# Part Test - 7 (JEE Advanced 2021)

## 72 Questions MCQ Test National Level Test Series for JEE Advanced 2020 | Part Test - 7 (JEE Advanced 2021)

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Attempt Part Test - 7 (JEE Advanced 2021) | 72 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study National Level Test Series for JEE Advanced 2020 for JEE Exam | Download free PDF with solutions
QUESTION: 1

### The spring block system lies on a smooth horizontal surface. The free end of the light spring is being pulled towards right with constant speed v0 = 2m/s. At t = 0 sec, the spring of spring constant k = 100 N/cm is unstretched and the block has a speed 1 m/s to left. The maximum

Solution:

In the frame (inertial w.r.t earth) of free end of spring, the initial velocity of
block is 3 m/s to left and the spring unstretched .

QUESTION: 2

### The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume V/4  in the cube as shown. This chamber is completely filled by m kg of water.  Water leaks through a hole H. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is :

Solution:

Let h be the height of water surface, finally

QUESTION: 3

### Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal  force F and 2F are applied on the two blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is :

Solution:

QUESTION: 4

A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 11 m/s2 downwards. The time after which the coin will strike the lift is :

Solution:

Relative to lift initial velocity and acceleration of coin are 0 m/s and 1 m/s2 upwards

QUESTION: 5

The extension in a uniform rod of length l, mass m, cross section radius r and young’s modulus Y when it is suspended at one of its end is :

Solution:

QUESTION: 6

A spherical soap bubble of radius 1.0 cm is formed inside another of radius 2 cm. If a single soap bubble is formed which maintains the same pressure difference as inside the smaller and outside the larger bubble, the radius of this bubble is -

Solution:

QUESTION: 7

Block ‘ A ‘ is hanging from a vertical spring and is at rest. Block ‘ B ‘ strikes the block ‘A’ with velocity ‘ v ‘ and sticks to it. Then the value of  ‘ v ‘ for which the spring just attains natural length is :

Solution:

The initial extension in spring is x0 =  mg/k
Just after collision of B with A the speed of combined mass is v/2 .
For the spring to just attain natural length the combined mass must rise up by x0 = mg/k(sec fig.) and comes to rest.

QUESTION: 8

A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane. Equal volumes of two liquids of densities ρ and σ (ρ > σ) fill half of the tube as shown. θ is the angle which the radius passing through the interface makes with the vertical.

Solution:

Pressure at 'A' from both side must balance. Figure is self–explanatory.

*Multiple options can be correct
QUESTION: 9

A ball tied to the end of a string swings in a vertical circle under the influence of gravity

Solution:

As shown in the figure , at A

Here, at is maximum because at other place, at = gcos θ . at and ac will be equal at some point. ac will have extreme
value at lowest points and there at = 0.

*Multiple options can be correct
QUESTION: 10

A particle moves along the X-axis as
x = u(t – 2) + a(t – 2)2

Solution:

*Multiple options can be correct
QUESTION: 11

Which of the following is not possible ?

Solution:

The pressure at any point can never have different values. Hence (A) & (D) are not possible. (Calculate the pressures at points A & D from both their left and right)

In case of insufficient length of capillary tube the shape of meniscus is as below :

*Multiple options can be correct
QUESTION: 12

Two particle P and Q are in motion under gravity. Then :

Solution:

QUESTION: 13

Statment–1 : Steady flow can be maintained in compressible liquids.

Statment–2 : Steady flow means all physical parameters related to the flow being constant with time though they may vary with position.

Solution:

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

QUESTION: 14

Statment–1 : In a circular motion, the force must be directed perpendicular to the velocity all the time.

Statment–2 : A centripetal force is required to provide the centripetal acceleration in a circular motion.

Solution:

Only in uniform circular motion force must be directed perpendicular to the velocity all the time. However in non-uniform circular motion, force will always not be perpendicular to velocity. Hence statement-1 is false.

QUESTION: 15

Statment–1 : The centre of gravity and centre of mass of a body may be at different positions.

Statment–2 : If the mass is uniformly distributed then only the centre of mass and centre of gravity of a body will be at the same place.

Solution:

Centre of mass and centre of gravity of a uniformly distributed mass shall coincide if acceleration due to gravity at all point of mass distribution is same. Hence statement-2 is false.

QUESTION: 16

Statment-1 : Consider an object that floats in water but sinks in oil. When the object floats in water, half of it is submerged. If we slowly pour oil on top of water till it completely covers the object, the object moves up.

Statment-2 :As the oil is poured in the situation of statement-1,  pressure inside the water will increase everywhere resulting in an increase in upward force on the object.

Solution:

As the oil is poured till it covers the object completely, pressure in water at all points keeps on increasing.  As a result upward force on object exerted by water increases and the object moves up for the given duration. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

QUESTION: 17

Consider a star and two planet system. The star has mass M. The planets  A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S1 is a geostationary satellite of planet A and S2 is a geostationary satellite of planet B. Neglect interaction between A and B, S1 and S2, S1 and B & S2 and A.

Q. If the time period of the satellite in geostationary orbit of planet A is T, then its time period in geostationary orbit of planet B is :

Solution:

The time in which the planet rotates about its axis is not given for either planet.

QUESTION: 18

Consider a star and two planet system. The star has mass M. The planets  A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S1 is a geostationary satellite of planet A and S2 is a geostationary satellite of planet B. Neglect interaction between A and B, S1 and S2, S1 and B & S2 and A.

Q. If the radius of the geostationary orbit in planet A is given by  then the time in which the geostationary satellite will complete 1 revolution is
I. 1 planet year = time in which planet revolves around the star
II. 1 planet day = time in which planet revolves about its axis.

Solution:

QUESTION: 19

Consider a star and two planet system. The star has mass M. The planets  A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S1 is a geostationary satellite of planet A and S2 is a geostationary satellite of planet B. Neglect interaction between A and B, S1 and S2, S1 and B & S2 and A.

Q. If planet A and B, both complete one revolution about their own axis in the same time, then the energy needed to transfer the satellite of mass m0 from planet A to planet B is -

Solution:

The energy of any geostationary satellite is the sum of kinetic energy of satellite, interaction energy of satellite and its own planet and interaction energy of satellite and star. Both planets have same mass and
same length of day. Geostationary satellite - planet system will have same interaction energy in either planet. Also kinetic energy of both satellites will be same. But the satellite-sun system will account for the energy
difference.

QUESTION: 20

In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.

Q. If the particle is released at the origin then :

Solution:

If the particle is released at the origin, it will try to go in the direction of force. Here  is positive and
hence force is negative, as a result it will move towards – ve x-axis.

QUESTION: 21

In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.

Q. If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :

Solution:

When the particle is released at x = 2 + Δ ; it will reach the point of least possible potential energy (–15J) where it will have maximum kinetic energy.

QUESTION: 22

In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.

Q. x = – 5 m and x = 10 m positions of the particle are respectively of

Solution:

QUESTION: 23

A particle is taken to a distance r (> R) from centre of the earth. R is radius of the earth. It is given velocity V which is perpendicular to . With the given values of V in column I you have to match the values of total energy and path of particle in column II. Here 'G' is the gravitational constant and 'M' is the mass of the earth.

Solution:

QUESTION: 24

An arrangement of the pipes is shown in the figure. The flow of water (incompressible and nonviscous) through the pipes is steady in nature. Three sections of the pipe are marked in which section 1 and section 2 are at same horizontal level, while being at a greater height than section 3. Correctly match order of the different physical parameter with the options given.

Column I                                    Column II
(A) volume flow rate                        (p) same everywhere
(B) kinetic energy per unit mass       (q) same at 2 and 3
(C) pressure in the sections.            (r) maximum at 1
(D) flow speed in sections               (s) minimum at 1
(t) maximum at 3

Solution:

Use equation of continuity and concept "pressure is greater at lower and broader section

QUESTION: 25

For 10 minute each, at 0 ºC, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 4 litre capacity. The resulting pressure is 2.8 atm and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas?

Solution:

QUESTION: 26

A 0.200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced to
copper (I) by iodine.   2Cu2+ + 4I– ——? 2 CuI + I2

If 20.0mL of 0.10 M Na2S2O3  is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu = 63.5 g/mole)

Solution:

QUESTION: 27

An ideal gaseous sample at initial state i (P0 , V0,T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2 = K1  and in second process the equation of the process is PV = K2. Then

Solution:

Work done in isothermal process will be more than PV2 = const, process whatever be the value of K1 and K2 as is shown in the diagram

QUESTION: 28

Given that T=298K

Now on the basis of above data which of the following predictions will be most appropriate under the standard conditions and reversible reaction.

Solution:

QUESTION: 29

An acid HA (Ka = 10–5) reacts with NaOH at 298 K. What would be the value of the rate constant of the reverse reaction at the same temperature if the rate constant of the forward reaction is 10–11 mol–1 L sec–1 ?

Solution:

QUESTION: 30

In a sample of H-atom electrons make transition from 5th excited state to ground state, producing all possible types of photons, then number of lines in infrared region are

Solution:

5 lines belong to ultra violet region and 4 lines are in visible region and rest are in infrared region.

QUESTION: 31

When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :

Solution:

Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.

QUESTION: 32

For the reversible reaction N2(g) + 3H2(g) 2NH3(g) at 500°C, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atm. The corresponding value of Kc, with concentration in mol L–1 is

Solution:

*Multiple options can be correct
QUESTION: 33

For the process H2O(l) (1 bar, 373 K)  H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is:

Solution:

*Multiple options can be correct
QUESTION: 34

0.1 M solution of KI reacts with excess of H2SO4 and KIO3 solutions, according to equation
5I + IO3 + 6H→3I2 + 3H2O ; which of the following statement is correct

Solution:

*Multiple options can be correct
QUESTION: 35

Equal volumes of following solutions are mixed, in which case the pH of  resulting solution will be average value of pH of two solutions.

Solution:

(D) pH = 5 CH3COOH and pH = 9 NH3 (aq), both must be of equal concentrations as pKa = pKb. Hence pH = pOH only if concentrations are equal so on mixing equal volume we will get CH3COONH4 salt solution and its pH is given is given by

*Multiple options can be correct
QUESTION: 36

Identify the correct statement (s)

Solution:

QUESTION: 37

Statement-1 : Solubility of BaSO4 in 0.1 M Na2SO4 is 10–9 M hence its KSP is 10–18

Statement-2 : Because for BaSO4 KSP = (s)2 [ symbols have their usual meanings]

Solution:

QUESTION: 38

Statement-1 : In a container containing gas ‘A’ at temperature 400 K, some more gas A gas at temperature 300 K is introduced. The pressure of the system increases.

Statement-2 : Increase in gaseous particles increases the number of collisions among the molecules, hence the pressure increases

Solution:

QUESTION: 39

Statement-1 : e/m ratio in case of anode ray experiment is different for different gases.

Statement-2 : The ion of gases formed after the ejection of electron are different if gas is different.

Solution:

Specific charge depends on mass of ion, which is different for different  gases.

QUESTION: 40

Statement-1 : Graphite can be converted into diamond by application of very high pressure and temperature, using a suitable catalyst.

Statement-2 : Graphite is thermodynamically more stable and less dense than diamond.

Solution:

QUESTION: 41

The enthalpy in the process HCl + nH2O → HCl in n moles of H2O
where n is the number of moles of water, is called the integral heat of solution. When n is large enough that
continued addition of water does not increase the heat of solution, one simply writes
HCl + aq → HCl (aq)
The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the
process
HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O
is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is
the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on
the composition of the solution as shown in the figure.

Q, Integral heat of solution for the following step is :
HCl + 5H2O → HCl (5H2O)

Solution:

From the graph.

QUESTION: 42

The enthalpy in the process HCl + nH2O → HCl in n moles of H2O
where n is the number of moles of water, is called the integral heat of solution. When n is large enough that
continued addition of water does not increase the heat of solution, one simply writes
HCl + aq → HCl (aq)
The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the
process
HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O
is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is
the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on
the composition of the solution as shown in the figure.

Q, What is the approximate enthalpy change for the reaction HCl + aq →HCl (aq)

Solution:

From the graph.

QUESTION: 43

The enthalpy in the process HCl + nH2O → HCl in n moles of H2O
where n is the number of moles of water, is called the integral heat of solution. When n is large enough that
continued addition of water does not increase the heat of solution, one simply writes
HCl + aq → HCl (aq)
The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the
process
HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O
is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is
the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on
the composition of the solution as shown in the figure.

Q, Approximate value of differential heat of that solution in which 1 mole of HCl is dissolved in 6 mole of water is:

Solution:

From the graph the slope of line shown in graph

QUESTION: 44

The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:
I2O5(s) + 5CO(g) → 5CO2(g) + I2(g) .
The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

Q. What must be the milligrams of CO in 1 L of the original gas sample?

Solution:

QUESTION: 45

The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:
I2O5(s) + 5CO(g) → 5CO2(g) + I2(g) .
The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

Q. What is the weight ( in mg) of iodine produced due to the reaction with I2O5(s) when it reacted with CO(g) in 20 L sample of gas?

Solution:

QUESTION: 46

The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:
I2O5(s) + 5CO(g) → 5CO2(g) + I2(g) .
The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

Q. In the above reaction (given in the comprehension) CO(g) is acting as

Solution:

QUESTION: 47

Solution:

QUESTION: 48

Solution:

QUESTION: 49

Solution:

QUESTION: 50

Solution:

QUESTION: 51

If |x|, |x – 1|, |x + 1| are first three terms of an A.P., then sum of its first 10 terms is equal to

Solution:

QUESTION: 52

Value of    is equal to

Solution:

QUESTION: 53

If (1 + x + x2)n = a0 + a1x + a2x2 + ..... + a2nx2n, then a0 + a3 + a6 + ....... is equal to

Solution:

QUESTION: 54

A bell rings at an interval of every 2 minutes, second bell rings at every 5 minutes, 3rd bell rings at every 6 minutes and 4th ring in every 8 minutes. In a span of 8 hours, how many times all 4 bells be ringing simultaneously

Solution:

QUESTION: 55

The value of the expression

(ω is the cube root of unity)

Solution:

QUESTION: 56

If α0 , α1 , α2 ....αn-1 be the n, nth roots of the unity, then the value of is equal to

Solution:

*Multiple options can be correct
QUESTION: 57

If 0 < a < b < c and the roots α, β of the equation ax2 + bx + c = 0 are imaginary, then -

Solution:

*Multiple options can be correct
QUESTION: 58

If both roots of the equation x2 – 2ax + a2 – 1 = 0 lie between – 3 and 4 and [a] denotes then intergral part of a, then [a] can be

Solution:

*Multiple options can be correct
QUESTION: 59

If |z1 + z2|2  = |z1|2 + |z2|2 , then

Solution:

*Multiple options can be correct
QUESTION: 60

If (1 + x)n = C0 + C1x + .... + Cnxn , where n is a positive integer, then

Solution:

QUESTION: 61

Statement-1 : Equation (x – a)3 + (x – b)3 + (x – c)3 = 0 has exactly one real root.

Statement-2 :  then y = f(x) meets x - axis only at one point.

Solution:

QUESTION: 62

Statement-1 :  is integer.

Statement-2 :  is divisible by n if n and r are co-prime.

Solution:

QUESTION: 63

Statement-1 : a, b, c are three distinct real numbers and ω  1 is a cube root of unity, then

Statement-2 : If z1, z2  0 are complex numbers, then

Solution:

QUESTION: 64

Statement-1 : If f : {a, b, c, d, e} → {a, b, c, d, e}, then number of onto function such that f(x)  x for each
x € {a, b, c, d, e}, is equal to 44.

Statement-2 : The number of dearrangement for n object is n!

Solution:

QUESTION: 65

If two line segments joining z1 and z2 ; z3 and z4 are parallel, then and if they are
perpendicular, then

Q. If the vertices of a triangle ABC are 0, z1, z2, then the orthocentre of ΔABC satisfies

Solution:

QUESTION: 66

If two line segments joining z1 and z2 ; z3 and z4 are parallel, then and if they are
perpendicular, then

A is point whose affix is i, B is a point whose affix is 1 + i  and C is a point on x-axis. If  |AC| + |BC|  is minimum, then the equation of  BC  is ( i =   )

Solution:

The line BC is the straight line joining the points whose affix is – i and the point B. Let Z be any point on the line,then equation of the line is the given by

QUESTION: 67

If two line segments joining z1 and z2 ; z3 and z4 are parallel, then and if they are
perpendicular, then

A point A is taken on a circle |z| = 3 whose real part is 3. It travels anticlockwise along the arc which subtends an angle of π/3  at the centre to reach at  B. OB is extended to P such that OP = 3 OB. Then the affix of
the foot of perpendicular of  P  on the real axis is

Solution:

QUESTION: 68

Let as consider the expression f(x) = sin2x – (a – 1) sin x + 2 (a – 3)

Q. If x  [0, π] and f(x) = 0 have exactly one real root, then the values of a lie in

Solution:

Since there is exactly one root in [0, π] which is possible only when x = π/2

QUESTION: 69

Let as consider the expression f(x) = sin2x – (a – 1) sin x + 2 (a – 3)

Q. f(x) = 0 have two real roots in (0, π), if a

Solution:

QUESTION: 70

Let as consider the expression f(x) = sin2x – (a – 1) sin x + 2 (a – 3)

Q.  then range of a is a subset of

Solution:

QUESTION: 71

Solution:

QUESTION: 72

If  N = 23 . 3 . 52 . 7, then

Column – I                                                                                                 Column – II

(A)    Number of odd proper divisors of  N is                                                  (p)    15

(B)    Number of even proper divisors not divisible by 3 is                               (q)    11

(C)    If sum of all the divisors which are not divisible by 7,                             (r)    14
is m, then the value of m/124 is

(D)    If sum of all the divisors which are divisible by 10,                                 (s)    18
is K, then   is

Solution:

(A) N = 2α . 3β . 5γ . 7δ
α can take the value 0 ; β can take the values 0, 1 ; γ can take the values 0, 1, 2 and δ can take the
values 0, 1
Number of odd proper divisors = 11
(B) N = 2α . 3β . 5γ . 7δ
α can take the values 1, 2, 3 ; β can take the value 0 ; γ can take the values 0, 1, 2 and δ can take
the values 0, 1
Number of even proper divisors = 18
(C) Required sum = (20 + 21 + 22 + 23) (30 + 31) (50 + 51 + 52) = 1860
(D) Required sum = (21 + 22 + 23) (30 + 31) (51 + 52) (70 + 71) = 13440

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