JEE Advanced Mock Test - 5 (Paper II) - JEE MCQ

# JEE Advanced Mock Test - 5 (Paper II) - JEE MCQ

Test Description

## 54 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Advanced Mock Test - 5 (Paper II)

JEE Advanced Mock Test - 5 (Paper II) for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Advanced Mock Test - 5 (Paper II) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 5 (Paper II) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 5 (Paper II) below.
 1 Crore+ students have signed up on EduRev. Have you?
*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 1

### Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier calipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 1

1 main scale division (M.S.D) = 1/8cm

5 Vernier scale division (V.S.D) = 4 M.S.D.

1 V.S.D. = 4/5M.S.D.

Least count of Vernier scale (L.C.) = 1 M.S.D.−1 V.S.D.

= 1 M.S.D.− 4/5M.S.D

For option A and B,

If the pitch of the screw gauge is twice the least count of the Vernier calipers then pitch = 2 × L.C. of Vernier scale

= 1/20cm = 0.05 cm

Hence, least count of screw gauge = Pitch/100 = 0.05/100 cm

= 0.0005 cm = 0.005 mm

For option C and D,

Least count of linear scale of screw gauge,

= 2 × 1/40 = 1/20 cm = 0.05 cm

Pitch = 2 × 1 / 20 = 110cm = 1mm

Least count of screw gauge = 1mm/100 = 0.01 mm

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 2

### If a body is projected with speed lesser than escape velocity:

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 2

We know that the escape velocity from the earth surface,

And the orbital velocity of satellite for revolving around the earth near the earth surface,

According to question body has projected with velocity v < ve,="" the="" followings="" possibilities="" can="" be="" />

(i) If body is projected vertically upward then at maximum height, velocity will be zero and due to effect of earth's gravity it will fall down in straight path.

(ii) If body is projected obliquely from the surface of the earth, then at maximum height only its vertical velocity will be zero, but due to horizontal velocity it will follow the parabolic path.

(iii) If body is projected from the earth surface such that at maximum height, its horizontal velocity becomes equal to orbital velocity of satellite as v0 < v="" />< />e, then body may orbit in elliptical path.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 3

### A listener is at rest with respect to the source of the sound. The wind starts blowing along the line, joining the source and the observer. Which of the following quantities do not change?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 3

According to Doppler's effect, when both the source and listener are at rest, and the wind alone is blowing, then there is no change in the frequency of the sound. We also know that the frequency is related to the time period. So, the time period also does not change.

Doppler's effect is the phenomenon of change in the apparent frequency of the sound due to relative motion between the source of the sound and the listener.

According to Doppler's expression of the apparent frequency,

Where w is the speed of the wind.

f' is the apparent frequency.

f is the original frequency.

v is the speed of sound.

uo is the velocity of the observer.

us is the velocity of source,

w is the speed of the wind.

When the wind is blowing along with the line joining, then the formula of the apparent frequency is,

But, according to the question, the source and the observer are at rest. So,

f' = f

According to Doppler's effect, when both the source and listener are at rest and the wind alone is blowing, then there is no change in the frequency of the sound.

Also, we know that the frequency is related to the time period, so the time period also does not change.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 4

A particle is in linear simple harmonic motion between two points. A and B,10 cm apart (figure), take the direction from A to B as the positive direction. Choose the correct statement(s).

AO = OB = 5 cm

BC = 8 cm

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 4

Situation at different positions is as shown in the figure:

When the particle is at 3 cm away from A and going towards B:

Direction is velocity is along the the direction of motion. Hence, velocity is towards AB i.e. velocity is positive.

In SHM, acceleration is always towards mean position (O) . Hence, in this case, accelertion is positive.

As force F = ma has same direction as that of acceleration, so force is also positive.

When the particle is at C and going towards B:

Clearly, velocity is positive in this case.

When the particle is 4 cm away from B going towards A:

In this case, velocity is negative and acceleration is towards mean position, i.e., acceleration is negative. Hence, force is also negative.

When the particle is at B:

Acceleration and force are towards mean position i.e. they are negative.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 5

500 g of mercury is poured into a bent tube whose right arm forms an angle of θ = 30° with the vertical. Cross-sectional area of tube is 0.6 cm2. Then,

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 5
Let a liquid of density ‘ρ’ be filled in a bent tube of cross-sectional area A. Let the liquid be depressed in the vertical arm of tube by distance 'y'. The vertical rise of liquid in the slant portion of tube = y cosθ . The vertical difference between liquid levels in two arms of bent tube = y + y cosθ = (1 + cosθ )y. The weight of liquid in column (1 + cosθ )y in two arms of bent tube will provide a restoring force to the liquid.

So, restoring force on the liquid,

F = -[Weight of liquid column of height (1 + cosθ )y]

= -(1 + cosθ )yρg A

= -(1 + cosθ )Aρgy ...(1)

On removing the force, liquid is left free and executes simple harmonic motion in the bent tube.

Equation (1) can be compared with the equation, F = -ky.

We have, spring factor, k = (1 + cosθ )Aρg

Here, Inertia factor = Mass of mercury, m = 500 g

Therefore, period of oscillation,

= 1.15 sec.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 6

The dipoles are joined along the axis of magnets as shown in the figure.

Which of the following is/are true?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 6

U = Potential energy of system

As U is negative, so force is attractive.

U is positive, so force is repulsive.

U = Potential energy of system

As U is negative, so force is attractive.

As U is positive, so force is repulsive.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 7

In a DC circuit, for C and R in series, decay of charge and q-t graph are shown.

C1 and C2 are two capacitors. Then,

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 7

q0 = EC [E → Applied voltage]

The capacitance C1 > C2

Smaller the time constant, faster the discharging

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 8

Directions: The question has 4 choices, out of which ONE OR MORE is correct.

The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 8

(1) p-V graph is not a rectangular hyperbola. Therefore, process A-B is not isothermal.

(2) In process BCD, the product pV (therefore temperature and internal energy) is decreasing. Further, volume is decreasing.

Hence, work done is also negative. Hence, Q will be negative or heat will flow out of the gas.

(3) WABC = positive

(4) For clockwise cycle on p-V diagram with P on y-axis, net work done is positive.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 9

The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio is m50 / m25 is

(Round off upto 2 decimal places)

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 9

Magnification, m = f / f + u

According to Cartesian sign conventions;

m25 = 20/20 - 25 = -4 , m50 = 20/20 - 50 = -2/3

∴ m50 / m25 = 2/12 = 0.167 = 0.17

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 10

A wire of length l = 6 ± 0.06 cm and radius r = 0.5 ± 0.005 cm and mass m = 0.3 ± 0.003 g. Maximum percentage error in density is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 10

Density,

Take log and differentiate for errors.

Or

Or

= 4

∴ Percentage error in density = 4%

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 11

Three objects A, B and C

are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 11

The law of conservation of momentum is valid for all collisions.

For elastic collision, conservation of energy is also valid.

u1 = u2 = (v1 - v2)

In this case, the second ball is at rest.

Initially, u1 = 9 ms-1, u2 = 0

mA will recoil back and 2m (B) will go forward.

B is moving with 6 ms-1 to make an inelastic collision with C.

Conservation of momentum gives

2m x 6 + m x 0 = (2m + m) vc

∴ vC = 12/3 = 4 ms-1

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 12

In a capillary tube of length 20 cm and bore diameter 1 mm1 mm, a very thin glass rod of diameter 0.5 mm is coaxially arranged. The arrangement is dipped in water for which angle of contact with glass is 0°. What is the difference in length of capillary raise if the tube is vertical in a case and is at 30° to the horizontal in other? Express the magnitude in cmcm. Surface tension of water is 0.075 N m−1

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 12
For capillary raise,

⇒ h = 6 cm

and at 30°,

∴ difference, Δℓ = 6 cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 13

An uncharged parallel plate capacitor having its lower end fixed and upper end is attached with spring having spring constant K. Upper plate is in equilibrium before switch is closed. After switch is closed, the condition on the potential of battery so that the system can acquire new equilibrium position without discharging is . Then p + q + r ? [ pp and qq are smallest possible positive integers and V is potential difference across the battery. Ignore gravity]

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 13

Let in new equilibrium the upper plate is displaced by x units.

If we get real x less than L then only new equilibrium is achieved.

Here, force between plates of capacitor = spring force

, where, x = extension in spring and A is area of the plate.

We know that, after connecting the switch, Q = C'V

So,

Maximum value of Kx(L − x)2

We get K(L − x)2 + 2Kx(L − x)(−1) = 0

x = L/3

Maximum value of

So for real x,

Hence, the value of p + q + r = 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 14

A point object O is placed on the principal axis of a convex lens of focal length 10cm at 12 cm from the lens. When object is displaced 1mm1mm along the principal axis magnitude of displacement of image is x1. When the lens is displaced by 1mm1mm perpendicular to the principal axis displacement of image is x2 in magnitude. The value of x1 / x2 is ____

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 14

We are given that :-

The distance of the object from the lens u = −12

Focal length of lens f = 10

We have to find the value of x1 / x2

We know that lens formula is

Where uu is the distance of the object from the lens, v is the distance of the image from the lens and f is the focal length.

Then,

The level of magnification is proportional to the ratio of v and u.

where v is image distance and u is object distance

According to given object is displaced 1mm along the principal axis magnitude of displacement of image is x1.

So,

x1 = (m2)(1mm)

The lens is displaced by 1mm perpendicular to the principal axis displacement of image is x2 in magnitude.

So,

x2 = (m)(1mm)

Hence, the value of x1/x2 = 5

JEE Advanced Mock Test - 5 (Paper II) - Question 15

Answer the following by appropriately matching the lists based on the information given:

List I includes devices and List II includes processes related to the devices.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 15

Bimetallic strip is a temperature-sensitive electrical contact used in some thermostats, consisting of two bands of different metals joined lengthwise. When heated, the metals expand at different rates, causing the strip to bend. The bimetallic strip consists of two thin strips of different metals, each having different coefficients of thermal expansion.

(A) → (s)

Steam engine is a machine that converts the heat energy of steam into mechanical energy.

(B) → (q)

Incandescent lamp converts electricity into light by heating a filament, using electric current, until it emits electromagnetic radiation. As current passes through the filament, its high resistance causes its temperature to rise until it glows.

JEE Advanced Mock Test - 5 (Paper II) - Question 16

Answer the following by appropriately matching the lists based on the information given:

List I includes devices and List II includes processes related to the devices.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 16

(C) → (p), (q)

Electric fuse is an electrical safety device that operates to provide overcurrent protection of an electrical circuit. Its essential component is a metal wire or strip that melts when too much current flows through it, thereby interrupting the current.

(D) → (r), (q)

JEE Advanced Mock Test - 5 (Paper II) - Question 17

A train is travelling on a straight horizontal track with a constant acceleration of 2 m s-2 across a bridge over a river. When the velocity of the train is 25 m s-1, a man inside one of the cars throws a stone horizontally out of a window in a direction perpendicular to direction of motion of the train with a speed of 5 m s-1 relative to himself. In the absence of air resistance the stone hits the water at point P, 40 m in a horizontal direction from track. Consider point of projection as origin and Cartesian coordinate system as shown in diagram. (Take g = 10 m s−2)

Based on above information answer the following questions.

At the instant when stone hits the water, the coordinate of car window from which stone is projected, is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 17
The distance travelled by car window from where the stone is projected in x - direction is

=264 m

So, required coordinate is (264 m, 0, 0)

JEE Advanced Mock Test - 5 (Paper II) - Question 18

A composite spherical shell is made up of two materials having thermal conductivities K and 2K respectively as shown in the diagram. The temperature at the innermost surface is maintained at T whereas the temperature at the outermost surface is maintained at 10 T. A, B, C and D are four points in the outer material such that AB = BC = CD. Now answer the following questions.

Out of the segments AB, BC and CD the magnitude of the temperature difference will be maximum across

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 18
Consider shell with cross-sectional thickness equal to AB, BC and CA respectively. Since, heat flow (heat current) is same through all of them, therefore temperature difference will be maximum across the shell with maximum resistance which will be AB.
*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 19

Acetone and carbon disulphide form binary liquid solution showing positive deviation from Raoult's law. If the normal boiling point (Tb) of pure acetone is more than that of pure carbon disulphide, then the correct statement(s) is/are:

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 19

Statements (1) and (2) are correct because for a solution with a positive deviation from Raoult's law, the boiling point of the mixture is lower than that of either of the components.

Statement (3) is correct because if the mole fraction of the lower boiling component is decreased, the boiling point will increase.

Only statement (4) is incorrect as the components of azeotropic mixtures cannot be separated completely.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 20

On being strongly heated, which of the following substances will give a gas that turns lime water milky?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 20

SO2 and CO2 gases can turn lime water milky due to formation of CaCO3.

SO2 + Ca(OH)2 →CaSO3 + H2O

CO2 + Ca(OH)2 → CaCO3 + H2O

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 21

Which of the following facts about sucrose is/are true?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 21

Sucrose is not a reducing sugar as the reducing aldehyde group of glucose in it is not in a free state. So, (1) is correct.

On hydrolysis with dilute acid, sucrose produces glucose and fructose.

Since glucose shows reducing property, the hydrolysis product reduces Fehling`s solution. So, (2) is correct.

Sucrose is dextrarotatory with specific rotating of + 66.50. On hydrolysis, it produces D (+) glucose and D (-) fructose. Since D (-) fructose has greater specific rotation than D (+) glucose, the resulting mixture is laevorotatory. So, (3) is correct.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 22

Which of the following comparisons of ionic radii is/are correct?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 22

Ti4+ > Mn7+ is correct as Ti > Mn because of the greater nuclear charge of Mn.

The trend in the same oxidation state of the cations remains the same.

Mn+7 is definitely smaller because of the greater nuclear pull.

35CI- = 37CI- is true because the nuclear charge is the same in the isotopes.

K+ < CI- is true as for the isoelectronic species, the cation is smaller than the anion.

P3+ > P5+ is ture as for homonuclear cations, the cation in the higher oxidation state is smaller.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 23

In the electrolysis of alumina, cryolite is added to :

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 23
Because of high melting point (2050oC), pure alumina cannot be electrolysed. Hence a mixture of alumina, cryolite (m.p. 1000oC) and calcium fluoride(to lower the temperature of the melt) is electrolysed at about 900oC. The function of cryolite is to increase the electrical conductivity of the electrolyte, and to lower the temperature of the melt.
*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 24

Which of the following are crossed aldol products in the given reaction?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 24

Proceed reverse to check the self and cross product.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 25

Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 25

According to emission photoelectric effect

E of incident radiation = hc/λ

= 9.94 × 10–18 J

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 26

Alcohols are -

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 26

Acidic strength ∝ Conjugate base stability (Anion)

(i) R-alkyl group acidic character

Conjugate base

In alcohol due to +I effect of alkyl group (R) So in alcohol is less as comapred to conjuate base of water(OH)

In ethyne -ve charge carbon Csp

In alcohol -ve charge on O -ve on more e negative atom is more satble as alcohol arer more acidic as compared to ethyne

(ii) In alcohol and ethyne conjuate base are R−O CH ≡ C− in charge on Csp R−O−R−O -ve charge on oxygen and -ve charge on more electro negative atom is more stable hence alcohol are more acidic then ethyne.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 27

In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 27

Given, CV = 2.5 kJ K-1 = 2500 J K-1

ΔT = T2 - T1 = 298.45 - 298 = 0.45 K

ΔH due to combustion of 3.5 g gas = CV x ΔT = 2500 x 0.45 = 1125 J

Given, molecular weight of gas = 28.

28 1 mole

Hence, ΔH due to combustion of 1 mole of gas

= 28 = 9000 J

∴ Δ H in kJ mol-1 = 9kJ mol-1.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 28

The dissociation constant of a substituted benzoic acid at 25°C is 1.0 x 10-4. The pH of a 0.01 M solution of its sodium salt is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 28

Given, Ka = 1 x 10-4

∴ pKa = -log (1 x 10-4) = 4

C = 0.01 M

Since the solution contains a salt of weak acid and strong base,

pH = 7 + 1/2 pKa +1/2log C

= 7 + 1/2 x 4 + 1/2 x log (0.01) = 9 + 1/2 x (-2) = 8

pH of solution = 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 29

The value of n in the molecular formula BenAI2Si6O18 is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 29
BenAI2Si6O18: It is Beryl, with formula Be3AI2Si6O18, where, n = 3. It is a blue coloured gemstone. It is an aluminosilicate with Be as impurity.
*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 30

Calculate molecular diameter (in nanometer) for a gas if its molar excluded volume is 3.2π ml. Give the answer by multiplying with 10. ((Take NA= 6.0 × 1023 )

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 30

Molar excluded volume

[b = 3.2π]

r = 10−8cm.

Molecular diameter for gas (2r) = 2 × 10−8cm

Multiply by 10 we get,

10 × 2 × 10−8cm = 2 × 10−7cm = 2nm

The volume that is not available for free movement of particles is called excluded volume. Excluded volume is zero for ideal gas.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 31

Find the number of native ores out of the given ores:

pyrolusite, chromite, siderite, cassiterite, calamine, argentite, limestone, chalcopyrite.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 31
An igneous or sedimentary rock that contains a metal element in its pure state is called native ore.

Pyrolusite - MnO2

Chromite - FeCr2O4

Siderite - FeCO3

Cassiterite - SnO2

Calamine - ZnCO3

Argentite - Ag2S

Limestone - CaCO3

Chalcopyrite - CuFeS2

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 32

Of the following carbonyl compounds, how many would give aldol condensation reaction?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 32
The carbonyl compounds in which α-H atom/atoms are present with respect to carbonyl group shows inter/intra molecular aldol condensation reactions. In the given question only 4 compounds have that.
JEE Advanced Mock Test - 5 (Paper II) - Question 33

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 includes dipole moments, hybridisations and shapes of various molecules.

List - 2 includes bond lengths, bond angles and bond energies of orbitals participating in hybridisation.

For PCl5, the correct combination is:

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 33

Hybridisation and structure of PCl5:

The bond angles in PCl5 are not equivalent.

• Three bonds lie in one plane at an angle of 120° to one another as in a triangular planar arrangement (named equatorial bond).
• Rest two lie above and below the equatorial plane, both making an angle of 90° called axial bonds.

Since axial bonds are larger than equatorial bonds, geometry of PCl5 is not symmetrical.

The net dipole moment of PCl5 molecule is zero, as there is no lone pair of electrons on the central atom, i.e. P. Three P-Cl bonds are at an angle of 120° from each which cancels each other's dipole moment and 2 P-Cl bonds are perpendicular to plane which are opposite to each other and hence nullify the effect.

JEE Advanced Mock Test - 5 (Paper II) - Question 34

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 includes alkali metals of group 1 in the modern periodic table.

List - 2 gives characteristic colours of the elements present in List - 1.

Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 34
The colours observed during the flame test result from the excitement of the electrons caused by the increased temperature. The electrons "jump" from their ground state to a higher energy level. As they return to their ground state, they emit visible light.

Alkali metals and their salts impart a characteristic colour to the flame.

Lithium: Crimson red

Sodium: Golden yellow

Potassium: Pale violet

Rubidium: Violet

JEE Advanced Mock Test - 5 (Paper II) - Question 35

The binaphthol (Bnp) is:

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 35

Compound is not having chiral carbon but still is optically active due to non-super impossible mirror image.

No plane of symmetry

cannot be a meso compound

• Chiral Centre

(Binol – No chiral carbon)
• Axial chirality = Chiral carbon is not the only criteria to have optical activity.

JEE Advanced Mock Test - 5 (Paper II) - Question 36

Disaccharides are carbohydrates those contain two monosaccharides molecules, each in the hemiacetal form, joined together by the elimination of a water between two hydroxyl groups. Dehydration involves the anomeric carbon of one monosaccharide and may or may not involve the anomeric carbon of the other monosaccharide when the hemiacetal hydroxyl group on an anomeric carbon is involved in a dehydration, the resulting product is an acetal (in common) and glycoside (in carbohydrate).

Sucrose is a non-reducing sugar (while its hydrolysis products glucose and fructose are reducing sugars) because :

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 36

Sucrose is having structure as

In above structure there is no free carbonyl centre of an aldehydic group present ; It is a non-reducing sugar.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 37

If the roots of the equation x3 + bx2 + cx − 1 = 0 form an increasing G.P., then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 37

Let the roots of the given equation be a/r, a, ar where a > 0 & r > 1, then

From equation (iii)iii, we get a3 = 1⇒ a = 1

From equation (i), we get

Also, from equation (ii), we get

From equations (iv) & (v), we get

−b = c ⇒ b + c = 0

As

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 38

Let f(t) = lnt. Then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 38

Let z = x(9x − 4)lnx

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 39

A real valued function f(x) is such that

where [x] and {x} denote integral and fractional part of x. Then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 39

Right hand limit at x = 0,

Left hand limit at x = 0,

=

Now,

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 40

Let , where [.] denotes the greatest integer function. Then f(x) is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 40

Given,

∵[x − π] = Integer = n (let) ∀ x ∈ R

⇒ sin(π[x − π]) = sin(nπ) = 0 ∀ x ∈ R

and 1 + [x2] ≠ 0

So, f(x) = 0

Hence, f(x) is continuous and differentiable for all x ∈ R.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 41

Directions: The question has four choices, out of which ONE or MORE is correct.

If f(x) = (t - 2)(t - 3) dt for all x ∈ (0, ∞), then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 41

f′(x) has maxima at x = 2 (∵ f′(x) changes sign from +ve to -ve)

f′(x) has minima at x = 3 (∵ f′(x) changes sign from -ve to +ve)

Also, f(x) is decreasing in (2, 3). [∵ f′(x) < 0]

f′(x) = 0 for x = 2 and x = 3

So, by Rolle's theorem, there exists c ϵ(2, 3) for which f"(c) = 0.

Options A, B, C and D are correct.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 42

f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 42

Since f(x) has local maxima at x = -1 and f(x) has local minima at x =0

Agin integrating both sides we get

Using (1), (2) and (3) we get

Using number line rule

∴ f(x) is increasing for [1, 2√5] and f(x) has local maximum at x = -1 and f(x) has local minimum at x = 1

Hence options (b) and (c) are correct.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 43

Let P(x1, y1) and Q(x2, y2) y1 < 0, y2 < 0 be the end points of the latus rectum of the ellipse x2 + 4y2 = 4 Find the equation(s) of parabola(s) with latus rectum PQ.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 43

The ellipse x2 + 4y2 = 4 has 2 and 1 as semi-major and semi-minor axes and eccentricity = √3/2

Thus, the ends of the latus rectum are:

Considering only

Now, PQ = 2√3

Thus, the coordinates of the vertex of the parabolas are

and the corresponding equations are:

i.e.

*Multiple options can be correct
JEE Advanced Mock Test - 5 (Paper II) - Question 44

Directions: The question has four choices, out of which ONE or MORE is correct.

or every integer n, let an and bn be real numbers. Let function f : R → R be given by:

, for all integers n.

If f is continuous, then which of the following hold(s) for all n?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 44

L.H.L

=

R.H.L

=

Now check for continuity at x = 2n + 1

For Continuity

Options (b) and (d) are correct.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 45

What is the number of points in (–∞, ∞) for which x2 – x sinx – cosx = 0?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 45

Given f(x) = x2- x sin x- cos x = 0

f'(x) = 2x - x cosx = x(2 - cosx)

We know that maximum value of cosx is 1, so x(2 - cosx) > 0.

Therefore, when x > 0, f(x) is increasing and when x < 0,="" f(x)="" is="" />

For f(0) = -1

There exists a solution for (-∞, 0) and (∞, 0).

Therefore, using intermediate value theorem, it will cut the x-axis at 2 points.

So, there are total 2 solutions.

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 46

Let P be a variable point on the ellipse with foci F1 and F2. If A be the area of triangle PF1F2, then maximum value of A is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 46

Let p be (x, y) and the foci F1 and F2 be (-ae, 0) and (ae, 0).

So A is maximum when x = 0

So maximum of A = abe

=

A.T.Q.

a2 = 10, b2 = 1 and

e =

Maximum =

= 3

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 47

L1: 2x + y = 50 and L2: y = mx + 1 are two lines. A point (x, y) is said to be integral point if x, y ∈ I.

The greatest integral value of m for which the point of intersection of L1 and L2 has integral coordinates is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 47

2x + y = 50

y = mx + 1

Using the value of y = mx + 1,

2x + mx + 1 = 50

x(2 + m) = 49

So, x coordinate of intersection point of

For x to be an integer, 2 + m should be divisible by 49, so m can be 47, 5, -1, or -51.

So, mgreatest = 47 and mleast = -51

Greatest value of m = 47

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 48

Let an ordered pair A be defined as A(x, y) where x ∈ prime number, such that x < 10 and y ∈ natural numbers and y ≤ 10.. If the probability that the ordered pair A satisfies the relation x2 − 3y2 = 1 is P then 60P equals.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 48

Given, x∈ prime numbers and x < 10="" and="" y∈="" natural="" numbers="" and="" y="" ≤="" />

Also given that A(x, y).

Total no. of order pair n(A) = 4 × 10 = 40

Let, E be the set of all order pair (x, y) which satisfy x2 − 3y2 = 1.

⇒ 60P = 3

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 49

Let and where , and I3 be the identity matrix of order 3. If the determinant of the matrix is αω2, then the value of α is equal to _________.

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 49
Let

Multiplying with P on both sides

Now,

Det(A − I)=(w2 + w + w) + 7(−w) + w3 = −6w

Det((A − I))2 = 36w2

⇒ α = 36

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 (Paper II) - Question 50

If x5 − x3 + x = a, where x > 0, then the maximum value of 2a − x6 is equal to

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 50

Given, x5 − x3 + x = a, where x > 0.

Here, x5,− x3, x is a G.P. whose first term is x5 and common ratio −1 / x2

Apply AM ≥ GM on x & 1/x

Hence, the maximum value of 2a − x6 is equal to 1.

JEE Advanced Mock Test - 5 (Paper II) - Question 51

Match the equations in Column I with the values in Column II.

Which of the following is the correct option?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 51

Explanation of A

3sin2x - 7sinx + 2

Number of real solution = 6 in [0, 5π]

So, A → Q

Explanation of B:

So, B → S

JEE Advanced Mock Test - 5 (Paper II) - Question 52

Match the integrals in Column I with the values in Column II.

Which of the following is the correct option?

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 52

Let A =

Put x = tanθ ⇒ dx = sec2θdθ

∴ A = 2 dθ =

Let B =

Put x = sinθ ⇒ dx = cosθdθ

∴ B =

C =

JEE Advanced Mock Test - 5 (Paper II) - Question 53

The distinct points A(0, 0), B(0, 1), C(1, 0) and D(2a, 3a) are concyclic, then

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 53

Since, B and C are the ends of diameter as ∠BAC is 90°

∴ Equation of circle is

Now, point D(2a, 3a) satisfies this equation

JEE Advanced Mock Test - 5 (Paper II) - Question 54

The region represented by the inequality |2Z − 3i| < |3Z − 2i| is

Detailed Solution for JEE Advanced Mock Test - 5 (Paper II) - Question 54

Let Z = x + iy, and

the given inequality is equivalent to

Above inequality is the exterior of the unit circle with its centre at (0, 0) or Z = 0.

## Mock Tests for JEE Main and Advanced 2024

357 docs|148 tests
In this test you can find the Exam questions for JEE Advanced Mock Test - 5 (Paper II) solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Advanced Mock Test - 5 (Paper II), EduRev gives you an ample number of Online tests for practice

### Up next

 Test | 54 ques
 Test | 54 ques

## Mock Tests for JEE Main and Advanced 2024

357 docs|148 tests

### Up next

 Test | 54 ques
 Test | 54 ques