’ Snell's law on surface AB

Then for net deviation to be double, the incident ray on side
A' B' of second prism should make angles i or e with normal.
Hence the angle between the given then will be 2e or i + e.

Part Test - 4 (JEE Advanced) - Question 6

Light of wavelength 4000 Å is incident at small angle on a prism of apex angle 4º. The prism has nv = 1.5 & nr = 1.48. The angle of dispersion produced by the prism in this light is :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 6

Part Test - 4 (JEE Advanced) - Question 7

Two plane mirrors are joined together as shown in figure. Two point objects O1 and O2 are placed symmetrically such that AO1 = AO2. The image of the two objects is common if :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 7

Part Test - 4 (JEE Advanced) - Question 8

Monochromatic light rays parallel to x-axis strike a convex lens AB. If the lens oscillates such that AB tilts upto a small angle q (in radian) on either side of y-axis, then the amplitude of oscillation of image will be (f = focal length of the lens) :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 8

From the figure :

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 9

A particle is moving towards a fixed convex mirror. The image also moves. If Vi = speed of image and VO = speed of the object, then

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*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 10

The speed of light in the material of a plano convex lens is 2 × 108 m/s and its greatest thickness is 3 mm. If the aperture diameter of the lens in 6.0 cm then :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 10

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 11

In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images formed by a convex lens placed between them is 4.84.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 11

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 12

Which of the following statements is/are correct about the refraction of light from a plane surface when light ray is incident in denser medium. [C is critical angle]

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 12

Part Test - 4 (JEE Advanced) - Question 13

Statement-1 : A beam of white light enters the curved surface of a semicircular piece of glass along the normal. The incoming beam is moved clockwise (so that the angle q increases), such that the beam always enters along the normal to the curved side. Just before the refracted beam disappears, it becomes predominantly red.

Statement-2 : The index of refraction for light at the red end of the visible spectrum is more than at the violet end

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 13

by cauchy's formula
: a & b are constant of medium. The index of refraction for light at the red end
of the visible spectrum is lesser than at the violet end.Hence statement -2 is false

Part Test - 4 (JEE Advanced) - Question 14

Statement-1 : A point object moves near the principal axis of a fixed spherical mirror along a straight line. Then the image formed by the spherical mirror also moves along a straight line.

Statement-2 : For an incident ray on a fixed spherical mirror there is a fixed reflected ray. If a point object moves along this incident ray, its image will always lie on the given reflected ray. Further an incident ray may  be drawn from the moving point object in its direction of velocity towards the mirror.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 14

Draw an incident ray on the mirror and trace the corresponding reflected ray. If a point object moves along this ray,its image will always lies on the traced reflected ray. Hence when a point object moves near the principal axisof a fixed spherical mirror along a straight line, then its image formed by the spherical mirror also moves along
a straight line.Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Part Test - 4 (JEE Advanced) - Question 15

Paragraph for Question Nos. 15

A monochromatic point source S of wavelength l = 5000√2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along y-axis at horizontal distance D = 1 metre from point source. The refractive index of liquid is √2 .  ( Neglect partial reflection of rays from source S at liquid-air interface.)

Q. The y-coordinates of second order bright formed on the screen is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 15

The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflected back (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
So Interference pattern is formed in region OP.Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0

Part Test - 4 (JEE Advanced) - Question 16

Paragraph for Question Nos. 15

A monochromatic point source S of wavelength l = 5000 √2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along y-axis at horizontal distance D = 1 metre from point source. The refractive index of liquid is√2 .  ( Neglect partial reflection of rays from source S at liquid-air interface.)

Q. The region on screen where interference pattern is formed lies in

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 16

The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflectedback (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
Interference pattern is formed in region OP.
Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0.

Part Test - 4 (JEE Advanced) - Question 17

Paragraph for Question Nos. 15

A monochromatic point source S of wavelength l = 5000√2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along y-axis at horizontal distance D = 1 metre from point source. The refractive index of liquid is√2 .  ( Neglect partial reflection of rays from source S at liquid-air interface.)

Q. The minimum horizontal distance of screen from fixed source such that at least one bright is formed on screen

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 17

The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflected back (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
Interference pattern is formed in region OP.
Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0.

Part Test - 4 (JEE Advanced) - Question 18

Matrix - Match Type

A white light ray is incident on a glass prism, and it create four refracted rays A, B, C and D. Match the refracted rays with the colours given (1 & D are rays due to total internal reflection.):

Ray            Colour
(A)    A        (p)    red
(B)    B        (q)    green
(C)    C        (r)    yellow
(D)    D        (s)    blue

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 18

By snell's law n =  and n = So at inclined face angular incidence is same for all ray and refraction is from denser to rarer. So emergent angle will be less for red.

*Answer can only contain numeric values
Part Test - 4 (JEE Advanced) - Question 19

An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1 and m2 and placed at the end of the tube. m1 & m2 are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014 Hz. The light reflected from m1 and m2 forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1 & m2.
The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1 & m2 is b × 10-3 m then find ab/40.

.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 19

*Answer can only contain numeric values
Part Test - 4 (JEE Advanced) - Question 20

In the figure shown light of wave length l = 5000 Å is incident on the slits (in a horizontal fixed plane) S1 and
S2 separated by distance d = 1 mm. A horizontal screen ‘S’ is released from rest from initial distance
D0 = 1 m from the plane of the slits. Taking origin at O and positive x and y axis as shown, at t = 2 seconds;
(Use g = 10 m/s2) velocity is 10n j ˆ and acceleration is m j ˆ of central maxima, find the value of [mm/25].

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 20

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Part Test - 4 (JEE Advanced) - Question 21

A is a thin walled sphere at rest made up of glass. The radius of the sphere is 1m and it is filled with a
transparent liquid of refractive index m. S is the luminous source moving directly towards a plane
vertical mirror M. A fish in the liquid is moving towards S. The eye of the fish and S are collinear &
perpendicular to mirror M. At the moment S is 3m away from the centre of the sphere, fish observes
that image of S due to reflection is moving with speed of 13m/s. If speed of the fish relative to the
sphere is 10m/s and m = 1.5 then find the speed of the source at that instant. The system is placed
in air.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 21

*Answer can only contain numeric values
Part Test - 4 (JEE Advanced) - Question 22

A light ray parallel to the principal axis is incident (as shown in the figure) on a thin planoconvex lens
with radius of curvature of its curved part equal to 10 cm. Assuming that the refractive index of the
material of the lens is 4/3 and medium on both sides of the lens is air, find the distance of the point from
the lens where this ray meets the principal axis. Find your answer in the form X/7 cm and fill value
of X.

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Part Test - 4 (JEE Advanced) - Question 23

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Part Test - 4 (JEE Advanced) - Question 24

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Part Test - 4 (JEE Advanced) - Question 25

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Part Test - 4 (JEE Advanced) - Question 26

On vigorous oxidation of (CH3)2C=CH—CH2 —CH3 by permanganate solution gives :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 26

Part Test - 4 (JEE Advanced) - Question 27

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Part Test - 4 (JEE Advanced) - Question 28

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Part Test - 4 (JEE Advanced) - Question 29

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Part Test - 4 (JEE Advanced) - Question 30

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 30

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Part Test - 4 (JEE Advanced) - Question 31

But-1-ene is formed in reaction/s.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 31

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 32

Which of the following reactions give alkylation product :

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 32

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 33

Among the following which Statement is /are correct

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 33

In (A) acid cannot give iodoform reaction
(B) lacks in α-hydrogen. It give cannizzaro. (C) It can give haloform reaction because of CCl3 group.

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 34

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Part Test - 4 (JEE Advanced) - Question 35

Statement-1 : When benzene diazonium chloride is heated with H2O,  PhOH is formed and rate of this ArSN reaction can be increased by adding +m group at para position of the .

Statement-2 :  + m group generally increases the stability of carbocation.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 35

Due to resonance or +m group, rate of ArSN does not increases

Part Test - 4 (JEE Advanced) - Question 36

Statement-1 : Nitration of aniline can be conveniently done by protecting the amino group by acetylation.

Statement-2 : Acetylation decrease the electron density in the benzene ring and thereby prevents oxidation of benzene ring.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 36

Part Test - 4 (JEE Advanced) - Question 37

Q. The product T is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 37

Part Test - 4 (JEE Advanced) - Question 38

Q. The product 'U' is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 38

Part Test - 4 (JEE Advanced) - Question 39

Q. The product mixture ‘V’ does not have.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 39

Part Test - 4 (JEE Advanced) - Question 40

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 40

(A) If there is no α-hydrogen atom than aldehyde gives cannizaro reactions.
(B) & (C) If α-hydrogen atoms are presents and the at least one –CH3 groups are presents then show aldol and
haloform reaction.
(D) By oxidation followed by haloform reaction.

*Answer can only contain numeric values
Part Test - 4 (JEE Advanced) - Question 41

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 41

*Answer can only contain numeric values
Part Test - 4 (JEE Advanced) - Question 42

Consider the following conversion   In order to carry out this conversion, five distinct reagents numbered 1 to 5 are to be used in successive
steps I to V. Reagents are numbred as Indicate the reagents that need to be used in successive steps I to V by indicating their number in successive
five boxes as seen below.

(For eg. If you think that reagent 2 is required in step I, reagent 4 is requried in step II, reagent 5 in step III,
reagent 1 in step IV and reagent 3 in step V, then your answer will be

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 42

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Part Test - 4 (JEE Advanced) - Question 43

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 43

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Part Test - 4 (JEE Advanced) - Question 44

Consider product C, F, G & step VII and follow the given instructions to fill the box
Where 'x' 'y' 'z' and 'w' are numerals If C can undergo cannizzaro reaction, write 1 in place of x, if it can undergo reformatsky reaction write 2 in place of x, if it can undergo michael addition write 3 in place of x.
If 'F' cannot show stereoisomerism write 0 (zero) in place of y, if F can show geometrical isomerism write 1
in place of y, if F can show optical isomerism write 2 in place y If G cannot show stereoisomeris write 0 in place of Z, if G can show geometrical isomerism write 1 in place of Z, if G cannot show optical isomerism write 2 in place Z. If step VII is neither a regiospecific nor stereospecific reaction write 0 in place of w, if step VII is stereospecific reaction then write 1 & if step VII is regosepecific reaction then write 2 in place of w.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 44

Part Test - 4 (JEE Advanced) - Question 45

Let f be a differentiable function satisfying f(xy) = f(x) . f(y),

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 45

Part Test - 4 (JEE Advanced) - Question 46

Let f'(x) is non- decreasing function for which f'(0) = 0, f(0) = 5 and f''(x) – f'(x)

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 46

Part Test - 4 (JEE Advanced) - Question 47

if then number of solutions of the equation (where {.} represents fractional part function)

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 47

clear from the graph

Part Test - 4 (JEE Advanced) - Question 48

Number of solutions for x between 3 and 15 if  where [.] represents greatest integer
function, is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 48

Part Test - 4 (JEE Advanced) - Question 49

Area bounded by y = f -1(x), x = 0, y =   & y =  , where f(x) = x + sin x, is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 49

Part Test - 4 (JEE Advanced) - Question 50

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Part Test - 4 (JEE Advanced) - Question 51

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 51

Part Test - 4 (JEE Advanced) - Question 52

Compute the area of the figure bounded by straight lines x = 0, x = 2 and the curves y = 2x and y = 2x – x2  is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 52

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 53

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 53

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 54

If f(x) = ae2x + bex + cx satisfies the conditions f(0) = – 1, f'(ln 2) = 31,

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 54

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 55

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 55

*Multiple options can be correct
Part Test - 4 (JEE Advanced) - Question 56

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 56

Part Test - 4 (JEE Advanced) - Question 57

STATEMENT-1 : The area bounded by the curve |x| + |y| = a (a > 0) is 2a2 and area bounded
|px + qy| + |qx – py| = a, where p2 + q2 = 1, is also 2a2.
STATEMENT-2 : Since αx + βy = 0 is perpendicular to βx – αy = 0, we can take one as x-axis and another
as y-axis and therefore the area bounded by |αx + βy| + |βx – αy| = a is 2a2 for all α, β€ R,
α  0, β  0.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 57

STATEMENT-1 :

Then the axis get rotated through an angle θ,

the equation of the given curve becomes |U| + |V| = a the area bounded = 2a2.
so statement-1 is true

STATEMENT-2 : The equation of the curve is

Part Test - 4 (JEE Advanced) - Question 58

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 58

STATEMENT 1 :

STATEMENT 2 :

(where [.] represents greater integer function)

Part Test - 4 (JEE Advanced) - Question 59

Consider the function defined implicitly by the equation  Where [x] denotes the greatest integer function

Q. The area of the region bounded by the curve & the line x = – 1 is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 59

Part Test - 4 (JEE Advanced) - Question 60

Consider the function defined implicitly by the equation  . Where [x] denotes the greatest integer function

Q. Line x = 0 divides the region mention above in two parts. The ratio of area of left hand side of line to right hand side of line is

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 60

Part Test - 4 (JEE Advanced) - Question 61

Consider the function defined implicitly by the equation  Where [x] denotes the greatest integer function

Q.

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Part Test - 4 (JEE Advanced) - Question 62

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 62

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Part Test - 4 (JEE Advanced) - Question 63

given that |a| > 1, then find the value of λ .

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 63

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Part Test - 4 (JEE Advanced) - Question 64

then find the value of λ.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 64

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Part Test - 4 (JEE Advanced) - Question 65

Solution of the differential equation  then find the value of λ.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 65

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Part Test - 4 (JEE Advanced) - Question 66

The area of the figure bounded by a curve, the x-axis, and two ordinates, one of which is constant, the other
variable, is equal to the ratio of the cube of the variable ordinate to the variable abscissa and equation of curve
is (λy2 –x2)3 = Cx2 , then find the value of λ.

Detailed Solution for Part Test - 4 (JEE Advanced) - Question 66

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