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# Part Test - 8 (JEE Advanced 2021)

## 72 Questions MCQ Test National Level Test Series for JEE Advanced 2020 | Part Test - 8 (JEE Advanced 2021)

Description
This mock test of Part Test - 8 (JEE Advanced 2021) for JEE helps you for every JEE entrance exam. This contains 72 Multiple Choice Questions for JEE Part Test - 8 (JEE Advanced 2021) (mcq) to study with solutions a complete question bank. The solved questions answers in this Part Test - 8 (JEE Advanced 2021) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Part Test - 8 (JEE Advanced 2021) exercise for a better result in the exam. You can find other Part Test - 8 (JEE Advanced 2021) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

Solution:

QUESTION: 3

### Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to :

Solution:

QUESTION: 4

A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is :

Solution:

QUESTION: 5

Four infinite ladder network containing identical resistances of R each, are combined as shown in figure. The equivalent resistance between A and B is RAB and between A and C is RAC. Then the value of  is :

Solution:

Let the equivalent resistance of one infinite ladder be x. Then the complete network reduces to

QUESTION: 6

Two small balls, each having equal positive charge Q are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole set-up is transferred to a satellite in orbit around the earth, the tension in equilibrium in each string is equal to

Solution:

A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the tension

QUESTION: 7

In a practical wheat stone bridge circuit as shown, when one more resistance of 100 ? is connected in parallel with unknown resistance ' x ', then ratio l1/l2 becomes '2'. l1 is balance length. AB is a uniform wire. Then value of ' x ' must be :

Solution:

QUESTION: 8

Loop A of radius (r << R) moves towards loop B with a constant velocity V in such a way that their  planes are always parallel. What is the distance between the two loops (x) when the induced emf in loop A is maximum

Solution:

*Multiple options can be correct
QUESTION: 9

The figure shows, a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10.

Solution:

*Multiple options can be correct
QUESTION: 10

A single circular loop of wire with radius 0.02 m carries a current of  8.0 A. It is placed at the centre of a solenoid that has length 0.65 m, radius 0.080 m and 1300 turns.

Solution:

*Multiple options can be correct
QUESTION: 11

In front of an earthed conductor a point charge + q is placed as shown in figure :

Solution:

Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside the conductor. Field due to only q is non-zero.

*Multiple options can be correct
QUESTION: 12

In the figure shown the key is switched on at t = 0. Let I1 and I2 be the currents through inductors having self inductances L1 & L2 at any time t respectively. The magnetic energy stored in the inductors 1 and 2 be U1 and U2. Then  at any instant of time is :

Solution:

QUESTION: 13

Statement – 1
Four point charges q1, q2, q3 and q4 are as shown in figure. The flux over the shown Gaussian surface depends only on charges q1 and q2.

Statement – 2
Electric field at all points on Gaussian surface depends only on charges q1 and q2.

Solution:

Statement ? is true directly from Gauss Theorem.
Statement 2 is false. Electric field at any point an Gaussian surface depends on all four charges.
Statement-1 is True, Statement-2 is False

QUESTION: 14

Statement 1  : A direct uniformly distributed current flows through a solid long metallic cylinder along its length. It produces magnetic field only outside the cylinder .

Statement 2 : A thin long cylindrical tube carrying uniformly distributed current along its length does not produce a magnetic field inside it. Moreover, a solid cylinder can be supposed to be made up of many thin cylindrical tubes.

Solution:

The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement-1
is false

QUESTION: 15

Statement–1 : Magnitude of potential difference across the terminals of a non-ideal battery in a circuit cannot be greater than its emf.

Statement–2 : When a current of magnitude I is passing through a battery of emf E and internal resistance r as shown, the magnitude of potential difference (V) across the battery is given by V = E– I r

Solution:

Statement-1 is obviously false if the current is sent in opposite direction given in figure of statement-2

QUESTION: 16

Statement–1 : No electric current will be present within a region having uniform and constant magnetic field.

Statement–2 : Within a region of uniform and constant magnetic field , the path integral of magnetic field  along any closed path is zero. Hence from Ampere circuital law

(where the given terms have usual meaning), no current can be present within a region having uniform and constant magnetic field.

Solution:

along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1.

QUESTION: 17

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The value of emf E1 is :

Solution:

QUESTION: 18

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The resistance R1 has value :

Solution:

QUESTION: 19

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The resistance R2 is equal to :

Solution:

QUESTION: 20

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. Which wire has the greatest radius ?

Solution:

QUESTION: 21

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. Which wire has the greatest magnitude of the magnetic field on the surface ?

Solution:

QUESTION: 22

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. The current density in wire a is

Solution:

QUESTION: 23

Match the following :

The following table gives the lengths of four copper rods at the same temperature, their diameters, and the potential differences between their ends.

Correctly match the physical quantities mentioned in the left column with the rods as marked.
(A) Rod 1            (p) Greatest Drift speed of the electrons.
(B) Rod 2            (q) Greatest Current
(C) Rod 3            (r) Second greatest current
(D) Rod 4            (s) Greatest Electric field
(t) greatest resistance

Solution:

QUESTION: 24

A  square loop of uniform conducting wire is as shown in figure. A current-I ( in amperes) enters the loop from one end and exits the loop from opposite end as shown in figure. The length of one side of square loop is l metre. The wire has uniform cross section area and uniform linear mass density. In four situations of column-I, the loop is subjected to four different uniform and constant magnetic field. Under the conditions of column-I, match the column-I with corresponding results of column-II ( Bo in column I is a positive nonzero constant)

Solution:

QUESTION: 25

In an f.c.c. crystal, which of the following shaded planes contains the following type of arrangement of atoms?

Solution:

Shown arrangement is hexagonally closed pack plane and these type of planes are arranged perpendicular to
body diagonal of fcc unit cell as shown.

QUESTION: 26

In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are :

(Atom numbered 3 is face centre of front face).

Solution:

Atoms along one edge or at corners do not touch each other in fcc cell.

QUESTION: 27

Electrolysis of a solution of HSO4- ions produces S2O8– –. Assuming 75% current efficiency, what  current  should  be  employed  to  achieve  a  production  rate  of 1 mole of S2O8- - per hour ?

Solution:

QUESTION: 28

You are given the following cell at 298 K,  = – 0.76 V. Which of the following amounts of NaOH (equivalent weight = 40) will just make the pH of cathodic compartment to be equal to 7.0 :

Solution:

QUESTION: 29

For the following parallel chain reaction

what will be that value of overall half-life of A in minutes?

Solution:

QUESTION: 30

Decomposition of A follows first order kinetics by the following equation.
4A(g) ——> B(g)  +  2C(g)

If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A ? (Assume only A is present initially)

Solution:

QUESTION: 31

Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following against an external pressure of 1000 Torr:

Solution:

QUESTION: 32

A complex containing K+, Pt(IV) and Cl is 100% ionised. Given that i = 3. Thus, complex is :

Solution:

and in [PtCl6]2-ion oxidation state of Pt is +4.

*Multiple options can be correct
QUESTION: 33

A current of 2.68 A is passed for one hour through an aqueous solution of CuSO4 using copper electrodes . Select the correct statement(s) from the following :

Solution:

Increase in mass of cathode = decrease in mass of Anode =

*Multiple options can be correct
QUESTION: 34

Decomposition of 3 A(g) ——> 2 B(g) + 2C(g) follows first order kinetics. Initially only A is present in the container. Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively. Which of the following is true.

Solution:

*Multiple options can be correct
QUESTION: 35

Which of the following is/are correct for an ideal binary solution of two volatile liquids (eg. benzene and toluene)?

Solution:

*Multiple options can be correct
QUESTION: 36

Which of the following is a mismatch :

Solution:

QUESTION: 37

Statement-1 : Time taken for the completion of 75% of a Ist order reaction is double than its t½.

Statement-2 : Time taken for completion of any fraction of a Ist order reaction is proportional to the extent of reaction completed.

Solution:

Statement-1 is correct but statement-2 is incorrect. The time taken for completion of any fraction of a Ist order
reaction is a fixed maximum value for a particular reaction.

QUESTION: 38

Statement-1 : The difference in the boiling points of equimolar solution of HCl and HF decreases as their molarity is decreased.

Statement-2 : The extent of dissociation decreases steadily with increasing dilution.

Solution:

Statement-1 is correct because As molarity decreases degree of dissociation of both HCI and HF will increase and approach unity.
Statement-2 is false because degree of dissociation increases with dilution.

QUESTION: 39

Statement-1 : A gas with higher critical temperature is absorbed more than a gas with lower critical temperature on the same absorbent.

Statement-2 :    Higher critical temperature implies that the gas is more easily liquifiable.

Solution:

Both the statements are individually correct but they are not related.

QUESTION: 40

Statement-1 : Specific conductance decreases with dilution whereas equivalent conductance increases.

Statement-2 : On dilution number of ions per millilitre decreases but total number of ions increases considerably.

Solution:

Total number of ions will increase slightly on dilution (Not considerably).

QUESTION: 41

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. What is the molar conductance of NaCl at infinite dilution ?

Solution:

QUESTION: 42

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. What is the cell constant of the conductivity cell (C) ?

Solution:

QUESTION: 43

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. If the cell (C) is filled with 5 x 10–3 (N) Na2SO4 the obserbed resistance was 400 ohm. What is the molar conductance of Na2SO4 ?

Solution:

QUESTION: 44

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product.

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The vant Hoff factor will be [ if we start with one mole of X ]

Solution:

QUESTION: 45

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product.

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The colligative properties observed will be

Solution:

(A) During association

QUESTION: 46

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product.

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The equilibrium constant for the process can be expressed as

Solution:

QUESTION: 47

Solution:

QUESTION: 48

For A + B → C in column - II the graphs given can be from any of these four types.
(p) –   Vs time (x axis) (q) t1/2 Vs initial concentration (x axis)         (s) Conc. Vs time (x axis) (t) log k Vs 1/Temperature (x axis)

Match the graphs in Column–II for the given order of reactions in Column – I

Solution:

(A) First order :– t1/2 is constant. (p)
concentration and rate decrease exponentially.(q)
Arhennius equation is applicable for every order reaction (t)
(B) Zero order :- Rate is constant. (p)
Concentration decreases linearly with t. (r)

Arhennius equation is applicable for every order reaction (t)

(C) Second order :-

Arhennius equation is applicable for every order reaction (t)

(D) Same as (A).

QUESTION: 49

Two roads OA and OB intersect at an angle of 60º.  A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.

Solution:

Let the distance between two cars after t seconds be s. Distance covered by first car in t second is 20t meter, so its distance from O will be (800 – 20t) meter and distance covered by second car will be 25t meter.

QUESTION: 50

If f(x) satisfies x + |f (x)| = 2f (x) then f–1 (x) satisfies

Solution:

QUESTION: 51

Range of the function y = sin  + cos  is

Solution:

QUESTION: 52

Let f : R → R and g : R → R be two one - one and onto functions such that they are mirror images of each other about the line y = 0, then h (x) = f(x) + g(x) is

Solution:

f : R → R & g : R →R be two one-one onto functions such that f & g are mirror images of each other about line                                            y = 0. It means one is –ve of the other
i.e. f(x) = – g(x)
⇒ f(x) + g(x) = 0
⇒ h(x) = 0

h(x) is not onto as well as not one-one

QUESTION: 53

The function f (x) = (x2 – 1) |x2 – 3x + 2| + cos |x| is not differentiable at x =

Solution:

QUESTION: 54

The function f (x) = (x – [x]) sin p x is, (where [.] denotes greatest integer function)

Solution:

f(x) is continuous at every integer & also f(x) is continuous for any other real number. So f(x) is continuous for all real numbers.To check differentiability

QUESTION: 55

If f(x) = x3 + ax2 + ax + x(tanθ + cotθ) is increasing for all real x and if    then

Solution:

QUESTION: 56

For what values of a, m and b, Lagrange's mean value theorem is applicable to the function f(x) for  x [0, 2]

Solution:

*Multiple options can be correct
QUESTION: 57

does not have critical points if 'a' equal to

Solution:

*Multiple options can be correct
QUESTION: 58

Let f(x) = Then f(x) is strictly increasing in R for

Solution:

*Multiple options can be correct
QUESTION: 59

Solution:

*Multiple options can be correct
QUESTION: 60

Which of the following is/are true

Solution:

If f is continuous at a point. Then |f| will also continuous at that point.
If f is differentiable then f|f| will also be differentiable

QUESTION: 61

Statement-1 : If a 4th degree polynomial function has four distinct real roots, then its graph have exactly  two inflection points.

Statement-2 : If the equation f"(x) = 0 has distinct real roots, then the equation f(x) = 0  has atleast four real roots.

Solution:

If a 4th degree polynomial function has four distinct real roots, then its graph will be like following

From graph we can say curve has 2 point of inflection.
Statement 1 is true
Statement 2 is not true.

QUESTION: 62

Statement-1 : Let 0 < a < b. Then Rolle's theorem is applicable to the function f(x) =  in [a,b] and Rolle's constant is the GM of a and b.

Statement-2 : All the three conditions of Rolle's theorem are satisfied by the above function f(x).

Solution:

QUESTION: 63

Statement-1    All points of intersection of y = f (x) & y = f – 1 (x) lies on y = x only.

Statement-2    If point P (α, β)  lies on y = f (x) then Q (β, α) lies on y = f – 1 (x).

Solution:

Pt. of intersection of y = f(x) & y = f-1(x) lies on y = x. Also they may not lie on y = x. Infact points which does not
lies on y = x are even in numbers.
and statement (2) is true.

QUESTION: 64

Statement-1  where [.] represent greatest integer function.
Statement-2

Solution:

QUESTION: 65

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.

Answer the following questions on the basis of this method

Q. The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 is

Solution:

QUESTION: 66

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.

Answer the following questions on the basis of this method

Q. The complete solution set of the equation |x2 – x| + |x + 3| = |x2 – 2x – 3| is

Solution:

QUESTION: 67

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.

Answer the following questions on the basis of this method

Q. The solution set belonging to (0, 2π) of the equation |sin x – cos x| = |sin x| + |cosx| is

Solution:

QUESTION: 68

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. In the following functions which one is a “two-one” function :-

Solution:

QUESTION: 69

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. Let f(x) = {x} be the fractional part function. For what domain is the function “two-one”?

Solution:

From graph it is clear that any horizontal line cut the graph at either one or two points between

QUESTION: 70

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. The values of ‘a’ for which the function f(x) = x4 – ax2 is a “two-one” function are

Solution:

QUESTION: 71

Solution:

QUESTION: 72

Solution:

(A)

(B)

(C)

(D)