JEE Advanced Mock Test - 4 (Paper I) - JEE MCQ

# JEE Advanced Mock Test - 4 (Paper I) - JEE MCQ

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## 54 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Advanced Mock Test - 4 (Paper I)

JEE Advanced Mock Test - 4 (Paper I) for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Advanced Mock Test - 4 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 4 (Paper I) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 4 (Paper I) below.
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JEE Advanced Mock Test - 4 (Paper I) - Question 1

### When liquid medicine of density p is to be put in the eye it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r << R)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 1
The vertical force due to the surface tension on the drop is,

JEE Advanced Mock Test - 4 (Paper I) - Question 2

### The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.For a = 0, the value of d (maximum value of p as shown in the figure) is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 2
As the charge density is

The charge inside the nucleus is

JEE Advanced Mock Test - 4 (Paper I) - Question 3

### Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x << l. Find the rate dq/dt with which the charge leaks off each sphere if their approach velocity varies as v = a/√x , where a is a constant.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 3

JEE Advanced Mock Test - 4 (Paper I) - Question 4

A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m which is stretched under a load by 0.7 mm. Calculate the load, given that the Young's modulus of elasticity for copper is 1.1 × 1011 N m–2 and that for steel is 2 × 1011 N m–2.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 4
Here, r = 3.0/2 = 1.5mm = 1.5 × 10−3m

LC = 2.2 m, Ls = 1.6 m

ΔLc + ΔLs = 0.7 × 10–3m ...........(1)

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

A stepped cylinder, with thread wound around smaller diametre, is released from rest and the cylinder moves down. Then,

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

Acceleration of the point of contact, ac = acm + αR

Velocity of point of contact is non-zero. Hence, rolling with slipping will occur and friction will act upward.

For translational motion, mg sinθ - T - μ mg cosθ = ma -----I

For rotational motion, torque about CM of cylinder

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 6

When a body of mass M is attached to lower end of wire (of length L) whose upper end is fixed, then the elongation of wire is l. Which of the following statements regarding this is/are correct?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 6
Loss in gratiational potential energy of M is Mgl, as M falls down by l.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 7

Positive point charges q1 and q2 are moving with velocities v1 and v2 as shown in the given figure. Mark the correct statements.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 7
Using Biot Savart law, magnetic field at location of q2 due a1,

directed into plane of paper

Force experienced by q2 due to magnetic field of q1,

F = q2v2

The direction of force can be calculated using left hand rule, hence direction of force is towards left.

The charge q1 is placed on line along which q2 moves. Hence, magnetic field at location as q1 due to q2 is zero.

Thus, magnetic force experienced by q1 is zero.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 8

Directions: Mark out the correct statement(s) about wave speed and particle velocity for the transverse travelling mechanical wave on string.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 8
Wave speed depends on properties of medium

Velocity of particle, y = sin (ωt- kx)

vP = dy/dt = -ωA cos (ωt - kx)

∴ Velocity of particle depends upon amplitude and frequency of wave.

Velocity of particle is different for different particles at a particular instant.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 9

An elliptical cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 9
Under electrostatic condition, all points lying on the conductor are at same potential. Therefore, potential at A = potential at B. Hence, Option (potential at A = potential at B) is correct. From Gauss theorem, total flux through the surface of the cavity will be q/ε0.
*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 10

It is observed that only 0.39% of the original radioactive sample remains un decayed after eight hours. Hence

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 10
From the law of radioactivity,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 11

Consider the situation shown in figure and select the correct statement from the following.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 11

Since, all pulleys and threads are light and pulleys are frictionless, the tensions in threads passing over B and C are equal i.e., TB = TC. Suppose pulley C moves down with acceleration a, then acceleration of B is a upward.

Writing newton's 2nd law equation for various blocks.

It means C moves up and B moves down with acceleration a = g/11

T = 2TB = 2TC

Solving the various equation, we get T = 12.46 g

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 12

In the figure shown AB is rod of length 30 cm and area of cross-section 1 cm2 and thermal conductivity 336 S.I. units. The ends A and B are maintained at temperature 20°C and 40°C respectively. A point C of this rod is connected to a box D, containing ice at 0°C, through a highly conducting wire of negligible heat capacity. Find the rate at which ice melts in the box. [Assume latent heat of fusion for ice Lice = 80 cal g−1]

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 12
Thermal resistance of

Temperature of C

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 13

A binary star consists of two stars A (mass 2.2Ms) and B (mass 11MS), where Ms is the mass of the Sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 13
Let stars A and B be rotating about their centre of mass with angular velocity ω.

Let the distance of stars A and B from the centre of mass be rA and r,, respectively, as shown in the figure.

Total angular momentum of the binary stars about the centre of mass is

Angular momentum of star B about the centre of mass is

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 14

When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 14
In series,

Rate of heat produced in R is

In parallel,

Rate of heat produced in R is

According to the given problem, J1 = 2.25J2

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 15

A large glass slab (μ = 5/4) of thickness 6 cm is placed over a point source of light on a plane surface. There is a bright circular patch of light on the top surface of the slab with radius R cm. What is the value of R?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 15
In the figure, C represents the critical angle.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 16

A mirror of radius of curvature 20 cm and an object which is placed at a distance of 15 cm are both moving with velocities 1 m s−1 and 10 m s−1 as shown in diagram. The velocity of the image at this situation is 9vP. Find vP.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 16

So, the image will move with velocity 45 cm s−1.

Hence, vP = 5 cm s−1

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 17

In the shown wire frame, each side of a square (the smallest square) has a resistance 2 Ω. The equivalent resistance of the circuit between the points A and B is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 17

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 18

A thin string of negligible mass, natural length L has Young's modulus Y. The string hangs from roof with masses m1 and m2 as shown in the figure. If mass m2 is removed gently, the mass m1 is just able to bounce back up to point O. Find the ratio m2/m1. (string will not obstruct the motion of mass m1 and system is initially in equilibrium)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 18
Using condition of equilibrium

JEE Advanced Mock Test - 4 (Paper I) - Question 19

Directions: The following question is based on the paragraph given below.

The noble gases have closed-shell electronic configuration and are monatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to give XeO3. The compound can also be prepared using XeF6 as the starting compound. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

The chemical nature of the compounds XeF4 and XeF6 is expected to be

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 19
The oxidation state of Xe in XeF4 is +4 and in XeF6 is +6. These oxidation states of Xe are displayed only with elements with high electronegativity, like oxygen and fluorine. Since, Xe will have a tendency to get reduced, the nature of fluorides of xenon is expected to be oxidising.
JEE Advanced Mock Test - 4 (Paper I) - Question 20

(1R, 3S)-Cis-1-Bromo-3-methyl cyclohexane. The product formed in the reaction is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 20
In the presence of a polar aprotic solvent, like acetone, the mechanism followed will be SN2.

Walden inversion takes place at C1, where -Br is substituted by -OH.

Hence, the product formed will be (1S, 3S)-Trans-3-methyl cyclohexanol

JEE Advanced Mock Test - 4 (Paper I) - Question 21

What is the magnetic moment of coordination compound formed during brown ring test?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 21
Brown ring test When a freshly prepared FeSO4 solution is added to aqueous solution of NO3 ion followed by addition of concentrated H2SO4 the brown ring is observed at junction between two liquids. This colour is due to charge transfer oxidation state of iron in this complex is + l. In this state Fe+, has 3 unpaired electrons, and hence, the magnetic moment will be: √3(3+2) = 3.87 BM

JEE Advanced Mock Test - 4 (Paper I) - Question 22

The equilibrium constant K for the reaction 2HI(g) ⇌ H2 (g) + I2(g) at room temperature is 2.85 and that at 698 K is 1.4 × 10–2. This implies that -

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 22
KC decreases with temperature i. e, reaction is exothermic or energy of HI is more than H2 and I2 or HI is less stable than H2, I2.
*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 23

Consider the following graph:

From this graph, it is clear that

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 23
More substituted alkenes are more stable due to more number of alpha hydrogens (hyperconjugation) According to given graph, trans-2-butene is more stable than 1-butene by

30.3 kcal – 27.6 kcal = 2.7 kcal,

or 2.7 kcal x 4.2 = 11 kJ

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 24

Which of the following oxides can act both as a reducing agent as well as an oxidising agent?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 24
In SO2, the oxidation state of S is +4. It can vary its oxidation state between -2 and +6 in its compounds and act both as an oxidising agent as well as a reducing agent.

In MnO2, the oxidation state of Mn is +4. It can vary its oxidation state between +2 and +7 in its compounds and act both as an oxidising agent as well as a reducing agent.

In CrO, the oxidation state of Cr is +2. It can vary its oxidation state between +1 and +6 in its compounds and act both as an oxidising agent as well as a reducing agent.

In Al2O3, the oxidation state of Al is +3. It can vary its oxidation state between +1 and +3 in its compounds and act only as an oxidising agent.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 25

Select the correct elimination product(s).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 25
NaI in acetone gives anti-elimination of HBr.

In case of anti-elimination,

(i) syn form gives trans isomer

(ii) anti form gives cis isomer

Hence, both (2) and (3) are correct answers.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 26

Among the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any) is/are

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 26
If all the atoms (except H) present are either sp2 or sp hybridised, then all the atoms will lie in a plane. Along C—C single bond, conformations are possible in butadiene in which all the atoms may not lie in the same plane. Hence, option 1 is false. AIlenes with even number of double bonds are non-planar. So, option 4 is also false.
*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 27

Amongst the following, the species having at least one unpaired electrons is/are

[Note - Use Molecular Orbital Theory to be valid in all the options]

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 27
As per MOT Theory

(One unpaired electron) F−2

As per MOT Theory

(Two unpaired electron) O2

(One unpaired electron) (XeF) ⇒ 15 valence electrons.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 28

Which of the following statements are correct ?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 28
A reaction in which different stereoisomers produce different products or act at different rates is called a stereospecific reaction. A reaction in which a given substrate produces diastereoisomeric products in different amounts and where one diastereomer predominates very much over the other, is called stereoselective reaction. If the replacement of one group at an achiral centre by a new substituent generates a chiral centre, the original molecule is said to be enantiotropic. The E and Z system of naming geometrical isomers, is based on the CIP sequence rule.

In All statements, compounds show optical isomerism.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 29

The species observed in the following sequence of reaction:

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 29
In presence of strong base carbanion intermediate is produced which acts as a nucleophile and attacks on the nearest carbon having chlorine atom. Cyclic ketone is opened to produce ester. The path of the reaction is

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 30

The correct order of resonance energies of the compounds

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 30
Greater delocalisation of π-electrons,

⇒ more resonance energy

Homocyclic aromatic are more stable than heterocyclic aromatic.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 31

The formula of the magnesium salt of a monobasic acid is MgA2.nH2O. (HA is the formula of the acid.) 1 gram of the salt on strong heating leaves behind 0.2 gram of MgO. Given that the molecule mass of the acid is 62, what is the value of n?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 31

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 32

How many structural isomers of ester having molecular formula C6H12O2 are possible with parent name as methanoate?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 32
In all of these structures, methanoate carboxylic acid part is fixed and structural isomers of the alcohol part of the ester are formed.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 33

For real gases, the graph of PV v/s p at constant temperature is not linear. So, d/P or W/V.P will not be independent of P. (W→ Mass of gas, V → Volume, P → Pressure) y-intercept of the graph d/P (g/atm-L) v/s P (atm) at 360 K is:

[Given: Molar mass of gas = 60 g/m,

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 33
For y-intercept

(∵ The gas will behave like an ideal gas)

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 34

How many isomers of 'x' C8H10 when reacts with hot alkaline KMnO4 gives only aromatic dicarboxylic acid? How many isomers of 'y' C4H8 when reacts with hot alkaline KMnO4 to give carbondioxide? What is the sum of 'x' and 'y'?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 34

C4H, has three structural isomeric alkenes. 2-butene, 1-butene and Isobutene. Among these 1-butene and isobutene gives carbon dioxide with hot alkaline KMnO4.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 35

A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 35
Molecular weight of B

Thus the increase in weight due to addition of Br2 is

= 374.22 − 54 = 320.22

Hence Br atom in the compound B is =320.22/80 = 4

Molecular weight of C is

So no. of H atom increased is 4 so a must be alkyne hence possible structure of the compound is:

Molecular weight of alkyl halide is 109 so it is ethyl bromide.

D. CH3−CH2−C ≡ C−CH2CH3

E.

So molecular weight of E is 114.

So x + y + z = 6

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 36

Calculate the number of different hydrocarbons possible, when a mixture of sodium ethanoate and sodium propanoate undergoes kolbe electrolysis.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 36
In the Kolbe's electrolysis hydrocarbons formed at anode.

Reactions at anode:

Hence, the possible hydrocarbons formed are

JEE Advanced Mock Test - 4 (Paper I) - Question 37

and then x + y + z equals

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 37

Taking dot product with we get

JEE Advanced Mock Test - 4 (Paper I) - Question 38

Out of 3n consecutive integers, three are selected at random. Find the probability that their sum is divisible by 3.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 38
Suppose the sequence of 3n consecutive integers begin with integer m.

Then the 3n consecutive integers are:

m, m + 1, m + 2,…………, m + (3n - 1)

Out of these integers, 3 integers can be chosen in 3nC3 ways.

Let us divide these 3n consecutive integers in the groups G1, G2 and G3 as follows:

G1: m, m + 3. m + 6, …….., m + (3n - 3)

G2: m + 1, m + 4, m + 7, ……., m + (3n - 2)

G3: m + 2, m + 5, m + 8, …….., m + (3n - 1)

The sum of 3 integers chosen from the given 3n integers will be divisible by 3 if either all the three integers are chosen from the same group or one integer is chosen from each group. The number of ways that the three integers are from the same group is and the number of ways that the integers are from different groups is .

So, number of ways in which the sum of three integers is divisible by 3 =

Hence, required probability

JEE Advanced Mock Test - 4 (Paper I) - Question 39

If SK be the perpendicular from the focus S on the tangent at any point P on the ellipse then locus of the foot of the perpendicular K is equal to

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 39
Let us assume that P(a cosθ, b sinθ) is any point on the given ellipse

Equation of tangent at any point (a cosθ, b sinθ) on the ellipse is equal to

A line perpendicular to Eq.(i) and passing through focus (ae, 0) will be

By eliminating θ from both the eqs. (i) and (ii) [by squaring and adding (i) & (ii)], we get the locus of K as x2 + y2 = a2 i.e., the auxiliary circle of the ellipse.

JEE Advanced Mock Test - 4 (Paper I) - Question 40

The value of where [.] denotes the greatest integer function is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 40
Let x = 4m + r is an integer & 0 < r="" />< />

∴ Given expression = 4m = [x]

Similarly, we can discuss it when rr belongs to (1, 2), (2, 3) and (3, 4).

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 41

The equations of two ellipses are and where p is a parameter. The locus of the points of intersection of both the ellipses is a set of curves comprising

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 41
Let the point of intersections is (h, k). Therefore,

x = ± y represents two straight lines

x2 + y2 = 1 represents a circle

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 42

The internal bisector of ∠A of a triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and side AB at F. If a, b and c represent the sides of ΔABC , then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 42

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - 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. The equation(s) of parabolas with latus rectum PQ is/are

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 43

a = 2 and b = 1

b2 = a2 (1 - e2)

or, 1 = 4(1 - e2)

(given y1 and y2 less than 0)

Co-ordinates of midpoint of

Length of latus rectum of parabola = 2√3

Two parabolas are possible whose vertices are

Only second and third option satisfy the equation of latus rectum of parabola.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 44

The vector(s) which is/are coplanar with vectors and perpendicular to the vector is/are

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 44

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 45

If ∫ x log(1 + x2) dx = ϕ(x) log(1 + x2) + ψ(x), then :

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 45
I = ∫ x log(1 + x2)dx

Let's assume log(1 + x2) as a first function and xx as a second function.

Now, by applying the formula of integration by parts we get,

So, substituting this in above equation we get ,

Putting the value of t we get,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 46

An ellipse intersects the hyperbola 2x2 − 2y2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 46
Eccentricity of the hyperbola is 2√2 as it is a rectangular hyperbola so eccentricity ee of the ellipse is 1/√2

Let the equation of the ellipse be

So, the equation of the ellipse is

x2 +2y2 = a2x2 + 2y2 = a2

Let (x1, y1) be a point of intersection of the ellipse and the hyperbola, then

Equations of the tangents at (x1, y1) to the two conics are

Since the two conics intersect orthogonally

and from (1), we get

Hence, the equation of the ellipse is x2 + 2y2 = 2 and its focus is

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 47

For a positive integer n, let a(n) Then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 47
We have

Next,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper I) - Question 48

A circle of radius 4cm is inscribed in ΔABC, which touches the side BC at D, if BD = 6cm and DC = 8cm, then which of the following options are correct ?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 48
Given, inradius of ΔABC = 4cm,

s − b = 6 and s − c = 8

(i) r = (s − a) tan A/2 = (s − b) tan B/2

= (s − c) tan C/2

⇒ tan B/2 = 2/3

and tan C/2 = 1/2

Since, all are less than 1

Hence, the triangle is necessarily acute angled triangle.

(ii) Solved in option (i).

⇒ 2s = 3a

⇒ 2s = 42

∴ Perimeter = 42

(iv) Δ = Area of ΔABC = r × s = 84 sq cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 49

Directions: The answer to the following question is a single digit integer ranging from 0 to 9.

The harmonic mean of the roots of the equation (5 + √2) x2 - (4 + √5) x + 8 + 2√5 = 0 is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 49
Let α,β be the roots of given quadratic equation. Then,

Let H be the harmonic mean between α and β, then

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 50

Directions: The answer to the following question is a single-digit integer ranging from 0 to 9.

If the system of equations x - ky - z = 0,

kx - y - z = 0,

x + y - z = 0

has a non-zero solution, then possible values of k are ±A. Calculate the value of A.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 50
Since the given system has non-zero solution, therefore

Applying C1 → C1 - C2, C2 → C2 + C3

⇒ 2(k + 1) - (k + 1)2 = 0

⇒ (k + 1) (2 - k - 1) = 0

⇒ k = ± 1

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 51

Directions: The answer to the question is a single-digit integer, ranging from 0 to 9.

Consider the parabola y2 = 8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point on the parabola, and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum.

Then Δ12 is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 51
Given: Equation of the parabola is: y2 = 8x ⇒ a = 2

The endpoints of its latus rectum are (a, 2a) and (a, -2a).

Points are (2, 4) and (2, -4).

Area of triangle

The equation of the tangent at (2, 4) is given by y = x + 2. ...(1)

The equation of the tangent at (2, -4) is given by -y = x + 2. ...(2)

And the equation of the tangent at (0.5, 2) is given by y = 2x + 1. ...(3)

Intersection points of (1), (2) and (3) are: (-2, 0), (-1, -1) and (1, 3)

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 52

The integer n for which is a finite non-zero number is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 52

to exist we must have n − 3 = 0 ⇒ n = 3.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 53

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 53
Given equation is

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper I) - Question 54

The total number of local maxima and local minima of the function

Detailed Solution for JEE Advanced Mock Test - 4 (Paper I) - Question 54

The graph y = f(x) is as shown in the figure, clearly there is one local maximum at x = −1 and one local minima at x = 0.

∴ Total number of local maxima or minima = 2.

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