JEE Advanced Mock Test - 6


54 Questions MCQ Test National Level Test Series for JEE Advanced 2020 | JEE Advanced Mock Test - 6


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This mock test of JEE Advanced Mock Test - 6 for JEE helps you for every JEE entrance exam. This contains 54 Multiple Choice Questions for JEE JEE Advanced Mock Test - 6 (mcq) to study with solutions a complete question bank. The solved questions answers in this JEE Advanced Mock Test - 6 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this JEE Advanced Mock Test - 6 exercise for a better result in the exam. You can find other JEE Advanced Mock Test - 6 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

A tank of uniform cross-section is completely filled with ice. The height of ice is H and its mass is m. When the entire ice melts, the work done by gravity is  

ice = 0.9 gm/cc, ρwater = 1 g/cc and g represents acceleration due to gravity) 

Solution:

H height of frozen water becomes 0.9 H height of liquid on melting.  
Thus Ui = m g(H/2) = 0.5 mgH,  Uf = mg(0.9H/2) = 0.45 mgH  
WG = Ui – Uf = 0.5mgH − 0.45mgH = 0.05 mgH 

QUESTION: 2

A hydrogen like species having atomic number Z = 2, in ground state, is excited by means of electromagnetic radiation of frequency 1.315 × 1016 Hz. How many spectral lines will be observed in the emission spectrum? 

(Planck’s constant h = 4.14 × 10-15 eVs) 

Solution:

E = hν = 4.14 × 10-15 × 1.315 × 1016 = 54.4 eV  
Energy of the given species in ground state is 

Ionization energy of the given species is = |E1| = 54.4 eV  
Thus the electron will get ionized and therefore become free.
So no spectral lines will be observed. 

QUESTION: 3

Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity ‘g’ at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is  

Solution:

Apparent value of g at pole is equal to

QUESTION: 4

A positive charge q is projected from origin with a velocity along positive x-axis in a region having uniform magnetic field directed towards negative y-axis. If T is the time period of circular motion then the velocity vector of charge q at some instant t where 

Solution:


The charge executes circular motion in xz plane with its center lying on the negative z-axis. At time t where the charge has completed (3/4)th of the circle and so its velocity vector is pointing towards 1st quadrant in x-z plane.  
⇒ vx is positive, vz is positive and vy is zero.  

QUESTION: 5

Four charges q1, q2, q3 and q4 are placed at the positions as shown in the figure, given q1 +q2+ q3 + q4 = 0 . The electric field on z-axis  

Solution:

As q1 +q2 + q3 + q4 = 0 , z axis is an equipotential line. Thus electric field is either zero or perpendicular to the z-axis. When q2 = q4 and q3 = q1 electric field is zero at origin and in all other cases it is perpendicular to z-axis. 

QUESTION: 6

For the circuit as shown in the figure q1 and q2 be the charges on 3μF and 6μF capacitors respectively, then 

Solution:

At any time t the current distribution in the circuit will be as shown. If i2 is the current going out of upper loop through 3 μF capacitor then current i2 must also come back to this through 6μF capacitor to make the current in that loop i1.Thus the current through both the capacitors will be same at all time instants. Thus they will have equal charge. 

*Answer can only contain numeric values
QUESTION: 7

A ball is released from position A and drops 10 m before striking a smooth incline. The coefficient of .  If the time taken by the ball to strike the incline again is t then find the value of t2 [in (second)2]. (g = 10 m/s2


Solution:


Just after collision component of velocity perpendicular to the incline is

Component of g perpendicular to incline is  

∴  time after which the ball strikes the incline again is  

*Answer can only contain numeric values
QUESTION: 8

Two identical beads P and Q of mass 1 kg each are connected by an inextensible massless string and they can slide along the two arms AB and BC of a rigid smooth wire frame in vertical plane. If the system is released from rest and vQ is the speed of bead Q when they have both moved by a distance of 0.1 m then find the value of (in m/s). (g = 10 m/s2


Solution:


Velocity of P and Q along the string should be same.  


From conservation of mechanical energy, decrease in potential energy of Q = increase in kinetic energy of both  


*Answer can only contain numeric values
QUESTION: 9

A solid uniform sphere rotating about its axis with kinetic energy Eo is gently placed on a rough horizontal plane. The coefficient of friction on the plane varies from point to point. After some time, the sphere begins pure rolling with total kinetic energy equal to E. Then find the value of 


Solution:

From conservation of angular momentum about point of contact  


*Answer can only contain numeric values
QUESTION: 10

A variable voltage V = 2t is applied across an inductor of inductance L = 2H as shown in figure. Then find the rate at which magnetic potential energy stored in the inductor is increasing at t = 1 s (in J/s). Take the current through the inductor at t = 0 as zero. 


Solution:


Potential energy stored in the inductor is  

QUESTION: 11

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.  

Q. 

The whole mixture in the calorimeter becomes ice if  

Solution:

For whole mixture to become ice tf ≤ 0oC  
Let tf = −to  then  
From heat lost = heat gained we get

QUESTION: 12

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.  

Q. 

The whole mixture in the calorimeter becomes water if  

Solution:

For whole mixture to convert to water tf ≥ 0oC  
Let tf = to then  
From heat lost = heat gained we get
 

QUESTION: 13

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.

Q. 

Water equivalent of calorimeter is  

Solution:

Let water equivalent of calorimeter be mo then  moC2 = mC1 ⇒ 

QUESTION: 14

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

The spring constant k is  

Solution:

When capacitor was uncharged, for equilibrium of lower plate of mass m   kxo = mg   where xo is the compression in the spring.   When voltage source is connected and lower plate again reaches equilibrium then compression in the spring is xo – (d1 – d2).  
Electric force between capacitor plates is 

QUESTION: 15

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

The maximum voltage Vm for a given k for which an equilibrium exists is  

Solution:



QUESTION: 16

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

When lower plate is slightly displaced about equilibrium position, time period T of small oscillations is  

Solution:

Let the lower plate be displaced upward by small distance x, then FBD of the plate is as shown. 



Net restoring force acting on the lower plate is 






Comparing with standard equation F = −Kx for SHM, we get

*Answer can only contain numeric values
QUESTION: 17

Two thin symmetrical lenses of different nature have equal radii of curvature R = 20 cm. The lenses are placed in contact and then immersed in water. The focal length of the system is found to be 24 cm. If the refractive indices of the two lenses are μ1 and μ2 respectively, then find the magnitude of 9(μ1 - μ2). Refractive index of water is 4/3.  


Solution:


where  are the focal lengths of the two lenses in water  

QUESTION: 18

A certain weak acid has a dissociation constant 1.0 x 10-4. The equilibrium constant for its reaction with strong base is  

Solution:

QUESTION: 19

The major product formed in the following reaction is 

Solution:

QUESTION: 20

Identify the major product of the following reaction: 

Solution:

This is nucleophilic addition followed by dehydration. 

QUESTION: 21

The oxidation state of molybdenum in [(η7-tropylium)Mo(CO)3]+ is 

Solution:

Tropylim cation has one positive charges and CO is neutral ligand. 

QUESTION: 22

In metal-olefin interaction, the extent of increase in metal ⎯→ olefin π-back-donation would 

Solution:
*Answer can only contain numeric values
QUESTION: 23

The highest oxidation state of an element in the following compound that behaves as an acid in H2SO4 is  

AcOH, HNO2,  HNO3, H2O, HClO, HClO4


Solution:

HClO4 is stronger acid than H2SO4 

*Answer can only contain numeric values
QUESTION: 24

How many unpaired electrons are present in O2 molecule? 


Solution:
*Answer can only contain numeric values
QUESTION: 25

One mole of Pb3O4 is separately reacted with excess of HCl and HNO3. The difference in moles of HCl and HNO3 is 


Solution:

Pb3O4 is a mixture of PbO2 and PbO. Only PbO reacted with HNO3 since Pb4+ is a very good oxidizing agent. HCl is a good reducing agent that can reduces Pb4+ into Pb2+.  
Pb3O4 + 8HCl ⎯⎯→ 3PbCl2 + Cl2 + 4H2O  
Pb3O4 + 4HNO3 ⎯⎯→ 2Pb(NO3)2 + PbO2 + 2H2

*Answer can only contain numeric values
QUESTION: 26

How many moles of phenyl hydrazine are used in the formation of osazone from glucose? 


Solution:


*Answer can only contain numeric values
QUESTION: 27

How many S – S bonds are present in (SO3)3


Solution:

QUESTION: 28

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

for an indefinitely dilute boundary then dμ1 is 

Solution:
QUESTION: 29

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

Which of the following solution has highest osmotic pressure? 

Solution:
QUESTION: 30

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

Which of the following is correct?

Solution:
QUESTION: 31

Thionyl chloride can be synthesized by chlorinating SO2 using PCl5. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2-dimethoxypropane.

Q. 

Consider the following reaction 


The compound X is 

Solution:

SO2 + PCl5 ⎯→ SOCl2 + POCl3 

QUESTION: 32

Thionyl chloride can be synthesized by chlorinating SO2 using PCl5. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2-dimethoxypropane.

Q. 

In the preparation of anhydrous FeCl3 from hexahydrated salt, SOCl2 acts as 

Solution:

SOCl2 acts as dehydrating agent. 

QUESTION: 33

Thionyl chloride can be synthesized by chlorinating SO2 using PCl5. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2-dimethoxypropane.

Q. 

In the following reaction  

FeCl3.6H2O + 6MeC(OH)(OH)Me ⎯⎯→FeCl3 + Y + Z

Solution:

FeCl3.6H2O + 6MeC(OH)(OH)Me ⎯⎯→ FeCl3 + 6 Acetone + 12 MeOH 

QUESTION: 34

Each question has four statements (A, B, C and D) given in Columns I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II.

Solution:

Nylon-6,6 is a condensation product of hexamethylenediamine and adipic acid, it is a poly amide fiber polymer. SBR is the addition polymer of styrene and butadiene; it is a synthetic rubber polymer which is also known as Buna-S.  Nylon-6 is a polyamide polymer of Caprolactam, it if also an example of fiber polymer. PTFE (Polytetrafluro ethylene) is the addition polymer of Tetrafluoroethylene. 

QUESTION: 35

Each question has four statements (A, B, C and D) given in Columns I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II.

Solution:
*Answer can only contain numeric values
QUESTION: 36

How many lone pair of electrons at Xe atom are present in XeOF4


Solution:

QUESTION: 37

Matrices of order 3 × 3 are formed by using the elements of the set A = {−3, −2, −1, 0, 1, 2, 3}, then probability that matrix is either symmetric or skew symmetric is  

Solution:

Probability that matrix is symmetric
Again that matrix is skew symmetric
One matrix containing all elements = 0; is common in both type of matrices.  
∴  Required probability

QUESTION: 38

Solution:

QUESTION: 39

Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of  

Solution:


Let P(x, y) be the point on the curve passing through the origin O(0, 0), and let PN and PM be the lines parallel to the x- and y-axes, respectively. If the equation of the curve is y = y(x), the area POM equals and the 


This solution represents a parabola. We will get a similar result if we had started instead with 2(PON) = POM. 

QUESTION: 40

 is neither an even function nor an odd function.   {0}, where [.] is greatest integer function.  
S3: f(x) = sgn (x) and g(x) = sgn(sgn(x)) is not a pair of identical function.  
S4: The sum of two non periodic function is always a non periodic function.  

Solution:
*Answer can only contain numeric values
QUESTION: 41

If the number of subsets X of {1, 2, 3, …. 10} such that X contains at least two elements and no two elements of X differ by 1 is K, then sum of digits of K is equal to 


Solution:

Every allowable k-elements subset corresponds to a way of choosing k out of a row of 10 objects so that no two are adjacents, remove (k − 1) unselected objects, one from each gap.  
This establishes, for each k ≥ 2 , a one to one correspondence between allowable subsets of {1, 2, …… 10} containing k elements, and the number of ways of choosing k out (10 − k+ 1) objects. It follows that there are  

Allowable subsets. 

*Answer can only contain numeric values
QUESTION: 42

A rectangle, HOMF is constructed with sides HO = 11 and OM = 5. The triangle ABC has orthocenter H, circumcentre O, M is the mid point of BC and F is foot of altitude from A. If length of BC = l , then l/7 is equal to


Solution:

The centroid G of triangle is collinear with H and O, and G lies two thirds of way from A to M. Therefore H is two thirds of the way from A to F, so AF = 3 × OM = 15.  
Since the triangle BFH and AFC are similar, hence,  

⇒  BF. FC = FH.AF = 75.
Now,  BC2 = (BF + FC)2 = (BF – FC)2 + 4BF.FC  
But,  FC – BF = (FM + MC) – (BM – FM) = 2FM = 2 HO = 22


*Answer can only contain numeric values
QUESTION: 43

If the smallest integer with exactly 24 divisors is N, then N/40 is equal to  


Solution:

If n is the required number and


Then divisor of n : T(n) = (α1 + 1)(α2 + 2)…….(αk + 1)  
But 24 can be written as the product of 2 or 3 or 4 factors.  
Corresponding to each factorization; we can get a smallest composite number.  

∴  The smallest number having 24 divisors is 360.  

*Answer can only contain numeric values
QUESTION: 44

In a certain town, the probability that, all the outgoing telephone lines are jammed, in the telephone exchange is 1/4. The probability that the customer will attempt to telephone is 3/20. If a customer telephones and it fails to get connected, the probability that he would replace his telephone by a cell phone is 3/4. If P is the probability that the customer replaces the telephone by a cell phone, when the outgoing lines of the exchange are jammed then  is equal to

(where [ ]  denotes the greatest integer function)  


Solution:

Let A : be the event that the outgoing telephone lines are jammed.  
B : be the event that the customer attempt telephone.  
C : be the event that the customer replaces the telephone by a cell phone due to the failure caused by the jamming of telephone lines. 

Here A and B are independent events, but A∩B and C are mutually dependent.   Required probability

*Answer can only contain numeric values
QUESTION: 45

The terms of  are all integers (where a, x > 0). If K is the least composite odd integral value of ‘a’, then K/5 is equal to  


Solution:


∴  The terms of the AP are :which are integers
∴  a = 5n2,   n∈N
for n = 1, a = 5 which is not composite  
for n = 2, a = 20 which is composite but not odd.  
for n = 3, a = 45 which is the least composite odd.  

QUESTION: 46

Let z1 and  z2 be complex numbers such that and the roots α and β of x2 + z1x + z2 + m = 0 for some complex number m satisfies 

Q.

The locus of the complex number m is a curve  

Solution:


QUESTION: 47

Let z1 and  z2 be complex numbers such that and the roots α and β of x2 + z1x + z2 + m = 0 for some complex number m satisfies 

Q. 

The maximum value of |m| is  

Solution:


QUESTION: 48

Let z1 and  z2 be complex numbers such that and the roots α and β of x2 + z1x + z2 + m = 0 for some complex number m satisfies 

Q. 

The minimum value of |m| is  

Solution:


QUESTION: 49

If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as Mean-Value Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i)  If f increases and has a concave graph in the interval [a, b] then

(ii) If f increases and has a convex graph in the interval [a, b] then  

Q. 

Using Mean-Value Theorem, the best upper bound of​ 

Solution:

QUESTION: 50

If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as Mean-Value Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i)  If f increases and has a concave graph in the interval [a, b] then

(ii) If f increases and has a convex graph in the interval [a, b] then  

Q. 

Using (i) or (ii) (above),  the best upper bound of 

Solution:

QUESTION: 51

If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as Mean-Value Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i)  If f increases and has a concave graph in the interval [a, b] then

(ii) If f increases and has a convex graph in the interval [a, b] then  

Q. 

Using (iii) (above), the best upper bound of

Solution:

QUESTION: 52

Match the following : 


Solution:

(A) Period of sin2 π x is 1 and period of x – [x] is equal to 1   Thus period of f(x) is 1. 


⇒ f(x) takes 6 integer values  
⇒ k = 6.  
(C) f ′(x) = 3x 2+ 2mx + m
for f(x) to be invertible it must be one–one

QUESTION: 53

Match the following :



Solution:





Hence any point on L1 and L2 can be A (λ, λ −1, λ) and B(2μ - 1, μ, μ)  

*Answer can only contain numeric values
QUESTION: 54

In a Δ A BC, A ≡ (α, β ) , B ≡ (1, 2), C ≡ (2, 3) and point A lies on the line y = 2x + 1 where α,β∈ integer. Area of triangle ABC is Δ such that [Δ] = 2, where [.] denotes greatest integer function. Find number of all possible coordinates of A.


Solution:



Since ‘α’ is a n integer  
⇒ α = −5, −4, 4, 5  
∴  A  ≡ (−5, − 9 ) , ( −4, −7 ) , ( 4, 9) , (5,11)

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